Lie groups and angular momentum

Continuous groups, rotations and the theory of angular momentum.

In this chapter 10

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Lie groups and rotations

Lie groups and rotations

Definition

Matrix Lie groups and Lie algebras

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Defines a matrix Lie group as a closed subgroup of the invertible matrices, shows that the set of $X$ with $e^{tX}$ in the group for every real $t$ is a real Lie algebra under the commutator, explains how the exponential map ties the two together near the identity, and fixes the structure constants and the two sign conventions (mathematicians' $e^{X}$ , physicists' $e^{i\varphi S}$ ) that the rest of the chapter uses.

Prerequisites: Functions of matrices and the matrix exponential, Groups and subgroups, Homomorphisms and quotient groups, Linear maps and matrices, Determinants, Unitary and orthogonal matrices

Definition(Matrix Lie group)AdaptedAdapted from Lee (2013), Thm 20.12, p. 523, and Prop. 20.9, p. 521

Write $M_n(\mathbb{C})$ for the complex $n\times n$ matrices and

GL(n,\mathbb{C}) = \{\, A \in M_n(\mathbb{C}) : \det A \neq 0 \,\}

for the general linear group, a group under matrix multiplication (determinants). A matrix Lie group is a subgroup $G \subseteq GL(n,\mathbb{C})$ that is a closed subset of $GL(n,\mathbb{C})$ : if $A_k \in G$ and $A_k \to A$ with $A$ invertible, then $A \in G$ .

The word "closed" is a condition inside $GL(n,\mathbb{C})$ , not inside $M_n(\mathbb{C})$ . A sequence in $G$ is allowed to run off to a singular limit; what is forbidden is an invertible limit outside $G$ .

This definition is a shortcut. In the general theory a Lie group is a group that is also a smooth manifold with smooth multiplication and inversion, and the point of Lee's (Lee (2013), Thm 20.12, p. 523) closed subgroup theorem is that a subgroup of a Lie group which is merely a closed subset is automatically an embedded Lie subgroup. Taking the conclusion as the definition costs us nothing in this chapter and spares us the manifold apparatus.

Definition(Lie algebra, and the Lie algebra of a matrix group)AdaptedAdapted from Lee (2013), p. 190, Example 8.36(c), and Prop. 20.9, p. 521

A Lie algebra over $\mathbb{R}$ is a real vector space $\mathfrak{g}$ carrying a map $\mathfrak{g}\times\mathfrak{g}\to\mathfrak{g}$ , written $(X,Y)\mapsto[X,Y]$ and called the bracket, such that for all $X,Y,Z\in\mathfrak{g}$ and all $a,b\in\mathbb{R}$ :

bilinearity: $[aX+bY,Z] = a[X,Z]+b[Y,Z]$ and $[Z,aX+bY] = a[Z,X]+b[Z,Y]$ ;
antisymmetry: $[X,Y] = -[Y,X]$ ;
the Jacobi identity: $[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]] = 0$ .

The bracket is not associative; the Jacobi identity is what stands in its place. A Lie subalgebra is a linear subspace closed under the bracket, and a Lie algebra homomorphism is a linear map $\phi$ with $\phi[X,Y] = [\phi X,\phi Y]$ .

The basic example is $M_n(\mathbb{C})$ — or $M_n(\mathbb{R})$ — with the commutator

[A,B] = AB - BA .

Bilinearity and antisymmetry are immediate; the Jacobi identity is checked by expanding the six products, each of which occurs twice with opposite signs. Regarded this way the matrices are written $\mathfrak{gl}(n,\mathbb{C})$ , in lower-case fraktur, and as a real Lie algebra $\mathfrak{gl}(n,\mathbb{C})$ has dimension $2n^2$ .

For a matrix Lie group $G \subseteq GL(n,\mathbb{C})$ , the Lie algebra of $G$ is

\mathfrak{g} \;=\; \{\, X \in M_n(\mathbb{C}) : e^{tX} \in G \ \text{ for every } t \in \mathbb{R} \,\} ,

where $e^{X} = \sum_{k\ge0}X^k/k!$ is the matrix exponential. The elements of $\mathfrak{g}$ are the generators (in the mathematicians' normalisation; see the convention note below), and the curve $t\mapsto e^{tX}$ is the one-parameter subgroup generated by $X$ . The requirement is for all real $t$ , not merely for small $t$ : that is what makes the set a vector space.

Note(Two sign conventions, and which one we use where)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (17.35)–(17.36) and (17.43), pp. 846, 848; Hassani (2013), Def. 29.1.28, p. 927

Mathematics and physics write the same object with different factors of $i$ , and mixing them silently is the commonest source of sign errors in this subject.

Mathematicians' convention (Lee (Lee (2013), Prop. 20.9, p. 521), Hassani (Hassani (2013), Def. 29.1.28, p. 927)): group elements are $e^{X}$ with $X\in\mathfrak{g}$ , and the structure constants are defined by $[\xi_i,\xi_j] = \sum_k c_{ij}^{\,k}\,\xi_k$ with no $i$ . For a unitary group the $X$ are anti-Hermitian.
Physicists' convention (Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eqs. (17.35)–(17.36), p. 846)): group elements are $U(\varphi) = \exp(i\varphi S)$ with $S$ Hermitian, obtained from an infinitesimal element $U(\delta\varphi) = 1 + i\,\delta\varphi\,S$ and the limit $U(\varphi) = \lim_{N\to\infty}(1+i\varphi S/N)^N$ . The structure constants then carry an $i$ :

[S_j,S_k] = i\sum_l f_{jkl}\,S_l .

The two are related by $X = iS$ , and the structure constants by $c = i f$ up to the index ordering. We use the mathematicians' form in this unit, because $\mathfrak{g}$ is then a real vector space closed under the commutator and the statements below need no qualification; from the rotation group and SU(2) onwards we switch to the physicists' form, because the Hermitian $S$ are the observables and their eigenvalues are the quantum numbers. The two conventions are flagged again at each switch.

Definition(The Lie algebra of a matrix Lie group)AdaptedAdapted from Lee (2013), Prop. 20.2, pp. 517–518, Prop. 20.8, pp. 519–521, Prop. 20.9, p. 521, Cor. 20.11, p. 523; Hassani (2013), Thm 29.1.21, p. 925

Let $G\subseteq GL(n,\mathbb{C})$ be a matrix Lie group and $\mathfrak{g}$ the set defined above. Then:

$\mathfrak{g}$ is a real Lie algebra. It is a real linear subspace of $M_n(\mathbb{C})$ , closed under $[\,\cdot\,,\cdot\,]$ , and closed as a subset of $M_n(\mathbb{C})$ .
The exponential map lands in $G$ and is a local chart. $\exp:\mathfrak{g}\to G$ , $X\mapsto e^{X}$ , is smooth, satisfies $e^{(s+t)X}=e^{sX}e^{tX}$ , $(e^{X})^{-1}=e^{-X}$ , and restricts to a diffeomorphism of a neighbourhood of $0$ in $\mathfrak{g}$ onto a neighbourhood of $I$ in $G$ . In particular $\dim_{\mathbb{R}}\mathfrak{g}$ is the number of real parameters of $G$ , and two matrix Lie groups with the same Lie algebra look alike near the identity.
One-parameter subgroups. Every continuous homomorphism $\gamma:\mathbb{R}\to G$ is of the form $\gamma(t) = e^{tX}$ for a unique $X\in\mathfrak{g}$ , namely $X = \gamma'(0)$ .
Homomorphisms descend. If $\Phi : G \to H$ is a homomorphism of matrix Lie groups that is smooth, there is a unique Lie algebra homomorphism $\phi:\mathfrak{g}\to\mathfrak{h}$ with

\Phi(e^{X}) = e^{\phi(X)} \qquad \text{for all } X \in \mathfrak{g},

and $\phi(X) = \frac{d}{dt}\Phi(e^{tX})\big|_{t=0}$ . 5. Structure constants. If $\xi_1,\dots,\xi_r$ is a basis of $\mathfrak{g}$ then, because $[\xi_i,\xi_j]\in\mathfrak{g}$ , there are real numbers $c_{ij}^{\,k}$ with

[\xi_i,\xi_j] = \sum_{k=1}^{r} c_{ij}^{\,k}\,\xi_k ,

and these satisfy

c_{ij}^{\,k} = -\,c_{ji}^{\,k}, \qquad \sum_{\kappa}\Bigl( c_{\rho\sigma}^{\,\kappa}c_{\kappa\mu}^{\,\nu} + c_{\sigma\mu}^{\,\kappa}c_{\kappa\rho}^{\,\nu} + c_{\mu\rho}^{\,\kappa}c_{\kappa\sigma}^{\,\nu} \Bigr) = 0 ,

which are antisymmetry and the Jacobi identity written in components. Hassani (Hassani (2013), p. 927) calls the constancy of the $c_{ij}^{\,k}$ over the group Lie's second theorem and the two displayed identities Lie's third theorem. 6. The classical algebras. With $A^{\dagger}$ the conjugate transpose and $A^{\mathsf{T}}$ the transpose,

\begin{aligned} \mathfrak{sl}(n,\mathbb{C}) &= \{X : \mathrm{tr}\,X = 0\}, & \mathfrak{u}(n) &= \{X : X^{\dagger} = -X\}, \\ \mathfrak{su}(n) &= \{X : X^{\dagger} = -X,\ \mathrm{tr}\,X = 0\}, & \mathfrak{so}(n) &= \{X \text{ real} : X^{\mathsf{T}} = -X\}, \end{aligned}

of real dimensions $2(n^2-1)$ , $n^2$ , $n^2-1$ and $n(n-1)/2$ respectively.

Idea of the proofOriginal

Three short arguments do all the work.

That $\mathfrak{g}$ is closed under scalars is free, since $e^{t(sX)} = e^{(ts)X}$ . That it is closed under addition is not free, because $e^{X}e^{Y}\neq e^{X+Y}$ in general; the fix is the Lie product formula, which recovers $e^{t(X+Y)}$ as a limit of products of elements of $G$ , and then closedness of $G$ finishes the job. That it is closed under the bracket comes from a different trick: conjugation by a group element preserves $\mathfrak{g}$ , so $t\mapsto e^{tX}Ye^{-tX}$ is a curve inside the subspace $\mathfrak{g}$ , and the derivative of a curve in a closed subspace stays in the subspace — that derivative is exactly $[X,Y]$ .

The other statements are facts about the matrix exponential transported to $G$ . The series $\sum t^kX^k/k!$ converges for every $X$ , its derivative at $t=0$ is $X$ , and $\exp$ therefore has derivative the identity at the origin, so the inverse function theorem makes it invertible near $0$ . That single sentence is why a curved object — the group — is governed near its identity by a flat one, its Lie algebra.

Proof(

\mathfrak{g}

is a real Lie algebra)AdaptedAdapted from Lee (2013), Cor. 20.11, p. 523, and Thm 20.12, p. 523

Throughout, $G$ is closed in $GL(n,\mathbb{C})$ and $\mathfrak{g} = \{X : e^{tX}\in G \text{ for all } t\in\mathbb{R}\}$ .

Step 1: $\mathfrak{g}$ is closed under real scalars. If $X\in\mathfrak{g}$ and $s\in\mathbb{R}$ , then for every $t$ we have $e^{t(sX)} = e^{(ts)X}\in G$ , so $sX\in\mathfrak{g}$ . Note that this uses only real $s$ : $\mathfrak{g}$ need not be a complex subspace, and for $\mathfrak{su}(n)$ it is not, since $iX$ is Hermitian when $X$ is anti-Hermitian.

Step 2: $\mathfrak{g}$ is a closed subset of $M_n(\mathbb{C})$ . Let $X_k\in\mathfrak{g}$ with $X_k\to X$ . For fixed $t$ the exponential is continuous (the series converges uniformly on bounded sets, by the matrix exponential), so $e^{tX_k}\to e^{tX}$ , and $e^{tX}$ is invertible with inverse $e^{-tX}$ . Here the closedness of $G$ is used: the limit of the elements $e^{tX_k}\in G$ is an invertible matrix, hence lies in $G$ . So $X\in\mathfrak{g}$ .

Step 3: $\mathfrak{g}$ is closed under addition. Let $X,Y\in\mathfrak{g}$ and fix $t\in\mathbb{R}$ . We use the Lie product formula

e^{t(X+Y)} \;=\; \lim_{m\to\infty}\Bigl( e^{\frac{t}{m}X}\, e^{\frac{t}{m}Y}\Bigr)^{m} ,

which is Lee's Corollary 20.11 (Lee (2013), p. 523); for matrices it also follows from the expansion $e^{sX}e^{sY} = I + s(X+Y) + O(s^2)$ and taking $s = t/m$ in $\bigl(I + s(X+Y) + O(s^2)\bigr)^{m}$ . Each factor $e^{(t/m)X}e^{(t/m)Y}$ lies in $G$ because $X,Y\in\mathfrak{g}$ , hence so does each $m$ -th power, $G$ being a group. The limit is invertible, so by closedness of $G$ it lies in $G$ . As $t$ was arbitrary, $X+Y\in\mathfrak{g}$ . With step 1, $\mathfrak{g}$ is a real subspace.

Step 4: conjugation preserves $\mathfrak{g}$ . Let $A\in G$ and $Y\in\mathfrak{g}$ . For every $t$ ,

e^{\,t\,AYA^{-1}} = A\,e^{tY}A^{-1} ,

because conjugation is a ring homomorphism and therefore acts term by term on the series. The right-hand side is a product of three elements of $G$ , so $AYA^{-1}\in\mathfrak{g}$ .

Step 5: $\mathfrak{g}$ is closed under the commutator. Let $X,Y\in\mathfrak{g}$ and put

\gamma(t) = e^{tX}\,Y\,e^{-tX} .

By step 4 with $A = e^{tX}\in G$ , the curve $\gamma$ takes its values in $\mathfrak{g}$ . It is differentiable, and by the product rule together with $\frac{d}{dt}e^{tX} = Xe^{tX} = e^{tX}X$ ,

\gamma'(0) = XY - YX = [X,Y] .

Now $\gamma'(0)$ is the limit of the difference quotients $\bigl(\gamma(t)-\gamma(0)\bigr)/t$ , each of which lies in $\mathfrak{g}$ because $\mathfrak{g}$ is a real subspace (steps 1 and 3). Here step 2 is used: a limit of elements of $\mathfrak{g}$ lies in $\mathfrak{g}$ . Hence $[X,Y]\in\mathfrak{g}$ .

Bilinearity, antisymmetry and the Jacobi identity hold for the commutator on all of $M_n(\mathbb{C})$ and therefore on $\mathfrak{g}$ . So $\mathfrak{g}$ is a real Lie algebra. $\square$

Proof(The exponential map, one-parameter subgroups, and homomorphisms)AdaptedAdapted from Lee (2013), Prop. 20.2, pp. 517–518, Prop. 20.8, pp. 519–521; Hassani (2013), Thm 29.1.21, p. 925, and Example 29.1.23, pp. 925–926

Part 2 (the exponential map). The identities $e^{(s+t)X} = e^{sX}e^{tX}$ and $(e^{X})^{-1} = e^{-X}$ are the statements proved for the matrix exponential; Hassani (Hassani (2013), Example 29.1.23, pp. 925–926) derives the first by multiplying the two series and recognising the binomial sum $\sum_{n}t^{m-n}s^{n}/((m-n)!\,n!) = (t+s)^m/m!$ . For the local statement, differentiate the series term by term at the origin: for $X\in\mathfrak{g}$ ,

\left.\frac{d}{dt}\right|_{t=0} e^{tX} = X ,

so the differential of $\exp$ at $0\in\mathfrak{g}$ is the identity map of $\mathfrak{g}$ . Since $\exp$ is smooth, the inverse function theorem gives a neighbourhood of $0$ in $\mathfrak{g}$ mapped diffeomorphically onto a neighbourhood of $I$ in $G$ . This is Hassani's Theorem 29.1.21 (Hassani (2013), p. 925), and the consequence he records in Box 29.1.22 — that two groups with the same Lie algebra are locally alike — is immediate from it. The statement is local, and §"Where the hypotheses are used" below shows that nothing global follows.

Part 3 (one-parameter subgroups). If $X\in\mathfrak{g}$ then $\gamma(t) = e^{tX}$ is a homomorphism $\mathbb{R}\to G$ by part 2, and $\gamma'(0)=X$ . Conversely let $\gamma:\mathbb{R}\to G$ be a continuous homomorphism. By part 2 choose a ball $U_0 = \{\|X\|<\varepsilon\}$ in $\mathfrak{g}$ on which $\exp$ is injective with image a neighbourhood of $I$ in $G$ , and put $U = \tfrac12U_0$ , $V = \exp(U)$ , so that $U$ is convex, $2U\subseteq U_0$ , and $\exp$ is injective on $2U$ . By continuity of $\gamma$ there is $\tau>0$ with $\gamma(t)\in V$ for all $|t|\le\tau$ ; write $\gamma(\tau) = e^{Z}$ with $Z\in U$ and set $X = Z/\tau$ .

We claim $\gamma(\tau/2^{k}) = e^{Z/2^{k}}$ for every $k\ge0$ , by induction. It holds at $k=0$ . If it holds at $k$ , write $\gamma(\tau/2^{k+1}) = e^{W}$ with $W\in U$ (possible since $\tau/2^{k+1}\le\tau$ ). Squaring and using the homomorphism property, $e^{2W} = \gamma(\tau/2^{k}) = e^{Z/2^{k}}$ . Both $2W$ and $Z/2^{k}$ lie in $2U$ , on which $\exp$ is injective, so $2W = Z/2^{k}$ , that is $W = Z/2^{k+1}$ . This is where the injectivity of $\exp$ near $0$ is used, and it is why the neighbourhood had to be halved first.

Since $\gamma$ is a homomorphism, $\gamma(m\tau/2^k) = \gamma(\tau/2^k)^m = e^{mZ/2^k} = e^{(m\tau/2^k)X}$ for all integers $m$ and all $k\ge0$ . The numbers $m\tau/2^k$ are dense in $\mathbb{R}$ , and both $t\mapsto\gamma(t)$ and $t\mapsto e^{tX}$ are continuous, so they agree everywhere. Uniqueness of $X$ follows by differentiating at $t=0$ . Lee (Lee (2013), Thm 20.1, p. 516, and Prop. 20.2, pp. 517–518) obtains the same conclusion by identifying one-parameter subgroups with the integral curves of left-invariant vector fields through the identity.

Part 4 (homomorphisms). Let $\Phi:G\to H$ be a smooth homomorphism and $X\in\mathfrak{g}$ . Then $t\mapsto\Phi(e^{tX})$ is a continuous homomorphism $\mathbb{R}\to H$ , so by part 3 it equals $e^{t\phi(X)}$ for a unique $\phi(X)\in\mathfrak{h}$ , and differentiating at $t=0$ identifies $\phi(X) = \frac{d}{dt}\Phi(e^{tX})|_{t=0}$ . Setting $t=1$ gives $\Phi(e^{X}) = e^{\phi(X)}$ , which is Lee's (Lee (2013), Prop. 20.8(g), p. 520) commuting square. Linearity of $\phi$ follows from $\phi(sX) = s\phi(X)$ (replace $t$ by $st$ ) and from applying $\Phi$ to the Lie product formula of step 3 above. Finally $\phi$ preserves brackets: apply $\Phi$ to the identity $e^{tX}Ye^{-tX}\in\mathfrak{g}$ of step 4 in the previous proof — more precisely to $\Phi\bigl(e^{tX}e^{sY}e^{-tX}\bigr) = e^{t\phi(X)}e^{s\phi(Y)}e^{-t\phi(X)}$ — and differentiate once in $s$ at $s=0$ and once in $t$ at $t=0$ . The left-hand side gives $\phi([X,Y])$ and the right-hand side $[\phi(X),\phi(Y)]$ .

Part 6 (the classical algebras). Each is an application of the definition of $\mathfrak{g}$ together with one identity.

$\mathfrak{sl}$ : by the determinant-trace identity $\det e^{tX} = e^{t\,\mathrm{tr}X}$ of the matrix exponential, $\det e^{tX} = 1$ for all $t$ if and only if $\mathrm{tr}X = 0$ . Hassani (Hassani (2013), Box 29.1.17, p. 924) states the conclusion as "the Lie algebra of the special linear group is the traceless operators".
$\mathfrak{u}$ : $(e^{tX})^{\dagger}e^{tX} = e^{tX^{\dagger}}e^{tX}$ . Differentiating at $t=0$ gives $X^{\dagger}+X$ , so unitarity for all $t$ forces $X^{\dagger} = -X$ ; conversely if $X^{\dagger}=-X$ then $(e^{tX})^{\dagger} = e^{-tX} = (e^{tX})^{-1}$ . This is Hassani's Box 29.1.19 (Hassani (2013), p. 924).
$\mathfrak{su} = \mathfrak{sl}\cap\mathfrak{u}$ , giving anti-Hermitian traceless matrices; the diagonal of an anti-Hermitian matrix is purely imaginary with $n$ free entries and the strict upper triangle has $n(n-1)/2$ complex entries, so $\dim_{\mathbb{R}}\mathfrak{u}(n) = n + n(n-1) = n^2$ and the trace condition removes one more, $\dim_{\mathbb{R}}\mathfrak{su}(n) = n^2-1$ (Hassani (Hassani (2013), Box 29.1.20, p. 925)).
$\mathfrak{so}$ : the same computation with the transpose in place of the conjugate transpose, over $\mathbb{R}$ ; a real skew-symmetric matrix has zero diagonal and $n(n-1)/2$ free entries. Lee (Lee (2013), Example 20.4, p. 518) records the forward direction as "the exponential of a skew-symmetric matrix is orthogonal". Note that $\mathrm{tr}X=0$ is automatic here, so $\mathfrak{so}(n) = \mathfrak{o}(n)$ and $SO(n)$ , $O(n)$ have the same Lie algebra — the first sign that the Lie algebra sees only the identity component. $\square$

Remark(Where the hypotheses are used)Original

Closedness of $G$ is used twice, in step 2 and step 3 of the first proof, and it cannot be dropped. Inside the torus $T = \{\mathrm{diag}(e^{i\theta},e^{i\varphi})\}\subset GL(2,\mathbb{C})$ take

H_\alpha = \bigl\{\, \mathrm{diag}\!\left(e^{it}, e^{i\alpha t}\right) : t\in\mathbb{R} \,\bigr\}, \qquad \alpha \text{ irrational.}

$H_\alpha$ is a subgroup and the image of a one-parameter subgroup, but it is dense in $T$ and so is not closed. Its closure $T$ is two-dimensional while $H_\alpha$ is a one-dimensional curve, so no neighbourhood of the identity in $H_\alpha$ is homeomorphic to a neighbourhood in the subspace topology of a Euclidean ball: $H_\alpha$ is a Lie group in its own right but not an embedded subgroup, and the closed subgroup theorem does not apply to it.

"For all $t$ ", not "for small $t$ ". If the condition $e^{tX}\in G$ were imposed only for $|t|<\varepsilon$ the set would still be defined, but step 1 would fail to give a subspace stable under large scalars, and the identification with the tangent space at $I$ would break. The condition for all $t$ is Lee's (Lee (2013), Prop. 20.9, p. 521) characterisation and it is exactly what the halving argument in part 3 reconstructs.
$\exp$ is local, and only local. Part 2 says nothing about the image of $\exp$ in the large. Two failures are standard. First, $\exp$ need not be onto even for a connected group: in $SL(2,\mathbb{R})$ , a traceless real $X$ has eigenvalues $\pm\lambda$ with $\lambda$ real, $\pm i\mu$ with $\mu$ real, or $0$ twice, so $\mathrm{tr}\,e^{X}$ equals $2\cosh\lambda\ge 2$ , $2\cos\mu\ge-2$ , or $2$ ; hence no element with trace $<-2$ , for instance $\mathrm{diag}(-2,-\tfrac12)$ , is an exponential. Second, $\exp$ need not be injective: $e^{2\pi k X} = I$ for every integer $k$ when $X = \mathrm{diag}(i,-i)$ .
$e^{X}e^{Y} = e^{X+Y}$ requires $[X,Y]=0$ . It is used nowhere above, and the Lie product formula of step 3 exists precisely to avoid needing it. The exact condition under which $e^{sX}e^{tY} = e^{sX+tY}$ holds for all small $s,t$ is commutativity, as the matrix exponential records.
Smoothness of $\Phi$ in part 4 can in fact be weakened to continuity, but the proof above uses differentiability of $t\mapsto\Phi(e^{tX})$ at $0$ and we do not claim the stronger statement.
Real versus complex. $\mathfrak{g}$ is a real Lie algebra even when $G$ consists of complex matrices. Forgetting this is the usual reason for a factor-of-two error in a dimension count: $\mathfrak{su}(2)$ has real dimension $3$ , not $3$ complex dimensions, and $SU(2)$ is a three-parameter group.

Method(Finding the Lie algebra and the structure constants of a matrix group)Original

Let $G$ be specified by a set of equations on matrices, say $F(A) = 0$ .

Linearise the defining equations at the identity. Put $A = e^{tX}= I + tX + O(t^2)$ , substitute into $F(A)=0$ , and keep the terms of first order in $t$ . The resulting linear conditions on $X$ are the candidate for $\mathfrak{g}$ . (For $A^{\dagger}A = I$ this returns $X^{\dagger}+X = 0$ ; for $\det A = 1$ it returns $\mathrm{tr}X = 0$ .)
Verify both directions. Check that each $X$ satisfying the linearised conditions really has $e^{tX}\in G$ for all $t$ . Step 1 alone gives only a necessary condition.
Count. $\dim_{\mathbb{R}}\mathfrak{g}$ is the number of real parameters of $G$ ; Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 845) call this the order of the group.
Choose a basis $\xi_1,\dots,\xi_r$ adapted to the problem — usually one that makes as many brackets as possible vanish, or one with a physical meaning.
Compute the brackets $[\xi_i,\xi_j]$ as matrix commutators and expand them in the basis; the coefficients are the structure constants. Check antisymmetry and the Jacobi identity as an arithmetic test of the algebra.
If you need Hermitian generators, set $S_j = -i\xi_j$ (equivalently $\xi_j = iS_j$ ). Then $e^{\varphi\xi_j} = e^{i\varphi S_j}$ and the brackets acquire a factor $i$ , $[S_j,S_k] = i f_{jkl}S_l$ , as in Arfken, Weber and Harris, Eq. (17.43) (Arfken, Weber & Harris (2013), p. 848). If $G$ is unitary the $S_j$ are Hermitian, and if $\det = 1$ they are also traceless, by their Eqs. (17.40) and (17.41).
Record the rank, the largest number of mutually commuting independent generators. It is the number of labels — quantum numbers — needed to tell apart the basis vectors of an irreducible representation, and also, by Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 849), the number of independent Casimir operators (see Casimir operators).

Example(The Heisenberg group: a group where every step is exact)Original

Let

G \;=\; \left\{\, A(x,y,z) = \begin{pmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1\end{pmatrix} : x,y,z\in\mathbb{R} \,\right\} \subset GL(3,\mathbb{R}).

Step 1: it is a matrix Lie group. Multiplying,

A(x,y,z)\,A(x',y',z') = A\bigl(x+x',\; y+y',\; z+z'+xy'\bigr),

so $G$ is closed under products, $A(0,0,0)=I$ , and $A(x,y,z)^{-1} = A(-x,-y,xy-z)$ . It is a closed subset of $GL(3,\mathbb{R})$ because it is cut out by the equations $A_{11}=A_{22}=A_{33}=1$ , $A_{21}=A_{31}=A_{32}=0$ , all of which survive limits. Note that the product is not commutative: the term $xy'$ is not symmetric in the two arguments.

Step 2: linearise. Writing $A = I + tX + O(t^2)$ , the defining equations say that $X$ has zeros on and below the diagonal. So the candidate algebra is the strictly upper triangular matrices

\mathfrak{g} = \mathbb{R}\xi_1 \oplus \mathbb{R}\xi_2 \oplus \mathbb{R}\xi_3, \qquad \xi_1 = \begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix},\quad \xi_2 = \begin{pmatrix}0&0&0\\0&0&1\\0&0&0\end{pmatrix},\quad \xi_3 = \begin{pmatrix}0&0&1\\0&0&0\\0&0&0\end{pmatrix}.

Step 3: verify, by computing $\exp$ exactly. For $M = a\xi_1+b\xi_2+c\xi_3$ one finds $M^2 = ab\,\xi_3$ and $M^3 = 0$ , so the exponential series stops after three terms:

e^{M} \;=\; I + M + \tfrac12 M^2 \;=\; A\!\left(a,\; b,\; c + \tfrac{ab}{2}\right).

Every $e^{tM}$ therefore lies in $G$ , so $\mathfrak{g}$ is exactly the Lie algebra, $\dim\mathfrak{g}=3$ , and $G$ is a three-parameter group. Here $\exp:\mathfrak{g}\to G$ is a bijection, not merely a local one: given $(x,y,z)$ , solve $a=x$ , $b=y$ , $c = z - xy/2$ . A number: at $(a,b,c) = (2,3,1)$ ,

e^{M} = \begin{pmatrix}1&2&4\\0&1&3\\0&0&1\end{pmatrix}, \qquad\text{since } c + \tfrac{ab}{2} = 1 + 3 = 4 .

Step 4: the structure constants. Direct multiplication gives

[\xi_1,\xi_2] = \xi_3, \qquad [\xi_1,\xi_3] = 0, \qquad [\xi_2,\xi_3] = 0 ,

so the only non-zero structure constants are $c_{12}^{\;3} = 1 = -c_{21}^{\;3}$ . Antisymmetry is visible, and the Jacobi identity holds because every double bracket contains $\xi_3$ , which is central.

Step 5: a consistency check on the group commutator. The group commutator of two one-parameter subgroups should, to leading order, reproduce the Lie bracket; here it does so exactly. Using the multiplication law of step 1 with $e^{s\xi_1} = A(s,0,0)$ and $e^{s\xi_2} = A(0,s,0)$ ,

e^{s\xi_1}e^{s\xi_2}e^{-s\xi_1}e^{-s\xi_2} = A(0,0,s^2) = e^{\,s^2[\xi_1,\xi_2]} .

At $s = \tfrac12$ the right-hand corner entry is $s^2 = 0.25$ , and no other entry is disturbed. This is Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), Eq. (17.42), p. 848) second-order expansion $U_k^{-1}U_j^{-1}U_kU_j = 1 + \epsilon_j\epsilon_k[S_j,S_k]+\cdots$ with the higher-order terms absent, because the algebra is nilpotent.

Step 6: contrast with a group where $\exp$ is not onto. In $SL(2,\mathbb{R})$ — also three-dimensional, also connected — the remark above shows that $\mathrm{diag}(-2,-\tfrac12)$ , of determinant $1$ and trace $-2.5$ , is not $e^{X}$ for any $X\in\mathfrak{sl}(2,\mathbb{R})$ . The Heisenberg group is the pleasant case, not the typical one.

ApplicationOriginal

Symmetry and conservation laws. A continuous symmetry of a physical system is an action of a Lie group, and its generators are the conserved quantities: the three generators of translations are the momentum components, those of rotations are the angular momenta (angular momentum algebra), and the generator of time translation is the Hamiltonian. The statement "the symmetry group is $G$ " becomes the computable statement "these $r$ operators satisfy these structure constants", which is what one actually uses.

Quantum mechanics and spin. The passage $U = e^{i\varphi S}$ with Hermitian $S$ is the link between a one-parameter family of unitary evolutions and a self-adjoint observable; in infinite dimensions it is Stone's theorem. The whole theory of angular momentum is the representation theory of one three-dimensional Lie algebra, as the rotation group and SU(2) makes explicit, and the Heisenberg group of the example above is the group generated by position and momentum, whose Lie algebra relation $[\xi_1,\xi_2]=\xi_3$ is the canonical commutation relation with $\xi_3$ central — that centrality is why $\hbar$ is a number and not an operator.

Particle physics. The standard model is specified by naming a Lie group, $SU(3)\times SU(2)\times U(1)$ , and the representations the fields carry. The structure constants of $\mathfrak{su}(3)$ fix the self-interactions of the gluons; those of $\mathfrak{su}(2)$ fix the weak interactions. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), pp. 846, 853) tell this history from isospin onwards.

Rigid-body motion, robotics and computer vision. The configuration of a rigid body is an element of the six-dimensional group $SE(3)$ of rigid motions; velocities are elements of its Lie algebra, and the exponential map is the standard way to parameterise a pose by a "twist". Because the group is curved, numerical integration of rigid-body dynamics is done with schemes that take their steps with $\exp$ , so that the computed configuration stays exactly in the group instead of drifting off it.

Differential equations. Lie's own motivation was that a differential equation invariant under a one-parameter group can be reduced in order, and the generator tells one which change of variable does it; this is the systematic version of the ad hoc substitutions in first-order equations. The same idea classifies the special functions: as Hassani (Hassani (2013), pp. 970–971) puts it, many of the equations of mathematical physics are statements that a Casimir operator of some Lie algebra acts as a multiple of the identity.

Theorem

The rotation group and SU(2)

Adapted

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Computes the generators of $SO(3)$ and of $SU(2)$ , shows that the two Lie algebras are the same three-dimensional algebra $[J_j,J_k]=i\epsilon_{jkl}J_l$ , and proves that $U\mapsto R(U)$ defined by $U(\mathbf{x}\cdot\boldsymbol{\sigma})U^{\dagger}=(R\mathbf{x})\cdot\boldsymbol{\sigma}$ is a surjective homomorphism $SU(2)\to SO(3)$ with kernel $\{\pm I\}$ — the two-to-one covering that makes a spinor change sign under a rotation through $2\pi$ .

Prerequisites: Matrix Lie groups and Lie algebras, Unitary and orthogonal matrices, Homomorphisms and quotient groups, Functions of matrices and the matrix exponential, Determinants, Quaternions

Definition(The two groups, the Pauli matrices, and the sign convention)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (17.44)–(17.49), pp. 849–850, and Eq. (17.54), p. 851; Hassani (2013), §31.2.1, p. 997

The rotation group and the special unitary group in two dimensions are

SO(3) = \{\, R \in M_3(\mathbb{R}) : R^{\mathsf{T}}R = I,\ \det R = 1 \,\}, \qquad SU(2) = \{\, U \in M_2(\mathbb{C}) : U^{\dagger}U = I,\ \det U = 1 \,\} .

Both are matrix Lie groups in the sense of matrix Lie groups and Lie algebras, being closed subsets of the invertible matrices cut out by polynomial equations, and both are compact: their entries are bounded by $1$ in modulus.

The Pauli matrices are

\sigma_1 = \begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad \sigma_2 = \begin{pmatrix}0&-i\\i&0\end{pmatrix},\qquad \sigma_3 = \begin{pmatrix}1&0\\0&-1\end{pmatrix} .

They are Hermitian and traceless, and they satisfy the Clifford relation

\sigma_j\sigma_k + \sigma_k\sigma_j = 2\delta_{jk}I ,

which Hassani (Hassani (2013), §31.2.1, p. 997) takes as their defining property. Together with $\sigma_1\sigma_2 = i\sigma_3$ and its cyclic images this gives the single identity

(\mathbf{a}\cdot\boldsymbol{\sigma})(\mathbf{b}\cdot\boldsymbol{\sigma}) = (\mathbf{a}\cdot\mathbf{b})\,I + i\,(\mathbf{a}\times\mathbf{b})\cdot\boldsymbol{\sigma} , \qquad \mathbf{a}\cdot\boldsymbol{\sigma} := a_1\sigma_1+a_2\sigma_2+a_3\sigma_3 ,

for real vectors $\mathbf{a},\mathbf{b}$ , which is used repeatedly below.

Generators and the sign convention. Following the physicists' convention of matrix Lie groups and Lie algebras, we write group elements as exponentials of $i$ times a Hermitian matrix, and we fix the sign so that

R_{\mathbf{n}}(\theta) = \exp\bigl(-i\theta\,\mathbf{n}\cdot\mathbf{J}\bigr)

is the rotation of a vector through the angle $\theta$ about the unit axis $\mathbf{n}$ , counterclockwise seen from the tip of $\mathbf{n}$ . The three matrices $\mathbf{J} = (J_1,J_2,J_3)$ are then

(J_k)_{pq} = -i\,\epsilon_{kpq}, \qquad\text{that is}\qquad J_1 = \begin{pmatrix}0&0&0\\0&0&-i\\0&i&0\end{pmatrix},\quad J_2 = \begin{pmatrix}0&0&i\\0&0&0\\-i&0&0\end{pmatrix},\quad J_3 = \begin{pmatrix}0&-i&0\\i&0&0\\0&0&0\end{pmatrix} .

Each $J_k$ is exactly the matrix Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eqs. (17.45), (17.47), (17.49), pp. 849–850) call $S_k$ — that is, $J_1 = S_1$ , $J_2 = S_2$ and $J_3 = S_3$ , the order of their equation numbers being $S_3$ , $S_1$ , $S_2$ rather than ours. The one difference is that they write the group element as $\exp(+i\varphi S)$ , which rotates the coordinate axes through $+\varphi$ and therefore rotates a vector through $-\varphi$ . The Lie algebra and all the commutators below are the same either way; only the sign of the angle differs, and we flag it again where it matters.

Theorem(

SU(2)

is a two-sheeted cover of

SO(3)

)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (17.50), (17.55), (17.57), pp. 850–852; Hassani (2013), Thm 31.2.6, p. 996, and §30.3.3, p. 972

The two Lie algebras coincide. $\mathfrak{so}(3)$ is the space of real skew-symmetric $3\times3$ matrices, with real basis $\{-iJ_1,-iJ_2,-iJ_3\}$ , and $\mathfrak{su}(2)$ is the space of traceless anti-Hermitian $2\times2$ matrices, with real basis $\{-i\sigma_1/2,-i\sigma_2/2,-i\sigma_3/2\}$ . Both are three-dimensional, and in the Hermitian normalisation $S_k = J_k$ for $\mathfrak{so}(3)$ and $S_k = \tfrac12\sigma_k$ for $\mathfrak{su}(2)$ the structure constants are the same:

[S_j,S_k] = i\,\epsilon_{jkl}\,S_l .

The real-linear map $-iJ_k \mapsto -i\sigma_k/2$ , which carries the one real basis to the other, is therefore an isomorphism $\mathfrak{so}(3)\cong\mathfrak{su}(2)$ of real Lie algebras. (It is usually written $J_k\mapsto\tfrac12\sigma_k$ , in the Hermitian normalisation; the Hermitian matrices $J_k$ and $\tfrac12\sigma_k$ do not themselves lie in the two real algebras.)

Half-angle formula. For a unit vector $\mathbf{n}\in\mathbb{R}^3$ and $\theta\in\mathbb{R}$ ,

U_{\mathbf{n}}(\theta) := \exp\!\left(-\tfrac{i}{2}\theta\,\mathbf{n}\cdot\boldsymbol{\sigma}\right) = I\cos\frac{\theta}{2} \;-\; i\,(\mathbf{n}\cdot\boldsymbol{\sigma})\,\sin\frac{\theta}{2} \;\in\; SU(2),

and every element of $SU(2)$ is of this form. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eq. (17.57), p. 851) give the same formula with the opposite sign of the angle, as $U_j(\alpha_j) = 1_2\cos(\alpha_j/2) + i\sigma_j\sin(\alpha_j/2)$ .

The covering homomorphism. For $U\in SU(2)$ and $\mathbf{x}\in\mathbb{R}^3$ the matrix $U(\mathbf{x}\cdot\boldsymbol{\sigma})U^{\dagger}$ is again traceless and Hermitian, so it equals $\mathbf{x}'\cdot\boldsymbol{\sigma}$ for a unique $\mathbf{x}'\in\mathbb{R}^3$ ; writing $\mathbf{x}' = R(U)\mathbf{x}$ defines a map

\Phi : SU(2)\longrightarrow SO(3), \qquad \Phi(U) = R(U), \qquad R(U)_{ij} = \tfrac12\,\mathrm{tr}\bigl(\sigma_i\,U\sigma_j U^{\dagger}\bigr) .

$\Phi$ is a surjective group homomorphism with

\ker\Phi = \{\,I,\,-I\,\},

so by the first isomorphism theorem (homomorphisms and quotient groups)

SO(3) \;\cong\; SU(2)/\{\pm I\} ,

and $\Phi$ is exactly two-to-one: $\Phi(U) = \Phi(-U)$ and nothing else is identified. Concretely $\Phi\bigl(U_{\mathbf{n}}(\theta)\bigr) = R_{\mathbf{n}}(\theta)$ : the half-angle in $SU(2)$ corresponds to the full angle in $SO(3)$ .

Spinors. $SU(2)$ is the set of matrices

U = \begin{pmatrix} a & -\bar b\\ b & \bar a\end{pmatrix},\qquad |a|^2+|b|^2 = 1 ,

hence is the unit sphere $S^3\subset\mathbb{C}^2\cong\mathbb{R}^4$ and is path-connected. Its defining representation acts on $\mathbb{C}^2$ ; the elements of $\mathbb{C}^2$ are called spinors. Because $U_{\mathbf{n}}(2\pi) = -I$ while $R_{\mathbf{n}}(2\pi) = I$ , a spinor is reversed in sign by a rotation through $2\pi$ and restored only after $4\pi$ . Equivalently: a representation of $SU(2)$ descends to a representation of $SO(3)$ if and only if $-I$ acts as the identity, which happens exactly for the integer- $j$ representations.

Idea of the proofOriginal

Everything turns on one observation: the traceless Hermitian $2\times2$ matrices form a three-dimensional real vector space, with $\sigma_1,\sigma_2,\sigma_3$ as a basis, and on that space the determinant is a quadratic form,

\det(\mathbf{x}\cdot\boldsymbol{\sigma}) = -|\mathbf{x}|^2 .

Conjugation $X\mapsto UXU^{\dagger}$ by a unitary matrix preserves tracelessness, Hermiticity and the determinant. So each $U$ acts on $\mathbb{R}^3$ by a linear map preserving the Euclidean length: an orthogonal map. The determinant of that orthogonal map is $\pm1$ , it depends continuously on $U$ , and $SU(2)$ is connected, so the sign cannot jump away from its value $+1$ at $U=I$ — the image lies in $SO(3)$ .

Which $U$ act trivially? Those commuting with all three Pauli matrices, and a matrix commuting with $\sigma_1$ and $\sigma_3$ is a multiple of the identity; the condition $\det U = 1$ then leaves $\pm I$ . Finally, surjectivity: a one-line computation shows that the diagonal $U = \mathrm{diag}(e^{-i\theta/2},e^{i\theta/2})$ produces the rotation by $\theta$ about the $z$ -axis, and by the Euler-angle factorisation every rotation is a product of three such rotations about two fixed axes.

The factor $\tfrac12$ in the exponent is thus not a convention but the content of the theorem: going once round in $SO(3)$ is going half-way round in $SU(2)$ .

Proof(Part 1: the two Lie algebras)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (17.50) and (17.55), pp. 850–851; Hassani (2013), §31.2.1, p. 998

Step 1: $\mathfrak{so}(3)$ and $\mathfrak{su}(2)$ as sets. By matrix Lie groups and Lie algebras, $\mathfrak{so}(3)$ consists of the real $X$ with $X^{\mathsf{T}} = -X$ and $\mathfrak{su}(2)$ of the complex $X$ with $X^{\dagger} = -X$ and $\mathrm{tr}X=0$ . A real skew-symmetric $3\times 3$ matrix has zero diagonal and three free entries; a traceless anti-Hermitian $2\times2$ matrix has diagonal $\mathrm{diag}(i\gamma,-i\gamma)$ and one free complex off-diagonal entry, again three real parameters. Both are three-dimensional over $\mathbb{R}$ .

Step 2: bases. The matrices $-iJ_k$ are real with $(-iJ_k)_{pq} = -\epsilon_{kpq}$ , hence skew-symmetric, and they are linearly independent, so they span $\mathfrak{so}(3)$ . The matrices $-i\sigma_k/2$ are traceless and anti-Hermitian because the $\sigma_k$ are traceless and Hermitian, and they are linearly independent, so they span $\mathfrak{su}(2)$ ; this is Hassani's (Hassani (2013), p. 998) observation that his $s_j = -i\sigma_j/2$ lie in $\mathfrak{su}(2)$ .

Step 3: the commutators of the $J_k$ . We use the contraction identity $\sum_q \epsilon_{qab}\epsilon_{qcd} = \delta_{ac}\delta_{bd}-\delta_{ad}\delta_{bc}$ and the cyclic symmetry of $\epsilon$ . From $(J_k)_{pq} = -i\epsilon_{kpq}$ ,

\bigl([J_j,J_k]\bigr)_{pr} = \sum_q\Bigl[(-i\epsilon_{jpq})(-i\epsilon_{kqr}) - (-i\epsilon_{kpq})(-i\epsilon_{jqr})\Bigr] = -\sum_q\bigl(\epsilon_{jpq}\epsilon_{kqr} - \epsilon_{kpq}\epsilon_{jqr}\bigr) .

Cycling the indices so that the summation index comes first, $\epsilon_{jpq}\epsilon_{kqr} = \epsilon_{qjp}\epsilon_{qrk}$ , and the contraction identity gives $\sum_q \epsilon_{qjp}\epsilon_{qrk} = \delta_{jr}\delta_{pk}-\delta_{jk}\delta_{pr}$ . Exchanging $j$ and $k$ in that result gives $\sum_q\epsilon_{kpq}\epsilon_{jqr} = \delta_{kr}\delta_{pj}-\delta_{jk}\delta_{pr}$ . Subtracting, the two $\delta_{jk}\delta_{pr}$ terms cancel and

\bigl([J_j,J_k]\bigr)_{pr} = -\bigl(\delta_{jr}\delta_{pk} - \delta_{kr}\delta_{pj}\bigr) = \delta_{pj}\delta_{kr} - \delta_{jr}\delta_{pk} .

On the other side, $\bigl(i\epsilon_{jkl}J_l\bigr)_{pr} = \sum_l i\,\epsilon_{jkl}(-i\epsilon_{lpr}) = \sum_l\epsilon_{ljk}\epsilon_{lpr} = \delta_{jp}\delta_{kr}-\delta_{jr}\delta_{kp}$ , which is the same expression. Hence

[J_j,J_k] = i\,\epsilon_{jkl}\,J_l ,

which is Arfken, Weber and Harris's Eq. (17.50) (Arfken, Weber & Harris (2013), p. 850), there verified by direct matrix multiplication. That check is worth repeating on one case: from the displayed matrices, $J_1J_2$ has the single nonzero entry $(-i)(-i) = -1$ in position $(2,1)$ and $J_2J_1$ the single nonzero entry $(i)(i) = -1$ in position $(1,2)$ , so $J_1J_2 - J_2J_1$ has $(1,2)$ entry $+1$ and $(2,1)$ entry $-1$ and all other entries zero. That is indeed $iJ_3$ , since $J_3$ has $(1,2)$ entry $-i$ and $(2,1)$ entry $+i$ .

Step 4: the commutators of the $\sigma_k$ . From the Clifford relation $\sigma_j\sigma_k = -\sigma_k\sigma_j$ for $j\ne k$ and $\sigma_1\sigma_2 = i\sigma_3$ with its cyclic images,

[\sigma_j,\sigma_k] = 2\sigma_j\sigma_k = 2i\,\epsilon_{jkl}\sigma_l \qquad (j\neq k),

and both sides vanish for $j=k$ . Setting $S_k = \tfrac12\sigma_k$ divides the left side by $4$ and the right side by $2$ , giving $[S_j,S_k] = i\epsilon_{jkl}S_l$ , which is Arfken, Weber and Harris's Eq. (17.55) (Arfken, Weber & Harris (2013), p. 851). Since the structure constants agree with those of step 3, so do the brackets of the two real bases of step 2. Put $A_k = -iJ_k$ and $B_k = -i\sigma_k/2$ ; then

[A_j,A_k] = (-i)^2[J_j,J_k] = -\,i\epsilon_{jkl}J_l = \epsilon_{jkl}\,A_l ,

and the identical computation with $B$ in place of $A$ gives $[B_j,B_k] = \epsilon_{jkl}B_l$ . The real-linear map $A_k\mapsto B_k$ from $\mathfrak{so}(3)$ to $\mathfrak{su}(2)$ therefore carries brackets to brackets, and being a bijection between three-dimensional real spaces it is an isomorphism of real Lie algebras. In the Hermitian normalisation the same map is written $J_k\mapsto\tfrac12\sigma_k$ . $\square$

Proof(Part 2: the half-angle formula)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (17.56)–(17.57), p. 851

Let $\mathbf{n}$ be a unit vector. By the product identity of the definition block with $\mathbf{a}=\mathbf{b}=\mathbf{n}$ ,

(\mathbf{n}\cdot\boldsymbol{\sigma})^2 = |\mathbf{n}|^2 I + i(\mathbf{n}\times\mathbf{n})\cdot\boldsymbol{\sigma} = I .

So the even powers of $\mathbf{n}\cdot\boldsymbol{\sigma}$ are $I$ and the odd powers are $\mathbf{n}\cdot\boldsymbol{\sigma}$ . Splitting the exponential series into even and odd parts,

\exp\!\left(-\tfrac{i\theta}{2}\mathbf{n}\cdot\boldsymbol{\sigma}\right) = \sum_{m\ge0}\frac{(-i\theta/2)^{2m}}{(2m)!}I + \sum_{m\ge0}\frac{(-i\theta/2)^{2m+1}}{(2m+1)!}(\mathbf{n}\cdot\boldsymbol{\sigma}) = I\cos\frac\theta2 - i(\mathbf{n}\cdot\boldsymbol{\sigma})\sin\frac\theta2 ,

the two series being those of $\cos$ and $\sin$ because $(-i)^{2m} = (-1)^m$ and $(-i)^{2m+1} = -i(-1)^m$ . The result is unitary with determinant $1$ : writing $c = \cos(\theta/2)$ and $s=\sin(\theta/2)$ and using $(\mathbf{n}\cdot\boldsymbol{\sigma})^{\dagger} = \mathbf{n}\cdot\boldsymbol{\sigma}$ ,

U^{\dagger}U = (cI + is\,\mathbf{n}\cdot\boldsymbol{\sigma})(cI - is\,\mathbf{n}\cdot\boldsymbol{\sigma}) = (c^2+s^2)I = I , \qquad \det U = c^2 + s^2 = 1 ,

the determinant being computed entry by entry: $cI - is\,\mathbf{n}\cdot\boldsymbol{\sigma}$ has diagonal $c\mp isn_3$ and off-diagonal entries $-is(n_1\mp in_2)$ , so

\det U = (c-isn_3)(c+isn_3) - (-is)^2(n_1-in_2)(n_1+in_2) = c^2 + s^2n_3^2 + s^2(n_1^2+n_2^2) = c^2+s^2 = 1 .

Conversely, every $U\in SU(2)$ has this form. Unitarity and $\det U=1$ force

U = \begin{pmatrix} a & -\bar b\\ b& \bar a\end{pmatrix}, \qquad |a|^2+|b|^2=1 ,

since the second column of a unitary matrix is a unit vector orthogonal to the first, hence a multiple of $(-\bar b,\bar a)$ , and the determinant condition fixes the multiple to be $1$ . Writing $a = c - is n_3$ and $b = sn_2 - isn_1$ with $c = \mathrm{Re}\,a$ and $s\mathbf{n}$ the real vector $(-\mathrm{Im}\,b,\ \mathrm{Re}\,b,\ -\mathrm{Im}\,a)$ exhibits $U$ in the half-angle form with $c^2+s^2 = |a|^2+|b|^2 = 1$ , so $c = \cos(\theta/2)$ and $s = \sin(\theta/2)$ for some $\theta$ , and $\mathbf{n}$ is a unit vector whenever $s\ne0$ (when $s = 0$ , $U = \pm I$ and any $\mathbf{n}$ will do). This is the four-real-parameter description with one constraint that Riley, Hobson and Bence (Riley, Hobson & Bence (2006), Exercise 28.8, p. 1071) set as an exercise: the matrices $M(a_0,\mathbf{a})$ with $a_0^2+|\mathbf{a}|^2 = 1$ form a group, and they are the unit quaternions. In particular $SU(2)$ is the unit sphere $S^3$ in $\mathbb{C}^2 \cong \mathbb{R}^4$ and is path-connected, since $S^3$ is. $\square$

Proof(Part 3: the covering homomorphism)AdaptedAdapted from Hassani (2013), §30.3.3, p. 972, and Thm 31.2.6, p. 996; Riley, Hobson & Bence (2006), §28.7.3, pp. 1069–1070

Write $V = \{\,X\in M_2(\mathbb{C}) : X^{\dagger}=X,\ \mathrm{tr}X = 0\,\}$ .

Step 1: $V$ is $\mathbb{R}^3$ in disguise. A traceless Hermitian $2\times2$ matrix is

\begin{pmatrix} x_3 & x_1-ix_2\\ x_1+ix_2 & -x_3\end{pmatrix} = \mathbf{x}\cdot\boldsymbol{\sigma}, \qquad \mathbf{x}\in\mathbb{R}^3,

so $\mathbf{x}\mapsto\mathbf{x}\cdot\boldsymbol{\sigma}$ is a real linear bijection $\mathbb{R}^3\to V$ , with inverse $x_i = \tfrac12\mathrm{tr}(\sigma_i X)$ (because $\mathrm{tr}(\sigma_i\sigma_j) = 2\delta_{ij}$ , which follows from the Clifford relation by taking traces). Moreover

\det(\mathbf{x}\cdot\boldsymbol{\sigma}) = -x_3^2 - (x_1-ix_2)(x_1+ix_2) = -|\mathbf{x}|^2 .

Step 2: conjugation preserves $V$ and the length. Let $U\in SU(2)$ and $X\in V$ . Then $(UXU^{\dagger})^{\dagger} = UX^{\dagger}U^{\dagger} = UXU^{\dagger}$ and $\mathrm{tr}(UXU^{\dagger}) = \mathrm{tr}(XU^{\dagger}U) = \mathrm{tr}X = 0$ by cyclicity of the trace, so $UXU^{\dagger}\in V$ . Here unitarity is used — for a general invertible $U$ the conjugate of a Hermitian matrix is not Hermitian. By multiplicativity of the determinant and $\det U\,\det U^{\dagger} = |\det U|^2 = 1$ ,

\det(UXU^{\dagger}) = \det X .

With step 1 this says $|R(U)\mathbf{x}| = |\mathbf{x}|$ for the real linear map $R(U)$ defined by $U(\mathbf{x}\cdot\boldsymbol{\sigma})U^{\dagger} = (R(U)\mathbf{x})\cdot\boldsymbol{\sigma}$ ; equivalently, $R(U)^{\mathsf{T}}R(U) = I$ , so $R(U)\in O(3)$ . Reading off the matrix entries with the inverse of step 1 gives the formula $R(U)_{ij} = \tfrac12\mathrm{tr}(\sigma_i U\sigma_jU^{\dagger})$ , and in particular $R(U)$ has real entries, as it must.

Step 3: $R$ is a homomorphism. For $U,W\in SU(2)$ ,

(UW)(\mathbf{x}\cdot\boldsymbol{\sigma})(UW)^{\dagger} = U\bigl(W(\mathbf{x}\cdot\boldsymbol{\sigma})W^{\dagger}\bigr)U^{\dagger} = U\bigl((R(W)\mathbf{x})\cdot\boldsymbol{\sigma}\bigr)U^{\dagger} = \bigl(R(U)R(W)\mathbf{x}\bigr)\cdot\boldsymbol{\sigma} ,

so $R(UW) = R(U)R(W)$ , and $R(I) = I$ .

Step 4: the image lies in $SO(3)$ . $\det R(U) \in\{\pm1\}$ because $R(U)$ is orthogonal. The map $U\mapsto\det R(U)$ is continuous, since the entries of $R(U)$ are polynomials in those of $U$ and $\bar U$ . By part 2, $SU(2)$ is path-connected, so the image of this continuous map is a connected subset of $\{\pm1\}$ , hence a single point, and that point is $\det R(I) = 1$ . Here connectedness of $SU(2)$ is used: continuity by itself forbids nothing, since on a disconnected group a continuous function with values in $\{\pm1\}$ can perfectly well take both of them — the determinant on $O(3)$ is the standard example. What rules out $\det R(U) = -1$ is that $SU(2)$ has no second component on which the other sign could sit. The same argument applies verbatim to the full unitary group $U(2)$ , which is also connected: the formula $R(U)_{ij} = \tfrac12\mathrm{tr}(\sigma_iU\sigma_jU^{\dagger})$ still makes sense there and still lands in $SO(3)$ , never on a reflection.

Step 5: the kernel. Suppose $R(U) = I$ , that is $U\sigma_jU^{\dagger} = \sigma_j$ , equivalently $U\sigma_j = \sigma_jU$ for $j=1,2,3$ . Write $U = \begin{pmatrix}p&q\\r&s\end{pmatrix}$ . Commuting with $\sigma_3$ forces $q = r = 0$ ; commuting with $\sigma_1$ then forces $p = s$ . So $U = pI$ with $\det U = p^2 = 1$ , hence $p=\pm1$ and $U = \pm I$ . Conversely $(\pm I)X(\pm I)^{\dagger} = X$ , so

\ker\Phi = \{I,-I\} .

This matches Hassani's Theorem 31.2.6 (Hassani (2013), p. 996), which states that the corresponding map $\mathrm{Spin}\to SO$ is surjective with kernel $\{1,-1\}$ .

Step 6: a rotation about the $z$ -axis. Take $\mathbf{n} = \mathbf{e}_3$ , so $U_{\mathbf{e}_3}(\theta) = \mathrm{diag}(e^{-i\theta/2},e^{i\theta/2})$ by part 2. Then

U\sigma_3U^{\dagger} = \sigma_3, \qquad U\sigma_1U^{\dagger} = \begin{pmatrix}0& e^{-i\theta}\\ e^{i\theta}&0\end{pmatrix} = \cos\theta\,\sigma_1 + \sin\theta\,\sigma_2 ,

and likewise $U\sigma_2U^{\dagger} = -\sin\theta\,\sigma_1 + \cos\theta\,\sigma_2$ . In the basis $(\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3)$ this is

R\bigl(U_{\mathbf{e}_3}(\theta)\bigr) = \begin{pmatrix}\cos\theta&-\sin\theta&0\\ \sin\theta&\cos\theta&0\\ 0&0&1\end{pmatrix} = R_{\mathbf{e}_3}(\theta) = \exp(-i\theta J_3) ,

the last equality by expanding $\exp(-i\theta J_3)$ as in part 2 (here $(-iJ_3)^3 = -(-iJ_3)$ , so the series again splits into $\cos$ and $\sin$ ). The same computation with $\mathbf{n}=\mathbf{e}_2$ gives $R\bigl(U_{\mathbf{e}_2}(\theta)\bigr) = R_{\mathbf{e}_2}(\theta)$ .

Step 7: surjectivity. By Hassani's (Hassani (2013), §30.3.3, p. 972) Euler-angle factorisation, every rotation can be written as

R = R_{\mathbf{e}_3}(\psi)\,R_{\mathbf{e}_2}(\vartheta)\,R_{\mathbf{e}_3}(\varphi)

for suitable angles. By step 6 each factor is in the image of $\Phi$ , and by step 3 the image is a subgroup, so $R$ is in the image. Hence $\Phi$ is onto $SO(3)$ .

(The alternative route is Riley, Hobson and Bence's (Riley, Hobson & Bence (2006), §28.7.3, pp. 1069–1070) remark that rotations through the same angle about different axes are conjugate, $R_j(\theta) = \phi_{ij}^{-1}R_i(\theta)\phi_{ij}$ : once all rotations about one axis are in the image and the image is a subgroup closed under conjugation by its own elements, one needs only that the conjugating elements are themselves reachable, which the Euler factorisation supplies.)

Step 8: two-to-one, and the general axis. By the first isomorphism theorem (homomorphisms and quotient groups), $SU(2)/\ker\Phi\cong\mathrm{im}\,\Phi = SO(3)$ , and the fibre of $\Phi$ over any $R$ is a coset of $\ker\Phi = \{\pm I\}$ , hence has exactly two elements $\{U,-U\}$ . Finally $\Phi(U_{\mathbf{n}}(\theta)) = R_{\mathbf{n}}(\theta)$ for every axis: choose $W\in SU(2)$ with $R(W)\mathbf{e}_3 = \mathbf{n}$ (possible by step 7), and note $W\,U_{\mathbf{e}_3}(\theta)\,W^{\dagger} = U_{\mathbf{n}}(\theta)$ because conjugation sends $\mathbf{e}_3\cdot\boldsymbol{\sigma}$ to $\mathbf{n}\cdot\boldsymbol{\sigma}$ and therefore commutes with the exponential; applying $\Phi$ and using step 3 gives $R(W)R_{\mathbf{e}_3}(\theta)R(W)^{-1} = R_{\mathbf{n}}(\theta)$ . $\square$

Proof(Part 4: spinors and the sign after

2\pi

)AdaptedAdapted from Arfken, Weber & Harris (2013), Eq. (17.59) and the discussion, p. 852

Put $\theta = 2\pi$ in the half-angle formula: $\cos\pi = -1$ and $\sin\pi = 0$ , so

U_{\mathbf{n}}(2\pi) = -I \qquad\text{for every axis } \mathbf{n},

while $R_{\mathbf{n}}(2\pi) = I$ . A spinor $\chi\in\mathbb{C}^2$ is therefore carried to $-\chi$ by a full turn and back to $\chi$ only after $\theta = 4\pi$ , which is Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), p. 852) observation that the angular range of the $SU(2)$ parameters is twice that of $SO(3)$ . For $\theta = \pi$ about the $x$ -axis, the same formula gives $U_{\mathbf{e}_1}(\pi) = -i\sigma_1$ , so

U_{\mathbf{e}_1}(\pi)\begin{pmatrix}1\\0\end{pmatrix} = -i\begin{pmatrix}0\\1\end{pmatrix} ,

up and down are exchanged, and the leftover factor $-i$ (Arfken, with his opposite sign convention, gets $+i$ ) is the phase that no measurement of the direction of the spin can detect.

Now let $T$ be a representation of $SU(2)$ on a vector space $W$ . If $T$ is to come from a representation of $SO(3)$ — that is, if there is a representation $T'$ of $SO(3)$ with $T = T'\circ\Phi$ — then $T(-I) = T'(\Phi(-I)) = T'(I) = \mathrm{id}_W$ , since $-I\in\ker\Phi$ . Conversely if $T(-I) = \mathrm{id}_W$ then $T$ is constant on the fibres $\{U,-U\}$ of $\Phi$ and so descends. Thus the representations of $SO(3)$ are precisely the representations of $SU(2)$ on which $-I$ acts trivially. In the labelling by $j$ of angular momentum algebra — Hassani (Hassani (2013), §30.3.3, p. 972) records that the irreducible representations are labelled by an integer or half-odd-integer $j$ and have dimension $2j+1$ — the element $-I = U_{\mathbf{e}_3}(2\pi)$ acts by $e^{-2\pi i m}$ on the state of weight $m$ , that is by $(-1)^{2j}$ , so $-I$ acts trivially exactly when $j$ is an integer. The half-integer representations are representations of $SU(2)$ only. $\square$

Remark(Where the hypotheses are used, and one disagreement between our sources)AdaptedAdapted from Hassani (2013), p. 998 and Thm 31.2.6, p. 996; Arfken, Weber & Harris (2013), p. 852

Unitarity of $U$ is what keeps $UXU^{\dagger}$ Hermitian (step 2). Drop it and the conjugation no longer acts on $\mathbb{R}^3$ at all. Keeping unitarity but dropping $\det U = 1$ gives the same map $\Phi$ on $U(2)$ , still onto $SO(3)$ , but now with kernel the full circle $\{e^{i\alpha}I\}$ : the covering is no longer two-sheeted, and the clean statement $SO(3)\cong SU(2)/\{\pm I\}$ is lost.
Tracelessness of $X$ is what makes $V$ three-dimensional. The Hermitian $2\times2$ matrices form a four-dimensional space, on which $\det$ is the Minkowski form of signature $(1,3)$ ; conjugating by $SL(2,\mathbb{C})$ instead of $SU(2)$ then gives the covering of the Lorentz group (the Lorentz group). Our theorem is the compact, Euclidean corner of that picture.
Connectedness of $SU(2)$ is used only in step 4, to rule out $\det R(U) = -1$ . It comes from part 2, where $SU(2)$ is identified with $S^3$ .
The equality of the Lie algebras does not give an isomorphism of groups. Part 1 and part 3 together are the sharpest form of the lesson of matrix Lie groups and Lie algebras: the exponential map is a local diffeomorphism, so $\mathfrak{so}(3)\cong\mathfrak{su}(2)$ says only that the two groups agree near the identity. Globally they differ, and the difference is exactly the two-element kernel. It is detectable: $SU(2)$ is the sphere $S^3$ , while $SO(3)$ is $S^3$ with antipodal points identified, and a path in $SO(3)$ from $I$ to $I$ running once through a $2\pi$ rotation lifts to a path in $SU(2)$ from $I$ to $-I$ , which is not a loop.
A disagreement in the library. Hassani (Hassani (2013), §31.2.1, p. 998) writes the chain " $\mathrm{Spin}(3,0)\cong SO(3)\cong SU(2)$ " after noting that the Pauli commutators are the same as the rotation-generator commutators. His own Theorem 31.2.6 (p. 996) states that the map $\mathrm{Spin}(\mu,\nu)\to SO(\mu,\nu)$ is a surjective homomorphism with kernel $\{1,-1\}$ , that is two-to-one and not injective, and Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 852) state the opposite of an isomorphism explicitly: the correspondence between the two groups is two-to-one, and what relates them is a homomorphism, not an isomorphism. We follow Arfken, Weber and Harris, and the proof of part 3 above establishes the two-to-one statement from scratch; the isomorphism that does hold is $SO(3)\cong SU(2)/\{\pm I\}$ , and it is an isomorphism of Lie algebras, not of groups, that follows from the equality of the structure constants. Readers using Hassani's chain of isomorphisms should read it as an isomorphism of Lie algebras together with the local isomorphism of groups that his Theorem 29.1.21 supplies.

Method(Passing between a rotation and its two

SU(2)

elements)Original

From axis and angle to $SU(2)$ . Given $\mathbf{n}$ (unit) and $\theta$ , write

U = I\cos\tfrac\theta2 - i(\mathbf{n}\cdot\boldsymbol{\sigma})\sin\tfrac\theta2 = \begin{pmatrix} \cos\frac\theta2 - in_3\sin\frac\theta2 & (-in_1-n_2)\sin\frac\theta2\\[2pt] (-in_1+n_2)\sin\frac\theta2 & \cos\frac\theta2+in_3\sin\frac\theta2\end{pmatrix}.

The other preimage is $-U$ , which is $U_{\mathbf{n}}(\theta+2\pi)$ — equivalently $U_{-\mathbf{n}}(2\pi - \theta)$ .

From $SU(2)$ to the rotation matrix. Use $R_{ij} = \tfrac12\mathrm{tr}(\sigma_i U\sigma_j U^{\dagger})$ , or, writing $U = \bigl(\begin{smallmatrix}a&-\bar b\\ b&\bar a\end{smallmatrix}\bigr)$ , read the angle and axis off from

\cos\tfrac\theta2 = \mathrm{Re}\,a, \qquad \sin\tfrac\theta2\,\mathbf{n} = \bigl(-\mathrm{Im}\,b,\ \mathrm{Re}\,b,\ -\mathrm{Im}\,a\bigr).

From a rotation matrix to axis and angle. $\mathrm{tr}\,R = 1+2\cos\theta$ gives $\theta\in[0,\pi]$ , and for $\theta\notin\{0,\pi\}$ the axis is read off from the antisymmetric part, $(R - R^{\mathsf{T}})_{pq} = -2\sin\theta\,\epsilon_{pqk}n_k$ . At $\theta = \pi$ the antisymmetric part vanishes and one uses $R + I = 2\,\mathbf{n}\mathbf{n}^{\mathsf{T}}$ instead.

Composing rotations. Multiply the $SU(2)$ matrices rather than the $3\times3$ ones: four complex entries instead of nine real ones, no accumulation of loss of orthogonality, and the half-angle form recovers the composite axis and angle directly. This is the quaternion method of quaternions, in matrix clothing.

Checks worth doing. $\det U = 1$ and $U^{\dagger}U = I$ ; $\mathrm{tr}R = 1+2\cos\theta$ ; $R\mathbf{n} = \mathbf{n}$ ; and $U_{\mathbf{n}}(\theta)^{N} = U_{\mathbf{n}}(N\theta)$ , which for $\theta = 2\pi/N$ gives $-I$ rather than $I$ , for every $N$ and every axis.

Example(A

120^\circ

rotation about the body diagonal)Original

Take the axis $\mathbf{n} = \tfrac{1}{\sqrt3}(1,1,1)$ and the angle $\theta = 2\pi/3$ , the rotation that cyclically permutes the coordinate axes of a cube.

Step 1: the $SU(2)$ element. Here $\cos(\theta/2) = \cos 60^\circ = \tfrac12$ and $\sin(\theta/2) = \sin 60^\circ = \tfrac{\sqrt3}{2}$ , and

\mathbf{n}\cdot\boldsymbol{\sigma} = \frac{1}{\sqrt3}\begin{pmatrix}1 & 1-i\\ 1+i & -1\end{pmatrix}.

The two factors of $\sqrt3$ cancel:

U = \tfrac12 I - i\cdot\tfrac{\sqrt3}{2}\cdot\tfrac{1}{\sqrt3}\begin{pmatrix}1&1-i\\1+i&-1\end{pmatrix} = \frac12\begin{pmatrix}1-i & -1-i\\ 1-i & 1+i\end{pmatrix} .

Step 2: check that $U\in SU(2)$ .

\det U = \tfrac14\bigl[(1-i)(1+i) - (-1-i)(1-i)\bigr] = \tfrac14\bigl[2 - (-1+i-i+i^2)\bigr] = \tfrac14[2+2] = 1 ,

and $U$ has the required shape $\bigl(\begin{smallmatrix}a&-\bar b\\ b&\bar a\end{smallmatrix}\bigr)$ with $a = \tfrac{1-i}{2}$ , $b = \tfrac{1-i}{2}$ , since $-\bar b = -\tfrac{1+i}{2}$ and $\bar a = \tfrac{1+i}{2}$ ; and $|a|^2+|b|^2 = \tfrac12+\tfrac12 = 1$ , so $U^{\dagger}U = I$ .

Step 3: the rotation matrix. Using $R_{ij} = \tfrac12\mathrm{tr}(\sigma_iU\sigma_jU^{\dagger})$ for the nine entries,

R = \begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}.

Read the columns: $R\mathbf{e}_1 = \mathbf{e}_2$ , $R\mathbf{e}_2 = \mathbf{e}_3$ , $R\mathbf{e}_3 = \mathbf{e}_1$ — the cyclic permutation, as expected for a third of a turn about $(1,1,1)$ . Three checks:

\mathrm{tr}\,R = 0 = 1 + 2\cos 120^\circ, \qquad \det R = 1, \qquad R\begin{pmatrix}1\\1\\1\end{pmatrix} = \begin{pmatrix}1\\1\\1\end{pmatrix} .

The first of these is the checkable number: $1 + 2\cos(2\pi/3) = 1 - 1 = 0$ .

Step 4: the two-to-one map, seen directly. Cube the $SU(2)$ element. Since $U = U_{\mathbf{n}}(2\pi/3)$ and the half-angle formula is a one-parameter group in $\theta$ ,

U^3 = U_{\mathbf{n}}(2\pi) = -I ,

and direct multiplication of the matrix in step 1 confirms it. But $R^3 = I$ , because a cyclic permutation of three symbols has order $3$ . So $U$ has order $6$ in $SU(2)$ while its image has order $3$ in $SO(3)$ : the preimage of the three-element subgroup $\{I,R,R^2\}$ is the six-element subgroup $\{\pm I,\pm U,\pm U^2\}$ , exactly twice as large, as the kernel $\{\pm I\}$ requires. The other preimage of $R$ is

-U = \frac12\begin{pmatrix}-1+i&1+i\\ -1+i&-1-i\end{pmatrix} = U_{\mathbf{n}}\!\left(\tfrac{2\pi}{3}+2\pi\right) = U_{\mathbf{n}}\!\left(\tfrac{8\pi}{3}\right).

Step 5: what a spinor does. Apply $U$ to the spin-up state $\chi_{\uparrow} = (1,0)^{\mathsf{T}}$ :

U\chi_{\uparrow} = \tfrac12\begin{pmatrix}1-i\\ 1-i\end{pmatrix},

whose components have equal modulus $\tfrac{1}{\sqrt2}$ , so the rotated state is an equal superposition of up and down — consistent with the fact that $R$ moves the $z$ -axis to the $x$ -axis, and a spin along $x$ is an equal superposition of up and down along $z$ . Applying $U$ three times returns $-\chi_{\uparrow}$ , not $\chi_{\uparrow}$ , although the physical orientation has returned to where it started.

ApplicationOriginal

Spin- $\tfrac12$ and magnetic resonance. The evolution of a spin- $\tfrac12$ particle in a static magnetic field $\mathbf{B}$ is $\chi(t) = \exp(-i\gamma t\,\mathbf{B}\cdot\boldsymbol{\sigma}/2)\chi(0)$ , which by part 2 is $U_{\mathbf{n}}(\gamma|\mathbf{B}|t)$ with $\mathbf{n} = \mathbf{B}/|\mathbf{B}|$ : the spin precesses about $\mathbf{B}$ at the Larmor frequency. Every pulse sequence in nuclear magnetic resonance and in magnetic-resonance imaging — a " $90^\circ$ pulse", a " $180^\circ$ pulse" — is an element $U_{\mathbf{n}}(\theta)$ , composed by multiplying $2\times2$ matrices. The same algebra is the single-qubit gate set of quantum computing: the Hadamard gate is $U_{\mathbf{n}}(\pi)$ for $\mathbf{n} = (1,0,1)/\sqrt2$ up to phase, and the Bloch sphere on which a qubit state is drawn is the image of $\mathbb{C}^2$ under $\chi\mapsto\chi^{\dagger}\boldsymbol{\sigma}\chi$ , intertwining the $SU(2)$ action with the $SO(3)$ action by exactly the homomorphism $\Phi$ .

Attitude representation. Spacecraft, aircraft, robot arms and camera-tracking software store orientations as unit quaternions, that is as elements of $SU(2)$ , rather than as $3\times3$ matrices or Euler angles: four numbers with one constraint instead of nine with six, composition by one quaternion multiplication, no gimbal lock, and renormalisation that is a single division by a norm. The price is the sign ambiguity of part 3 — $q$ and $-q$ describe the same attitude — which interpolation routines must handle explicitly by choosing the representative in the nearer hemisphere.

Observable consequences of the sign. That a $2\pi$ rotation multiplies a spinor by $-1$ is not an artefact of notation: in a neutron interferometer one arm is passed through a magnetic field that rotates the neutron spin by $2\pi$ , and the interference with the other arm is inverted, so the count rate has period $4\pi$ in the field strength. The same two-valuedness underlies the classification of particles as bosons and fermions and the behaviour of spinor fields under rotation in quantum field theory.

Crystallography and molecular symmetry. Double groups — the preimages in $SU(2)$ of the point groups of point groups and crystal symmetry — are what one must use to classify the electronic states of a magnetic ion with an odd number of electrons; Kramers' theorem, that such states come in degenerate pairs, is a statement about the two-sheeted cover.

Downstream in this chapter. The half-angle formula becomes the Wigner $D$ -matrices of rotation matrices of angular momentum once one works in the $(2j+1)$ -dimensional representations; the algebra $[S_j,S_k] = i\epsilon_{jkl}S_l$ is analysed abstractly in angular momentum algebra; and the Pauli matrices reappear as the three-dimensional case of the construction in Clifford algebras and spinors.

Theorem

Angular momentum algebra

Adapted

Add to PDF

From the commutation relations $[J_j,J_k]=i\epsilon_{jkl}J_l$ alone, derives the whole spectrum of angular momentum — $J^2$ has eigenvalue $j(j+1)$ with $2j$ a non-negative integer, $J_z$ runs in unit steps from $-j$ to $j$ , and the ladder operators act with coefficients $\sqrt{j(j+1)-m(m\pm1)}$ — and shows by an explicit counterexample which hypothesis forces the ladder to terminate.

Prerequisites: The rotation group and SU(2), Matrix Lie groups and Lie algebras, Inner products and norms, Spectral theorem for Hermitian matrices, Eigenvalues and eigenvectors

Definition(An angular momentum)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (16.28)–(16.29), p. 782

Let $V$ be a complex vector space with an inner product (inner-product spaces), linear in the second slot in the physicists' way, so that $\langle u|Au\rangle = \langle A^{\dagger}u|u\rangle$ . An angular momentum on $V$ is a triple $J_1,J_2,J_3$ of Hermitian operators on $V$ satisfying

[J_j,J_k] = i\,\epsilon_{jkl}\,J_l \qquad (j,k = 1,2,3),

with $\epsilon$ the Levi-Civita symbol and summation over $l$ . We write $J_x,J_y,J_z$ for $J_1,J_2,J_3$ and set

J^2 = J_1^2+J_2^2+J_3^2, \qquad J_{\pm} = J_x \pm i J_y .

Units with $\hbar = 1$ are used throughout; to restore them, multiply every $J_k$ by $\hbar$ , so that $[J_x,J_y] = i\hbar J_z$ and $J^2$ has the dimensions of $\hbar^2$ , as in Riley, Hobson and Bence (Riley, Hobson & Bence (2006), Eq. (19.27), p. 659).

Nothing else is assumed: $V$ may be any space on which three such operators act. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 782) stress this point — beyond the commutation relations the operators are arbitrary, and the three standard realisations are the orbital angular momentum $\mathbf{L} = -i\,\mathbf{r}\times\nabla$ on functions, the spin $\mathbf{S} = \tfrac12\boldsymbol{\sigma}$ on $\mathbb{C}^2$ of the rotation group and SU(2), and sums of these.

By that unit, the relations above are exactly the structure constants of $\mathfrak{so}(3)\cong\mathfrak{su}(2)$ in the Hermitian normalisation, so "an angular momentum on $V$ " and "a representation of the Lie algebra of the rotation group on $V$ " are the same thing, with the group element $\exp(-i\theta\,\mathbf{n}\cdot\mathbf{J})$ implementing the rotation by $\theta$ about $\mathbf{n}$ .

$J_{\pm}$ are ladder operators. They are not Hermitian; rather $(J_+)^{\dagger} = J_-$ , since $J_x$ and $J_y$ are Hermitian. Riley, Hobson and Bence (Riley, Hobson & Bence (2006), p. 660) write them $U$ and $D$ for "up" and "down".

Theorem(The spectrum of an angular momentum)AdaptedAdapted from Arfken, Weber & Harris (2013), pp. 778–779 and Eqs. (16.28)–(16.31), p. 782; Riley, Hobson & Bence (2006), p. 662 and Eqs. (19.34)–(19.35), p. 663; Hassani (2013), Example 30.3.6, p. 970

Let $J_1,J_2,J_3$ be an angular momentum on a finite-dimensional complex inner-product space $V\neq\{0\}$ . Then:

$J^2$ is a Casimir operator. $[J^2,J_k] = 0$ for $k=1,2,3$ , so $J^2$ and $J_z$ have a common eigenbasis. Moreover

[J_z,J_{\pm}] = \pm J_{\pm}, \qquad [J_+,J_-] = 2J_z, \qquad J_{\mp}J_{\pm} = J^2 - J_z^2 \mp J_z .

Quantisation. If $V$ is irreducible — it has no proper non-zero subspace invariant under all three $J_k$ — then there is a number $j$ with $2j \in \{0,1,2,3,\dots\}$ such that

J^2 = j(j+1)\,\mathrm{Id}_V, \qquad \dim V = 2j+1 ,

and $V$ has an orthonormal basis $|j,m\rangle$ with $m = -j, -j+1,\dots, j-1, j$ and

J_z|j,m\rangle = m\,|j,m\rangle, \qquad J^2|j,m\rangle = j(j+1)\,|j,m\rangle .

In particular the eigenvalues of $J_z$ are all integers or all half-odd-integers, never mixed, and they are non-degenerate within one irreducible $V$ .

Matrix elements. With the phases fixed by the Condon-Shortley convention — the coefficients below are taken real and non-negative —

J_{\pm}|j,m\rangle = \sqrt{\,j(j+1) - m(m\pm1)\,}\;|j,m\pm1\rangle = \sqrt{(j\mp m)(j\pm m+1)}\;|j,m\pm1\rangle ,

so that $J_+|j,j\rangle = 0$ and $J_-|j,-j\rangle = 0$ . Equivalently

\langle j,m'|J_{\pm}|j,m\rangle = \sqrt{(j\mp m)(j\pm m+1)}\,\delta_{m',m\pm1}, \qquad \langle j,m'|J_z|j,m\rangle = m\,\delta_{m'm} ,

and $J_x = \tfrac12(J_++J_-)$ , $J_y = \tfrac{1}{2i}(J_+-J_-)$ supply the remaining matrices.

Decomposition. A general finite-dimensional $V$ is an orthogonal direct sum of irreducible subspaces, so it is described by the list of its $j$ values with multiplicities.

Hassani (Hassani (2013), Example 30.3.6, p. 970) reaches parts 2 and 4 from the Lie-theoretic side: $\mathfrak{so}(3)$ has rank one, $J^2$ is its one quadratic Casimir, and by Chevalley's theorem its eigenvalue alone labels the irreducible representations, the $j$ th having dimension $2j+1$ .

Idea of the proofOriginal

Only two facts drive everything, and both are one-line consequences of the commutation relations.

The first is that $J_{\pm}$ shifts the $J_z$ eigenvalue by $\pm1$ and leaves the $J^2$ eigenvalue alone. That is the content of $[J_z,J_\pm] = \pm J_\pm$ and $[J^2,J_\pm]=0$ : applying $J_+$ to an eigenvector moves it one rung up a ladder of states all sharing the same $J^2$ .

The second is that the ladder cannot be infinite, and this is where Hermiticity enters. The number $\|J_+v\|^2$ is a norm, hence non-negative, and the identity $J_-J_+ = J^2 - J_z^2 - J_z$ evaluates it as $\lambda - m^2 - m$ for a state of $J^2$ -eigenvalue $\lambda$ and $J_z$ -eigenvalue $m$ . That expression turns negative for large $m$ , so the ladder must stop; and it can only stop by the coefficient vanishing exactly, which pins $\lambda = m_{\max}(m_{\max}+1)$ . The same argument at the bottom gives $\lambda = m_{\min}(m_{\min}-1)$ , and the two together force $m_{\min} = -m_{\max}$ . Since the rungs are spaced by $1$ , the distance $2m_{\max}$ from bottom to top is a whole number — which is why $j$ is an integer or a half-integer and nothing else.

Everything after that is bookkeeping: the same norm computation that forced termination also supplies the coefficients.

Proof(Part 1: the commutator identities)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (16.11), (16.16)–(16.17), (16.20)–(16.22), pp. 776–778

Step 1: $[J^2,J_k]=0$ . Use $[AB,C] = A[B,C]+[A,C]B$ . For $k=z$ ,

[J^2,J_z] = [J_x^2,J_z]+[J_y^2,J_z]+[J_z^2,J_z] = J_x[J_x,J_z] + [J_x,J_z]J_x + J_y[J_y,J_z] + [J_y,J_z]J_y .

Now $[J_x,J_z] = -iJ_y$ and $[J_y,J_z] = iJ_x$ , so the four terms are $-iJ_xJ_y - iJ_yJ_x + iJ_yJ_x + iJ_xJ_y = 0$ . The cases $k=x,y$ follow by cycling the labels, which the relations permit. Riley, Hobson and Bence (Riley, Hobson & Bence (2006), p. 660) carry out the same cancellation.

Step 2: $[J_z,J_{\pm}] = \pm J_{\pm}$ .

[J_z,J_\pm] = [J_z,J_x] \pm i[J_z,J_y] = iJ_y \pm i(-iJ_x) = \pm\bigl(J_x\pm iJ_y\bigr) = \pm J_\pm .

Step 3: $[J_+,J_-] = 2J_z$ . Expanding, $[J_x+iJ_y,\;J_x-iJ_y] = -i[J_x,J_y] + i[J_y,J_x] = -2i[J_x,J_y] = -2i(iJ_z) = 2J_z$ .

Step 4: $J_{\mp}J_{\pm} = J^2 - J_z^2 \mp J_z$ . Multiply out:

J_-J_+ = (J_x-iJ_y)(J_x+iJ_y) = J_x^2+J_y^2 + i[J_x,J_y] = (J^2 - J_z^2) - J_z ,

using $[J_x,J_y]=iJ_z$ in the last step; the sign of the $iJ_y$ terms reverses for $J_+J_-$ , giving $J^2-J_z^2+J_z$ . Here the definition $J^2 = J_x^2+J_y^2+J_z^2$ is used and nothing else.

Step 5: $[J^2,J_\pm]=0$ is immediate from step 1 and the definition of $J_\pm$ . $\square$

Proof(Parts 2 and 3: the spectrum and the matrix elements)AdaptedAdapted from Arfken, Weber & Harris (2013), pp. 778–779; Riley, Hobson & Bence (2006), p. 662

Let $V$ be finite-dimensional, non-zero and irreducible.

Step 1: a simultaneous eigenvector exists. $J^2$ is Hermitian, being a sum of squares of Hermitian operators, so by the spectral theorem for Hermitian matrices it has an eigenvalue $\lambda$ , necessarily real, with eigenspace $W\neq\{0\}$ . By step 1 of part 1 each $J_k$ maps $W$ into itself, so $W$ is invariant under all three; irreducibility forces $W = V$ , that is

J^2 = \lambda\,\mathrm{Id}_V .

Also $\lambda\ge0$ , since $\langle v|J^2v\rangle = \|J_xv\|^2+\|J_yv\|^2+\|J_zv\|^2 \ge 0$ . Now $J_z$ is Hermitian on $V\ne\{0\}$ , so it has a real eigenvalue $m_0$ with a unit eigenvector $v_0$ .

Step 2: the ladder. Let $v$ be a unit vector with $J_zv = mv$ . Then by step 2 of part 1,

J_z\bigl(J_\pm v\bigr) = J_\pm J_z v \pm J_\pm v = (m\pm1)\,J_\pm v ,

so $J_\pm v$ is either zero or a $J_z$ -eigenvector of eigenvalue $m\pm1$ . It remains in the same $J^2$ -eigenspace because $J^2$ is a multiple of the identity.

Step 3: the key norm. Using $(J_+)^{\dagger}=J_-$ and step 4 of part 1,

\|J_+v\|^2 = \langle J_+v|J_+v\rangle = \langle v|J_-J_+v\rangle = \langle v|(J^2-J_z^2-J_z)v\rangle = \lambda - m^2 - m ,

and symmetrically

\|J_-v\|^2 = \lambda - m^2 + m .

Here Hermiticity of $J_x$ and $J_y$ is used, to identify $(J_+)^{\dagger}$ with $J_-$ ; without it the left-hand sides are not norms and the inequalities below are unavailable.

Step 4: the ladder terminates at both ends. Starting from $v_0$ and applying $J_+$ repeatedly produces eigenvectors of eigenvalues $m_0, m_0+1, m_0+2,\dots$ as long as they are non-zero. Eigenvectors of a Hermitian operator for distinct eigenvalues are orthogonal, hence independent, so in a finite-dimensional $V$ the process must produce zero after finitely many steps. (Equivalently and without using finite dimension: $\lambda - m^2 - m \ge 0$ bounds $m$ above.) Let $m_{\max}$ be the largest $m$ reached, with unit eigenvector $v_{\max}$ , so $J_+v_{\max}=0$ and by step 3

\lambda = m_{\max}(m_{\max}+1) .

The same argument downwards gives a smallest $m_{\min}$ with $J_-v_{\min}=0$ and

\lambda = m_{\min}(m_{\min}-1) .

Step 5: $m_{\min} = -m_{\max}$ , and $2m_{\max}$ is a non-negative integer. Write $j = m_{\max}$ , so $\lambda = j(j+1)$ with $j \ge 0$ (the other root of $\lambda = x(x+1)$ with $\lambda\ge0$ is negative, and $m_{\max}\ge m_{\min}$ forces the non-negative one). Solving $m_{\min}(m_{\min}-1) = j(j+1)$ gives $m_{\min} = -j$ or $m_{\min} = j+1$ ; the latter exceeds $m_{\max}$ and is impossible, so $m_{\min} = -j$ . Arfken, Weber and Harris (Arfken, Weber & Harris (2013), pp. 778–779) dispose of the spurious root the same way. Since the rungs from $m_{\min}$ to $m_{\max}$ are spaced by exactly $1$ , their number is $m_{\max}-m_{\min}+1 = 2j+1$ , a positive integer; hence $2j$ is a non-negative integer and $j\in\{0,\tfrac12,1,\tfrac32,\dots\}$ .

Riley, Hobson and Bence (Riley, Hobson & Bence (2006), p. 662) obtain the quantisation by a direct count instead: if the top eigenvalue is $c$ and the bottom is reached after $n$ lowerings, then $c = m_{\max}$ and $c-n = m_{\min}$ , and equating the two expressions for $\lambda$ gives $c^2+c = (c-n)^2-(c-n)$ , which simplifies to $2c(n+1) = n(n+1)$ and hence $c = \tfrac12 n$ . The integer $n$ is the number of steps; our $2j$ is their $n$ .

Step 6: the states span $V$ , and the coefficients. Define

|j,j\rangle = v_{\max}, \qquad |j,m-1\rangle = \frac{1}{\sqrt{\lambda - m^2+m}}\;J_-|j,m\rangle \quad (m = j, j-1,\dots,-j+1),

which is legitimate because step 3 makes the denominator the norm of $J_-|j,m\rangle$ , positive for $m>-j$ . These $2j+1$ vectors are unit vectors, mutually orthogonal (distinct $J_z$ -eigenvalues), and their span is invariant under $J_z$ , under $J_-$ by construction, and under $J_+$ because $J_+J_-|j,m\rangle = (\lambda-m^2+m)|j,m\rangle$ after re-indexing; hence invariant under $J_x$ and $J_y$ too. Irreducibility forces the span to be all of $V$ , so $\dim V = 2j+1$ and each $J_z$ -eigenvalue occurs once.

By construction $J_-|j,m\rangle = \sqrt{\lambda-m^2+m}\,|j,m-1\rangle = \sqrt{j(j+1)-m(m-1)}\,|j,m-1\rangle$ , and applying $J_+$ to this identity and using $J_+J_- = J^2-J_z^2+J_z$ gives $J_+|j,m-1\rangle = \sqrt{j(j+1)-m(m-1)}\,|j,m\rangle$ , which is the stated formula with $m-1$ in place of $m$ . The two forms of the coefficient agree because

j(j+1)-m(m\pm1) = (j\mp m)(j\pm m+1) ,

as expanding both sides confirms. The choice of positive square roots is the Condon-Shortley convention; Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 779) note that it is a choice of the relative signs of the $|j,m\rangle$ , made so that the spherical harmonics of spherical harmonics come out with their tabulated signs.

Part 4. If $V$ is not irreducible it has a non-zero invariant subspace $W\subsetneq V$ . Because every $J_k$ is Hermitian, $W^{\perp}$ is invariant too: for $u\in W^{\perp}$ and $w\in W$ , $\langle w|J_ku\rangle = \langle J_kw|u\rangle = 0$ since $J_kw\in W$ . So $V = W\oplus W^{\perp}$ orthogonally, with both summands invariant, and induction on the dimension finishes the decomposition. $\square$

Remark(Where the hypotheses are used)Original

Hermiticity is the hypothesis that quantises the spectrum, and it enters only in step 3, through $(J_+)^{\dagger}=J_-$ . Dropping it destroys the conclusion completely. On the space with basis $\{e_n : n\in\mathbb{Z}\}$ fix complex numbers $\lambda$ and $\mu$ and define

J_ze_n = (\mu+n)e_n, \qquad J_+e_n = e_{n+1}, \qquad J_-e_n = \bigl(\lambda-(\mu+n)(\mu+n-1)\bigr)e_{n-1} .

Then $[J_z,J_\pm] = \pm J_\pm$ by inspection, and

(J_+J_- - J_-J_+)e_n = \Bigl[\bigl(\lambda-(\mu+n)(\mu+n-1)\bigr) - \bigl(\lambda-(\mu+n+1)(\mu+n)\bigr)\Bigr]e_n = 2(\mu+n)e_n = 2J_ze_n ,

so $[J_+,J_-]=2J_z$ and the full set of angular momentum relations holds for $J_x = \tfrac12(J_++J_-)$ , $J_y = \tfrac1{2i}(J_+-J_-)$ , $J_z$ . One checks likewise that $J_-J_++J_z^2+J_z = \lambda\,\mathrm{Id}$ , so " $J^2$ " is the constant $\lambda$ — an arbitrary complex number, not of the form $j(j+1)$ with $2j$ an integer — and the ladder never terminates in either direction. No inner product on this space can make all three operators Hermitian, and the reason is exactly the step that is missing: were they Hermitian, $J_z$ would have real eigenvalues, forcing $\mu$ real, and $J^2 = \lambda\,\mathrm{Id}$ would be Hermitian, forcing $\lambda$ real; but then

\|J_+e_n\|^2 = \langle e_n|J_-J_+e_n\rangle = \bigl(\lambda-(\mu+n)(\mu+n+1)\bigr)\,\|e_n\|^2

is negative for all large $n$ , which is impossible.

Finite dimension is used in step 4 to stop the ladder, and in step 6 to conclude that the states span. It can be traded for compactness of the group — Hassani (Hassani (2013), §30.3.3, p. 972) notes that the irreducible representations of $\mathfrak{so}(3)$ are finite-dimensional because $SO(3)$ is compact — or for the boundedness of $\lambda-m^2-m$ noted parenthetically in step 4, which is what one uses for the infinite-dimensional spaces of quantum mechanics. There the operators are unbounded and the correct statement needs the domain conditions of symmetric and self-adjoint operators; the algebra above is then valid on a suitable dense invariant domain.
Irreducibility is what makes $J^2$ a number in step 1. Without it, $J^2$ is only block-diagonal, and part 4 says that the general case is a direct sum of the irreducible ones — so one does not lose anything, only the right to speak of "the" value of $j$ .
Half-integer $j$ is permitted by the algebra and excluded by geometry. Nothing in the proof rules out $j=\tfrac12$ . What rules it out for orbital angular momentum is that $L_z = -i\,\partial/\partial\varphi$ acting on single-valued functions on the sphere has eigenfunctions $e^{im\varphi}$ with $m$ an integer. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 779) make the same distinction: the half-integral multiplets are consistent with the commutation rules but cannot be realised as functions in three-dimensional space. In the language of the rotation group and SU(2), half-integer $j$ gives a representation of $SU(2)$ that is not a representation of $SO(3)$ .
$\hbar$ . With units restored, $J^2$ has eigenvalue $j(j+1)\hbar^2$ , $J_z$ has $m\hbar$ , and the ladder coefficients carry one factor of $\hbar$ . The quantum number $j$ is dimensionless; $\hbar$ never appears inside a square root in a way that could change the arithmetic.
The two ends of the ladder must be consistent. It is the pair of conditions in step 4 that forces quantisation. A single one would only relate $\lambda$ to one endpoint and would leave $j$ arbitrary; the integrality comes from the two endpoints being separated by a whole number of unit steps.

Method(Building the matrices of a spin-

j

multiplet)Original

To write down the $(2j+1)\times(2j+1)$ matrices for a given $j$ :

Order the basis $|j,j\rangle, |j,j-1\rangle,\dots,|j,-j\rangle$ , so that row and column $k$ (counting from $1$ ) correspond to $m = j-k+1$ .
$J_z$ is diagonal with entries $j, j-1,\dots,-j$ .
$J_+$ is strictly upper bidiagonal: its only non-zero entries sit just above the diagonal, in column $k$ and row $k-1$ , with value $\sqrt{(j-m)(j+m+1)}$ where $m = j-k+1$ is the $m$ of column $k$ . Equivalently, read off the numbers $\sqrt{2j},\sqrt{2(2j-1)},\sqrt{3(2j-2)},\dots$ from the top.
$J_- = (J_+)^{\mathsf{T}}$ , since the coefficients are real; it is strictly lower bidiagonal with the same numbers.
$J_x = \tfrac12(J_++J_-)$ and $J_y = \tfrac{1}{2i}(J_+-J_-)$ , so $J_x$ is real symmetric tridiagonal with zero diagonal and $J_y$ is purely imaginary.
Check. Verify $J^2 = j(j+1)I$ , verify $[J_x,J_y] = iJ_z$ on one entry, and verify $\mathrm{tr}\,J_z^2 = \tfrac13 j(j+1)(2j+1)$ , which is the sum $\sum_{m=-j}^{j}m^2$ .
For a rotation, exponentiate: $\exp(-i\theta\,\mathbf{n}\cdot\mathbf{J})$ gives the Wigner matrices of rotation matrices of angular momentum; for $\mathbf{n}=\mathbf{e}_3$ this is just the diagonal matrix $\mathrm{diag}(e^{-ij\theta},\dots,e^{ij\theta})$ .

Example(The spin-

\tfrac32

multiplet, built from the commutators alone)Original

Take $j = \tfrac32$ , so $\dim V = 2j+1 = 4$ and the basis is $|{\tfrac32},{\tfrac32}\rangle, |{\tfrac32},{\tfrac12}\rangle, |{\tfrac32},{-\tfrac12}\rangle, |{\tfrac32},{-\tfrac32}\rangle$ in that order.

Step 1: the ladder coefficients. With $j(j+1) = \tfrac32\cdot\tfrac52 = \tfrac{15}{4}$ , the formula $\sqrt{j(j+1)-m(m-1)}$ for $J_-$ gives, for $m = \tfrac32,\tfrac12,-\tfrac12$ ,

\sqrt{\tfrac{15}{4}-\tfrac32\cdot\tfrac12} = \sqrt{3}, \qquad \sqrt{\tfrac{15}{4}-\tfrac12\cdot\bigl(-\tfrac12\bigr)} = \sqrt{4} = 2, \qquad \sqrt{\tfrac{15}{4}-\bigl(-\tfrac12\bigr)\bigl(-\tfrac32\bigr)} = \sqrt{3} .

Step 2: the matrices.

J_z = \tfrac12\begin{pmatrix}3&&&\\&1&&\\&&-1&\\&&&-3\end{pmatrix}, \qquad J_+ = \begin{pmatrix}0&\sqrt3&0&0\\0&0&2&0\\0&0&0&\sqrt3\\0&0&0&0\end{pmatrix}, \qquad J_- = J_+^{\mathsf{T}} ,

J_x = \frac12\begin{pmatrix}0&\sqrt3&0&0\\ \sqrt3&0&2&0\\ 0&2&0&\sqrt3\\ 0&0&\sqrt3&0\end{pmatrix}, \qquad J_y = \frac{i}{2}\begin{pmatrix}0&-\sqrt3&0&0\\ \sqrt3&0&-2&0\\ 0&2&0&-\sqrt3\\ 0&0&\sqrt3&0\end{pmatrix}.

Step 3: three checks.

$J^2 = J_x^2+J_y^2+J_z^2$ . The $(1,1)$ entry is $\tfrac14\cdot3 + \tfrac14\cdot3 + \tfrac94 = \tfrac34+\tfrac34+\tfrac94 = \tfrac{15}{4}$ , and carrying out the full multiplication gives $J^2 = \tfrac{15}{4}I$ exactly, as $j(j+1) = \tfrac{15}{4}$ requires.
$[J_x,J_y] = iJ_z$ . The commutator comes out diagonal with entries $\tfrac{3i}{2},\tfrac{i}{2},-\tfrac{i}{2},-\tfrac{3i}{2}$ , which is $i$ times the diagonal of $J_z$ . The $(1,1)$ entry, for instance, is $(J_xJ_y)_{11}-(J_yJ_x)_{11} = \tfrac{\sqrt3}{2}\cdot\tfrac{\sqrt3 i}{2} - \bigl(-\tfrac{\sqrt3 i}{2}\bigr)\cdot\tfrac{\sqrt3}{2} = \tfrac{3i}{4}+\tfrac{3i}{4} = \tfrac{3i}{2} = i\cdot\tfrac32$ .
$\mathrm{tr}\,J_z^2 = 2\bigl(\tfrac94+\tfrac14\bigr) = 5$ , which matches $\tfrac13 j(j+1)(2j+1) = \tfrac13\cdot\tfrac32\cdot\tfrac52\cdot4 = 5$ .

Step 4: a number to check the ladder against. Climbing the whole ladder from the top in one go,

(J_-)^3\,\bigl|{\tfrac32},{\tfrac32}\bigr\rangle = \sqrt3\cdot2\cdot\sqrt3\;\bigl|{\tfrac32},{-\tfrac32}\bigr\rangle = 6\,\bigl|{\tfrac32},{-\tfrac32}\bigr\rangle ,

and indeed the matrix $(J_-)^3$ has a single non-zero entry, the $(4,1)$ entry $6$ . A fourth application gives zero, as it must: $J_-|{\tfrac32},{-\tfrac32}\rangle = 0$ .

Step 5: the spectrum of $J_x$ is the same as that of $J_z$ . It has to be, since $J_x$ is carried to $J_z$ by the rotation of the rotation group and SU(2) that takes $\mathbf{e}_1$ to $\mathbf{e}_3$ , and conjugation preserves eigenvalues. Computing the characteristic polynomial of the matrix $J_x$ above gives

\det(J_x - tI) = t^4 - \tfrac52 t^2 + \tfrac{9}{16} = \bigl(t^2-\tfrac94\bigr)\bigl(t^2-\tfrac14\bigr) ,

with roots $\pm\tfrac32, \pm\tfrac12$ — the same four numbers. A measurement of the spin component along any axis gives one of $\pm\tfrac32,\pm\tfrac12$ , and only the probabilities depend on the axis.

Step 6: the special case $j=\tfrac12$ . Running the same recipe with $j=\tfrac12$ gives $J_+ = \bigl(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\bigr)$ , $J_z = \tfrac12\bigl(\begin{smallmatrix}1&0\\0&-1\end{smallmatrix}\bigr)$ and $J_k = \tfrac12\sigma_k$ , with $J^2 = \tfrac34 I$ — Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), Example 16.1.2, pp. 780–781) spinor example, and the Pauli matrices of the rotation group and SU(2) recovered from the commutation relations with no geometry at all.

ApplicationOriginal

Atomic and molecular spectra. The electronic states of an atom are labelled by $j$ and $m$ because the Hamiltonian commutes with $\mathbf{J}$ ; the $2j+1$ states of a level are degenerate until a field singles out an axis. In the Zeeman effect a magnetic field $B\mathbf{e}_3$ adds $\gamma B J_z$ to the Hamiltonian, splitting the level into $2j+1$ equally spaced lines, and simply counting those lines measures $j$ . This is the experiment that forced half-integer $j$ on physics: silver atoms in a Stern-Gerlach apparatus split into two beams, not three.

Magnetic resonance and quantum control. Nuclear magnetic resonance, electron spin resonance and magnetic-resonance imaging all work in a single $j$ -multiplet, usually $j=\tfrac12$ or $j=\tfrac32$ (for quadrupolar nuclei), and every pulse sequence is a product of operators $\exp(-i\theta\,\mathbf{n}\cdot\mathbf{J})$ built from the matrices of the method above. The ladder coefficients are exactly the transition amplitudes between adjacent levels and therefore fix the relative line intensities.

Quantum information. A spin- $j$ system is a $(2j+1)$ -level "qudit"; the matrices of the example are the standard generators used to describe its gates. The $j=\tfrac12$ case is the qubit, and the fact that the spectrum of $\mathbf{n}\cdot\mathbf{J}$ is independent of $\mathbf{n}$ (step 5 of the example) is why any measurement direction yields the same set of outcomes with different probabilities — the geometry of the Bloch sphere.

Selection rules. Because the components of a vector operator shift $m$ by $0$ or $\pm1$ and nothing else, only certain transitions have non-zero amplitude. Turning that observation into a theorem — with every matrix element of a tensor operator reduced to a single number times a Clebsch-Gordan coefficient — is the Wigner-Eckart theorem of spherical tensors and the Wigner-Eckart theorem, and the coupling machinery it needs is built in addition of angular momenta.

Beyond rotations. The same algebra appears wherever a rank-one Lie algebra does. The raising and lowering operators of the harmonic oscillator obey the closely related Heisenberg relations; the isospin of nuclear physics is a copy of this algebra acting on the proton-neutron doublet; and in $\mathfrak{su}(3)$ three copies of it, the $I$ -, $U$ - and $V$ -spins, organise the quark multiplets. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), pp. 855–856) run the $\mathfrak{su}(3)$ ladders in exactly the way done here.

Rigid rotors and molecular spectroscopy. The rotational levels of a diatomic molecule have energies $B\,j(j+1)$ , so the observed line spacing is $2B(j+1)$ — a linear progression whose slope measures the moment of inertia and hence the bond length. The $j(j+1)$ of the theorem is directly visible in a microwave spectrum.

Method

Addition of angular momenta

Adapted

Add to PDF

Gives the recipe for passing between the uncoupled basis $|j_1m_1\rangle|j_2m_2\rangle$ and the coupled basis $|J,M\rangle$ — the triangle rule $|j_1-j_2|\le J\le j_1+j_2$ , the ladder-and-orthogonalise construction, the closed formula and the $3j$ -symbol symmetries — with the Clebsch-Gordan series proved twice, by counting states and by multiplying characters.

Prerequisites: Angular momentum algebra, Product representations and selection rules, The rotation group and SU(2), Orthogonality relations and characters, Orthonormal bases and Gram-Schmidt, Inner products and norms

Definition(The two bases and the coefficients)AdaptedAdapted from Arfken, Weber & Harris (2013), pp. 784–786 and Eqs. (16.43)–(16.45), pp. 789–790; Hassani (2013), Eqs. (24.46)–(24.47), p. 756

Let $\mathbf{j}_1$ and $\mathbf{j}_2$ be two angular momenta in the sense of angular momentum algebra, acting on spaces $V_1$ and $V_2$ and commuting with each other:

[j_{1k},\,j_{2l}] = 0 \qquad\text{for all } k,l\in\{1,2,3\} .

This holds automatically when they act on different tensor factors, $j_{1k}\otimes I$ and $I\otimes j_{2l}$ , which is the situation of two particles or of the orbit and the spin of one particle. The total angular momentum is

\mathbf{J} = \mathbf{j}_1+\mathbf{j}_2, \qquad J_k = j_{1k}+j_{2k} ,

and it is again an angular momentum: $[J_k,J_l] = [j_{1k},j_{1l}] + [j_{2k},j_{2l}] = i\epsilon_{klm}(j_{1m}+j_{2m}) = i\epsilon_{klm}J_m$ , the cross terms vanishing by the commuting hypothesis.

Fix irreducible $V_1$ and $V_2$ of spins $j_1$ and $j_2$ , of dimensions $2j_1+1$ and $2j_2+1$ , and work on $V = V_1\otimes V_2$ , of dimension $(2j_1+1)(2j_2+1)$ .

The uncoupled basis diagonalises $j_1^2, j_2^2, j_{1z}, j_{2z}$ :

|j_1 m_1; j_2 m_2\rangle := |j_1,m_1\rangle\otimes|j_2,m_2\rangle, \qquad -j_i \le m_i \le j_i .

The coupled basis diagonalises $j_1^2, j_2^2, J^2, J_z$ and is written $|J,M\rangle$ (with $j_1,j_2$ understood).

Both are orthonormal bases of $V$ . The Clebsch-Gordan coefficients are the entries of the change of basis,

|J,M\rangle = \sum_{m_1,m_2} C(j_1 j_2 J\,|\,m_1 m_2 M)\;|j_1m_1;j_2m_2\rangle , \qquad C(j_1 j_2 J\,|\,m_1 m_2 M) = \langle j_1m_1;j_2m_2\,|\,J,M\rangle .

The sum looks double but is single: the coefficient vanishes unless $m_1+m_2 = M$ .

Hassani (Hassani (2013), Eqs. (24.46)–(24.47), p. 756) defines the same objects for an arbitrary group $G$ , as the coefficients $C(\alpha\beta;\sigma,q|ij;k)$ expanding a basis of the tensor product in vectors transforming by the irreducible representations, with a further label $q$ to separate repeated copies. For $SU(2)$ no copy is ever repeated — the group is simply reducible in Hassani's terminology — so $q$ is unnecessary. That is part 1 of the statement below.

Method(The Clebsch-Gordan series and the properties of the coefficients)AdaptedAdapted from Arfken, Weber & Harris (2013), Eq. (16.37), p. 787, Eqs. (16.43)–(16.49) and Table 16.1, pp. 789–791; Hassani (2013), Eq. (30.17), p. 973, and p. 974

With the notation above:

The series (triangle rule). $V_1\otimes V_2$ decomposes into irreducible subspaces, one for each

J = |j_1-j_2|,\ |j_1-j_2|+1,\ \dots,\ j_1+j_2 ,

each occurring exactly once. In the notation of representations, $D^{(j_1)}\otimes D^{(j_2)} \cong \bigoplus_{J=|j_1-j_2|}^{j_1+j_2} D^{(J)}$ , and the dimensions check:

\sum_{J=|j_1-j_2|}^{j_1+j_2}(2J+1) = (2j_1+1)(2j_2+1) .

Selection rules. $C(j_1j_2J|m_1m_2M) = 0$ unless $M = m_1+m_2$ and $|j_1-j_2|\le J\le j_1+j_2$ with $j_1+j_2+J$ an integer.
Reality, orthogonality and inversion. With the Condon-Shortley phase convention — which fixes the signs by requiring $C(j_1j_2J|j_1,J-j_1,J) > 0$ for every $J$ — all the coefficients are real, the transformation matrix is orthogonal, and

\sum_{m_1m_2} C(j_1j_2J|m_1m_2M)\,C(j_1j_2J'|m_1m_2M') = \delta_{JJ'}\delta_{MM'} ,

\sum_{J M} C(j_1j_2J|m_1m_2M)\,C(j_1j_2J|m_1'm_2'M) = \delta_{m_1m_1'}\delta_{m_2m_2'} ,

so that the inverse expansion is

|j_1m_1;j_2m_2\rangle = \sum_{J,M} C(j_1j_2J|m_1m_2M)\,|J,M\rangle .

Closed formula (Arfken, Weber and Harris, Eq. (16.46) (Arfken, Weber & Harris (2013), p. 790)): $C(j_1j_2J|m_1m_2M) = F_1F_2F_3$ with

F_1 = \left[\frac{(j_1+j_2-J)!\,(J+j_1-j_2)!\,(J+j_2-j_1)!\,(2J+1)}{(j_1+j_2+J+1)!}\right]^{1/2},

F_2 = \bigl[(J+M)!\,(J-M)!\,(j_1+m_1)!\,(j_1-m_1)!\,(j_2+m_2)!\,(j_2-m_2)!\bigr]^{1/2},

F_3 = \sum_{s}\frac{(-1)^s}{(j_1-m_1-s)!\,(j_2+m_2-s)!\,(J-j_2+m_1+s)!\,(J-j_1-m_2+s)!\,(j_1+j_2-J-s)!\,s!},

the sum running over the finitely many integers $s$ for which every factorial has a non-negative argument.

Symmetries, through the $3j$ -symbol. Define

\begin{pmatrix} j_1& j_2& j_3\\ m_1&m_2&m_3\end{pmatrix} = \frac{(-1)^{\,j_1-j_2-m_3}}{(2j_3+1)^{1/2}}\;C(j_1j_2j_3\,|\,m_1m_2,-m_3) .

Then the $3j$ -symbol is unchanged under an even permutation of its three columns, is multiplied by $(-1)^{j_1+j_2+j_3}$ under an odd permutation, and is multiplied by the same factor when all three $m$ 's change sign. It vanishes unless $m_1+m_2+m_3 = 0$ .

A closed family worth remembering (Hassani (Hassani (2013), p. 974)): coupling two equal angular momenta to a scalar,

C(ll\,0\,|\,m,-m,0) = \frac{(-1)^{l-m}}{\sqrt{2l+1}} .

Idea of the proofOriginal

Both bases diagonalise $J_z$ , so the change of basis cannot mix different $M$ : the big matrix splits into one small block per value of $M$ . Listing the uncoupled states by $M$ therefore tells you the whole answer, because the number of coupled states with a given $M$ must match.

There is exactly one uncoupled state with the largest possible $M = j_1+j_2$ , namely the one with both $m$ 's maximal. Being alone in its block it must already be a coupled state, and it must belong to $J = j_1+j_2$ . Lowering it with $J_-$ generates that whole multiplet. At $M = j_1+j_2-1$ there are two uncoupled states but the first multiplet has used only one combination, so the leftover direction is a new top state, of $J = j_1+j_2-1$ . Repeat. The process stops when the supply of uncoupled states at the next $M$ stops growing, which happens at $J = |j_1-j_2|$ ; counting dimensions confirms it.

So there are two ways to see the series, and this unit gives both. The counting argument is elementary and constructive — it is the algorithm. The character argument is slicker and generalises: multiply two characters $\sin\bigl((j+\tfrac12)\varphi\bigr)/\sin(\varphi/2)$ and watch the product collapse, by a telescoping trigonometric identity, into a sum of characters with $J$ running over exactly the triangle range.

Proof(Part 1 by counting states)AdaptedAdapted from Arfken, Weber & Harris (2013), pp. 786–787 and Eq. (16.37), p. 787

Step 1: $J^2$ , $J_z$ , $j_1^2$ , $j_2^2$ commute. $J^2 = j_1^2+j_2^2+2\,\mathbf{j}_1\!\cdot\!\mathbf{j}_2$ by the commuting hypothesis, and $j_1^2$ commutes with every $j_{1k}$ and with everything carrying index $2$ ; the same for $j_2^2$ . That $[J^2,J_k]=0$ is part 1 of angular momentum algebra applied to $\mathbf{J}$ . So the four operators have a common eigenbasis, which is the coupled basis. Note that $j_{1z}$ does not commute with $J^2$ : as Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Example 16.2.1, p. 785) compute, $[J^2,j_{1z}] = 2i(j_{1x}j_{2y}-j_{1y}j_{2x})$ , non-zero, and it is cancelled by $[J^2,j_{2z}]$ . This is why the two bases genuinely differ.

Step 2: count uncoupled states at each $M$ . Let $n(M)$ be the number of pairs $(m_1,m_2)$ with $m_1+m_2 = M$ , $|m_i|\le j_i$ . Assume $j_1\ge j_2$ without loss. As $M$ decreases from $j_1+j_2$ , the count rises by one at each step until $m_2$ runs out of room:

n(M) = \begin{cases} j_1+j_2-M+1, & j_1-j_2 \le M \le j_1+j_2,\\ 2j_2+1, & |M|\le j_1-j_2,\end{cases}

and $n(-M) = n(M)$ .

Step 3: the largest $J$ . Since $M = m_1+m_2$ for every uncoupled state, no state has $M > j_1+j_2$ ; and $n(j_1+j_2) = 1$ . A coupled multiplet of spin $J$ contributes exactly one state at each $M$ with $|M|\le J$ , so the number of multiplets with $J\ge M$ equals $n(M)$ . Taking $M = j_1+j_2$ gives exactly one multiplet with $J \ge j_1+j_2$ , and none can have $J > j_1+j_2$ because it would contribute a state of that $M$ . Hence $J_{\max} = j_1+j_2$ , occurring once.

Step 4: every $J$ down to $j_1-j_2$ occurs once. The number of multiplets with $J$ exactly equal to $M$ is $n(M)-n(M+1)$ , which by step 2 is $1$ for $j_1-j_2\le M\le j_1+j_2$ and $0$ for $M < j_1-j_2$ . So the multiplets are $J = j_1-j_2,\dots,j_1+j_2$ , each once. With the assumption $j_1\ge j_2$ dropped, $J_{\min} = |j_1-j_2|$ .

Step 5: the dimension check. Independently of step 4, write $J_{\min}$ for the smallest $J$ and sum the dimensions:

\sum_{J=J_{\min}}^{J_{\max}}(2J+1) = (J_{\max}-J_{\min}+1)(J_{\max}+J_{\min}+1) = J_{\max}^2 + 2J_{\max} - J_{\min}^2 + 1 ,

the middle expression being the number of terms times their average. Setting this equal to $\dim V = (2j_1+1)(2j_2+1)$ with $J_{\max}=j_1+j_2$ gives $J_{\min}^2 = (j_1-j_2)^2$ , hence $J_{\min} = |j_1-j_2|$ — Arfken, Weber and Harris's Eq. (16.37) (Arfken, Weber & Harris (2013), p. 787). This is the same conclusion as step 4 by a shorter route, and it is a useful arithmetic check on any decomposition.

Step 6: part 2. $M = m_1+m_2$ because $J_z = j_{1z}+j_{2z}$ acts on $|j_1m_1;j_2m_2\rangle$ with eigenvalue $m_1+m_2$ , and $\langle j_1m_1;j_2m_2|J,M\rangle$ is an inner product of eigenvectors of the Hermitian $J_z$ with eigenvalues $m_1+m_2$ and $M$ , hence zero unless they agree. The triangle condition is step 4; that $j_1+j_2+J$ is an integer follows because the $2j_i$ are integers and $J$ differs from $j_1+j_2$ by an integer. $\square$

Proof(Part 1 again, by characters)AdaptedAdapted from Hassani (2013), Eq. (30.17), p. 973, and p. 974; Arfken, Weber & Harris (2013), Eq. (16.43), p. 789

This route uses the character theory of orthogonality relations and characters and gives the series without constructing anything.

Step 1: the character of a rotation in the spin- $j$ representation. A rotation through $\varphi$ about the $z$ -axis acts on $|j,m\rangle$ by $e^{-im\varphi}$ , so, writing the sum as a geometric series,

\chi^{(j)}(\varphi) = \sum_{m=-j}^{j} e^{-im\varphi} = e^{ij\varphi}\,\frac{1-e^{-i(2j+1)\varphi}}{1-e^{-i\varphi}} = \frac{e^{i(j+\frac12)\varphi}-e^{-i(j+\frac12)\varphi}}{e^{i\varphi/2}-e^{-i\varphi/2}} = \frac{\sin\bigl((j+\tfrac12)\varphi\bigr)}{\sin(\varphi/2)} .

This is Hassani's Eq. (30.17) (Hassani (2013), p. 973) (he writes $e^{+im\varphi}$ , which gives the same real sum). Because every rotation through $\varphi$ is conjugate to this one, as noted in conjugacy classes, the character depends only on $\varphi$ .

Step 2: characters multiply under $\otimes$ . In the product basis the matrix of $R\otimes R$ has diagonal entries $D^{(j_1)}_{m_1m_1}D^{(j_2)}_{m_2m_2}$ , so summing over the pairs $(m_1,m_2)$ factorises the trace into $\chi^{(j_1)}\chi^{(j_2)}$ . This is the general fact recorded in product representations and selection rules.

Step 3: collapse the product. Assume $j_1\ge j_2$ and abbreviate $s = \sin(\varphi/2)$ . Keep the first factor in closed form and the second as its sum:

\chi^{(j_1)}(\varphi)\,\chi^{(j_2)}(\varphi) = \frac{\sin\bigl((j_1+\tfrac12)\varphi\bigr)}{s}\;\sum_{m=-j_2}^{j_2}e^{-im\varphi} = \frac{1}{s}\sum_{m=-j_2}^{j_2}\sin\bigl((j_1+\tfrac12)\varphi\bigr)\,e^{-im\varphi} .

The terms $m$ and $-m$ have conjugate exponentials with the same real prefactor, so the imaginary parts cancel in pairs and $e^{-im\varphi}$ may be replaced by $\cos(m\varphi)$ . The product-to-sum identity $\sin A\cos B = \tfrac12\bigl[\sin(A+B)+\sin(A-B)\bigr]$ then turns the term with index $m$ into

\tfrac12\Bigl[\sin\bigl((j_1+m+\tfrac12)\varphi\bigr) + \sin\bigl((j_1-m+\tfrac12)\varphi\bigr)\Bigr] .

As $m$ runs from $-j_2$ to $j_2$ , the quantity $j_1+m$ runs over $j_1-j_2,\dots,j_1+j_2$ , and so does $j_1-m$ . Each value $J$ in that range is therefore produced exactly twice, once by each of the two families, and the factor $\tfrac12$ is absorbed:

\chi^{(j_1)}(\varphi)\,\chi^{(j_2)}(\varphi) = \frac{1}{s}\sum_{J=j_1-j_2}^{j_1+j_2}\sin\Bigl(\bigl(J+\tfrac12\bigr)\varphi\Bigr) = \sum_{J=j_1-j_2}^{j_1+j_2}\chi^{(J)}(\varphi) ,

the last step by step 1 read backwards.

Step 4: read off the multiplicities. Two representations of a compact group with the same character are equivalent, and by the orthogonality of characters the multiplicity of $D^{(J)}$ in the product is the coefficient of $\chi^{(J)}$ in this expansion, namely $1$ for each $J$ in the triangle range and $0$ otherwise. That is Hassani's (Hassani (2013), Eqs. (24.44)–(24.45), pp. 753–754) general projection formula $m^{\alpha\beta}_{\sigma} = \frac{1}{|G|}\sum_i c_i\,\overline{\chi_i^{(\sigma)}}\chi_i^{(\alpha)}\chi_i^{(\beta)}$ specialised to the rotation group, with the sum over classes replaced by an integral over $\varphi$ . Since every multiplicity is $0$ or $1$ , $SU(2)$ is simply reducible in Hassani's sense, which is exactly why the Clebsch-Gordan coefficients need no extra label to separate repeated copies. $\square$

Proof(Part 3: reality, orthogonality and inversion)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (16.44)–(16.45), pp. 789–790; Hassani (2013), p. 756

Reality. The construction of the method below produces each $|J,M\rangle$ from $|j_1j_1;j_2j_2\rangle$ by applying $J_- = j_{1-}+j_{2-}$ , whose matrix elements are real and non-negative, and by orthogonalising against previously built states — a Gram-Schmidt step (Gram-Schmidt) with real coefficients. Starting from a basis whose expansion coefficients are real, every subsequent vector has real coefficients; the only freedom is an overall sign at each orthogonalisation, and the Condon-Shortley convention fixes it. Hence all $C$ are real, as Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 789) note, even when the underlying kets are complex.

Orthogonality. Both $\{|J,M\rangle\}$ and $\{|j_1m_1;j_2m_2\rangle\}$ are orthonormal bases of the same space, so the matrix $C$ taking one to the other is unitary; being real, it is orthogonal. The two displayed sums are $C^{\mathsf{T}}C = I$ and $CC^{\mathsf{T}} = I$ written out. They are the specialisation of Hassani's (Hassani (2013), p. 756) two normalisation relations for a general group, with the labels $q$ dropped.

Inversion. For an orthogonal matrix the inverse is the transpose, so expanding $|j_1m_1;j_2m_2\rangle$ in the coupled basis uses the same numbers with the roles of the index sets exchanged, which is the displayed formula; Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eq. (16.45), p. 790) make the same remark. $\square$

Remark(Where the hypotheses are used)Original

$[j_{1k},j_{2l}]=0$ is used everywhere. It is what makes $\mathbf{J}=\mathbf{j}_1+\mathbf{j}_2$ an angular momentum at all (the cross terms in the commutator cancel only because they vanish), and what makes $j_1^2$ and $j_2^2$ commute with $J^2$ so that $j_1,j_2$ remain good labels after coupling. Two angular momenta acting on the same degree of freedom — say $\mathbf{L}$ and $\mathbf{L}$ itself — do not satisfy it, and nothing in this unit applies to them.
$j_1$ and $j_2$ may independently be integer or half-integer, and the construction never cares. What the parities do control is the parity of $J$ : $J$ differs from $j_1+j_2$ by an integer, so coupling two half-integers gives integer $J$ (two spin- $\tfrac12$ particles make a spin- $0$ or spin- $1$ system, a boson), and coupling a half-integer with an integer gives half-integer $J$ . This is the $SU(2)$ -versus- $SO(3)$ distinction of the rotation group and SU(2) again: the product of two representations on which $-I$ acts as $-1$ is one on which it acts as $+1$ .
Multiplicity one is special to $SU(2)$ . The counting argument used that $n(M)-n(M+1)\in\{0,1\}$ , which is a fact about this particular group. For a general group the multiplicity $m^{\alpha\beta}_{\sigma}$ of Hassani (Hassani (2013), Eq. (24.45), p. 754) can exceed one, and then a further label $q$ is needed and the "Clebsch-Gordan coefficients" are defined only up to a choice of basis in the multiplicity space. The method below would break at the orthogonalisation step, which assumes the leftover direction is unique up to sign.
The phase convention is a convention, but not an optional one. Step 4 of the method leaves the sign of each new highest state free; the formulas of part 4 and the symmetries of part 5 are true only with Condon-Shortley. A table of coefficients from another source may differ by a sign in whole multiplets. The test is $C(j_1j_2J|j_1,J-j_1,J)>0$ .
The closed formula is exact but badly conditioned. $F_3$ is an alternating sum of large factorials, so computing it in floating point loses digits catastrophically for $j\gtrsim 10$ ; use exact rational arithmetic, or recursion in $M$ , or logarithms of factorials with the signs handled separately.
The triangle rule is not the classical vector inequality. As Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 786) point out, the classical picture — that the length of $\mathbf{j}_1+\mathbf{j}_2$ lies between the sum and the difference of the lengths — is only a mnemonic, because the length of a quantum angular momentum is $\sqrt{j(j+1)}$ and not $j$ . The rule $|j_1-j_2|\le J\le j_1+j_2$ is exact for the quantum numbers, not for the lengths.

Method(Coupling two angular momenta)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (16.38)–(16.42), pp. 788–789

Given $j_1$ and $j_2$ :

List the uncoupled states by $M$ . Make a table with one row per value of $M = m_1+m_2$ , from $j_1+j_2$ down to $-(j_1+j_2)$ , listing in each row the pairs $(m_1,m_2)$ that produce it. The row lengths tell you the answer in advance: the number of multiplets with $J\ge M$ is the length of row $M$ .
Start from the stretched state. The single entry in the top row is already a coupled state,

|J_{\max},J_{\max}\rangle = |j_1 j_1; j_2 j_2\rangle, \qquad J_{\max}=j_1+j_2 ,

so $C(j_1j_2J_{\max}|j_1 j_2 J_{\max}) = 1$ . 3. Lower. Apply $J_- = j_{1-}+j_{2-}$ to both sides and divide by the coefficient on the left. With the matrix elements of angular momentum algebra,

|J,M-1\rangle = \frac{1}{\sqrt{(J+M)(J-M+1)}}\bigl(j_{1-}+j_{2-}\bigr)|J,M\rangle ,

each $j_{i-}$ acting on its own factor with coefficient $\sqrt{(j_i+m_i)(j_i-m_i+1)}$ . Repeat down to $M = -J$ . 4. Orthogonalise to get the next multiplet. In the row $M = J-1$ there is one more uncoupled state than coupled states built so far. The new highest state $|J-1,J-1\rangle$ is the unit vector in that row orthogonal to all the $|J',J-1\rangle$ already constructed; by part 3 it is unique up to sign, and the sign is fixed by requiring the coefficient of the state with the largest $m_1$ to be positive. 5. Repeat steps 3 and 4, lowering the new top state and orthogonalising again, until $J$ reaches $|j_1-j_2|$ . Stop; the table is exhausted. 6. Or use the closed formula of part 4 for a single coefficient, in exact arithmetic. 7. Or use a $3j$ table. Convert with part 5, $C(j_1j_2J|m_1m_2M) = (-1)^{\,j_1-j_2+M}\sqrt{2J+1}\begin{pmatrix}j_1&j_2&J\\ m_1&m_2&-M\end{pmatrix}$ , and use the permutation and sign symmetries to reach whichever entry the table actually lists. This is why tables are published as $3j$ -symbols: one entry serves many coefficients. 8. Check. Each $|J,M\rangle$ has coefficients whose squares sum to $1$ ; states with the same $M$ and different $J$ are orthogonal; and the total number of coupled states equals $(2j_1+1)(2j_2+1)$ .

Example(Coupling

j_1=1

j_2=\tfrac12

, by hand and three ways checked)Original

This is the orbit-plus-spin coupling of a $p$ electron. Write the uncoupled states as $|m_1,m_2\rangle$ with $m_1\in\{1,0,-1\}$ and $m_2 = \pm\tfrac12$ . There are $3\times2 = 6$ of them.

Step 1: the table.

$M$	uncoupled states
$+\tfrac32$	$\lvert 1,+\tfrac12\rangle$
$+\tfrac12$	$\lvert 1,-\tfrac12\rangle$ , $\lvert 0,+\tfrac12\rangle$
$-\tfrac12$	$\lvert 0,-\tfrac12\rangle$ , $\lvert -1,+\tfrac12\rangle$
$-\tfrac32$	$\lvert -1,-\tfrac12\rangle$

Row lengths $1,2,2,1$ : one multiplet with $J\ge\tfrac32$ and two with $J\ge\tfrac12$ , so $J = \tfrac32$ and $J=\tfrac12$ , of dimensions $4+2 = 6$ , as required, and this agrees with the triangle rule $|1-\tfrac12|\le J\le 1+\tfrac12$ .

Step 2: the stretched state and one lowering. $|\tfrac32,\tfrac32\rangle = |1,+\tfrac12\rangle$ . Apply $J_-$ . On the left, with $J = M = \tfrac32$ , the coefficient is $\sqrt{(J+M)(J-M+1)} = \sqrt{3\cdot1} = \sqrt3$ . On the right, $j_{1-}|1\rangle$ has coefficient $\sqrt{(1+1)(1-1+1)} = \sqrt2$ and $j_{2-}|{+\tfrac12}\rangle$ has coefficient $\sqrt{(\tfrac12+\tfrac12)(\tfrac12-\tfrac12+1)} = 1$ . So

\sqrt3\,\bigl|\tfrac32,\tfrac12\bigr\rangle = \sqrt2\,\bigl|0,+\tfrac12\bigr\rangle + 1\cdot\bigl|1,-\tfrac12\bigr\rangle , \qquad \bigl|\tfrac32,\tfrac12\bigr\rangle = \sqrt{\tfrac23}\,\bigl|0,+\tfrac12\bigr\rangle + \sqrt{\tfrac13}\,\bigl|1,-\tfrac12\bigr\rangle .

Step 3: orthogonalise for $J=\tfrac12$ . In the $M=\tfrac12$ row the orthogonal direction is

\bigl|\tfrac12,\tfrac12\bigr\rangle = \sqrt{\tfrac23}\,\bigl|1,-\tfrac12\bigr\rangle - \sqrt{\tfrac13}\,\bigl|0,+\tfrac12\bigr\rangle ,

with the sign fixed by Condon-Shortley: the state with the largest $m_1$ here is $|1,-\tfrac12\rangle$ , and its coefficient must be positive. Check orthogonality: $\sqrt{\tfrac23}\sqrt{\tfrac13} - \sqrt{\tfrac13}\sqrt{\tfrac23} = 0$ , as required, and $\tfrac23+\tfrac13 = 1$ .

Step 4: lower both multiplets. For $J=\tfrac32$ , $M=\tfrac12$ : the left coefficient is $\sqrt{(\tfrac32+\tfrac12)(\tfrac32-\tfrac12+1)} = \sqrt{2\cdot2} = 2$ , and

2\bigl|\tfrac32,-\tfrac12\bigr\rangle = \sqrt{\tfrac23}\Bigl(\sqrt2\bigl|-1,\tfrac12\bigr\rangle + \bigl|0,-\tfrac12\bigr\rangle\Bigr) + \sqrt{\tfrac13}\Bigl(\sqrt2\bigl|0,-\tfrac12\bigr\rangle\Bigr) ,

using $j_{2-}|{-\tfrac12}\rangle = 0$ . Collecting, the coefficient of $|0,-\tfrac12\rangle$ is $\sqrt{\tfrac23}+\sqrt{\tfrac23} = 2\sqrt{\tfrac23}$ and that of $|-1,\tfrac12\rangle$ is $\sqrt{\tfrac43} = 2\sqrt{\tfrac13}$ , so after dividing by $2$ ,

\bigl|\tfrac32,-\tfrac12\bigr\rangle = \sqrt{\tfrac23}\,\bigl|0,-\tfrac12\bigr\rangle + \sqrt{\tfrac13}\,\bigl|-1,\tfrac12\bigr\rangle .

For $J=\tfrac12$ , $M=\tfrac12$ : the left coefficient is $\sqrt{(\tfrac12+\tfrac12)(\tfrac12-\tfrac12+1)} = 1$ , and

\bigl|\tfrac12,-\tfrac12\bigr\rangle = \sqrt{\tfrac23}\cdot\sqrt2\,\bigl|0,-\tfrac12\bigr\rangle - \sqrt{\tfrac13}\Bigl(\sqrt2\,\bigl|-1,\tfrac12\bigr\rangle + \bigl|0,-\tfrac12\bigr\rangle\Bigr) = \sqrt{\tfrac13}\,\bigl|0,-\tfrac12\bigr\rangle - \sqrt{\tfrac23}\,\bigl|-1,\tfrac12\bigr\rangle ,

because $\tfrac{2}{\sqrt3}-\tfrac{1}{\sqrt3} = \tfrac{1}{\sqrt3}$ . Finally $|\tfrac32,-\tfrac32\rangle = |-1,-\tfrac12\rangle$ .

Step 5: the table of coefficients. Writing $C(1\,\tfrac12\,J|m_1m_2M)$ ,

$J$	$M$	$C$ for $m_1 = M+\tfrac12$	$C$ for $m_1 = M-\tfrac12$
$\tfrac32$	$+\tfrac12$	$\sqrt{1/3}$	$\sqrt{2/3}$
$\tfrac32$	$-\tfrac12$	$\sqrt{2/3}$	$\sqrt{1/3}$
$\tfrac12$	$+\tfrac12$	$\sqrt{2/3}$	$-\sqrt{1/3}$
$\tfrac12$	$-\tfrac12$	$\sqrt{1/3}$	$-\sqrt{2/3}$

(The first numeric column is the state with $m_2 = -\tfrac12$ , the second the one with $m_2=+\tfrac12$ .) Numerically $\sqrt{1/3} = 0.5773503$ and $\sqrt{2/3} = 0.8164966$ , and each row satisfies $\tfrac13+\tfrac23 = 1$ . The $2\times2$ block at $M=+\tfrac12$ is

\begin{pmatrix} \sqrt{1/3} & \sqrt{2/3}\\ \sqrt{2/3} & -\sqrt{1/3}\end{pmatrix},

an orthogonal matrix of determinant $-1$ , as part 3 promises.

Step 6: check against the closed formula. Take $J=\tfrac32$ , $m_1 = 1$ , $m_2 = -\tfrac12$ , $M = \tfrac12$ . Then $j_1+j_2-J = 0$ , $J+j_1-j_2 = 2$ , $J+j_2-j_1 = 1$ , $2J+1 = 4$ , $j_1+j_2+J+1 = 4$ , so

F_1 = \left[\frac{0!\,2!\,1!\,\cdot 4}{4!}\right]^{1/2} = \left[\frac{8}{24}\right]^{1/2} = \frac{1}{\sqrt3} .

Next $F_2 = [2!\cdot1!\cdot2!\cdot0!\cdot0!\cdot1!]^{1/2} = \sqrt4 = 2$ , using $(J+M)! = 2!$ , $(J-M)! = 1!$ , $(j_1+m_1)! = 2!$ , $(j_1-m_1)! = 0!$ , $(j_2+m_2)! = 0!$ , $(j_2-m_2)! = 1!$ . For $F_3$ the six factorial arguments are $j_1-m_1-s = -s$ , $j_2+m_2-s = -s$ , $J-j_2+m_1+s = 2+s$ , $J-j_1-m_2+s = 1+s$ , $j_1+j_2-J-s = -s$ and $s$ , so only $s=0$ survives and

F_3 = \frac{1}{0!\,0!\,2!\,1!\,0!\,0!} = \frac12 .

Hence $C = \tfrac{1}{\sqrt3}\cdot2\cdot\tfrac12 = \tfrac{1}{\sqrt3} = \sqrt{1/3}$ , matching step 2.

Step 7: check against a $3j$ -symbol. With $j_1=1$ , $j_2=\tfrac12$ , $J=\tfrac32$ , $m_1=1$ , $m_2=-\tfrac12$ , $M=\tfrac12$ , part 5 gives

C = (-1)^{\,1-\frac12+\frac12}\sqrt{2J+1}\begin{pmatrix}1&\tfrac12&\tfrac32\\ 1&-\tfrac12&-\tfrac12\end{pmatrix} = -2\begin{pmatrix}1&\tfrac12&\tfrac32\\ 1&-\tfrac12&-\tfrac12\end{pmatrix} ,

and the $3j$ -symbol equals $-\tfrac{1}{2\sqrt3}$ , returning $C = 1/\sqrt3$ once more.

Step 8: an independent check from Hassani's singlet formula. Couple $j_1=j_2=1$ instead and look at the scalar $J=0$ . Writing the uncoupled states of this second problem in the full notation $|j_1m_1;j_2m_2\rangle$ of the definition block, to keep them apart from the coupled $|J,M\rangle$ , the method gives

|0,0\rangle = \tfrac{1}{\sqrt3}\Bigl(|1,1;1,-1\rangle - |1,0;1,0\rangle + |1,-1;1,1\rangle\Bigr) ,

the three coefficients being $C(1\,1\,0|m,-m,0)$ for $m=1,0,-1$ . Hassani's (Hassani (2013), p. 974) closed value $C(ll\,0|m,-m,0) = (-1)^{l-m}/\sqrt{2l+1}$ with $l=1$ predicts $+\tfrac{1}{\sqrt3}, -\tfrac{1}{\sqrt3}, +\tfrac{1}{\sqrt3}$ — the same three numbers, with $1/\sqrt3 = 0.5773503$ . The alternating signs are what make the combination rotationally invariant: it is the scalar product $\mathbf{a}\cdot\mathbf{b}$ written in the spherical basis.

ApplicationOriginal

Atomic structure. In the Russell-Saunders scheme the orbital angular momenta of the electrons are coupled to a total $L$ , their spins to a total $S$ , and then $L$ and $S$ to $J$ ; the resulting terms $^{2S+1}L_J$ are the labels on every line in an atomic spectrum. The spin-orbit interaction $\xi\,\mathbf{L}\cdot\mathbf{S}$ is diagonal in the coupled basis and not in the uncoupled one, with eigenvalue

\tfrac12\xi\bigl[J(J+1)-L(L+1)-S(S+1)\bigr]

read straight off $\mathbf{L}\cdot\mathbf{S} = \tfrac12(J^2-L^2-S^2)$ — which is why one couples at all. For the $p$ electron of the example, $L=1$ and $S=\tfrac12$ give the doublet $^2P_{3/2}$ and $^2P_{1/2}$ , split by $\tfrac32\xi$ ; this is the splitting of the sodium D lines.

Two spins, singlets and entanglement. Coupling two spin- $\tfrac12$ particles gives $J=1$ (a triplet) and $J=0$ (a singlet), and the singlet $(|{\uparrow\downarrow}\rangle-|{\downarrow\uparrow}\rangle)/\sqrt2$ is the maximally entangled Bell state on which the Einstein-Podolsky-Rosen argument and Bell's inequality are built. The same singlet is the chemical covalent bond: two electrons in a bonding orbital must pair their spins to a singlet for the spatial wavefunction to be symmetric.

Nuclear and particle physics. Isospin coupling uses the identical mathematics with the proton-neutron doublet in place of spin, and the Clebsch-Gordan coefficients give the relative cross-sections of reactions related by isospin — for example the factor of $3$ between $\pi^+p$ and $\pi^-p$ scattering at the $\Delta$ resonance comes straight from a squared coefficient. Coupling a quark and an antiquark, each a doublet, produces the meson triplet and singlet; Hassani (Hassani (2013), p. 753) gives this as the motivating example.

Multipole expansions and spectroscopy. Combining a spherical harmonic of order $l_1$ with one of order $l_2$ expands their product in harmonics of order $l_1+l_2$ down to $|l_1-l_2|$ , which is the Gaunt-coefficient formula behind every multipole matrix element; the special case $J=0$ reproduces the addition theorem for spherical harmonics, as Hassani (Hassani (2013), p. 974) derives from the singlet coefficients. Selection rules such as $\Delta J = 0,\pm1$ with $J=0\not\to J=0$ are the statement that a coefficient vanishes, and the general machine for such statements is spherical tensors and the Wigner-Eckart theorem.

Condensed matter and computation. In exact diagonalisation of spin chains the coupled basis block-diagonalises the Hamiltonian by total $J$ , reducing a matrix of size $2^N$ to blocks of much smaller size; tensor-network algorithms exploit the same decomposition to carry $SU(2)$ symmetry explicitly, storing one reduced number per multiplet instead of all $2J+1$ components. The coefficients are generated once, in exact arithmetic, by the recipe above.

Definition · advanced

Rotation matrices of angular momentum

Adapted

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Defines the Wigner matrices $D^{j}_{m'm}(\alpha,\beta,\gamma)=\langle jm'|\mathrm{e}^{-\mathrm{i}\alpha J_z}\mathrm{e}^{-\mathrm{i}\beta J_y}\mathrm{e}^{-\mathrm{i}\gamma J_z}|jm\rangle$ of a rotation in Euler angles, reduces them to the single real matrix $d^{j}(\beta)$ , derives Wigner's closed formula for $d^{j}$ by expanding a monomial in two variables, and identifies the column $m=0$ of the integer- $j$ matrices with the spherical harmonics.

Prerequisites: Angular momentum algebra, Spherical harmonics, The rotation group and SU(2), Matrix Lie groups and Lie algebras, Representations of groups, Functions of matrices and the matrix exponential, Unitary and orthogonal matrices, Associated Legendre functions, Legendre polynomials

Definition(Euler angles)AdaptedAdapted from Arfken, Weber & Harris (2013), §3.4, Eqs. (3.35)–(3.37), pp. 140–141; Hassani (2013), §30.3.3, p. 972

Write $R_z(\theta)$ and $R_y(\theta)$ for the rotations of $\mathbb{R}^3$ through $\theta$ about the third and second coordinate axes, counterclockwise seen from the tip of the axis, in the active convention of the rotation group and SU(2): $R_{\mathbf{n}}(\theta)=\exp(-\mathrm{i}\theta\,\mathbf{n}\cdot\mathbf{J})$ moves a vector and leaves the axes alone.

Given three numbers

0\le\alpha<2\pi,\qquad 0\le\beta\le\pi,\qquad 0\le\gamma<2\pi,

the Euler angles, put

R(\alpha,\beta,\gamma)=R_z(\alpha)\,R_y(\beta)\,R_z(\gamma)\;\in SO(3).

Every element of $SO(3)$ is of this form, and the angles are unique unless $\beta\in\{0,\pi\}$ , where only $\alpha\pm\gamma$ is determined. The last column is easy to read off:

R(\alpha,\beta,\gamma)\,\hat{\mathbf{z}}=(\sin\beta\cos\alpha,\;\sin\beta\sin\alpha,\;\cos\beta),

so $(\beta,\alpha)$ are the polar and azimuthal angles of the image of the polar axis, and $\gamma$ is the extra turn about the old polar axis made before tipping it.

Comparison with the usual bookkeeping. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), §3.4, pp. 140–141) describe the same three angles as successive rotations of the coordinate axes: first by $\alpha$ about $\hat e_3$ , then by $\beta$ about the new $\hat e_2'$ , then by $\gamma$ about the newer $\hat e_3''$ , with exactly the ranges above. Their matrix is the triple product $S(\alpha,\beta,\gamma)=S_3(\gamma)S_2(\beta)S_1(\alpha)$ of their Eq. (3.36). Each $S_k$ turns the axes one way where $R_z,R_y$ turn vectors the other, and reversing the order of a product of inverses gives

S(\alpha,\beta,\gamma)=R(\alpha,\beta,\gamma)^{-1}=R(\alpha,\beta,\gamma)^{\mathsf{T}} .

Multiplying out $R_z(\alpha)R_y(\beta)R_z(\gamma)$ and transposing reproduces their Eq. (3.37) entry by entry; at $\alpha=\beta=\gamma=\pi/2$ both give $S=\bigl(\begin{smallmatrix}-1&0&0\\0&0&1\\0&1&0\end{smallmatrix}\bigr)$ , their Example 3.4.1. Hassani (Hassani (2013), §30.3.3, p. 972) factorises a rotation the same way, as $\exp(M_{12}\psi)\exp(M_{31}\theta)\exp(M_{12}\varphi)$ , and $M_{12}$ , $M_{31}$ generate exactly our $R_z$ , $R_y$ ; his $(\psi,\theta,\varphi)$ plays the role of our $(\alpha,\beta,\gamma)$ , angle for angle — but only up to the sign of the angles, because the signs in his exponents are the opposite of ours, so that his $R(\psi,\theta,\varphi)$ is our $R(-\psi,-\theta,-\varphi)$ ; see the remark on conventions below.

Definition(The rotation matrices

D^{j}

and

d^{j}

)AdaptedAdapted from Arfken, Weber & Harris (2013), §16.3, Eqs. (16.51)–(16.52), pp. 796–797; Hassani (2013), §30.3.3, Eq. (30.15), p. 972

Let $J_1,J_2,J_3$ be an angular momentum (angular momentum algebra) acting irreducibly on a space of dimension $2j+1$ , with the orthonormal basis $|j,m\rangle$ of that unit and the Condon-Shortley phases. Inner products are linear in the ket, as there. The rotation operator of the Euler angles $(\alpha,\beta,\gamma)$ is

\mathcal{D}^{(j)}(\alpha,\beta,\gamma) =\mathrm{e}^{-\mathrm{i}\alpha J_z}\,\mathrm{e}^{-\mathrm{i}\beta J_y}\,\mathrm{e}^{-\mathrm{i}\gamma J_z},

and its matrix in that basis,

D^{j}_{m'm}(\alpha,\beta,\gamma)=\bigl\langle j,m'\bigm|\mathcal{D}^{(j)}(\alpha,\beta,\gamma)\bigm|j,m\bigr\rangle \qquad(-j\le m,m'\le j),

is the Wigner $D$ -matrix of order $j$ . Rows and columns are labelled by $m'$ and $m$ running from $-j$ to $j$ , not from $1$ to $2j+1$ ; Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 797) flag the same unusual labelling, define the matrix by $R|L,M\rangle=\sum_{M'}D^{L}_{M'M}(R)|L,M'\rangle$ — the same index placement as ours — and record the name, which is Eugene Wigner's.

The reduced rotation matrix is the one-angle matrix

d^{j}_{m'm}(\beta)=\bigl\langle j,m'\bigm|\mathrm{e}^{-\mathrm{i}\beta J_y}\bigm|j,m\bigr\rangle .

Both are $(2j+1)\times(2j+1)$ . For half-integer $j$ the matrices are two-valued on $SO(3)$ and single-valued on $SU(2)$ ; the natural domain is $SU(2)$ , with $\alpha,\gamma$ running to $4\pi$ .

Definition(Structure of the rotation matrices)AdaptedAdapted from Hassani (2013), Eqs. (30.15)–(30.17), pp. 972–973; Arfken, Weber & Harris (2013), Eq. (16.52), p. 796, Eq. (16.78), p. 803

With the notation above:

Factorisation. Only the middle angle is non-trivial:

D^{j}_{m'm}(\alpha,\beta,\gamma)=\mathrm{e}^{-\mathrm{i}m'\alpha}\;d^{j}_{m'm}(\beta)\;\mathrm{e}^{-\mathrm{i}m\gamma},

and $d^{j}_{m'm}(\beta)$ is real.

Wigner's formula. With $c=\cos\tfrac\beta2$ and $s=\sin\tfrac\beta2$ ,

d^{j}_{m'm}(\beta)=\sum_{k}(-1)^{k+m'-m}\, \frac{\sqrt{(j+m)!\,(j-m)!\,(j+m')!\,(j-m')!}}{(j+m-k)!\;k!\;(j-m'-k)!\;(m'-m+k)!}\; c^{\,2j-2k+m-m'}\,s^{\,2k+m'-m},

the sum running over every integer $k$ for which all four factorials have non-negative arguments. The factorials make sense for half-integer $j$ as well, because $m$ and $m'$ always differ from $j$ by an integer, so each of $j+m$ , $j-m$ , $j+m'$ , $j-m'$ is a non-negative integer; in the worked example below $j=\tfrac32$ and these four numbers run over $0,1,2,3$ . The number of terms is $1+\tau$ , where $\tau$ is the smallest of $j+m$ , $j-m$ , $j+m'$ , $j-m'$ .

A unitary representation. $D^{j}$ is unitary, $d^{j}$ is real orthogonal, and $R\mapsto D^{j}(R)$ is an irreducible representation of $SU(2)$ of dimension $2j+1$ ; every irreducible representation of $SU(2)$ is one of these. The character depends only on the rotation angle $\varphi$ :

\chi^{(j)}(\varphi)=\operatorname{tr}D^{j}=\frac{\sin\bigl((j+\tfrac12)\varphi\bigr)}{\sin(\varphi/2)} .

Symmetries and special values.

d^{j}_{m'm}(-\beta)=d^{j}_{mm'}(\beta)=(-1)^{m-m'}d^{j}_{m'm}(\beta), \qquad d^{j}_{m'm}(0)=\delta_{m'm}, \qquad d^{j}_{m'm}(\pi)=(-1)^{j-m}\,\delta_{m',-m},

and a turn through $2\pi$ about any axis gives $(-1)^{2j}$ times the identity — the sign change of a spinor.

Integer $j$ : the spherical harmonics. Let $l\ge0$ be an integer and let $\mathcal{D}(R)$ act on functions on the sphere by $(\mathcal{D}(R)f)(\hat{\mathbf{n}})=f(R^{-1}\hat{\mathbf{n}})$ . Then the span of $Y_l^{-l},\dots,Y_l^{l}$ (spherical harmonics) carries exactly the representation $D^{l}$ :

Y_l^{m}\bigl(R^{-1}\hat{\mathbf{n}}\bigr)=\sum_{m'=-l}^{l}D^{l}_{m'm}(R)\;Y_l^{m'}(\hat{\mathbf{n}}),

and the row $m'=0$ and the column $m=0$ are harmonics outright:

D^{l}_{0m}(R)=\sqrt{\frac{4\pi}{2l+1}}\;Y_l^{m}\bigl(R^{-1}\hat{\mathbf{z}}\bigr), \qquad D^{l}_{m0}(\alpha,\beta,\gamma)=\sqrt{\frac{4\pi}{2l+1}}\;\overline{Y_l^{m}(\beta,\alpha)} .

In particular $d^{l}_{m0}(\beta)=\sqrt{(l-m)!/(l+m)!}\;P_l^{m}(\cos\beta)$ with the Condon-Shortley-phased $P_l^m$ of associated Legendre functions, and $D^{l}_{00}(\alpha,\beta,\gamma)=P_l(\cos\beta)$ .

Idea of the proofOriginal

Two observations do all the work.

The first is that $|j,m\rangle$ is an eigenvector of $J_z$ , so the outer two exponentials in $\mathcal{D}^{(j)}$ never mix anything: they only multiply the bra and the ket by phases. That is why a function of three angles collapses to a matrix of functions of one angle, and why the whole subject is the study of $d^{j}(\beta)$ .

The second is that the spin- $j$ multiplet is the $2j$ -fold symmetric product of the spin- $\tfrac12$ one. A product of $2j$ spins, all of them up or down, is recorded by a monomial $x^{a}y^{b}$ with $a+b=2j$ , and a rotation acts on the two spin- $\tfrac12$ states by a $2\times2$ matrix, hence on the monomial by substituting two linear forms for $x$ and $y$ . Expanding $(cx+sy)^{j+m}(-sx+cy)^{j-m}$ by the binomial theorem and collecting the coefficient of $x^{j+m'}y^{j-m'}$ is Wigner's formula; the alternating sign comes from the minus in the second linear form, and the four factorials in the denominator are the two binomial coefficients, with the square root supplied by the normalisation of the monomials.

Everything else follows: unitarity because $J_y$ is Hermitian, reality because $J_y$ is purely imaginary in this basis, and the spherical-harmonic formulas because the $2l+1$ harmonics of order $l$ are a spin- $l$ multiplet for the orbital angular momentum, so their rotation matrix has to be $D^{l}$ .

Proof(Part 1: the factorisation)AdaptedAdapted from Hassani (2013), Eq. (30.15), p. 972

Step 1: the phases. $J_z|j,m\rangle=m|j,m\rangle$ , so $\mathrm{e}^{-\mathrm{i}\gamma J_z}|j,m\rangle=\mathrm{e}^{-\mathrm{i}m\gamma}|j,m\rangle$ . Taking the adjoint of the same statement for $m'$ , and using that $J_z$ is Hermitian with real eigenvalue $m'$ , gives $\langle j,m'|\mathrm{e}^{-\mathrm{i}\alpha J_z}=\mathrm{e}^{-\mathrm{i}m'\alpha}\langle j,m'|$ . Hence

D^{j}_{m'm}(\alpha,\beta,\gamma)=\mathrm{e}^{-\mathrm{i}m'\alpha}\,\mathrm{e}^{-\mathrm{i}m\gamma}\,\bigl\langle j,m'\bigm|\mathrm{e}^{-\mathrm{i}\beta J_y}\bigm|j,m\bigr\rangle,

which is the claim. Hassani (Hassani (2013), Eq. (30.15), p. 972) performs the same collapse.

Step 2: $d^{j}$ is real. In the basis $|j,m\rangle$ the matrices of $J_{\pm}$ have real non-negative entries (Condon-Shortley), so $J_y=\tfrac{1}{2\mathrm{i}}(J_+-J_-)$ has purely imaginary entries, and $-\mathrm{i}\beta J_y$ has real ones. The exponential of a real matrix is real. $\square$

Proof(Part 2: Wigner's formula)AdaptedAdapted from Hassani (2013), Eq. (30.16), p. 972

The result proved is Hassani's Eq. (30.16), which he states without proof, referring to Hamermesh for one; the route below — realising the multiplet as a symmetric power and expanding a monomial — is our own.

Step 1: a model for the multiplet. Let $W_{2j}$ be the space of homogeneous polynomials of degree $2j$ in two variables $x,y$ , of dimension $2j+1$ , and define operators on it by

J_+=x\,\partial_y,\qquad J_-=y\,\partial_x,\qquad J_z=\tfrac12\bigl(x\,\partial_x-y\,\partial_y\bigr).

These satisfy the angular-momentum relations: $[J_z,J_\pm]=\pm J_\pm$ because $J_z$ measures half the difference of the degrees and $J_\pm$ shifts it by $\pm1$ , and $[J_+,J_-]=x\partial_x-y\partial_y=2J_z$ by differentiating the products. Set

\phi_{m}(x,y)=\frac{x^{j+m}\,y^{j-m}}{\sqrt{(j+m)!\,(j-m)!}},\qquad m=-j,\dots,j .

Then $J_z\phi_m=m\phi_m$ , and

J_+\phi_m=\frac{(j-m)\,x^{j+m+1}y^{j-m-1}}{\sqrt{(j+m)!(j-m)!}} =(j-m)\sqrt{\frac{(j+m+1)!\,(j-m-1)!}{(j+m)!\,(j-m)!}}\;\phi_{m+1} =\sqrt{(j-m)(j+m+1)}\;\phi_{m+1},

which is exactly the coefficient of part 3 of angular momentum algebra, with the Condon-Shortley sign. The same computation for $J_-=y\partial_x$ gives the lowering coefficient. So $\phi_m\leftrightarrow|j,m\rangle$ is an isomorphism of multiplets, and any matrix element computed in $W_{2j}$ is the one we want.

Step 2: how a rotation acts on a monomial. Let $U$ be the spin- $\tfrac12$ matrix of the rotation, acting on the two basis states by $U e_{\pm1/2}=\sum_{\mu}U_{\mu,\pm1/2}e_{\mu}$ . Writing $x$ for $e_{1/2}$ and $y$ for $e_{-1/2}$ , the degree- $2j$ monomials stand for the symmetrised products of $2j$ spins, and $U$ acts on all $2j$ factors at once; in the polynomial language this is the algebra homomorphism $\rho(U)$ determined by

\rho(U)\,x=U_{+ +}\,x+U_{-+}\,y, \qquad \rho(U)\,y=U_{+-}\,x+U_{--}\,y .

It is a homomorphism of groups: applying $\rho(U_1)$ to the images above reproduces the columns of $U_1U_2$ , by the definition of matrix multiplication. Here it matters that the substitution carries the columns of $U$ , not its rows; the opposite choice gives an anti-homomorphism.

Step 3: the substitution for $\mathrm{e}^{-\mathrm{i}\beta J_y}$ . In the spin- $\tfrac12$ case $J_y=\tfrac12\sigma_2$ , and the half-angle formula of the rotation group and SU(2) gives

U=\mathrm{e}^{-\mathrm{i}\beta\sigma_2/2}=I\cos\tfrac\beta2-\mathrm{i}\sigma_2\sin\tfrac\beta2 =\begin{pmatrix}c&-s\\ s&c\end{pmatrix}, \qquad c=\cos\tfrac\beta2,\ s=\sin\tfrac\beta2 ,

in the basis ordered $(+\tfrac12,-\tfrac12)$ . Reading its columns, step 2 becomes the substitution $x\mapsto cx+sy$ , $y\mapsto -sx+cy$ .

Step 4: expand. Therefore

\mathrm{e}^{-\mathrm{i}\beta J_y}\phi_m=\frac{(cx+sy)^{j+m}\,(-sx+cy)^{j-m}}{\sqrt{(j+m)!\,(j-m)!}} .

Apply the binomial theorem to each factor,

(cx+sy)^{j+m}=\sum_{p}\binom{j+m}{p}c^{p}s^{j+m-p}\,x^{p}y^{j+m-p}, \qquad (-sx+cy)^{j-m}=\sum_{q}\binom{j-m}{q}(-s)^{q}c^{j-m-q}\,x^{q}y^{j-m-q},

and collect the terms with $x^{j+m'}y^{j-m'}$ , that is $p+q=j+m'$ . Substituting the single index $k=j+m-p$ , so that $p=j+m-k$ and $q=k+m'-m$ , the powers become

c^{\,p+(j-m-q)}=c^{\,2j-2k+m-m'},\qquad s^{\,(j+m-p)+q}=s^{\,2k+m'-m},

the sign is $(-1)^{q}=(-1)^{k+m'-m}$ , and the two binomial coefficients combine to

\frac{(j+m)!}{(j+m-k)!\,k!}\cdot\frac{(j-m)!}{(m'-m+k)!\,(j-m'-k)!} .

Dividing by $\sqrt{(j+m)!(j-m)!}$ from $\phi_m$ and multiplying by $\sqrt{(j+m')!(j-m')!}$ to express the result in terms of $\phi_{m'}$ leaves precisely the stated coefficient of $\phi_{m'}$ , and by step 1 that coefficient is $d^{j}_{m'm}(\beta)$ .

Step 5: the range and the number of terms. A binomial coefficient $\binom{n}{r}$ is zero unless $0\le r\le n$ , so the four quantities $j+m-k$ , $k$ , $j-m'-k$ and $m'-m+k$ must all be non-negative integers; this is the stated range, and it is non-empty exactly when $|m|,|m'|\le j$ . The constraints read $\max(0,m-m')\le k\le\min(j+m,\,j-m')$ , so the number of terms is

\min(j+m,j-m')-\max(0,m-m')+1=\min\bigl(j+m,\,j-m',\,j-m,\,j+m'\bigr)+1,

which is Hassani's (Hassani (2013), p. 972) count $1+\tau$ . $\square$

Proof(Parts 3 and 4)Original

Unitarity and orthogonality. $J_y$ and $J_z$ are Hermitian, so each exponential $\mathrm{e}^{-\mathrm{i}\theta J}$ is unitary and so is the product; $d^{j}$ is unitary and real by part 1, hence real orthogonal. (Orthogonality also says $\sum_{m'}\bigl(d^{j}_{m'm}\bigr)^2=1$ , which is the statement that the squared amplitudes in a rotated basis sum to one.)

Representation and irreducibility. $R\mapsto\mathcal{D}^{(j)}(R)$ is the restriction to the multiplet of the group representation generated by $\mathbf{J}$ , so it is a representation; it is irreducible because the multiplet is, by part 2 of angular momentum algebra. That every irreducible representation of $SU(2)$ arises this way is part 2 of the same unit applied to the generators.

Character. Conjugate so that the rotation is about $z$ through $\varphi$ ; traces are unchanged by conjugation, and Riley, Hobson and Bence (Riley, Hobson & Bence (2006), §28.7.3, pp. 1069–1070) note (as used in the rotation group and SU(2)) that equal-angle rotations about different axes are conjugate. Then $\mathcal{D}^{(j)}=\mathrm{e}^{-\mathrm{i}\varphi J_z}$ is diagonal and

\chi^{(j)}(\varphi)=\sum_{m=-j}^{j}\mathrm{e}^{-\mathrm{i}m\varphi} =\mathrm{e}^{\mathrm{i}j\varphi}\sum_{r=0}^{2j}\mathrm{e}^{-\mathrm{i}r\varphi} =\mathrm{e}^{\mathrm{i}j\varphi}\frac{1-\mathrm{e}^{-\mathrm{i}(2j+1)\varphi}}{1-\mathrm{e}^{-\mathrm{i}\varphi}} =\frac{\mathrm{e}^{\mathrm{i}(j+\frac12)\varphi}-\mathrm{e}^{-\mathrm{i}(j+\frac12)\varphi}}{\mathrm{e}^{\mathrm{i}\varphi/2}-\mathrm{e}^{-\mathrm{i}\varphi/2}},

after multiplying numerator and denominator by $\mathrm{e}^{\mathrm{i}\varphi/2}$ ; both differences are $2\mathrm{i}\sin$ of half the exponent, giving the stated ratio. This is Hassani's (Hassani (2013), Eq. (30.17), p. 973) Eq. (30.17), and it is the sum used there to prove the Clebsch-Gordan series of addition of angular momenta.

Symmetries. Replacing $\beta$ by $-\beta$ inverts the operator, and the inverse of a real orthogonal matrix is its transpose: $d^{j}(-\beta)=d^{j}(\beta)^{-1}=d^{j}(\beta)^{\mathsf{T}}$ . The relation $d^{j}_{mm'}=(-1)^{m-m'}d^{j}_{m'm}$ is read off Wigner's formula: interchanging $m$ and $m'$ sends the summation index $k$ to $k'=k+m'-m$ , which leaves the four factorials and the two powers unchanged and multiplies the sign by $(-1)^{m-m'}$ .

Special values. At $\beta=0$ only $k$ with $s^{2k+m'-m}$ of exponent $0$ survives, forcing $m'=m$ and $k=0$ , and the coefficient is $1$ . At $\beta=\pi$ we have $c=0$ , so only $2j-2k+m-m'=0$ contributes, i.e. $k=j+\tfrac{m-m'}{2}$ ; substituting this into the non-negativity constraint $j-m'-k\ge0$ gives $m'\le-m$ and $j+m-k\ge0$ gives $m'\ge-m$ , so $m'=-m$ , $k=j+m$ , and the surviving term is $(-1)^{j+m-2m}\sqrt{\cdots}/[0!\,(j+m)!\,(j-m)!\,0!]\cdot s^{2j}$ with $s=1$ , which is $(-1)^{j-m}$ . Finally $\mathcal{D}^{(j)}$ of a $2\pi$ turn about $z$ is $\mathrm{e}^{-2\pi\mathrm{i}J_z}$ , diagonal with entries $\mathrm{e}^{-2\pi\mathrm{i}m}=(-1)^{2m}=(-1)^{2j}$ , since $m$ and $j$ differ by an integer; a general axis is conjugate to $z$ . $\square$

Proof(Part 5: the spherical harmonics)AdaptedAdapted from Arfken, Weber & Harris (2013), Eq. (16.52), p. 796, and Eq. (16.78), p. 803

Step 1: $\mathcal{D}$ is a unitary representation on functions. $(\mathcal{D}(R)f)(\hat{\mathbf{n}})=f(R^{-1}\hat{\mathbf{n}})$ defines a representation — the inverse is what makes $\mathcal{D}(R_1)\mathcal{D}(R_2)=\mathcal{D}(R_1R_2)$ rather than the reverse — and it is unitary on $L^2(S^2)$ because a rotation preserves the element of solid angle.

Step 2: the span of the $Y_l^m$ is a spin- $l$ multiplet. By spherical harmonics the space $\mathcal{H}_l=\operatorname{span}\{Y_l^{m}\}$ is $(2l+1)$ -dimensional and invariant under all rotations, which is Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), Eq. (15.147), p. 763) Eq. (15.147). Its generators are the orbital angular momentum $\mathbf{L}=-\mathrm{i}\,\mathbf{r}\times\nabla$ , for which $L^2Y_l^m=l(l+1)Y_l^m$ and $L_zY_l^m=mY_l^m$ with the $2l+1$ values of $m$ each occurring once. So $\mathcal{H}_l$ is an irreducible multiplet of spin $l$ , and the $Y_l^m$ are its basis $|l,m\rangle$ — with the correct phases, because the Condon-Shortley factor $(-1)^m$ carried by $P_l^m$ is put there exactly so that $L_\pm$ acts with positive coefficients, as the remark on conventions in spherical harmonics records. Hence the matrix of $\mathcal{D}(R)$ on $\mathcal{H}_l$ is $D^{l}(R)$ :

Y_l^{m}(R^{-1}\hat{\mathbf{n}})=\bigl(\mathcal{D}(R)Y_l^m\bigr)(\hat{\mathbf{n}})=\sum_{m'}D^{l}_{m'm}(R)\,Y_l^{m'}(\hat{\mathbf{n}}).

Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eq. (16.52), p. 796) write the same identity as $Y_l^m(R\Omega)=\sum_{m'}D^l_{m'm}(R)Y_l^{m'}(\Omega)$ , their $R\Omega$ being the new coordinates of the old point, which is our $R^{-1}\hat{\mathbf{n}}$ .

Step 3: evaluate at the pole. Put $\hat{\mathbf{n}}=\hat{\mathbf{z}}$ . Only $m'=0$ survives on the right, because $Y_l^{m'}(0,\varphi)=\sqrt{(2l+1)/4\pi}\,\delta_{m'0}$ (spherical harmonics, special values). Hence

D^{l}_{0m}(R)=\sqrt{\frac{4\pi}{2l+1}}\;Y_l^{m}\bigl(R^{-1}\hat{\mathbf{z}}\bigr).

Step 4: transpose it. $D^{l}$ is unitary and $D^{l}(R)^{-1}=D^{l}(R^{-1})$ , so $D^{l}_{m0}(R)=\overline{D^{l}_{0m}(R^{-1})}$ . Applying step 3 to $R^{-1}$ ,

D^{l}_{m0}(R)=\sqrt{\frac{4\pi}{2l+1}}\;\overline{Y_l^{m}\bigl(R\hat{\mathbf{z}}\bigr)} =\sqrt{\frac{4\pi}{2l+1}}\;\overline{Y_l^{m}(\beta,\alpha)},

the last step by the formula for $R(\alpha,\beta,\gamma)\hat{\mathbf{z}}$ in the first definition. This is Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), Eq. (16.78), p. 803) lemma, which they state as $D^{l}_{m0}(R)=\sqrt{4\pi/(2l+1)}\,Y_l^m(\Omega_1)^*$ for a coordinate rotation $R$ carrying the direction $\Omega_1$ to the pole — and a coordinate rotation carrying $\Omega_1$ to the pole is our active $R$ carrying the pole to $\Omega_1$ , so $\Omega_1=(\beta,\alpha)$ and the two statements coincide.

Step 5: read off $d^{l}_{m0}$ and $D^{l}_{00}$ . Setting $\gamma=0$ in part 1, $D^{l}_{m0}(\alpha,\beta,0)=\mathrm{e}^{-\mathrm{i}m\alpha}d^{l}_{m0}(\beta)$ . Inserting the definition of $Y_l^m$ , whose only complex factor is $\mathrm{e}^{\mathrm{i}m\alpha}$ ,

\mathrm{e}^{-\mathrm{i}m\alpha}d^{l}_{m0}(\beta) =\sqrt{\frac{4\pi}{2l+1}}\sqrt{\frac{2l+1}{4\pi}\frac{(l-m)!}{(l+m)!}}\;P_l^{m}(\cos\beta)\,\mathrm{e}^{-\mathrm{i}m\alpha},

and cancelling the phase gives $d^{l}_{m0}(\beta)=\sqrt{(l-m)!/(l+m)!}\,P_l^m(\cos\beta)$ . At $m=0$ this is $P_l(\cos\beta)$ , and by part 1 the two outer phases are then trivial, so $D^{l}_{00}=P_l(\cos\beta)$ . $\square$

Remark(Where the hypotheses are used, and the conventions)Original

Irreducibility of the multiplet is what makes $D^{j}$ a square matrix of a single size and the label $j$ meaningful. On a reducible space the same construction gives a block-diagonal matrix, one block per irreducible summand, by part 4 of angular momentum algebra.
The Condon-Shortley phases are used twice and are not cosmetic. They make $d^{j}$ real (step 2 of the first proof): with arbitrary phases $|j,m\rangle\to\mathrm{e}^{\mathrm{i}\chi_m}|j,m\rangle$ the matrix picks up $\mathrm{e}^{\mathrm{i}(\chi_{m'}-\chi_m)}$ and is no longer real, and Wigner's formula is simply false. They are used again in step 2 of the last proof, to identify $Y_l^m$ with $|l,m\rangle$ rather than with some rephasing of it.
Hermiticity of $J_y$ and $J_z$ gives unitarity; it is the hypothesis behind $\sum_{m'}|D^{j}_{m'm}|^2=1$ , that is, behind the amplitudes being probabilities.
The ranges of the Euler angles matter only for uniqueness, not for the formulas. At $\beta=0$ the two outer rotations are about the same axis and only $\alpha+\gamma$ is defined; this is the familiar coordinate degeneracy of Euler angles, and it is why numerical work near $\beta=0$ or $\pi$ is better done with quaternions (quaternions) than with angles.
Half-integer $j$ needs $SU(2)$ . Part 4 gives $\mathcal{D}^{(j)}(2\pi)=(-1)^{2j}$ , so for half-integer $j$ the assignment $R\mapsto D^{j}(R)$ is not a function on $SO(3)$ at all: it is a genuine representation only of the double cover, as the rotation group and SU(2) explains. The $3\times3$ rotation matrix of the first definition is $D^{1}$ written in the Cartesian rather than the spherical basis.
Sign conventions. We put $\mathcal{D}=\mathrm{e}^{-\mathrm{i}\alpha J_z}\mathrm{e}^{-\mathrm{i}\beta J_y}\mathrm{e}^{-\mathrm{i}\gamma J_z}$ , to match the active convention $R_{\mathbf{n}}(\theta)=\exp(-\mathrm{i}\theta\,\mathbf{n}\cdot\mathbf{J})$ of the rotation group and SU(2). Hassani (Hassani (2013), Eq. (30.15), p. 972) writes the operator with the opposite sign in all three exponents, $\mathrm{e}^{\mathrm{i}J_z\psi}\mathrm{e}^{\mathrm{i}J_y\theta}\mathrm{e}^{\mathrm{i}J_z\varphi}$ , so that $D^{(j)}_{m'm}=\mathrm{e}^{\mathrm{i}(m'\psi+m\varphi)}d^{(j)}_{m'm}(\theta)$ ; correspondingly his coefficient $\phi(j,m,m';\mu)$ in Eq. (30.16) carries $(-1)^{\mu}$ where ours carries $(-1)^{k+m'-m}$ . The two are consistent: by part 4, dropping the factor $(-1)^{m'-m}$ turns $d^{j}_{m'm}(\beta)$ into $d^{j}_{m'm}(-\beta)$ , which is exactly the effect of reversing the sign of the angle in the exponent. Reading his angles as ours, $\mathcal{D}^{\text{Hassani}}(\psi,\theta,\varphi)=\mathcal{D}(-\psi,-\theta,-\varphi)$ . Nothing below depends on the choice provided one keeps it; what is not safe is to mix a $D$ -matrix from one book with a Clebsch-Gordan coefficient from another without checking.
A further convention. Many texts define the Euler angles by rotating about $x'$ rather than $y'$ in the middle step. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Exercise 3.4.1, p. 142) record the dictionary: that convention's $(\varphi,\theta,\psi)$ equals our $(\alpha+\pi/2,\ \beta,\ \gamma-\pi/2)$ , which leaves $D^{j}(R)$ itself unchanged — it is the same physical rotation in two parametrisations — and changes only the reduced matrix, which for that convention is $d^{x,j}_{m'm}(\theta) = \langle jm'\rvert\mathrm{e}^{-\mathrm{i}\theta J_x}\lvert jm\rangle = \mathrm{e}^{+\mathrm{i}(m'-m)\pi/2}\,d^{j}_{m'm}(\theta)$ , since $\mathrm{e}^{-\mathrm{i}\theta J_x} = \mathrm{e}^{+\mathrm{i}\pi J_z/2}\,\mathrm{e}^{-\mathrm{i}\theta J_y}\,\mathrm{e}^{-\mathrm{i}\pi J_z/2}$ in the sign convention used here. The phase is $\mathrm{e}^{+\mathrm{i}(m'-m)\pi/2}$ and not its conjugate: at $j=\tfrac12$ , $m'=-\tfrac12$ , $m=+\tfrac12$ , $\theta=\pi/2$ one has $d^{1/2}_{-1/2,1/2} = 1/\sqrt2$ while $\langle-\tfrac12\rvert\mathrm{e}^{-\mathrm{i}(\pi/2)J_x}\lvert\tfrac12\rangle = -\mathrm{i}/\sqrt2$ , a ratio of $-\mathrm{i}$ . So $d^{x,j}$ is not real, while $d^{j}$ is.

Method(Computing

D^{j}

for a given

j

and rotation)Original

Get the Euler angles. If the rotation is given as an axis $\mathbf{n}$ and angle $\theta$ , or as a $3\times3$ matrix $R$ , read $\beta=\arccos R_{33}$ , and then $\alpha=\operatorname{atan2}(R_{23},R_{13})$ and $\gamma=\operatorname{atan2}(R_{32},-R_{31})$ from the last column and last row. (Check against $R\hat{\mathbf{z}}$ , whose polar angles must be $(\beta,\alpha)$ .) If $\beta=0$ or $\pi$ , set $\gamma=0$ and put the whole turn into $\alpha$ .
Build $d^{j}(\beta)$ . Either exponentiate — form the $(2j+1)\times(2j+1)$ matrix $J_y$ by the recipe of angular momentum algebra and compute $\exp(-\mathrm{i}\beta J_y)$ — or use Wigner's formula entry by entry. Exponentiating is better for large $j$ : the closed formula alternates in sign and loses accuracy by cancellation.
Fill in the top-left quadrant only. The symmetries of part 4, together with $d^{j}_{-m',-m}=(-1)^{m'-m}d^{j}_{m'm}$ , reduce the work by a factor of about four. That last identity comes straight out of Wigner's formula: reindex the sum by $k\mapsto k+m'-m$ , which permutes the four denominator factorials among themselves, leaves the powers of $c$ and $s$ untouched, and changes the overall sign by $(-1)^{m'-m}$ .
Attach the phases: $D^{j}_{m'm}=\mathrm{e}^{-\mathrm{i}m'\alpha}d^{j}_{m'm}(\beta)\mathrm{e}^{-\mathrm{i}m\gamma}$ .
Check. Verify $D^{j}D^{j\dagger}=I$ ; verify one column sum $\sum_{m'}\bigl(d^{j}_{m'm}\bigr)^2=1$ (the sum is over the row index $m'$ at fixed column $m$ ); verify $\operatorname{tr}D^{j}$ against the character formula of part 3 evaluated at the rotation angle $\varphi$ , where $\cos\tfrac\varphi2=\cos\tfrac\beta2\cos\tfrac{\alpha+\gamma}{2}$ .
For integer $j$ , check the $m=0$ column against $d^{l}_{m0}(\beta)=\sqrt{(l-m)!/(l+m)!}\,P_l^m(\cos\beta)$ , which costs nothing and catches phase errors at once.

Example(The spin-

\tfrac32

rotation matrix, and a quarter turn)Original

Step 1: the matrix from Wigner's formula. Take $j=\tfrac32$ , basis ordered $m=\tfrac32,\tfrac12,-\tfrac12,-\tfrac32$ , and write $c=\cos\tfrac\beta2$ , $s=\sin\tfrac\beta2$ . The corner entry has $m=m'=\tfrac32$ , so $j+m-k=3-k$ , $j-m'-k=-k$ and $m'-m+k=k$ force $k=0$ and leave

d^{3/2}_{3/2,3/2}=\frac{\sqrt{3!\,0!\,3!\,0!}}{3!\,0!\,0!\,0!}\,c^{3}=c^{3}.

Running through the other fifteen entries the same way gives

d^{3/2}(\beta)= \begin{pmatrix} c^{3} & -\sqrt3\,c^{2}s & \sqrt3\,cs^{2} & -s^{3}\\[2pt] \sqrt3\,c^{2}s & c\,(3c^{2}-2) & s\,(3s^{2}-2) & \sqrt3\,cs^{2}\\[2pt] \sqrt3\,cs^{2} & s\,(2-3s^{2}) & c\,(3c^{2}-2) & -\sqrt3\,c^{2}s\\[2pt] s^{3} & \sqrt3\,cs^{2} & \sqrt3\,c^{2}s & c^{3} \end{pmatrix}.

Two checks: the top row and left column are related by $d_{mm'}=(-1)^{m-m'}d_{m'm}$ (signs $+,+,+$ down the first column against $-,+,-$ across the first row, the two lists differing by $(-1)^{m-m'}=-1,+1,-1$ as required), and the first column of squares sums to

c^{6}+3c^{4}s^{2}+3c^{2}s^{4}+s^{6}=(c^{2}+s^{2})^{3}=1 .

Step 2: a quarter turn. At $\beta=\pi/2$ , $c=s=1/\sqrt2$ , so $c^3=s^3=2^{-3/2}=\tfrac{\sqrt2}{4}$ , $\sqrt3c^2s=\sqrt3\cdot2^{-3/2}=\tfrac{\sqrt6}{4}$ , and $3c^2-2=\tfrac32-2=-\tfrac12$ , giving $c(3c^2-2)=-\tfrac{1}{2\sqrt2}=-\tfrac{\sqrt2}{4}$ . Hence

d^{3/2}(\tfrac\pi2)=\frac14 \begin{pmatrix} \sqrt2 & -\sqrt6 & \sqrt6 & -\sqrt2\\ \sqrt6 & -\sqrt2 & -\sqrt2 & \sqrt6\\ \sqrt6 & \sqrt2 & -\sqrt2 & -\sqrt6\\ \sqrt2 & \sqrt6 & \sqrt6 & \sqrt2 \end{pmatrix}, \qquad \bigl(d^{3/2}_{m'm}(\tfrac\pi2)\bigr)^{2}= \begin{pmatrix} \tfrac18&\tfrac38&\tfrac38&\tfrac18\\ \tfrac38&\tfrac18&\tfrac18&\tfrac38\\ \tfrac38&\tfrac18&\tfrac18&\tfrac38\\ \tfrac18&\tfrac38&\tfrac38&\tfrac18 \end{pmatrix}.

Every row and every column of the table of squares sums to $\tfrac18+\tfrac38+\tfrac38+\tfrac18=1$ , as orthogonality demands. The physical reading: a spin- $\tfrac32$ nucleus prepared with $m=+\tfrac32$ along $z$ and then measured along an axis at $90^{\circ}$ gives $m'=+\tfrac32$ with probability $\tfrac18$ , $m'=+\tfrac12$ with probability $\tfrac38$ , $m'=-\tfrac12$ with probability $\tfrac38$ and $m'=-\tfrac32$ with probability $\tfrac18$ — a binomial $(1,3,3,1)/8$ , which is no accident: the model of step 1 of the second proof realises the spin- $\tfrac32$ state as three independent spin- $\tfrac12$ coin flips, each giving heads with probability $c^2=\tfrac12$ .

Step 3: a composition check. Squaring the matrix of step 2 must give $d^{3/2}(\pi)$ , and by part 4 that is the antidiagonal matrix whose entry in column $m$ is $(-1)^{3/2-m}$ : reading the columns from $m=\tfrac32$ across to $m=-\tfrac32$ , the antidiagonal is $+1,-1,+1,-1$ . (Reading the same four numbers down the rows, from $m'=\tfrac32$ to $m'=-\tfrac32$ , reverses the list to $-1,+1,-1,+1$ , so it matters which way one reads.) Taking the $(1,4)$ entry of the square,

\tfrac{1}{16}\bigl(\sqrt2\cdot(-\sqrt2)+(-\sqrt6)\cdot\sqrt6+\sqrt6\cdot(-\sqrt6)+(-\sqrt2)\cdot\sqrt2\bigr) =\tfrac{1}{16}(-2-6-6-2)=-1 ,

and the $(1,1)$ entry,

\tfrac1{16}\bigl(\sqrt2\cdot\sqrt2+(-\sqrt6)\cdot\sqrt6+\sqrt6\cdot\sqrt6+(-\sqrt2)\cdot\sqrt2\bigr)=\tfrac1{16}(2-6+6-2)=0 ,

both as required.

Step 4: the spherical-harmonic check at $j=l=2$ . Part 5 predicts $d^{2}_{10}(\beta)=\sqrt{1!/3!}\,P_2^{1}(\cos\beta)$ . With the Condon-Shortley convention $P_2^{1}(x)=-3x\sqrt{1-x^{2}}$ , so $P_2^1(\cos\beta)=-3\cos\beta\sin\beta=-\tfrac32\sin2\beta$ and

d^{2}_{10}(\beta)=\frac{1}{\sqrt6}\Bigl(-\frac32\sin2\beta\Bigr)=-\sqrt{\frac{3}{8}}\;\sin2\beta .

Wigner's formula gives the same thing: with $j=2$ , $m=0$ , $m'=1$ the constraints $2-k\ge0$ , $k\ge0$ , $1-k\ge0$ , $1+k\ge0$ allow $k=0,1$ , and

d^{2}_{10}=(-1)^{1}\frac{\sqrt{2!\,2!\,3!\,1!}}{2!\,0!\,1!\,1!}c^{3}s+(-1)^{2}\frac{\sqrt{2!\,2!\,3!\,1!}}{1!\,1!\,0!\,2!}c\,s^{3} =-\sqrt6\,c^{3}s+\sqrt6\,cs^{3} =-\sqrt6\,cs\,(c^{2}-s^{2}),

and $cs=\tfrac12\sin\beta$ , $c^{2}-s^{2}=\cos\beta$ , so this is $-\tfrac{\sqrt6}{2}\sin\beta\cos\beta=-\tfrac{\sqrt6}{4}\sin2\beta=-\sqrt{3/8}\,\sin2\beta$ , since $\sqrt6/4=\sqrt{6/16}=\sqrt{3/8}$ . The two routes agree, which tests the phase conventions of both this unit and spherical harmonics at once.

ApplicationOriginal

Magnetic resonance. Every pulse in an NMR or ESR sequence is a rotation of the spin multiplet, so a sequence is a product of $D^{j}$ matrices and the observed signal is a matrix element of that product. For $j=\tfrac12$ the $90^{\circ}$ pulse is $d^{1/2}(\pi/2)=\tfrac{1}{\sqrt2}\bigl(\begin{smallmatrix}1&-1\\1&1\end{smallmatrix}\bigr)$ ; for the quadrupolar nuclei of solid-state NMR ( $^{23}$ Na with $j=\tfrac32$ , the case worked in the example above, and $^{27}$ Al with $j=\tfrac52$ ) the line intensities are the squares computed as in the example.

Rotational spectroscopy. The wavefunctions of a symmetric-top molecule are literally rotation matrices: $\psi_{JKM}(\alpha,\beta,\gamma)\propto\overline{D^{J}_{MK}(\alpha,\beta,\gamma)}$ , with $K$ the angular momentum along the molecular symmetry axis and $M$ that along the laboratory axis. Microwave selection rules, line strengths and Stark patterns are then integrals of products of three $D$ matrices.

Angular distributions. In a scattering or decay experiment the laboratory frame and the frame picked out by the beam differ by a rotation, so every measured angular distribution is a sum $\sum_{m}|{\textstyle\sum_{m'}}D^{j}_{m'm}a_{m'}|^2$ . The $(1,3,3,1)/8$ pattern of the example is the simplest instance: it is what a spin- $\tfrac32$ resonance produces when observed at right angles to its polarisation axis.

Orientational averaging. Averaging a tensor property over all orientations of a molecule — in a liquid, a powder or a glass — means integrating products of $D$ matrices over the group, and the orthogonality of the $D^{j}$ turns those integrals into Clebsch-Gordan coefficients. This is the machinery behind powder line shapes, Raman depolarisation ratios and the theory of liquid crystals, where the order parameter is the average of $D^{2}_{00}=P_2(\cos\beta)$ .

Products of harmonics. The lemma of part 5 is Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), pp. 803–806) route to the product formula

Y_{l_1}^{m_1}(\Omega)\,Y_{l_2}^{m_2}(\Omega)=\sum_{L}\sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2L+1)}}\,C(l_1l_2L\,|\,000)\,C(l_1l_2L\,|\,m_1m_2,m_1{+}m_2)\,Y_{L}^{m_1+m_2}(\Omega),

and hence to the integral of three harmonics of the same argument, their Eq. (16.84); it is non-zero only when $|l_1-l_2|\le l_3\le l_1+l_2$ , $m_3=m_1+m_2$ , and $l_1+l_2+l_3$ is even. Those integrals are the angular part of every two-electron Coulomb matrix element in atomic structure, and they are the reason a closed subshell is spherically symmetric. The coefficients $C$ are those of addition of angular momenta, and the general statement behind the pattern is the Wigner-Eckart theorem.

Geodesy and climate. The same matrices re-express a spherical-harmonic expansion of a field on the Earth in a rotated frame. Gravity-field models, geomagnetic models and the spectral dynamical cores of atmosphere models all need this when a pole is moved or a grid is rotated, and they use $D^{l}$ (there usually written as "Wigner rotation of spherical-harmonic coefficients") for degrees $l$ into the thousands — which is why the stable route of step 2 of the method, exponentiating rather than summing the closed formula, matters in practice.

Theorem · advanced

Spherical tensors and the Wigner-Eckart theorem

Adapted

Add to PDF

Defines a spherical tensor operator of rank $k$ by the transformation law $\mathcal{D}(R)T^{k}_{q}\mathcal{D}(R)^{-1}=\sum_{q'}D^{k}_{q'q}(R)T^{k}_{q'}$ , and proves that all $(2k+1)(2j+1)(2j'+1)$ of its matrix elements between two angular-momentum multiplets are one number — the reduced matrix element — times a Clebsch-Gordan coefficient, so that every dependence on the three magnetic quantum numbers is pure geometry.

Prerequisites: Addition of angular momenta, Angular momentum algebra, Rotation matrices of angular momentum, The rotation group and SU(2), Representations of groups, Schur's lemma, Product representations and selection rules, Spherical harmonics, Inner products and norms

Definition(Spherical tensor operator)AdaptedAdapted from Hassani (2013), Definition 24.8.4 and Eq. (24.48), p. 756; Arfken, Weber & Harris (2013), §16.3, Eqs. (16.51)–(16.52), pp. 796–797

Let $J_1,J_2,J_3$ be an angular momentum on a complex inner-product space $V$ (angular momentum algebra), with $\hbar=1$ and the inner product linear in the ket, as there. Write

\mathcal{D}(R)=\exp\bigl(-\mathrm{i}\theta\,\mathbf{n}\cdot\mathbf{J}\bigr)

for the unitary operator implementing the rotation $R=R_{\mathbf{n}}(\theta)$ , and $D^{k}_{q'q}(R)$ for the Wigner rotation matrices.

A family of $2k+1$ operators $T^{k}_{-k},\dots,T^{k}_{k}$ on $V$ , indexed by $q=-k,\dots,k$ with $2k$ a non-negative integer, is an irreducible spherical tensor operator of rank $k$ if

\mathcal{D}(R)\,T^{k}_{q}\,\mathcal{D}(R)^{-1}\;=\;\sum_{q'=-k}^{k} D^{k}_{q'q}(R)\;T^{k}_{q'} \qquad\text{for every rotation }R .

The $2k+1$ components rotate among themselves by the same matrix, and with the same placement of indices, by which the kets $|k,q\rangle$ of a spin- $k$ multiplet rotate. Hassani (Hassani (2013), Definition 24.8.4, p. 756) takes exactly this as the definition of an irreducible set of operators for a general group $G$ , with $T^{(\alpha)}_{ji}(g)$ in place of $D^{k}_{q'q}(R)$ ; Arfken, Weber and Harris (Arfken, Weber & Harris (2013), pp. 796–797) supply the name — the objects are spherical because the rotation matrices act on them, and irreducible because a multiplet of fixed $k$ has no smaller family that rotates within itself.

A rank-zero tensor is a single operator with $\mathcal{D}S\mathcal{D}^{-1}=S$ , that is, one commuting with all rotations; Hassani calls it a scalar operator, the word recording that it has only the one component.

Infinitesimal form. Differentiate the defining law along $R=R_{\mathbf{n}}(t)$ at $t=0$ . The left side gives $-\mathrm{i}[\mathbf{n}\cdot\mathbf{J},T^{k}_{q}]$ and the right side gives the matrix of $-\mathrm{i}\,\mathbf{n}\cdot\mathbf{J}$ in the spin- $k$ multiplet acting on the index $q$ , so the map $T\mapsto[J_a,T]$ acts on the label $q$ exactly as $J_a$ acts on $|k,q\rangle$ . The equivalent conditions are therefore

[J_z,\,T^{k}_{q}]=q\,T^{k}_{q}, \qquad [J_{\pm},\,T^{k}_{q}]=\sqrt{k(k+1)-q(q\pm1)}\;T^{k}_{q\pm1},

with the same coefficients as the ladder operators acting on $|k,q\rangle$ . We use the two forms interchangeably.

Two families to keep in mind.

Spherical harmonics as multiplication operators. On functions on the sphere, with $(\mathcal{D}(R)f)(\hat{\mathbf{n}})=f(R^{-1}\hat{\mathbf{n}})$ , let $T^{k}_{q}$ be multiplication by $Y_k^{q}$ (spherical harmonics). Then $\mathcal{D}(R)T^{k}_{q}\mathcal{D}(R)^{-1}$ is multiplication by $Y_k^{q}(R^{-1}\hat{\mathbf{n}})$ , and the transformation rule of that unit, $Y_k^{q}(R^{-1}\hat{\mathbf{n}})=\sum_{q'}D^{k}_{q'q}(R)Y_k^{q'}(\hat{\mathbf{n}})$ , is the defining law. So $\{Y_k^{q}\}$ is a rank- $k$ tensor; this is the case Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eq. (16.52), p. 796) write down first.
Vector operators. A triple $V_x,V_y,V_z$ with $[J_a,V_b]=\mathrm{i}\epsilon_{abc}V_c$ becomes a rank-one tensor in the spherical basis

V_{+1}=-\frac{V_x+\mathrm{i}V_y}{\sqrt2},\qquad V_{0}=V_z,\qquad V_{-1}=\frac{V_x-\mathrm{i}V_y}{\sqrt2} .

The three signs are the ones that make the commutators above come out, and they are the Condon-Shortley signs: the same recombination of $\hat{\mathbf{e}}_x,\hat{\mathbf{e}}_y,\hat{\mathbf{e}}_z$ is Arfken, Weber and Harris's (Arfken, Weber & Harris (2013), Eq. (16.87), p. 810) basis $\mathbf{k}_{\pm1},\mathbf{k}_0$ , which they obtain by diagonalising the matrix $K_3$ of their Eq. (16.86) and then fixing the phases by Condon-Shortley. In this basis the scalar product of two vector operators is

\mathbf{A}\cdot\mathbf{B}=\sum_{q=-1}^{1}(-1)^{q}A_{q}B_{-q},

as multiplying out the three terms confirms. Family 1 at $k=1$ is the special case $V_q\propto Y_1^{q}$ : the unit vector $\hat{\mathbf{r}}$ has spherical components $\hat r_q=\sqrt{4\pi/3}\;Y_1^{q}$ , as substituting $Y_1^{0}=\sqrt{3/4\pi}\,z/r$ and $Y_1^{\pm1}=\mp\sqrt{3/8\pi}\,(x\pm\mathrm{i}y)/r$ confirms.

Theorem(The Wigner-Eckart theorem)AdaptedAdapted from Hassani (2013), Eqs. (24.50)–(24.51) and the scalar-operator corollary, pp. 757–758

Let $V$ be a complex inner-product space carrying a unitary representation of the rotations, decomposed into mutually orthogonal irreducible multiplets with orthonormal Condon-Shortley bases $|\alpha,j,m\rangle$ , where $\alpha$ collects whatever labels besides $j$ are needed to tell two multiplets of the same $j$ apart. Let $\{T^{k}_{q}\}$ be a spherical tensor operator of rank $k$ on $V$ . Then there is a single number $\langle\alpha',j'\,\|\,T^{k}\,\|\,\alpha,j\rangle$ , depending on $\alpha,\alpha',j,j',k$ and not on $m,m',q$ , such that

\bigl\langle \alpha',j',m'\bigm| T^{k}_{q}\bigm|\alpha,j,m\bigr\rangle = C\bigl(k\,j\,j'\,\big|\,q\,m\,m'\bigr)\;\bigl\langle\alpha',j'\,\|\,T^{k}\,\|\,\alpha,j\bigr\rangle ,

with $C(j_1j_2J|m_1m_2M)=\langle j_1m_1;j_2m_2|J,M\rangle$ the Clebsch-Gordan coefficients of addition of angular momenta. The number is the reduced matrix element.

Three consequences are read off from the selection rules obeyed by the coefficient:

Selection rules. The matrix element vanishes unless

m'=m+q \qquad\text{and}\qquad |j-k|\le j'\le j+k .

The second is the triangle condition; in particular a rank- $k$ tensor connects $j$ to at most $2k+1$ values of $j'$ , and $j=j'=0$ is connected only by $k=0$ .

Counting. The left-hand side is a table of $(2k+1)(2j+1)(2j'+1)$ complex numbers; the right-hand side is one number times a table of Clebsch-Gordan coefficients fixed once and for all by the group. All the physics of the operator sits in the reduced matrix element, all the dependence on orientation in the coefficient. Ratios of matrix elements within one pair of multiplets are therefore pure numbers depending on nothing about the operator but its rank.
Scalar operators. For $k=0$ the coefficient is $C(0jj'|0mm')=\delta_{jj'}\delta_{mm'}$ , so a scalar operator has no matrix elements between multiplets of different $j$ , and inside one multiplet it is a multiple of the identity, the multiple depending on $\alpha,\alpha',j$ but not on $m$ . Hassani (Hassani (2013), p. 758) records this corollary and notes it is also a direct consequence of Schur's lemma.

The general-group statement. Hassani's (Hassani (2013), Eq. (24.51), p. 758) Eq. (24.51) is the same identity for an arbitrary group $G$ : for an operator $A^{(\alpha)}_i$ transforming by the $i$ th row of the irreducible representation $T^{(\alpha)}$ , and vectors transforming by $T^{(\beta)}$ and $T^{(\gamma)}$ ,

\bigl\langle\phi^{(\gamma)}_m\bigm|A^{(\alpha)}_i\bigm|\phi^{(\beta)}_j\bigr\rangle =\sum_{q} C^{*}\bigl(\alpha\beta;\gamma,q\,\big|\,ij;m\bigr)\;\bigl\langle\phi^{(\gamma)}\,\|\,A^{(\alpha)}\,\|\,\phi^{(\beta)}\bigr\rangle_{q},

with one reduced matrix element for each copy $q$ of $T^{(\gamma)}$ inside $T^{(\alpha)}\otimes T^{(\beta)}$ . For $SU(2)$ no copy is ever repeated — the group is simply reducible in Hassani's sense — the sum has one term, and the coefficients are real, which is the displayed form above.

Idea of the proofOriginal

The theorem is the statement that an operator and a ket are the same kind of object as far as rotations can tell.

Read the defining law of a tensor operator and the transformation law of a ket side by side: $T^{k}_{q}$ carries the index $q$ exactly as $|k,q\rangle$ carries its magnetic label, and $|\alpha,j,m\rangle$ carries $m$ . So the family of vectors $T^{k}_{q}|\alpha,j,m\rangle$ , as $q$ and $m$ run, rotates precisely like the product basis $|k,q\rangle\otimes|j,m\rangle$ . Everything known about that product — above all, how it splits into multiplets of total angular momentum $J$ — therefore applies to it unchanged, and the splitting is performed by the Clebsch-Gordan coefficients.

Once the family is sorted into multiplets, ask for its overlap with one fixed multiplet $|\alpha',j',m'\rangle$ . Overlaps between two multiplets are governed by Schur's lemma: an overlap table commuting with the rotation matrices on both sides is zero unless the two multiplets have the same $j$ , and is a multiple of the identity table when they do. That multiple is the reduced matrix element, and the rest — all of the $m,m',q$ dependence — is the Clebsch-Gordan coefficient used in the sorting.

So there is nothing to compute. The content is that rotations cannot distinguish "which component of the operator" from "which state of the multiplet", and the only invariant left over is a single number.

ProofAdaptedAdapted from Hassani (2013), Eqs. (24.49)–(24.51), pp. 757–758

Fix $\alpha,j,\alpha',j',k$ throughout. We follow Hassani's route — transport the vectors $T^{k}_{q}|\alpha,j,m\rangle$ into the coupled basis and then show that the overlap with a fixed multiplet cannot depend on the magnetic label — but replace his appeal to the orthogonality relations between basis vectors of irreducible representations with a direct use of Schur's lemma.

Step 1: the image vectors rotate like a tensor product. Put $w_{qm}=T^{k}_{q}|\alpha,j,m\rangle$ . Insert $\mathcal{D}^{-1}\mathcal{D}$ and use the two transformation laws in turn:

\mathcal{D}(R)\,w_{qm} =\bigl(\mathcal{D}(R)T^{k}_{q}\mathcal{D}(R)^{-1}\bigr)\,\mathcal{D}(R)\,|\alpha,j,m\rangle =\sum_{q'}\sum_{m'} D^{k}_{q'q}(R)\,D^{j}_{m'm}(R)\;w_{q'm'} .

The first factor is the definition of a rank- $k$ tensor, the second the definition of $D^{j}$ . The coefficient $D^{k}_{q'q}D^{j}_{m'm}$ is exactly the matrix of $R$ in the product representation on $|k,q\rangle\otimes|j,m\rangle$ , which is the product representation studied there. This is Hassani's Eq. (24.49).

Step 2: sort the image vectors into multiplets. The Clebsch-Gordan coefficients $C(kjJ|qmM)$ are the entries of a real orthogonal matrix taking the product basis to the coupled basis. Apply the same orthogonal matrix to the $w_{qm}$ and set

\Psi^{J}_{M}=\sum_{q,m} C\bigl(k\,j\,J\,\big|\,q\,m\,M\bigr)\;w_{qm}, \qquad\text{so that}\qquad w_{qm}=\sum_{J,M} C\bigl(k\,j\,J\,\big|\,q\,m\,M\bigr)\;\Psi^{J}_{M} ,

the inversion being the orthogonality of the coefficients. Because the coefficients depend only on the group and step 1 says the $w_{qm}$ transform exactly like the product basis, the sorted vectors transform like the coupled basis:

\mathcal{D}(R)\,\Psi^{J}_{M}=\sum_{M'} D^{J}_{M'M}(R)\;\Psi^{J}_{M'} .

Here $J$ runs over $|j-k|,\dots,j+k$ only, and $M$ over $-J,\dots,J$ ; outside that range no coefficient exists. Nothing says the $\Psi^{J}_{M}$ are non-zero or independent — they may all vanish — and nothing in what follows needs them to be.

Step 3: project onto the target multiplet. Let $P$ be the orthogonal projection of $V$ onto the span of $|\alpha',j',m'\rangle$ , $m'=-j',\dots,j'$ . Since $V$ is an orthogonal sum of multiplets and the representation is unitary, both that span and its orthogonal complement are rotation-invariant, so $P\mathcal{D}(R)=\mathcal{D}(R)P$ . Define a linear map $\Phi$ from the abstract spin- $J$ space with basis $|J,M\rangle$ into the target multiplet by

\Phi\bigl(|J,M\rangle\bigr)=P\,\Psi^{J}_{M} .

By step 2 and the commuting of $P$ with rotations, $\Phi$ intertwines: $\Phi\circ\mathcal{D}^{(J)}(R)=\mathcal{D}(R)\circ\Phi$ for every $R$ , since both sides send $|J,M\rangle$ to $\sum_{M'}D^{J}_{M'M}(R)P\Psi^{J}_{M'}$ .

Step 4: Schur's lemma. Both the spin- $J$ space and the target multiplet are irreducible under the rotations, by the definition of a multiplet. Schur's lemma therefore gives: $\Phi=0$ unless the two representations are equivalent, which for $SU(2)$ means $J=j'$ ; and when $J=j'$ , $\Phi$ is a multiple of the identification $|j',M\rangle\mapsto|\alpha',j',M\rangle$ of the two copies of the spin- $j'$ representation. Writing that multiple $c$ ,

\bigl\langle\alpha',j',m'\bigm|\Psi^{J}_{M}\bigr\rangle=\bigl\langle\alpha',j',m'\bigm|P\,\Psi^{J}_{M}\bigr\rangle=c\;\delta_{Jj'}\,\delta_{Mm'} ,

the first equality because $|\alpha',j',m'\rangle$ lies in the range of the self-adjoint projection $P$ . The number $c$ depends on $\alpha,j,k,\alpha',j'$ alone — it was produced before any magnetic label was chosen.

Step 5: assemble. Take the inner product of the second display of step 2 with $|\alpha',j',m'\rangle$ and use step 4:

\bigl\langle\alpha',j',m'\bigm|T^{k}_{q}\bigm|\alpha,j,m\bigr\rangle =\sum_{J,M} C\bigl(k\,j\,J\,\big|\,q\,m\,M\bigr)\,c\,\delta_{Jj'}\delta_{Mm'} = c\;C\bigl(k\,j\,j'\,\big|\,q\,m\,m'\bigr) .

Naming $c=\langle\alpha',j'\|T^{k}\|\alpha,j\rangle$ gives the theorem. If $j'$ lies outside $|j-k|\le j'\le j+k$ the sum is empty and the matrix element is zero, which is the triangle rule; and $C(kjj'|qmm')=0$ unless $m'=q+m$ , which is the other selection rule. Part 3 follows by putting $k=0$ , where the only coefficient is $C(0jj'|0mm')=\delta_{jj'}\delta_{mm'}$ . $\square$

Remark(Where the hypotheses are used)Original

The full tensor law, not one commutator. Step 1 needs all $2k+1$ components and the exact transformation matrix. An operator obeying only $[J_z,A]=0$ need not be a scalar operator: inside a fixed multiplet take $A=|j,m_0\rangle\langle j,m_0|$ for one chosen $m_0$ . It commutes with $J_z$ , and all its matrix elements are diagonal — but they are $\delta_{m m_0}$ , not a constant, so the conclusion of part 3 fails. What fails in the hypothesis is the commutator with $J_{\pm}$ : $[J_+,A]\neq0$ , because $A$ is not rotation-invariant. The moral is that the vanishing of one commutator fixes only the $m'=m+q$ rule; the triangle rule and the single reduced matrix element need the rest.

Irreducibility, on both sides. Step 4 uses Schur's lemma twice over, and it applies only to irreducible representations. If the target space were a reducible block — say two multiplets of the same $j'$ handled together, without the label $\alpha'$ — an intertwiner need not be a multiple of the identity, and the $m$ -independence would be lost. That is precisely why the statement carries $\alpha'$ : one reduced matrix element per pair of multiplets, not per pair of $j$ values.

Unitarity and orthogonality of the decomposition. Step 3 needs $P$ to commute with the rotations, which came from the complement of a multiplet being invariant too. For a non-unitary representation an invariant subspace need not have an invariant complement, and the projection argument collapses.

Simple reducibility. The single reduced matrix element is special to $SU(2)$ . In $k\otimes j$ each $J$ between $|j-k|$ and $j+k$ occurs exactly once, so Schur's lemma in step 4 produces one constant. For a group where an irreducible representation occurs $n\ge2$ times in a product, step 2 produces $n$ independent families $\Psi^{\gamma,q}_{m}$ and step 4 gives one constant for each: that is the label $q$ and the sum in Hassani's (Hassani (2013), Eq. (24.51), p. 758) general form. Writing the $SU(2)$ form for such a group would be a false statement, not merely an imprecise one.

Conventions, and how to detect the wrong one. Three choices enter.

Normalisation. We follow Hassani and put no extra factor on the reduced matrix element. Many texts in atomic physics define it with a $1/\sqrt{2j'+1}$ , so that their reduced matrix element is $\sqrt{2j'+1}$ times ours. Only ratios of matrix elements within a fixed pair of multiplets, such as those in the example below, are convention-free.
Coupling order. We couple the operator rank first, $C(kjj'|qmm')$ , matching Hassani's $C(\alpha\beta;\gamma|ij;m)$ with $\alpha$ the operator's representation. Coupling $j$ first costs the symmetry factor $(-1)^{k+j-j'}$ of addition of angular momenta, which changes the sign of the reduced matrix element for odd $k+j-j'$ but not of any matrix element.
Phases. The displayed form assumes Condon-Shortley phases on the multiplets and on the tensor components, which is what makes the coefficients real so that no complex conjugate is needed.

A sharp test of which normalisation a table uses: for a scalar operator our form forces the reduced matrix element to equal the matrix element, since $C(0jj|0mm)=1$ , whereas the $1/\sqrt{2j'+1}$ convention makes it $\sqrt{2j+1}$ times larger.

Method(Computing with the theorem)Original

To obtain all matrix elements of a rank- $k$ tensor between the multiplets $(\alpha,j)$ and $(\alpha',j')$ :

Check the rank. Verify $[J_z,T^{k}_{q}]=qT^{k}_{q}$ and $[J_{\pm},T^{k}_{q}]=\sqrt{k(k+1)-q(q\pm1)}\,T^{k}_{q\pm1}$ for the family at hand. For a vector operator, pass to the spherical basis first.
Apply the selection rules. If $j'$ fails $|j-k|\le j'\le j+k$ , every matrix element is zero and there is nothing to do. Otherwise only the pairs with $m'=m+q$ survive, leaving at most $\min(2j+1,2j'+1)$ entries for each value of $q$ .
Compute one matrix element directly, choosing the easiest: usually a stretched one, $m=j$ and $q=k$ , or a diagonal one at $m=m'=0$ . Any honest method will do — an integral, an explicit matrix, a known commutator.
Divide out the coefficient. The reduced matrix element is $\langle\alpha',j'\|T^{k}\|\alpha,j\rangle=\langle\alpha',j',m'|T^{k}_{q}|\alpha,j,m\rangle / C(kjj'|qmm')$ , computed at the pair used in step 3. Pick a pair whose coefficient is not zero.
Multiply back. Every other matrix element is that number times its own Clebsch-Gordan coefficient.
Check. Recompute one further matrix element directly and compare, or verify the sum rule

\sum_{q=-k}^{k}\ \sum_{m=-j}^{j}\bigl|\langle\alpha',j',m+q|T^{k}_{q}|\alpha,j,m\rangle\bigr|^{2}=(2j'+1)\,\bigl|\langle\alpha',j'\|T^{k}\|\alpha,j\rangle\bigr|^{2},

which follows from the theorem and the orthogonality of the Clebsch-Gordan coefficients, every $m'$ being hit exactly once. Its content is what Hassani (Hassani (2013), Problem 24.24, p. 760) asks for in Problem 24.24: averaging $|\langle\cdot|T^{k}_{q}|\cdot\rangle|^{2}$ over the component and state labels returns $|\langle\alpha',j'\|T^{k}\|\alpha,j\rangle|^{2}$ up to a fixed factor.

Example(A rank-two tensor on a

p

multiplet)Original

Take $V=L^2(S^2)$ with the orbital angular momentum, the multiplet $l=1$ spanned by $Y_1^{m}$ , and the rank-two tensor $T^{2}_{q}=Y_2^{q}$ acting as multiplication. The matrix elements are integrals of three spherical harmonics,

\bigl\langle 1,m'\bigm|Y_2^{q}\bigm|1,m\bigr\rangle=\int_{S^2}\overline{Y_1^{m'}}\;Y_2^{q}\;Y_1^{m}\;\mathrm{d}\Omega .

Step 1: the selection rules. With $k=2$ , $j=j'=1$ , the triangle condition $|1-2|\le1\le1+2$ holds, so the block does not vanish, and $m'=m+q$ leaves nine non-zero entries out of the $5\times3\times3=45$ in the table.

Step 2: one matrix element. Take $q=m=m'=0$ . With $Y_1^{0}=\sqrt{3/4\pi}\cos\theta$ and $Y_2^{0}=\sqrt{5/16\pi}\,(3\cos^2\theta-1)$ , substituting $t=\cos\theta$ and integrating out the azimuth gives

\bigl\langle 1,0\bigm|Y_2^{0}\bigm|1,0\bigr\rangle =2\pi\cdot\frac{3}{4\pi}\sqrt{\frac{5}{16\pi}}\int_{-1}^{1}t^2\,(3t^2-1)\,\mathrm{d}t =\frac32\sqrt{\frac{5}{16\pi}}\Bigl(\frac65-\frac23\Bigr) =\frac{1}{\sqrt{5\pi}} .

Numerically $1/\sqrt{5\pi}=0.2523133$ .

Step 3: the reduced matrix element. The coefficient at this entry is $C(2\,1\,1|0\,0\,0)=-\sqrt{2/5}=-0.6324555$ , so

\bigl\langle 1\,\|\,Y_2\,\|\,1\bigr\rangle=\frac{1/\sqrt{5\pi}}{-\sqrt{2/5}}=-\frac{1}{\sqrt{2\pi}}=-0.3989423 .

The general formula $\langle l'\|Y_k\|l\rangle=\sqrt{(2k+1)(2l+1)/4\pi(2l'+1)}\;C(k\,l\,l'|0\,0\,0)$ agrees: $\sqrt{5\cdot3/(4\pi\cdot3)}\cdot(-\sqrt{2/5})=-1/\sqrt{2\pi}$ .

Step 4: the whole block. Multiplying the reduced matrix element by the coefficients $C(2\,1\,1|q\,m\,m')$ , written $C$ in the table, gives

$q$	$m$	$m'$	$C$	matrix element
$0$	$\pm1$	$\pm1$	$+\sqrt{1/10}=0.3162278$	$-1/\sqrt{20\pi}=-0.1261566$
$0$	$0$	$0$	$-\sqrt{2/5}=-0.6324555$	$+1/\sqrt{5\pi}=+0.2523133$
$+1$	$0$	$+1$	$-\sqrt{3/10}=-0.5477226$	$+\sqrt{3/20\pi}=+0.2185097$
$+1$	$-1$	$0$	$+\sqrt{3/10}=+0.5477226$	$-\sqrt{3/20\pi}=-0.2185097$
$+2$	$-1$	$+1$	$+\sqrt{3/5}=+0.7745967$	$-\sqrt{3/10\pi}=-0.3090194$

together with the entries at $-q,-m,-m'$ , which repeat the same numbers.

Step 5: the check. Integrating $\overline{Y_1^{1}}\,Y_2^{0}\,Y_1^{1}$ directly over the sphere returns $-1/\sqrt{20\pi}=-0.1261566$ , and $\overline{Y_1^{1}}\,Y_2^{2}\,Y_1^{-1}$ returns $-\sqrt{3/10\pi}=-0.3090194$ ; each agrees with the table to the last digit shown. The sum rule of the method is a second check: adding the squares of all nine entries gives $0.0954930+0.1909859+0.1909859=0.4774648$ , and $(2l'+1)|\langle1\|Y_2\|1\rangle|^{2}=3/2\pi=0.4774648$ . The first ratio is worth pausing on: the diagonal of $Y_2^{0}$ on the three states $m=1,0,-1$ is in the ratio $-1:2:-1$ , exactly $C(211|0mm)$ , and no property of $Y_2$ beyond its rank entered. This is why the quadrupole splitting of a $p$ level has the pattern it has.

Comparison with Arfken, Weber and Harris. Their Eq. (16.84), reached by an entirely different route — the addition theorem, then the product formula for two harmonics of the same argument — reads

\int \overline{Y_{l_3}^{m_3}}\,Y_{l_1}^{m_1}\,Y_{l_2}^{m_2}\,\mathrm{d}\Omega =\sqrt{\frac{(2l_1+1)(2l_2+1)}{4\pi(2l_3+1)}}\;C(l_1l_2l_3|0\,0\,0)\;C(l_1l_2l_3|m_1m_2m_3) ,

which is the Wigner-Eckart theorem for this case with the reduced matrix element written out. Their three conditions on p. 806 — the triangle condition, $m_3=m_1+m_2$ , and $l_1+l_2+l_3$ even — are our two selection rules plus one more, the parity rule, which is extra information: it comes from $C(l_1l_2l_3|000)$ vanishing for odd $l_1+l_2+l_3$ and is invisible to the Wigner-Eckart theorem, which knows nothing about reflections. Checking the eight cases $(l_3m_3;l_1m_1;l_2m_2)$ used above and in the table against direct integration confirms the formula with $2l_3+1$ in the denominator.

Note(A printing slip in the source)Original

Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eq. (16.84), p. 805) print $4\pi(2L+1)$ in the denominator of the square root of Eq. (16.84), carrying over the summation index $L$ from their Eq. (16.83); by the point the integral is taken, the surviving term is $L=l_3$ . We write $2l_3+1$ , which is what the derivation on that page produces and what direct integration confirms to seven digits in the eight cases checked in the example above.

ApplicationOriginal

Selection rules without a single integral. The leading coupling of an atom to light is the electric dipole operator $e\mathbf{r}$ , a vector, hence a rank-one tensor. The theorem gives at once $\Delta m\in\{0,\pm1\}$ and $|\Delta j|\le1$ with $j=j'=0$ forbidden, for any atom and any state, before any wavefunction is written down; the whole atomic structure is buried in one radial number per pair of levels. The next multipole, the electric quadrupole, is rank two, and its rules are $|\Delta j|\le2$ with $0\to0$ and $\tfrac12\to\tfrac12$ forbidden. Observed line intensities inside one multiplet are then ratios of squared Clebsch-Gordan coefficients, as in the $-1:2:-1$ pattern of the example.

The projection theorem and the Landé factor. Let $\mathbf{V}$ be any vector operator and work inside one multiplet, $\alpha'=\alpha$ and $j'=j$ . Both $V_q$ and $J_q$ are rank-one tensors between the same two multiplets, so by the theorem their matrix elements share the same Clebsch-Gordan table and differ only by the ratio of reduced matrix elements:

\bigl\langle\alpha,j,m'\bigm|V_q\bigm|\alpha,j,m\bigr\rangle=\lambda\;\bigl\langle\alpha,j,m'\bigm|J_q\bigm|\alpha,j,m\bigr\rangle, \qquad \lambda=\frac{\langle\alpha,j\|V\|\alpha,j\rangle}{\langle\alpha,j\|J\|\alpha,j\rangle} ,

one constant for all $m,m',q$ . To identify $\lambda$ , contract with $\mathbf{J}$ using $\mathbf{A}\cdot\mathbf{B}=\sum_q(-1)^qA_qB_{-q}$ : since each $J_{-q}$ keeps us inside the multiplet, inserting the resolution of the identity on that multiplet is legitimate, and

\langle\mathbf{J}\cdot\mathbf{V}\rangle=\lambda\,\langle\mathbf{J}\cdot\mathbf{J}\rangle=\lambda\,j(j+1), \qquad\text{so}\qquad \lambda=\frac{\langle\mathbf{J}\cdot\mathbf{V}\rangle}{j(j+1)} .

Only the component of $\mathbf{V}$ along $\mathbf{J}$ survives inside a multiplet — the perpendicular part averages away, as the old vector model asserted without proof.

Apply this to the magnetic moment of an atom, $\boldsymbol{\mu}=-\mu_B(\mathbf{L}+g_s\mathbf{S})$ with $\mathbf{J}=\mathbf{L}+\mathbf{S}$ ; taking the electron $g$ -factor to be exactly $g_s=2$ this is $-\mu_B(\mathbf{J}+\mathbf{S})$ . The Landé $g$ -factor is defined by $\boldsymbol{\mu}\to-g_J\mu_B\mathbf{J}$ inside the multiplet, so

g_J=1+\frac{\langle\mathbf{J}\cdot\mathbf{S}\rangle}{j(j+1)} =1+\frac{j(j+1)+s(s+1)-l(l+1)}{2j(j+1)} ,

the last step from $\mathbf{L}^2=(\mathbf{J}-\mathbf{S})^2$ . For the level $^2P_{3/2}$ ( $l=1$ , $s=\tfrac12$ , $j=\tfrac32$ ): $\mathbf{J}\cdot\mathbf{S}=\tfrac12(3.75+0.75-2)=1.25$ and $j(j+1)=3.75$ , giving

g_J=1+\frac{1.25}{3.75}=\frac43=1.3333\ldots ,

while its partner $^2P_{1/2}$ ( $j=\tfrac12$ ) has $\mathbf{J}\cdot\mathbf{S}=\tfrac12(0.75+0.75-2)=-0.25$ and $g_J=1-\tfrac{0.25}{0.75}=\tfrac23=0.6666\ldots$ . These two numbers are what the anomalous Zeeman splittings of the sodium D lines measure, and they follow from the theorem and arithmetic alone, with no atomic wavefunction anywhere in the derivation.

Definition · advanced

The Lorentz group

Adapted

Add to PDF

Defines $O(3,1)$ as the matrices preserving $\eta=\mathrm{diag}(1,-1,-1,-1)$ , separates its four pieces by the signs of $\det\Lambda$ and $\Lambda^{0}{}_{0}$ , computes the six generators and the brackets $[J_j,J_k]=\mathrm{i}\epsilon_{jkl}J_l$ , $[J_j,K_k]=\mathrm{i}\epsilon_{jkl}K_l$ , $[K_j,K_k]=-\mathrm{i}\epsilon_{jkl}J_l$ , factorises every proper orthochronous transformation uniquely as a boost times a rotation, and builds the two-to-one covering $SL(2,\mathbb{C})\to SO^{+}(3,1)$ .

Prerequisites: Matrix Lie groups and Lie algebras, The rotation group and SU(2), Homomorphisms and quotient groups, Functions of matrices and the matrix exponential, Determinants, Quadratic forms and the Rayleigh quotient, Unitary and orthogonal matrices

Definition(Minkowski space and the Lorentz group)AdaptedAdapted from Hassani (2013), Example 23.2.2 item 6, pp. 706–707, and Example 29.2.10 with Eqs. (29.38)–(29.39), pp. 940–941; Arfken, Weber & Harris (2013), Eq. (17.74), p. 865

Write a point of space-time as a column $x=(x^{0},x^{1},x^{2},x^{3})^{\mathsf{T}}\in\mathbb{R}^4$ with $x^{0}=ct$ , and give $\mathbb{R}^4$ the symmetric bilinear form

\langle x,y\rangle_{\eta}=x^{\mathsf{T}}\eta\,y, \qquad \eta=\operatorname{diag}(1,-1,-1,-1), \qquad \langle x,x\rangle_{\eta}=(x^{0})^{2}-(x^{1})^{2}-(x^{2})^{2}-(x^{3})^{2} .

This is Minkowski space. A vector is timelike, null or spacelike as $\langle x,x\rangle_{\eta}$ is positive, zero or negative.

The Lorentz group is the set of real $4\times4$ matrices preserving the form,

O(3,1)=\bigl\{\,\Lambda\in M_4(\mathbb{R}) : \Lambda^{\mathsf{T}}\eta\,\Lambda=\eta \,\bigr\} ,

a subgroup of $GL(4,\mathbb{R})$ because the defining condition is preserved by products and inverses. Its elements are Lorentz transformations. Taking $\eta=\operatorname{diag}(1,-1,\dots,-1)$ to $\operatorname{diag}(\underbrace{-1,\dots,-1}_{p},\underbrace{1,\dots,1}_{n-p})$ gives the general pseudo-orthogonal group $O(p,n-p)$ of Hassani (Hassani (2013), Example 23.2.2 item 6, p. 706), of which $O(3,1)$ is the case $n=4$ , $p=3$ ; see also the classical matrix groups for the definite case $O(n)$ .

Two conventions, stated once. Hassani numbers the three spatial coordinates $1,2,3$ with $\eta_{ii}=-1$ and the time coordinate last, then records in a footnote that he relabels it $0$ , which is what we do; his invariant on p. 940 is our $\langle x,x\rangle_{\eta}$ exactly. Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eq. (17.74), p. 865) write the same $\operatorname{diag}(1,-1,-1,-1)$ and warn that the opposite overall sign is equally correct provided it is used consistently; we never use it. Indices are raised and lowered with $\eta$ , and we write $\Lambda^{\mu}{}_{\nu}$ for the entry in row $\mu$ , column $\nu$ , so that $x'^{\mu}=\Lambda^{\mu}{}_{\nu}x^{\nu}$ .

Rotations sit inside. For $R\in SO(3)$ the matrix $\widehat R=\bigl(\begin{smallmatrix}1&0\\0&R\end{smallmatrix}\bigr)$ satisfies $\widehat R^{\mathsf{T}}\eta\widehat R=\eta$ , so the rotation group embeds in $O(3,1)$ as the transformations fixing the time axis.

Definition(Boosts and rapidity)AdaptedAdapted from Arfken, Weber & Harris (2013), Eqs. (17.69)–(17.72), pp. 863–864; Hassani (2013), pp. 942–943

For a rapidity $\rho\in\mathbb{R}$ the boost along the third axis is

B_3(\rho)= \begin{pmatrix} \cosh\rho&0&0&\sinh\rho\\ 0&1&0&0\\ 0&0&1&0\\ \sinh\rho&0&0&\cosh\rho \end{pmatrix},

and $B_{\hat{\mathbf{n}}}(\rho)=\widehat R\,B_3(\rho)\,\widehat R^{-1}$ for any $R\in SO(3)$ carrying $\hat{\mathbf{z}}$ to $\hat{\mathbf{n}}$ . A pure Lorentz transformation, or boost, is a transformation of this form. Writing $\boldsymbol{\beta}=\hat{\mathbf{n}}\tanh\rho$ , so that

\beta=|\boldsymbol{\beta}|=\tanh\rho=\frac{v}{c}, \qquad \gamma=\cosh\rho=\frac{1}{\sqrt{1-\beta^{2}}}, \qquad \sinh\rho=\beta\gamma,

the boost is given in block form by

B(\boldsymbol{\beta})= \begin{pmatrix} \gamma & \gamma\,\boldsymbol{\beta}^{\mathsf{T}}\\[2pt] \gamma\,\boldsymbol{\beta} & I_3+(\gamma-1)\dfrac{\boldsymbol{\beta}\boldsymbol{\beta}^{\mathsf{T}}}{|\boldsymbol{\beta}|^{2}} \end{pmatrix},

a symmetric matrix, as Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 865) note every single boost must be. The parametrisation by $\rho$ rather than $v$ is their Eq. (17.71) (Arfken, Weber & Harris (2013), p. 863); they call $\rho$ the rapidity and point out that $\tanh\rho<1$ for every finite $\rho$ , which is why $c$ bounds $v$ and is never attained. Along one fixed direction the boosts form a one-parameter group, $B_{\hat{\mathbf{n}}}(\rho')B_{\hat{\mathbf{n}}}(\rho)=B_{\hat{\mathbf{n}}}(\rho+\rho')$ — rapidities add where velocities do not, which is their Eq. (17.72). Hassani (Hassani (2013), pp. 942–943) builds the same object as the $2\times2$ block $\bigl(\begin{smallmatrix}\cosh\xi&-\sinh\xi\\-\sinh\xi&\cosh\xi\end{smallmatrix}\bigr)$ sitting in one time and one space index, with $\xi=\tanh^{-1}(v/c)$ .

Active or passive. Ours is the active convention: $B_3(\rho)$ moves a point, and a particle at rest at the origin acquires velocity $+v\hat{\mathbf{z}}$ . Both sources write the matrix the other way round, as the change of coordinates to a frame moving at $+v$ : Arfken's $U(\rho)$ of Eq. (17.70) and Hassani's block carry the minus signs off the diagonal. The two differ by the sign of $\rho$ ,

B_3(\rho)=U(\rho)^{-1}=U(-\rho),

and nothing below depends on the choice provided it is not mixed.

Definition(Structure of the Lorentz group)AdaptedAdapted from Hassani (2013), Example 29.2.10 with Eq. (29.40), pp. 940–941, Eqs. (29.41)–(29.42), p. 943, and §30.3.3, pp. 974–975; Arfken, Weber & Harris (2013), §17.8, pp. 863–865

Let $\Lambda\in O(3,1)$ and write $a_{00}=\Lambda^{0}{}_{0}$ .

Two discrete invariants. $(\det\Lambda)^{2}=1$ and

a_{00}^{2}-\sum_{i=1}^{3}(\Lambda^{i}{}_{0})^{2}=1, \qquad a_{00}^{2}-\sum_{i=1}^{3}(\Lambda^{0}{}_{i})^{2}=1 ,

so $|a_{00}|\ge1$ and in particular $a_{00}\neq0$ . $\Lambda$ is proper if $\det\Lambda=+1$ and orthochronous if $a_{00}\ge1$ .

Four pieces. The pair $(\operatorname{sgn}\det\Lambda,\operatorname{sgn}a_{00})$ is multiplicative, so the proper orthochronous transformations form a subgroup

SO^{+}(3,1)=\{\Lambda\in O(3,1):\det\Lambda=1,\ \Lambda^{0}{}_{0}\ge1\}

of index four in $O(3,1)$ , the other three cosets being represented by space inversion $P=\operatorname{diag}(1,-1,-1,-1)$ , time reversal $T=\operatorname{diag}(-1,1,1,1)$ and $PT=-I$ . Hassani (Hassani (2013), p. 941) names the orthochronous and the proper orthochronous transformations and records that the proper orthochronous ones form a subgroup each of whose elements is joined to the identity by a continuous path; $P$ , $T$ and $PT$ are of course not among them. The converse is ours and follows from part 1: $\det$ and $\operatorname{sgn}\Lambda^{0}{}_{0}$ are continuous on $O(3,1)$ — the second because $a_{00}$ never vanishes there — and take only the values $\pm1$ , so each of the four pieces is open and closed in $O(3,1)$ and nothing outside $SO^{+}(3,1)$ can be joined to $I$ by a path. Hence $SO^{+}(3,1)$ is exactly the identity component.

Polar decomposition. Every $\Lambda\in SO^{+}(3,1)$ factors

\Lambda=B(\boldsymbol{\beta})\,\widehat R

with a unique boost $B(\boldsymbol{\beta})$ and a unique rotation $R\in SO(3)$ , namely $\gamma=\Lambda^{0}{}_{0}$ and $\gamma\beta_i=\Lambda^{i}{}_{0}$ . Consequently $SO^{+}(3,1)$ is six-dimensional and connected, and the boosts, which do not form a subgroup, are a set of coset representatives for $SO(3)$ in it.

The Lie algebra. $\mathfrak{so}(3,1)=\{X\in M_4(\mathbb{R}):X^{\mathsf{T}}\eta+\eta X=0\}$ has dimension six, with basis the antisymmetric family of Hassani's mixed-index generators (Hassani (2013), the unnumbered line above Eq. (29.41), p. 943),

(M_{\mu\nu})^{\alpha}{}_{\beta}=\delta^{\alpha}_{\nu}\,\eta_{\mu\beta}-\delta^{\alpha}_{\mu}\,\eta_{\nu\beta}, \qquad M_{\mu\nu}=-M_{\nu\mu},

which is his Eq. (29.41) with the first index raised by $\eta$ (he prints the all-lower-index form $(M_{\mu\nu})_{\alpha\beta}=\eta_{\mu\beta}\eta_{\nu\alpha}-\eta_{\nu\beta}\eta_{\mu\alpha}$ as the numbered equation). They obey his Eq. (29.42),

[M_{\mu\nu},M_{\kappa\lambda}]=\eta_{\mu\kappa}M_{\nu\lambda}-\eta_{\mu\lambda}M_{\nu\kappa}+\eta_{\nu\lambda}M_{\mu\kappa}-\eta_{\nu\kappa}M_{\mu\lambda}.

In the physicists' Hermitian-style normalisation

J_i=-\tfrac{\mathrm{i}}{2}\epsilon_{ijk}M_{jk}, \qquad K_i=\mathrm{i}\,M_{0i}\qquad(i,j,k=1,2,3),

so that $\widehat R_{\hat{\mathbf{n}}}(\theta)=\exp(-\mathrm{i}\theta\,\hat{\mathbf{n}}\cdot\mathbf{J})$ and $B_{\hat{\mathbf{n}}}(\rho)=\exp(-\mathrm{i}\rho\,\hat{\mathbf{n}}\cdot\mathbf{K})$ , the brackets are

[J_j,J_k]=\mathrm{i}\,\epsilon_{jkl}J_l, \qquad [J_j,K_k]=\mathrm{i}\,\epsilon_{jkl}K_l, \qquad [K_j,K_k]=-\,\mathrm{i}\,\epsilon_{jkl}J_l .

The three $J_i$ are Hermitian and the three $K_i$ are anti-Hermitian, which is the algebraic record of the group being non-compact. Neither family lies in $\mathfrak{so}(3,1)$ itself, which was defined just above as a set of real matrices: each of $J_i,K_i$ is $\mathrm{i}$ times a real matrix, and it is the $-\mathrm{i}J_i$ — real and antisymmetric — that span over $\mathbb{R}$ the $\mathfrak{so}(3)$ of the embedded rotations, while the $-\mathrm{i}K_i$ are real symmetric and give the three remaining real directions.

Two commuting copies of $\mathfrak{su}(2)$ , after complexifying. Put

N^{\pm}_{i}=\tfrac12\bigl(J_i\pm\mathrm{i}K_i\bigr).

Then $[N^{\pm}_j,N^{\pm}_k]=\mathrm{i}\epsilon_{jkl}N^{\pm}_l$ and $[N^{+}_j,N^{-}_k]=0$ . The two sets are not real combinations of $J$ and $K$ , so this is an isomorphism of the complexified algebra, $\mathfrak{so}(3,1)\otimes\mathbb{C}\cong\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$ , not a splitting of the real one. The two quantities $\mathbf{J}^2-\mathbf{K}^2$ and $\mathbf{J}\cdot\mathbf{K}$ commute with all six generators and are the Casimir invariants, so the irreducible representations are labelled by a pair $(j_1,j_2)$ of spins. The splitting, the Casimir pair and that labelling are Hassani's (Hassani (2013), §30.3.3, pp. 974–975), in a notation and normalisation different from ours and without the word complexified; the remark below gives the dictionary and says where we part from him.

The covering group $SL(2,\mathbb{C})$ . For $x\in\mathbb{R}^4$ set

X(x)=x^{0}I+\mathbf{x}\cdot\boldsymbol{\sigma} =\begin{pmatrix}x^{0}+x^{3}&x^{1}-\mathrm{i}x^{2}\\x^{1}+\mathrm{i}x^{2}&x^{0}-x^{3}\end{pmatrix},

a bijection from $\mathbb{R}^4$ onto the Hermitian $2\times2$ matrices with $\det X(x)=\langle x,x\rangle_{\eta}$ . For $A\in SL(2,\mathbb{C})$ the map $X\mapsto AXA^{\dagger}$ preserves Hermiticity and the determinant, so there is a unique $\Lambda(A)\in O(3,1)$ with

A\,X(x)\,A^{\dagger}=X\bigl(\Lambda(A)\,x\bigr), \qquad \Lambda(A)^{\mu}{}_{\nu}=\tfrac12\operatorname{tr}\bigl(\sigma_{\mu}A\,\sigma_{\nu}A^{\dagger}\bigr)

where $\sigma_0=I$ and $\sigma_1,\sigma_2,\sigma_3$ are the Pauli matrices, with no index raising intended. The map

\Lambda:SL(2,\mathbb{C})\longrightarrow SO^{+}(3,1)

is a surjective group homomorphism with kernel $\{\pm I\}$ , so $SO^{+}(3,1)\cong SL(2,\mathbb{C})/\{\pm I\}$ and the covering is two-to-one. On the two generating families

\Lambda\Bigl(\exp\bigl(-\tfrac{\mathrm{i}}{2}\theta\,\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}\bigr)\Bigr)=\widehat R_{\hat{\mathbf{n}}}(\theta), \qquad \Lambda\Bigl(\exp\bigl(\tfrac12\rho\,\hat{\mathbf{n}}\cdot\boldsymbol{\sigma}\bigr)\Bigr)=B_{\hat{\mathbf{n}}}(\rho) ,

so the restriction to $SU(2)$ is exactly the covering $SU(2)\to SO(3)$ of the rotation group and SU(2), and the boosts come from the Hermitian positive-definite elements.

This part is ours, not our sources': neither book covers $SO^{+}(3,1)$ by $SL(2,\mathbb{C})$ , and the proof below is the $SU(2)\to SO(3)$ construction of the rotation group and SU(2) with "traceless Hermitian" widened to "Hermitian". The nearest statement in our library is Hassani's Theorem 31.2.6 (Hassani (2013), p. 996), that $\mathrm{Spin}(\mu,\nu)\to SO(\mu,\nu)$ is a surjective homomorphism with kernel $\{1,-1\}$ ; it is stated for the full $SO(\mu,\nu)$ rather than for the identity component, and it is not accompanied there by the identification of $\mathrm{Spin}(3,1)$ with $SL(2,\mathbb{C})$ , so we prove what we need from scratch.

Idea of the proofOriginal

Three observations organise everything.

The first is that $\eta$ has one plus sign and three minuses, and that asymmetry alone produces the discrete structure. Reading the condition $\Lambda^{\mathsf{T}}\eta\Lambda=\eta$ in its $(0,0)$ entry says that the first column is a unit timelike vector: its time part beats its space part by exactly one. A unit timelike vector cannot have zero time component, so the sign of $\Lambda^{0}{}_{0}$ can never change continuously — and that sign, together with the sign of the determinant, cuts the group into four pieces.

The second is that a unit timelike vector is precisely what a boost is built from. Every $\Lambda$ sends the time axis somewhere, and wherever it sends it, there is a boost doing the same thing; dividing that boost out leaves a transformation fixing the time axis, which can only be a rotation. That is the whole of the polar decomposition, and it says that the Lorentz group is the rotations plus three more directions, one per boost parameter.

The third is that the traceless Hermitian $2\times2$ matrices form a three-dimensional real vector space, and adding the real scalars gives all $2\times2$ Hermitian matrices, four-dimensional over $\mathbb{R}$ — and on that four-dimensional space the determinant is exactly the Minkowski form. The construction that covers $SO(3)$ by $SU(2)$ therefore runs again verbatim with "traceless Hermitian" widened to "Hermitian", "unitary" widened to "unimodular", and $U\cdot U^{\dagger}$ left unchanged. Unitarity was what kept the trace — the time coordinate — fixed; dropping it is exactly what lets the time coordinate move, and the boosts are what appears.

Proof(Parts 1 and 2: the four pieces)AdaptedAdapted from Hassani (2013), Eqs. (29.38)–(29.40), pp. 940–941

Step 1: the determinant. Taking determinants in $\Lambda^{\mathsf{T}}\eta\Lambda=\eta$ and using $\det\Lambda^{\mathsf{T}}=\det\Lambda$ gives $(\det\Lambda)^{2}\det\eta=\det\eta$ , and $\det\eta=-1\neq0$ , so $(\det\Lambda)^{2}=1$ . This is Hassani's Eq. (29.38) (Hassani (2013), p. 940).

Step 2: the two relations on $a_{00}$ . The $(0,0)$ entry of $\Lambda^{\mathsf{T}}\eta\Lambda=\eta$ reads $\sum_{\mu}\Lambda^{\mu}{}_{0}\,\eta_{\mu\mu}\,\Lambda^{\mu}{}_{0}=\eta_{00}=1$ , that is $a_{00}^{2}-\sum_i(\Lambda^{i}{}_{0})^{2}=1$ . Since $\eta^{2}=I$ , the defining condition also gives $\Lambda\eta\Lambda^{\mathsf{T}}=\eta$ : right-multiplying $\Lambda^{\mathsf{T}}\eta\Lambda=\eta$ by $\Lambda^{-1}\eta$ turns the left side into $\Lambda^{\mathsf{T}}$ and the right into $\eta\Lambda^{-1}\eta$ , so

\Lambda^{\mathsf{T}}=\eta\,\Lambda^{-1}\eta, \qquad\text{equivalently}\qquad \Lambda^{-1}=\eta\,\Lambda^{\mathsf{T}}\eta ,

and substituting that second form into the identity $\Lambda\,\Lambda^{-1}\eta=\eta$ , with $\eta^{2}=I$ once more, gives $\Lambda\eta\Lambda^{\mathsf{T}}=\eta$ . Its $(0,0)$ entry is the second relation. These are Hassani's Eqs. (29.39)–(29.40) (Hassani (2013), pp. 940–941). Both force $a_{00}^{2}\ge1$ .

Step 3: orthochronicity is multiplicative. Let $\Lambda,\Lambda'$ be orthochronous, $a=\Lambda^{0}{}_{0}\ge1$ and $b=\Lambda'^{0}{}_{0}\ge1$ . Put $u_i=\Lambda^{0}{}_{i}$ and $v_i=\Lambda'^{i}{}_{0}$ , so that by step 2, $|\mathbf{u}|^{2}=a^{2}-1$ and $|\mathbf{v}|^{2}=b^{2}-1$ . Then

(\Lambda\Lambda')^{0}{}_{0}=\Lambda^{0}{}_{\mu}\Lambda'^{\mu}{}_{0}=ab+\mathbf{u}\cdot\mathbf{v} \;\ge\; ab-|\mathbf{u}||\mathbf{v}| = ab-\sqrt{(a^{2}-1)(b^{2}-1)} ,

by Cauchy-Schwarz. The right-hand side is positive, because $a^{2}b^{2}-(a^{2}-1)(b^{2}-1)=a^{2}+b^{2}-1\ge1>0$ . Here both relations of step 2 are used, one for each factor. So $(\Lambda\Lambda')^{0}{}_{0}>0$ ; and part 1, applied to the Lorentz transformation $\Lambda\Lambda'$ , gives $|(\Lambda\Lambda')^{0}{}_{0}|\ge1$ , so the entry is in fact $\ge1$ and the product is orthochronous. Since $a_{00}$ never vanishes, the same computation applied to $\Lambda^{-1}=\eta\Lambda^{\mathsf{T}}\eta$ (step 2) — whose $(0,0)$ entry is again $a_{00}$ — shows the orthochronous transformations are closed under inverses.

Step 4: four cosets. $\det$ is multiplicative, so $\Lambda\mapsto(\operatorname{sgn}\det\Lambda,\operatorname{sgn}\Lambda^{0}{}_{0})$ is a homomorphism onto $\{\pm1\}\times\{\pm1\}$ with kernel $SO^{+}(3,1)$ ; it is onto because $P$ , $T$ and $PT$ realise the three non-trivial values, as their diagonal entries show. By the first isomorphism theorem the kernel has index four. That every element of the kernel is joined to the identity by a continuous path is Hassani's (Hassani (2013), p. 941) statement, which part 3 makes concrete by writing each one as a boost times a rotation and joining both factors to $I$ ; the converse, that nothing else can be, is the continuity argument of part 2 above, so the kernel is the identity component. $\square$

Proof(Part 3: boost times rotation)Original

Step 1: read off the boost. Let $\Lambda\in SO^{+}(3,1)$ and let $c=\Lambda e_0$ be its first column, $c=(\gamma,\gamma\boldsymbol{\beta})$ with $\gamma=\Lambda^{0}{}_{0}\ge1$ and $\gamma\beta_i=\Lambda^{i}{}_{0}$ . Step 2 of the previous proof gives $\gamma^{2}(1-|\boldsymbol{\beta}|^{2})=1$ , so $|\boldsymbol{\beta}|<1$ and $\gamma=(1-|\boldsymbol{\beta}|^{2})^{-1/2}$ , exactly the relation the boost $B(\boldsymbol{\beta})$ requires. Its first column is $(\gamma,\gamma\boldsymbol{\beta})=c$ as well. (For $\boldsymbol{\beta}=\mathbf{0}$ read $B=I$ .)

Step 2: what is left fixes the time axis. Put $\widehat R:=B(\boldsymbol{\beta})^{-1}\Lambda$ . It lies in $SO^{+}(3,1)$ , being a product of two such, and $\widehat R e_0=B(\boldsymbol{\beta})^{-1}c=e_0$ by step 1, so its first column is $\delta^{\mu}_{0}$ . The first row vanishes off the diagonal too: the $(0,j)$ entry of $\widehat R^{\mathsf{T}}\eta\widehat R=\eta$ reads

\sum_{\mu}\widehat R^{\mu}{}_{0}\,\eta_{\mu\mu}\,\widehat R^{\mu}{}_{j}=\eta_{0j}=0\qquad(j=1,2,3),

and the first factor is $\delta^{\mu}_{0}$ , leaving $\widehat R^{0}{}_{j}=0$ . So $\widehat R=\bigl(\begin{smallmatrix}1&0\\0&R\end{smallmatrix}\bigr)$ , the remaining entries of $\widehat R^{\mathsf{T}}\eta\widehat R=\eta$ collapse to $-R^{\mathsf{T}}R=-I_3$ , and $\det R=\det\widehat R=1$ : $R\in SO(3)$ .

Step 3: uniqueness. In any factorisation $\Lambda=B(\boldsymbol{\beta}')\widehat R'$ the first column is $B(\boldsymbol{\beta}')e_0=(\gamma',\gamma'\boldsymbol{\beta}')$ , so $\gamma'=\gamma$ and $\boldsymbol{\beta}'=\boldsymbol{\beta}$ , whence $\widehat R'=\widehat R$ .

Step 4: dimension and connectedness. The boosts are parametrised by the open unit ball $|\boldsymbol{\beta}|<1$ in $\mathbb{R}^3$ and the rotations by the connected three-dimensional $SO(3)$ , so step 3 is a bijection $SO^{+}(3,1)\leftrightarrow\{|\boldsymbol{\beta}|<1\}\times SO(3)$ , continuous both ways. Hence $\dim SO^{+}(3,1)=3+3=6$ and $SO^{+}(3,1)$ is connected, both factors being so. Here orthochronicity is used: without $\gamma\ge1$ the first column could be $(-\gamma,\dots)$ and no boost would match it. $\square$

Proof(Part 4: the generators and their brackets)AdaptedAdapted from Hassani (2013), Eqs. (29.41)–(29.42), p. 943, and §30.3.3, pp. 974–975

Step 1: the algebra as a matrix condition. Differentiating $\Lambda(t)^{\mathsf{T}}\eta\Lambda(t)=\eta$ at $t=0$ along a curve through the identity gives $X^{\mathsf{T}}\eta+\eta X=0$ , that is, $\eta X$ is antisymmetric (matrix Lie groups and Lie algebras). The antisymmetric $4\times4$ matrices form a space of dimension $\binom42=6$ , and $X\mapsto\eta X$ is a bijection onto it, so $\dim\mathfrak{so}(3,1)=6$ . Conversely $\exp$ of such an $X$ lies in $O(3,1)$ , since $\mathrm{e}^{tX^{\mathsf{T}}}\eta\,\mathrm{e}^{tX}$ has vanishing $t$ -derivative.

Step 2: a basis. Hassani's matrices $(M_{\mu\nu})^{\alpha}{}_{\beta}=\delta^{\alpha}_{\nu}\eta_{\mu\beta}-\delta^{\alpha}_{\mu}\eta_{\nu\beta}$ satisfy $\eta M_{\mu\nu}$ antisymmetric and $M_{\mu\nu}=-M_{\nu\mu}$ , and the six with $\mu<\nu$ are independent, hence a basis. Spelling out two of them,

M_{03}=\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\1&0&0&0\end{pmatrix}, \qquad M_{12}=\begin{pmatrix}0&0&0&0\\0&0&1&0\\0&-1&0&0\\0&0&0&0\end{pmatrix} ,

and exponentiating: $\exp(\rho M_{03})=B_3(\rho)$ is the boost of the definition, since $M_{03}^{2}$ is the projector on the $(0,3)$ plane so the series sums to $\cosh$ and $\sinh$ ; and $\exp(\theta M_{12})$ is the rotation by $-\theta$ about the third axis, since $M_{12}^{2}$ is minus the projector on the $(1,2)$ plane.

Step 3: the covariant brackets. Contracting the index $\sigma$ ,

(M_{\mu\nu}M_{\kappa\lambda})^{\alpha}{}_{\beta} =\bigl(\delta^{\alpha}_{\nu}\eta_{\mu\sigma}-\delta^{\alpha}_{\mu}\eta_{\nu\sigma}\bigr)\bigl(\delta^{\sigma}_{\lambda}\eta_{\kappa\beta}-\delta^{\sigma}_{\kappa}\eta_{\lambda\beta}\bigr) =\delta^{\alpha}_{\nu}\eta_{\mu\lambda}\eta_{\kappa\beta} -\delta^{\alpha}_{\nu}\eta_{\mu\kappa}\eta_{\lambda\beta} -\delta^{\alpha}_{\mu}\eta_{\nu\lambda}\eta_{\kappa\beta} +\delta^{\alpha}_{\mu}\eta_{\nu\kappa}\eta_{\lambda\beta}.

Subtracting the same expression with $(\mu\nu)$ and $(\kappa\lambda)$ exchanged gives eight terms, which pair off two at a time: the terms carrying $\eta_{\mu\kappa}$ are $-\delta^{\alpha}_{\nu}\eta_{\mu\kappa}\eta_{\lambda\beta}+\delta^{\alpha}_{\lambda}\eta_{\mu\kappa}\eta_{\nu\beta}=\eta_{\mu\kappa}(M_{\nu\lambda})^{\alpha}{}_{\beta}$ , and the three other pairs assemble the same way. Hence

[M_{\mu\nu},M_{\kappa\lambda}]=\eta_{\mu\kappa}M_{\nu\lambda}-\eta_{\mu\lambda}M_{\nu\kappa}+\eta_{\nu\lambda}M_{\mu\kappa}-\eta_{\nu\kappa}M_{\mu\lambda},

which is Hassani's Eq. (29.42). We have also checked the identity entry by entry for all $4^{4}$ index choices.

Step 4: the physicists' basis. With $J_i=-\tfrac{\mathrm{i}}{2}\epsilon_{ijk}M_{jk}$ and $K_i=\mathrm{i}M_{0i}$ , each bracket follows from step 3 by contracting with Levi-Civita symbols and using $\eta_{jk}=-\delta_{jk}$ for spatial indices and $\eta_{00}=1$ . Explicitly, $J_3=-\mathrm{i}M_{12}$ so $\exp(-\mathrm{i}\theta J_3)=\exp(-\theta M_{12})$ is the rotation by $+\theta$ , matching the convention of the rotation group; and $K_3=\mathrm{i}M_{03}$ so $\exp(-\mathrm{i}\rho K_3)=\exp(\rho M_{03})=B_3(\rho)$ . The three relations are

[J_j,J_k]=\mathrm{i}\epsilon_{jkl}J_l, \qquad [J_j,K_k]=\mathrm{i}\epsilon_{jkl}K_l, \qquad [K_j,K_k]=-\mathrm{i}\epsilon_{jkl}J_l ,

verified matrix by matrix for all nine pairs in each line. The middle one says that $\mathbf{K}$ is a vector under rotations; the last one, with its minus sign, is where the signature enters — in $\mathfrak{so}(4)$ the sign is $+$ . Hermiticity: $M_{jk}$ is real antisymmetric so $J_i=-\tfrac{\mathrm{i}}{2}\epsilon_{ijk}M_{jk}$ is Hermitian; $M_{0i}$ is real symmetric so $K_i=\mathrm{i}M_{0i}$ is anti-Hermitian.

Step 5: part 5. The splitting is Hassani's (Hassani (2013), §30.3.3, pp. 974–975), who forms it from his real generators; we do it in the Hermitian normalisation of step 4. Expanding and using $(\pm\mathrm{i})^{2}=-1$ ,

[N^{\pm}_j,N^{\pm}_k]=\tfrac14\Bigl([J_j,J_k]-[K_j,K_k]\pm\mathrm{i}\bigl([J_j,K_k]+[K_j,J_k]\bigr)\Bigr).

The first two brackets contribute $\mathrm{i}\epsilon_{jkl}J_l-(-\mathrm{i}\epsilon_{jkl}J_l)=2\mathrm{i}\epsilon_{jkl}J_l$ , and since $[K_j,J_k]=-[J_k,K_j]=-\mathrm{i}\epsilon_{kjl}K_l=\mathrm{i}\epsilon_{jkl}K_l$ the last two contribute $2\mathrm{i}\epsilon_{jkl}K_l$ . Hence

[N^{\pm}_j,N^{\pm}_k]=\tfrac14\bigl(2\mathrm{i}\epsilon_{jkl}J_l\pm\mathrm{i}\cdot2\mathrm{i}\epsilon_{jkl}K_l\bigr) =\tfrac{\mathrm{i}}{2}\epsilon_{jkl}\bigl(J_l\pm\mathrm{i}K_l\bigr)=\mathrm{i}\epsilon_{jkl}N^{\pm}_l .

For the mixed bracket the same expansion gives $\tfrac14\bigl([J_j,J_k]+[K_j,K_k]-\mathrm{i}[J_j,K_k]+\mathrm{i}[K_j,J_k]\bigr)$ , in which the first pair is $\mathrm{i}\epsilon_{jkl}J_l-\mathrm{i}\epsilon_{jkl}J_l=0$ and the second is $-\mathrm{i}\cdot\mathrm{i}\epsilon_{jkl}K_l+\mathrm{i}\cdot\mathrm{i}\epsilon_{jkl}K_l=0$ : the two families commute. $\square$

Proof(Part 6: the covering by

SL(2,\mathbb{C})

)Original

This is the argument of the rotation group and SU(2) with "traceless Hermitian" relaxed to "Hermitian".

Step 1: $X$ is the right dictionary. A $2\times2$ matrix is Hermitian exactly when it is $x^{0}I+\mathbf{x}\cdot\boldsymbol{\sigma}$ for real $x^{0},\mathbf{x}$ , since $I,\sigma_1,\sigma_2,\sigma_3$ are Hermitian and a real basis of that four-dimensional real space. Expanding the $2\times2$ determinant of the displayed matrix,

\det X(x)=(x^{0}+x^{3})(x^{0}-x^{3})-(x^{1}-\mathrm{i}x^{2})(x^{1}+\mathrm{i}x^{2})=(x^{0})^{2}-|\mathbf{x}|^{2}=\langle x,x\rangle_{\eta} .

The inverse dictionary is $x^{0}=\tfrac12\operatorname{tr}X$ and $x^{i}=\tfrac12\operatorname{tr}(\sigma_iX)$ , from $\operatorname{tr}(\sigma_i\sigma_j)=2\delta_{ij}$ and $\operatorname{tr}\sigma_i=0$ .

Step 2: the action. For $A\in SL(2,\mathbb{C})$ , $(AXA^{\dagger})^{\dagger}=AX^{\dagger}A^{\dagger}=AXA^{\dagger}$ , so $AXA^{\dagger}$ is Hermitian, and $\det(AXA^{\dagger})=|\det A|^{2}\det X=\det X$ . By step 1 there is a unique real linear $\Lambda(A)$ with $AX(x)A^{\dagger}=X(\Lambda(A)x)$ , and it preserves $\langle\cdot,\cdot\rangle_{\eta}$ — a quadratic form determines its bilinear form by polarisation — so $\Lambda(A)\in O(3,1)$ . The entry formula follows by applying $x^{\mu}=\tfrac12\operatorname{tr}(\sigma^{\mu}X)$ to $A\sigma_{\nu}A^{\dagger}$ . Composition: $(AB)X(AB)^{\dagger}=A(BXB^{\dagger})A^{\dagger}$ , so $\Lambda(AB)=\Lambda(A)\Lambda(B)$ .

Step 3: the image. Both generating families are direct computations. For $A=\exp(-\tfrac{\mathrm{i}}{2}\theta\hat{\mathbf{n}}\cdot\boldsymbol{\sigma})\in SU(2)$ we have $A^{\dagger}=A^{-1}$ , so $\operatorname{tr}(AXA^{\dagger})=\operatorname{tr}X$ and $x^{0}$ is untouched, while on the traceless part the action is the rotation $R_{\hat{\mathbf{n}}}(\theta)$ of the rotation group and SU(2); hence $\Lambda(A)=\widehat R_{\hat{\mathbf{n}}}(\theta)$ . For $A=\exp(\tfrac12\rho\,\hat{\mathbf{n}}\cdot\boldsymbol{\sigma})$ , take $\hat{\mathbf{n}}=\hat{\mathbf{z}}$ (the general case follows by conjugating with the previous family): $A=\operatorname{diag}(\mathrm{e}^{\rho/2},\mathrm{e}^{-\rho/2})$ is Hermitian, and $AX A^{\dagger}$ multiplies the diagonal entries $x^{0}\pm x^{3}$ of $X$ by $\mathrm{e}^{\pm\rho}$ , which is

x'^{0}=x^{0}\cosh\rho+x^{3}\sinh\rho,\qquad x'^{3}=x^{0}\sinh\rho+x^{3}\cosh\rho,

that is $B_3(\rho)$ . By part 3 every element of $SO^{+}(3,1)$ is a boost times a rotation, so $\Lambda$ is onto $SO^{+}(3,1)$ . It lands in no other piece: every $A\in SL(2,\mathbb{C})$ is $\mathrm{e}^{h}U$ with $h$ traceless Hermitian and $U\in SU(2)$ — take $\mathrm{e}^{h}=(AA^{\dagger})^{1/2}$ , which is positive-definite Hermitian with determinant $|\det A|=1$ , and $U=\mathrm{e}^{-h}A$ — so $SL(2,\mathbb{C})$ is connected, being a product of two connected sets; $\Lambda$ is continuous and $\Lambda(I)=I$ , so the image lies in the connected component of the identity.

Step 4: the kernel. Suppose $AXA^{\dagger}=X$ for every Hermitian $X$ . Taking $X=I$ gives $AA^{\dagger}=I$ , so $A\in SU(2)$ ; then $A\sigma_iA^{\dagger}=\sigma_i$ for each $i$ , and the kernel computation of the rotation group and SU(2) gives $A=\pm I$ . Both are in $SL(2,\mathbb{C})$ , so the kernel is $\{\pm I\}$ and the homomorphism is two-to-one; the isomorphism $SO^{+}(3,1)\cong SL(2,\mathbb{C})/\{\pm I\}$ is the first isomorphism theorem. $\square$

Remark(Where the hypotheses are used)Original

The signature, not just the non-degeneracy. The bound $|\Lambda^{0}{}_{0}|\ge1$ of part 1 comes from $\eta$ having exactly one positive entry: the relation $a_{00}^{2}=1+\sum_i(\Lambda^{i}{}_{0})^{2}$ puts all the squares on one side only because the three space-space entries of $\eta$ share a sign. For a form of signature $(2,2)$ the same $(0,0)$ entry reads $a_{00}^{2}+a_{10}^{2}-a_{20}^{2}-a_{30}^{2}=1$ , which does not keep $a_{00}$ away from zero — the rotation mixing the two positive directions drives it through $0$ — so there is no orthochronous/antichronous dichotomy of this shape, and the second discrete label of $O(2,2)$ has to be built from the determinant of its $2\times2$ time block instead. One time direction is what makes the sign of $\Lambda^{0}{}_{0}$ meaningful, and with it the statement that an observer's future stays the future. The minus sign in $[K_j,K_k]=-\mathrm{i}\epsilon_{jkl}J_l$ has the same origin, and flipping it gives $\mathfrak{so}(4)\cong\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ as a real direct sum.

Both signs are needed, separately. Neither invariant alone cuts out the identity component. $PT=-I$ has $\det=+1$ yet $\Lambda^{0}{}_{0}=-1$ , so "proper" does not imply "connected to the identity"; $P=\operatorname{diag}(1,-1,-1,-1)$ has $\Lambda^{0}{}_{0}=1$ yet $\det=-1$ , so "orthochronous" does not either. This is why the physically relevant group carries two decorations, $S$ and $+$ .

Orthochronicity in the polar decomposition. Step 4 of that proof flags it: a boost has $\gamma\ge1$ by construction, so an antichronous $\Lambda$ can never be written as boost times rotation. It is $T$ times one that can.

Non-compactness, and why there are no finite-dimensional unitary representations. The $K_i$ are anti-Hermitian in the defining representation, so $\exp(-\mathrm{i}\rho\hat{\mathbf{n}}\cdot\mathbf{K})$ grows like $\mathrm{e}^{|\rho|}$ and the matrix entries of $SO^{+}(3,1)$ are unbounded. No change of basis repairs this, but the reason is subtler than a clash of Hermiticity. Suppose a finite-dimensional representation of the algebra were unitary, so that all six of $J_i,K_i$ act as Hermitian matrices. The three brackets raise no objection on their own: for Hermitian $A,B$ the commutator $[A,B]$ is anti-Hermitian, so each of $[J_j,J_k]=\mathrm{i}\epsilon_{jkl}J_l$ and $[K_j,K_k]=-\mathrm{i}\epsilon_{jkl}J_l$ says only that $J_l$ is Hermitian, which was the assumption. What Hermiticity really does is act on the splitting of part 5: it gives

(N^{+}_i)^{\dagger}=\tfrac12\bigl(J_i+\mathrm{i}K_i\bigr)^{\dagger}=\tfrac12\bigl(J_i-\mathrm{i}K_i\bigr)=N^{-}_i ,

so the adjoint carries each $\mathfrak{sl}(2,\mathbb{C})$ factor onto the other one — and the other one commutes with it. Now use $N^{+}_3=-\mathrm{i}[N^{+}_1,N^{+}_2]$ together with the trace identity $\operatorname{tr}\bigl([A,B]C\bigr)=\operatorname{tr}\bigl(A[B,C]\bigr)$ , valid for any three matrices:

\operatorname{tr}\bigl(N^{+}_3(N^{+}_3)^{\dagger}\bigr) =\operatorname{tr}\bigl(N^{+}_3N^{-}_3\bigr) =-\mathrm{i}\operatorname{tr}\bigl([N^{+}_1,N^{+}_2]\,N^{-}_3\bigr) =-\mathrm{i}\operatorname{tr}\bigl(N^{+}_1\,[N^{+}_2,N^{-}_3]\bigr)=0 .

The left-hand side is the sum of $|\cdot|^{2}$ over the entries of $N^{+}_3$ — this is where finite-dimensionality is used, since the trace has to exist — so $N^{+}_3=0$ , and cycling the indices gives $N^{+}_1=N^{+}_2=0$ and then $N^{-}_i=(N^{+}_i)^{\dagger}=0$ . Hence $J_i=N^{+}_i+N^{-}_i=0$ and $K_i=-\mathrm{i}(N^{+}_i-N^{-}_i)=0$ : the representation of the algebra is trivial, and since $SO^{+}(3,1)$ is connected (part 3), the representation of the group is trivial too. This is the special case for $\mathfrak{so}(3,1)$ of the standard fact that a connected non-compact simple Lie group has no non-trivial finite-dimensional unitary representation. The unitary representations of physical interest are therefore infinite-dimensional, and the finite-dimensional ones — the ones that carry fields — are not unitary.

The sign in $[K,K]$ is what does the work. Running the previous paragraph on $\mathfrak{so}(4)$ must fail, because $\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ has unitary representations in every dimension, and it does fail at exactly the step the sign controls. With $[K_j,K_k]=+\mathrm{i}\epsilon_{jkl}J_l$ the two commuting families are the real combinations $A^{\pm}_i=\tfrac12(J_i\pm K_i)$ , not $\tfrac12(J_i\pm\mathrm{i}K_i)$ , so Hermitian $J,K$ give $(A^{+}_i)^{\dagger}=A^{+}_i$ instead of $(A^{+}_i)^{\dagger}=A^{-}_i$ : the adjoint fixes each factor rather than swapping them. The trace then reads $\operatorname{tr}\bigl(A^{+}_3(A^{+}_3)^{\dagger}\bigr)=-\mathrm{i}\operatorname{tr}\bigl(A^{+}_1[A^{+}_2,A^{+}_3]\bigr)=\operatorname{tr}\bigl((A^{+}_1)^{2}\bigr)$ , with no commuting partner to annihilate it and no conclusion to draw. The two-dimensional example $A^{+}_i=\tfrac12\sigma_i$ , $A^{-}_i=0$ , that is $J_i=K_i=\tfrac12\sigma_i$ , satisfies the $\mathfrak{so}(4)$ brackets with both families Hermitian.

Complexified, not real. Part 5 is an isomorphism of complexifications. The combinations $N^{\pm}=\tfrac12(J\pm\mathrm{i}K)$ require the imaginary unit, and no real change of basis splits $\mathfrak{so}(3,1)$ : it is a simple real Lie algebra, isomorphic to $\mathfrak{sl}(2,\mathbb{C})$ regarded as a six-dimensional real algebra. Saying " $\mathfrak{so}(3,1)=\mathfrak{su}(2)\oplus\mathfrak{su}(2)$ " without the word "complexified" is a false statement, and it is the statement that would be true for $\mathfrak{so}(4)$ .

Hassani's notation in part 5, and where we leave him. Hassani (Hassani (2013), §30.3.3, pp. 974–975) reaches the same splitting from the real generators $\mathbf{M}=(M_{23},M_{31},M_{12})$ and $\mathbf{N}=(M_{01},M_{02},M_{03})$ , which in our normalisation are $\mathrm{i}\mathbf{J}$ and $-\mathrm{i}\mathbf{K}$ . His two families $\tfrac{1}{2\mathrm{i}}(\mathbf{M}+\mathrm{i}\mathbf{N})$ and $\tfrac{1}{2\mathrm{i}}(\mathbf{M}-\mathrm{i}\mathbf{N})$ — which he calls $\mathbf{J}$ and $\mathbf{K}$ , our symbols for something else — are exactly our $\mathbf{N}^{-}$ and $\mathbf{N}^{+}$ ; we checked all four identifications on the matrices of part 4. His Casimir pair $\mathbf{M}^{2}-\mathbf{N}^{2}$ , $\mathbf{M}\cdot\mathbf{N}$ is our $\mathbf{K}^{2}-\mathbf{J}^{2}$ , $\mathbf{J}\cdot\mathbf{K}$ : the same two invariants, the first with the opposite overall sign (in the defining representation his is $-3I$ where ours is $+3I$ ). Two further differences. His brackets are written for real generators and so carry no factor $\mathrm{i}$ , and he keeps that form for the two complex families as well, where his own relations for $\mathbf{M}$ and $\mathbf{N}$ give the factor back; ours is the Hermitian normalisation, in which $[N^{\pm}_j,N^{\pm}_k]=\mathrm{i}\epsilon_{jkl}N^{\pm}_l$ , and the difference is recorded in verification/discrepancies.md. And he calls $\mathfrak{so}(3,1)$ the direct sum of the two algebras without the qualifier complexified; by the previous paragraph that reading is the one we do not take, which is why part 5 says "after complexifying" and why this unit states the isomorphism for $\mathfrak{so}(3,1)\otimes\mathbb{C}$ . The labelling of the irreducible representations by a pair of spins, which the application uses, is his and is unaffected.

One boost is symmetric; two need not be. $B(\boldsymbol{\beta})$ is symmetric, as both sources note. A product of two boosts in different directions is not symmetric, so by part 3 it is not a boost: the leftover rotation is unavoidable. The worked example computes it.

Method(Identifying and decomposing a Lorentz transformation)Original

Given a real $4\times4$ matrix $\Lambda$ :

Is it Lorentz? Compute $\Lambda^{\mathsf{T}}\eta\Lambda$ and compare with $\eta$ . Sixteen entries, ten of them independent by symmetry.
Which piece? Read $\det\Lambda$ and $\Lambda^{0}{}_{0}$ . The pair of signs names the coset: $(+,+)$ is $SO^{+}(3,1)$ , $(-,+)$ is $P\,SO^{+}(3,1)$ , $(+,-)$ is $PT\,SO^{+}(3,1)$ , $(-,-)$ is $T\,SO^{+}(3,1)$ . If the signs are not $(+,+)$ , multiply by whichever of $P$ , $T$ , $PT$ restores them and continue with the result.
Is it a pure boost? A boost is symmetric. If $\Lambda\neq\Lambda^{\mathsf{T}}$ , it is not one, and a rotation is hiding in it.
Extract the boost. From the first column set $\gamma=\Lambda^{0}{}_{0}$ and $\beta_i=\Lambda^{i}{}_{0}/\gamma$ ; check $\gamma=(1-|\boldsymbol{\beta}|^{2})^{-1/2}$ as an arithmetic test of step 1. Build $B(\boldsymbol{\beta})$ from the block formula, and the rapidity is $\rho=\operatorname{artanh}|\boldsymbol{\beta}|$ .
Extract the rotation. $\widehat R=B(\boldsymbol{\beta})^{-1}\Lambda$ , whose first row and column must come out $(1,0,0,0)$ ; the lower block is $R\in SO(3)$ , and its angle is $\cos\theta=\tfrac12(\operatorname{tr}R-1)$ with axis the eigenvector of eigenvalue $1$ .
Lift to $SL(2,\mathbb{C})$ if spinors are wanted. $A=\pm\exp\bigl(\tfrac12\rho\,\hat{\boldsymbol{\beta}}\cdot\boldsymbol{\sigma}\bigr)\exp\bigl(-\tfrac{\mathrm{i}}{2}\theta\,\hat{\mathbf{m}}\cdot\boldsymbol{\sigma}\bigr)$ , with $\hat{\mathbf{m}}$ the rotation axis. The two signs are the whole content of the double cover; neither is preferred.
Check. Recover $\Lambda$ as $\tfrac12\operatorname{tr}(\sigma_{\mu}A\sigma_{\nu}A^{\dagger})$ , or simply re-multiply $B(\boldsymbol{\beta})\widehat R$ .

Example(Two perpendicular boosts, and the rotation they leave behind)Original

Boost along the first axis with $\beta_1=3/5$ , then along the second with $\beta_2=4/5$ , so that

\gamma_1=\tfrac54,\qquad \gamma_2=\tfrac53,\qquad \rho_1=\operatorname{artanh}\tfrac35=0.6931472,\qquad \rho_2=\operatorname{artanh}\tfrac45=1.0986123 .

Step 1: the composite. Multiplying $B(\beta_2\hat{\mathbf{e}}_2)\,B(\beta_1\hat{\mathbf{e}}_1)$ out,

\Lambda= \begin{pmatrix} \tfrac{25}{12}&\tfrac54&\tfrac43&0\\[2pt] \tfrac34&\tfrac54&0&0\\[2pt] \tfrac53&1&\tfrac53&0\\[2pt] 0&0&0&1 \end{pmatrix}, \qquad \Lambda^{\mathsf{T}}\eta\Lambda=\eta,\qquad \det\Lambda=1,\qquad \Lambda^{0}{}_{0}=\tfrac{25}{12}=2.0833333 .

It is proper and orthochronous — but it is not symmetric ( $\Lambda^{0}{}_{1}=\tfrac54$ against $\Lambda^{1}{}_{0}=\tfrac34$ ), so by part 3 and the last remark it is not a pure boost.

Step 2: extract the boost. The first column gives $\gamma=\tfrac{25}{12}$ and

\boldsymbol{\beta}=\frac{1}{\gamma}\Bigl(\tfrac34,\tfrac53,0\Bigr)=\Bigl(\tfrac{9}{25},\tfrac45,0\Bigr), \qquad |\boldsymbol{\beta}|^{2}=\frac{81}{625}+\frac{16}{25}=\frac{481}{625}=0.7696 ,

and indeed $(1-481/625)^{-1/2}=(144/625)^{-1/2}=25/12$ , confirming the consistency relation of step 1 of that proof. Note in passing that $\gamma=\gamma_1\gamma_2$ , which is special to perpendicular boosts. With $(\gamma-1)/|\boldsymbol{\beta}|^{2}=(13/12)(625/481)=625/444$ the block formula gives

B(\boldsymbol{\beta})= \begin{pmatrix} \tfrac{25}{12}&\tfrac34&\tfrac53&0\\[2pt] \tfrac34&\tfrac{175}{148}&\tfrac{15}{37}&0\\[2pt] \tfrac53&\tfrac{15}{37}&\tfrac{211}{111}&0\\[2pt] 0&0&0&1 \end{pmatrix} = \begin{pmatrix} 2.0833333&0.75&1.6666667&0\\ 0.75&1.1824324&0.4054054&0\\ 1.6666667&0.4054054&1.9009009&0\\ 0&0&0&1 \end{pmatrix},

symmetric and $\eta$ -orthogonal, as checked directly.

Step 3: the leftover rotation. Inverting and multiplying,

\widehat R=B(\boldsymbol{\beta})^{-1}\Lambda= \begin{pmatrix} 1&0&0&0\\ 0&\tfrac{35}{37}&-\tfrac{12}{37}&0\\ 0&\tfrac{12}{37}&\tfrac{35}{37}&0\\ 0&0&0&1 \end{pmatrix} ,

a rotation about the third axis through the angle $\theta$ with

\cos\theta=\tfrac{35}{37}=0.9459459, \qquad \sin\theta=\tfrac{12}{37}=0.3243243, \qquad \tan\theta=\tfrac{12}{35},

\theta=0.3302974\ \text{rad}=18.9246^{\circ} .

The Pythagorean triple $(12,35,37)$ is an accident of the chosen velocities, not of the construction. This angle is the Wigner rotation; accumulated around a closed orbit it is the Thomas precession that Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 864) identify as the origin of the spin-orbit factor in atomic fine structure, and its appearance here is their statement that non-collinear boosts compose into a boost plus a rotation.

Step 4: the same number from $SL(2,\mathbb{C})$ . Repeating the calculation in the covering group, $A=\exp(\tfrac12\rho_2\sigma_2)\exp(\tfrac12\rho_1\sigma_1)$ has $\det A=1$ , and $\Lambda(A)$ computed from $\tfrac12\operatorname{tr}(\sigma_{\mu}A\sigma_{\nu}A^{\dagger})$ reproduces the matrix $\Lambda$ of step 1 entry for entry. Its polar decomposition $A=HU$ has

U=\begin{pmatrix}\mathrm{e}^{-\mathrm{i}\theta/2}&0\\0&\mathrm{e}^{\mathrm{i}\theta/2}\end{pmatrix}, \qquad \cos\tfrac{\theta}{2}=0.9863939 ,

and $2\arccos(0.9863939)=0.3302974$ : the same angle, as it must be, and the half-angle is the visible sign that $SL(2,\mathbb{C})$ is the double cover.

Step 5: a general formula, checked. For two perpendicular boosts the construction above gives

\tan\theta=\frac{\gamma_1\gamma_2\beta_1\beta_2}{\gamma_1+\gamma_2} ,

which at our values is $\dfrac{(25/12)(12/25)}{35/12}=\dfrac{12}{35}$ , as found. Repeating steps 1–3 numerically at $(\beta_1,\beta_2)=(\tfrac12,\tfrac12)$ gives $\theta=0.1433476$ against $\arctan\bigl((4/3)(1/4)/(4/\sqrt3)\bigr)=0.1433476$ , and at $(\tfrac{9}{10},\tfrac13)$ gives $0.2142566$ from both routes.

ApplicationOriginal

Electromagnetism is a two-index Lorentz tensor. The electric and magnetic fields are not separately Lorentz objects; the antisymmetric field tensor $F^{\mu\lambda}$ of Arfken, Weber and Harris (Arfken, Weber & Harris (2013), Eqs. (17.77)–(17.79), p. 866) holds both, with $\mathbf{E}$ in the time row and $c\mathbf{B}$ in the spatial block, and it transforms with one copy of $\Lambda$ on each index. Their Eq. (17.79) carries an overall factor $\varepsilon_0$ , which we drop: a constant multiple of $F$ transforms in exactly the same way, and the factor cancels in the ratios of components and in the two invariants below. Their component rules, Eqs. (17.82)–(17.83), are written in the passive convention: there $v$ is the velocity, along the third axis, of the transformed coordinate system relative to the original, so the matrix used is Arfken's $U(\rho)$ and not our $B_3(\rho)$ . As printed they read

E'_x=\gamma(E_x-vB_y),\quad E'_y=\gamma(E_y+vB_x),\quad E'_z=E_z,

B'_x=\gamma\bigl(B_x+\tfrac{v}{c^{2}}E_y\bigr),\quad B'_y=\gamma\bigl(B_y-\tfrac{v}{c^{2}}E_x\bigr),\quad B'_z=B_z .

Since $B_3(\rho)=U(-\rho)$ , the active rules are these with $v$ replaced by $-v$ ; we keep the active convention declared above, so every sign below is the $-v$ one.

Take a lab field that is purely electric, $\mathbf{E}=(E_0,0,0)$ and $\mathbf{B}=\mathbf{0}$ , and boost it actively along the third axis with $\beta=3/5$ , $\gamma=5/4$ , $\beta\gamma=3/4$ . Evaluating $F'=\Lambda F\Lambda^{\mathsf{T}}$ with $\Lambda=B_3(\operatorname{artanh}\tfrac35)$ and $F$ the matrix of Eq. (17.79) gives

\mathbf{E}'=(1.25\,E_0,0,0),\qquad c\mathbf{B}'=(0,+0.75\,E_0,0) ,

a magnetic field out of nothing. (In Arfken's passive convention the same arithmetic returns $cB'_y=-0.75\,E_0$ ; the magnitude is the one he prints and only the sign tracks the convention.) The two Lorentz invariants of $F$ survive the change:

|\mathbf{E}'|^{2}-c^{2}|\mathbf{B}'|^{2}=(1.25^{2}-0.75^{2})E_0^{2}=1.0\,E_0^{2}=|\mathbf{E}|^{2}-c^{2}|\mathbf{B}|^{2}, \qquad \mathbf{E}'\cdot\mathbf{B}'=0=\mathbf{E}\cdot\mathbf{B} .

Because $|\mathbf{E}|^{2}-c^{2}|\mathbf{B}|^{2}>0$ is invariant, no boost can remove this electric field, and no boost can turn a light wave (where both invariants vanish) into a static field. The coordinate-free statement of the same facts is in the Hodge star and Maxwell's equations.

Spin, and why half-integers are allowed. Part 6 says that the symmetry group acting on quantum states is $SL(2,\mathbb{C})$ , not $SO^{+}(3,1)$ : a state need only return to itself up to a phase, so a double-valued representation of the Lorentz group is a single-valued representation of its cover. The two-component objects on which $SL(2,\mathbb{C})$ acts by $A$ and by $(A^{\dagger})^{-1}$ are the two kinds of Weyl spinor, and part 5 labels every finite-dimensional representation by the pair of spins $(j_1,j_2)$ of the two commuting $\mathfrak{su}(2)$ factors: $(0,0)$ is a scalar, $(\tfrac12,0)$ and $(0,\tfrac12)$ the Weyl spinors, their sum the Dirac spinor of Clifford algebras and spinors, and $(\tfrac12,\tfrac12)$ the four-vector — which is the defining representation of this unit, consistent with its Casimir value $\mathbf{J}^{2}-\mathbf{K}^{2}=2\bigl(j_1(j_1+1)+j_2(j_2+1)\bigr)I=3I$ , verified directly from the matrices of part 4.

What the four pieces mean physically. Electromagnetism and gravity are invariant under all of $O(3,1)$ ; the weak interaction is not, violating $P$ and $CP$ and so distinguishing the cosets. Hassani's classification of the five regions around an event — inside and on the future and past light cones, and "elsewhere" — is preserved exactly by the orthochronous transformations, which is the group-theoretic content of causality: no observer related to another by a proper orthochronous transformation disagrees about which of two causally connected events came first.

Note(What lies just outside)AdaptedAdapted from Hassani (2013), p. 707 and Eqs. (29.43)–(29.45), p. 943; Arfken, Weber & Harris (2013), p. 862

Adjoining the four translations gives the Poincaré group $P(3,1)$ , the inhomogeneous Lorentz group: Hassani (Hassani (2013), Eq. (29.43), p. 943) represents its elements as $5\times5$ matrices $\bigl(\begin{smallmatrix}\Lambda&u\\0&1\end{smallmatrix}\bigr)$ and obtains the algebra

[M_{\mu\nu},P_{\kappa}]=\eta_{\mu\kappa}P_{\nu}-\eta_{\nu\kappa}P_{\mu}, \qquad [P_{\mu},P_{\nu}]=0,

alongside Eq. (29.42). Arfken, Weber and Harris (Arfken, Weber & Harris (2013), p. 862) reserve "homogeneous Lorentz group" for the group of this unit when the distinction matters. The Poincaré algebra is not semisimple, so its invariants are not built by the recipe of Casimir operators, which treats that case separately; the two invariants that replace the quadratic one there are what the words mass and spin name.

Reference table

The classical matrix groups

Adapted

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Tabulates the defining condition, real dimension, compactness, number of connected components and Lie algebra of GL(n), SL(n), O(n), SO(n), U(n), SU(n) and Sp(2n,R), with a single constraint-counting argument that produces every dimension and proofs of every entry in the other columns.

Prerequisites: Matrix Lie groups and Lie algebras, Determinants, Unitary and orthogonal matrices, Functions of matrices and the matrix exponential, Orthonormal bases and Gram-Schmidt, Spectral theorem for Hermitian matrices, Compactness, Connectedness and simple connectivity

Definition(The seven families)AdaptedAdapted from Hassani (2013), Box 23.1.7 p. 705, Example 23.2.2 and Boxes 23.2.3–23.2.4, pp. 706–707; Lee (2013), Examples 7.18, 7.27–7.30, pp. 158, 166–168, and Problem 22-3, p. 591

Write $M(n,\mathbb{F})$ for the $n \times n$ matrices over $\mathbb{F} = \mathbb{R}$ or $\mathbb{C}$ , $A^{\mathsf T}$ for the transpose, $A^{\dagger} = \overline{A}^{\mathsf T}$ for the conjugate transpose, and $I$ for the identity matrix. All of the following are groups under matrix multiplication, and each is a matrix Lie group.

General linear. $\mathrm{GL}(n,\mathbb{F}) = \{A \in M(n,\mathbb{F}) : \det A \neq 0\}$ , the invertible matrices. Hassani defines it first as the group $\mathrm{GL}(\mathcal{V})$ of invertible endomorphisms of a vector space $\mathcal{V}$ , and writes $\mathrm{GL}(n,\mathbb{C})$ for $\mathrm{GL}(\mathbb{C}^n)$ .
Special linear. $\mathrm{SL}(n,\mathbb{F}) = \{A : \det A = 1\}$ . Since $\det$ is a homomorphism into the multiplicative group $\mathbb{F}^{\times}$ , this is the kernel of $\det$ and therefore a normal subgroup of $\mathrm{GL}(n,\mathbb{F})$ .
Orthogonal and special orthogonal. $\mathrm{O}(n) = \{A \in M(n,\mathbb{R}) : A^{\mathsf T}A = I\}$ , equivalently the real matrices preserving the Euclidean dot product, equivalently those whose columns are orthonormal (see unitary and orthogonal matrices); and $\mathrm{SO}(n) = \mathrm{O}(n) \cap \mathrm{SL}(n,\mathbb{R})$ , the rigid rotations of $\mathbb{R}^n$ .
Unitary and special unitary. $\mathrm{U}(n) = \{A \in M(n,\mathbb{C}) : A^{\dagger}A = I\}$ , the matrices preserving the Hermitian inner product of $\mathbb{C}^n$ , and $\mathrm{SU}(n) = \mathrm{U}(n) \cap \mathrm{SL}(n,\mathbb{C})$ .
Symplectic. Let $n \ge 1$ and put
$J = \begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix} \in M(2n,\mathbb{R}), \qquad J^{\mathsf T} = -J, \qquad J^2 = -I_{2n} .$
Then $\mathrm{Sp}(2n,\mathbb{R}) = \{A \in M(2n,\mathbb{R}) : A^{\mathsf T}J A = J\}$ : the real matrices preserving the antisymmetric bilinear form $\omega(x,y) = x^{\mathsf T}Jy$ on $\mathbb{R}^{2n}$ . This is Hassani's definition. Lee instead takes $J$ with $n$ copies of $\bigl(\begin{smallmatrix} 0 & 1 \\ -1 & 0\end{smallmatrix}\bigr)$ down the diagonal; the two matrices differ by a permutation of coordinates, so the two groups are conjugate inside $\mathrm{GL}(2n,\mathbb{R})$ and everything tabulated below is the same for both.

The Lie algebra $\mathfrak{g}$ of such a group $G$ is, as in matrix Lie groups and Lie algebras, the set of $X$ with $e^{tX} \in G$ for all real $t$ ; it is a real vector space, and $\dim_{\mathbb{R}} G = \dim_{\mathbb{R}} \mathfrak{g}$ . Throughout, dimension means dimension as a real manifold, so that $\mathrm{GL}(n,\mathbb{C})$ , an open subset of $M(n,\mathbb{C}) \cong \mathbb{R}^{2n^2}$ , has dimension $2n^2$ and not $n^2$ .

Note(The catalogue)AdaptedAdapted from Lee (2013), Examples 7.18, 7.27–7.30 and Problem 8-29, pp. 158, 166–168, 203–204, and Propositions 21.34–21.35, pp. 558–559

Every entry is proved or cited below. "Components" is the number of connected components; a group with one component is connected.

Group	Defining condition	$\dim_{\mathbb{R}}$	Compact	Components	Lie algebra
$\mathrm{GL}(n,\mathbb{R})$	$\det A \neq 0$	$n^2$	no	$2$	$\mathfrak{gl}(n,\mathbb{R})$ : all real $X$
$\mathrm{GL}(n,\mathbb{C})$	$\det A \neq 0$	$2n^2$	no	$1$	$\mathfrak{gl}(n,\mathbb{C})$ : all complex $X$
$\mathrm{SL}(n,\mathbb{R})$	$\det A = 1$	$n^2 - 1$	no	$1$	$\operatorname{tr}X = 0$ , $X$ real
$\mathrm{SL}(n,\mathbb{C})$	$\det A = 1$	$2n^2 - 2$	no	$1$	$\operatorname{tr}X = 0$ , $X$ complex
$\mathrm{O}(n)$	$A^{\mathsf T}A = I$	$n(n-1)/2$	yes	$2$	$X^{\mathsf T} + X = 0$
$\mathrm{SO}(n)$	$A^{\mathsf T}A = I$ , $\det A = 1$	$n(n-1)/2$	yes	$1$	$X^{\mathsf T} + X = 0$
$\mathrm{U}(n)$	$A^{\dagger}A = I$	$n^2$	yes	$1$	$X^{\dagger} + X = 0$
$\mathrm{SU}(n)$	$A^{\dagger}A = I$ , $\det A = 1$	$n^2 - 1$	yes	$1$	$X^{\dagger} + X = 0$ , $\operatorname{tr}X = 0$
$\mathrm{Sp}(2n,\mathbb{R})$	$A^{\mathsf T}JA = J$	$n(2n+1)$	no	$1$	$X^{\mathsf T}J + JX = 0$

The algebras carry the names $\mathfrak{sl}$ , $\mathfrak{o}$ , $\mathfrak{u}$ , $\mathfrak{su}$ , $\mathfrak{sp}$ matching their groups; Lee records the first five of them in Problem 8-29. Because $\mathrm{SO}(n)$ is an open subgroup of $\mathrm{O}(n)$ and $\mathrm{GL}^{+}(n,\mathbb{R})$ is an open subgroup of $\mathrm{GL}(n,\mathbb{R})$ , each has the same Lie algebra and the same dimension as the larger group: passing to the determinant-one part removes a component, not a direction. For $\mathrm{U}(n) \to \mathrm{SU}(n)$ it does remove a direction, because $\det$ on $\mathrm{U}(n)$ takes every value on the unit circle rather than only $\pm 1$ .

Two further columns are worth recording, for the small cases that occur most often in physics. The rank is the largest number of mutually commuting independent generators, which Arfken, Weber and Harris introduce because it is the number of labels (quantum numbers) a basis of a representation needs.

Group	$\dim$	Rank	Familiar description
$\mathrm{SO}(2) \cong \mathrm{U}(1)$	$1$	$1$	the circle; rotations of the plane; phase factors $e^{i\theta}$
$\mathrm{SO}(3)$	$3$	$1$	rotations of space (see the rotation group and SU(2))
$\mathrm{SU}(2)$	$3$	$1$	the unit quaternions; the three-sphere $S^3$ (see quaternions)
$\mathrm{SU}(3)$	$8$	$2$	colour and flavour symmetry; eight generators
$\mathrm{Sp}(2,\mathbb{R}) = \mathrm{SL}(2,\mathbb{R})$	$3$	$1$	area-preserving linear maps of the plane

The identification $\mathrm{Sp}(2,\mathbb{R}) = \mathrm{SL}(2,\mathbb{R})$ is immediate from the definition: for $n = 1$ , $J = \bigl(\begin{smallmatrix} 0 & 1 \\ -1 & 0\end{smallmatrix}\bigr)$ and $A^{\mathsf T}JA = (\det A)\,J$ for every $2 \times 2$ real $A$ , so the condition $A^{\mathsf T}JA = J$ says exactly $\det A = 1$ . The dimensions agree: $n(2n+1) = 3 = n^2 - 1$ at $n = 1$ and $n = 2$ respectively.

Idea of the proofOriginal

Every group in the table except the two general linear ones is cut out of the matrices by an equation $\Phi(A) = c$ , where $\Phi$ is a quadratic or polynomial map. Two observations turn that into all of the dimension column. First, differentiating the equation at $A = I$ gives a linear condition on the tangent direction $X$ , and that linear condition is the Lie algebra. Second, the number of independent equations is the dimension of the space $\Phi$ lands in, provided $\Phi$ is onto a neighbourhood in that space — which one checks by showing the derivative at $I$ is surjective. So each dimension is a subtraction: the dimension of the matrices minus the dimension of the space of constraints.

Compactness is then a two-line matter of whether the defining equation bounds the entries, and connectedness comes from factorising a group element into a piece one can shrink to the identity along an obvious path and a piece lying in a group already known to be connected.

Proof(The dimension column)AdaptedAdapted from Lee (2013), Example 7.18(c),(e), p. 158, and Examples 7.27 and 7.29, pp. 166–168

Throughout, $\mathfrak{g}$ denotes the set of $X$ with $e^{tX} \in G$ for all $t \in \mathbb{R}$ , and we use the fact from matrix Lie groups and Lie algebras that $\mathfrak{g}$ is the kernel of the derivative at $I$ of the defining constraint, and that $\dim G = \dim_{\mathbb{R}}\mathfrak{g}$ .

Step 1 (the two general linear groups). $\det$ is a polynomial in the entries, hence continuous, so $\mathrm{GL}(n,\mathbb{F})$ is the preimage of $\mathbb{F}\setminus\{0\}$ : an open subset of $M(n,\mathbb{F})$ . An open subset has the dimension of the whole space, which is $n^2$ for $\mathbb{F} = \mathbb{R}$ and $2n^2$ for $\mathbb{F} = \mathbb{C}$ counted over $\mathbb{R}$ . There is no constraint to differentiate, and $\mathfrak{gl}(n,\mathbb{F})$ is all of $M(n,\mathbb{F})$ . This is Lee's Example 1.27.

Step 2 (special linear). Write $\Phi(A) = \det A$ , a homomorphism $\mathrm{GL}(n,\mathbb{F}) \to \mathbb{F}^{\times}$ with $\mathrm{SL}(n,\mathbb{F}) = \Phi^{-1}(1)$ . Differentiating at the identity, Jacobi's formula gives $\left.\tfrac{d}{dt}\right|_{t=0}\det(I + tX) = \operatorname{tr}X$ , so the tangent condition is $\operatorname{tr}X = 0$ and

\mathfrak{sl}(n,\mathbb{F}) = \{X \in M(n,\mathbb{F}) : \operatorname{tr}X = 0\} .

The trace is a non-zero linear functional, so it has a one-dimensional image over $\mathbb{F}$ : one real condition when $\mathbb{F} = \mathbb{R}$ and two when $\mathbb{F} = \mathbb{C}$ . Hence $\dim \mathrm{SL}(n,\mathbb{R}) = n^2 - 1$ and $\dim_{\mathbb{R}}\mathrm{SL}(n,\mathbb{C}) = 2n^2 - 2$ , which are Lee's counts, there obtained from the surjectivity of $\det$ . That $\operatorname{tr}X = 0$ is also sufficient for membership follows from $\det e^{tX} = e^{t\operatorname{tr}X}$ , recorded in the matrix exponential.

Step 3 (orthogonal). Let $\Phi(A) = A^{\mathsf T}A$ . Since $\Phi(A)^{\mathsf T} = \Phi(A)$ , the map lands in the symmetric matrices $\mathrm{Sym}(n)$ , of dimension $n(n+1)/2$ — one free entry on and above the diagonal. Differentiating along $A = I + tX$ ,

\left.\frac{d}{dt}\right|_{t=0}(I + tX)^{\mathsf T}(I + tX) = X^{\mathsf T} + X ,

so $\mathfrak{o}(n) = \{X : X^{\mathsf T} + X = 0\}$ , the antisymmetric matrices. The derivative $X \mapsto X^{\mathsf T}+X$ is onto $\mathrm{Sym}(n)$ : given $S$ symmetric, $X = \tfrac12 S$ has $X^{\mathsf T}+X = S$ . So the constraint is exactly $n(n+1)/2$ independent conditions, and

\dim \mathrm{O}(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2} ,

which also counts the strictly-upper-triangular entries of an antisymmetric matrix directly. $\mathrm{SO}(n)$ is an open subset of $\mathrm{O}(n)$ , by Step 6 below, so it has the same dimension and the same Lie algebra. This is Lee's Examples 7.27 and 7.28.

Step 4 (unitary). Let $\Phi(A) = A^{\dagger}A$ , which lands in the Hermitian matrices $\mathrm{Herm}(n)$ . A Hermitian matrix has $n$ real diagonal entries and $n(n-1)/2$ complex entries above the diagonal, so $\dim_{\mathbb{R}}\mathrm{Herm}(n) = n + 2 \cdot n(n-1)/2 = n^2$ . The derivative at $I$ is $X \mapsto X^{\dagger} + X$ , giving

\mathfrak{u}(n) = \{X \in M(n,\mathbb{C}) : X^{\dagger} + X = 0\} ,

the anti-Hermitian matrices; and it is onto $\mathrm{Herm}(n)$ , since $X = \tfrac12 H$ works for Hermitian $H$ . Counting over $\mathbb{R}$ ,

\dim \mathrm{U}(n) = 2n^2 - n^2 = n^2 .

Step 5 (special unitary). For $A \in \mathrm{U}(n)$ , $|\det A| = 1$ , because $1 = \det(A^{\dagger}A) = \overline{\det A}\,\det A$ . So on $\mathrm{U}(n)$ the extra condition $\det A = 1$ is one real condition, not two: it pins the phase of a number already known to lie on the unit circle. The tangent version is $\operatorname{tr}X = 0$ , and for anti-Hermitian $X$ the trace is purely imaginary, again one real condition. Hence

\mathfrak{su}(n) = \{X : X^{\dagger}+X = 0,\ \operatorname{tr}X = 0\}, \qquad \dim \mathrm{SU}(n) = n^2 - 1 .

Step 6 (symplectic). Let $\Phi(A) = A^{\mathsf T}JA$ . Then $\Phi(A)^{\mathsf T} = A^{\mathsf T}J^{\mathsf T}A = -\Phi(A)$ , so $\Phi$ lands in the antisymmetric $2n \times 2n$ matrices, of dimension $\tfrac12 (2n)(2n-1) = n(2n-1)$ . Differentiating at $I$ ,

\left.\frac{d}{dt}\right|_{t=0}(I+tX)^{\mathsf T}J(I+tX) = X^{\mathsf T}J + JX ,

so $\mathfrak{sp}(2n,\mathbb{R}) = \{X : X^{\mathsf T}J + JX = 0\}$ . The image of $X \mapsto X^{\mathsf T}J + JX$ is antisymmetric, since $(X^{\mathsf T}J + JX)^{\mathsf T} = J^{\mathsf T}X + X^{\mathsf T}J^{\mathsf T} = -(JX + X^{\mathsf T}J)$ , and it is onto the antisymmetric matrices: given $C$ with $C^{\mathsf T} = -C$ , put $X = -\tfrac12 JC$ . Then, using $J^{\mathsf T} = -J$ and $J^2 = -I$ ,

X^{\mathsf T}J + JX = -\tfrac12 C^{\mathsf T}J^{\mathsf T}J - \tfrac12 J^2 C = -\tfrac12 (-C)(-J)J + \tfrac12 C = \tfrac12 C + \tfrac12 C = C .

Therefore

\dim \mathrm{Sp}(2n,\mathbb{R}) = (2n)^2 - n(2n-1) = 4n^2 - 2n^2 + n = n(2n+1) . \qquad\square

Proof(The compactness column)AdaptedAdapted from Lee (2013), Example 7.27, p. 167 (the closed-and-bounded argument for O(n))

Each group is a subset of a finite-dimensional real vector space, so by Heine–Borel compactness is closedness together with boundedness.

Step 1 (the four compact ones). Each of $\mathrm{O}(n)$ , $\mathrm{SO}(n)$ , $\mathrm{U}(n)$ , $\mathrm{SU}(n)$ is defined by equations between continuous functions of the entries, hence is closed. For boundedness, the condition $A^{\dagger}A = I$ (or $A^{\mathsf T}A = I$ in the real case) says that the columns $a_1,\dots,a_n$ are orthonormal, so every entry has modulus at most $1$ and the Frobenius norm is

\|A\|_F^2 = \sum_{j=1}^{n}\|a_j\|^2 = n .

The group therefore sits inside a sphere of radius $\sqrt{n}$ and is bounded. This is Lee's argument for $\mathrm{O}(n)$ ; it uses only orthonormality of the columns, so it applies verbatim with $\dagger$ in place of $\mathsf T$ . $\mathrm{SO}(n)$ and $\mathrm{SU}(n)$ are closed subsets of compact sets, hence compact.

Step 2 (the five non-compact ones). Each fails boundedness, and exhibiting one unbounded curve in the group is enough.

$\mathrm{SL}(n,\mathbb{F})$ and hence $\mathrm{GL}(n,\mathbb{F})$ contain $\operatorname{diag}(t, t^{-1}, 1, \dots, 1)$ for every $t > 0$ , whose norm tends to infinity with $t$ .
$\mathrm{Sp}(2n,\mathbb{R})$ contains $Z_t = \operatorname{diag}(t,\dots,t,\,t^{-1},\dots,t^{-1})$ , with $n$ entries of each, for every $t>0$ : writing $Z_t$ in the block form $\bigl(\begin{smallmatrix} tI & 0 \\ 0 & t^{-1}I\end{smallmatrix}\bigr)$ ,
$Z_t^{\mathsf T}JZ_t = \begin{pmatrix} tI & 0 \\ 0 & t^{-1}I\end{pmatrix}\begin{pmatrix} 0 & I \\ -I & 0\end{pmatrix}\begin{pmatrix} tI & 0 \\ 0 & t^{-1}I\end{pmatrix} = \begin{pmatrix} 0 & I \\ -I & 0 \end{pmatrix} = J ,$
the factors $t$ and $t^{-1}$ cancelling in each off-diagonal block. Again $\|Z_t\| \to \infty$ .

$\mathrm{GL}(n,\mathbb{R})$ and $\mathrm{GL}(n,\mathbb{C})$ are not closed either: $\operatorname{diag}(s,1,\dots,1) \to \operatorname{diag}(0,1,\dots,1)$ as $s \to 0$ , and the limit is singular. $\square$

Proof(The components column)AdaptedAdapted from Lee (2013), Propositions 21.34 and 21.35 with the QR argument, pp. 558–559

Step 1 (quoted). Lee's Proposition 21.34 proves that $\mathrm{SO}(n)$ , $\mathrm{U}(n)$ and $\mathrm{SU}(n)$ are connected for every $n \ge 1$ , by induction along the fibrations $\mathrm{SO}(n)/\mathrm{SO}(n-1) \cong S^{n-1}$ and $\mathrm{U}(n)/\mathrm{U}(n-1) \cong \mathrm{SU}(n)/\mathrm{SU}(n-1) \cong S^{2n-1}$ . His Proposition 21.35 proves that $\mathrm{GL}(n,\mathbb{R})$ has exactly the two components $\mathrm{GL}^{\pm}(n,\mathbb{R})$ of positive and of negative determinant. We quote both.

Step 2 ( $\mathrm{O}(n)$ has two components). From $A^{\mathsf T}A = I$ and $\det A^{\mathsf T} = \det A$ we get $(\det A)^2 = 1$ , so $\det A = \pm 1$ and $\mathrm{O}(n)$ is the disjoint union of the two relatively open sets $\mathrm{O}^{\pm}(n)$ on which the sign is fixed. The first is $\mathrm{SO}(n)$ , connected by Step 1. Fix any $R \in \mathrm{O}^{-}(n)$ , for instance the reflection $\operatorname{diag}(-1,1,\dots,1)$ ; then $A \mapsto RA$ is a homeomorphism of $\mathrm{SO}(n)$ onto $\mathrm{O}^{-}(n)$ , since it is continuous with continuous inverse $A \mapsto R^{-1}A$ and it reverses the sign of the determinant. So $\mathrm{O}^{-}(n)$ is connected too, and the two sets are the components. (This also shows $\mathrm{SO}(n)$ is open in $\mathrm{O}(n)$ , used in Step 3 of the dimension proof.)

Step 3 ( $\mathrm{GL}(n,\mathbb{C})$ is connected). Let $A, B \in \mathrm{GL}(n,\mathbb{C})$ and consider $p(z) = \det\bigl(A + z(B-A)\bigr)$ , a polynomial in the single complex variable $z$ . It is not identically zero, because $p(0) = \det A \neq 0$ ; so it has at most $n$ roots. The complex plane minus finitely many points is path-connected, and it contains $0$ and $1$ , so there is a path $z(s)$ from $0$ to $1$ avoiding all roots. Then $s \mapsto A + z(s)(B-A)$ is a path from $A$ to $B$ inside $\mathrm{GL}(n,\mathbb{C})$ . The contrast with the real case is exactly that $\mathbb{R}\setminus\{\text{finitely many points}\}$ is not connected, which is why $\det$ can separate $\mathrm{GL}(n,\mathbb{R})$ and cannot separate $\mathrm{GL}(n,\mathbb{C})$ .

Step 4 ( $\mathrm{SL}(n,\mathbb{R})$ and $\mathrm{SL}(n,\mathbb{C})$ are connected). We follow Lee's QR route, with the path chosen to keep the determinant equal to $1$ rather than merely positive. Let $A$ have $\det A = 1$ . Gram–Schmidt applied to the columns of $A$ gives $A = QR$ with $Q$ orthogonal (unitary in the complex case) and $R$ upper triangular with real positive diagonal entries $r_1,\dots,r_n$ . Taking determinants, $\det Q \cdot \prod_j r_j = 1$ ; the product is real and positive and $|\det Q| = 1$ , so $\prod_j r_j = 1$ and $\det Q = 1$ . Thus $Q \in \mathrm{SO}(n)$ , respectively $Q \in \mathrm{SU}(n)$ . Now define, for $0 \le t \le 1$ ,

(R_t)_{jj} = r_j^{\,1-t}, \qquad (R_t)_{jk} = (1-t)\,R_{jk} \quad (j < k), \qquad (R_t)_{jk} = 0 \quad (j > k) .

Each $R_t$ is upper triangular with positive diagonal, $\det R_t = \prod_j r_j^{\,1-t} = 1^{\,1-t} = 1$ , and $t \mapsto R_t$ is continuous with $R_0 = R$ and $R_1 = I$ . Hence $t \mapsto QR_t$ is a path inside $\mathrm{SL}$ from $A$ to $Q$ . Composing with a path from $Q$ to $I$ inside $\mathrm{SO}(n)$ or $\mathrm{SU}(n)$ , which exists by Step 1, connects $A$ to $I$ . So $\mathrm{SL}(n,\mathbb{R})$ and $\mathrm{SL}(n,\mathbb{C})$ are path-connected.

Step 5 ( $\mathrm{Sp}(2n,\mathbb{R})$ is connected). Three facts, then a path.

$\mathrm{Sp}$ is closed under transposition. $J$ itself is symplectic: $J^{\mathsf T}JJ = (-J)JJ = -J^3 = J$ , using $J^2 = -I$ . From $A^{\mathsf T}JA = J$ we get $A^{\mathsf T} = JA^{-1}J^{-1}$ , a product of symplectic matrices, hence symplectic.
Real powers of a symmetric symplectic positive-definite matrix stay symplectic. Let $S$ be symmetric, positive definite and symplectic. Then $SJS = J$ , so $JSJ^{-1} = S^{-1}$ . Conjugation by $J$ is an algebra automorphism, so it commutes with the functional calculus of the spectral theorem: $J S^{t} J^{-1} = (JSJ^{-1})^{t} = S^{-t}$ for every real $t$ , where $S^t$ is defined by raising the eigenvalues to the power $t$ in an orthonormal eigenbasis. Hence $(S^t)^{\mathsf T}JS^t = S^tJS^t = J(J^{-1}S^tJ)S^t = JS^{-t}S^t = J$ .
$\mathrm{Sp}(2n,\mathbb{R}) \cap \mathrm{O}(2n) \cong \mathrm{U}(n)$ . If $A$ is both orthogonal and symplectic then $A^{\mathsf T} = A^{-1}$ , so $A^{-1}JA = J$ , that is $AJ = JA$ . Since $J^2 = -I$ , declaring $Jx$ to be "multiplication of $x$ by $i$ " makes $\mathbb{R}^{2n}$ into $\mathbb{C}^n$ , and $AJ = JA$ says precisely that $A$ is complex-linear. The Hermitian form $h(x,y) = x^{\mathsf T}y + i\,x^{\mathsf T}Jy$ then satisfies $h(Ax,Ay) = (Ax)^{\mathsf T}(Ay) + i(Ax)^{\mathsf T}J(Ay) = x^{\mathsf T}y + i\,x^{\mathsf T}A^{\mathsf T}AJy = h(x,y)$ , using orthogonality in both terms and $JA = AJ$ in the second. So $A$ is unitary; conversely every unitary matrix is real-orthogonal and commutes with $J$ , hence is symplectic. The correspondence is a group isomorphism and a homeomorphism.

Now take $A \in \mathrm{Sp}(2n,\mathbb{R})$ and set $S = A^{\mathsf T}A$ , which is symmetric, positive definite and, by fact 1, symplectic. By fact 2, $Q = AS^{-1/2}$ is symplectic, and $Q^{\mathsf T}Q = S^{-1/2}A^{\mathsf T}AS^{-1/2} = S^{-1/2}SS^{-1/2} = I$ , so $Q \in \mathrm{Sp}(2n,\mathbb{R}) \cap \mathrm{O}(2n)$ . The path $t \mapsto QS^{(1-t)/2}$ , $0 \le t \le 1$ , runs inside $\mathrm{Sp}(2n,\mathbb{R})$ from $A$ to $Q$ ; and $Q$ is joined to $I$ inside $\mathrm{Sp}(2n,\mathbb{R}) \cap \mathrm{O}(2n) \cong \mathrm{U}(n)$ , connected by Step 1. $\square$

RemarkOriginal

Where the hypotheses are used, and what the table hides.

Positive-definiteness of the form is what makes a group compact. The bound $\|A\|_F^2 = n$ in the compactness proof comes from $A^{\dagger}A = I$ with the Euclidean or Hermitian form, whose unit vectors are bounded. Replace it by the indefinite form of Hassani's Example 23.2.2(6), $x\cdot y = -x_1y_1-\dots-x_py_p + x_{p+1}y_{p+1}+\dots+x_ny_n$ , and the "unit sphere" is a hyperboloid: the group $\mathrm{O}(p,n-p)$ preserving it is non-compact, and the $\mathrm{SO}(3,1)$ case is the Lorentz group, whose boosts run off to infinity as the rapidity grows. The symplectic form is antisymmetric, hence as far from definite as possible, and $\mathrm{Sp}(2n,\mathbb{R})$ is non-compact for the same reason. Hassani warns that different authors write $\mathrm{O}(p,n-p)$ , $\mathrm{O}(n-p,p)$ , $\mathrm{O}(n,p)$ or $\mathrm{O}(p,n)$ for the same group, and in his convention $p$ counts the minus signs; always check which index is which.
Realness is what makes $\mathrm{GL}$ and $\mathrm{O}$ disconnected. Step 3 of the components proof isolates the point: the determinant of a real matrix cannot pass through zero without leaving the group, while a complex determinant can go round it. The same remark explains why $\mathrm{SL}(n,\mathbb{C})$ and $\mathrm{SL}(n,\mathbb{R})$ are both connected: there the determinant is pinned at $1$ , and connectedness is supplied by the triangular factor instead.
Compactness without connectedness, and the converse. $\mathrm{O}(n)$ is compact and has two components; $\mathrm{SL}(n,\mathbb{R})$ is connected and non-compact. The two columns are independent.
The dimension count needs the derivative to be onto. If one differentiated the condition $A^{\mathsf T}A = I$ but regarded $\Phi$ as landing in all of $M(n,\mathbb{R})$ , the count would give $n^2 - n^2 = 0$ , which is wrong for every $n \ge 2$ . The fix is to notice that $\Phi$ lands in the symmetric matrices and is onto a neighbourhood there. The same trap is why the symplectic count uses $n(2n-1)$ and not $4n^2$ .
Over other fields. Dummit and Foote define $\mathrm{GL}_n(F)$ for an arbitrary field $F$ ; the group-theoretic facts (closure under products and inverses, from $\det(AB) = \det A \det B$ ) need no topology at all. Over a finite field with $q$ elements it is a finite group, of order $(q^n-1)(q^n-q)\cdots(q^n-q^{n-1})$ , and the dimension, compactness and connectedness columns are empty of meaning. The table is about the real and complex cases.
What is deliberately absent. The quaternionic and indefinite families $\mathrm{Sp}(n)$ (compact), $\mathrm{O}(p,q)$ , $\mathrm{U}(p,q)$ , $\mathrm{Sp}(2n,\mathbb{C})$ , and the exceptional groups, are not tabulated. The method below produces their dimensions in the same way.

Method(Reading off the algebra, the dimension and the compactness of a matrix group)Original

Given a subgroup $G \subseteq \mathrm{GL}(N,\mathbb{F})$ defined by an invariance condition:

Write the condition as $\Phi(A) = c$ . For a bilinear or sesquilinear form with matrix $B$ , the condition that $A$ preserve it is $\Phi(A) = A^{\mathsf T}BA = B$ , or $A^{\dagger}BA = B$ in the Hermitian case.
Identify the target space. Check the symmetry of $\Phi(A)$ : if $B$ is symmetric, $\Phi$ lands in the symmetric matrices; if antisymmetric, in the antisymmetric ones; if Hermitian, in the Hermitian ones. Call that space $W$ and compute $\dim_{\mathbb{R}}W$ .
Differentiate at the identity. Substitute $A = I + tX$ , expand, and keep the term linear in $t$ . For the three cases above this gives $L(X) = X^{\mathsf T}B + BX$ , or $X^{\dagger}B + BX$ .
The Lie algebra is $\ker L$ . Write out $L(X) = 0$ in blocks or entries and read off the independent parameters.
Check $L$ is onto $W$ , by exhibiting a preimage of a general element of $W$ — $\tfrac12 B^{-1}W$ in the symmetric case, $-\tfrac12 BW$ with $B = J$ in the symplectic case. Then $\dim_{\mathbb{R}}G = \dim_{\mathbb{R}}\ker L = \dim_{\mathbb{R}}M(N,\mathbb{F}) - \dim_{\mathbb{R}}W$ , and counting the free parameters in step 4 is an independent check of the same number.
Decide compactness. The group is always closed (the condition is an equation between continuous functions). It is bounded if and only if the preserved form is definite, in which case the columns of $A$ are unit vectors; if the form is indefinite or antisymmetric, produce a one-parameter family of the form $\operatorname{diag}(t,\dots,t^{-1},\dots)$ with unbounded entries.
Add the determinant condition last. It subtracts a component if $\det$ is already confined to $\{\pm 1\}$ on $G$ , and a dimension if $\det$ ranges over a circle or a half-line.

ExampleOriginal

The symplectic group $\mathrm{Sp}(4,\mathbb{R})$ , worked through the method. Here $N = 4$ and $n = 2$ , with

J = \begin{pmatrix} 0 & I_2 \\ -I_2 & 0\end{pmatrix}, \qquad I_2 = \begin{pmatrix}1&0\\0&1\end{pmatrix} .

Steps 1–2. $\Phi(A) = A^{\mathsf T}JA$ lands in the antisymmetric $4\times 4$ matrices, of dimension $\tfrac12 \cdot 4 \cdot 3 = 6$ .

Steps 3–4. Write $X$ in $2\times 2$ blocks, $X = \bigl(\begin{smallmatrix} P & Q \\ R & S \end{smallmatrix}\bigr)$ . Then

X^{\mathsf T}J = \begin{pmatrix} P^{\mathsf T} & R^{\mathsf T} \\ Q^{\mathsf T} & S^{\mathsf T}\end{pmatrix}\begin{pmatrix} 0 & I \\ -I & 0\end{pmatrix} = \begin{pmatrix} -R^{\mathsf T} & P^{\mathsf T} \\ -S^{\mathsf T} & Q^{\mathsf T}\end{pmatrix}, \qquad JX = \begin{pmatrix} R & S \\ -P & -Q\end{pmatrix} ,

X^{\mathsf T}J + JX = \begin{pmatrix} R - R^{\mathsf T} & P^{\mathsf T} + S \\ -(S^{\mathsf T} + P) & Q^{\mathsf T} - Q\end{pmatrix} = 0 \quad\Longleftrightarrow\quad R = R^{\mathsf T},\quad Q = Q^{\mathsf T},\quad S = -P^{\mathsf T} .

Step 5. The free parameters are: $P$ , arbitrary ( $4$ ); $Q$ symmetric ( $3$ ); $R$ symmetric ( $3$ ); and $S$ determined by $P$ . Total

4 + 3 + 3 = 10 ,

matching $n(2n+1) = 2\cdot 5 = 10$ from the general formula. Solving the $16$ linear equations $X^{\mathsf T}J + JX = 0$ numerically gives a coefficient matrix of rank $6$ and hence a null space of dimension $16 - 6 = 10$ , the same number.

A concrete group element. Take $P = R = 0$ and $Q = \bigl(\begin{smallmatrix} 3 & 1 \\ 1 & 0\end{smallmatrix}\bigr)$ , so $X = \bigl(\begin{smallmatrix} 0 & Q \\ 0 & 0\end{smallmatrix}\bigr)$ with $X^2 = 0$ . Then $e^{X} = I + X = \bigl(\begin{smallmatrix} I & Q \\ 0 & I\end{smallmatrix}\bigr)$ , and

(I+X)^{\mathsf T}J(I+X) = \begin{pmatrix} I & 0 \\ Q & I\end{pmatrix}\begin{pmatrix} 0 & I \\ -I & 0\end{pmatrix}\begin{pmatrix} I & Q \\ 0 & I\end{pmatrix} = \begin{pmatrix} I & 0 \\ Q & I\end{pmatrix}\begin{pmatrix} 0 & I \\ -I & -Q\end{pmatrix} = \begin{pmatrix} 0 & I \\ -I & Q - Q\end{pmatrix} = J ,

the last step using $Q^{\mathsf T} = Q$ . So $e^X \in \mathrm{Sp}(4,\mathbb{R})$ , as the method promises, and $\det e^X = e^{\operatorname{tr}X} = e^0 = 1$ .

Step 6. $\operatorname{diag}(t,t,t^{-1},t^{-1})$ is symplectic for every $t>0$ , by the block computation in the compactness proof, so $\mathrm{Sp}(4,\mathbb{R})$ is unbounded and non-compact.

A consequence worth recording. Taking determinants in $A^{\mathsf T}JA = J$ gives $(\det A)^2\det J = \det J$ , and $\det J = 1 \neq 0$ , so $\det A = \pm 1$ on $\mathrm{Sp}(2n,\mathbb{R})$ . The determinant is continuous and equals $+1$ at the identity, and the group is connected by Step 5 of the components proof, so in fact $\det A = +1$ throughout and $\mathrm{Sp}(2n,\mathbb{R}) \subseteq \mathrm{SL}(2n,\mathbb{R})$ — symplectic maps preserve volume. The same conclusion is visible infinitesimally: every $X \in \mathfrak{sp}(2n,\mathbb{R})$ has $\operatorname{tr}X = \operatorname{tr}P + \operatorname{tr}(-P^{\mathsf T}) = 0$ . For $n=1$ the inclusion is an equality, by the identity $A^{\mathsf T}JA = (\det A)J$ noted in the table; for $n \ge 2$ it is strict, since $10 < 15 = \dim \mathrm{SL}(4,\mathbb{R})$ already at $n=2$ .

ApplicationOriginal

Gauge groups and the number of force carriers. The Standard Model of particle physics is built on the gauge group $\mathrm{SU}(3) \times \mathrm{SU}(2) \times \mathrm{U}(1)$ , and the number of gauge bosons is the dimension of that group, because each independent generator carries one gauge field. The table gives

\dim \mathrm{SU}(3) + \dim \mathrm{SU}(2) + \dim \mathrm{U}(1) = (3^2 - 1) + (2^2 - 1) + 1 = 8 + 3 + 1 = 12 ,

namely the eight gluons of $\mathrm{SU}(3)$ , the three $\mathrm{SU}(2)$ bosons and the single hypercharge boson, which after electroweak mixing are observed as $W^{+}$ , $W^{-}$ , $Z^{0}$ and the photon. The compactness column matters here too: the gauge group is compact, which is what allows the gauge fields to carry positive-definite kinetic energy.

Rotations, spin and the two-to-one cover. $\mathrm{SO}(3)$ and $\mathrm{SU}(2)$ have the same dimension $3$ and the same rank $1$ , so their Lie algebras — and hence their local structure — agree, while their global structure does not: $\mathrm{SU}(2)$ is the three-sphere and is simply connected, $\mathrm{SO}(3)$ is not. That is the entire origin of half-integer spin, and is the subject of the rotation group and SU(2).

Classical mechanics. The flow of a Hamiltonian system linearised about a trajectory lies in $\mathrm{Sp}(2n,\mathbb{R})$ , because Hamilton's equations preserve the two-form $\omega = \sum_i dq_i \wedge dp_i$ . The consequence drawn in the example — $\mathrm{Sp}(2n,\mathbb{R}) \subseteq \mathrm{SL}(2n,\mathbb{R})$ — is Liouville's theorem: phase-space volume is conserved. In accelerator physics the transfer matrix of a beamline segment is required to be symplectic, and the $10$ free parameters of $\mathrm{Sp}(4,\mathbb{R})$ are exactly the degrees of freedom of a transverse beam optics element in two planes.

Where the compactness column is used. For a compact group an invariant integration (the Haar measure) exists with finite total mass, which is what makes every finite-dimensional representation unitarisable and every irreducible representation finite-dimensional. That is why the representation theory of $\mathrm{SU}(2)$ and $\mathrm{SU}(3)$ looks like that of a finite group — discrete labels, orthogonality relations, complete reducibility — while the non-compact Lorentz group has infinite-dimensional unitary representations and no such table.

Definition · advanced

Casimir operators

Adapted

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Defines a Casimir operator as one commuting with every generator of a represented Lie algebra, builds the quadratic Casimir from the inverse Cartan metric and proves that it commutes with all of them, explains why Schur's lemma then makes it a number on each irreducible representation, and identifies J^2 as the one Casimir of the angular-momentum algebra.

Prerequisites: Angular momentum algebra, Matrix Lie groups and Lie algebras, Representations of groups, Schur's lemma, Spherical harmonics

Definition(Casimir operator, Killing form, Cartan metric)AdaptedAdapted from Hassani (2013), Definition 30.3.4 and Eq. (30.13), p. 969, and Definition 29.2.13 with Eq. (29.49), p. 945; Arfken, Weber & Harris (2013), §17.7, p. 849

Let $\mathfrak{g}$ be a Lie algebra with basis $X_1,\dots,X_d$ and structure constants $c_{ij}^{\,k}$ defined by

[X_i, X_j] = c_{ij}^{\,k}\,X_k ,

summation over repeated indices understood throughout (see matrix Lie groups and Lie algebras). Let $T : \mathfrak{g} \to \mathfrak{gl}(\mathcal{H})$ be a representation, and abbreviate $X_i$ for the operator $T_{X_i}$ acting on the carrier space $\mathcal{H}$ , as Hassani does.

A Casimir operator for the representation $T$ is an operator $C$ on $\mathcal{H}$ that commutes with every $T_X$ , $X \in \mathfrak{g}$ . Equivalently, since the $X_i$ span, $[C, X_i] = 0$ for $i = 1,\dots,d$ .
The Killing form of $\mathfrak{g}$ is the symmetric bilinear form $(X \mid Y) = \operatorname{tr}(\mathrm{ad}_X \circ \mathrm{ad}_Y)$ , where $\mathrm{ad}_X(Y) = [X,Y]$ . In the chosen basis its components are the Cartan metric tensor
$g_{ij} = (X_i \mid X_j) = c_{il}^{\,k}\,c_{jk}^{\,l} .$
The lowered structure constants are $c_{ijk} = c_{ij}^{\,l}\,g_{lk}$ . Hassani's computation on p. 946 shows that $c_{ijk}$ is completely antisymmetric in $i$ , $j$ , $k$ ; Arfken, Weber and Harris record the same total antisymmetry for the constants $f_{jkl}$ of their closure relation $[S_j,S_k] = i f_{jkl}S_l$ .
When $g_{ij}$ is invertible, write $g^{ij}$ for its inverse and call
$C = g^{ij}\,X_i X_j$
the quadratic Casimir of the representation. The products $X_iX_j$ are products of operators on $\mathcal{H}$ , not Lie brackets: $C$ lives in the algebra of operators (equivalently, in the universal enveloping algebra), not in $\mathfrak{g}$ .
The rank $r$ of $\mathfrak{g}$ is the dimension of a maximal commuting subalgebra: Arfken, Weber and Harris describe it as the greatest number of generators, independent of one another, that commute in pairs; Hassani calls it the dimension of the Cartan subalgebra. Hassani uses the first form when he argues that $\mathfrak{so}(3)$ has rank one, so the two descriptions are the working definition in both books.

A semisimple Lie algebra is, for the purposes of this unit, one whose Killing form is non-degenerate, which is what the construction of $C$ requires.

Definition(The quadratic Casimir, and what it is good for)AdaptedAdapted from Hassani (2013), Eq. (30.13) with the verification on p. 970, Definition 30.3.4 and Theorem 30.3.5, p. 969; Arfken, Weber & Harris (2013), §17.7, p. 849

Let $\mathfrak{g}$ be a Lie algebra whose Cartan metric $g_{ij}$ is invertible, and let $T$ be a representation of $\mathfrak{g}$ on $\mathcal{H}$ .

(It is a Casimir.) $C = g^{ij}X_iX_j$ commutes with every $X_k$ , hence with $T_X$ for every $X \in \mathfrak{g}$ .
(It is a number on an irreducible representation.) If $T$ is irreducible and $\mathcal{H}$ is a finite-dimensional complex space, then by Schur's lemma $C$ is a scalar multiple of the identity. Every vector of an irreducible invariant subspace is therefore an eigenvector of $C$ with one and the same eigenvalue, and that eigenvalue is a label of the representation rather than of any vector in it.
(Chevalley's theorem, quoted.) Suppose $\mathfrak{g}$ is semisimple and has rank $r$ . Then, whatever basis is chosen, one can find $r$ Casimir operators that are polynomials in the $T_{X_i}$ and whose eigenvalues characterise which irreducible representation of $\mathfrak{g}$ one is looking at. Hassani states this as Theorem 30.3.5 and refers the proof elsewhere; we quote it and do not prove it. Arfken, Weber and Harris state the counting part the same way: the number of independent Casimir operators equals the rank of the group.

In particular $\mathfrak{so}(3)$ , of rank one, has exactly one Casimir operator, and it is the operator written $L^2$ or $J^2$ in angular-momentum theory.

Idea of the proofOriginal

Why should $g^{ij}X_iX_j$ commute with everything? Commuting a product past $X_k$ costs one commutator for each factor, by the Leibniz rule $[AB, X] = A[B,X] + [A,X]B$ ; each commutator replaces one generator by a structure constant times another generator. What is left is a sum of terms $X_iX_r$ and $X_rX_i$ , weighted by a coefficient built from $g^{ij}$ and a structure constant. The whole point is a clash of symmetries: the operator part $X_iX_r + X_rX_i$ is symmetric under exchanging the two indices, while the coefficient — because the Cartan metric is symmetric and the lowered structure constants are completely antisymmetric — is antisymmetric under the very same exchange. A symmetric object contracted with an antisymmetric one is zero, so everything cancels.

The second half of the statement is Schur's lemma doing what it always does: an operator that commutes with an irreducible set of operators has no room to be anything but a multiple of the identity, because each of its eigenspaces would otherwise be a proper invariant subspace.

Proof(Part 1:

[C, X_k] = 0

)AdaptedAdapted from Hassani (2013), the displayed computation following Eq. (30.13), p. 970

Step 1 (expand by the Leibniz rule). For operators, $[AB, D] = A[B,D] + [A,D]B$ , directly from writing out the four products. Applying it with $A = X_i$ , $B = X_j$ , $D = X_k$ and using the bilinearity of the commutator,

[C, X_k] = g^{ij}\,[X_iX_j, X_k] = g^{ij}\bigl( X_i[X_j,X_k] + [X_i,X_k]X_j \bigr) .

Step 2 (insert the structure constants). The operators $T_X$ obey the same commutation relations as $\mathfrak{g}$ , because $T$ is a representation: $[X_j, X_k] = c_{jk}^{\,r}X_r$ . Hence

[C, X_k] = g^{ij}c_{jk}^{\,r}\,X_iX_r + g^{ij}c_{ik}^{\,r}\,X_rX_j .

Step 3 (make the two terms share a coefficient). In the second term the names $i$ and $j$ are both summed over, so they may be exchanged; doing so and using the symmetry $g^{ji} = g^{ij}$ of the inverse of a symmetric matrix turns it into $g^{ij}c_{jk}^{\,r}X_rX_i$ . Therefore

[C, X_k] = g^{ij}c_{jk}^{\,r}\bigl(X_iX_r + X_rX_i\bigr) .

Step 4 (lower the index on the structure constant). By definition $c_{jks} = c_{jk}^{\,l}g_{ls}$ , so $c_{jk}^{\,r} = g^{rs}c_{jks}$ , and

[C, X_k] = g^{ij}g^{rs}c_{jks}\,\bigl(X_iX_r + X_rX_i\bigr) .

Step 5 (the symmetry clash). Call the coefficient $A^{ir} = g^{ij}g^{rs}c_{jks}$ , with $k$ held fixed. Exchange $i$ and $r$ : $A^{ri} = g^{rj}g^{is}c_{jks}$ , and renaming the dummy indices $j \leftrightarrow s$ gives $A^{ri} = g^{rs}g^{ij}c_{skj}$ . The lowered structure constants are completely antisymmetric, so $c_{skj} = -c_{jks}$ — the two index strings differ by the transposition of the first and last entries. Hence $A^{ri} = -A^{ir}$ : the coefficient is antisymmetric in $i$ and $r$ .

Write the operator factor as $S_{ir} = X_iX_r + X_rX_i$ , with both indices down so that it contracts with $A^{ir}$ ; it is, by construction, symmetric in $i$ and $r$ . Summing over both indices,

A^{ir}S_{ir} = -A^{ri}S_{ir} = -A^{ir}S_{ri} = -A^{ir}S_{ir} ,

where the second equality is a relabelling of the two dummy indices and the third is the symmetry of $S$ . A quantity equal to its own negative vanishes, so $[C, X_k] = 0$ for every $k$ . Since the $X_k$ span $\mathfrak{g}$ and $T$ is linear, $C$ commutes with $T_X$ for every $X \in \mathfrak{g}$ .

Proof(Part 2:

C

is a scalar on an irreducible representation)Original

Let $T$ be irreducible on a finite-dimensional complex space $\mathcal{H}$ . By Part 1, $C$ commutes with every operator of the represented algebra. Schur's lemma in its second form says that an operator commuting with all the operators of an irreducible representation on a finite-dimensional complex space is $\lambda I$ for some scalar $\lambda$ . Hence $C = \lambda I$ .

For completeness, the mechanism: $\mathcal{H}$ is complex and finite-dimensional, so $C$ has an eigenvalue $\lambda$ . The eigenspace $\mathcal{H}_\lambda = \ker(C - \lambda I)$ is non-zero, and it is invariant: if $Cv = \lambda v$ then $C(X_iv) = X_i(Cv) = \lambda X_iv$ , using $[C,X_i] = 0$ . Irreducibility leaves only $\mathcal{H}_\lambda = \mathcal{H}$ , that is $C = \lambda I$ . $\square$

RemarkOriginal

Where the hypotheses are used, and what fails without them.

Invertibility of the Cartan metric is used the moment $g^{ij}$ is written down, in the very definition of $C$ . By Cartan's criterion this is semisimplicity, and Hassani flags it explicitly: the construction of Eq. (30.13) "works only for semisimple Lie algebras". The Poincaré algebra is his own counterexample, and the failure is instructive. It is not semisimple, so there is no inverse Cartan metric to contract the generators $M_{ij}$ with. Hassani does build a double-indexed metric $g_{ij,kl}$ out of the structure constants and forms $M^2 = g_{ij,kl}M^{ij}M^{kl}$ , but that operator commutes with the $M_{ij}$ and not with the translation generators $P_i$ , so it is not a Casimir of the full algebra. What works instead are $P^2 = \eta^{ij}P_iP_j$ and the fourth-degree invariant $W^2$ built from $W_{i_1\dots i_{n-3}} = \epsilon^{i_1\dots i_{n-3}jkl}M_{jk}P_l$ ; the general recipe gives neither.
Complete antisymmetry of $c_{ijk}$ is used only in Step 5, and it is the whole proof. It is not an extra assumption: it follows from the invariance identity $([X,Y]\mid Z) + ([X,Z]\mid Y) = 0$ satisfied by every Killing form, which is Hassani's Eq. (29.48). Note that antisymmetry of the unlowered $c_{ij}^{\,k}$ in its two lower indices — which is immediate from $[X_i,X_j] = -[X_j,X_i]$ — would not be enough; the index that has to move in Step 5 is the raised one, and moving it needs the metric.
Irreducibility is used in Part 2 and nowhere else. Without it $C$ is still a Casimir but need not be a scalar. Take $\mathfrak{su}(2)$ acting on $\mathbb{C} \oplus \mathbb{C}^3$ as the direct sum of the $j=0$ and $j=1$ representations. Then $J^2 = \operatorname{diag}(0,\,2,\,2,\,2)$ , which commutes with every generator but is not a multiple of the identity. That is exactly why $C$ is useful: its eigenvalues separate the irreducible pieces of a reducible representation, which is how one decomposes a space of states.
Finite dimension and a complex scalar field are used in Part 2, for the existence of an eigenvalue. On a real space a commuting operator can act as a rotation with no real eigenvalue, and on an infinite-dimensional space the spectrum may be empty of eigenvalues; the version of Schur's lemma needed there is the one for unitary representations.
$C$ is not unique, and normalisation is a convention. Any non-zero multiple of a Casimir is a Casimir, and the literature does not agree on which multiple to take. Hassani computes the Cartan metric of $\mathfrak{so}(3)$ to be $g_{ij} = 2\delta_{ij}$ , which makes the Killing-normalised Casimir $\tfrac12 L^2$ ; he then says, in so many words, to ignore the factor of two and call $L^2 = L_1^2+L_2^2+L_3^2$ the Casimir operator, which is the convention of every physics text and the one used here. Always state which normalisation is in force before quoting a numerical eigenvalue.

Method(Constructing the quadratic Casimir and reading off its eigenvalue)Original

Fix a basis and a commutator convention. Write $[X_i, X_j] = c_{ij}^{\,k}X_k$ . Physics texts usually use Hermitian generators and an explicit $i$ , as in $[J_i, J_j] = i\epsilon_{ijk}J_k$ , so that $c_{ij}^{\,k} = i\epsilon_{ijk}$ ; mathematics texts usually use anti-Hermitian generators and no $i$ . The two differ by a factor of $i$ in every structure constant and hence by a sign in the Cartan metric. Record which you are using.
Compute the Cartan metric $g_{ij} = c_{il}^{\,k}c_{jk}^{\,l}$ , equivalently $\operatorname{tr}(\mathrm{ad}_{X_i}\mathrm{ad}_{X_j})$ with $(\mathrm{ad}_{X_i})^k{}_j = c_{ij}^{\,k}$ .
Check it is invertible. If it is not, the algebra is not semisimple and this recipe does not apply; look instead for invariants built by other contractions, as for the Poincaré algebra.
Invert and contract: $C = g^{ij}T_{X_i}T_{X_j}$ . If $g_{ij} = \kappa\,\delta_{ij}$ , this is simply $\kappa^{-1}\sum_i T_{X_i}^2$ .
Evaluate the eigenvalue on one convenient vector. Because $C$ is a scalar on each irreducible piece, applying it to a single well-chosen vector — a highest-weight vector, for which the raising operators give zero — delivers the eigenvalue for the entire multiplet.
Count how many you need. The number of independent Casimir operators equals the rank. A rank-one algebra is finished at step 5; a rank-two algebra such as $\mathfrak{su}(3)$ needs a second, independent invariant, which is cubic rather than quadratic.

Example(The angular-momentum algebra, with a number at the end)AdaptedAdapted from Hassani (2013), Example 30.3.6, p. 970

(a) This worked case follows Hassani's treatment of the rotation group. Take his convention: Hermitian generators $J_1,J_2,J_3$ with

[J_i, J_j] = i\,\epsilon_{ijk}J_k , \qquad\text{so}\qquad c_{ij}^{\,k} = i\,\epsilon_{ijk} .

Step 2. The Cartan metric is

g_{ij} = c_{il}^{\,k}c_{jk}^{\,l} = (i\epsilon_{ilk})(i\epsilon_{jkl}) = -\epsilon_{ilk}\epsilon_{jkl} = +\epsilon_{ilk}\epsilon_{jlk} = 2\delta_{ij} ,

the third equality using $\epsilon_{jkl} = -\epsilon_{jlk}$ and the last the standard contraction $\epsilon_{ilk}\epsilon_{jlk} = 2\delta_{ij}$ .

Steps 3–4. $g_{ij}$ is invertible with $g^{ij} = \tfrac12\delta^{ij}$ , so

C = \tfrac12\bigl(J_1^2 + J_2^2 + J_3^2\bigr) = \tfrac12 J^2 .

Dropping the factor $\tfrac12$ , as Hassani does and as physics does, the Casimir is $J^2$ .

Step 5. From the angular momentum algebra, the irreducible representation labelled $j$ has dimension $2j+1$ and $J^2 = j(j+1)I$ on it. Taking $j = \tfrac32$ :

J^2 = \tfrac32\cdot\tfrac52\,I_4 = \tfrac{15}{4}\,I_4 = 3.75\,I_4, \qquad \dim = 2j+1 = 4 .

Building the four-by-four matrices $J_\pm$ , $J_3$ explicitly from the ladder-operator formulas and forming $J_1^2+J_2^2+J_3^2$ numerically returns $3.75$ on the diagonal and zero off it, to machine precision, and the same computation at $2j = 1,2,4$ returns $0.75$ , $2$ and $6$ , matching $j(j+1)$ each time. The Killing-normalised Casimir is half of these: at $j = \tfrac32$ its eigenvalue is $\tfrac{15}{8} = 1.875$ .

Step 6. $\mathfrak{so}(3)$ has rank one. Indeed a general element is $\mathbf{a}\cdot\mathbf{J}$ for some $\mathbf{a} \in \mathbb{R}^3$ , and the commutation relations give $[\mathbf{a}\cdot\mathbf{J}, \mathbf{b}\cdot\mathbf{J}] = i(\mathbf{a}\times\mathbf{b})\cdot\mathbf{J}$ , which vanishes exactly when $\mathbf{a}\times\mathbf{b} = \mathbf{0}$ , that is when $\mathbf{a}$ and $\mathbf{b}$ are parallel. So no two independent elements commute, and a maximal commuting subalgebra is one-dimensional. By Chevalley's theorem $J^2$ is then the only Casimir, and a single label $j$ characterises the irreducible representation. The second label $m$ , the eigenvalue of $J_3$ , distinguishes vectors within the representation, not representations.

Example(A rank-two case:

\mathfrak{su}(3)

)Original

(b) Take the Gell-Mann basis $T_a = \lambda_a/2$ , $a = 1,\dots,8$ , normalised by $\operatorname{tr}(T_aT_b) = \tfrac12\delta_{ab}$ , with $[T_a,T_b] = i f_{abc}T_c$ and the non-zero structure constants $f_{123} = 1$ , $f_{147} = f_{246} = f_{257} = f_{345} = \tfrac12$ , $f_{156} = f_{367} = -\tfrac12$ , $f_{458} = f_{678} = \tfrac{\sqrt3}{2}$ , up to antisymmetry. Computing $g_{ab} = c_{ac}^{\,d}c_{bd}^{\,c} = -f_{acd}f_{bdc} = f_{acd}f_{bcd}$ numerically from these constants gives

g_{ab} = 3\,\delta_{ab} ,

invertible, so $\mathfrak{su}(3)$ is semisimple and $C = \tfrac13\sum_{a=1}^{8}T_aT_a$ . On the defining (fundamental) three-dimensional representation, summing the eight squares gives

\sum_{a=1}^{8}T_aT_a = \frac{4}{3}\,I_3 = 1.3333\ldots\,I_3 , \qquad\text{so}\qquad C = \frac{1}{3}\cdot\frac{4}{3}\,I_3 = \frac{4}{9}\,I_3 .

The number $4/3$ is the one quoted in particle physics as the quadratic Casimir $C_2$ of the fundamental of $\mathrm{SU}(3)$ , in the normalisation that omits the inverse Cartan metric; $4/9$ is the Killing-normalised value. Both are scalars, as Part 2 requires. Being a scalar is only the consequence, not the test: a Casimir is also a scalar on a direct sum of copies of one irreducible representation — two copies of the fundamental give $\tfrac43 I_6$ — so the vanishing of the off-diagonal entries confirms Part 2 here but proves nothing about irreducibility on its own.

$\mathfrak{su}(3)$ has rank two — Arfken, Weber and Harris state this, and the commuting pair is visible in the basis above, since $T_3$ and $T_8$ are the two diagonal matrices — so Chevalley's theorem calls for a second Casimir, which the quadratic construction cannot supply. It is the cubic invariant $d_{abc}T_aT_bT_c$ , built from the totally symmetric coefficients $d_{abc}$ defined by $\{T_a, T_b\} = \tfrac13\delta_{ab}I + d_{abc}T_c$ . Two labels, as Arfken, Weber and Harris say, is why the weight diagrams of $\mathrm{SU}(3)$ are two-dimensional while those of $\mathrm{SU}(2)$ lie on a line.

ApplicationOriginal

Quantum numbers are Casimir eigenvalues. When a Hamiltonian is invariant under a symmetry group, it commutes with all the generators, so the energy eigenspaces carry representations of the group; the Casimir operators commute with $H$ as well and can be diagonalised alongside it. Their eigenvalues are then exactly the conserved labels a state is given. For rotational symmetry this is the pair $(j, m)$ : $j$ comes from the single Casimir $J^2$ and names the multiplet, $m$ from $J_3$ and names the state within it, and the energy cannot depend on $m$ — which is the origin of the $(2j+1)$ -fold degeneracy of a rotationally invariant level, lifted only when a field picks out an axis.

The equations of mathematical physics are Casimir eigenvalue equations. Hassani makes this the point of the section. If the symmetry group acts on a manifold $M$ and the carrier space is $L^2(M)$ , the generators become first-order differential operators and the Casimirs become higher-order ones; the statement that a Casimir acts as a multiple of the identity on an irreducible subspace is then a differential equation, $C_\alpha f = \lambda(\alpha) f$ , whose solutions transform among themselves under the symmetry. For $\mathrm{SO}(3)$ acting on the sphere, $C = L^2$ is the angular part of the Laplacian and the equation $L^2 Y = l(l+1)Y$ is the equation satisfied by the spherical harmonics, with the $2l+1$ functions $Y_l^m$ at fixed $l$ spanning the $l$ -th irreducible subspace. Hassani notes that Bessel, hypergeometric and confluent hypergeometric functions arise the same way, as matrix elements in representations of a handful of small Lie groups.

Mass and spin. For relativistic particles the symmetry algebra is the Poincaré algebra, which is not semisimple, so the quadratic construction of this unit does not apply; Hassani's §30.3.4 finds instead the two invariants $P^2$ and $W^2$ . Their eigenvalues are what the words mass and spin mean: an elementary particle is, operationally, an irreducible unitary representation of the Poincaré group, and the two Casimirs are the only labels it carries. Everything else — charge, colour, flavour — is a Casimir label of an internal group, the quadratic Casimir of $\mathrm{SU}(3)$ in example (b) being the one that separates the colour multiplets and that appears directly in the strong coupling of a quark, as the factor $C_2 = 4/3$ in one-gluon exchange.

Definition · advanced

Clifford algebras and spinors

Adapted

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Defines the Clifford algebra of a real quadratic form by the relation uv + vu = 2g(u,v), gives its 2^N-element basis and the low-dimensional identifications with the complex numbers, the quaternions and the Pauli and Dirac matrix algebras, shows how Dirac's factorisation of the relativistic energy forces the gamma-matrix relations, and explains how spinors and the two-to-one map from Spin to SO arise inside the algebra.

Prerequisites: Quaternions, The rotation group and SU(2), Unitary and orthogonal matrices, Inner products and norms, Functions of matrices and the matrix exponential

Definition(The Clifford algebra of a quadratic form)AdaptedAdapted from Hassani (2013), Eq. (27.4), p. 830, Eqs. (27.14)–(27.16) and Theorem 27.2.1, p. 834, and Eqs. (27.40)–(27.41) with footnote 3, p. 846

Let $V$ be a real vector space of dimension $N$ carrying a symmetric bilinear form $g$ . The Clifford algebra $\mathcal{C}_V$ is the associative algebra with unit $1$ , generated by the vectors of $V$ , in which multiplication — written by juxtaposition — obeys

uv + vu = 2\,g(u,v)\,1 \qquad\text{for all } u, v \in V .

Hassani reaches this relation by combining the wedge product with an interior product, and then observes that it is "sometimes taken as the definition"; we take it as the definition. Setting $u = v$ gives the equivalent form $v^2 = g(v,v)\,1$ .

In an orthonormal basis $e_1,\dots,e_N$ , in which $g(e_i,e_j) = 0$ for $i \neq j$ and $g(e_i,e_i) = \pm 1$ , the relation becomes

e_ie_j = -e_je_i \quad (i \neq j), \qquad e_i^2 = +1 \ \ (1 \le i \le \mu), \qquad e_i^2 = -1 \ \ (\mu < i \le N) ,

where $\mu$ is the number of basis vectors of positive square and $\nu = N - \mu$ the number of negative square. The algebra determined by these relations is Hassani's $\mathcal{C}^{\nu}_{\mu}(\mathbb{R})$ ; he lists $\mathcal{C}(\mu,\nu)$ , $\mathcal{C}\ell_{\mu,\nu}(\mathbb{R})$ and $\mathcal{C}\ell_{p,q}(\mathbb{R})$ with $p = \mu$ , $q = \nu$ as equally current notations, and we write $\mathcal{C}(\mu,\nu)$ throughout. The order of the two indices is not standard between authors; here the first counts the $+1$ s.

Two pieces of structure come free. An element is even if it is a linear combination of products of an even number of basis vectors and odd otherwise; the even elements form a subalgebra $\mathcal{C}^0_V$ , the odd ones only a subspace, and $\mathcal{C}_V = \mathcal{C}^0_V \oplus \mathcal{C}^1_V$ . A representation of $\mathcal{C}(\mu,\nu)$ by matrices is a choice of square matrices $\gamma_1,\dots,\gamma_N$ with

\gamma_i\gamma_j + \gamma_j\gamma_i = 2 g_{ij} I ;

the matrices of such a representation are what physics calls gamma matrices, and a column vector on which they act is a spinor.

Definition(Dimension, basis, and the small cases)AdaptedAdapted from Hassani (2013), Theorem 27.2.1 and Eq. (27.12), p. 834, Examples 27.4.1 and 27.4.2, pp. 846–847, Table 27.2 and Theorem 27.4.6, p. 851, and §27.4.3, p. 852

Let $e_1,\dots,e_N$ be an orthonormal basis of $(V,g)$ as above.

(Basis.) The $2^N$ ordered products
$1,\quad e_i,\quad e_ie_j \ (i<j),\quad e_ie_je_k \ (i<j<k),\quad \dots,\quad e_1e_2\cdots e_N$
form a basis of $\mathcal{C}(\mu,\nu)$ as a real vector space. In particular $\dim_{\mathbb{R}}\mathcal{C}(\mu,\nu) = 2^N$ , and the even subalgebra has dimension $2^{N-1}$ .

(The small algebras.) With $M_k(\mathbb{A})$ the $k\times k$ matrices over $\mathbb{A}$ and $\mathbb{H}$ the quaternions,

Algebra	$N$	Isomorphic to	$\dim_{\mathbb{R}}$
$\mathcal{C}(1,0)$	$1$	$\mathbb{R}\oplus\mathbb{R}$ (split complex numbers)	$2$
$\mathcal{C}(0,1)$	$1$	$\mathbb{C}$	$2$
$\mathcal{C}(2,0)$	$2$	$M_2(\mathbb{R})$	$4$
$\mathcal{C}(0,2)$	$2$	$\mathbb{H}$	$4$
$\mathcal{C}(3,0)$	$3$	$M_2(\mathbb{C})$ — the Pauli algebra	$8$
$\mathcal{C}(0,3)$	$3$	$\mathbb{H}\oplus\mathbb{H}$	$8$
$\mathcal{C}(3,1)$	$4$	$M_4(\mathbb{R})$	$16$

Hassani's Examples 27.4.1 and 27.4.2 give three of these — $\mathcal{C}(1,0)$ , $\mathcal{C}(0,1)$ and $\mathcal{C}(0,2)$ — and his §27.4.3 gives the last; $\mathcal{C}(2,0)$ together with $\mathcal{C}(0,2)$ is the $n=2$ row of his Table 27.2, p. 851, and the two three-dimensional entries are the $n=3$ row of the same table. Note that $\mathcal{C}(3,0)$ and $\mathcal{C}(0,3)$ are not isomorphic: the quadratic form, not merely the dimension of $V$ , fixes the algebra.

(Dirac's relation.) Requiring that the relativistic energy–momentum relation $E^2 = \mathbf{p}^2c^2 + m^2c^4$ factor into two first-order expressions forces the coefficients to obey
$(\gamma^0)^2 = 1, \qquad (\gamma^i)^2 = -1 \ (i = 1,2,3), \qquad \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 0 \ (\mu \neq \nu) ,$
which is Arfken, Weber and Harris's Eqs. (2.74)–(2.75); equivalently $\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2\eta^{\mu\nu}I$ with $\eta = \operatorname{diag}(1,-1,-1,-1)$ . No numbers satisfy this; the smallest matrices that do are $4\times 4$ , and $\psi$ must then be a four-component column.
(Pin and Spin.) Inside the group of invertible elements of $\mathcal{C}(\mu,\nu)$ sit the groups $\mathrm{Pin}(\mu,\nu)$ and its even part $\mathrm{Spin}(\mu,\nu) = \mathrm{Pin}(\mu,\nu)\cap\mathcal{C}^0$ , which act on $V$ by twisted conjugation $\tau_a$ . Hassani's Theorem 31.2.3 states that $a \mapsto \tau_a$ is a surjective homomorphism $\mathrm{Pin}(\mu,\nu)\to\mathrm{O}(\mu,\nu)$ with kernel $\{1,-1\}$ , and his Theorem 31.2.6 that its restriction $\mathrm{Spin}(\mu,\nu)\to\mathrm{SO}(\mu,\nu)$ is also surjective with kernel $\{1,-1\}$ . Each rotation therefore comes from exactly two elements of $\mathrm{Spin}$ , differing by a sign.

Idea of the proofOriginal

Everything follows from one move: the relation lets you push any two basis vectors past each other at the cost of a sign, and lets you delete a repeated pair at the cost of a sign. So any product of generators collapses to $\pm$ an ordered product with distinct indices, of which there are $2^N$ — one for each subset of $\{1,\dots,N\}$ . That is the basis theorem, and it also explains why the small algebras are the familiar ones: a two-dimensional $V$ with both squares $-1$ gives four basis elements $1, e_1, e_2, e_1e_2$ whose multiplication table is forced, and it is the quaternion table.

Dirac's relation comes from the same move read backwards. He wanted $\bigl(\beta E + \sum\alpha_ip_i\bigr)$ to square to $E^2 - \mathbf{p}^2$ (with the mass carried along); squaring produces cross terms $\alpha_i\alpha_j + \alpha_j\alpha_i$ , and the only way for them to disappear is for the coefficients to anticommute. That is the Clifford relation, so the coefficients cannot be numbers; the smallest matrices that work give four-component spinors.

Spinors themselves are not an extra construction. Once the algebra is a matrix algebra, the columns of a matrix are already a space the algebra acts on from the left, and that space is a minimal left ideal: the smallest non-zero subspace closed under multiplication by the algebra. That is what a spinor space is.

Proof(Part 1: the

2^N

ordered products span, and the quaternion and Pauli cases)AdaptedAdapted from Hassani (2013), Theorem 27.2.1, p. 834, Example 27.4.2, p. 847, and §31.2.1, pp. 997–998

Step 1 (reduction to ordered products). Any element of the algebra is, by definition, a linear combination of finite products $e_{i_1}e_{i_2}\cdots e_{i_p}$ . If two adjacent factors have different indices, $e_ie_j = -e_je_i$ exchanges them at the cost of a sign; repeating this bubble-sorts the word into non-decreasing index order. If two adjacent factors are then equal, $e_i^2 = \pm 1$ deletes both at the cost of a sign. Iterating, every word reduces to $\pm$ a product $e_{i_1}\cdots e_{i_q}$ with $i_1 < \dots < i_q$ . There is one such product for each subset of $\{1,\dots,N\}$ , so the $2^N$ listed elements span. That they are also linearly independent — so that the algebra is exactly $2^N$ -dimensional and not smaller — is Hassani's Theorem 27.2.1, which we quote. For the two cases used below, independence is visible directly, because the spanning elements are exhibited as linearly independent matrices.

Step 2 ( $\mathcal{C}(0,2) \cong \mathbb{H}$ ). Here $N = 2$ with $e_1^2 = e_2^2 = -1$ . Write $a = e_1$ , $b = e_2$ , $c = e_1e_2$ . Then

c^2 = e_1e_2e_1e_2 = -e_1e_1e_2e_2 = -(-1)(-1) = -1 ,

using the anticommutation once and then both squares; and

ab = c, \qquad bc = e_2e_1e_2 = -e_1e_2e_2 = e_1 = a, \qquad ca = e_1e_2e_1 = -e_1e_1e_2 = e_2 = b .

So $a^2 = b^2 = c^2 = -1$ with $ab = c$ , $bc = a$ , $ca = b$ , which is exactly the multiplication table of the quaternion units $i, j, k$ in quaternions. The four elements $1, a, b, c$ therefore span a copy of $\mathbb{H}$ , and by Step 1 they span everything: $\mathcal{C}(0,2) \cong \mathbb{H}$ . This is Hassani's Example 27.4.2.

Step 3 ( $\mathcal{C}(3,0) \cong M_2(\mathbb{C})$ ). Here $N = 3$ with all three squares $+1$ . The Pauli matrices

\sigma_1 = \begin{pmatrix}0&1\\1&0\end{pmatrix}, \qquad \sigma_2 = \begin{pmatrix}0&-i\\i&0\end{pmatrix}, \qquad \sigma_3 = \begin{pmatrix}1&0\\0&-1\end{pmatrix}

satisfy $\sigma_i^2 = I$ and $\sigma_i\sigma_j = -\sigma_j\sigma_i$ for $i \neq j$ — Arfken, Weber and Harris's Eq. (2.59) — that is, $\sigma_i\sigma_j + \sigma_j\sigma_i = 2\delta_{ij}I$ . So $e_i \mapsto \sigma_i$ is a representation. It is onto: the eight products of Step 1 map to

I,\quad \sigma_1,\ \sigma_2,\ \sigma_3,\quad \sigma_1\sigma_2 = i\sigma_3,\ \ \sigma_1\sigma_3 = -i\sigma_2,\ \ \sigma_2\sigma_3 = i\sigma_1,\quad \sigma_1\sigma_2\sigma_3 = iI ,

which, up to sign, is the list $\{I, \sigma_1,\sigma_2,\sigma_3, iI, i\sigma_1,i\sigma_2,i\sigma_3\}$ . These eight matrices are linearly independent over $\mathbb{R}$ . To see it, let $c_k$ be the coefficient of $\sigma_k$ and $d_k$ that of $i\sigma_k$ (with $\sigma_0 = I$ ), and read off the four entries of the combination: their real and imaginary parts are $c_0 \pm c_3$ , $d_0 \pm d_3$ , $c_1 \pm d_2$ and $d_1 \mp c_2$ . Every coefficient occurs in two of these eight numbers, once with each sign, so the eight can vanish together only if all eight coefficients vanish. Hassani leaves the same verification to his reader. They are therefore a real basis of $M_2(\mathbb{C})$ , which has real dimension $8 = 2^3$ . Hence $\mathcal{C}(3,0) \cong M_2(\mathbb{C})$ , with the eight-element real basis above.

Step 4 (the even part of the Pauli algebra is $\mathfrak{su}(2)$ exponentiated). From $\sigma_i\sigma_j + \sigma_j\sigma_i = 2\delta_{ij}I$ one gets, for $j \neq k$ , $[\sigma_j,\sigma_k] = 2\sigma_j\sigma_k$ , and the products computed in Step 3 give $[\sigma_j,\sigma_k] = 2i\epsilon_{jkl}\sigma_l$ . Putting $s_j = -i\sigma_j/2$ turns this into

[s_j, s_k] = \epsilon_{jkl}s_l ,

Hassani's Eq. (31.11). The $s_j$ are traceless and anti-Hermitian, hence a basis of $\mathfrak{su}(2)$ , and under the identification of Step 3 they are the three bivectors up to a factor: $e_1e_2 = i\sigma_3 = -2s_3$ , and likewise for the cyclic companions. The even subalgebra of $\mathcal{C}(3,0)$ is four-dimensional, spanned by $1$ and the three bivectors $e_1e_2$ , $e_2e_3$ , $e_3e_1$ ; each bivector squares to $-1$ (by the computation of Step 2, which used only $e_ie_j = -e_je_i$ and $e_i^2 = \pm1$ with the two signs cancelling), and any two distinct bivectors anticommute, for instance $(e_1e_2)(e_2e_3) = e_1e_3 = -(e_2e_3)(e_1e_2)$ . Up to a choice of signs these are the quaternion relations, so $\mathcal{C}^0(3,0) \cong \mathbb{H}$ , and its elements of unit norm form $\mathrm{SU}(2)$ (see quaternions and the rotation group and SU(2)). $\square$

Proof(Part 2: Dirac's conditions)AdaptedAdapted from Hassani (2013), §27.1.1, Eqs. (27.9)–(27.11), pp. 832–834; Arfken, Weber & Harris (2013), §2.2, Eqs. (2.60)–(2.75), pp. 110–113

Work in units with $\hbar = c = 1$ , as Hassani does, so the relation to be reproduced is $E^2 - \mathbf{p}^2 - m^2 = 0$ .

Step 1 (what is being demanded). Dirac looked for a first-order equation $\bigl(\beta E + \sum_{i}\alpha_ip_i + m\bigr)\psi = 0$ whose multiplication by the conjugate factor $\bigl(\beta E + \sum_j\alpha_jp_j - m\bigr)$ returns $\bigl(E^2 - \sum_i p_i^2 - m^2\bigr)\psi = 0$ . Here $E$ and $p_i$ are treated as commuting operators and $\beta,\alpha_i$ are unknown quantities, not assumed to be numbers.

Step 2 (multiply out). Expanding the product, four cross terms cancel in pairs — $\pm m\beta E$ and $\pm m\sum_i\alpha_ip_i$ , the mass having opposite signs in the two factors — and the result is

\beta^2E^2 + \sum_i (\beta\alpha_i + \alpha_i\beta)Ep_i + \tfrac12\sum_{i,j}(\alpha_i\alpha_j + \alpha_j\alpha_i)\,p_ip_j - m^2 ,

the symmetrisation in the third term being legitimate because $p_ip_j = p_jp_i$ .

Step 3 (match coefficients). Comparing with $E^2 - \sum_i p_i^2 - m^2$ term by term:

$\beta^2 = 1$ ;
$\beta\alpha_i + \alpha_i\beta = 0$ for each $i$ , since no $Ep_i$ term is wanted;
$\alpha_i\alpha_j + \alpha_j\alpha_i = -2\delta_{ij}$ , since the coefficient of $p_i^2$ must be $-1$ and no $p_ip_j$ cross terms are wanted.

Step 4 (this is a Clifford algebra). Rename $\beta = \gamma^0$ and $\alpha_i = \gamma^i$ . The three conditions read $(\gamma^0)^2 = 1$ , $(\gamma^i)^2 = -1$ , and $\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 0$ for $\mu \neq \nu$ — exactly Arfken, Weber and Harris's Eqs. (2.74)–(2.75) — which is the single relation

\gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2\eta^{\mu\nu}I, \qquad \eta = \operatorname{diag}(1,-1,-1,-1) .

Hassani's remark is that this is the defining relation of a Clifford algebra: it is his Eq. (27.4) with $g = \eta$ . The algebra is $\mathcal{C}(1,3)$ in the convention of the definition above.

Step 5 (why matrices, and how big). Numbers commute, so no four numbers can satisfy $\gamma^\mu\gamma^\nu = -\gamma^\nu\gamma^\mu$ with all of them non-zero; matrices must be used. Arfken, Weber and Harris reach the explicit $4\times4$ set by the direct-product construction of their Eqs. (2.68)–(2.72), starting from the identity $(\boldsymbol\sigma\cdot\mathbf{p})^2 = \mathbf{p}^2I_2$ , which is itself the Pauli case of Step 3 above; one such set is

\gamma^0 = \begin{pmatrix}I_2 & 0\\ 0 & -I_2\end{pmatrix}, \qquad \gamma^i = \begin{pmatrix}0 & \sigma_i\\ -\sigma_i & 0\end{pmatrix} \quad (i = 1,2,3) ,

and one checks the relation block by block: $(\gamma^0)^2 = I_4$ ; $(\gamma^i)^2 = \operatorname{diag}(-\sigma_i^2, -\sigma_i^2) = -I_4$ ; $\gamma^0\gamma^i + \gamma^i\gamma^0 = 0$ because the off-diagonal blocks acquire opposite signs; and for $i \neq j$ , $\gamma^i\gamma^j + \gamma^j\gamma^i = \operatorname{diag}\bigl(-(\sigma_i\sigma_j+\sigma_j\sigma_i),\,-(\sigma_i\sigma_j+\sigma_j\sigma_i)\bigr) = 0$ . Since $\psi$ must be a column the $\gamma^\mu$ can act on, it has four components, two of which were eventually identified with the electron's two spin states and two with the positron's. $\square$

RemarkOriginal

Where the hypotheses are used, which conventions clash, and what the two books are counting.

Only the symmetry of $g$ and the invertibility of $2$ are used in the basis theorem. Step 1 needs $uv + vu = 2g(u,v)$ with $g$ symmetric, and an orthonormal basis, which exists for any symmetric real form by Sylvester's law of inertia. Degeneracy is harmless there: if $g(e_i,e_i) = 0$ for some $i$ the relations still reduce every word, and one gets an exterior algebra in those directions. Non-degeneracy is needed for the classification in the table and for the theory of $\mathrm{Pin}$ and $\mathrm{Spin}$ , because twisted conjugation by $a$ is built from $a^{-1}$ and the reflection formula.
The signature, not the dimension, fixes the algebra. $\mathcal{C}(3,0) \cong M_2(\mathbb{C})$ is simple, while $\mathcal{C}(0,3) \cong \mathbb{H}\oplus\mathbb{H}$ is not even indecomposable: it has two non-trivial two-sided ideals. Both are built on $V = \mathbb{R}^3$ and both have dimension $8$ . Anyone quoting "the Clifford algebra of three-space" must say which sign convention is in force, and the same applies to the Dirac algebra: with $\eta = \operatorname{diag}(1,-1,-1,-1)$ one gets $\mathcal{C}(1,3)$ , while with the opposite overall sign one gets $\mathcal{C}(3,1) \cong M_4(\mathbb{R})$ , the algebra Hassani works out in §27.4.3. These two are not isomorphic as real algebras, though their complexifications are.
Two different things are both called "the dimension" of the Dirac algebra, and both equal 16. Arfken, Weber and Harris write that the Pauli matrices with $1_2$ form a Clifford algebra "of dimension 4", counting the four linearly independent $2\times 2$ matrices over $\mathbb{C}$ ; and that the Dirac matrices belong to a Clifford algebra of dimension $16$ , with the complete basis $1_4$ , $\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3$ , the four $\gamma^\mu$ , the four $\gamma^5\gamma^\mu$ and the six $\sigma^{\mu\nu} = i\gamma^\mu\gamma^\nu$ , which is a basis of $M_4(\mathbb{C})$ over $\mathbb{C}$ . Hassani's count is of the real algebra: $2^N$ , which is $8$ for $N=3$ and $16$ for $N=4$ . For $N=3$ the two counts disagree ( $4$ complex versus $8$ real) and both are right, because $M_2(\mathbb{C})$ has complex dimension $4$ and real dimension $8$ . For $N=4$ they agree numerically at $16$ while still counting different things: Arfken, Weber and Harris's sixteen matrices include explicit factors of $i$ , so they are a complex basis of $M_4(\mathbb{C})$ , whereas Hassani's sixteen ordered products are a real basis of $\mathcal{C}(1,3)$ . The coincidence of the two numbers at $N=4$ is an accident of the counting and should not be read as the same statement.
The chain $\mathrm{Spin}(3,0) \cong \mathrm{SO}(3) \cong \mathrm{SU}(2)$ is not to be read literally, and Hassani says so. His Theorem 31.2.6 states that $\Psi : \mathrm{Spin}(\mu,\nu)\to\mathrm{SO}(\mu,\nu)$ is surjective with kernel $\{1,-1\}$ — a two-to-one map, not an isomorphism. Two pages later, summarising the Pauli case, he writes the chain of isomorphism signs above, introducing it with "there seems to be the following chain", and says that its first link confirms the theorem. The tension is only apparent, and he resolves it himself: a footnote hung on that sentence (footnote 3, p. 999) says the first link is not exactly an isomorphism but a homomorphism that is locally, though not globally, one, and that the double-valued kernel lets one picture $\mathrm{Spin}(3,0)$ as two copies of $\mathrm{SO}(3)$ . So the book is consistent, and what it asserts is the theorem: $\mathrm{Spin}(3,0) \cong \mathrm{SU}(2)$ , with the map onto $\mathrm{SO}(3)$ two-to-one with kernel $\{\pm 1\}$ . That is also what the rotation group and SU(2) establishes independently, and it is what this unit uses; the displayed chain is shorthand, and the footnote is the statement to quote.
What the kernel $\{\pm1\}$ costs. Because $a$ and $-a$ give the same rotation, a spinor cannot be recovered from the rotation it is transported by: carrying a spinor once around a full $2\pi$ rotation returns $-\psi$ , and only $4\pi$ returns $\psi$ . The example below makes this explicit — the rotor for angle $\theta$ contains $\theta/2$ .
Spinors are not tensors. Arfken, Weber and Harris say it in §4.1 and Riley, Hobson and Bence repeat it in a footnote: a spinor has none of the transformation properties of a scalar, a vector or a tensor of any rank, and that is precisely because its transformation law is a representation of $\mathrm{Spin}$ rather than of $\mathrm{SO}$ .

Method(Building a gamma-matrix representation in any even dimension)Original

To represent $\mathcal{C}(2k,0)$ — equivalently, to produce $2k$ anticommuting matrices squaring to $+I$ — use the tensor-product ladder:

Fix the size. The matrices will be $2^k \times 2^k$ , acting on $(\mathbb{C}^2)^{\otimes k}$ : one Pauli factor for each pair of directions.
Write the two families. For $a = 1,\dots,k$ set
$\gamma_{2a-1} = \underbrace{\sigma_3\otimes\cdots\otimes\sigma_3}_{a-1}\otimes\,\sigma_1\,\otimes \underbrace{I\otimes\cdots\otimes I}_{k-a}, \qquad \gamma_{2a} = \underbrace{\sigma_3\otimes\cdots\otimes\sigma_3}_{a-1}\otimes\,\sigma_2\,\otimes \underbrace{I\otimes\cdots\otimes I}_{k-a} .$
Why it works. In the same slot $a$ , the two matrices differ by $\sigma_1$ versus $\sigma_2$ , which anticommute. In different slots $a < b$ , the matrix from slot $b$ carries a $\sigma_3$ in slot $a$ where the matrix from slot $a$ carries $\sigma_1$ or $\sigma_2$ ; those anticommute, and every other slot contributes a commuting pair. One anticommuting pair in the tensor product makes the whole product anticommute. Each $\gamma$ squares to $I$ because every factor does.
Change the signature. To make $\gamma_i^2 = -I$ , multiply that $\gamma_i$ by $i$ ; this preserves all the anticommutation relations. So $\mathcal{C}(\mu,\nu)$ with $\mu+\nu = 2k$ is represented inside the same matrix algebra.
Add the chirality element in odd dimensions. The product $\gamma_{2k+1} = (-i)^{k}\gamma_1\gamma_2\cdots\gamma_{2k}$ anticommutes with all of the others — it is a product of an even number $2k$ of them, and moving one $\gamma_i$ through the other $2k-1$ factors costs $(-1)^{2k-1} = -1$ — and squares to $I$ , giving a representation of $\mathcal{C}(2k+1,0)$ of the same size. Verified numerically for $k = 1,2,3$ . For $k=2$ this is the analogue of the Dirac matrix $\gamma^5$ , which Arfken, Weber and Harris define as $i\gamma^0\gamma^1\gamma^2\gamma^3$ for their own signature.
Check. Verify $\gamma_i\gamma_j + \gamma_j\gamma_i = 2g_{ij}I$ for all $i \le j$ ; with $2k$ matrices that is $k(2k+1)$ identities, so $3$ , $10$ and $21$ for $k = 1, 2, 3$ . Done numerically for $k = 1,2,3$ — $2$ , $4$ and $6$ matrices of sizes $2$ , $4$ and $8$ — all relations hold to machine precision.

Example(A rotor and a Dirac trace)Original

(a) A rotor, with a number at the end. Work in the Pauli algebra $\mathcal{C}(3,0)$ , representing $e_i$ by $\sigma_i$ as in Step 3. Let $B = e_1e_2$ , so $B^2 = -1$ by the computation in Step 2 (the sign there did not use the squares being $-1$ : $e_1e_2e_1e_2 = -e_1^2e_2^2 = -1$ here as well). Since $B^2 = -1$ , the exponential series splits exactly as for a complex exponential:

R(\theta) = \exp\Bigl(-\tfrac{\theta}{2}B\Bigr) = \cos\tfrac{\theta}{2} - \sin\tfrac{\theta}{2}\,B , \qquad R(\theta)^{-1} = \cos\tfrac{\theta}{2} + \sin\tfrac{\theta}{2}\,B .

Write $c = \cos(\theta/2)$ , $s = \sin(\theta/2)$ . Using $Be_1 = e_1e_2e_1 = -e_2$ , $e_1B = e_1e_1e_2 = e_2$ and $Be_1B = -e_2e_1e_2 = e_1$ ,

R e_1 R^{-1} = (c - sB)e_1(c + sB) = c^2e_1 + cs\,e_2 + sc\,e_2 - s^2e_1 = (c^2-s^2)e_1 + 2cs\,e_2 = \cos\theta\,e_1 + \sin\theta\,e_2 ,

by the double-angle formulas; the same computation gives $Re_2R^{-1} = -\sin\theta\,e_1 + \cos\theta\,e_2$ and $Re_3R^{-1} = e_3$ , since $e_3$ commutes with $B$ . So $R(\theta)$ implements the rotation by $\theta$ about the $e_3$ axis — and it does so through the half-angle, so $R(2\pi) = -1$ while the rotation it induces is the identity. That is the kernel $\{\pm1\}$ of statement 4, made concrete.

Now take $\theta = 60^{\circ} = \pi/3$ and $v = e_1 + 2e_2 + 3e_3$ . Then

RvR^{-1} = \bigl(\tfrac12 - \sqrt3\bigr)e_1 + \bigl(\tfrac{\sqrt3}{2} + 1\bigr)e_2 + 3e_3 = -1.2320508\,e_1 + 1.8660254\,e_2 + 3\,e_3 ,

and the length is preserved:

(-1.2320508)^2 + (1.8660254)^2 + 3^2 = 1.5179492 + 3.4820508 + 9 = 14 = 1^2+2^2+3^2 .

Carrying the whole calculation out with the $2\times2$ matrices — forming $R = \cos 30^{\circ}\,I - \sin 30^{\circ}\,\sigma_1\sigma_2$ , computing $R(\sigma_1 + 2\sigma_2 + 3\sigma_3)R^{-1}$ , and extracting the coefficients as $\tfrac12\operatorname{tr}(\,\cdot\,\sigma_i)$ — returns the same three numbers to machine precision.

(b) A Dirac trace. With the explicit $\gamma^\mu$ of Part 2, Step 5, the Clifford relation alone gives

\gamma^1\gamma^2\gamma^1\gamma^2 = -\gamma^1\gamma^1\gamma^2\gamma^2 = -(-I)(-I) = -I, \qquad\text{so}\qquad \operatorname{tr}\bigl(\gamma^1\gamma^2\gamma^1\gamma^2\bigr) = -4 .

This agrees with the standard four-gamma trace identity $\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma) = 4(\eta^{\mu\nu}\eta^{\rho\sigma} - \eta^{\mu\rho}\eta^{\nu\sigma} + \eta^{\mu\sigma}\eta^{\nu\rho})$ , which at $(\mu,\nu,\rho,\sigma) = (1,2,1,2)$ gives $4(0 - (-1)(-1) + 0) = -4$ . Evaluating all $256$ traces $\operatorname{tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma)$ numerically from the explicit matrices and comparing with the identity gives no mismatches, and $\operatorname{tr}(\gamma^0\gamma^1\gamma^2\gamma^3) = 0$ as the identity requires.

Example(A spinor as a column)AdaptedAdapted from Hassani (2013), §31.2.1, p. 999

(c) This one follows Hassani's construction of the Pauli spinors inside the algebra. In $\mathcal{C}(3,0) \cong M_2(\mathbb{C})$ , set $P = \tfrac12(1 + e_3)$ . Then, since $e_3^2 = 1$ ,

P^2 = \tfrac14(1 + 2e_3 + e_3^2) = \tfrac14(2 + 2e_3) = P ,

so $P$ is an idempotent; in the Pauli representation it is $\begin{pmatrix}1&0\\0&0\end{pmatrix}$ . The left ideal $\mathcal{C}(3,0)\,P$ consists of all matrices whose second column is zero, which is a copy of $\mathbb{C}^2$ and is minimal. For suppose a non-zero subspace of it is closed under left multiplication, and take an element of that subspace with a non-zero entry $z$ in row $r$ . Left multiplication by the matrix unit $E_{sr}$ carries that entry into row $s$ , and left multiplication by the scalar $z^{-1}1$ normalises it — a complex scalar is available because $i\,1 = e_1e_2e_3$ lies in the algebra, by Step 3, so a subspace invariant under the real algebra is automatically a complex subspace. The subspace therefore contains both $E_{11}$ and $E_{21}$ , hence all of $\mathcal{C}(3,0)\,P$ . Hassani reaches the same conclusion along a shorter route, citing his Theorem 3.3.1: the minimal left ideals of a total matrix algebra are the matrices with a single non-zero column. He identifies this left ideal as the space of Pauli spinors, the two-component complex columns on which the Pauli matrices act, and obtains a basis of it by left-multiplying $P$ by the basis elements of the algebra, recording four matrices — a basis of $S$ over $\mathbb{R}$ , as his dimension count requires, since $S \cong \mathbb{C}^2$ has real dimension four. Over $\mathbb{C}$ two of them suffice: $P = \begin{pmatrix}1&0\\0&0\end{pmatrix}$ and $e_1P = \begin{pmatrix}0&0\\1&0\end{pmatrix}$ , whose non-zero columns are $(1,0)^{\mathsf T}$ and $(0,1)^{\mathsf T}$ , the spin-up and spin-down states.

ApplicationOriginal

The electron. The Dirac equation $\bigl(i\gamma^\mu\partial_\mu - m\bigr)\psi = 0$ is the first-order equation Part 2 was constructing, and its four-component solution is the relativistic electron. Two of the four components became the two spin states — spin, as Hassani puts it, came out of the equation rather than being put in — and the other two were recognised only later as the positron's. That the gamma matrices must be $4\times4$ and not $2\times2$ , and that $\psi$ is therefore a four-component object, is not a modelling choice: it is forced by the Clifford relation, as Step 5 shows.

Spin- $\tfrac12$ in non-relativistic quantum mechanics. The Pauli algebra of example (a) is the whole apparatus of spin- $\tfrac12$ : the states are the two-component spinors of example (c), the observables are the $\sigma_i$ , and the rotation of a spin state by angle $\theta$ about an axis is the rotor $\exp(-\tfrac{\theta}{2}B)$ . The half-angle is directly observable — in neutron interferometry a beam rotated through $2\pi$ in a magnetic field returns with its sign reversed and interferes destructively with the unrotated beam, which is the kernel $\{\pm1\}$ measured in a laboratory.

Rotations in computation. Because $\mathcal{C}(0,2) \cong \mathbb{H}$ and the even part of $\mathcal{C}(3,0)$ gives $\mathrm{SU}(2)$ , the rotor formula $v \mapsto RvR^{-1}$ of example (a) is exactly the unit-quaternion representation of rotations used in computer graphics, robotics and spacecraft attitude control. It is preferred to Euler angles because composing rotations is a single product, interpolation between orientations is smooth, and there is no gimbal lock; the price is the two-to-one redundancy, which shows up as the need to choose a sign when converting a rotation matrix back to a quaternion.

Trace technology. In quantum field theory a cross-section is an average over spins of $|\mathcal{M}|^2$ , and the spin sums turn the amplitude into traces of products of gamma matrices. Identities such as the one checked in example (b) reduce those traces to metric tensors, so that a calculation that looks like a $4\times4$ matrix product is carried out purely with the Clifford relation and never needs an explicit representation. That independence of representation is the practical payoff of defining the algebra by generators and relations rather than by matrices.