Irreducible Representations of SU(2) and Angular Momentum

Angular momentum in quantum mechanics is not solved — it is classified. You never integrate a differential equation to learn that a spin can be $\tfrac12$ but never $\tfrac13$ ; you read it straight off the algebra of rotations. The three generators $J_x,J_y,J_z$ close under commutation into the Lie algebra $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ , and every physically realizable multiplet is an irreducible representation of that algebra. The whole list of allowed quantum numbers — the total magnitude $j$ , the projection $m$ — falls out of three demands: that $J^2$ and $J_z$ be simultaneously diagonalizable, that the raising and lowering operators shift $m$ by one unit, and that the resulting ladder be finite. This is representation theory doing the work of dynamics. The payoff is a complete, closed classification. Each irrep $D^{(j)}$ is $(2j+1)$ -dimensional, uniquely fixed by a single non-negative half-integer $j$ , and every rotation acts on it through the exponential $\exp(-\tfrac{i}{\hbar}\theta\,\mathbf n\cdot\mathbf J)$ . The split between integer and half-integer $j$ is not algebraic but topological: $\mathrm{SU}(2)$ is the simply connected double cover of the rotation group $\mathrm{SO}(3)$ , so integer- $j$ irreps descend to genuine rotations while half-integer ones — the spinors — remember the covering and flip sign under a $2\pi$ turn. From this one structure spill the Stern–Gerlach doublet, atomic selection rules, isospin multiplets, the qubit, and the $4\pi$ periodicity that neutron interferometers have measured. This article assembles it end to end: the algebra, the ladder quantization, the explicit matrices, the characters, and the Clebsch–Gordan rule for adding two of them.

Key takeaways

The rotation generators satisfy $[J_i,J_j]=i\hbar\,\epsilon_{ijk}J_k$ ; the finite-dimensional irreducible representations of this $\mathfrak{su}(2)$ algebra are labelled by one non-negative half-integer $j$ and have dimension $2j+1$ .
Inside each irrep the joint eigenstates of $J^2$ and $J_z$ obey $J^2\to\hbar^2 j(j+1)$ and $J_z\to\hbar m$ with $m=-j,\dots,+j$ in unit steps.
The ladder operators $J_\pm=J_x\pm iJ_y$ raise and lower $m$ ; demanding the ladder terminate at top and bottom is what quantizes $j$ to $\{0,\tfrac12,1,\tfrac32,\dots\}$ .
$\mathrm{SU}(2)$ is the simply connected double cover of $\mathrm{SO}(3)$ : integer- $j$ irreps are true $\mathrm{SO}(3)$ representations, half-integer (spinor) irreps flip sign under $2\pi$ and need $4\pi$ to return.
Tensor products decompose by the Clebsch–Gordan rule $D^{(j_1)}\otimes D^{(j_2)}=\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}D^{(j)}$ — the algebra of adding angular momenta.
Key idea: Diagonalize the Casimir $J^2$ and $J_z$ , then use $[J_z,J_\pm]=\pm\hbar J_\pm$ to climb the ladder; finiteness of the ladder forces $j\in\{0,\tfrac12,1,\tfrac32,\dots\}$ and fixes the entire $(2j+1)$ -dimensional irrep.

Where this lives today

The $\mathrm{SU}(2)$ irreps are load-bearing in modern physics. Every qubit is a state in the $j=\tfrac12$ irrep, and every single-qubit gate is a finite $\mathrm{SU}(2)$ rotation $e^{-i\theta\,\mathbf n\cdot\boldsymbol\sigma/2}$ on the Bloch sphere. In magnetic resonance — NMR and clinical MRI — radiofrequency pulses are literally $\mathrm{SU}(2)$ rotations of nuclear spins, and the spin-echo that sharpens every image is a sequence of $\pi$ and $\pi/2$ rotations from this same group. In particle physics the same mathematics reappears as isospin: proton and neutron form a spin- $\tfrac12$ doublet under an internal $\mathrm{SU}(2)$ , the pions an $I=1$ triplet, and cross-section ratios follow from Clebsch–Gordan coefficients exactly as for spin. Most strikingly, the $4\pi$ periodicity of spinors was confirmed by direct experiment: in 1975 neutron-interferometry groups (Rauch and Zeilinger in Vienna; Werner, Colella, Overhauser and Eagen in the United States) rotated a neutron's spin through controlled angles and watched the interference pattern return only after a full $4\pi$ — the spinor's double-cover behaviour, measured exactly as the representation theory demands.

The physical picture

Think of $J_z$ as a ruler measuring the projection of angular momentum onto a chosen axis, and of $J_\pm$ as gears that click that reading up or down by one unit of $\hbar$ . The Casimir $J^2$ is the total magnitude, and Schur's lemma guarantees it is blind to which rung you stand on: commuting with every generator, on an irreducible multiplet it can only be a number, the same $\hbar^2 j(j+1)$ for every state in the tower. So a spin- $j$ system is a rigid ladder of $2j+1$ equally spaced rungs sharing one conserved total. The first figure draws it for $j=2$ and puts real numbers on the gears.

Figure

The spin-2 ladder: J_+ and J_− step m by ±1 unit and annihilate the top and bottom rungs; the ladder coefficients are ℏ·2, ℏ√6, ℏ√6, ℏ·2, and the Casimir J² = 6ℏ² is shared by all five states

The rungs tell you nothing about a subtle global fact: what happens when you rotate the whole system by a full turn. For integer $j$ a $2\pi$ rotation is the identity; for half-integer $j$ it is minus the identity — the state returns with a reversed sign, and only $4\pi$ restores it. This is the fingerprint of the double cover: quantum rotations are represented not by $\mathrm{SO}(3)$ but by its two-sheeted cover $\mathrm{SU}(2)$ , and half-integer spin lives on the extra sheet. The second figure shows the two clocks running at different rates — a vector home after $2\pi$ , a spinor only after $4\pi$ .

Figure

Two turns to come home: an SO(3) vector observable follows cos θ and returns after 2π, but the SU(2) spinor amplitude follows cos(θ/2) and needs a full 4π — at θ = 2π it has flipped to −1, so a 2π rotation acts as −I on a spinor and only 4π restores +I

The core equations

Everything starts from one line. The rotation generators do not commute — the exact way they fail to is the algebra — and fixing that structure determines the entire representation theory.

[J_i,J_j]=i\hbar\,\epsilon_{ijk}J_k \qquad (1)

where: $J_i$ — Hermitian angular-momentum (rotation) generators, $i=x,y,z$ ; $\epsilon_{ijk}$ — totally antisymmetric Levi-Civita symbol; $\hbar$ — reduced Planck constant (set to $1$ in many texts). Reading: the commutator of two generators is the third, so the three form a closed three-dimensional Lie algebra, $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ . The total-squared magnitude commutes with every generator, so by Schur's lemma it is a constant on any irrep — the single number that labels the multiplet.

J^2=J_x^2+J_y^2+J_z^2,\qquad [J^2,J_i]=0 \qquad (2)

where: $J^2$ — quadratic Casimir operator; it commutes with all generators and acts as a scalar within each irrep. Reading: $\{J^2,J_z\}$ is the maximal commuting set (the Cartan subalgebra plus its Casimir). The remaining two generators are repackaged into raising and lowering combinations — non-Hermitian shift operators, the machinery that moves you along the ladder.

J_\pm=J_x\pm iJ_y,\qquad [J_z,J_\pm]=\pm\hbar J_\pm,\qquad [J_+,J_-]=2\hbar J_z \qquad (3)

where: $J_\pm$ — raising/lowering operators, non-Hermitian with $J_-=J_+^\dagger$ . Reading: $[J_z,J_\pm]=\pm\hbar J_\pm$ says $J_\pm$ carries one unit of $J_z$ , so it steps $m$ up or down by one. Diagonalizing the commuting pair simultaneously gives the two quantum numbers every physicist quotes.

J^2|j,m\rangle=\hbar^2 j(j+1)|j,m\rangle,\qquad J_z|j,m\rangle=\hbar m|j,m\rangle \qquad (4)

where: $j$ — total angular-momentum (spin) quantum number, $j\in\{0,\tfrac12,1,\dots\}$ ; $m$ — magnetic quantum number, $m\in\{-j,\dots,+j\}$ . Reading: the magnitude $\hbar\sqrt{j(j+1)}$ is strictly larger than the maximum projection $\hbar j$ — the vector never fully aligns, a purely quantum effect. Acting with $J_\pm$ produces the neighbouring state, and normalization fixes the coefficient exactly (up to the standard real-positive phase).

J_\pm|j,m\rangle=\hbar\sqrt{j(j+1)-m(m\pm1)}\;|j,m\pm1\rangle \qquad (5)

where: $\sqrt{j(j+1)-m(m\pm1)}$ — the ladder coefficient, from the norm of $J_\pm|j,m\rangle$ (Condon–Shortley convention). Reading: it vanishes at $m=+j$ for $J_+$ and at $m=-j$ for $J_-$ — exactly how the ladder closes off at both ends. A finite rotation is the exponential of the generators — the Wigner $D$ -matrix attached to each irrep.

D^{(j)}(\mathbf n,\theta)=\exp\!\Big(-\tfrac{i}{\hbar}\,\theta\,\mathbf n\cdot\mathbf J\Big) \qquad (6)

where: $\mathbf n$ — unit rotation axis; $\theta$ — rotation angle; $\mathbf J$ — the vector of generators in the spin- $j$ representation. Reading: this is the $(2j+1)\times(2j+1)$ unitary matrix by which a rotation acts on a spin- $j$ multiplet. Counting the rungs gives the dimension, the single most-used fact about the irrep.

\dim D^{(j)}=2j+1 \qquad (7)

where: $2j+1$ — the number of allowed $m$ values, hence the size of the multiplet. Reading: $j=0\to1$ , $j=\tfrac12\to2$ , $j=1\to3$ , $j=\tfrac32\to4$ , and so on. The trace of the rotation matrix is a class function — depending only on the angle, not the axis — the fingerprint that tells irreps apart.

\chi^{(j)}(\theta)=\sum_{m=-j}^{j}e^{im\theta}=\frac{\sin\!\big((2j+1)\theta/2\big)}{\sin(\theta/2)} \qquad (8)

where: $\theta$ — rotation angle (class parameter); $\chi^{(j)}$ — character, the trace of $D^{(j)}$ . Reading: this Dirichlet kernel is orthonormal against the $\mathrm{SU}(2)$ Haar (Weyl) measure, which is what lets character algebra decompose any representation uniquely. Finally, combining two systems multiplies their state spaces, and the product reduces to a direct sum of irreps — the addition of angular momenta.

D^{(j_1)}\otimes D^{(j_2)}=\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}D^{(j)} \qquad (9)

where: $\otimes$ — tensor (Kronecker) product of representations; $\oplus$ — direct sum (block-diagonal decomposition into irreps). Reading: two spins combine into everything from $|j_1-j_2|$ up to $j_1+j_2$ in integer steps, each appearing exactly once.

Derivation AQuantization of j and m from a finite ladder

The goal is to prove, using only the commutation relations and the positivity of norms, that $J^2$ has eigenvalue $\hbar^2 j(j+1)$ , that $m$ runs from $-j$ to $+j$ in unit steps, and that $2j$ must be a non-negative integer — so $j\in\{0,\tfrac12,1,\tfrac32,\dots\}$ , with no differential equation solved anywhere. Fix an irrep, and let $|m\rangle$ be a normalized $J_z$ eigenstate with eigenvalue $\hbar m$ and shared $J^2$ eigenvalue $\hbar^2\lambda$ .

Step 1. Apply $J_z$ to $J_\pm|m\rangle$ and use $[J_z,J_\pm]=\pm\hbar J_\pm$ : $J_z(J_\pm|m\rangle)=(J_\pm J_z\pm\hbar J_\pm)|m\rangle=\hbar(m\pm1)(J_\pm|m\rangle)$ . So $J_\pm|m\rangle$ is again a $J_z$ eigenstate, shifted by one unit. Why: this is what makes $J_\pm$ a ladder — it turns one rung into the next, generating the whole multiplet from any single state.

Step 2. Bound the ladder. Since $J_x^2+J_y^2=J^2-J_z^2$ is a sum of squares of Hermitian operators, its expectation is non-negative: $\langle m|J^2-J_z^2|m\rangle=\hbar^2(\lambda-m^2)\ge0$ , so $m^2\le\lambda$ . Why: an infinite ladder would give unbounded $|m|$ with negative $J_x^2+J_y^2$ , impossible in an inner-product space — finiteness is forced by positivity, not assumed.

Step 3. Because the ladder is bounded there is a top state with $J_+|m_{\max}\rangle=0$ and a bottom with $J_-|m_{\min}\rangle=0$ . Name $m_{\max}=j$ . Using $J_-J_+=J^2-J_z^2-\hbar J_z$ , the vanishing norm gives $0=\|J_+|j\rangle\|^2=\hbar^2[\lambda-j(j+1)]$ , so $\lambda=j(j+1)$ . Why: the top rung is where the algebra pins the Casimir — the $\lambda$ labelling the whole multiplet is read off the one state $J_+$ annihilates.

Step 4. Repeat at the bottom with $J_+J_-=J^2-J_z^2+\hbar J_z$ : $0=\|J_-|m_{\min}\rangle\|^2=\hbar^2[\lambda-m_{\min}(m_{\min}-1)]$ , so $\lambda=m_{\min}(m_{\min}-1)$ . Matching to $\lambda=j(j+1)$ forces $m_{\min}=-j$ . The ladder is symmetric about zero. Why: top and bottom must give the same Casimir, and the only consistent solution places the floor at $-j$ , exactly opposite the ceiling.

Step 5. The ladder steps from $-j$ to $+j$ in unit increments, so $j-(-j)=2j$ must be a non-negative integer. Therefore $j\in\{0,\tfrac12,1,\tfrac32,\dots\}$ and each irrep holds $2j+1$ states. Why: this is the whole quantization — half-integer spin is allowed because $2j$ , not $j$ , must be an integer, which opens the door to spinors. *Verified: reproduces the standard spectrum of Sakurai Ch. 3 and Arfken Ch. 17; eigenvalue $j(j+1)$ , unit spacing, and $2j+1$ multiplicity all match.*

Go deeper · Derivation B · 1 minThe spin one-half representation and the SU(2) to SO(3) double cover

Derivation BThe spin one-half representation and the SU(2) to SO(3) double cover

The goal is to build the smallest nontrivial irrep from the Pauli matrices, exponentiate it into a finite rotation, and expose the global structure separating $\mathrm{SU}(2)$ from $\mathrm{SO}(3)$ — that two group elements $\pm U$ map to one physical rotation, so the spinor picks up a sign under $2\pi$ and returns only after $4\pi$ . Take $j=\tfrac12$ , where the generators are $J_i=\tfrac{\hbar}{2}\sigma_i$ with the Pauli matrices $\sigma_i$ .

Step 1. Verify the algebra. Using $[\sigma_i,\sigma_j]=2i\epsilon_{ijk}\sigma_k$ , one finds $\big[\tfrac{\hbar}{2}\sigma_i,\tfrac{\hbar}{2}\sigma_j\big]=\tfrac{\hbar^2}{4}\cdot2i\epsilon_{ijk}\sigma_k=i\hbar\,\epsilon_{ijk}\big(\tfrac{\hbar}{2}\sigma_k\big)$ , exactly equation (1). Why: the Pauli matrices are a faithful two-dimensional realization of $\mathfrak{su}(2)$ , so everything derived abstractly must hold here concretely.

Step 2. Exponentiate. Because $(\mathbf n\cdot\boldsymbol\sigma)^2=I$ , the Taylor series resums to $e^{-i\theta(\mathbf n\cdot\boldsymbol\sigma)/2}=\cos\tfrac{\theta}{2}\,I-i\sin\tfrac{\theta}{2}\,(\mathbf n\cdot\boldsymbol\sigma)$ , a generic $\mathrm{SU}(2)$ element. Why: the half-angle $\theta/2$ is the whole story — the seed of the spinor's slow clock, running at half a vector's rate.

Step 3. Map down to $\mathrm{SO}(3)$ . For any $U\in\mathrm{SU}(2)$ , the conjugation $U(\mathbf x\cdot\boldsymbol\sigma)U^\dagger=(R\mathbf x)\cdot\boldsymbol\sigma$ defines a rotation $R\in\mathrm{SO}(3)$ ; but $U$ and $-U$ give the same $R$ (the sign cancels in the two-sided product). The map is two-to-one with kernel $\{\pm I\}$ , so $\mathrm{SO}(3)\cong\mathrm{SU}(2)/\mathbb{Z}_2$ . Why: this is the double cover made algebraic — the two sheets of $\mathrm{SU}(2)$ collapse onto one $\mathrm{SO}(3)$ , and the lost information is the spinor sign.

Step 4. Read off the periodicity. Along $\hat z$ , $D^{(1/2)}(\hat z,\theta)=\mathrm{diag}(e^{-i\theta/2},e^{+i\theta/2})$ , so $D^{(1/2)}(\hat z,2\pi)=-I$ and $D^{(1/2)}(\hat z,4\pi)=+I$ . A $2\pi$ rotation returns every vector to itself but multiplies the spinor by $-1$ . Why: this sign is not a bookkeeping artifact — neutron interferometry has measured it directly, confirming that half-integer spin lives on the double cover. *Verified: Pauli-matrix algebra and the rotation formula are standard (Sakurai Ch. 3); the kernel $\{\pm I\}$ matches $\mathrm{SU}(2)/\mathbb{Z}_2\cong\mathrm{SO}(3)$ .*

Go deeper · Derivation C · 1 minThe Clebsch-Gordan series from character multiplication

Derivation CThe Clebsch-Gordan series from character multiplication

The goal is to derive the addition rule $D^{(j_1)}\otimes D^{(j_2)}=\bigoplus_{j=|j_1-j_2|}^{j_1+j_2}D^{(j)}$ without computing a single Clebsch–Gordan coefficient, using one fact: the character of a tensor product is the product of characters, and characters form an orthonormal basis, so the decomposition is unique. Two systems of spins $j_1,j_2$ build a product space of character $\chi^{(j_1)}(\theta)\,\chi^{(j_2)}(\theta)$ .

Step 1. Write each character as a weight sum and multiply: $\big(\sum_{m_1=-j_1}^{j_1}e^{im_1\theta}\big)\big(\sum_{m_2=-j_2}^{j_2}e^{im_2\theta}\big)=\sum_M c_M\,e^{iM\theta}$ , with $M=m_1+m_2$ running from $-(j_1+j_2)$ to $+(j_1+j_2)$ . Why: the total projection is additive, so the combined weights are pairwise sums — the only bookkeeping is how many pairs land on each $M$ .

Step 2. Count $c_M=\#\{(m_1,m_2):m_1+m_2=M\}$ . From the top, $M=j_1+j_2$ occurs once, $M=j_1+j_2-1$ twice, and so on — rising by one per step until it plateaus, a symmetric trapezoid. Why: the shape of $c_M$ encodes which irreps are present, since each $D^{(j)}$ contributes one weight at every $M$ from $-j$ to $+j$ .

Step 3. Peel off irreps from the top. The highest weight is $M=j_1+j_2$ , appearing once, so $D^{(j_1+j_2)}$ must be present exactly once; subtract its character $\chi^{(j_1+j_2)}$ . The remainder now has top weight $M=j_1+j_2-1$ , so $D^{(j_1+j_2-1)}$ appears once — subtract again. Why: because characters are linearly independent, matching top weights one at a time is a deterministic algorithm that cannot double-count or miss an irrep.

Step 4. Iterate until the top weight reaches $|j_1-j_2|$ , where the plateau ends and the recursion terminates, giving $\sum_{j=|j_1-j_2|}^{j_1+j_2}\chi^{(j)}$ — equation (9). As a check, dimensions balance: $(2j_1+1)(2j_2+1)=\sum(2j+1)$ . Why: the dimension identity is an independent arithmetic test — for $j_1=j_2=1$ it reads $3\times3=9=5+3+1$ , confirming $1\otimes1=2\oplus1\oplus0$ . *Verified: the dimension identity holds for all tested $(j_1,j_2)$ — $(1,1)\to9=5+3+1$ , $(\tfrac12,\tfrac12)\to4=3+1$ , $(\tfrac12,1)\to6=4+2$ ; matches Arfken Ch. 17.*

Limiting cases

Special values of $j$ recover the most familiar objects in physics, and the classical limit shows how the discrete cone melts into a smooth vector.

Case	Result	Meaning
Trivial / scalar	$j=0,\ \dim=1$	The invariant singlet; every generator acts as zero. It is the rotational scalar that appears as the singlet in $\tfrac12\otimes\tfrac12$ and as the dot product in $1\otimes1$ .
Fundamental spinor	$j=\tfrac12,\ J_i=\tfrac{\hbar}{2}\sigma_i$	The defining two-dimensional representation of $\mathrm{SU}(2)$ ; the qubit, the electron spin, the object that flips sign under $2\pi$ .
Adjoint / vector	$j=1,\ \dim=3$	Descends to the defining representation of $\mathrm{SO}(3)$ on ordinary $3$ -vectors; the orbital $\ell=1$ states and the vector (dipole) operator.
Classical	$j\to\infty,\ \hbar\to0,\ \hbar j=L$ fixed	$\sqrt{j(j+1)}\to j$ , the spacing $\hbar$ shrinks to zero, and the discrete cone of $J_z$ values becomes a continuous classical angular-momentum vector of length $L$ .

Common trapConfusing SU(2) with SO(3), and reading J² as ℏ²j²

Two traps travel together. First, the magnitude of angular momentum is $\hbar\sqrt{j(j+1)}$ , not $\hbar j$ : the eigenvalue of $J^2$ is $\hbar^2 j(j+1)$ , and the extra $+j$ comes straight from non-commutativity, $[J_x,J_y]=i\hbar J_z$ . It is why a spin can never point exactly along $z$ — the maximum projection $\hbar j$ is strictly below the length $\hbar\sqrt{j(j+1)}$ . Second, do not identify the symmetry group of quantum rotations with $\mathrm{SO}(3)$ . Because quantum states are rays (defined up to a phase), the true group is the universal cover $\mathrm{SU}(2)$ — and that is the entire reason half-integer spin exists. Integer- $j$ irreps happen to descend to honest $\mathrm{SO}(3)$ representations, but the half-integer ones represent $\mathrm{SO}(3)$ only projectively, single-valued only on $\mathrm{SU}(2)$ where the $2\pi$ sign flip is perfectly consistent.

Worked examples

1 · Building the spin-1 representation from scratch (fully worked)

Construct the $3\times3$ matrices $J_x,J_y,J_z$ for $j=1$ in the $\{|1,1\rangle,|1,0\rangle,|1,-1\rangle\}$ basis, verify that $J^2=2\hbar^2 I$ , and evaluate the character at $\theta=\pi/2$ .

Step 1. $J_z$ is diagonal with eigenvalues $\hbar m$ : $J_z=\hbar\,\mathrm{diag}(1,0,-1)$ . The ladder coefficients from equation (5) with $j=1$ are $\hbar\sqrt{1\cdot2-m(m\pm1)}$ : for both $m=0\to1$ and $m=-1\to0$ the coefficient is $\hbar\sqrt2$ . So $J_+$ has the two nonzero entries $\langle1,1|J_+|1,0\rangle=\langle1,0|J_+|1,-1\rangle=\hbar\sqrt2$ , giving

J_+=\hbar\sqrt2\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix},\qquad J_-=J_+^\dagger=\hbar\sqrt2\begin{pmatrix}0&0&0\\1&0&0\\0&1&0\end{pmatrix}.

Step 2. Recombine into Hermitian generators with $J_x=\tfrac12(J_++J_-)$ and $J_y=\tfrac{1}{2i}(J_+-J_-)$ :

J_x=\frac{\hbar}{\sqrt2}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix},\qquad J_y=\frac{\hbar}{\sqrt2}\begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix}.

Step 3. Square and add. Direct multiplication gives $J_x^2+J_y^2=\hbar^2\,\mathrm{diag}(1,2,1)$ and $J_z^2=\hbar^2\,\mathrm{diag}(1,0,1)$ , so $J^2=\hbar^2\,\mathrm{diag}(2,2,2)=\mathbf{2\hbar^2 I}$ , matching $\hbar^2 j(j+1)=\hbar^2\cdot1\cdot2$ exactly — a scalar, as Schur's lemma demands.

Step 4. The character is $\chi^{(1)}(\theta)=e^{i\theta}+1+e^{-i\theta}=1+2\cos\theta$ . At $\theta=\pi/2$ : $\chi^{(1)}(\pi/2)=1+2\cos(\pi/2)=\mathbf{1}$ . Cross-check with the closed form $\sin(3\theta/2)/\sin(\theta/2)=\sin(3\pi/4)/\sin(\pi/4)=1$ . The spin-1 generators satisfy $J^2=2\hbar^2 I$ and $\chi^{(1)}(\pi/2)=1$ .

2 · Adding two spin-½ — the triplet and singlet

Combine two spin- $\tfrac12$ particles. Show that the four-dimensional product space splits as $D^{(1)}\oplus D^{(0)}$ , and write the four combined states explicitly.

Step 1. The product basis $\{|{\uparrow\uparrow}\rangle,|{\uparrow\downarrow}\rangle,|{\downarrow\uparrow}\rangle,|{\downarrow\downarrow}\rangle\}$ has dimension $2\times2=4$ . The Clebsch–Gordan rule (9) predicts $\tfrac12\otimes\tfrac12=1\oplus0$ , dimensions $3+1=4$ . With $\mathbf S=\mathbf S^{(1)}+\mathbf S^{(2)}$ , the total $S_z$ eigenvalue is $m=m_1+m_2$ .

Step 2. The top state is unique: $|1,1\rangle=|{\uparrow\uparrow}\rangle$ . Apply $S_-=S_-^{(1)}+S_-^{(2)}$ : with $S_-|{\uparrow}\rangle=\hbar|{\downarrow}\rangle$ we get $S_-|{\uparrow\uparrow}\rangle=\hbar(|{\downarrow\uparrow}\rangle+|{\uparrow\downarrow}\rangle)$ . Since $S_-|1,1\rangle=\hbar\sqrt2\,|1,0\rangle$ , normalizing gives $|1,0\rangle=\tfrac{1}{\sqrt2}(|{\uparrow\downarrow}\rangle+|{\downarrow\uparrow}\rangle)$ .

Step 3. Continue down: $|1,-1\rangle=|{\downarrow\downarrow}\rangle$ , completing the triplet ( $j=1$ , three states, symmetric under exchange). The one remaining orthogonal $m=0$ combination is the singlet $|0,0\rangle=\tfrac{1}{\sqrt2}(|{\uparrow\downarrow}\rangle-|{\downarrow\uparrow}\rangle)$ , antisymmetric under exchange.

Step 4. Check magnitudes: on the triplet $S^2=2\hbar^2$ , on the singlet $S^2=0$ . The Clebsch–Gordan coefficients are just $\pm\tfrac{1}{\sqrt2}$ and $1$ . The four product states reorganize as a spin-1 triplet plus a spin-0 singlet: $4=3+1$ , the singlet being the rotational scalar $\tfrac{1}{\sqrt2}(|{\uparrow\downarrow}\rangle-|{\downarrow\uparrow}\rangle)$ .

Physics of the portrait

The representation theory of $\mathrm{SU}(2)$ is where abstract group theory became a working tool of the physicist — and no one did more to force that marriage than Eugene Wigner, who turned the group of rotations into a machine for predicting spectra.

Origins · 1 minEugene Wigner and how this was discovered

Eugene Wigner

1902–1995 · group theory and the quantum theory of angular momentum

Eugene Paul Wigner, a Hungarian-born theorist and one of the great mathematical minds of twentieth-century physics, made the machinery of group representations the native language of angular momentum. His 1931 book Gruppentheorie und ihre Anwendung auf die Quantenmechanik der Atomspektren showed that the multiplet structure of atomic spectra, the selection rules, and the classification of states are all consequences of rotational symmetry — of the irreducible representations of $\mathrm{SU}(2)$ and $\mathrm{SO}(3)$ — rather than of any particular Hamiltonian. From this program came the tools every practitioner still uses: the Wigner $D$ -matrices $D^{(j)}(\mathbf n,\theta)$ that rotate a spin- $j$ multiplet, the $3j$ symbols that tidy Clebsch–Gordan bookkeeping, and above all the Wigner–Eckart theorem, which factors any tensor matrix element into a geometric Clebsch–Gordan coefficient times a single reduced matrix element. The ground had been laid by others — Élie Cartan's 1913 spinor representations, Wolfgang Pauli's 1927 two-component spinor and $\sigma$ matrices, Hermann Weyl's systematic representation theory — but it was Wigner who insisted that symmetry, not force, dictates what quantum numbers a system can carry, and who made the double cover $\mathrm{SU}(2)\to\mathrm{SO}(3)$ a computational reality. He shared the 1963 Nobel Prize for the symmetry principles that organize atomic and particle physics alike. His legacy is a way of thinking: given a symmetry, classify its irreps first, and much of the physics is already determined.

History

1913

— Cartan discovers the spinor representations of the rotation algebra

1925

— Uhlenbeck and Goudsmit propose the spinning electron

1927

— Pauli introduces the two-component spinor and the σ matrices

1928

— Weyl systematizes group representations for quantum mechanics

1931

— Wigner applies SU(2) irreps to atomic spectra and states the Wigner-Eckart theorem

Going further

Texts. Sakurai & Napolitano, Modern Quantum Mechanics, 3rd ed., Ch. 3 (theory of angular momentum) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 17 · Stone & Goldbart, Mathematics for Physics, Ch. 16 (Lie groups and representations) · Georgi, Lie Algebras in Particle Physics, 2nd ed., Ch. 3 · Hassani, Mathematical Physics, Ch. 24–25 (group representations) · Nakahara, Geometry, Topology and Physics, Ch. 5.

Provenance. The algebra, the eigenvalues $\hbar^2 j(j+1)$ and $\hbar m$ , the ladder coefficient $\hbar\sqrt{j(j+1)-m(m\pm1)}$ , the dimension $2j+1$ , the character $\sin((2j+1)\theta/2)/\sin(\theta/2)$ , and the Clebsch–Gordan series are standard and were recomputed against Sakurai Ch. 3, Arfken Ch. 17, and Stone & Goldbart Ch. 16. The spin-1 matrices, $J^2=2\hbar^2 I$ , $\chi^{(1)}(\pi/2)=1$ , the triplet/singlet construction, and the dimension identities $(2j_1+1)(2j_2+1)=\sum(2j+1)$ [checked at $(1,1)\to9=5+3+1$ , $(\tfrac12,\tfrac12)\to4=3+1$ , $(\tfrac12,1)\to6=4+2$ ] were verified directly. The Pauli relations and the kernel $\{\pm I\}$ of $\mathrm{SU}(2)\to\mathrm{SO}(3)$ are elementary. No non-standard numbers are used.

Builds on: Lie Algebras and the Exponential Map · Lie Groups · Reducible and Irreducible Representations · Schur's Lemmas · The Rotation Group SO(3). Leads to: Tensor Products and Clebsch–Gordan Decomposition · Characters and Orthogonality Theorems · Group Homomorphisms and Quotient Groups.

Ask the tutor: Ask me to build the explicit spin- $\tfrac32$ matrices, walk through any Clebsch–Gordan table by hand, or show exactly how the $4\pi$ neutron-interferometry sign flip is measured.

Why is the eigenvalue of J² equal to ℏ²j(j+1) and not ℏ²j²?

Because $J^2$ is not just $J_z^2$ — it also contains the transverse pieces $J_x^2+J_y^2$ , which do not vanish even on the top state. Write $J^2=J_z^2+\tfrac12(J_+J_-+J_-J_+)$ and evaluate on $|j,j\rangle$ , using $J_+|j,j\rangle=0$ and the identity $J_-J_+=J^2-J_z^2-\hbar J_z$ . This yields $J^2=\hbar^2(j^2+j)=\hbar^2 j(j+1)$ . The extra $+j$ is the direct imprint of non-commutativity $[J_x,J_y]=i\hbar J_z$ : because $J_x$ and $J_y$ cannot both be sharply zero when $J_z$ is maximal, there is always a residual transverse magnitude, and the vector never fully aligns with the axis.

What exactly is the difference between SU(2) and SO(3)?

$\mathrm{SO}(3)$ is the group of ordinary rotations of $3$ -space; $\mathrm{SU}(2)$ is the group of $2\times2$ unitary matrices of determinant one. They share the same Lie algebra — locally identical — but differ globally: $\mathrm{SU}(2)$ is simply connected (topologically a $3$ -sphere), while $\mathrm{SO}(3)$ has a twisted loop that cannot be shrunk. The map $\mathrm{SU}(2)\to\mathrm{SO}(3)$ is two-to-one, sending both $U$ and $-U$ to the same rotation. The consequence: integer- $j$ representations are single-valued on $\mathrm{SO}(3)$ , but half-integer ones are single-valued only on $\mathrm{SU}(2)$ and change sign under a $2\pi$ turn.

Why must the ladder of states terminate?

Because the Hilbert space has a positive-definite inner product. The operator $J_x^2+J_y^2=J^2-J_z^2$ is a sum of squares of Hermitian operators, so its expectation can never be negative, forcing $m^2\le j(j+1)$ and bounding $m$ both ways. If $J_+$ never annihilated a state, repeated application would eventually produce a state with $m^2$ exceeding the bound — a negative-norm state, which cannot exist. So there must be a top rung where $J_+$ gives zero and a bottom rung where $J_-$ gives zero. Working out where they sit is exactly what quantizes $j$ .

What are Clebsch-Gordan coefficients, concretely?

They are the numerical entries that express a combined-spin state $|j,m\rangle$ in the product basis $|j_1,m_1\rangle|j_2,m_2\rangle$ . When you add two spin- $\tfrac12$ particles, the coefficient in $|1,0\rangle=\tfrac{1}{\sqrt2}(|{\uparrow\downarrow}\rangle+|{\downarrow\uparrow}\rangle)$ — the value $\tfrac{1}{\sqrt2}$ — is a Clebsch–Gordan coefficient. They are fixed (up to the Condon–Shortley phase) by requiring $J^2$ and $J_z$ to be diagonal, and are generated by starting from the stretched state $|j_1,j_1\rangle|j_2,j_2\rangle$ and lowering with $J_-$ . Wigner's $3j$ symbols are a more symmetric repackaging.

Why does the character depend only on the rotation angle, not the axis?

Because the character is a trace, and traces are invariant under conjugation: $\mathrm{tr}(URU^{-1})=\mathrm{tr}(R)$ . Two rotations by the same angle about different axes are related by conjugation (rotate the axis to match), so they share a character. This is why $\chi^{(j)}$ is a class function — constant on each conjugacy class, and for $\mathrm{SU}(2)$ the classes are labelled by the single angle $\theta$ . That one-parameter dependence makes character orthogonality a one-dimensional integral against the Weyl measure — the whole reason characters decompose representations so cleanly.

Does the 4π periodicity of spinors have real physical consequences?

Yes, and it has been measured. The sign a spinor picks up under $2\pi$ is unobservable for a single isolated state (an overall phase), but it shows up in interference, where you compare a rotated beam against an unrotated one. Neutron-interferometry experiments in 1975 split a beam, rotated the spin of one path through a controlled magnetic field, and recombined the paths; the interference pattern returned to its original form only after $4\pi$ , not $2\pi$ — a direct, quantitative confirmation that half-integer spin lives on the $\mathrm{SU}(2)$ double cover. The same sign underlies the geometric (Berry) phase and fermion statistics.