2.2.2 Spin Representations#

Prompts

What distinguishes spin from orbital angular momentum? Why can spin be half-integer?
Derive the spinor rotation formula \(\hat{R}(\boldsymbol{n}, \theta) = \cos(\theta/2)\,I - \mathrm{i}\sin(\theta/2)\,(\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}})\). Show that a \(2\pi\) rotation gives \(-I\) for spin-1/2.
Write the \(3\times 3\) matrix representations for spin-1 operators and verify they satisfy the angular momentum commutation relations.
How does the Wigner D-matrix generalize spinor rotations to arbitrary spin-\(j\)?

Lecture Notes#

Overview#

Spin is intrinsic angular momentum — a purely quantum degree of freedom with no classical analog. Unlike orbital angular momentum \(\hat{\boldsymbol{L}} = \hat{\boldsymbol{r}} \times \hat{\boldsymbol{p}}\), spin does not arise from motion through space. Yet it obeys the same commutation relations and quantization rules derived in §2.2.1. This section builds the explicit matrix representations for spin-1/2, spin-1, and general spin-\(j\), and introduces the striking fact that spinors transform differently from classical vectors under rotations.

Spin as Intrinsic Angular Momentum#

Definition: Spin

Spin is an intrinsic angular momentum carried by particles. It satisfies \([\hat{S}_i, \hat{S}_j] = \mathrm{i}\hbar\epsilon_{ijk}\hat{S}_k\) but is not associated with spatial motion (\(\hat{\boldsymbol{S}} \neq \hat{\boldsymbol{r}} \times \hat{\boldsymbol{p}}\)). The spin quantum number \(s\) is fixed for each particle species.

The spin-statistics theorem (proved in relativistic QFT) connects spin to particle statistics:

Spin	Statistics	Examples
Integer (\(s = 0, 1, 2, \ldots\))	Bosons	Photons (\(s{=}1\)), pions (\(s{=}0\)), gravitons (\(s{=}2\))
Half-integer (\(s = \tfrac{1}{2}, \tfrac{3}{2}, \ldots\))	Fermions	Electrons (\(s{=}\tfrac{1}{2}\)), quarks (\(s{=}\tfrac{1}{2}\)), \(\Omega^-\) (\(s{=}\tfrac{3}{2}\))

Discussion

The spin-statistics connection — integer spin means bosons, half-integer means fermions — has no simple non-relativistic proof. Why should the rotation properties of a single particle dictate the behavior of many identical particles? What would go wrong physically if spin-1/2 particles obeyed Bose statistics?

Spin-1/2 in Detail#

The simplest nontrivial case. Basis states and Pauli matrices were introduced in §1.1.3; here we view them as the \(j = 1/2\) angular momentum representation.

Spin-1/2 Basis and Operators

Basis states: \(\vert\uparrow\rangle = \vert\tfrac{1}{2}, +\tfrac{1}{2}\rangle\) and \(\vert\downarrow\rangle = \vert\tfrac{1}{2}, -\tfrac{1}{2}\rangle\). The spin operators are \(\hat{S}_i = \frac{\hbar}{2}\hat{\sigma}^i\):

\[\begin{split} \hat{\sigma}^x = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \hat{\sigma}^y = \begin{pmatrix} 0 & -\mathrm{i} \\ \mathrm{i} & 0 \end{pmatrix}, \quad \hat{\sigma}^z = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \end{split}\]

Key properties: \((\hat{\sigma}^i)^2 = I\), \(\hat{\sigma}^i \hat{\sigma}^j = \delta_{ij} I + \mathrm{i}\epsilon_{ijk}\hat{\sigma}^k\), eigenvalues \(\pm 1\).

Derivation: Spin-1/2 Matrices from Ladder Operators

For \(j=\tfrac12\), use basis

\[ \vert\uparrow\rangle=\left\vert\tfrac12,+\tfrac12\right\rangle, \qquad \vert\downarrow\rangle=\left\vert\tfrac12,-\tfrac12\right\rangle. \]

From ladder action,

\[\begin{split} \begin{split} \hat J_+\vert\downarrow\rangle&=\hbar\vert\uparrow\rangle, \qquad \hat J_+\vert\uparrow\rangle=0,\\ \hat J_-\vert\uparrow\rangle&=\hbar\vert\downarrow\rangle, \qquad \hat J_-\vert\downarrow\rangle=0. \end{split} \end{split}\]

So in ordered basis \(\{\vert\uparrow\rangle,\vert\downarrow\rangle\}\),

\[\begin{split} \hat J_+=\hbar\begin{pmatrix}0&1\\0&0\end{pmatrix}, \qquad \hat J_-=\hbar\begin{pmatrix}0&0\\1&0\end{pmatrix}. \end{split}\]

Then

\[\begin{split} \hat J_x=\frac12(\hat J_++\hat J_-)=\frac{\hbar}{2}\begin{pmatrix}0&1\\1&0\end{pmatrix} =\frac{\hbar}{2}\sigma^x, \end{split}\]

\[\begin{split} \hat J_y=\frac{1}{2\mathrm{i}}(\hat J_+-\hat J_-)=\frac{\hbar}{2}\begin{pmatrix}0&-\mathrm{i}\\\mathrm{i}&0\end{pmatrix} =\frac{\hbar}{2}\sigma^y. \end{split}\]

Also, from \(\hat J_z\vert\uparrow\rangle=\frac{\hbar}{2}\vert\uparrow\rangle\) and \(\hat J_z\vert\downarrow\rangle=-\frac{\hbar}{2}\vert\downarrow\rangle\),

\[\begin{split} \hat J_z=\frac{\hbar}{2}\begin{pmatrix}1&0\\0&-1\end{pmatrix} =\frac{\hbar}{2}\sigma^z. \end{split}\]

Hence \(\hat J_i=\frac{\hbar}{2}\sigma^i\), i.e. the \(j=\tfrac12\) ladder representation is exactly the Pauli-matrix representation. \(\checkmark\)

Spinor Rotations#

A rotation by angle \(\theta\) about axis \(\boldsymbol{n}\) acts on a spin-1/2 state via:

Spinor Rotation Formula

(47)#\[ \hat{R}(\boldsymbol{n}, \theta) = \mathrm{e}^{-\mathrm{i}\theta\,\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}/2} = \cos\frac{\theta}{2}\,I - \mathrm{i}\sin\frac{\theta}{2}\,(\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}) \]

The \(2\pi\) Mystery

A \(2\pi\) rotation gives \(\hat{R}(\boldsymbol{n}, 2\pi) = -I\): every spinor picks up a factor of \(-1\). A full \(4\pi\) rotation is needed to return to the original state. This is the hallmark of spin-1/2 — spinors are not vectors.

Physically: the overall sign is not observable for a single particle, but it becomes observable in interference experiments and is essential for the fermion sign in many-body wavefunctions.

Derivation: \(2\pi\) Rotation of a Spinor

Set \(\theta = 2\pi\) in the rotation formula:

\[ \hat{R}(\boldsymbol{n}, 2\pi) = \cos\pi\,I - \mathrm{i}\sin\pi\,(\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}) = -I \]

For \(4\pi\): \(\hat{R}(\boldsymbol{n}, 4\pi) = \cos 2\pi\,I - \mathrm{i}\sin 2\pi\,(\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}) = +I\). \(\checkmark\)

Mathematically, SU(2) is the double cover of SO(3): two elements of SU(2) (\(\pm I\)) map to the same rotation (identity) in SO(3).

Spin-1#

For \(j = 1\), three basis states: \(\vert 1, +1\rangle\), \(\vert 1, 0\rangle\), \(\vert 1, -1\rangle\).

Spin-1 Operators

\[\begin{split} \hat{S}_z = \hbar\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}, \quad \hat{S}_x = \frac{\hbar}{\sqrt{2}}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}, \quad \hat{S}_y = \frac{\hbar}{\sqrt{2}}\begin{pmatrix} 0 & -\mathrm{i} & 0 \\ \mathrm{i} & 0 & -\mathrm{i} \\ 0 & \mathrm{i} & 0 \end{pmatrix} \end{split}\]

Ladder operators:

\[\begin{split} \hat{S}_+ = \hbar\begin{pmatrix} 0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2} \\ 0 & 0 & 0 \end{pmatrix}, \quad \hat{S}_- = \hbar\begin{pmatrix} 0 & 0 & 0 \\ \sqrt{2} & 0 & 0 \\ 0 & \sqrt{2} & 0 \end{pmatrix} \end{split}\]

Under a \(2\pi\) rotation, \(\hat{R}(\boldsymbol{n}, 2\pi) = +I\) for spin-1 (integer spin states are true vectors). The rotation matrix is \(\hat{D}^{(1)}(\boldsymbol{n}, \theta) = \mathrm{e}^{-\mathrm{i}\theta\,\boldsymbol{n}\cdot\hat{\boldsymbol{S}}/\hbar}\), a \(3\times 3\) unitary matrix that reduces to the familiar SO(3) rotation matrices.

Derivation: Spin-1 Matrices from Ladder Operators

From the general formula \(\hat{J}_+ \vert j, m\rangle = \hbar\sqrt{(j-m)(j+m+1)}\vert j, m+1\rangle\) with \(j = 1\):

\[ \hat{S}_+ \vert 1, -1\rangle = \hbar\sqrt{2}\,\vert 1, 0\rangle, \quad \hat{S}_+ \vert 1, 0\rangle = \hbar\sqrt{2}\,\vert 1, 1\rangle, \quad \hat{S}_+ \vert 1, 1\rangle = 0 \]

Reading off the matrix elements in the basis \(\{\vert 1, 1\rangle, \vert 1, 0\rangle, \vert 1, -1\rangle\}\) gives the \(3\times 3\) matrices above. Then \(\hat{S}_x = \frac{1}{2}(\hat{S}_+ + \hat{S}_-)\) and \(\hat{S}_y = \frac{1}{2\mathrm{i}}(\hat{S}_+ - \hat{S}_-)\). \(\checkmark\)

General Spin-\(j\)#

Irreducible Representations

For spin quantum number \(j\), the \((2j+1)\) basis states \(\{\vert j, m\rangle : m = -j, \ldots, j\}\) form an irreducible representation of SU(2). The rotation operator is:

(48)#\[ \hat{D}^{(j)}(\boldsymbol{n}, \theta) = \mathrm{e}^{-\mathrm{i}\theta\,\boldsymbol{n}\cdot\hat{\boldsymbol{J}}/\hbar} \]

Its matrix elements \(D^{(j)}_{m'm}(\boldsymbol{n}, \theta) = \langle j, m' \vert \hat{D}^{(j)} \vert j, m\rangle\) are the Wigner D-matrices.

The Wigner D-matrices are the complete set of matrix representations for rotations. For \(j = 1/2\), they reduce to the \(2\times 2\) spinor rotation matrices; for \(j = 1\), to the \(3\times 3\) SO(3) rotation matrices. No closed-form derivation is needed — they are tabulated and computed numerically.

Note: \(2\pi\) Rotation for Integer vs Half-Integer Spin

For a full \(2\pi\) rotation, integer-spin and half-integer-spin representations differ by a sign.

For spin-\(\tfrac12\),

\[ \hat{R}_{1/2}(\boldsymbol{n},2\pi)=-\hat{I}. \]

For spin-1,

\[ \hat{R}_{1}(\boldsymbol{n},2\pi)=+\hat{I}. \]

So fermionic spinors acquire a minus sign under self-rotation, while bosonic vectors return with no phase.

Mathematically, this is the statement that SU(2) double-covers SO(3): two elements \(U\) and \(-U\) in SU(2) correspond to the same physical rotation in SO(3).

Physically, this matches the spin-statistics sign structure: bosons carry exchange sign \(+1\), fermions carry exchange sign \(-1\).

Summary#

Spin is intrinsic angular momentum with no classical analog; the spin quantum number \(s\) is fixed per particle species.
Spin-statistics theorem: integer spin \(\to\) bosons; half-integer spin \(\to\) fermions.
Spin-1/2: Pauli matrices, spinor rotations \(\hat{R} = \mathrm{e}^{-\mathrm{i}\theta\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}/2}\), and the \(2\pi\) sign flip distinguish spinors from vectors.
Spin-1: \(3\times 3\) matrices; \(2\pi\) rotation returns to identity (true vectors).
General spin-\(j\): \((2j+1)\)-dimensional irreducible representation; Wigner D-matrices describe rotations.
Spinor wavefunctions combine spatial and spin degrees of freedom: \(\Psi(\boldsymbol{r})\) is a \((2s+1)\)-component field.

Homework#

1. Show that the Pauli matrices satisfy \((\hat{\sigma}^i)^2 = I\), \(\hat{\sigma}^i\hat{\sigma}^j = \delta_{ij}I + \mathrm{i}\epsilon_{ijk}\hat{\sigma}^k\), and \(\mathrm{Tr}(\hat{\sigma}^i) = 0\).

2. Find the eigenvalues and eigenstates of \(\hat{S}_x = \frac{\hbar}{2}\hat{\sigma}^x\). Express the eigenstates \(\vert+\rangle\) and \(\vert-\rangle\) as linear combinations of \(\vert\uparrow\rangle\) and \(\vert\downarrow\rangle\).

3. Compute the rotation operator \(\hat{R}(\hat{\boldsymbol{x}}, \pi/2) = \mathrm{e}^{-\mathrm{i}\pi\hat{\sigma}^x/4}\) and apply it to \(\vert\uparrow\rangle\). Verify the result is an equal superposition of spin-up and spin-down along \(z\).

4. Show that \(\hat{R}(\boldsymbol{n}, 2\pi) = -I\) for spin-1/2, and \(\hat{R}(\boldsymbol{n}, 4\pi) = +I\). Explain why the \(-1\) phase is unobservable for a single particle but becomes physical in interference experiments.

5. For spin-1, verify \([\hat{S}_x, \hat{S}_y] = \mathrm{i}\hbar\hat{S}_z\) using the \(3\times 3\) matrices. Compute \(\hat{S}_+\vert 1, -1\rangle\) and confirm \(\hat{S}_+\vert 1, -1\rangle = \hbar\sqrt{2}\,\vert 1, 0\rangle\).

6. For general spin-\(j\), explain why the representation is \((2j+1)\)-dimensional and irreducible. What does “irreducible” mean physically?