2.2.2 Spin Representations#

Prompts

What distinguishes spin from orbital angular momentum? Why can spin be half-integer?
What does the spinor rotation formula \(\hat{D}^{(1/2)}(\boldsymbol{n}, \theta) = \cos(\theta/2)\,I - \mathrm{i}\sin(\theta/2)\,(\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}})\) tell us about spin-1/2? Why does a \(2\pi\) rotation give \(-I\)?
How are the \(3\times 3\) matrix representations for spin-1 operators constructed? Why must they satisfy the angular momentum commutation relations?
How does the Wigner D-matrix generalize spinor rotations to arbitrary spin-\(j\)? What patterns emerge as \(j\) increases?

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

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

(45)#\[ \hat{D}^{(1/2)}(\boldsymbol{n}, \theta) = \mathrm{e}^{-\mathrm{i}\theta\,\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}/2} = \cos\frac{\theta}{2}\,I - \mathrm{i}\sin\frac{\theta}{2}\,(\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}) \]

The \(2\pi\) Mystery

A \(2\pi\) rotation gives \(\hat{D}^{(1/2)}(\boldsymbol{n}, 2\pi) = -I\): 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:

\[ \hat{D}^{(1/2)}(\boldsymbol{n}, 2\pi) = \cos\pi\,I - \mathrm{i}\sin\pi\,(\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}) = -I \]

For \(4\pi\): \(\hat{D}^{(1/2)}(\boldsymbol{n}, 4\pi) = \cos 2\pi\,I - \mathrm{i}\sin 2\pi\,(\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}) = +I\). \(\checkmark\)

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

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

Under a \(2\pi\) rotation, \(\hat{D}^{(1)}(\boldsymbol{n}, 2\pi) = +I\) 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.

Note: Basis Dependence of Spin-1 Matrices

The explicit \(3\times 3\) matrices depend on basis choice. In the standard \(\{\vert 1,+1\rangle, \vert 1,0\rangle, \vert 1,-1\rangle\}\) basis we wrote \(\hat{S}_a\) above; in the Cartesian basis \(\{\vert x\rangle,\vert y\rangle,\vert z\rangle\}\) they take a different matrix form.

Define

\[\begin{split} \begin{split} \begin{pmatrix} \vert 1,+1\rangle \\ \vert 1,0\rangle \\ \vert 1,-1\rangle \end{pmatrix} &= \hat{U} \begin{pmatrix} \vert x\rangle \\ \vert y\rangle \\ \vert z\rangle \end{pmatrix},\\ \hat{U} &= \begin{pmatrix} \tfrac{\mathrm{i}}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} & 0 \\ 0 & 0 & -\mathrm{i} \\ -\tfrac{\mathrm{i}}{\sqrt{2}} & \tfrac{1}{\sqrt{2}} & 0 \end{pmatrix}. \end{split} \end{split}\]

Using spin operators directly, the transformed matrices are

\[ \hat{S}_a^{(xyz)} = \hat{U}^\dagger \hat{S}_a \hat{U}. \]

To distinguish this Cartesian-basis form from the earlier basis, define the dimensionless generators

\[ \hat{L}_a \equiv \frac{\hat{S}_a^{(xyz)}}{\hbar}. \]

Then

\[\begin{split} \hat{L}_1 = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & -\mathrm{i} \\ 0 & \mathrm{i} & 0 \end{pmatrix},\quad \hat{L}_2 = \begin{pmatrix} 0 & 0 & \mathrm{i} \\ 0 & 0 & 0 \\ -\mathrm{i} & 0 & 0 \end{pmatrix},\quad \hat{L}_3 = \begin{pmatrix} 0 & -\mathrm{i} & 0 \\ \mathrm{i} & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}. \end{split}\]

So a rotation about the \(z\)-axis is

\[\begin{split} \hat{D}^{(1)}_z(\theta) = \mathrm{e}^{-\mathrm{i}\theta\hat{L}_3} = \begin{pmatrix} \cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1 \end{pmatrix}. \end{split}\]

Same representation, different basis: matrices change by similarity transform, physics does not.

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:

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

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.

Poll: Dimension of spin-1 representation

A spin-1 system has three eigenstates \(\vert 1\rangle, \vert 0\rangle, \vert -1\rangle\) of \(\hat{J}_z\). How large are the matrices representing \(\hat{J}_x, \hat{J}_y, \hat{J}_z\)?

(A) \(2 \times 2\) (like spin-1/2).

(B) \(3 \times 3\) (one row/column for each eigenstate).

(D) \(6 \times 6\) (one matrix per component per dimension).

Summary#

Spin is intrinsic angular momentum with fixed quantum number \(s\) per species; spin-statistics theorem connects integer spin to bosons, half-integer to fermions.
Spin-1/2: Pauli matrices and spinor rotations \(\hat{D}^{(1/2)} = \mathrm{e}^{-\mathrm{i}\theta\boldsymbol{n}\cdot\hat{\boldsymbol{\sigma}}/2}\) with \(2\pi\) sign flip (spinors \(\neq\) vectors).
Spin-1: \(3\times 3\) matrices; \(2\pi\) rotation returns to identity. General spin-\(j\): \((2j+1)\)-dimensional irreducible rep; Wigner D-matrices describe all rotations.
Spinor wavefunctions combine spatial and spin: \(\Psi(\boldsymbol{r})\) is a \((2s+1)\)-component field.

Homework#

1. Stern-Gerlach for spin-1. A beam of spin-1 atoms (say, \(\Omega^-\) baryons or any composite system with integer spin) passes through a Stern-Gerlach apparatus oriented along the \(z\)-axis. The magnetic moment couples to the spin via \(\hat H_{\mathrm{int}} = -\gamma\hat{\boldsymbol S}\cdot\boldsymbol B\), and the field gradient produces a force proportional to \(\hat S_z\).

(a) State the possible eigenvalues of \(\hat S_z\) for a spin-1 system, and identify how many distinct branches the beam splits into.

(b) For spin-1/2 the beam splits into two branches; for spin-1 it splits into three. Argue from the lecture’s general rule \(\dim = 2j+1\) that the number of branches in any Stern-Gerlach measurement directly reads off the value of \(j\).

(c) An initial spin-1 state \(\vert\psi(0)\rangle = a\vert 1,+1\rangle + b\vert 1,0\rangle + c\vert 1,-1\rangle\) (with \(\vert a\vert^2 + \vert b\vert^2 + \vert c\vert^2 = 1\)) enters the apparatus. State the probability that the atom emerges in each of the three branches.

(d) The beam-splitting can be inverted: by counting the three branches and measuring their relative intensities, one measures \(j\) (the existence of three rather than two branches) and the populations \(\vert a\vert^2, \vert b\vert^2, \vert c\vert^2\). Argue that a single-shot Stern-Gerlach experiment is therefore a complete \(\hat J_z\)-basis measurement of a spin-\(j\) atom — the natural generalisation of the qubit projective measurement.

2. Spin-1 rotation operator via polynomial truncation. For spin-1/2, the identity \((\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})^2 = \hat I\) truncates the Taylor series of \(\hat D^{(1/2)}(\boldsymbol{n}, \theta) = \mathrm{e}^{-\mathrm{i}\theta\boldsymbol{n}\cdot\hat{\boldsymbol\sigma}/2}\) into the spinor formula \(\cos(\theta/2)\hat I - \mathrm{i}\sin(\theta/2)(\boldsymbol{n}\cdot\hat{\boldsymbol\sigma})\).

For spin-1 the analogous truncation requires one more power. Take \(\boldsymbol{n} = \boldsymbol{e}_x\), where \(\boldsymbol{e}_a\) denotes the unit vector along the \(a\)-axis.

(a) Using the spin-1 matrix \(\hat J_x = \tfrac{\hbar}{\sqrt 2}\begin{pmatrix}0&1&0\\ 1&0&1\\ 0&1&0\end{pmatrix}\) from 2.2.1 P1, verify by explicit matrix multiplication that \(\hat J_x^3 = \hbar^2\hat J_x\) (in particular, the cube reduces back to the first power, not to a higher polynomial). Conclude that the eigenvalues of \(\hat J_x\) are \(\hbar\cdot m\) with \(m \in \{+1, 0, -1\}\) — the cube relation \(\lambda^3 = \lambda\) holds because \(\lambda \in \{+1, 0, -1\}\) solves it.

(b) Use \(\hat J_x^3 = \hbar^2\hat J_x\) to truncate the Taylor series \(\hat D^{(1)}(\boldsymbol{e}_x, \theta) = \mathrm{e}^{-\mathrm{i}\theta\hat J_x/\hbar}\). Show

\[ \hat D^{(1)}(\boldsymbol{e}_x, \theta) = \hat I - \mathrm{i}\sin\theta\,\frac{\hat J_x}{\hbar} + (\cos\theta - 1)\,\frac{\hat J_x^2}{\hbar^2}. \]

(c) Evaluate at \(\theta = 2\pi\) to show \(\hat D^{(1)}(\boldsymbol{e}_x, 2\pi) = +\hat I\). Contrast with the spin-1/2 result \(\hat D^{(1/2)}(\boldsymbol{e}_x, 2\pi) = -\hat I\) (Problem 4 below). What property of \(j\) produces the sign difference?

3. Spinor rotation by ninety degrees about x. Compute the rotation operator \(\hat{D}^{(1/2)}(\boldsymbol{e}_x, \pi/2) = \mathrm{e}^{-\mathrm{i}\pi\hat{\sigma}^x/4}\) and apply it to \(\vert\uparrow\rangle\). Verify the result is an equal superposition of spin-up and spin-down along \(z\), and identify the resulting Bloch direction.

4. Spinor 2-pi rotation and neutron interferometry. Show that \(\hat{D}^{(1/2)}(\boldsymbol{n}, 2\pi) = -\hat{I}\) for spin-1/2, and \(\hat{D}^{(1/2)}(\boldsymbol{n}, 4\pi) = +\hat{I}\). Explain why the \(-1\) phase is unobservable for a single particle but becomes physical in interferometric experiments — describe how a neutron-interferometry setup detects it.

5. Spin-j rotation: integer vs half-integer. Generalise the \(2\pi\) rotation comparison to arbitrary spin.

(a) Compute \(\hat D^{(j)}(\boldsymbol{e}_z, 2\pi) = \mathrm{e}^{-2\pi\mathrm{i}\hat J_z/\hbar}\) in the \(\vert j, m\rangle\) basis. Show it is diagonal with entries \(\mathrm{e}^{-2\pi\mathrm{i}m}\).

(b) Evaluate \(\mathrm{e}^{-2\pi\mathrm{i}m}\) for integer \(m\) versus half-integer \(m\). Conclude that for integer \(j\), \(\hat D^{(j)}(\boldsymbol{e}_z, 2\pi) = +\hat I\), while for half-integer \(j\), \(\hat D^{(j)}(\boldsymbol{e}_z, 2\pi) = -\hat I\).

(c) By the rotation-axis-independence of the \(2\pi\) behaviour (Problem 4 for spin-1/2 and Problem 2 for spin-1), state the general formula

\[ \hat D^{(j)}(\boldsymbol n, 2\pi) = (-1)^{2j}\,\hat I. \]

(d) Connect to the spin-statistics theorem quoted in the lecture: integer spin ↔ bosons; half-integer spin ↔ fermions. Argue that the \((-1)^{2j}\) factor under a \(2\pi\) rotation is the single-particle signature of the multi-particle exchange statistics — the same minus sign that flips a fermion wavefunction under particle swap.

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

★ 7. Spin-1 time evolution. A spin-1 particle is prepared in \(\vert 1, +1\rangle\) and evolves under \(\hat{H} = \omega\hat{J}_x^2\) (set \(\hbar = 1\)).

(a) Write \(\hat{J}_x^2\) as a \(3\times 3\) matrix in the \(\{\vert 1,+1\rangle, \vert 1,0\rangle, \vert 1,-1\rangle\}\) basis. Show that \(\vert 1,0\rangle\) is an eigenstate, and find the two eigenvalues and eigenstates of \(\hat{J}_x^2\) in the \(\{\vert 1,+1\rangle, \vert 1,-1\rangle\}\) subspace.

(b) Decompose \(\vert 1,+1\rangle\) in the eigenbasis of \(\hat{J}_x^2\) and find \(\vert\psi(t)\rangle\). Show that \(P_{+1}(t) = \cos^2(\omega t/2)\), \(P_{-1}(t) = \sin^2(\omega t/2)\), and \(P_0 = 0\) at all times.

(c) At \(t = \pi/(2\omega)\) the state has \(P_{+1} = P_{-1} = 1/2\). One measures \(\hat{J}_z^2\) and obtains the outcome \(1\) (i.e. \(m^2 = 1\)). What is the post-measurement state? Compute \(\langle\hat{J}_z\rangle\).