Lecture Notes#

Angular Momentum Representation Theory#

Because \([\hat J^2,\hat J_z]=0\), we can choose simultaneous eigenstates and build irreducible multiplets algebraically.

Common Eigenstates of \(\hat J^2\) and \(\hat J_z\).

Simultaneous Eigenstates

Let \(\vert j,m\rangle\) satisfy

(41)#\[\begin{split} \begin{split} \hat J^2\vert j,m\rangle &= \hbar^2 j(j+1)\vert j,m\rangle,\\ \hat J_z\vert j,m\rangle &= \hbar m\vert j,m\rangle. \end{split} \end{split}\]

At this stage, \(j,m\in\mathbb{R}\) are labels; quantization is derived below.

Ladder Operator Action.

Ladder Action on \(\vert j,m\rangle\)

(42)#\[\begin{split} \begin{split} \hat J_+\vert j,m\rangle &= \hbar\sqrt{j(j+1)-m(m+1)}\,\vert j,m+1\rangle,\\ \hat J_-\vert j,m\rangle &= \hbar\sqrt{j(j+1)-m(m-1)}\,\vert j,m-1\rangle. \end{split} \end{split}\]

So \(\hat J_+\) raises \(m\) by one and \(\hat J_-\) lowers \(m\) by one.

Derivation: Ladder-Action Coefficients

From \([\hat J_z,\hat J_\pm]=\pm\hbar\hat J_\pm\) and \([\hat J^2,\hat J_\pm]=0\), the states \(\hat J_\pm\vert j,m\rangle\) lie in the same \(j\) multiplet with \(m\to m\pm1\).

For normalization, use (40):

\[ \|\hat J_+\vert j,m\rangle\|^2 =\langle j,m\vert\hat J_-\hat J_+\vert j,m\rangle =\hbar^2\bigl[j(j+1)-m(m+1)\bigr], \]

\[ \|\hat J_-\vert j,m\rangle\|^2 =\langle j,m\vert\hat J_+\hat J_-\vert j,m\rangle =\hbar^2\bigl[j(j+1)-m(m-1)\bigr]. \]

Taking square roots yields (42).

Angular Momentum Quantization.

Quantization Rules

(43)#\[ j\in\left\{0,\tfrac12,1,\tfrac32,2,\ldots\right\}, \qquad m\in\{-j,-j+1,\ldots,j\}. \]

For each \(j\), the multiplet dimension is \(2j+1\).

Proof: Quantum bootstrap argument

The quantum bootstrap is an algebraic method: constrain quantum numbers by positivity of norms. For any operator \(\hat A\) and state \(\vert\psi\rangle\),

\[ \|\hat A\vert\psi\rangle\|^2\ge 0. \]

Applied to angular momentum, this principle fixes the allowed values of \(j\) and \(m\).

From (41), take simultaneous eigenstates \(\vert j,m\rangle\) of \(\hat J^2\) and \(\hat J_z\).

(1) Positivity constraints. Apply positivity to \(\hat A=\hat J_+\) and \(\hat A=\hat J_-\) on \(\vert j,m\rangle\):

\[ \|\hat J_+\vert j,m\rangle\|^2\ge0, \qquad \|\hat J_-\vert j,m\rangle\|^2\ge0. \]

Using (42), these become

(44)#\[\begin{split} \begin{split} \text{(raising)}\;& j(j+1)-m(m+1)\ge 0,\\ \text{(lowering)}\;& j(j+1)-m(m-1)\ge 0. \end{split} \end{split}\]

So \(m\) is bounded: \(-j\le m\le j\).

(2) Ladder unit-step structure. From \([\hat J_z,\hat J_\pm]=\pm\hbar\hat J_\pm\), each ladder action changes \(m\) by exactly one unit. Therefore all \(m\) values in one multiplet differ by integers.

(3) Finite-dimensional irreducibility. In a finite-dimensional irrep, the ladder must terminate at top and bottom states:

\[ \hat J_+\vert j,m_{\max}\rangle=0, \qquad \hat J_-\vert j,m_{\min}\rangle=0. \]

By (42), vanishing coefficients imply

\[ m_{\max}=j, \qquad m_{\min}=-j. \]

Hence the ladder runs

\[ m=-j,-j+1,\ldots,j, \]

with \(N=2j+1\) states.

Since \(N\in\mathbb N\), we must have \(2j\in\mathbb N_0\). Therefore

\[ j\in\left\{0,\tfrac12,1,\tfrac32,\ldots\right\}, \qquad m\in\{-j,-j+1,\ldots,j\}. \]

So quantization follows from algebra + positivity + finite-dimensional irreducibility.

Integer vs. half-integer: Orbital angular momentum (\(\hat{\boldsymbol{L}}=\hat{\boldsymbol{r}}\times\hat{\boldsymbol{p}}\)) gives only integer \(\ell\) from single-valued wavefunctions, while spin admits half-integer \(j\).

Matrix Representations.

For fixed \(j\), the operators act on a \((2j+1)\)-dimensional space.

Spin-1/2: Pauli Matrices

For \(j=1/2\):

\[\begin{split} \hat J_z=\frac{\hbar}{2} \begin{pmatrix}1&0\\0&-1\end{pmatrix}, \quad \hat J_x=\frac{\hbar}{2} \begin{pmatrix}0&1\\1&0\end{pmatrix}, \quad \hat J_y=\frac{\hbar}{2} \begin{pmatrix}0&-\mathrm{i}\\\mathrm{i}&0\end{pmatrix}. \end{split}\]

Spin-1: 3\(\times\)3 Matrices

For \(j=1\):

\[\begin{split} \hat J_z=\hbar \begin{pmatrix}1&0&0\\0&0&0\\0&0&-1\end{pmatrix}, \quad \hat J_x=\frac{\hbar}{\sqrt2} \begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}, \quad \hat J_y=\frac{\hbar}{\sqrt2} \begin{pmatrix}0&-\mathrm{i}&0\\\mathrm{i}&0&-\mathrm{i}\\0&\mathrm{i}&0\end{pmatrix}. \end{split}\]

Discussion: integer vs half-integer angular momentum

Why do orbital and spin angular momentum components differ in their allowed quantum numbers? Orbital angular momentum must have integer \(\ell\) to ensure single-valued wavefunctions in physical space, while spin can have half-integer \(j\). Yet both satisfy the exact same commutation algebra \([\hat{J}_i, \hat{J}_j] = \mathrm{i}\hbar\epsilon_{ijk}\hat{J}_k\). Does this mean the algebra alone cannot distinguish integer from half-integer representations? What is the fundamental reason—is it topology, boundary conditions, or something deeper about the nature of rotation in quantum mechanics?

Poll: Ladder operator eigenvalue shift

Angular momentum operators satisfy \([\hat{J}_x, \hat{J}_y] = \mathrm{i}\hbar\hat{J}_z\). The raising operator is \(\hat{J}_+ = \hat{J}_x + \mathrm{i}\hat{J}_y\). If \(\hat{J}_z\vert j, m\rangle = \hbar m\vert j, m\rangle\), what is \(\hat{J}_z(\hat{J}_+\vert j, m\rangle)\) proportional to?

(A) \(\hat{J}_+\vert j, m\rangle\) with eigenvalue \(\hbar m\) (unchanged).

(B) \(\hat{J}_+\vert j, m\rangle\) with eigenvalue \(\hbar(m+1)\) (raised by 1).

(C) Zero (raising operators are orthogonal to Z eigenstates).

(D) \((\hat{J}_+)^2\vert j, m\rangle\) (squared).

Homework#

1. Spin-1 verification of the angular-momentum algebra. The lecture states that the commutation relations \([\hat J_i,\hat J_j] = \mathrm{i}\hbar\epsilon_{ijk}\hat J_k\) hold for any representation. The spin-1/2 case follows from the Pauli commutators (1.1.3). Here, verify the relation directly for the spin-1 representation, in the basis \(\{\vert 1,+1\rangle, \vert 1,0\rangle, \vert 1,-1\rangle\}\) where \(\hat J_z = \hbar\,\mathrm{diag}(1, 0, -1)\) and

(a) Compute \([\hat J_x, \hat J_y]\) by matrix multiplication and verify it equals \(\mathrm{i}\hbar\hat J_z\).

(b) Compute \(\hat J^2 = \hat J_x^2 + \hat J_y^2 + \hat J_z^2\) explicitly and confirm \(\hat J^2 = 2\hbar^2\hat I = \hbar^2 \cdot 1\cdot(1+1)\hat I\), i.e. the Casimir eigenvalue \(j(j+1) = 2\) for \(j = 1\).

(c) Compare with the spin-1/2 case from 1.1.3 (where \(\hat J^2 = \frac{3}{4}\hbar^2\hat I\)). State the general pattern: the Casimir eigenvalue is \(j(j+1)\hbar^2\) for every spin-\(j\) multiplet, set by \(\hat J^2\) alone.

2. j=3/2 multiplet by repeated lowering. Start from the stretched state \(\vert 3/2, 3/2\rangle\) and apply the lowering operator \(\hat J_-\) repeatedly to construct all four states of the spin-\(\tfrac{3}{2}\) multiplet.

(a) Verify that \(\hat J_+\vert 3/2, 3/2\rangle = 0\) using the ladder formula \(\hat J_+\vert j,m\rangle = \hbar\sqrt{(j-m)(j+m+1)}\vert j, m+1\rangle\).

(b) Compute \(\hat J_-\vert 3/2, m\rangle\) for \(m = 3/2, 1/2, -1/2, -3/2\), using \(\hat J_-\vert j,m\rangle = \hbar\sqrt{(j+m)(j-m+1)}\vert j, m-1\rangle\). State the coefficient in each step.

(c) Verify that \(\hat J_-\vert 3/2, -3/2\rangle = 0\) — the multiplet terminates at the bottom rung.

(d) The four states \(\{\vert 3/2, 3/2\rangle, \vert 3/2, 1/2\rangle, \vert 3/2, -1/2\rangle, \vert 3/2, -3/2\rangle\}\) form a \(4\)-dimensional representation. Argue from this construction that the dimension of a spin-\(j\) multiplet is \(2j+1\).

3. Ladder formula via Schwinger bosons. Recall the Schwinger boson construction from 2.1.3 Problem 10: spin operators \(\hat S_+ = \hat a^\dagger\hat b\), \(\hat S_- = \hat b^\dagger\hat a\), \(\hat S_z = \tfrac{1}{2}(\hat a^\dagger\hat a - \hat b^\dagger\hat b)\), with the Fock state \(\vert n_a, n_b\rangle\) identified with \(\vert s, m\rangle\) via \(s = (n_a + n_b)/2\) and \(m = (n_a - n_b)/2\). Use the bosonic algebra to derive the angular-momentum ladder formula.

(a) Express \(n_a\) and \(n_b\) in terms of \(s\) and \(m\).

(b) Apply \(\hat S_+ = \hat a^\dagger\hat b\) to \(\vert n_a, n_b\rangle\) and read off the coefficient using \(\hat a^\dagger\vert n_a\rangle = \sqrt{n_a+1}\vert n_a+1\rangle\) and \(\hat b\vert n_b\rangle = \sqrt{n_b}\vert n_b-1\rangle\). Show that

reproducing the lecture’s ladder formula (in units \(\hbar = 1\)).

(c) Argue that this derivation makes the ladder coefficient \(\sqrt{(s-m)(s+m+1)}\) automatic — it emerges directly from the bosonic normalisation factors \(\sqrt{n}\), with no separate calculation needed. Contrast with the standard approach (lecture), which derives the same coefficient from \(\hat J_+\hat J_- = \hat J^2 - \hat J_z^2 + \hbar\hat J_z\).

4. Ladder action: termination and verification. Use \(\hat J_+\vert j, m\rangle = \hbar\sqrt{(j-m)(j+m+1)}\,\vert j, m+1\rangle\) to explain why \(\hat J_+\vert 1,1\rangle = 0\). Then compute \(\hat J_-\vert 1, 0\rangle\) from the lowering formula and verify the result using the spin-1 matrix representation from Problem 1.

5. Transverse variance on an angular-momentum eigenstate. Compute \(\langle\hat J_x\rangle\), \(\langle\hat J_x^2\rangle\), and the variance \((\Delta\hat J_x)^2\) on a state \(\vert j, m\rangle\).

(a) Express \(\hat J_x = \tfrac{1}{2}(\hat J_+ + \hat J_-)\) and use it to compute \(\langle\hat J_x\rangle = \langle j, m\vert\hat J_x\vert j, m\rangle\). Explain in one sentence why the result is zero.

(b) Use the lecture’s ladder product identity \(\hat J_+\hat J_- + \hat J_-\hat J_+ = 2(\hat J^2 - \hat J_z^2)\) to compute \(\hat J_x^2 + \hat J_y^2 = \tfrac{1}{2}(\hat J_+\hat J_- + \hat J_-\hat J_+) = \hat J^2 - \hat J_z^2\). Conclude

(c) By the symmetry of the algebra under \(x \leftrightarrow y\) on a \(\hat J_z\)-eigenstate, \(\langle\hat J_x^2\rangle = \langle\hat J_y^2\rangle\). Conclude

(d) Identify the limiting values: at the stretched state \(m = \pm j\), \((\Delta\hat J_x)^2 = \hbar^2 j/2\) — not zero, even though the state is a \(\hat J_z\) eigenstate. Explain physically why even the most “extreme” \(\hat J_z\) eigenstate has some transverse uncertainty — the quantum AM vector cannot be perfectly aligned with \(\boldsymbol{e}_z\).

6. Robertson uncertainty for angular momentum. Apply the Robertson uncertainty relation from 1.2.2 to the pair \((\hat J_x, \hat J_y)\) on a state \(\vert j, m\rangle\).

(a) Using \([\hat J_x, \hat J_y] = \mathrm{i}\hbar\hat J_z\), write down the Robertson bound \(\Delta\hat J_x\cdot\Delta\hat J_y \geq \tfrac{1}{2}\vert\langle[\hat J_x, \hat J_y]\rangle\vert\).

(b) Evaluate both sides on the stretched state \(\vert j, j\rangle\). Use the variance from Problem 5 and the fact that \(\langle\hat J_z\rangle_{\vert j, j\rangle} = \hbar j\).

(c) Show that the Robertson inequality is saturated on the stretched state. Why is this the angular-momentum analogue of the minimum-uncertainty state \(\vert 0\rangle\) for \((\hat X, \hat Z)\) from 1.2.2 Problem 6?

(d) Compute the ratio \(\Delta\hat J_x/\langle\hat J_z\rangle\) on the stretched state and show that in the classical limit \(j \to \infty\), this ratio vanishes as \(1/\sqrt j\). The angular momentum becomes a sharp 3-vector in the classical limit, recovering the macroscopic notion of AM.

7. Vector model and the tilt angle. The state \(\vert j, m\rangle\) has \(\langle\hat J^2\rangle = \hbar^2 j(j+1)\), \(\langle\hat J_z\rangle = \hbar m\), and \(\langle\hat J_x\rangle = \langle\hat J_y\rangle = 0\). The vector model pictures the angular momentum as a classical 3-vector of length \(\vert\boldsymbol J\vert = \hbar\sqrt{j(j+1)}\) with \(z\)-component \(J_z = \hbar m\).

(a) Show that the angle \(\theta\) between \(\boldsymbol J\) and the \(z\)-axis satisfies

(b) For the stretched state \(\vert j, j\rangle\), find the minimum tilt angle \(\theta_{\min}\) in terms of \(j\), and evaluate it for \(j = 1/2\), \(j = 1\), and \(j = 100\).

(c) The quantum AM cannot be exactly aligned with \(\boldsymbol{e}_z\): even the stretched state has \(\theta_{\min} > 0\). Use Problem 5 to compute the transverse magnitude \(\sqrt{\langle\hat J_x^2 + \hat J_y^2\rangle} = \hbar\sqrt{j(j+1) - m^2}\) and identify it with the radius of a precession circle in the vector model.

(d) Compare with the classical prediction: for a classical AM vector of length \(\vert\boldsymbol J\vert\) tilted at angle \(\theta\), the transverse projection is \(\vert\boldsymbol J\vert\sin\theta\). Verify that the quantum result \(\hbar\sqrt{j(j+1) - m^2}\) equals \(\vert\boldsymbol J\vert\sin\theta\) using the value of \(\cos\theta\) from (a).

★ 8. Quantum bootstrap. Two operators \(\hat{\alpha}\) and \(\hat{\beta}\) satisfy the algebraic relations

where \([\hat{A}, \hat{B}] = \hat{A}\hat{B} - \hat{B}\hat{A}\) is the commutator, \(\{\hat{A}, \hat{B}\} = \hat{A}\hat{B} + \hat{B}\hat{A}\) is the anticommutator, and \(\hat{I}\) is the identity operator.

(a) Show that \(\hat{\beta}\) is Hermitian.

(b) Express \(\hat{\alpha}^\dagger\hat{\alpha}\) and \(\hat{\alpha}\hat{\alpha}^\dagger\) separately as linear combinations of \(\hat{\beta}\) and \(\hat{I}\).

(c) Show that \((\hat{\alpha}^\dagger)^2 \hat{\alpha}^2 = \hat{\beta}^2 - \hat{I}\) by expressing the left-hand side in terms of \(\hat{\beta}\). Using the positivity conditions \(\hat{\alpha}^\dagger\hat{\alpha} \geq 0\), \(\hat{\alpha}\hat{\alpha}^\dagger \geq 0\), and \((\hat{\alpha}^\dagger)^2\hat{\alpha}^2 \geq 0\), determine the feasible eigenvalues of \(\hat{\beta}\).

(d) Identify the angular momentum operators that \(\hat{\alpha}\) and \(\hat{\beta}\) correspond to (with \(\hbar = 1\)). What spin representation does this algebra describe? Write the \(2\times 2\) matrix representations of \(\hat{\alpha}\) and \(\hat{\beta}\) in the eigenbasis of \(\hat{\beta}\).