2.2.1 Angular Momentum Algebra#

Worked solutions for the homework problems in the 2.2.1 Angular Momentum Algebra lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

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

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

(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.

Solution.

(a) Compute the two products. Let \(a \equiv \hbar/\sqrt 2\). Then

\[\begin{split} \hat J_x\hat J_y = a^2 \begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0\end{pmatrix}\begin{pmatrix} 0&-\mathrm{i}&0\\ \mathrm{i}&0&-\mathrm{i}\\ 0&\mathrm{i}&0\end{pmatrix} = a^2\begin{pmatrix} \mathrm{i} & 0 & -\mathrm{i} \\ 0 & 0 & 0 \\ \mathrm{i} & 0 & -\mathrm{i}\end{pmatrix}, \end{split}\]

\[\begin{split} \hat J_y\hat J_x = a^2 \begin{pmatrix} 0&-\mathrm{i}&0\\ \mathrm{i}&0&-\mathrm{i}\\ 0&\mathrm{i}&0\end{pmatrix}\begin{pmatrix} 0&1&0\\ 1&0&1\\ 0&1&0\end{pmatrix} = a^2\begin{pmatrix} -\mathrm{i} & 0 & -\mathrm{i} \\ 0 & 0 & 0 \\ \mathrm{i} & 0 & \mathrm{i}\end{pmatrix}. \end{split}\]

Subtract:

\[\begin{split} [\hat J_x,\hat J_y] = a^2 \begin{pmatrix} 2\mathrm{i} & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -2\mathrm{i}\end{pmatrix} = 2\mathrm{i}a^2\,\mathrm{diag}(1, 0, -1) = \mathrm{i}\hbar^2\,\mathrm{diag}(1, 0, -1) = \mathrm{i}\hbar\,\hat J_z. \quad\checkmark \end{split}\]

(using \(2a^2 = 2\cdot\hbar^2/2 = \hbar^2\)). The spin-1 matrices reproduce the universal algebra.

(b) Compute \(\hat J_x^2\) and \(\hat J_y^2\):

\[\begin{split} \hat J_x^2 = a^2\begin{pmatrix} 1 & 0 & 1 \\ 0 & 2 & 0 \\ 1 & 0 & 1\end{pmatrix}, \qquad \hat J_y^2 = a^2\begin{pmatrix} 1 & 0 & -1 \\ 0 & 2 & 0 \\ -1 & 0 & 1\end{pmatrix}, \end{split}\]

\[ \hat J_z^2 = \hbar^2\,\mathrm{diag}(1, 0, 1) = 2a^2\,\mathrm{diag}(1, 0, 1). \]

Add term by term. Off-diagonals: \(\hat J_x^2 + \hat J_y^2\) cancels the \(\pm 1\) corners, leaving zero off-diagonal. Diagonal: \(a^2(1+1) + a^2(1+1) = 4a^2 = 2\hbar^2\) for entries \((1,1)\) and \((3,3)\); \(a^2(2) + a^2(2) = 4a^2 = 2\hbar^2\) for entry \((2,2)\); plus \(\hat J_z^2\) contributions \(2a^2 = \hbar^2, 0, \hbar^2\). Wait, let me redo:

\(\hat J_x^2 + \hat J_y^2 = a^2\begin{pmatrix}2 & 0 & 0\\ 0 & 4 & 0\\ 0 & 0 & 2\end{pmatrix} = \hbar^2\,\mathrm{diag}(1, 2, 1)\).

Add \(\hat J_z^2 = \hbar^2\,\mathrm{diag}(1, 0, 1)\):

\[ \hat J^2 = \hbar^2\,\mathrm{diag}(2, 2, 2) = 2\hbar^2\,\hat I = \hbar^2\cdot j(j+1)\,\hat I\quad\text{with } j(j+1) = 2. \quad\checkmark \]

So every spin-1 basis state has the same Casimir eigenvalue \(\hbar^2\cdot 2\), identifying \(j = 1\).

(c) For spin-1/2: \(\hat J^2 = \tfrac{\hbar^2}{4}((\hat\sigma^x)^2 + (\hat\sigma^y)^2 + (\hat\sigma^z)^2) = \tfrac{\hbar^2}{4}\cdot 3\hat I = \tfrac{3}{4}\hbar^2\hat I\), consistent with \(j(j+1) = \tfrac{1}{2}\cdot\tfrac{3}{2} = \tfrac{3}{4}\).

General pattern. On any spin-\(j\) multiplet, \(\hat J^2 = \hbar^2 j(j+1)\hat I_{2j+1}\). The Casimir labels the multiplet; the value \(j(j+1)\) is set by the algebra and is the same for every state in the multiplet (no \(m\)-dependence). The quantum spectrum of total angular momentum is therefore non-classical — a state with \(j = 1\) has \(\vert\hat{\boldsymbol J}\vert = \hbar\sqrt 2\), not \(\hbar\cdot 1\); this excess is the hallmark of quantum AM and underlies the vector-model picture of Problem 7.

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.

(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 \(2j+1\).

Solution.

(a) With \(j = 3/2\), \(m = 3/2\): \((j-m) = 0\), so \(\hat J_+\vert 3/2,3/2\rangle = 0\). ✓ The factor \((j-m)\) guarantees that the multiplet does not exceed \(m = j\) at the top.

(b) Apply the lowering formula step by step:

\[ \hat J_-\vert 3/2, 3/2\rangle = \hbar\sqrt{(3/2 + 3/2)(3/2 - 3/2 + 1)}\vert 3/2, 1/2\rangle = \hbar\sqrt{3\cdot 1}\,\vert 3/2, 1/2\rangle = \hbar\sqrt 3\,\vert 3/2, 1/2\rangle. \]

\[ \hat J_-\vert 3/2, 1/2\rangle = \hbar\sqrt{(3/2 + 1/2)(3/2 - 1/2 + 1)}\vert 3/2, -1/2\rangle = \hbar\sqrt{2\cdot 2}\,\vert 3/2, -1/2\rangle = 2\hbar\,\vert 3/2, -1/2\rangle. \]

\[ \hat J_-\vert 3/2, -1/2\rangle = \hbar\sqrt{(3/2 - 1/2)(3/2 + 1/2 + 1)}\vert 3/2, -3/2\rangle = \hbar\sqrt{1\cdot 3}\,\vert 3/2, -3/2\rangle = \hbar\sqrt 3\,\vert 3/2, -3/2\rangle. \]

The coefficients \(\hbar\sqrt 3, 2\hbar, \hbar\sqrt 3\) form a palindromic sequence — the ladder is symmetric under reflection \(m\to -m\), because \(\sqrt{(j+m)(j-m+1)} = \sqrt{(j+(-m+1))(j-(-m+1)+1)}\) — i.e. the lowering coefficient from \(m\) equals the raising coefficient from \(-m-1\).

(c) With \(m = -3/2\): \((j+m) = 0\), so \(\hat J_-\vert 3/2, -3/2\rangle = 0\). ✓ The factor \((j+m)\) guarantees that the multiplet does not fall below \(m = -j\).

(d) The multiplet runs from \(m = j\) down to \(m = -j\) in unit steps, giving the states \(m = j, j-1, j-2, \ldots, -j+1, -j\). The number of integer (or half-integer) steps from \(-j\) to \(j\) inclusive is \(2j + 1\). For \(j = 3/2\) this is \(4\); for \(j = 1\) it is \(3\); for \(j = 1/2\) it is \(2\) (the Bloch sphere). Every multiplet has dimension \(2j+1\) — the dimension of the irreducible representation of \(\mathfrak{su}(2)\) with Casimir \(\hbar^2 j(j+1)\).

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

\[ \hat S_+\vert s, m\rangle = \sqrt{(s-m)(s+m+1)}\,\vert s, m+1\rangle, \]

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 \(\hat J_+\hat J_- = \hat J^2 - \hat J_z^2 + \hbar\hat J_z\).

Solution.

(a) From \(s = (n_a + n_b)/2\) and \(m = (n_a - n_b)/2\):

\[ n_a = s + m, \qquad n_b = s - m. \]

(For consistency, \(s\) is a non-negative half-integer (set by \(N = n_a + n_b \ge 0\)) and \(m\) runs from \(-s\) to \(+s\) in unit steps — exactly the AM multiplet structure.)

(b) Apply \(\hat S_+ = \hat a^\dagger\hat b\) to \(\vert n_a, n_b\rangle\):

\[ \hat S_+\vert n_a, n_b\rangle = \sqrt{n_b}\,\sqrt{n_a+1}\,\vert n_a+1, n_b-1\rangle. \]

Under the identification \(\vert n_a, n_b\rangle \leftrightarrow \vert s, m\rangle\) with \(s\) fixed (since \(\hat S_+\) raises \(n_a\) by one and lowers \(n_b\) by one, preserving \(n_a + n_b = 2s\)) and \(m \to m + 1\) (since \(n_a - n_b\) increases by \(2\)):

\[ \hat S_+\vert s, m\rangle = \sqrt{(s-m)(s+m+1)}\,\vert s, m+1\rangle, \]

substituting \(n_b = s - m\) and \(n_a + 1 = s + m + 1\).

(c) The Schwinger construction makes the ladder formula structurally obvious: it is simply the product of two boson normalisation factors \(\sqrt{n_b}\cdot\sqrt{n_a+1}\). There is no need to compute commutators, evaluate \(\hat J^2 - \hat J_z^2\), or take square roots of positive operators — the algebra of \(\hat a, \hat b\) does all the work.

The conventional derivation (lecture) starts from \(\hat J_+\hat J_- + \hat J_z^2 = \hat J^2\) together with \([\hat J_z, \hat J_\pm] = \pm\hbar\hat J_\pm\), and constructs the coefficient \(\sqrt{j(j+1) - m(m\pm 1)} = \sqrt{(j\mp m)(j\pm m + 1)}\) via a careful norm calculation. The Schwinger route gives the same answer by reinterpreting AM as a system of two harmonic oscillators, where the same combinatorial factor emerges from straightforward Fock-space arithmetic. This is one of the deepest pieces of the Schwinger boson formalism: it trivialises the most computationally heavy part of angular-momentum theory.

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.

Solution.

Why \(\hat J_+\vert 1,1\rangle = 0\). Set \(j=1\), \(m=1\) in the raising formula. The coefficient \((j-m) = 0\):

\[ \hat J_+\vert 1,1\rangle = \hbar\sqrt{(1-1)(1+1+1)}\,\vert 1,2\rangle = 0. \]

The factor \((j-m)\) vanishes at \(m = j\), terminating the multiplet at the top rung. A nonzero \(\hat J_+\vert 1,1\rangle\) would be a state with \(m = 2 > j\), which doesn’t exist in the spin-1 multiplet. The ladder closes exactly at \(m = \pm j\).

Compute \(\hat J_-\vert 1, 0\rangle\). From \(\hat J_-\vert j,m\rangle = \hbar\sqrt{(j+m)(j-m+1)}\vert j, m-1\rangle\) with \(j = 1\), \(m = 0\):

\[ \hat J_-\vert 1,0\rangle = \hbar\sqrt{(1+0)(1-0+1)}\,\vert 1,-1\rangle = \hbar\sqrt 2\,\vert 1,-1\rangle. \]

Matrix verification. Construct \(\hat J_- = \hat J_x - \mathrm{i}\hat J_y\) from the spin-1 matrices in Problem 1. Computing,

\[\begin{split} \hat J_- = \frac{\hbar}{\sqrt 2}\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 1 & 0\end{pmatrix} - \mathrm{i}\,\frac{\hbar}{\sqrt 2}\begin{pmatrix} 0 & -\mathrm{i} & 0 \\ \mathrm{i} & 0 & -\mathrm{i} \\ 0 & \mathrm{i} & 0\end{pmatrix} = \frac{\hbar}{\sqrt 2}\begin{pmatrix} 0 & 0 & 0 \\ 2 & 0 & 0 \\ 0 & 2 & 0\end{pmatrix} = \hbar\sqrt 2\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{pmatrix}. \end{split}\]

Acting on \(\vert 1, 0\rangle = (0, 1, 0)^{\mathsf T}\):

\[\begin{split} \hat J_-\vert 1, 0\rangle = \hbar\sqrt 2\begin{pmatrix} 0\\ 0\\ 1\end{pmatrix} = \hbar\sqrt 2\,\vert 1, -1\rangle. \quad\checkmark \end{split}\]

The matrix calculation and the ladder formula agree exactly — and they must, because the lecture’s ladder formula was constructed to be consistent with the matrix representation of each multiplet.

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

\[ \langle\hat J_x^2\rangle + \langle\hat J_y^2\rangle = \hbar^2\bigl[j(j+1) - m^2\bigr]. \]

(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

\[ (\Delta\hat J_x)^2 = \langle\hat J_x^2\rangle - \langle\hat J_x\rangle^2 = \frac{\hbar^2}{2}\bigl[j(j+1) - m^2\bigr]. \]

(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 \(\boldsymbol{e}_z\).

Solution.

(a) Using \(\hat J_x = \tfrac{1}{2}(\hat J_+ + \hat J_-)\) and the ladder action \(\hat J_\pm\vert j,m\rangle \propto \vert j, m\pm 1\rangle\):

\[ \langle j, m\vert\hat J_x\vert j, m\rangle = \tfrac{1}{2}\langle j, m\vert(\hat J_+ + \hat J_-)\vert j, m\rangle. \]

Both \(\hat J_+\vert j,m\rangle\) and \(\hat J_-\vert j,m\rangle\) shift to states orthogonal to \(\vert j, m\rangle\) (different \(\hat J_z\) eigenvalue), so each matrix element vanishes. Hence \(\langle\hat J_x\rangle = 0\). The same argument gives \(\langle\hat J_y\rangle = 0\). Physically: a \(\hat J_z\) eigenstate has no preferred direction in the \(xy\)-plane (rotational symmetry about \(\boldsymbol{e}_z\)), so the transverse expectation values must vanish.

(b) Compute \(\hat J^2 = \hat J_x^2 + \hat J_y^2 + \hat J_z^2\), so \(\hat J_x^2 + \hat J_y^2 = \hat J^2 - \hat J_z^2\). Take the expectation on \(\vert j, m\rangle\):

\[ \langle\hat J_x^2\rangle + \langle\hat J_y^2\rangle = \langle\hat J^2\rangle - \langle\hat J_z^2\rangle = \hbar^2 j(j+1) - \hbar^2 m^2 = \hbar^2\bigl[j(j+1) - m^2\bigr]. \]

(c) The state \(\vert j, m\rangle\) has axial symmetry about \(\boldsymbol{e}_z\) — rotations about \(\boldsymbol{e}_z\) leave it invariant (up to a phase). This symmetry implies \(\langle\hat J_x^2\rangle = \langle\hat J_y^2\rangle\). So each equals half the sum:

\[ \langle\hat J_x^2\rangle = \frac{\hbar^2}{2}\bigl[j(j+1) - m^2\bigr]. \]

With \(\langle\hat J_x\rangle = 0\) from (a), the variance is exactly this:

\[ (\Delta\hat J_x)^2 = \frac{\hbar^2}{2}\bigl[j(j+1) - m^2\bigr]. \]

(d) At \(m = \pm j\):

\[ (\Delta\hat J_x)^2_{\text{stretched}} = \frac{\hbar^2}{2}\bigl[j(j+1) - j^2\bigr] = \frac{\hbar^2 j}{2}. \]

So \(\Delta\hat J_x = \hbar\sqrt{j/2}\) — strictly positive for any \(j > 0\). The quantum AM vector can never be perfectly aligned with \(\boldsymbol{e}_z\): even the stretched state has irreducible transverse uncertainty \(\Delta J_x \sim \hbar\sqrt j\). Geometrically, the vector tilts at an angle (Problem 7) from \(\boldsymbol{e}_z\), and the transverse uncertainty reflects rotational averaging around this tilted direction. Only in the classical limit \(j\to\infty\) does the relative uncertainty \(\Delta J_x/(\hbar j) \sim 1/\sqrt j \to 0\), recovering the sharp classical AM vector.

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.

Solution.

(a) The Robertson uncertainty relation reads

\[ \Delta\hat J_x\cdot\Delta\hat J_y \geq \tfrac{1}{2}\vert\langle[\hat J_x, \hat J_y]\rangle\vert = \tfrac{\hbar}{2}\vert\langle\hat J_z\rangle\vert. \]

The bound is state-dependent, with right-hand side proportional to \(\langle\hat J_z\rangle\) — it is largest on states with definite, large \(\hat J_z\).

(b) On \(\vert j, j\rangle\), \(\langle\hat J_z\rangle = \hbar j\), so the right-hand side is \(\tfrac{\hbar^2}{2}j\). From Problem 5(d),

\[ (\Delta\hat J_x)^2 = (\Delta\hat J_y)^2 = \frac{\hbar^2 j}{2}, \qquad \Delta\hat J_x = \Delta\hat J_y = \hbar\sqrt{j/2}. \]

Therefore

\[ \Delta\hat J_x\cdot\Delta\hat J_y = \frac{\hbar^2 j}{2}. \]

(c) The two sides are equal — the Robertson inequality is saturated: \(\Delta\hat J_x\cdot\Delta\hat J_y = \tfrac{\hbar}{2}\langle\hat J_z\rangle\). The stretched state is the angular-momentum analogue of the minimum-uncertainty state: it has the smallest possible transverse uncertainty consistent with the algebra (just as \(\vert 0\rangle\) does for \((\hat X, \hat Z)\) in 1.2.2 P6, saturating \(\Delta\hat X\cdot\Delta\hat Z \geq \tfrac{1}{2}\vert\langle\hat Y\rangle\vert = 0\)). The geometric picture: among all states in the spin-\(j\) multiplet, \(\vert j, j\rangle\) is the most strongly aligned with \(\boldsymbol{e}_z\), and its transverse uncertainty is the price the algebra exacts for that alignment.

(d) Ratio:

\[ \frac{\Delta\hat J_x}{\langle\hat J_z\rangle} = \frac{\hbar\sqrt{j/2}}{\hbar j} = \frac{1}{\sqrt{2j}}. \]

As \(j \to \infty\), this ratio vanishes as \(j^{-1/2}\). The classical limit: a macroscopic AM with \(j \sim 10^{34}\) (typical of a spinning top) has relative transverse uncertainty \(\sim 10^{-17}\) — undetectable. The classical sharp AM vector is the \(j\to\infty\) limit of the quantum stretched state, and the algebraic uncertainty bound — never zero at finite \(j\) — disappears smoothly into the macroscopic regime. This is the AM analogue of the harmonic-oscillator’s coherent-state classical limit (2.1.3 P3).

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

\[ \cos\theta = \frac{m}{\sqrt{j(j+1)}}. \]

(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).

Solution.

(a) Treating \(\boldsymbol J\) as a classical vector of length \(\hbar\sqrt{j(j+1)}\) with \(z\)-component \(\hbar m\):

\[ \cos\theta = \frac{J_z}{\vert\boldsymbol J\vert} = \frac{\hbar m}{\hbar\sqrt{j(j+1)}} = \frac{m}{\sqrt{j(j+1)}}. \]

(b) Stretched state: \(m = j\).

\[ \cos\theta_{\min} = \frac{j}{\sqrt{j(j+1)}} = \sqrt{\frac{j}{j+1}}. \]

Evaluations:

\(j = 1/2\): \(\cos\theta_{\min} = \sqrt{1/3} \approx 0.577\), so \(\theta_{\min} \approx 54.7^\circ\).
\(j = 1\): \(\cos\theta_{\min} = \sqrt{1/2} = 1/\sqrt 2\), so \(\theta_{\min} = 45^\circ\).
\(j = 100\): \(\cos\theta_{\min} = \sqrt{100/101} \approx 0.9950\), so \(\theta_{\min} \approx 5.71^\circ\).

The angle decreases monotonically toward \(0\) as \(j \to \infty\) — the stretched state aligns ever more sharply with \(\boldsymbol{e}_z\).

\[ \sqrt{\langle\hat J_x^2 + \hat J_y^2\rangle} = \hbar\sqrt{j(j+1) - m^2}. \]

In the vector model this is the radius of the circle traced by the perpendicular projection of \(\boldsymbol J\) — the “precession circle” of the semiclassical picture.

(d) Classical: \(\vert\boldsymbol J\vert\sin\theta = \hbar\sqrt{j(j+1)}\sin\theta\). Using \(\sin^2\theta = 1 - \cos^2\theta = 1 - m^2/[j(j+1)] = [j(j+1) - m^2]/[j(j+1)]\):

\[ \vert\boldsymbol J\vert\sin\theta = \hbar\sqrt{j(j+1)}\cdot\sqrt{\frac{j(j+1) - m^2}{j(j+1)}} = \hbar\sqrt{j(j+1) - m^2}, \]

exactly the quantum result. The vector model is internally consistent: \(\boldsymbol J\) has length \(\hbar\sqrt{j(j+1)}\), \(z\)-projection \(\hbar m\), and transverse projection \(\hbar\sqrt{j(j+1) - m^2}\), all in agreement with the quantum expectation values. The picture should be used with care — \(\hat J_x\) and \(\hat J_y\) are not simultaneously sharp, so the “precession circle” is a statistical/conceptual aid, not a literal trajectory — but it is the right intuitive bridge between the quantum spectrum and the macroscopic notion of a spinning object.

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

\[ [\hat{\alpha}, \hat{\beta}] = 2\hat{\alpha}, \quad [\hat{\alpha}^\dagger, \hat{\beta}] = -2\hat{\alpha}^\dagger, \quad [\hat{\alpha}^\dagger, \hat{\alpha}] = 2\hat{\beta}, \quad \{\hat{\alpha}^\dagger, \hat{\alpha}\} = 2\hat{I}, \]

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 \(\hat{\beta}\).

Solution.

(a) Take the Hermitian conjugate of \([\hat\alpha^\dagger, \hat\alpha] = 2\hat\beta\). The conjugate of a commutator: \([\hat A, \hat B]^\dagger = (\hat A\hat B)^\dagger - (\hat B\hat A)^\dagger = \hat B^\dagger\hat A^\dagger - \hat A^\dagger\hat B^\dagger = [\hat B^\dagger, \hat A^\dagger] = -[\hat A^\dagger, \hat B^\dagger]\). Applying:

\[ 2\hat\beta^\dagger = ([\hat\alpha^\dagger, \hat\alpha])^\dagger = -[\hat\alpha, \hat\alpha^\dagger] = [\hat\alpha^\dagger, \hat\alpha] = 2\hat\beta. \]

So \(\hat\beta^\dagger = \hat\beta\). Hermitian. ✓

(b) Add and subtract the commutator and anticommutator:

\[ [\hat\alpha^\dagger, \hat\alpha] + \{\hat\alpha^\dagger, \hat\alpha\} = 2\hat\alpha^\dagger\hat\alpha, \qquad \{\hat\alpha^\dagger, \hat\alpha\} - [\hat\alpha^\dagger, \hat\alpha] = 2\hat\alpha\hat\alpha^\dagger. \]

Substituting the given relations:

\[ \hat\alpha^\dagger\hat\alpha = \tfrac{1}{2}(2\hat\beta + 2\hat I) = \hat\beta + \hat I, \]

\[ \hat\alpha\hat\alpha^\dagger = \tfrac{1}{2}(2\hat I - 2\hat\beta) = \hat I - \hat\beta. \]

(c) Compute \((\hat\alpha^\dagger)^2\hat\alpha^2\) by inserting \(\hat\alpha^\dagger\hat\alpha = \hat\beta + \hat I\) in the middle:

\[ (\hat\alpha^\dagger)^2\hat\alpha^2 = \hat\alpha^\dagger(\hat\alpha^\dagger\hat\alpha)\hat\alpha = \hat\alpha^\dagger(\hat\beta + \hat I)\hat\alpha = \hat\alpha^\dagger\hat\beta\hat\alpha + \hat\alpha^\dagger\hat\alpha. \]

For the first term, use \([\hat\alpha^\dagger, \hat\beta] = -2\hat\alpha^\dagger\), i.e. \(\hat\alpha^\dagger\hat\beta = \hat\beta\hat\alpha^\dagger - 2\hat\alpha^\dagger\), then \(\hat\alpha^\dagger\hat\alpha = \hat\beta + \hat I\):

\[\begin{split} \begin{split} \hat\alpha^\dagger\hat\beta\hat\alpha &= (\hat\beta\hat\alpha^\dagger - 2\hat\alpha^\dagger)\hat\alpha \\ &= \hat\beta\hat\alpha^\dagger\hat\alpha - 2\hat\alpha^\dagger\hat\alpha \\ &= \hat\beta(\hat\beta + \hat I) - 2(\hat\beta + \hat I) \\ &= \hat\beta^2 - \hat\beta - 2\hat I. \end{split} \end{split}\]

Therefore

\[ (\hat\alpha^\dagger)^2\hat\alpha^2 = \hat\beta^2 - \hat\beta - 2\hat I + (\hat\beta + \hat I) = \hat\beta^2 - \hat I. \quad\checkmark \]

Feasible eigenvalues of \(\hat\beta\). Apply positivity to each operator in turn. Let \(\beta\) be an eigenvalue of \(\hat\beta\) (real, since \(\hat\beta\) is Hermitian).

\(\hat\alpha^\dagger\hat\alpha = \hat\beta + \hat I \geq 0\) forces \(\beta + 1 \geq 0\), i.e. \(\beta \geq -1\).
\(\hat\alpha\hat\alpha^\dagger = \hat I - \hat\beta \geq 0\) forces \(1 - \beta \geq 0\), i.e. \(\beta \leq +1\).
\((\hat\alpha^\dagger)^2\hat\alpha^2 = \hat\beta^2 - \hat I \geq 0\) forces \(\beta^2 \geq 1\), i.e. \(\beta \leq -1\) or \(\beta \geq +1\).

The intersection of all three is \(\beta = +1\) or \(\beta = -1\). Only two eigenvalues are feasible: \(\beta = \pm 1\). The algebra forces a two-dimensional representation.

(d) The relations \([\hat\beta, \hat\alpha] = -2\hat\alpha\) and \([\hat\beta, \hat\alpha^\dagger] = 2\hat\alpha^\dagger\) are exactly the standard ladder commutators \([\hat J_z, \hat J_\pm] = \pm\hbar\hat J_\pm\) with \(\hbar = 1\), \(\hat J_z = \hat\beta/2\), \(\hat J_- = \hat\alpha\), \(\hat J_+ = \hat\alpha^\dagger\). The values \(\beta = \pm 1\) correspond to \(m = \pm 1/2\) — exactly the spin-1/2 representation.

In the eigenbasis of \(\hat\beta\), label the states \(\vert+\rangle\) (with \(\beta = +1\), i.e. \(m = +1/2\)) and \(\vert-\rangle\) (with \(\beta = -1\)). Then \(\hat\beta = \mathrm{diag}(1, -1) = \hat\sigma^z\). The ladder operators act as \(\hat\alpha^\dagger\vert-\rangle = \vert+\rangle\) (raising) and \(\hat\alpha^\dagger\vert+\rangle = 0\) (top), with coefficient \(\sqrt{(\tfrac{1}{2} - m)(\tfrac{1}{2} + m + 1)}\) from Problem 4 — equal to \(\sqrt{1} = 1\) for \(m = -1/2\). Likewise \(\hat\alpha\vert+\rangle = \vert-\rangle\), \(\hat\alpha\vert-\rangle = 0\). So in the \(\{\vert+\rangle, \vert-\rangle\}\) basis,

\[\begin{split} \hat\alpha = \begin{pmatrix} 0 & 0 \\ 1 & 0\end{pmatrix} = \hat\sigma^-, \qquad \hat\alpha^\dagger = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix} = \hat\sigma^+, \qquad \hat\beta = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix} = \hat\sigma^z. \end{split}\]

The “quantum bootstrap” recovers the spin-1/2 Pauli operators from the algebra alone — no commutator-by-commutator construction, no requirement of a specific representation, just positivity bounds applied to algebraic relations. This is the operator-algebra version of the conformal-bootstrap approach in modern theoretical physics: spectrum and matrix elements pinned down entirely by symmetry constraints + positivity.