Angular Momentum and Spin Questions and Answers

Answer: B

The beam splits into 7 beams.

The magnitude of an angular momentum with quantum number \(j\) is \(\sqrt{j(j+1)}\,\hbar\). Setting

\[ j(j + 1) = 12 \Rightarrow j = 3 \]

In a Stern-Gerlach apparatus the beam splits according to the projection \(m_{j}\), which takes the \(2j + 1 = 7\) values \(-3, -2, \ldots, 3\).

3 is the value of \(j\) itself, 9 is \((j)^{2}\) or the count for \(j = 4\), and 25 is \(2j + 1\) with 12 taken as \(j\); none gives the number of \(m_{j}\) states.

Answer: C

The eigenvalues are \(L^{2} = 2\hbar^{2}\) and \(L_{z} = 0\).

Write \(z = r\cos\theta\), so

\[ \begin{aligned} \psi &= N r\cos\theta\, e^{-\alpha r^{2}} \\ &\propto r e^{-\alpha r^{2}}\, Y_{1}^{0}(\theta) \end{aligned} \]

The angular part is exactly \(Y_{1}^{0}\), so the state has \(\ell = 1\) and \(m = 0\). The eigenvalues are \(\ell(\ell + 1)\hbar^{2} = 2\hbar^{2}\) and \(m\hbar = 0\). The radial factor \(re^{-\alpha r^{2}}\) does not affect the angular momentum.

0 and 0 would need the spherically symmetric \(Y_{0}^{0}\) (no factor of \(z\)). The options with \(\hbar^{2}\) misuse \(\ell^{2}\hbar^{2}\) for \(L^{2}\), and \(\pm\hbar\) for \(L_{z}\) would need the \(x \pm iy\) combinations that go with \(Y_{1}^{\pm 1}\).

Answer: A

The expectation value is \(-\lambda\).

Write the dot product with raising and lowering operators:

\[ \begin{aligned} \vec{S}_{1}\cdot\vec{S}_{2} &= S_{1z}S_{2z} \\ &\quad + \tfrac{1}{2}(S_{1+}S_{2-} + S_{1-}S_{2+}) \end{aligned} \]

In the product state \(|\uparrow\rangle|\downarrow\rangle\), the flip terms turn the state into \(|\downarrow\rangle|\uparrow\rangle\) (or zero), which is orthogonal to it, so their expectation vanishes. Only the \(z\) term survives:

\[ \begin{aligned} \langle H \rangle &= \frac{4\lambda}{\hbar^{2}}\left(\frac{\hbar}{2}\right)\left(-\frac{\hbar}{2}\right) \\ &= -\lambda \end{aligned} \]

\(+\lambda\) is the value for parallel spins (the triplet \(|\uparrow\uparrow\rangle\)). \(-3\lambda\) would be the singlet, and \(\pm 2\lambda\) do not follow from the spin algebra.

Answer: D

The probability is \(\cos^{2}\frac{\theta}{2}\).

The spin-up state along a direction in the \(xy\)-plane at angle \(\phi\) from \(x\) is \(\frac{1}{\sqrt{2}}(|\uparrow\rangle + e^{i\phi}|\downarrow\rangle)\) in the \(z\) basis. For \(\hat{x}\) (\(\phi = 0\)) and \(\hat{x}'\) (\(\phi = \theta\)):

\[ \begin{aligned} \langle x'+|x+\rangle &= \tfrac{1}{2}(1 + e^{-i\theta}) \\ |\langle x'+|x+\rangle|^{2} &= \tfrac{1}{2}(1 + \cos\theta) = \cos^{2}\frac{\theta}{2} \end{aligned} \]

This is the general spin-1/2 rule: the overlap between spin-up states along two directions an angle \(\theta\) apart is \(\cos^{2}(\theta/2)\). Check: \(\theta = 0\) gives certainty and \(\theta = \pi\) gives zero. The options with \(\theta/4\) or an extra factor 1/2 fail these checks.

Answer: B

The minimum time is \(\frac{\pi}{4\omega}\).

The evolution operator for \(H = -\hbar\omega\sigma_{y}\) is \(U = e^{i\omega t\sigma_{y}} = \cos\omega t + i\sigma_{y}\sin\omega t\), and \(i\sigma_{y} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\). Acting on spin up:

\[ U\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \cos\omega t \\ -\sin\omega t \end{pmatrix} \]

This equals \(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ -1 \end{pmatrix}\) first when \(\cos\omega t = \sin\omega t\), that is \(\omega t = \frac{\pi}{4}\).

The spin precesses about \(y\) at angular rate \(2\omega\), so turning from \(+z\) to \(-x\) (90°) takes \(\frac{\pi/2}{2\omega} = \frac{\pi}{4\omega}\). \(\frac{\pi}{2\omega}\) would take it to spin down along \(z\), and \(\frac{\pi}{\omega}\) back to spin up.

Answer: A

The eigenvalues are \(0, \frac{A}{2}, -A\).

Let \(\vec{S}_{12} = \vec{S}_{1} + \vec{S}_{2}\), with \(s_{12} = 0\) or 1, and total spin \(\vec{S} = \vec{S}_{12} + \vec{S}_{3}\). Then

\[ \begin{aligned} \vec{S}_{12}\cdot\vec{S}_{3} = \frac{\hbar^{2}}{2}[&s(s+1) \\ &- s_{12}(s_{12}+1) - \tfrac{3}{4}] \end{aligned} \]

• \(s_{12} = 0\), \(s = \frac{1}{2}\): \(E = 0\).
• \(s_{12} = 1\), \(s = \frac{3}{2}\): \(E = \frac{A}{2}(\frac{15}{4} - 2 - \frac{3}{4}) = \frac{A}{2}\).
• \(s_{12} = 1\), \(s = \frac{1}{2}\): \(E = \frac{A}{2}(\frac{3}{4} - 2 - \frac{3}{4}) = -A\).

The sets containing \(-\frac{A}{2}\) or \(\pm\frac{3A}{2}\) come from using the wrong \(s(s+1)\) values or dropping the factor \(\frac{1}{2}\).

Answer: C

The trace is \(4i\).

The Pauli matrices satisfy \([\sigma_{x}, \sigma_{y}] = 2i\sigma_{z}\), and \(\sigma_{z}^{2} = I\):

\[ \begin{aligned} \sigma_{z}[\sigma_{x}, \sigma_{y}] &= 2i\,\sigma_{z}^{2} = 2i\,I \\ \text{Tr}(2i\,I) &= 2i \times 2 = 4i \end{aligned} \]

Directly: \(\sigma_{x}\sigma_{y} = \text{diag}(i, -i)\) and \(\sigma_{y}\sigma_{x} = \text{diag}(-i, i)\), so the commutator is \(\text{diag}(2i, -2i)\); multiplying by \(\sigma_{z}\) gives \(\text{diag}(2i, 2i)\). 2i forgets that the trace of the 2 by 2 identity is 2, and i and i/2 drop the factor 2 of the commutator as well.

Answer: A

The coefficients are \(c_{1} = 2/\sqrt{5}\) and \(c_{2} = 1/\sqrt{5}\).

Start from the stretched state \(|5/2, 5/2\rangle = |2, 2\rangle|1/2, 1/2\rangle\) and apply \(\hat{J}_{-} = \hat{J}_{1-} + \hat{J}_{2-}\) to both sides using the hint:

\[ \begin{aligned} \hat{J}_{-}|\tfrac{5}{2}, \tfrac{5}{2}\rangle &= \sqrt{\tfrac{35}{4} - \tfrac{15}{4}}\,|\tfrac{5}{2}, \tfrac{3}{2}\rangle = \sqrt{5}\,|\tfrac{5}{2}, \tfrac{3}{2}\rangle \\ \hat{J}_{1-}|2, 2\rangle &= \sqrt{6 - 2}\,|2, 1\rangle = 2|2, 1\rangle \\ \hat{J}_{2-}|\tfrac{1}{2}, \tfrac{1}{2}\rangle &= 1 \cdot |\tfrac{1}{2}, -\tfrac{1}{2}\rangle \end{aligned} \]

\[ \begin{aligned} \sqrt{5}\,|\tfrac{5}{2}, \tfrac{3}{2}\rangle = \ &2|2,1\rangle|\tfrac{1}{2},\tfrac{1}{2}\rangle \\ &+ |2,2\rangle|\tfrac{1}{2},-\tfrac{1}{2}\rangle \end{aligned} \]

This gives \(c_{1} = 2/\sqrt{5}\) and \(c_{2} = 1/\sqrt{5}\). The swapped pair belongs to the orthogonal state \(|3/2, 3/2\rangle\) (up to sign), equal weights do not follow from the ladder factors, and \(c_{1} = 0\) is not an eigenstate of \(J^{2}\).

Answer: A

The expectation value is \(-\hbar/6\).

Each \(\psi_{nlm}\) is an eigenstate of \(L_{z}\) with eigenvalue \(m\hbar\), and the probability of each is the squared coefficient. The coefficients are 4/6, -2/6, \(\sqrt{6}/6\) and \(-\sqrt{10}/6\), giving probabilities 16/36, 4/36, 6/36 and 10/36, which add to 1, so the state is normalized.

\[ \begin{aligned} \langle L_{z}\rangle &= \hbar\Big[0\cdot\tfrac{16}{36} + 1\cdot\tfrac{4}{36} \\ &\quad + 0\cdot\tfrac{6}{36} - 1\cdot\tfrac{10}{36}\Big] \\ &= -\frac{6\hbar}{36} = -\frac{\hbar}{6} \end{aligned} \]

Only the m = 1 and m = -1 components contribute. The value \(+\hbar/6\) has the sign reversed, and \(\pm 6\hbar\) forget to divide by the normalization 36.

Answer: C

The keyed count is 2.

Coupling two spin-1/2 angular momenta gives total j = 1 (a triplet, m = 1, 0, -1) and j = 0 (a singlet, m = 0), four states in all. Two of these have m = 0: the triplet state \(|1, 0\rangle\) = \((|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle)/\sqrt{2}\) and the singlet \(|0, 0\rangle\) = \((|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)/\sqrt{2}\).

Both are built as linear combinations of the same two product states \(|\uparrow\downarrow\rangle\) and \(|\downarrow\uparrow\rangle\), which is the sense of the question; as states they are orthogonal to each other. The m = 1 and m = -1 states are each a single product state, so the number of m = 0 states is 2, not zero, 1 or 4.