Angular Momentum and Spin Questions and Answers

Answer: B

The \( z \)-component of angular momentum is zero.

In cylindrical coordinates \( L_{z} = -i\hbar \frac{\partial}{\partial \phi} \). The function \( e^{ikz} \) depends only on \( z \), not on the azimuthal angle \( \phi \), so

\[ \begin{aligned} L_{z}\, e^{ikz} &= -i\hbar \frac{\partial}{\partial \phi} e^{ikz} \\ &= 0 \end{aligned} \]

The state is an eigenstate of \( L_{z} \) with eigenvalue 0. \( \hbar k \) is the trap: it is the eigenvalue of the linear momentum \( p_{z} \), not of the angular momentum. The imaginary values cannot be eigenvalues of a Hermitian operator.

Answer: D

The orbital angular momentum of a 2s electron is zero.

The magnitude of orbital angular momentum is \( \sqrt{l(l+1)}\,\frac{h}{2\pi} \), and an s electron has \( l = 0 \) whatever its principal quantum number, so the result is 0. The 2 in 2s is n, which does not enter.

\( h/2\pi \) is the Bohr-model value \( n\hbar \) for \( n = 1 \), which the quantum theory replaces. \( h\sqrt{2}/2\pi \) is the value for a p electron (\( l = 1 \)), and \( h/4\pi \) is the size of the spin component, not of orbital motion.

Answer: B

The Pauli matrices anti-commute.

Multiplying them both ways:

\[ \sigma_{x}\sigma_{y} = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} = i\sigma_{z} \]

\[ \sigma_{y}\sigma_{x} = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} = -i\sigma_{z} \]

So \(\sigma_{x}\sigma_{y} + \sigma_{y}\sigma_{x} = 0\): they anti-commute. They do not commute (their commutator is \(2i\sigma_{z}\), not zero), and they are not inverses of each other, since their product is \(i\sigma_{z}\), not the identity; each is its own inverse.

Answer: D

The probabilities are 1/3 and 2/3.

The probability of each outcome is the squared modulus of the corresponding component:

\[ \begin{aligned} P(+\hbar/2) &= \frac{|1+i|^{2}}{6} = \frac{2}{6} = \frac{1}{3} \\ P(-\hbar/2) &= \left(\frac{\sqrt{2}}{\sqrt{3}}\right)^{2} = \frac{2}{3} \end{aligned} \]

They add to 1, so the state is normalised. The trap is \(|1+i|^{2}\): it is 2, not 1 (which would give 1/6), and not \((1+i)^{2} = 2i\). Swapping the two values gives 2/3 and 1/3.

Answer: C

The commutator is \(\hbar(xp_{z} + p_{x}z)\).

Write \(\hat{A} = iL_{z}\) with \(L_{z} = xp_{y} - yp_{x}\), and use \([L_{z}, y] = -i\hbar x\), \([L_{z}, p_{y}] = -i\hbar p_{x}\), while \(L_{z}\) commutes with z and \(p_{z}\):

\[ \begin{aligned} [L_{z}, yp_{z}] &= [L_{z}, y]p_{z} = -i\hbar xp_{z} \\ [L_{z}, zp_{y}] &= z[L_{z}, p_{y}] = -i\hbar zp_{x} \\ [\hat{A}, \hat{B}] &= i(-i\hbar)(xp_{z} + zp_{x}) \\ &= \hbar(xp_{z} + p_{x}z) \end{aligned} \]

Since z and \(p_{x}\) commute, \(zp_{x} = p_{x}z\). The options with a minus sign in front come from dropping the factor i in \(\hat{A}\), and those with \(xp_{z} - p_{x}z\) from a sign slip in one of the two commutators.

Answer: B

The ratio \(I_{i} : I_{f}\) is 4 : 1.

An unpolarised beam of spin-1/2 particles splits equally along any axis, so \(S_{1}\), passing only spin +x, transmits half the beam. Those particles are in the state \(|+x\rangle\). In the y basis

\[ |\langle +y|+x\rangle|^{2} = \frac{1}{2} \]

so \(S_{2}\) passes half of what reaches it. The final intensity is \(\frac{1}{2}\times\frac{1}{2} = \frac{1}{4}\) of the initial, a ratio of 4 : 1.

2 : 1 counts only the first apparatus. 1 : 0 would follow if x and y spin states were orthogonal, but only opposite directions, such as +y and -y, are orthogonal for spin 1/2. 8 : 1 would need a third filter.

Answer: D

\([L_{+}, L_{-}] = 2\hbar L_{z}\).

Expand using \([L_{x}, L_{y}] = i\hbar L_{z}\):

\[ \begin{aligned} [L_{x} + iL_{y}, L_{x} - iL_{y}] &= -i[L_{x}, L_{y}] + i[L_{y}, L_{x}] \\ &= -2i[L_{x}, L_{y}] \\ &= -2i(i\hbar L_{z}) = 2\hbar L_{z} \end{aligned} \]

The commutator is not zero because \(L_{x}\) and \(L_{y}\) do not commute, and the cyclic algebra produces \(L_{z}\), not \(L_{x}\) or \(L_{y}\).

Answer: B

Spin-orbit coupling leads to the Anomalous Zeeman effect.

Spin-orbit coupling combines L and S into J, splitting levels into fine-structure components with different Landé g factors. In a weak magnetic field each J level splits into \(2J + 1\) sublevels spaced by \(g_{J}\mu_{B}B\), and because \(g_{J}\) differs between levels, the lines split into more than three components: the anomalous Zeeman effect.

The normal Zeeman effect (three lines) is seen when spin plays no part (singlet states). The Stark effect is due to an electric field, and hyperfine splitting comes from coupling with the nuclear spin.

Answer: D

The statement that is not correct is that, with spin-orbit interaction, \(\bar{L}^{2}\), \(\bar{S}^{2}\) and \(\bar{J}\) are not conserved.

Spin-orbit coupling is \(\xi(r)\bar{L}\cdot\bar{S} = \frac{\xi}{2}(J^{2} - L^{2} - S^{2})\). This commutes with \(L^{2}\), \(S^{2}\), \(J^{2}\) and the components of \(\bar{J}\), so those quantities remain conserved, while \(\bar{L}\) and \(\bar{S}\) separately are not, since the coupling transfers angular momentum between them.

So statements A and B are correct, and C is the standard form of the spin-orbit Hamiltonian (including the Thomas factor). Statement D claims the opposite of what the commutators show.

Answer: C

The probabilities are \(P_{1}^{x} = 1/4, P_{0}^{x} = 1/2, P_{-1}^{x} = 1/4\).

For spin 1, the eigenvectors of \(S_{x}\) written in the \(S_{z}\) basis \((|{+1}\rangle, |0\rangle, |{-1}\rangle)\) are

\[ \begin{aligned} |{+1}\rangle_{x} &= \tfrac{1}{2}(1, \sqrt{2}, 1) \\ |0\rangle_{x} &= \tfrac{1}{\sqrt{2}}(1, 0, -1) \\ |{-1}\rangle_{x} &= \tfrac{1}{2}(1, -\sqrt{2}, 1) \end{aligned} \]

The probability of each outcome is the square of the first component (the overlap with \(|S_{z}, +1\rangle\)): \((1/2)^{2} = 1/4\), \((1/\sqrt{2})^{2} = 1/2\) and \(1/4\). They add to 1, and the symmetry between \(+\hbar\) and \(-\hbar\) is expected because \(|S_{z}, +1\rangle\) has \(\langle S_{x} \rangle = 0\).

The equal split 1/3 each and the asymmetric splits fail that symmetry check.