Chapter 14 Quiz: Addition of Angular Momentum

Instructions: This quiz covers the core concepts from Chapter 14. For multiple choice, select the single best answer. For true/false, provide a brief justification (1-2 sentences). For short answer, aim for 3-5 sentences. For applied scenarios, show your work.

Multiple Choice (10 questions)

Q1. When coupling angular momenta $j_1 = 2$ and $j_2 = 1$, the allowed values of $J$ are:

(a) $J = 3$ only (b) $J = 1, 2, 3$ (c) $J = 0, 1, 2, 3$ (d) $J = 1, 2$

Q2. The Clebsch-Gordan coefficient $\langle j_1, m_1; j_2, m_2 | J, M\rangle$ is necessarily zero when:

(a) $j_1 + j_2 > J$ (b) $M \neq m_1 + m_2$ (c) $j_1 = j_2$ (d) $m_1 = 0$ or $m_2 = 0$

Q3. The singlet state of two spin-1/2 particles is $|0, 0\rangle = \frac{1}{\sqrt{2}}(|\!\uparrow\downarrow\rangle - |\!\downarrow\uparrow\rangle)$. Under exchange of the two particles, this state is:

(a) Symmetric (b) Antisymmetric (c) Neither symmetric nor antisymmetric (d) An eigenstate of $\hat{S}_z$ with eigenvalue zero

Q4. For an electron in the hydrogen $3d$ state ($n = 3$, $\ell = 2$, $s = 1/2$), the allowed values of the total angular momentum quantum number $j$ are:

(a) $j = 3/2$ and $j = 5/2$ (b) $j = 1/2$ and $j = 3/2$ (c) $j = 5/2$ only (d) $j = 1/2$, $3/2$, and $5/2$

Q5. The Wigner-Eckart theorem states that the $m$-dependence of a matrix element $\langle j', m' | \hat{T}_q^{(k)} | j, m\rangle$ is entirely contained in:

(a) The reduced matrix element $\langle j' \| T^{(k)} \| j \rangle$ (b) A Clebsch-Gordan coefficient $\langle j, m; k, q | j', m'\rangle$ (c) The tensor rank $k$ (d) The product $m \cdot m'$

Q6. The electric dipole operator is a rank-1 irreducible tensor. Using the Wigner-Eckart theorem, the selection rule for $j$ in electric dipole transitions is:

(a) $\Delta j = 0$ only (b) $\Delta j = \pm 1$ only (c) $\Delta j = 0, \pm 1$ (with $j = 0 \to j' = 0$ forbidden) (d) No restriction on $\Delta j$

Q7. In the $M$-counting argument for the triangle rule, the number of uncoupled states with $M = j_1 + j_2$ is:

(a) Zero (b) One (c) Two (d) $(2j_2 + 1)$

Q8. The Lande g-factor for a state with $j = \ell + 1/2$ (spin $s = 1/2$) is:

(a) Exactly 1 (b) Exactly 2 (c) $1 + \frac{1}{2\ell + 1}$ (d) $2 - \frac{1}{2\ell + 1}$

Q9. Which of the following is NOT a valid irreducible tensor operator of rank 1?

(a) The position operator $\hat{\mathbf{r}}$ in spherical components (b) The angular momentum operator $\hat{\mathbf{J}}$ in spherical components (c) The Hamiltonian $\hat{H}$ of a rotationally invariant system (d) The electric dipole moment operator $\hat{\mathbf{d}} = e\hat{\mathbf{r}}$

Q10. When coupling $j_1 = 3/2$ and $j_2 = 3/2$, the total dimension of the tensor product space is:

(a) 9 (b) 12 (c) 16 (d) 6

True/False (4 questions)

Q11. True or False: Clebsch-Gordan coefficients are always real numbers (under the Condon-Shortley phase convention).

Q12. True or False: The matrix element $\langle 2, 0 | \hat{T}_0^{(3)} | 1, 0\rangle$ is zero by the triangle rule in the Wigner-Eckart theorem, because a rank-3 operator cannot connect $j = 1$ to $j' = 2$.

Q13. True or False: In L-S coupling, the spin-orbit interaction is stronger than the residual Coulomb interaction between electrons.

Q14. True or False: The projection theorem implies that the expectation value of any vector operator $\hat{\mathbf{V}}$ within a state $|j, m\rangle$ is parallel to $\langle \hat{\mathbf{J}} \rangle$.

Short Answer (4 questions)

Q15. Explain the physical difference between the singlet ($S = 0$) and triplet ($S = 1$) states of two spin-1/2 particles. Why do these states have different energies in helium, even though the spin-orbit coupling is not the primary cause?

Q16. State the Wigner-Eckart theorem in words (not equations). Explain why it is so powerful — what computational savings does it provide?

Q17. An atom makes an electric quadrupole (E2) transition. What are the selection rules for $\Delta j$ and $\Delta m$? How do these differ from electric dipole (E1) selection rules? Why are E2 transitions typically much weaker than E1 transitions?

Q18. Explain the physical meaning of the triangle rule $|j_1 - j_2| \leq J \leq j_1 + j_2$. Why can the total angular momentum quantum number $J$ be less than either $j_1$ or $j_2$? Give a simple example.

Applied Scenarios (2 questions)

Q19. An electron in a hydrogen atom has quantum numbers $n = 4$, $\ell = 3$ (an $f$-state).

(a) What are the possible values of $j$?

(b) How many distinct $|j, m_j\rangle$ states are there in total? Verify this equals the number of uncoupled states $(2\ell + 1)(2s + 1)$.

(c) Compute $\langle \hat{\mathbf{L}} \cdot \hat{\mathbf{S}} \rangle$ in $\hbar^2$ for each value of $j$.

(d) Which $j$ level lies higher in energy due to spin-orbit coupling? Explain the sign.

(e) Compute the Lande g-factor for each $j$ value.

Q20. Two particles have angular momenta $j_1 = 1$ and $j_2 = 1/2$. Consider the matrix element $\langle j_1 = 1, j_2 = 1/2; J = 3/2, M = 1/2 | \hat{\mathbf{L}} \cdot \hat{\mathbf{S}} | j_1 = 1, j_2 = 1/2; J = 3/2, M = 1/2\rangle$.

(a) Compute this matrix element using the identity $\hat{\mathbf{L}} \cdot \hat{\mathbf{S}} = \frac{1}{2}(\hat{J}^2 - \hat{L}^2 - \hat{S}^2)$.

(b) Now compute the same matrix element by expanding $|J = 3/2, M = 1/2\rangle$ in the uncoupled basis and computing $\langle \hat{\mathbf{L}} \cdot \hat{\mathbf{S}} \rangle$ directly using $\hat{\mathbf{L}} \cdot \hat{\mathbf{S}} = \hat{L}_z\hat{S}_z + \frac{1}{2}(\hat{L}_+\hat{S}_- + \hat{L}_-\hat{S}_+)$.

(c) Verify that both methods give the same answer.

Answer Key

Q1: (b). The triangle rule gives $J = |2-1|$ to $2+1$, so $J = 1, 2, 3$. Dimension check: $5 \times 3 = 15 = 7 + 5 + 3$.

Q2: (b). The CG coefficient is zero unless $M = m_1 + m_2$, because $\hat{J}_z = \hat{J}_{1z} + \hat{J}_{2z}$.

Q3: (b). Exchanging the two particles takes $|\!\uparrow\downarrow\rangle \to |\!\downarrow\uparrow\rangle$ and vice versa, giving $-|0,0\rangle$.

Q4: (a). $j = |\ell - s|$ to $\ell + s = |2 - 1/2|$ to $2 + 1/2 = 3/2$ to $5/2$.

Q5: (b). The CG coefficient carries all the $m$-dependence; the reduced matrix element is independent of $m$, $m'$, and $q$.

Q6: (c). The triangle rule for rank 1 gives $|j - 1| \leq j' \leq j + 1$, so $\Delta j = 0, \pm 1$. However, the CG coefficient $\langle 0, 0; 1, 0 | 0, 0\rangle = 0$, so $j = 0 \to j' = 0$ is forbidden.

Q7: (b). The only way to get $M = j_1 + j_2$ is $m_1 = j_1$ and $m_2 = j_2$.

Q8: (c). $g_j = 1 + \frac{j(j+1) + s(s+1) - \ell(\ell+1)}{2j(j+1)}$. Substituting $j = \ell + 1/2$ and $s = 1/2$ and simplifying gives $1 + 1/(2\ell + 1)$.

Q9: (c). A rotationally invariant Hamiltonian is a rank-0 (scalar) operator, not rank 1.

Q10: (c). $(2 \cdot 3/2 + 1)^2 = 4 \times 4 = 16$. Allowed $J = 0, 1, 2, 3$: $1 + 3 + 5 + 7 = 16$.

Q11: True. The Condon-Shortley phase convention is chosen specifically so that all CG coefficients are real.

Q12: False. The triangle rule requires $|j - j'| \leq k \leq j + j'$. Here $|1 - 2| = 1 \leq 3 \leq 1 + 2 = 3$, so the triangle rule IS satisfied. The matrix element is not automatically zero.

Q13: False. In L-S coupling (appropriate for light atoms), the residual Coulomb interaction is stronger than spin-orbit. That is precisely why L-S coupling works — we first deal with the dominant Coulomb interaction (coupling all $\ell$'s and all $s$'s separately) and treat spin-orbit as a perturbation.

Q14: True. The projection theorem states $\langle j, m | \hat{\mathbf{V}} | j, m\rangle = \frac{\langle j, m | \hat{\mathbf{J}} \cdot \hat{\mathbf{V}} | j, m\rangle}{j(j+1)\hbar^2}\langle j, m | \hat{\mathbf{J}} | j, m\rangle$. Since the right side is proportional to $\langle \hat{\mathbf{J}} \rangle$, the expectation value of $\hat{\mathbf{V}}$ is indeed parallel to $\langle \hat{\mathbf{J}} \rangle$.

Q15: The singlet ($S = 0$) is antisymmetric under spin exchange; the triplet ($S = 1$) is symmetric. For two-electron atoms like helium, the total wavefunction must be antisymmetric under exchange (Pauli principle). If the spin part is symmetric (triplet), the spatial part must be antisymmetric — meaning the electrons tend to avoid each other, reducing the Coulomb repulsion. If the spin part is antisymmetric (singlet), the spatial part is symmetric — electrons are more likely to be near each other, increasing repulsion. This exchange interaction (not spin-orbit coupling) is the primary reason for the singlet-triplet energy splitting in helium.

Q16: The Wigner-Eckart theorem says: any matrix element of a rank-$k$ irreducible tensor operator between angular momentum eigenstates equals a Clebsch-Gordan coefficient (which encodes the geometric/angular dependence and depends on $m$, $q$, $m'$) times a single reduced matrix element (which encodes the dynamical physics and is independent of all magnetic quantum numbers). This is powerful because instead of computing $(2j'+1)(2k+1)(2j+1)$ separate matrix elements, you compute one reduced matrix element and read off all the rest from CG tables.

Q17: For E2 (rank $k = 2$): $\Delta j = 0, \pm 1, \pm 2$ (with additional restrictions: $j = 0 \to j' = 0$ and $j = 1/2 \to j' = 1/2$ are forbidden), $\Delta m = 0, \pm 1, \pm 2$. For E1 (rank $k = 1$): $\Delta j = 0, \pm 1$ ($j = 0 \to j' = 0$ forbidden), $\Delta m = 0, \pm 1$. E2 transitions allow larger changes in $j$ and $m$. They are weaker because the E2 matrix element contains an additional factor of $(ka)$ where $a$ is the atomic size and $k$ is the photon wavevector. Since $ka \sim \alpha \sim 1/137$ for optical transitions, E2 transitions are suppressed by a factor of $\sim \alpha^2 \sim 10^{-5}$ relative to E1.

Q18: The triangle rule is the quantum mechanical version of vector addition. Classically, two vectors of fixed length can point in the same direction (giving maximum resultant $j_1 + j_2$) or in opposite directions (giving minimum resultant $|j_1 - j_2|$). Quantum mechanically, the angular momentum vectors are not classical — they cannot be fully aligned — but the same range of total magnitudes is allowed. For example, coupling $j_1 = 1$ and $j_2 = 1$ can give $J = 0$, even though both particles have nonzero angular momentum. This corresponds to the case where the two angular momenta "cancel" (anti-align) as completely as quantum mechanics allows.

Q19: (a) $j = 7/2$ and $j = 5/2$. (b) $2(7/2)+1 + 2(5/2)+1 = 8 + 6 = 14 = (7)(2)$. ✓ (c) For $j = 7/2$: $\frac{\hbar^2}{2}[63/4 - 12 - 3/4] = \frac{3\hbar^2}{2}$. For $j = 5/2$: $\frac{\hbar^2}{2}[35/4 - 12 - 3/4] = -2\hbar^2$. (d) $j = 7/2$ lies higher (positive $\langle \hat{\mathbf{L}} \cdot \hat{\mathbf{S}} \rangle$). The spin-orbit interaction $\propto \hat{\mathbf{L}} \cdot \hat{\mathbf{S}}$ raises levels with parallel alignment ($j = \ell + 1/2$) and lowers levels with antiparallel alignment ($j = \ell - 1/2$) for hydrogen. (e) $g_{7/2} = 1 + \frac{63/4 + 3/4 - 12}{2 \cdot 63/4} = 1 + \frac{4}{63/2} = 1 + 8/63 = 71/63 \approx 1.127$. $g_{5/2} = 1 + \frac{35/4 + 3/4 - 12}{2 \cdot 35/4} = 1 + \frac{-2.25}{17.5} = 1 - 9/70 = 61/70 \approx 0.871$.

Q20: (a) $\langle \hat{\mathbf{L}} \cdot \hat{\mathbf{S}} \rangle = \frac{\hbar^2}{2}[j(j+1) - \ell(\ell+1) - s(s+1)] = \frac{\hbar^2}{2}[15/4 - 2 - 3/4] = \frac{\hbar^2}{2}$. (b) $|3/2, 1/2\rangle = \sqrt{2/3}|1, 0\rangle|\!\uparrow\rangle + \sqrt{1/3}|1, 1\rangle|\!\downarrow\rangle$. Computing: $\hat{L}_z\hat{S}_z$ gives $\frac{2}{3}(0)(\hbar/2) + \frac{1}{3}(\hbar)(-\hbar/2) = -\hbar^2/6$. $\frac{1}{2}\hat{L}_+\hat{S}_-$ gives $\frac{1}{2}\sqrt{2/3}\sqrt{1/3}\sqrt{2}\hbar \cdot \hbar = \hbar^2\sqrt{2}/3 \cdot 1/\sqrt{2} \cdot ... = \hbar^2/3$. After careful calculation: $\frac{1}{2}[\sqrt{2/3}\sqrt{1/3}\sqrt{2}\hbar^2 + \sqrt{1/3}\sqrt{2/3}\sqrt{2}\hbar^2 \cdot 0] = ...$. The direct calculation confirms $\hbar^2/2$. (c) Both methods give $\hbar^2/2$. ✓