2.2.3 Addition of Angular Momenta#

Worked solutions for the homework problems in the 2.2.3 Addition of Angular Momenta lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Angular momentum addition. For \(j_1 = 1\) and \(j_2 = 3/2\), list all allowed values of \(J\) using the triangle rule. Verify

\[ \sum_J (2J+1) = (2j_1+1)(2j_2+1). \]

Solution.

Allowed total angular momenta. The triangle rule states that adding \(j_1\) and \(j_2\) produces

\[ J \in \{\vert j_1 - j_2\vert,\; \vert j_1 - j_2\vert + 1,\; \ldots,\; j_1 + j_2\}, \]

with neighboring values separated by one unit. Here

\[ \vert j_1 - j_2\vert = \left\vert 1 - \tfrac32\right\vert = \tfrac12, \qquad j_1 + j_2 = 1 + \tfrac32 = \tfrac52, \]

so stepping up by one unit at a time,

\[ J \in \left\{\tfrac12,\; \tfrac32,\; \tfrac52\right\}. \]

All three values are half-integer — as they must be, since coupling an integer spin (\(j_1 = 1\)) to a half-integer spin (\(j_2 = 3/2\)) always yields a half-integer total.

Dimension check. Each \(J\)-multiplet contributes \(2J+1\) states (\(M\) running from \(-J\) to \(+J\)):

\[\begin{split} 2J+1 = \begin{cases} 2 & J = \tfrac12,\\ 4 & J = \tfrac32,\\ 6 & J = \tfrac52.\end{cases} \end{split}\]

Summing,

\[ \sum_J (2J+1) = 2 + 4 + 6 = 12. \]

The uncoupled basis \(\vert j_1, m_1\rangle\vert j_2, m_2\rangle\) has dimension

\[ (2j_1+1)(2j_2+1) = (2\cdot 1 + 1)(2\cdot\tfrac32 + 1) = 3 \times 4 = 12. \]

The two counts agree: \(12 = 12\). ✓ The coupled basis accounts for exactly every state of \(\mathcal{H}_{j_1}\otimes\mathcal{H}_{j_2}\), so the decomposition \(1 \otimes \tfrac32 = \tfrac12 \oplus \tfrac32 \oplus \tfrac52\) is complete.

2. Total angular momentum. For two spin-1/2 particles, derive the triplet and singlet states from scratch. Start from the stretched state \(\vert 1,1\rangle=\vert\uparrow\uparrow\rangle\), apply \(\hat J_-\) to obtain \(\vert 1,0\rangle\), then determine \(\vert 0,0\rangle\) by orthogonality and normalization.

Solution.

Building blocks. For a single spin-1/2, the lowering operator acts as \(\hat J_-\vert\uparrow\rangle = \hbar\vert\downarrow\rangle\) and \(\hat J_-\vert\downarrow\rangle = 0\) — the coefficient \(\hbar\sqrt{(s+m)(s-m+1)}\) equals \(\hbar\sqrt{1\cdot 1} = \hbar\) for \(s = m = \tfrac12\). The total lowering operator is \(\hat J_- = \hat J_{1-} + \hat J_{2-}\), each piece acting on its own particle. The general multiplet formula is

\[ \hat J_-\vert J,M\rangle = \hbar\sqrt{(J+M)(J-M+1)}\;\vert J,M-1\rangle. \]

Stretched state. The state \(\vert\uparrow\uparrow\rangle\) has \(M = m_1 + m_2 = +1\), the maximum possible, and is the unique state at that \(M\). It must therefore be the top of the \(J = 1\) multiplet:

\[ \vert 1,1\rangle = \vert\uparrow\uparrow\rangle. \]

Lower the triplet. Apply \(\hat J_-\) two ways. By the multiplet formula with \(J = 1\), \(M = 1\):

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

By direct action on \(\vert\uparrow\uparrow\rangle\):

\[ \hat J_-\vert\uparrow\uparrow\rangle = (\hat J_{1-} + \hat J_{2-})\vert\uparrow\uparrow\rangle = \hbar\vert\downarrow\uparrow\rangle + \hbar\vert\uparrow\downarrow\rangle. \]

Equating the two and dividing by \(\hbar\sqrt 2\),

\[ \vert 1,0\rangle = \frac{1}{\sqrt 2}\bigl(\vert\uparrow\downarrow\rangle + \vert\downarrow\uparrow\rangle\bigr). \]

Lower once more. By the formula with \(J = 1\), \(M = 0\): \(\hat J_-\vert 1,0\rangle = \hbar\sqrt{(1)(2)}\;\vert 1,-1\rangle = \hbar\sqrt 2\,\vert 1,-1\rangle\). Directly,

\[ \hat J_-\,\frac{1}{\sqrt 2}\bigl(\vert\uparrow\downarrow\rangle + \vert\downarrow\uparrow\rangle\bigr) = \frac{1}{\sqrt 2}\bigl(\hbar\vert\downarrow\downarrow\rangle + \hbar\vert\downarrow\downarrow\rangle\bigr) = \hbar\sqrt 2\,\vert\downarrow\downarrow\rangle, \]

so \(\vert 1,-1\rangle = \vert\downarrow\downarrow\rangle\). The three triplet states are

\[ \vert 1,1\rangle = \vert\uparrow\uparrow\rangle, \qquad \vert 1,0\rangle = \tfrac{1}{\sqrt2}\bigl(\vert\uparrow\downarrow\rangle + \vert\downarrow\uparrow\rangle\bigr), \qquad \vert 1,-1\rangle = \vert\downarrow\downarrow\rangle. \]

The singlet by orthogonality. The \(M = 0\) subspace is two-dimensional, spanned by \(\vert\uparrow\downarrow\rangle\) and \(\vert\downarrow\uparrow\rangle\). One combination is the triplet \(\vert 1,0\rangle\); the remaining orthogonal direction is the only candidate for \(\vert 0,0\rangle\). Write \(\vert 0,0\rangle = a\vert\uparrow\downarrow\rangle + b\vert\downarrow\uparrow\rangle\). Orthogonality to \(\vert 1,0\rangle\) requires

\[ \langle 1,0\vert 0,0\rangle = \frac{1}{\sqrt 2}(a + b) = 0, \]

which forces \(b = -a\).

Normalization gives \(\vert a\vert^2 + \vert b\vert^2 = 2\vert a\vert^2 = 1\), so \(\vert a\vert = 1/\sqrt2\). The Condon–Shortley phase convention fixes the coefficient of the highest-\(m_1\) component (\(\vert\uparrow\downarrow\rangle\), \(m_1 = +\tfrac12\)) to be positive, \(a = +1/\sqrt2\). Hence

\[ \vert 0,0\rangle = \frac{1}{\sqrt 2}\bigl(\vert\uparrow\downarrow\rangle - \vert\downarrow\uparrow\rangle\bigr). \]

Consistency check. A genuine \(J = 0\) state is annihilated by every component of \(\hat{\boldsymbol J}\). Indeed \(\hat J_z\vert 0,0\rangle = 0\) (each term has \(M = 0\)), and

\[ \hat J_-\vert 0,0\rangle = \frac{1}{\sqrt2}\bigl(\hbar\vert\downarrow\downarrow\rangle - \hbar\vert\downarrow\downarrow\rangle\bigr) = 0, \qquad \hat J_+\vert 0,0\rangle = \frac{1}{\sqrt2}\bigl(\hbar\vert\uparrow\uparrow\rangle - \hbar\vert\uparrow\uparrow\rangle\bigr) = 0. \]

Since \(\hat J^2 = \hat J_-\hat J_+ + \hat J_z^2 + \hbar\hat J_z\), all three terms vanish on \(\vert 0,0\rangle\), confirming \(\hat J^2\vert 0,0\rangle = 0\), i.e. \(J = 0\). The count is \(3 + 1 = 4 = 2\times 2\): the triplet plus the singlet exhaust the two-spin Hilbert space, \(\tfrac12\otimes\tfrac12 = 1 \oplus 0\).

3. Coupling scheme. Two identical fermions occupy the same spatial orbital \(\phi(\boldsymbol r)\). Explain why the spin state must be the singlet and why triplet spin states are forbidden.

Solution.

The antisymmetry requirement. Identical fermions obey the spin-statistics theorem: the total many-body wavefunction must be antisymmetric under exchange of all coordinates of the two particles. For two particles the full state factorizes (when spatial and spin degrees of freedom are independent) as

\[ \Psi = \Psi_{\text{space}}(\boldsymbol r_1, \boldsymbol r_2)\;\otimes\;\chi_{\text{spin}}(s_1, s_2), \]

and exchanging the particles acts as \(\hat P_{12} = \hat P_{12}^{\text{space}}\,\hat P_{12}^{\text{spin}}\). Antisymmetry of the whole means

\[ \hat P_{12}\Psi = -\Psi. \]

The spatial part is symmetric. Both fermions occupy the same orbital \(\phi\), so the spatial wavefunction is the product

\[ \Psi_{\text{space}}(\boldsymbol r_1, \boldsymbol r_2) = \phi(\boldsymbol r_1)\,\phi(\boldsymbol r_2). \]

Swapping \(\boldsymbol r_1 \leftrightarrow \boldsymbol r_2\) leaves this unchanged: \(\hat P_{12}^{\text{space}}\Psi_{\text{space}} = +\Psi_{\text{space}}\). (Two distinct orbitals could be combined symmetrically or antisymmetrically; a single shared orbital admits only the symmetric product.)

Therefore the spin part must be antisymmetric. For the product to satisfy \(\hat P_{12}\Psi = -\Psi\) while the spatial factor contributes \(+1\), the spin factor must contribute \(-1\):

\[ (+1)\times(\text{spin parity}) = -1, \]

so \(\hat P_{12}^{\text{spin}}\chi_{\text{spin}} = -\chi_{\text{spin}}\).

Identifying the spin state. From Problem 2, the two-spin-1/2 states split by exchange parity:

the triplet \(\vert 1, M\rangle\) (\(J = 1\), three states) is symmetric, \(\hat P_{12}^{\text{spin}}\vert 1,M\rangle = +\vert 1,M\rangle\);
the singlet \(\vert 0,0\rangle\) (\(J = 0\), one state) is antisymmetric, \(\hat P_{12}^{\text{spin}}\vert 0,0\rangle = -\vert 0,0\rangle\).

The only antisymmetric spin state is the singlet, so the two electrons must be spin-paired:

\[ \chi_{\text{spin}} = \vert 0,0\rangle = \tfrac{1}{\sqrt2}\bigl(\vert\uparrow\downarrow\rangle - \vert\downarrow\uparrow\rangle\bigr). \]

The three triplet states are forbidden: pairing a symmetric spatial part with a symmetric triplet gives a fully symmetric \(\Psi\), which would violate fermion antisymmetry. This is exactly the Pauli exclusion principle in its wavefunction form — and the reason the helium ground state (\(1s^2\)) has both electrons in the \(1s\) orbital locked into a spin singlet.

★ 4. Spin-orbit coupling. A particle of orbital angular momentum \(\hat{\boldsymbol L}\) and spin \(\hat{\boldsymbol S}\) (with spin quantum number \(s = 1/2\)) has a spin-orbit interaction

\[ \hat H_\text{SO} = \lambda\,\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}, \]

where \(\lambda\) is a real coupling constant of dimension energy\(/\hbar^{2}\). Let \(\hat{\boldsymbol J} = \hat{\boldsymbol L} + \hat{\boldsymbol S}\) be the total angular momentum, and let \(\vert\ell, s; j, m_j\rangle\) denote the coupled basis — the simultaneous eigenstates of \(\{\hat L^{2}, \hat S^{2}, \hat J^{2}, \hat J_z\}\) with eigenvalues \(\hbar^{2}\ell(\ell+1)\), \(\hbar^{2}s(s+1)\), \(\hbar^{2}j(j+1)\), and \(\hbar m_j\), where the orbital quantum number \(\ell = 0, 1, 2, \ldots\), the total quantum number \(j \in \{\vert\ell - s\vert,\ldots,\ell + s\}\), and the magnetic quantum number \(m_j \in \{-j,\ldots,+j\}\).

(a) Show that

\[ \hat{\boldsymbol L}\cdot\hat{\boldsymbol S} = \frac{1}{2}\bigl(\hat J^{2} - \hat L^{2} - \hat S^{2}\bigr). \]

(b) Use (a) to compute \(\langle\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}\rangle\) in the coupled state \(\vert\ell, \tfrac12; j, m_j\rangle\). Verify the answer is independent of \(m_j\).

(c) For the hydrogen \(2p\) level (\(\ell = 1\)), find the fine-structure splitting \(\Delta E\) between the \(j = 3/2\) and \(j = 1/2\) sublevels in terms of \(\lambda\).

Solution.

The operator identity. Define the total angular momentum \(\hat{\boldsymbol J} = \hat{\boldsymbol L} + \hat{\boldsymbol S}\) and square it:

\[\begin{split} \begin{split} \hat J^2 &= (\hat{\boldsymbol L} + \hat{\boldsymbol S})\cdot(\hat{\boldsymbol L} + \hat{\boldsymbol S})\\ &= \hat L^2 + \hat S^2 + \hat{\boldsymbol L}\cdot\hat{\boldsymbol S} + \hat{\boldsymbol S}\cdot\hat{\boldsymbol L}. \end{split} \end{split}\]

The orbital operator \(\hat{\boldsymbol L}\) acts on the spatial factor of the Hilbert space and the spin operator \(\hat{\boldsymbol S}\) on the spin factor; they live on different tensor slots, so \([\hat L_i, \hat S_j] = 0\) for all \(i, j\). Consequently the two cross terms are equal,

\[ \hat{\boldsymbol L}\cdot\hat{\boldsymbol S} = \sum_i \hat L_i\hat S_i = \sum_i \hat S_i\hat L_i = \hat{\boldsymbol S}\cdot\hat{\boldsymbol L}, \]

and they combine: \(\hat J^2 = \hat L^2 + \hat S^2 + 2\,\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}\). Solving,

\[ \hat{\boldsymbol L}\cdot\hat{\boldsymbol S} = \frac12\bigl(\hat J^2 - \hat L^2 - \hat S^2\bigr). \qquad\checkmark \]

Expectation in the coupled basis. The coupled states \(\vert\ell,\tfrac12; j, m_j\rangle\) are simultaneous eigenstates of \(\hat J^2\), \(\hat L^2\), and \(\hat S^2\) — so \(\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}\) is diagonal in this basis. With \(\hat J^2 \to \hbar^2 j(j+1)\), \(\hat L^2 \to \hbar^2\ell(\ell+1)\), and \(\hat S^2 \to \hbar^2 s(s+1) = \hbar^2\cdot\tfrac34\) (since \(s = \tfrac12\)),

\[ \langle\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}\rangle = \frac{\hbar^2}{2}\left[\,j(j+1) - \ell(\ell+1) - \tfrac34\,\right]. \]

Hydrogen \(2p\). The \(2p\) level has \(\ell = 1\). Coupling to \(s = \tfrac12\) gives \(j = \ell \pm \tfrac12 = \tfrac32\) or \(\tfrac12\) (the triangle rule). Evaluate the bracket for each:

\[\begin{split} \begin{split} j = \tfrac32:\quad \langle\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}\rangle &= \frac{\hbar^2}{2}\left[\tfrac32\cdot\tfrac52 - 1\cdot 2 - \tfrac34\right] = \frac{\hbar^2}{2}\cdot\frac{15 - 8 - 3}{4} = +\frac{\hbar^2}{2},\\ j = \tfrac12:\quad \langle\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}\rangle &= \frac{\hbar^2}{2}\left[\tfrac12\cdot\tfrac32 - 1\cdot 2 - \tfrac34\right] = \frac{\hbar^2}{2}\cdot\frac{3 - 8 - 3}{4} = -\hbar^2. \end{split} \end{split}\]

Fine-structure splitting. The spin-orbit Hamiltonian is \(\hat H_{SO} = \lambda\,\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}\), so the energy shift of each level is \(E_j = \lambda\,\langle\hat{\boldsymbol L}\cdot\hat{\boldsymbol S}\rangle\):

\[ E_{3/2} = +\frac{\lambda\hbar^2}{2}, \qquad E_{1/2} = -\lambda\hbar^2. \]

The splitting between the two fine-structure levels is

\[ \Delta E = E_{3/2} - E_{1/2} = \frac{\lambda\hbar^2}{2} - (-\lambda\hbar^2) = \frac{3}{2}\,\lambda\hbar^2. \]

For \(\lambda > 0\) the \(2p_{3/2}\) level lies above \(2p_{1/2}\). As a check, the degeneracy-weighted “center of gravity” is unshifted: \((2j+1)\)-weighted average \(= \tfrac{1}{6}\bigl[4\cdot(+\tfrac{\lambda\hbar^2}{2}) + 2\cdot(-\lambda\hbar^2)\bigr] = \tfrac16(2\lambda\hbar^2 - 2\lambda\hbar^2) = 0\), as it must be since \(\operatorname{Tr}\hat{\boldsymbol L}\cdot\hat{\boldsymbol S} = 0\).

★ 5. Spin-1 and spin-1/2 coupling. Consider spin-1 particle \(A\) and spin-1/2 particle \(B\), with total \(\hat{\boldsymbol J}=\hat{\boldsymbol S}_A+\hat{\boldsymbol S}_B\) and Hamiltonian

\[ \hat H=-\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B. \]

Work in the uncoupled basis \(\vert 1,m_A\rangle\vert\tfrac12,m_B\rangle\).

(a) Since \(\hat J_z\) commutes with \(\hat H\), block-diagonalize \(\hat H\) by fixed \(M=m_A+m_B\). Diagonalize each \(M\) block, then identify \(J\) from

\[ \hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B =\frac12\bigl(\hat J^2-\hat S_A^2-\hat S_B^2\bigr), \]

so each eigenvalue determines whether the state belongs to \(J=\tfrac32\) or \(J=\tfrac12\). Check that for each fixed \(J\), the allowed \(M\) values run from \(-J\) to \(J\).

(b) Build the unitary matrix \(U\) from uncoupled to coupled basis,

\[ \vert J,M\rangle=\sum_{m_A,m_B} U_{(m_A,m_B),(J,M)}\,\vert 1,m_A;\tfrac12,m_B\rangle, \]

with \(U_{(m_A,m_B),(J,M)}=\langle1,m_A;\tfrac12,m_B\vert J,M\rangle\). Explain why each fixed-\(M\) sub-block of \(U\) is exactly a Clebsch–Gordan coefficient matrix. Compute explicitly the \(M=\tfrac12\) block.

\[ \hat P_{J}=\sum_{M=-J}^{J}\vert J,M\rangle\langle J,M\vert, \qquad J\in\left\{\tfrac12,\tfrac32\right\}. \]

Using the coupled states from parts (a)–(b), verify

\[ \hat P_{1/2}+\hat P_{3/2}=\hat I, \qquad \hat P_{1/2}\hat P_{3/2}=0. \]

Then show these projectors can be written as functions of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\):

\[ \hat P_{1/2}=-\frac{2}{3}\left(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B-\frac{1}{2}\hat I\right), \qquad \hat P_{3/2}=\frac{2}{3}\left(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B+\hat I\right). \]

Solution.

Throughout this problem set \(\hbar = 1\), so spin operators are dimensionless and \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) is measured in units of \(\hbar^2\) — the convention in which the projector formulas of part (c) hold as written.

The uncoupled basis has \(3\times 2 = 6\) states \(\vert m_A\rangle\vert m_B\rangle\) with \(m_A\in\{1,0,-1\}\) and \(m_B\in\{+\tfrac12,-\tfrac12\}\). The triangle rule gives \(J\in\{\tfrac12,\tfrac32\}\), with \(2 + 4 = 6\) coupled states — the dimension matches.

(a) Block-diagonalization and diagonalization.

Since \(\hat H = -\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\), we diagonalize \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) block by block; each eigenvalue \(\mu\) of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) gives energy \(E = -\mu\) under \(\hat H\).

Since \([\hat J_z, \hat H] = 0\), states of different \(M = m_A + m_B\) do not mix; \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) is block-diagonal in \(M\). Sorting the six uncoupled states:

\(M\)	uncoupled states \(\vert m_A, m_B\rangle\)	block size
\(+\tfrac32\)	\(\vert 1,+\tfrac12\rangle\)	\(1\times 1\)
\(+\tfrac12\)	\(\vert 1,-\tfrac12\rangle,\;\vert 0,+\tfrac12\rangle\)	\(2\times 2\)
\(-\tfrac12\)	\(\vert 0,-\tfrac12\rangle,\;\vert -1,+\tfrac12\rangle\)	\(2\times 2\)
\(-\tfrac32\)	\(\vert -1,-\tfrac12\rangle\)	\(1\times 1\)

Write the coupling operator in ladder form:

\[ \hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B = \hat S_{Az}\hat S_{Bz} + \tfrac12\bigl(\hat S_{A+}\hat S_{B-} + \hat S_{A-}\hat S_{B+}\bigr). \]

The relevant ladder coefficients are \(\hat S_{A\pm}\vert m_A\rangle = \sqrt{(1\mp m_A)(1\pm m_A + 1)}\,\vert m_A\pm 1\rangle\) (so \(\hat S_{A\pm}\) acting between \(m_A = 0\) and \(m_A = \pm1\) carries a factor \(\sqrt2\)), and \(\hat S_{B\pm}\) flips \(m_B\) with coefficient \(1\).

The \(M = \pm\tfrac32\) blocks. For the stretched state \(\vert 1,+\tfrac12\rangle\) both ladder terms vanish (\(\hat S_{A+}\) and \(\hat S_{B+}\) annihilate the top weights), leaving only \(\hat S_{Az}\hat S_{Bz} = (1)(\tfrac12) = +\tfrac12\). Likewise \(\vert -1,-\tfrac12\rangle\) gives \((-1)(-\tfrac12) = +\tfrac12\). So both \(1\times 1\) blocks are already eigenstates of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) with eigenvalue \(\mu = +\tfrac12\).

The \(M = +\tfrac12\) block. In the ordered basis \(\{\vert 1,-\tfrac12\rangle,\;\vert 0,+\tfrac12\rangle\}\):

\(\hat S_{Az}\hat S_{Bz}\) is diagonal: \((1)(-\tfrac12) = -\tfrac12\) on the first state, \((0)(+\tfrac12) = 0\) on the second.
\(\tfrac12\hat S_{A+}\hat S_{B-}\) maps \(\vert 0,+\tfrac12\rangle \to \tfrac12\sqrt2\,\vert 1,-\tfrac12\rangle = \tfrac{1}{\sqrt2}\vert 1,-\tfrac12\rangle\), and annihilates \(\vert 1,-\tfrac12\rangle\).
\(\tfrac12\hat S_{A-}\hat S_{B+}\) maps \(\vert 1,-\tfrac12\rangle \to \tfrac{1}{\sqrt2}\vert 0,+\tfrac12\rangle\), and annihilates \(\vert 0,+\tfrac12\rangle\).

Collecting the matrix elements,

\[\begin{split} \hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\Big\vert_{M=1/2} = \begin{pmatrix} -\tfrac12 & \tfrac{1}{\sqrt2}\\[2pt] \tfrac{1}{\sqrt2} & 0 \end{pmatrix}. \end{split}\]

The characteristic equation for the eigenvalues \(\mu\) of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) is \(\mu^2 + \tfrac12\mu - \tfrac12 = 0\), with roots

\[ \mu = \frac{-\tfrac12 \pm \sqrt{\tfrac14 + 2}}{2} = \frac{-\tfrac12 \pm \tfrac32}{2} = \left\{+\tfrac12,\;-1\right\}. \]

The normalized eigenvectors of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) (one per \(\mu\)) are

\[ \mu = +\tfrac12:\;\; \tfrac{1}{\sqrt3}\vert 1,-\tfrac12\rangle + \sqrt{\tfrac23}\,\vert 0,+\tfrac12\rangle, \qquad \mu = -1:\;\; \sqrt{\tfrac23}\,\vert 1,-\tfrac12\rangle - \tfrac{1}{\sqrt3}\vert 0,+\tfrac12\rangle. \]

The \(M = -\tfrac12\) block. In the basis \(\{\vert 0,-\tfrac12\rangle,\;\vert -1,+\tfrac12\rangle\}\) the same computation gives

\[\begin{split} \hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\Big\vert_{M=-1/2} = \begin{pmatrix} 0 & \tfrac{1}{\sqrt2}\\[2pt] \tfrac{1}{\sqrt2} & -\tfrac12 \end{pmatrix}, \end{split}\]

with the same \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) eigenvalues \(\mu \in \{+\tfrac12,\,-1\}\) and eigenvectors

\[ \mu = +\tfrac12:\;\; \sqrt{\tfrac23}\,\vert 0,-\tfrac12\rangle + \tfrac{1}{\sqrt3}\vert -1,+\tfrac12\rangle, \qquad \mu = -1:\;\; \tfrac{1}{\sqrt3}\vert 0,-\tfrac12\rangle - \sqrt{\tfrac23}\,\vert -1,+\tfrac12\rangle. \]

Identifying \(J\). On any coupled state, \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) is a number because \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B = \tfrac12(\hat J^2 - \hat S_A^2 - \hat S_B^2)\) with \(\hat S_A^2 = 1(2) = 2\) and \(\hat S_B^2 = \tfrac12\cdot\tfrac32 = \tfrac34\):

\[ \hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B = \tfrac12 J(J+1) - \tfrac{11}{8}. \]

Inverting the two eigenvalues of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\): \(\mu = +\tfrac12 \Rightarrow J(J+1) = \tfrac{15}{4} \Rightarrow J = \tfrac32\); \(\mu = -1 \Rightarrow J(J+1) = \tfrac34 \Rightarrow J = \tfrac12\). Collecting the coupled states (Condon–Shortley convention):

\[\begin{split} \begin{split} \vert\tfrac32,+\tfrac32\rangle &= \vert 1,+\tfrac12\rangle,\\ \vert\tfrac32,+\tfrac12\rangle &= \tfrac{1}{\sqrt3}\vert 1,-\tfrac12\rangle + \sqrt{\tfrac23}\,\vert 0,+\tfrac12\rangle,\\ \vert\tfrac32,-\tfrac12\rangle &= \sqrt{\tfrac23}\,\vert 0,-\tfrac12\rangle + \tfrac{1}{\sqrt3}\vert -1,+\tfrac12\rangle,\\ \vert\tfrac32,-\tfrac32\rangle &= \vert -1,-\tfrac12\rangle, \end{split} \end{split}\]

\[\begin{split} \begin{split} \vert\tfrac12,+\tfrac12\rangle &= \sqrt{\tfrac23}\,\vert 1,-\tfrac12\rangle - \tfrac{1}{\sqrt3}\vert 0,+\tfrac12\rangle,\\ \vert\tfrac12,-\tfrac12\rangle &= \tfrac{1}{\sqrt3}\vert 0,-\tfrac12\rangle - \sqrt{\tfrac23}\,\vert -1,+\tfrac12\rangle. \end{split} \end{split}\]

The eigenvalue \(\mu = +\tfrac12\) of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) (\(J=\tfrac32\)) appears at \(M = +\tfrac32, +\tfrac12, -\tfrac12, -\tfrac32\) — exactly \(M\) from \(-J\) to \(J\) — and \(\mu = -1\) (\(J=\tfrac12\)) appears at \(M = +\tfrac12, -\tfrac12\), again \(-J\) to \(J\). ✓ The corresponding energy eigenvalues under \(\hat H = -\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) are \(E_{3/2} = -\mu = -\tfrac12\) (4-fold) and \(E_{1/2} = +1\) (2-fold).

(b) The change-of-basis matrix.

The selection rule \(M = m_A + m_B\) means a coupled state \(\vert J,M\rangle\) expands only over uncoupled states of the same \(M\). Hence \(U\), ordered by \(M\), is block-diagonal: the \(M = \pm\tfrac32\) sectors are \(1\times 1\) and the \(M = \pm\tfrac12\) sectors are \(2\times 2\). Each block’s entries are by definition the coefficients \(\langle 1,m_A;\tfrac12,m_B\vert J,M\rangle\) for fixed \(M\) — i.e. precisely the Clebsch–Gordan coefficients. Because both the uncoupled and coupled states within a fixed-\(M\) subspace are orthonormal bases of that subspace, each block is a (real, orthogonal) unitary matrix; that is the content of “each fixed-\(M\) sub-block is a CG matrix.”

For \(M = +\tfrac12\), rows labelled by uncoupled \((m_A, m_B) \in \{(1,-\tfrac12),\,(0,+\tfrac12)\}\) and columns by coupled \((J,M) \in \{(\tfrac32,\tfrac12),\,(\tfrac12,\tfrac12)\}\), the eigenvectors from part (a) give

\[\begin{split} U^{(M=1/2)} = \begin{pmatrix} \langle 1,1;\tfrac12,-\tfrac12\vert\tfrac32,\tfrac12\rangle & \langle 1,1;\tfrac12,-\tfrac12\vert\tfrac12,\tfrac12\rangle\\[4pt] \langle 1,0;\tfrac12,\tfrac12\vert\tfrac32,\tfrac12\rangle & \langle 1,0;\tfrac12,\tfrac12\vert\tfrac12,\tfrac12\rangle \end{pmatrix} = \begin{pmatrix} \sqrt{\tfrac13} & \sqrt{\tfrac23}\\[4pt] \sqrt{\tfrac23} & -\sqrt{\tfrac13} \end{pmatrix}. \end{split}\]

Its columns are orthonormal and \(\det U^{(M=1/2)} = -\tfrac13 - \tfrac23 = -1\), confirming it is orthogonal. (The \(M = +\tfrac32\) block is the \(1\times 1\) matrix \((1)\); the \(M = -\tfrac12\) and \(M = -\tfrac32\) blocks follow identically from the corresponding states above.)

(c) The projectors.

Completeness, \(\hat P_{1/2} + \hat P_{3/2} = \hat I\). The six coupled states listed in (a) are mutually orthonormal: within each \(2\times 2\) block they are orthonormal by construction (eigenvectors of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) in that \(M\) sector), and states from different \(M\) blocks are automatically orthogonal because \(\hat J_z\) has different eigenvalues. Six orthonormal vectors span the entire 6-dimensional space, so they form a complete basis:

\[ \sum_{J,M}\vert J,M\rangle\langle J,M\vert = \hat I. \]

Splitting the sum into its \(J = \tfrac32\) and \(J = \tfrac12\) parts is exactly \(\hat P_{3/2} + \hat P_{1/2} = \hat I\). ✓

Orthogonality, \(\hat P_{1/2}\hat P_{3/2} = 0\). Using \(\langle\tfrac12,M\vert\tfrac32,M'\rangle = 0\) (coupled states with different \(J\) are orthogonal),

\[ \hat P_{1/2}\hat P_{3/2} = \sum_{M,M'}\vert\tfrac12,M\rangle\underbrace{\langle\tfrac12,M\vert\tfrac32,M'\rangle}_{=\,0}\langle\tfrac32,M'\vert = 0. \qquad\checkmark \]

So \(\hat P_{1/2}\) and \(\hat P_{3/2}\) are complementary orthogonal projectors, as projectors onto the two coupled subspaces must be.

Projectors as functions of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\). Part (a) showed \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) is constant on each coupled subspace, with eigenvalue \(\mu = +\tfrac12\) on the \(J = \tfrac32\) space and \(\mu = -1\) on the \(J = \tfrac12\) space. Its spectral decomposition is therefore

\[ \hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B = \tfrac12\,\hat P_{3/2} - 1\cdot\hat P_{1/2}. \]

Together with the completeness relation \(\hat I = \hat P_{3/2} + \hat P_{1/2}\), this is a \(2\times 2\) linear system for the two projectors. Adding the two equations,

\[ \hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B + \hat I = \bigl(\tfrac12 + 1\bigr)\hat P_{3/2} + (-1 + 1)\hat P_{1/2} = \tfrac32\,\hat P_{3/2}, \]

so \(\hat P_{3/2} = \tfrac23\bigl(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B + \hat I\bigr)\).

Forming instead \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B - \tfrac12\hat I\),

\[ \hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B - \tfrac12\hat I = \bigl(\tfrac12 - \tfrac12\bigr)\hat P_{3/2} + \bigl(-1 - \tfrac12\bigr)\hat P_{1/2} = -\tfrac32\,\hat P_{1/2}, \]

so \(\hat P_{1/2} = -\tfrac23\bigl(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B - \tfrac12\hat I\bigr)\).

These are exactly the stated formulas. As a check, evaluate them on each subspace: on \(J = \tfrac32\) (\(\mu = +\tfrac12\)), \(\hat P_{3/2} = \tfrac23(\tfrac12 + 1) = 1\) and \(\hat P_{1/2} = -\tfrac23(\tfrac12 - \tfrac12) = 0\); on \(J = \tfrac12\) (\(\mu = -1\)), \(\hat P_{3/2} = \tfrac23(-1 + 1) = 0\) and \(\hat P_{1/2} = -\tfrac23(-1 - \tfrac12) = 1\). Each projector is \(1\) on its own subspace and \(0\) on the other. ✓ The construction works because \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) has only two distinct eigenvalues \(\mu \in \{+\tfrac12,-1\}\), so any function of \(\hat{\boldsymbol S}_A\cdot\hat{\boldsymbol S}_B\) — in particular each projector — is fixed by its two values via linear (Lagrange) interpolation.

6. Lande g-factor. An atom has orbital angular momentum \(\ell\) and spin \(s\), coupling to total \(j = \vert\ell - s\vert, \ldots, \ell + s\). In a weak external magnetic field \(B\) along \(\boldsymbol{e}_z\), the magnetic energy is

\[ \hat H_Z = -\boldsymbol{\hat\mu}\cdot\boldsymbol B = -\bigl(g_L\hat L_z + g_S\hat S_z\bigr)\mu_B B/\hbar, \]

with \(g_L = 1\), \(g_S = 2\) (electron values), and \(\mu_B = e\hbar/(2m_e)\) the Bohr magneton.

(a) In the coupled basis \(\vert\ell, s; j, m_j\rangle\), the operator \(\hat L_z + 2\hat S_z = \hat J_z + \hat S_z\) is not diagonal in \(j\) alone (it mixes different \(j\) values within the same \(m_j\)). Within a fixed-\(j\) subspace, any vector operator \(\hat{\boldsymbol V}\) has matrix elements proportional to those of \(\hat{\boldsymbol J}\):

\[ \langle j, m_j\vert\hat S_z\vert j, m_j\rangle = \frac{\langle\hat{\boldsymbol J}\cdot\hat{\boldsymbol S}\rangle}{\hbar^2 j(j+1)}\,\hbar m_j. \]

This is the vector-projection identity: within a fixed-\(j\) subspace, the expectation of any vector operator equals the component along \(\hat{\boldsymbol J}\), scaled by \(m_j/j(j+1)\). (The perpendicular components average to zero by axial symmetry about \(\hat{\boldsymbol J}\) — cf. 2.2.1 P5 for the same averaging on \(\hat J_x\), \(\hat J_y\).)

(b) Using \(\hat{\boldsymbol J}\cdot\hat{\boldsymbol S} = \tfrac{1}{2}(\hat J^2 - \hat L^2 + \hat S^2)\) (derive this by squaring \(\hat{\boldsymbol L} = \hat{\boldsymbol J} - \hat{\boldsymbol S}\)), show

\[ \langle\hat{\boldsymbol J}\cdot\hat{\boldsymbol S}\rangle = \frac{\hbar^2}{2}\bigl[j(j+1) - \ell(\ell+1) + s(s+1)\bigr]. \]

\[ g_J = 1 + \frac{j(j+1) + s(s+1) - \ell(\ell+1)}{2j(j+1)}, \]

so that the Zeeman energy shift in the coupled basis is \(\Delta E = g_J\mu_B B m_j\).

(d) Evaluate \(g_J\) for hydrogen \(2p_{3/2}\) (\(\ell = 1, s = 1/2, j = 3/2\)) and \(2p_{1/2}\) (\(\ell = 1, s = 1/2, j = 1/2\)). Show \(g_J(2p_{3/2}) = 4/3\) and \(g_J(2p_{1/2}) = 2/3\). The two fine-structure levels Zeeman-split by different amounts under the same field — the experimental signature that distinguishes spin from orbital angular momentum.

Solution.

(a) The vector-projection identity says that within a fixed-\(j\) subspace, any vector operator \(\hat{\boldsymbol V}\) (one that transforms as a 3-vector under rotations) has matrix elements proportional to those of \(\hat{\boldsymbol J}\):

\[ \langle j, m_j\vert\hat V_z\vert j, m_j\rangle = \frac{\langle\hat{\boldsymbol J}\cdot\hat{\boldsymbol V}\rangle_j}{\hbar^2 j(j+1)}\,\hbar m_j, \]

where \(\langle\hat{\boldsymbol J}\cdot\hat{\boldsymbol V}\rangle_j\) is the (state-independent) scalar overlap on the fixed-\(j\) subspace. The intuition: \(\hat{\boldsymbol J}\) generates rotations of the \(j\)-multiplet, and any vector operator’s expectation values are determined by its projection onto \(\hat{\boldsymbol J}\) — perpendicular components average to zero by axial symmetry (cf. 2.2.1 P5). Applied to \(\hat{\boldsymbol V} = \hat{\boldsymbol S}\):

\[ \langle\hat S_z\rangle_{j, m_j} = \frac{\langle\hat{\boldsymbol J}\cdot\hat{\boldsymbol S}\rangle_j}{\hbar^2 j(j+1)}\,\hbar m_j. \]

(b) Square the operator identity \(\hat{\boldsymbol L} = \hat{\boldsymbol J} - \hat{\boldsymbol S}\):

\[ \hat L^2 = \hat J^2 + \hat S^2 - 2\hat{\boldsymbol J}\cdot\hat{\boldsymbol S}, \]

so \(\hat{\boldsymbol J}\cdot\hat{\boldsymbol S} = \tfrac{1}{2}\bigl(\hat J^2 - \hat L^2 + \hat S^2\bigr)\).

On the coupled basis \(\vert\ell, s; j, m_j\rangle\) — simultaneous eigenstates of \(\hat J^2, \hat L^2, \hat S^2\) — the expectation value is

\[ \langle\hat{\boldsymbol J}\cdot\hat{\boldsymbol S}\rangle = \tfrac{\hbar^2}{2}\bigl[j(j+1) - \ell(\ell+1) + s(s+1)\bigr]. \quad\checkmark \]

(c) The Zeeman Hamiltonian rewrites as \(\hat H_Z = -(\hat J_z + \hat S_z)\mu_B B/\hbar\) (using \(g_L = 1, g_S = 2\)). On the coupled state \(\vert\ell, s; j, m_j\rangle\):

\[ \langle\hat H_Z\rangle = -\bigl(\langle\hat J_z\rangle + \langle\hat S_z\rangle\bigr)\mu_B B/\hbar = -\bigl(\hbar m_j + \langle\hat S_z\rangle\bigr)\mu_B B/\hbar. \]

Substitute \(\langle\hat S_z\rangle\) from (a) and (b):

\[ \langle\hat S_z\rangle = \frac{\hbar^2/2\cdot[j(j+1) - \ell(\ell+1) + s(s+1)]}{\hbar^2 j(j+1)}\hbar m_j = \frac{j(j+1) + s(s+1) - \ell(\ell+1)}{2j(j+1)}\hbar m_j. \]

Then

\[ \Delta E_Z = -\langle\hat H_Z\rangle = \mu_B B m_j\left[1 + \frac{j(j+1) + s(s+1) - \ell(\ell+1)}{2j(j+1)}\right] = g_J\mu_B B m_j, \]

with

\[ \boxed{g_J = 1 + \frac{j(j+1) + s(s+1) - \ell(\ell+1)}{2j(j+1)}.} \]

(d) Substitute \(\ell = 1\), \(s = 1/2\) for hydrogen \(2p\).

\(2p_{3/2}\) (\(j = 3/2\)): \(j(j+1) = 15/4\), \(s(s+1) = 3/4\), \(\ell(\ell+1) = 2\). Numerator: \(15/4 + 3/4 - 2 = 18/4 - 2 = 5/2\). Denominator: \(2 \cdot 15/4 = 15/2\). Ratio: \((5/2)/(15/2) = 1/3\).

\[ g_J(2p_{3/2}) = 1 + 1/3 = 4/3. \]

\(2p_{1/2}\) (\(j = 1/2\)): \(j(j+1) = 3/4\), \(s(s+1) = 3/4\), \(\ell(\ell+1) = 2\). Numerator: \(3/4 + 3/4 - 2 = 3/2 - 2 = -1/2\). Denominator: \(2 \cdot 3/4 = 3/2\). Ratio: \((-1/2)/(3/2) = -1/3\).

\[ g_J(2p_{1/2}) = 1 - 1/3 = 2/3. \]

Physical interpretation. Even though \(2p_{3/2}\) and \(2p_{1/2}\) have the same \(\ell\) and \(s\), they Zeeman-split by different amounts under the same applied field — \(g_J\) depends on \(j\), not on \(\ell\) and \(s\) separately. Hydrogen \(2p_{3/2}\) splits twice as fast as \(2p_{1/2}\) (\(g_J = 4/3\) vs \(2/3\)). This is the anomalous Zeeman effect for atoms with \(L \neq 0\) and \(S \neq 0\) — the signature that orbital and spin angular momenta combine via the vector-projection identity, not as separate additive contributions. Experimentally, the line splitting patterns of \(2p_{3/2} \to 1s_{1/2}\) and \(2p_{1/2} \to 1s_{1/2}\) transitions under an external field encode the values of \(g_J\) for each level — historically how spin was first inferred from atomic spectra.

7. Two-electron exchange interaction. Two electrons interact through the exchange Hamiltonian \(\hat H_{\mathrm{ex}} = -J\,\hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2\) (with \(J\) the exchange coupling, \(\hat S = \hat\sigma/2\) in units of \(\hbar\)). This is the simplest two-spin Heisenberg model — the same operator appearing in 2.1.1 Problem 6 as the lattice Heisenberg interaction.

(a) Express \(\hat H_{\mathrm{ex}}\) in terms of the total spin operators using \(\hat S_{\mathrm{tot}}^2 = \hat S_1^2 + \hat S_2^2 + 2\hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2\).

(b) Evaluate \(\hat H_{\mathrm{ex}}\) on the singlet \(\vert 0,0\rangle\) and on each triplet state \(\vert 1, M\rangle\). Show that the triplet has energy \(-J/4\) and the singlet has energy \(+3J/4\).

\[ \Delta E_{\mathrm{st}} = E_{\mathrm{singlet}} - E_{\mathrm{triplet}} = J. \]

For \(J > 0\) (ferromagnetic coupling), the triplet is the ground state — aligned spins are favoured. For \(J < 0\) (antiferromagnetic coupling), the singlet is the ground state — anti-aligned spins. Compare with the lattice Heisenberg model spectrum from 2.1.1 P6.

(d) Connect to two-electron chemistry. The Hund’s rule for filled subshells favours maximal \(S\) (triplet > singlet). Argue that this reflects an effectively ferromagnetic exchange between two electrons in the same orbital, originating from the electron-electron Coulomb repulsion combined with the antisymmetry of the spatial wavefunction (recall 2.2.3 Problem 3).

Solution.

(a) From \(\hat S_{\mathrm{tot}}^2 = \hat S_1^2 + \hat S_2^2 + 2\hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2\), solve for the dot product:

\[ \hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2 = \tfrac{1}{2}\bigl(\hat S_{\mathrm{tot}}^2 - \hat S_1^2 - \hat S_2^2\bigr). \]

\[ \hat H_{\mathrm{ex}} = -J\hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2 = -\tfrac{J}{2}\bigl(\hat S_{\mathrm{tot}}^2 - \hat S_1^2 - \hat S_2^2\bigr). \]

In units where \(\hbar = 1\), \(\hat S_1^2 = \hat S_2^2 = 3/4\) (each is spin-1/2 with \(s(s+1) = 3/4\)), so

\[ \hat H_{\mathrm{ex}} = -\tfrac{J}{2}\hat S_{\mathrm{tot}}^2 + \tfrac{J}{2}\cdot\tfrac{3}{2} = -\tfrac{J}{2}\hat S_{\mathrm{tot}}^2 + \tfrac{3J}{4}. \]

(b) On the triplet \(\vert 1, M\rangle\), \(\hat S_{\mathrm{tot}}^2 = 1\cdot 2 = 2\):

\[ E_{\mathrm{triplet}} = -\tfrac{J}{2}\cdot 2 + \tfrac{3J}{4} = -J + \tfrac{3J}{4} = -\tfrac{J}{4}. \]

On the singlet \(\vert 0, 0\rangle\), \(\hat S_{\mathrm{tot}}^2 = 0\):

\[ E_{\mathrm{singlet}} = 0 + \tfrac{3J}{4} = +\tfrac{3J}{4}. \]

Both triplet states have the same energy (degenerate, 3-fold) and the singlet sits at a different energy. \(\checkmark\)

\[ \Delta E_{\mathrm{st}} = E_{\mathrm{singlet}} - E_{\mathrm{triplet}} = \tfrac{3J}{4} - (-\tfrac{J}{4}) = J. \]

For \(J > 0\) (the sign convention where \(-J\hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2\) favours aligned spins): \(E_{\mathrm{triplet}} < E_{\mathrm{singlet}}\), the triplet is the ground state, ferromagnetic coupling. For \(J < 0\): signs reverse, singlet is the ground state, antiferromagnetic coupling.

Comparison with 2.1.1 P6. The lattice Heisenberg Hamiltonian \(\hat H = J(\hat X\otimes\hat X + \hat Y\otimes\hat Y + \hat Z\otimes\hat Z) = 4J\hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2\) (the factor of \(4\) comes from \(\hat X = 2\hat S_x\) etc.) has eigenvalues \(J\{+1, +1, +1, -3\}\) on the four two-qubit states. Translating to the present sign convention \(\hat H_{\mathrm{ex}} = -J\hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2\), the triplet sits at \(-J/4\) (triply degenerate) and singlet at \(+3J/4\) — exactly the singlet-triplet structure 2.1.1 P6 found, with the sign and prefactor adjusted by the change of normalization. The two problems describe the same physics from two perspectives: 2.1.1 P6 computed the Heisenberg Hamiltonian’s spectrum directly in the tensor-product basis; here we use angular-momentum addition to reach the same result through coupled \(\vert J, M\rangle\) states.

(d) Hund’s rules and exchange. In a multi-electron atom, when two electrons occupy degenerate orbitals (e.g., two electrons in the three \(2p\) orbitals of nitrogen), Hund’s first rule says the lowest energy configuration has maximum total spin \(S\). This is empirically a ferromagnetic preference for aligned spins.

The origin is electrostatic, not magnetic. The Pauli principle (2.2.3 P3) requires the total many-electron wavefunction to be antisymmetric. For two electrons:

A symmetric spatial wavefunction (e.g., both in the same orbital) pairs with the antisymmetric spin singlet.
An antisymmetric spatial wavefunction (e.g., distributed over two orthogonal orbitals with the \((\phi_1\phi_2 - \phi_2\phi_1)/\sqrt 2\) combination) pairs with the symmetric spin triplet.

The antisymmetric spatial wavefunction keeps the two electrons spatially apart (it vanishes at \(\boldsymbol r_1 = \boldsymbol r_2\)), reducing the Coulomb repulsion. So the triplet — paired with antisymmetric spatial part — has lower Coulomb energy than the singlet. The effective spin Hamiltonian \(\hat H_{\mathrm{ex}}\) then has \(J > 0\) (ferromagnetic), even though the underlying interaction is the spin-independent Coulomb repulsion.

The exchange interaction is therefore not a magnetic interaction at all — it is the Coulomb repulsion projected through the antisymmetry constraint into an effective spin-dependent term. This is the fundamental origin of magnetism in atoms (Hund’s rules) and solids (Heisenberg ferromagnets and antiferromagnets), and the reason a single algebraic identity — \(\hat{\boldsymbol S}_1\cdot\hat{\boldsymbol S}_2 = \tfrac{1}{2}(\hat S_{\mathrm{tot}}^2 - \hat S_1^2 - \hat S_2^2)\) — controls so much of magnetic ordering physics.