Past exam of the further quantum theory course of the University of Oxford / 2023 / 1

An irreducible spin representation is a nonzero Hilbert space with no proper nonzero closed invariant subspace under all three angular momentum operators. Equivalently, the associated unitary representation of SU(2) cannot be decomposed into two nonzero orthogonal invariant pieces. The irreducible possibilities are labelled by a spin $j\in \{0,\frac{1}{2},1,\frac{3}{2},\dots \}$ and have dimension $2j+1$.

Choose an orthonormal basis $|j,m_s⟩$, with $m_s=-j,-j+1,\dots ,j$. The angular momentum commutation relations imply $[{\mathbf{S}}^2,S_i]=0$, $[S_3,S_{\pm }]=\pm \hslash S_{\pm }$, $[S_{+},S_{-}]=2\hslash S_3$, and $S_{+}^{†}=S_{-}$. The complete action isIn particular, the spin raising operator annihilates $|j,j⟩$ and the spin lowering operator annihilates $|j,-j⟩$. The coefficient follows by taking the squared norm and using $S_{\mp }{S}_{\pm }={\mathbf{S}}^2-S_3^2\mp \hslash S_3$. Thus the ladder operators connect every member of the basis, while terminating at its two endpoints.

On the tensor product of spins $j_1$ and $j_2$, the combined angular momentum operator is $\mathbf{J}={\mathbf{S}}^{(1)}\otimes I+I\otimes {\mathbf{S}}^{(2)}$. The Clebsch-Gordan decomposition iswhere $J$ increases by one and each irreducible spin representation occurs once. The dimensions agree: ${\sum }_J(2J+1)=(2j_1+1)(2j_2+1)$.

The uncoupled orthonormal basis diagonalizes $S_3^{(1)},S_3^{(2)}$; the coupled orthonormal basis diagonalizes ${\mathbf{J}}^2,J_3$. The Clebsch-Gordan coefficients are their change-of-basis entries, with conventionThey vanish unless $m_1+m_2=M$. Since both bases are orthonormal, this change-of-basis matrix obeys unitarity. Overall phases of the coupled basis vectors are fixed by a convention and do not change the energy eigenvalues.

Write $|\uparrow ⟩,|\downarrow ⟩$ for the two $S_3$ eigenstates, of eigenvalues $\pm \hslash /2$. The spin-spin interaction can be evaluated usingFor $|\beta |\gg |\alpha |$, take $H_0=\beta (S_3^{(1)}-S_3^{(2)})/\hslash$ and $V=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$. The unperturbed energy eigenvalues are $\beta ,-\beta ,0,0$ on $|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩$, respectively. For an isolated normalized energy eigenstate $|n⟩$, first-order nondegenerate perturbation theory gives $\Delta E_n^{(1)}=⟨n|V|n⟩$. The two antiparallel spins have expectation $-\alpha /4$, soFor the zero-energy degenerate eigenspace, degenerate perturbation theory requires diagonalizing the restriction of $V$. On the ordered basis $(|\uparrow \uparrow ⟩,|\downarrow \downarrow ⟩)$ it is $(\alpha /4)I$: the ladder operators cannot take one parallel state to the other. Indeed neither parallel state couples to anything else under $H$, so the remaining two levels are exactly $\alpha /4$, retaining their twofold quantum degeneracy.

For $|\alpha |\gg |\beta |$, take $H_0=\alpha {\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}/{\hslash }^2$ and $V=\beta D$, where $D=(S_3^{(1)}-S_3^{(2)})/\hslash$. The spin-one-half triplet state basis and spin-one-half singlet state areThe identity ${\mathbf{S}}^{(1)}\cdot {\mathbf{S}}^{(2)}=({\mathbf{J}}^2-3{\hslash }^2/2)/2$ gives unperturbed energy eigenvalues $E_t^{(0)}=\alpha /4$ and $E_s^{(0)}=-3\alpha /4$. Direct action givesConsequently the restriction of $V$ to the entire triplet degenerate eigenspace is zero, so no first-order splitting is missed. The singlet expectation is also zero. In second-order nondegenerate perturbation theory, $\Delta E_n^{(2)}={\sum }_{r\ne n}|⟨r|V|n⟩{|}^2/(E_n^{(0)}-E_r^{(0)})$. Within the triplet, the corresponding second-order effective matrix is ${\beta }^2|t_0⟩⟨t_0|/\alpha$, already diagonal in this basis. Thus, through second order,The last two expressions have higher corrections beginning at $O({\beta }^4/|\alpha {|}^3)$, as the exact calculation in part (d) also verifies. For positive $\alpha$, coupling to the lower singlet raises the triplet's $M=0$ level and lowers the singlet; the signs are precisely the energy denominators above.

In the ordered product basis $(|\uparrow \uparrow ⟩,|\uparrow \downarrow ⟩,|\downarrow \uparrow ⟩,|\downarrow \downarrow ⟩)$, the Hamiltonian operator has matrixThe middle block has characteristic polynomial $(E+\alpha /4{)}^2-{\beta }^2-{\alpha }^2/4$. Hence the exact energy eigenvalues areThis is the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field, with its interaction and field parameters rescaled to the paper's units. For positive $\beta$ and $|\alpha |\ll \beta$, the Taylor expansion giveswhich agrees with the first-order result of part (c). For positive $\alpha$ and $|\beta |\ll \alpha$,These agree with the second-order triplet and singlet results. If $\alpha <0$, their association with the signs $\pm$ reverses; if $\beta <0$, the two antiparallel product-state labels reverse. The unordered exact spectrum and the perturbative expressions with their signed denominators remain valid.

Let $X$ exchange the two equal-spin factors, so $X^{†}=X=X^{-1}$. For the unperturbed Hamiltonian operator, addition of angular momentum givesFor nonzero $\alpha$, different $J$ have distinct energy eigenvalues, while each eigenspace has the $2J+1$ values of $M$. The stated symmetry of the Clebsch-Gordan coefficients gives the exchange parity of two coupled identical spins:However, $D=(S_3^{(1)}-S_3^{(2)})/\hslash$ changes sign: $XDX=-D$. For any two basis vectors in the same $J$ eigenspace, insertion of $X^2=I$ therefore givesThus the entire projected perturbation $P_J\beta D{P}_J$ is zero, including its off-diagonal matrix elements. Degenerate perturbation theory now proves that every first-order-in-$\beta$ energy correction vanishes. This is first-order cancellation for an exchange-odd perturbation; checking only the individual diagonal expectations would not by itself settle a degenerate problem.