Past exam of the advanced quantum physics course of the University of Cambridge / 2021 / Section B / 6

Inside the infinite square well, the orbital and spin operators act on different factors of the tensor product. Write $\mathbf{S}={\mathbf{S}}_1+{\mathbf{S}}_2$. The identityshows that the exchange interaction commutes with $S^2$ and $S_z$. The sum of the one-particle kinetic energies also commutes with them and with particle exchange. Thus one may use orbital quantum numbers together with the total-spin sector $S=0,1$ and projection $M=-S,\dots ,S$, with physical $S_z=\hslash M$.

For identical spin one-half particles the Pauli exclusion principle further restricts the states. The spin-one-half singlet state is antisymmetric in spin and must multiply a symmetric spatial wavefunction; a spin-one-half triplet state must multiply an antisymmetric spatial wavefunction. Thus $n_1,n_2$ label unordered orbital occupations in symmetry-adapted states, not arbitrary ordered product states. This gives the labels $|n_1,n_2,S,M⟩$, with the paper's last label expressed in units of $\hslash$.

LetThe normalized orbital energy eigenstates vanish outside the well and have one-particle energies $n^2\epsilon$. Define the normalized symmetric and antisymmetric combinations for $r\ne s$ byThe normalized spin functions are the singlet state ${\chi }_{00}=(|\uparrow \downarrow ⟩-|\downarrow \uparrow ⟩)/\sqrt{2}$ and the triplet states ${\chi }_{11}=|\uparrow \uparrow ⟩$, ${\chi }_{10}=(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$, ${\chi }_{1,-1}=|\downarrow \downarrow ⟩$. Their exchange energies are respectively $3J/4$ and $-J/4$.

For $0<J\ll \epsilon$, the three lowest distinct energy levels of the exchange-coupled spin-one-half particles in an infinite square well areThe spatially doubly occupied orbital can only occur with the singlet state. The next orbital sum is $1^2+2^2=5$, where either exchange symmetry is possible. The next unused sum is $2^2+2^2=8$, well separated under the small-coupling assumption. The paper's inequality is equivalent, up to a numerical factor, to $|J|\ll \epsilon$; the factor ${\hslash }^2$ in $J$ is essential because the spin operators carry units of angular momentum.

The sign of $\lambda$ is not explicitly specified. The table assumes the usual intended positive coupling in the displayed minus-sign convention. For $J<0$ with $|J|\ll \epsilon$, the same three sets of eigenfunctions occur, but the two rows at $5\epsilon$ exchange order. At $J=0$ those rows merge into a single energy level of multiplicity four; if three distinct energies are requested exactly at that point, the third is $8\epsilon$ with the single state ${\Phi }_{22}^{+}{\chi }_{00}$.

Each one-particle energy eigenstate has definite parity about the centre of the well, so $⟨n|x|n⟩=0$. Consequently, a doubly occupied orbital has $⟨x_1{x}_2⟩=0$. For distinct orbitals, the two off-diagonal exchange terms instead survive:The position matrix elements in a symmetric infinite square well, with the present width $a$, giveThe sign of $x_{12}$ depends on the orbital phase convention, while $D$ does not. The perturbation acts only on the spatial part and cannot mix orthogonal singlet states and triplet states. Within a triplet it is proportional to the identity on the spin factor. Thus first-order nondegenerate perturbation theory, with this trivial diagonal treatment of spin degeneracy, gives the shifts corresponding to the three rows in part (b):These formulae also apply for $J<0$, with the reordered levels described above. At $J=0$, they resolve the fourfold $5\epsilon$ level into a shifted triplet and singlet. The next doubly occupied orbital ${\Phi }_{22}^{+}$ still has zero first-order shift. The perturbative requirement is $|\alpha |a^2\ll \epsilon$ for mixing with remote orbital states; the spin-independent perturbation does not require division by the small singlet–triplet gap.

For distinguishable particles, the ordered spatial products are independent states, and neither singlet states nor triplet states are excluded from any orbital occupation. The unperturbed energy eigenvalues areFor $0<J\ll \epsilon$, the three lowest distinct energy levels arewhere $M=-1,0,1$. The first two levels do have zero first-order correction, since their spatial expectation value factorizes as $⟨1|x|1{⟩}^2=0$.

The last instruction in the original PDF is false for the third level. Although every diagonal product-state matrix element vanishes, the two orbital states have degenerate energy levels. Degenerate perturbation theory requires diagonalizing the perturbation within their span. For each $M$, its matrix in the ordered basis $(u_1{u}_2,u_2{u}_1)$ isThe eigenfunctions are ${\Phi }_{12}^{\pm }{\chi }_{1M}$, withThus the sixfold unperturbed level splits into two triplets. Its average shift is zero, and the diagonal expectation in an individual ordered product state is zero, but neither fact makes all its energy eigenvalue corrections zero. This is the degenerate orbital exchange splitting from a product perturbation; distinguishability allows both combinations rather than removing their coupling.

For negative $J$, the lowest two rows exchange order and the third unperturbed level is instead the twofold singlet orbital pair at $5\epsilon +3J/4$. It splits into ${\Phi }_{12}^{\pm }{\chi }_{00}$ with the same shifts $\pm \alpha D$, each nondegenerate. At $J=0$, the three lowest distinct unperturbed energies are $2\epsilon ,5\epsilon ,8\epsilon$, with multiplicities $4,8,4$. The first and third have zero first-order shift; the middle splits by $\pm \alpha D$, each fourfold. The printed zero-shift assertion therefore fails also in the zero-exchange limit.