Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 4 / 59

Let $H^{(0)}|n⟩=E_n^{(0)}|n⟩$ with a nondegenerate reference eigenvalue, and choose intermediate normalization $⟨n|n^{(1)}⟩=0$. The first-order equation isProjection on $|j⟩$, $j\ne n$, gives $⟨j|n^{(1)}⟩=H_{jn}^{(1)}/(E_n^{(0)}-E_j^{(0)})$. The second-order projected equation gives $E_n^{(2)}=⟨n|H^{(1)}|n^{(1)}⟩$, henceThe sum runs over a complete orthonormal reference basis except the reference state; continuum states require integration if present. Degenerate states with zero denominators require degenerate perturbation theory first. These are expansion coefficients in $E_n=E_n^{(0)}+\lambda E_n^{(1)}+{\lambda }^2{E}_n^{(2)}+\cdots$.

The electric potential is $\epsilon (x-a/2)$, so inside the boxThe infinite walls impose zero wavefunction at $0,a$. The normalized particle in a box states are ${\psi }_n(x)=\sqrt{2/a}\sin (n\pi x/a)$ and $E_n^{(0)}=n^2{h}^2/(8m{a}^2)$ for positive integers $n$.

The probability density $|{\psi }_n(x){|}^2$ is symmetric about $a/2$, while $x-a/2$ is antisymmetric. ThereforeThe use of a midpoint-zero electric potential matters: adding a constant potential would add the corresponding uniform energy shift.

Reflection $x\mapsto a-x$ preserves the walls and kinetic operator but sends $\lambda (x-a/2)$ to $-\lambda (x-a/2)$. The Hamiltonians at $\lambda$ and $-\lambda$ are unitarily equivalent, so each analytic nondegenerate energy branch obeys $E_n(\lambda )=E_n(-\lambda )$. Its Taylor expansion contains only even powers: all odd-order energy coefficients vanish, not merely the first-order expectation.

The midpoint-odd perturbation couples the ground state only to even-index states. Direct integration givesInsertion in the second-order expression yieldsIt is negative because all coupled states lie above the ground state. This is the coefficient of ${\lambda }^2$; the physical energy shift includes the additional factor $q^2{\epsilon }^2$.

With the supplied sum, the second-order approximation isFor an electron in the $1\mu \mathrm{m}$ box, $E_1^{(0)}=6.02\times 10^{-26}\mathrm{J}$ and the correction is $-4.61\times 10^{-27}\mathrm{J}$, giving $5.56\times 10^{-26}\mathrm{J}=3.47\times 10^{-7}\mathrm{eV}$.

For $10\mu \mathrm{m}$, the formal result is approximately $-4.61\times 10^{-23}\mathrm{J}=-2.88\times 10^{-4}\mathrm{eV}$. It is not a credible approximation: even the minimum electric potential energy in that box is only $-5\mu \mathrm{eV}$, whereas this answer is much lower. The ground-to-first-excited-state mixing estimate increases from $0.160$ to $160$ when the length grows tenfold. The shift grows as $a^4$ while the level spacing decreases as $a^{-2}$, so the larger box is far outside the perturbative regime.

The first excited unperturbed level is twofold degenerate, spanned by $|1,2⟩$ and $|2,1⟩$, with energy $5h^2/(8m{a}^2)$. The perturbation is $q{\epsilon }_Q(x-a/2)(y-a/2)$. Both diagonal matrix elements vanish, but the off-diagonal element does not:Diagonalizing the $2\times 2$ matrix with zero diagonal gives first-order shiftswith symmetric/antisymmetric combinations $(|1,2⟩\pm |2,1⟩)/\sqrt{2}$. Their order depends on the sign of $q{\epsilon }_Q$; the lower branch has shift $-|W_{12,21}|$. Applying nondegenerate perturbation theory to either product state would miss the splitting.