Past exam of the advanced quantum physics course of the University of Cambridge / 2022 / Section B / 5

The unperturbed infinite square well is invariant under the parity operator. Its one-dimensional energy eigenvalues are nondegenerate, so each energy eigenstate has definite parity. Its position expectation value therefore vanishes: $|{\varphi }_n(x){|}^2$ is even, while $x$ is odd. The first-order nondegenerate perturbation theory correction is consequentlyThe electric field does mix opposite-parity quantum states, and second-order nondegenerate perturbation theory contains two factors of $eEx$. Its leading contribution is thus proportional to $E^2$: this is a quadratic Stark effect. More generally, applying the parity operator changes $H(E)$ into $H(-E)$, so their corresponding energy eigenvalues agree. An analytic expansion about zero electric field contains only even powers of $E$. We assume $e|E|a$ is small compared with the unperturbed excitation scale.

Use zero-based labels to agree with ${\varphi }_0$ in the later parts. The normalized infinite square well energy eigenstates and energy eigenvalues areThe position matrix elements in a symmetric infinite square well, obtained by integrating the product of the two wavefunctions against $x$, areThe integral in the question uses positive integer labels; replacing them by $k+1,n+1$ is essential here. Only opposite-parity states, $k+n$ odd, contribute, as also follows from the parity operator selection rule.

Second-order nondegenerate perturbation theory now gives the requested infinite sum:In fully explicit form this isThe symbol $m$ here denotes the electron mass, not a summation index. For the ground state $n=0$, every denominator in the first form is negative, so the ground energy level moves downward. For excited states, contributions from lower and higher energy levels have opposite signs.

The electron's electric dipole moment is $\mathbf{d}=-e⟨x⟩\hat{\mathbf{x}}$. It vanishes in the unperturbed ground energy eigenstate. The first correction to the wavefunction is linear in the electric field and has the opposite parity; its interference with the unperturbed state therefore gives a nonzero position expectation value of first order.

In detail, the normalized ground quantum state, to the order needed, isThe normalization correction starts at second order and does not affect the linear term below. The two cross terms giveThe absence of an $E^2$ term follows from the parity operator: $⟨x⟩$ changes sign under field reversal. Comparing with the second-order nondegenerate perturbation theory energy sum yieldsThus the induced electric dipole moment points along the electric field, with positive electric polarizability. The induced dipole moment and quadratic energy shift obey $\delta {ϵ}_0^{(2)}=-{\alpha }_{\mathrm{pol}}{E}^2/2$. This factor of one half accounts for the energy needed to induce the dipole. As a check, the Hellmann–Feynman theorem gives $d_x=-\partial {ϵ}_0/\partial E$, with the same result.

Normalize the trial quantum state as $(|{\varphi }_0⟩+c|{\varphi }_1⟩)/\sqrt{1+c^2}$. With the phases used above, the off-diagonal position matrix element is $x_{01}=-32a/(9{\pi }^2)$. DefineThe diagonal perturbation matrix elements vanish by parity, so the Rayleigh quotient isThe stationarity equation is $W+\Delta c-W{c}^2=0$. The minimum continuously connected to the unperturbed ground energy eigenstate hasEquivalently, diagonalize the Hamiltonian operator matrix in this two-dimensional trial space. Its lower eigenvalue isThereforeThis retains exactly the $k=1$ term of the full ground-state second-order nondegenerate perturbation theory sum. Every omitted term is negative, so the full quadratic correction is more negative than this restricted one. This is consistent with the Rayleigh-Ritz variational principle: minimizing over a two-state subspace gives an upper bound on the exact ground energy. The bound applies to the unexpanded two-state minimum; a truncated series is to be compared order by order.