Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 4 / 30A (Principles of Quantum Mechanics)

The interaction picture removes the evolution generated by $H_0$: $|{\psi }_I(t)⟩=e^{i{H}_{0}t/\hslash }|{\psi }_S(t)⟩$. Differentiating the Schrodinger equation gives $i\hslash {\partial }_t|{\psi }_I⟩=\lambda V_I(t)|{\psi }_I⟩$, with $V_I=e^{i{H}_{0}t/\hslash }V(t)e^{-i{H}_{0}t/\hslash }$. The integral equation, iterated once, givesDistinct-energy eigenstates are orthogonal. Their transition amplitude is therefore $-i\lambda {\hslash }^{-1}{\int }_0^t⟨\varphi |V(s)|\chi ⟩e^{i(E^{\prime }-E)s/\hslash }ds+O({\lambda }^2)$. Squaring its modulus givesThe first-order amplitude produces a second-order probability.

For the isotropic quantum harmonic oscillator, parity makes $\int x_i{x}_3|\chi {|}^2{d}^{3}x=0$ for $i=1,2$. Rotational symmetry and the given radial integral give $\int x_3^2|\chi {|}^2{d}^{3}x=\hslash /(2m\omega )$. Thus $⟨{\varphi }_i|{\hat{x}}_3|\chi ⟩={\delta }_{i3}\sqrt{\hslash /(2m\omega )}$, and the energy gap is $\hslash \omega$. The time integral is ${\int }_0^{\infty }{e}^{(-\mu +i\omega )s}ds=(\mu -i\omega {)}^{-1}$. ConsequentlyThe perturbation picks the excitation polarized along the third coordinate, rather than an arbitrary member of the degenerate first excited level.