= Solution
The <interaction picture> removes the evolution generated by $H_0$: $|\psi_I(t)\rangle=e^{iH_0t/\hbar}|\psi_S(t)\rangle$. Differentiating the <Schrödinger equation> gives $i\hbar\partial_t|\psi_I\rangle=\lambda V_I(t)|\psi_I\rangle$, with $V_I=e^{iH_0t/\hbar}V(t)e^{-iH_0t/\hbar}$. The integral equation, iterated once, gives
$$
|\psi_I(t)\rangle=|\chi\rangle-\frac{i\lambda}{\hbar}\int_0^t V_I(s)|\chi\rangle\,ds+O(\lambda^2).
$$
Distinct-energy <eigenstates> are orthogonal. Their transition amplitude is therefore $-i\lambda\hbar^{-1}\int_0^t\langle\phi|V(s)|\chi\rangle e^{i(E'-E)s/\hbar}ds+O(\lambda^2)$. Squaring its modulus gives
$$
\boxed{P_{\chi\to\phi}(t)=\frac{\lambda^2}{\hbar^2}\left|\int_0^t\langle\phi|V(s)|\chi\rangle e^{i(E'-E)s/\hbar}ds\right|^2+O(\lambda^3).}
$$
The first-order amplitude produces a second-order probability.
For the isotropic <quantum harmonic oscillator>, parity makes $\int x_i x_3|\chi|^2d^3x=0$ for $i=1,2$. Rotational symmetry and the given radial integral give $\int x_3^2|\chi|^2d^3x=\hbar/(2m\omega)$. Thus $\langle\phi_i|\hat x_3|\chi\rangle=\delta_{i3}\sqrt{\hbar/(2m\omega)}$, and the energy gap is $\hbar\omega$. The time integral is $\int_0^\infty e^{(-\mu+i\omega)s}ds=(\mu-i\omega)^{-1}$. Consequently
$$
\boxed{P_{\chi\to\phi_3}=\frac{\lambda^2}{2\hbar m\omega(\mu^2+\omega^2)}+O(\lambda^3),\qquad P_{\chi\to\phi_1}=P_{\chi\to\phi_2}=0\text{ to this order}.}
$$
The perturbation picks the excitation polarized along the third coordinate, rather than an arbitrary member of the degenerate first excited level.
Back to article page
