A quantum system is prepared in the ground state $|0\rangle$ at time $t=0$. It is subjected to a time-varying Hamiltonian $H=H_{0}+\Delta(t)$. Show that, to first order in $\Delta(t)$, the system evolves as

$$|\psi(t)\rangle=\sum_{k} c_{k}(t) \mathrm{e}^{-i E_{k} t / \hbar}|k\rangle$$

where $H_{0}|k\rangle=E_{k}|k\rangle$ and

$$c_{k}(t)=\frac{1}{i \hbar} \int_{0}^{t}\left\langle k\left|\Delta\left(t^{\prime}\right)\right| 0\right\rangle \mathrm{e}^{i\left(E_{k}-E_{0}\right) t^{\prime} / \hbar} \mathrm{d} t^{\prime}$$

A large number of hydrogen atoms, each in the ground state, are subjected to an electric field

$$\mathbf{E}(t)=\left\{\begin{array}{lll} 0 & \text { for } & t<0 \\ \hat{\mathbf{z}} \mathcal{E}_{0} \exp (-t / \tau) & \text { for } & t>0 \end{array}\right.$$

where $\mathcal{E}_{0}$ is a constant. Show that the fraction of atoms found in the state $|n, \ell, m\rangle=$ $|2,1,0\rangle$ is, after a long time and to lowest non-trivial order in $\mathcal{E}_{0}$,

$$\frac{2^{15}}{3^{10}} \frac{a_{0}^{2} e^{2} \mathcal{E}_{0}^{2}}{\hbar^{2}\left(\omega^{2}+1 / \tau^{2}\right)}$$

where $\hbar \omega$ is the energy difference between the $|2,1,0\rangle$ and $|1,0,0\rangle$ states, and $e$ is the electron charge and $a_{0}$ the Bohr radius. What fraction of atoms lie in the $|2,0,0\rangle$ state?

[Hint: You may assume the hydrogenic wavefunctions

$$\langle\mathbf{r} \mid 1,0,0\rangle=\frac{2}{\sqrt{4 \pi}} \frac{1}{a_{0}^{3 / 2}} \exp \left(-\frac{r}{a_{0}}\right) \quad \text { and } \quad\langle\mathbf{r} \mid 2,1,0\rangle=\frac{1}{\sqrt{4 \pi}} \frac{1}{\left(2 a_{0}\right)^{3 / 2}} \frac{r}{a_{0}} \cos \theta \exp \left(-\frac{r}{2 a_{0}}\right)$$

and the integral

$$\int_{0}^{\infty} r^{m} \mathrm{e}^{-\alpha r} \mathrm{~d} r=\frac{m !}{\alpha^{m+1}}$$

for $m$ a positive integer.]