A particle of mass $m$ is initially trapped by the well with potential $V(x)=-V_{\delta} \delta(x)$, where $V_{\delta}>0$. From $t=0$ it is disturbed by the time-dependent potential $v(x, t)=-F x \mathrm{e}^{-\mathrm{i} \omega t} .$ Its subsequent wavefunction can be written
$$
|\psi\rangle=a(t) \mathrm{e}^{-\mathrm{i} E_{0} t / \hbar}|0\rangle+\int \mathrm{d} k\left\{b_{k}(t)|k, \mathrm{e}\rangle+c_{k}(t)|k, \mathrm{o}\rangle\right\} \mathrm{e}^{-\mathrm{i} E_{k} t / \hbar}
$$
where $E_{0}$ is the energy of the bound state $|0\rangle$ and $E_{k} \equiv \hbar^{2} k^{2} / 2 m$ and $|k, \mathrm{e}\rangle$ and $|k, \mathrm{o}\rangle$ are, respectively the even- and odd-parity states of energy $E_{k}$ (see Problem 5.17). Obtain the equations of motion
$$
\begin{aligned}
\mathrm{i} \hbar\left\{\dot{a}|0\rangle \mathrm{e}^{-\mathrm{i} E_{0} t / \hbar}+\int \mathrm{d} k\left(\dot{b}_{k}|k, \mathrm{e}\rangle+\dot{c}_{k}|k, \mathrm{o}\rangle\right) \mathrm{e}^{-\mathrm{i} E_{k} t / \hbar}\right\} \\
&=v\left\{a|0\rangle \mathrm{e}^{-\mathrm{i} E_{0} t / \hbar}+\int \mathrm{d} k\left(b_{k}|k, \mathrm{e}\rangle+c_{k}|k, \mathrm{o}\rangle\right) \mathrm{e}^{-\mathrm{i} E_{k} t / \hbar}\right\}
\end{aligned}
$$
Given that the free states are normalised such that $\left\langle k^{\prime}, \mathrm{o} \mid k, \mathrm{o}\right\rangle=\delta\left(k-k^{\prime}\right)$, show that to first order in $v, b_{k}=0$ for all $t$, and that
$$
c_{k}(t)=\frac{\mathrm{i} F}{\hbar}\langle k, \mathrm{o}|x| 0\rangle \mathrm{e}^{\mathrm{i} \Omega_{k} t / 2} \frac{\sin \left(\Omega_{k} t / 2\right)}{\Omega_{k} / 2}, \quad \text { where } \quad \Omega_{k} \equiv \frac{E_{k}-E_{0}}{\hbar}-\omega
$$
Hence show that at late times the probability that the particle has become free is
$$
P_{\mathrm{fr}}(t)=\left.\frac{2 \pi m F^{2} t}{\hbar^{3}} \frac{|\langle k, \mathrm{o}|x| 0\rangle|^{2}}{k}\right|_{\Omega_{k}=0}
$$
Given that from Problem $5.17$ we have
$$
\langle x \mid 0\rangle=\sqrt{K e}^{-K|x|} \quad \text { where } \quad K=\frac{m V_{\delta}}{\hbar^{2}} \quad \text { and } \quad\langle x \mid k, o\rangle=\frac{1}{\sqrt{\pi}} \sin (k x)
$$ show that
$$
\langle k, \mathrm{o}|x| 0\rangle=\sqrt{\frac{K}{\pi}} \frac{4 k K}{\left(k^{2}+K^{2}\right)^{2}}
$$
Hence show that the probability of becoming free is
$$
P_{\mathrm{fr}}(t)=\frac{8 \hbar F^{2} t}{m E_{0}^{2}} \frac{\sqrt{E_{\mathrm{f}} /\left|E_{0}\right|}}{\left(1+E_{\mathrm{f}} /\left|E_{0}\right|\right)^{4}}
$$
where $E_{\mathrm{f}}>0$ is the final energy. Check that this expression for $P_{\mathrm{fr}}$ is dimensionless and give a physical explanation of the general form of the energy-dependence of $P_{\mathrm{fr}}(t)$