The Hamiltonian for a quantum system in the Schrödinger picture is $H_{0}+\lambda V(t)$, where $H_{0}$ is independent of time and the parameter $\lambda$ is small. Define the interaction picture corresponding to this Hamiltonian and derive a time evolution equation for interaction picture states.

Suppose that $|\chi\rangle$ and $|\phi\rangle$ are eigenstates of $H_{0}$ with distinct eigenvalues $E$ and $E^{\prime}$, respectively. Show that if the system is in state $|\chi\rangle$ at time zero then the probability of measuring it to be in state $|\phi\rangle$ at time $t$ is

$$\frac{\lambda^{2}}{\hbar^{2}}\left|\int_{0}^{t} d t^{\prime}\left\langle\phi\left|V\left(t^{\prime}\right)\right| \chi\right\rangle e^{i\left(E^{\prime}-E\right) t^{\prime} / \hbar}\right|^{2}+O\left(\lambda^{3}\right)$$

Let $H_{0}$ be the Hamiltonian for an isotropic three-dimensional harmonic oscillator of mass $m$ and frequency $\omega$, with $\chi(r)$ being the ground state wavefunction (where $r=|\mathbf{x}|$ ) and $\phi_{i}(\mathbf{x})=(2 m \omega / \hbar)^{1 / 2} x_{i} \chi(r)$ being wavefunctions for the states at the first excited energy level $(i=1,2,3)$. The oscillator is in its ground state at $t=0$ when a perturbation

$$\lambda V(t)=\lambda \hat{x}_{3} e^{-\mu t}$$

is applied, with $\mu>0$, and $H_{0}$ is then measured after a very large time has elapsed. Show that to first order in perturbation theory the oscillator will be found in one particular state at the first excited energy level with probability

$$\frac{\lambda^{2}}{2 \hbar m \omega\left(\mu^{2}+\omega^{2}\right)},$$

but that the probability that it will be found in either of the other excited states is zero (to this order).

$\left[\right.$ You may use the fact that $\left.4 \pi \int_{0}^{\infty} r^{4}|\chi(r)|^{2} d r=\frac{3 \hbar}{2 m \omega} .\right]$