Define the interaction picture for a quantum mechanical system with Schrödinger picture Hamiltonian $H_{0}+V(t)$ and explain why the interaction and Schrödinger pictures give the same physical predictions for transition rates between eigenstates of $H_{0}$. Derive the equation of motion for the interaction picture states $|\overline{\psi(t)}\rangle$.

A system consists of just two states $|1\rangle$ and $|2\rangle$, with respect to which

$$H_{0}=\left(\begin{array}{cc} E_{1} & 0 \\ 0 & E_{2} \end{array}\right), \quad V(t)=\hbar \lambda\left(\begin{array}{cc} 0 & e^{i \omega t} \\ e^{-i \omega t} & 0 \end{array}\right)$$

Writing the interaction picture state as $\overline{\langle(t)}\rangle=a_{1}(t)|1\rangle+a_{2}(t)|2\rangle$, show that the interaction picture equation of motion can be written as

$$i \dot{a}_{1}(t)=\lambda e^{i \mu t} a_{2}(t), \quad i \dot{a}_{2}(t)=\lambda e^{-i \mu t} a_{1}(t)$$

where $\mu=\omega-\omega_{21}$ and $\omega_{21}=\left(E_{2}-E_{1}\right) / \hbar$. Hence show that $a_{2}(t)$ satisfies

$$\ddot{a}_{2}+i \mu \dot{a}_{2}+\lambda^{2} a_{2}=0 .$$

Given that $a_{2}(0)=0$, show that the solution takes the form

$$a_{2}(t)=\alpha e^{-i \mu t / 2} \sin \Omega t,$$

where $\Omega$ is a frequency to be determined and $\alpha$ is a complex constant of integration.

Substitute this solution for $a_{2}(t)$ into $(*)$ to determine $a_{1}(t)$ and, by imposing the normalization condition $\||\overline{\psi(t)}\rangle \|^{2}=1$ at $t=0$, show that $|\alpha|^{2}=\lambda^{2} / \Omega^{2}$.

At time $t=0$ the system is in the state $|1\rangle$. Write down the probability of finding the system in the state $|2\rangle$ at time $t$.