The Hamiltonian for a quantum system in the Schrödinger picture is

$$H_{0}+V(t)$$

where $H_{0}$ is independent of time. Define the interaction picture corresponding to this Hamiltonian and derive a time evolution equation for interaction picture states.

Let $|a\rangle$ and $|b\rangle$ be orthonormal eigenstates of $H_{0}$ with eigenvalues $E_{a}$ and $E_{b}$ respectively. Assume $V(t)=0$ for $t \leqslant 0$. Show that if the system is initially, at $t=0$, in the state $|a\rangle$ then the probability of measuring it to be the state $|b\rangle$ after a time $t$ is

$$\frac{1}{\hbar^{2}}\left|\int_{0}^{t} d t^{\prime}\left\langle b\left|V\left(t^{\prime}\right)\right| a\right\rangle e^{i\left(E_{b}-E_{a}\right) t^{\prime} / \hbar}\right|^{2}$$

to order $V(t)^{2}$.

Suppose a system has a basis of just two orthonormal states $|1\rangle$ and $|2\rangle$, with respect to which

$$H_{0}=E I, \quad V(t)=v t \sigma_{1}, \quad t \geqslant 0,$$

where

$$I=\left(\begin{array}{ll} 1 & 0 \\ 0 & 1 \end{array}\right), \quad \sigma_{1}=\left(\begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right)$$

Use $(*)$ to calculate the probability of a transition from state $|1\rangle$ to state $|2\rangle$ after a time $t$ to order $v^{2}$.

Show that the time dependent Schrödinger equation has a solution

$$|\psi(t)\rangle=\exp \left(-\frac{i}{\hbar}\left(E t I+\frac{1}{2} v t^{2} \sigma_{1}\right)\right)|\psi(0)\rangle$$

Calculate the transition probability exactly. Hence find the condition for the order $v^{2}$ approximation to be valid.