If $A$ and $B$ are operators which each commute with their commutator $[A, B]$, show that

$$F(\lambda)=e^{\lambda A} e^{\lambda B} e^{-\lambda(A+B)} \quad \text { satisfies } \quad F^{\prime}(\lambda)=\lambda[A, B] F(\lambda)$$

By solving this differential equation for $F(\lambda)$, deduce that

$$e^{A} e^{B}=e^{\frac{1}{2}[A, B]} e^{A+B}$$

The annihilation and creation operators for a harmonic oscillator of mass $m$ and frequency $\omega$ are defined by

$$a=\sqrt{\frac{m \omega}{2 \hbar}}\left(\hat{x}+\frac{i}{m \omega} \hat{p}\right), \quad a^{\dagger}=\sqrt{\frac{m \omega}{2 \hbar}}\left(\hat{x}-\frac{i}{m \omega} \hat{p}\right)$$

Write down an expression for the general normalised eigenstate $|n\rangle(n=0,1,2, \ldots)$ of the oscillator Hamiltonian $H$ in terms of the ground state $|0\rangle$. What is the energy eigenvalue $E_{n}$ of the state $|n\rangle ?$

Suppose the oscillator is now subject to a small perturbation so that it is described by the modified Hamiltonian $H+\varepsilon V(\hat{x})$ with $V(\hat{x})=\cos (\mu \hat{x})$. Show that

$$V(\hat{x})=\frac{1}{2} e^{-\gamma^{2} / 2}\left(e^{i \gamma a^{\dagger}} e^{i \gamma a}+e^{-i \gamma a^{\dagger}} e^{-i \gamma a}\right)$$

where $\gamma$ is a constant, to be determined. Hence show that to $O\left(\varepsilon^{2}\right)$ the shift in the ground state energy as a result of the perturbation is

$$\varepsilon e^{-\mu^{2} \hbar / 4 m \omega}-\varepsilon^{2} e^{-\mu^{2} \hbar / 2 m \omega} \frac{1}{\hbar \omega} \sum_{p=1}^{\infty} \frac{1}{(2 p) ! 2 p}\left(\frac{\mu^{2} \hbar}{2 m \omega}\right)^{2 p} .$$

[Standard results of perturbation theory may be quoted without proof.]