(a) If $A$ and $B$ are operators which each commute with their commutator $[A, B]$, show that $\left[A, e^{B}\right]=[A, B] e^{B}$. By considering

$$F(\lambda)=e^{\lambda A} e^{\lambda B} e^{-\lambda(A+B)}$$

and differentiating with respect to the parameter $\lambda$, show also that

$$e^{A} e^{B}=C e^{A+B}=e^{A+B} C$$

where $C=e^{\frac{1}{2}[A, B]}$.

(b) Consider a one-dimensional quantum system with position $\hat{x}$ and momentum $\hat{p}$. Write down a formula for the operator $U(\alpha)$ corresponding to translation through $\alpha$, calculate $[\hat{x}, U(\alpha)]$, and show that your answer is consistent with the assumption that position eigenstates obey $|x+\alpha\rangle=U(\alpha)|x\rangle$. Given this assumption, express the wavefunction for $U(\alpha)|\psi\rangle$ in terms of the wavefunction $\psi(x)$ for $|\psi\rangle$.

Now suppose the one-dimensional system is a harmonic oscillator of mass $m$ and frequency $\omega$. Show that

$$\psi_{0}(x-\alpha)=e^{-m \omega \alpha^{2} / 4 \hbar} \sum_{n=0}^{\infty}\left(\frac{m \omega}{2 \hbar}\right)^{n / 2} \frac{\alpha^{n}}{\sqrt{n !}} \psi_{n}(x),$$

where $\psi_{n}(x)$ are normalised wavefunctions with energies $E_{n}=\hbar \omega\left(n+\frac{1}{2}\right)$.

[Standard results for constructing normalised energy eigenstates in terms of annihilation and creation operators

$$a=\left(\frac{m \omega}{2 \hbar}\right)^{1 / 2}\left(\hat{x}+\frac{i}{m \omega} \hat{p}\right), \quad a^{\dagger}=\left(\frac{m \omega}{2 \hbar}\right)^{1 / 2}\left(\hat{x}-\frac{i}{m \omega} \hat{p}\right)$$

may be quoted without proof.]