For any given operators $A$ and $B$, show that $F(\lambda)=e^{\lambda A} B e^{-\lambda A}$ has derivative $F^{\prime}(\lambda)=e^{\lambda A}[A, B] e^{-\lambda A}$ and deduce an analogous formula for the $n$th derivative. Hence, by considering $F(\lambda)$ as a power series in $\lambda$, show that

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

A particle of unit mass in one dimension has position $\hat{x}$ and momentum $\hat{p}$ in the Schrödinger picture, and Hamiltonian

$$H=\frac{1}{2} \hat{p}^{2}-\alpha \hat{x},$$

where $\alpha$ is a constant. Apply $(*)$ to find the Heisenberg picture operators $\hat{x}(t)$ and $\hat{p}(t)$ in terms of $\hat{x}$ and $\hat{p}$, and check explicitly that $H(\hat{x}(t), \hat{p}(t))=H(\hat{x}, \hat{p})$.

A particle of unit mass in two dimensions has position $\hat{x}_{i}$ and momentum $\hat{p}_{i}$ in the Schrödinger picture, and Hamiltonian

$$H=\frac{1}{2}\left(\hat{p}_{1}^{2}+\hat{p}_{2}^{2}\right)-\beta\left(\hat{x}_{1} \hat{p}_{2}-\hat{x}_{2} \hat{p}_{1}\right)$$

where $\beta$ is a constant. Calculate the Heisenberg picture momentum components $\hat{p}_{i}(t)$ in terms of $\hat{p}_{i}$ and verify that $\hat{p}_{1}(t)^{2}+\hat{p}_{2}(t)^{2}$ is independent of time. Now consider the interaction picture corresponding to $H=H_{0}+V$ : show that if $H_{0}=\frac{1}{2}\left(\hat{p}_{1}^{2}+\hat{p}_{2}^{2}\right)$ then the interaction picture position operators are $\hat{x}_{i}+t \hat{p}_{i}$, and use this to find the Heisenberg picture position operators $\hat{x}_{i}(t)$ in terms of $\hat{x}_{i}$ and $\hat{p}_{i}$.

[Hint: If $\left[H_{0}, V\right]=0$ and $\bar{Q}(t)$ is an operator in the interaction picture, then the corresponding operator in the Heisenberg picture is $\left.Q(t)=e^{i t V / \hbar} \bar{Q}(t) e^{-i t V / \hbar} \cdot\right]$