 Solution

Let ${\mathrm{ad}}_A(B)=[A,B]$. Differentiating both exponentials and using that $A$ commutes with $e^{\lambda A}$ givesApplying the same calculation repeatedly gives $F^{(n)}(\lambda )=e^{\lambda A}{\mathrm{ad}}_A^n(B)e^{-\lambda A}$. Its Taylor series at zero, evaluated at one, proves the commutator expansion for exponential conjugationFor bounded operators the series converges in operator norm by the bound $(2\Vert A\Vert {)}^n\Vert B\Vert$. The quantum computations below hold on a common invariant domain, such as Schwartz functions, with the commutator series either terminating or closing on a finite-dimensional span.

Use the canonical commutation relations $[\hat{x},\hat{p}]=i\hslash$ and the Heisenberg picture $\hat{B}(t)=e^{itH/\hslash }\hat{B}{e}^{-itH/\hslash }$. For the first Hamiltonian,All further relevant commutators vanish, soThe explicit energy check isThe scalar shifts commute with all operators, so there is no ordering ambiguity.

For the two-dimensional system put $L={\hat{x}}_1{\hat{p}}_2-{\hat{x}}_2{\hat{p}}_1$ and $V=-\beta L$. Direct commutation gives $[L,{\hat{p}}_1]=i\hslash {\hat{p}}_2$, $[L,{\hat{p}}_2]=-i\hslash {\hat{p}}_1$. Thus the Heisenberg equations of motion are ${\dot{\hat{p}}}_1=\beta {\hat{p}}_2$, ${\dot{\hat{p}}}_2=-\beta {\hat{p}}_1$, whose solution, equivalently the resummed commutator series, isExpanding both squares cancels their cross terms and gives ${\hat{p}}_1(t{)}^2+{\hat{p}}_2(t{)}^2={\hat{p}}_1^2+{\hat{p}}_2^2$.

In the interaction picture generated by $H_0$, $[H_0,{\hat{x}}_i]=-i\hslash {\hat{p}}_i$ and $[H_0,{\hat{p}}_i]=0$. The terminating expansion therefore gives ${\overline{x}}_i(t)={\hat{x}}_i+t{\hat{p}}_i$. Also $[H_0,L]=0$, either by the momentum-square calculation or by direct commutation. Conjugating ${\overline{x}}_i$ by $e^{itV/\hslash }$ rotates position and momentum by the same matrix, since $[L,{\hat{x}}_1]=i\hslash {\hat{x}}_2$ and $[L,{\hat{x}}_2]=-i\hslash {\hat{x}}_1$. Hence