I am currently trying to prove that the two following commutator relationships are equivalent (for an operator $\hat{A}(s)$ that depends on a continuous parameter $s$), so if one holds the other one should hold as well:
$$0=\left[\frac{d\hat{A}(s)}{ds},\exp(\hat{A}(s))\right]$$
and
$$0=\left[\frac{d\hat{A}(s)}{ds},\hat{A}(s)\right].$$
Proof from the second relation to the first
That the second equation implies the first one is easy to see. Just start from the first equation and fill in the series expansion of the exponential, yielding:
$$0=\sum\limits_{n=0}^\infty\frac{1}{n!}\left[\frac{d\hat{A}(s)}{ds},\hat{A}^n(s)\right].$$ The above commutator can be expanded in terms of the second commutator, proving that if the second relation holds, the first one is implied.
Proof from the first equation to the second
The other way around is much harder, and it's here that I am stuck! I could re-enter the series and expand the first relation as:
$$0=\left[\frac{d\hat{A}(s)}{ds},\exp(\hat{A}(s))\right]=\left[\frac{d\hat{A}(s)}{ds},\hat{1}\right]+\left[\frac{d\hat{A}(s)}{ds},\hat{A}(s)\right]+\frac{1}{2}\left[\frac{d\hat{A}(s)}{ds},\hat{A}(s)^2\right]+…,$$
but I don't see how this might help me.
So my question is: do these relations truly imply each other? And if not, what are the conditions such that they do ? So, if one of the two is true, does this automatically imply the other?
Best Answer
Theorem. Let $B :D(B) \to H$, $A: D(A) \to H$ be densely defined self-adjoint operators in the Hilbert space $H$. Suppose that $e^{A}$ is bounded (it happens in particular if $\sigma(A)$ is bounded from above). $$Be^A\psi = e^AB\psi$$ for every $\psi \in D(B)$ is equivalent to the fact that $B$ commute with the spectral measure of $e^A$ which in turn is equivalent to the fact that $B$ commutes with the spectral measure of $A$. Therefore $BA\psi = AB\psi$ whenever both sides are defined.
Applying this theorem you prove (2) out of (1). I think the statement is valid also if $A$ and $B$ are closed and normal provided $e^A$ is bounded, but I do not have spare time to produce a proof.