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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?

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  • $\begingroup$ @ValterMoretti, I know that in general both are false. My question is, if one is true, will the other be true as well ? $\endgroup$ – Nick Mar 2 '16 at 10:55
  • $\begingroup$ You approach is too formal. At most it may make sense for matrices. However regarding (1) -> (2), assuming that all matrices are at least normal, the first requirement implies that there is a common basis of eigenvectors of $dA/ds$ and $exp(A(s))$. So both matrices can be written in diagonal form simultaneously. The matrix $A(s)$ is diagonal if $exp(A(s))$ is, because they have the same eigenvectors with eigenvalues related in a obvious way $\lambda$ and $\ln \lambda$. Thus (2) holds. $\endgroup$ – Valter Moretti Mar 2 '16 at 11:07
  • $\begingroup$ @ValterMoretti He is putting the derivative also in the second commutator (on the left) so it is not zero; still I suppose that the property is false in general (there are counterexamples if they commute only on a common dense domain of definition). If they are normal and either bounded or commuting in the sense of the spectral projections, then it is true by the spectral theorem. But of course you know that very well ;-) $\endgroup$ – yuggib Mar 2 '16 at 11:07
  • $\begingroup$ Yes, the spectral theorem immediately implies the thesis if $dA/ds$ and A are selfadjoint, because the former identity (paying attention to domains) implies that $dA/ds$ commutes with the spectral measure of $\exp(A(s))$ which is the same as the one of $A(s)$ with a change of variables. $\endgroup$ – Valter Moretti Mar 2 '16 at 11:10
  • $\begingroup$ To sum up: there is surely a problem with the second identity not implying the first in general, even for self-adjoint operators (there are explicit counterexamples for operators commuting on a common domain of essential self-adjointness). On the other hand, the first identity implies the second for self adjoint operators because of the spectral theorem (as @ValterMoretti explained). $\endgroup$ – yuggib Mar 2 '16 at 11:20
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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.

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  • $\begingroup$ just to be sure (since I am not very mathematical). It only goes for hermitian, bounded operators? So basically in quantum mechanics, what does this imply? Does it go for all (physically acceptable) operators, or are there exceptions ? $\endgroup$ – Nick Mar 2 '16 at 13:29
  • $\begingroup$ In QM observables are self-adjoint operators. Self-adjointness is a stronger version of Hermiticity. Almost all self-adjoint operators in QM are unbounded because the values the corresponding observables may attain for an unbounded set of reals and this is equivalent to unboundedness of the operators. $\endgroup$ – Valter Moretti Mar 2 '16 at 14:46
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Looking at you're equations I am quite sure A(s) is a unitary operator for all s? In that case Valter's answer applies, too, as the spectral theorem (in its general form) holds for all normal operators (and hence for self adjoint as well as for unitary operators). A nice script (however in German) about the spectral theory of unbound normal operators can be found here: SpectralTheory with the spectral theorem on pg. 68.

To be sure that the spectral theorem can be applied (on the certain domains of the operators) you should make sure that for A in Valter's formula it holds AA^*=A^*A (normality).

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