As in the title. Is it possible that $[A,B]\neq0$, but $[A,e^B]=0$? I tried expanding the exponential and using $[A,B^n]=\sum_k {n\choose k} B^{n-k}[A,B]B^k $ but this doesn't seem to give any insight.

I'm inclined to think the answer is yes, because a sum of terms being $0$ is a weaker requirement than each term being $0$, but I was wondering if there's a clearer way to see it.

EDIT: in light of lisyarus' answer, what if the matrices in question are hermitian and have real eigenvalues?

  • $\begingroup$ Another way of putting it: $A,B$ commutes means $A$ has to commute with the entire one-parameter subgroup $\exp(tB)$ generated by $B$. Commuting with just a discrete subgroup of this one-parameter is a much weaker requirement, especially when there are multiple one-parameter subgroups that contain this discrete subgroup. $\endgroup$ – user10354138 May 19 at 10:40
  • $\begingroup$ Is your formula for $[A,B^k]$ correct as stated? Think you’ve got some indices mixed up $\endgroup$ – Adam Higgins May 19 at 13:22
  • $\begingroup$ @AdamHiggins Thank you, edited, I had mixed up my $n$ and $k$. Now it should be correct assuming I can still do induction $\endgroup$ – user438666 May 19 at 13:29

Yes, this can happen.

Take $B$ to be diagonal, with entries $2\pi i k$ with different $k$ (so that $B$ is not a scalar matrix). Then, $\exp B = 1$, so it commutes with anything.

Now, since $B$ is not a scalar matrix, there is some $A$ that doesn't commute with $B$.

  • 3
    $\begingroup$ yep, so more examples in the same vein here math.stackexchange.com/questions/349180/… $\endgroup$ – zwim May 19 at 10:05
  • 1
    $\begingroup$ Interesting, thank you. What if $A$ and $B$ are hermitian, hence the eigenvalues are real? $\endgroup$ – user438666 May 19 at 12:02

Referring to your question in the comment to @lisyarus, if $A,B$ are hermitian then $\left[A,e^{B}\right]=0\Longrightarrow\left[A,B\right]=0$.

To see this, observe that two hermitian matrices commute iff they can be simultaneously diagonalized. Thus, if $\left[A,e^{B}\right]=0$ there exist a unitarian $U$ such that both $U^{\dagger}AU$ and $U^{\dagger}e^{B}U$ are diagonal. Now

$$U^{\dagger}BU=U^{\dagger}\ln e^{B}U=\ln\left(U^{\dagger}e^{B}U\right)$$

is also diagonal and therefore $\left[A,B\right]=0$. Note that indeed

$$B=\ln e^{B}$$

since in the diagonal basis this translates to $\lambda_{i}=\ln e^{\lambda_{i}}$ where $\lambda_{i}$ are the real eigenvalues. This fails for non-hermitian matrices because in that case $\ln$ is multivalued and not injective.


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