# Commutator of exponential operators

How do I compute the commutator $[a,e^{-iHt}]$, knowing that $[H,a]=-Ea$?

I tried by Taylor expanding the exponential, but I get $-iEta$ to first order, which seems wrong.

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mfb
Mentor
Why does it seem wrong?

The answer I was expecting to get is something like $e^{-iEt}$. This follows from the fact that I was trying to replicate the following calculation

$$e^{iHt}ae^{-iHt}=e^{-iEt}a$$

which is said to be justified by $[H,a]=-Ea$.

If I take my result to be true, I get

$$e^{iHt}ae^{-iHt}=e^{iHt}(e^{-iHt}a-iEta)=a-ie^{iHt}Eta$$

mfb
Mentor
The first order is not the final result...

Second order does the job....thanks.

mfb
Mentor
Second order does the job....thanks.
I am surprised by that, as the final result has all orders, and just ignoring orders means changing the value of things.

It does the job in the sense that with two orders I can get back two terms which I can combine back together to form an exponential by recognising they are the first two terms in its Taylor expansion. I thought, naively, that one order would have been enough, but the fact that the first commutator vanishes doesn't make it work.

mfb
Mentor
Ah, okay.