# Commutator of exponential operators

1. Jan 5, 2014

### gentsagree

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.

2. Jan 5, 2014

### Staff: Mentor

Why does it seem wrong?

3. Jan 5, 2014

### gentsagree

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

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

4. Jan 5, 2014

### Staff: Mentor

The first order is not the final result...

5. Jan 5, 2014

### gentsagree

Second order does the job....thanks.

6. Jan 5, 2014

### Staff: Mentor

I am surprised by that, as the final result has all orders, and just ignoring orders means changing the value of things.

7. Jan 5, 2014

### gentsagree

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.

8. Jan 5, 2014

Ah, okay.