# Commuting creation and annihilation operators

1. Oct 5, 2008

### Rettaw

Hello, I have the missfortune of having to calculate a commutator with some powers of the creation and the annihilation operators, something like:

$$\left[ a^m , (a^{\dagger})^n \right]$$

I have managed to derive
$$\left[ a^m , (a^{\dagger})^n \right]= m a^{m-1} \left[ a , a^{\dagger} \right]$$
(altought I should really have remebered that) but I don't know how to use that to calculate
the big thing other than by recursive application of it, and that's very messy.

Any suggestions?

2. Oct 5, 2008

### ismaili

(I guess you got a typo in your second equation. I corrected it in the way I thought.)
Actually, by directly repeating use of the eq(*), you will get
$$\left[ a^m , (a^{\dagger})^n \right] = nma^{m-1}(a^\dagger)^{n-1}\left[a,a^\dagger\right]$$.
What you only have to notice is, the commutator is a c-number.
Application of eq(*) to $$\left[ a^m , (a^{\dagger})^n \right]$$ is just of as many lines calculation as the derivation of eq(*).
Actually, the answer can be read off directly, but I think you should go into the calculation, and should not feel cumbersome please.

Cheers

3. Oct 11, 2008

### Rettaw

Yeah, you're right it's supposed to be only $$a^{\dagger}$$ and no powers of n.
Still, I'm not entirely convinced, the $$\left[ a , a^{\dagger} \right]$$ is indeed a c-number, but the $$\left[ a^m , a^{\dagger} \right]$$ is an operator, and when I expand the full $$\left[ a^m , (a^{\dagger})^n \right]$$ I get things that contain higher powes of the operators and thus I'm quite sure do not commute trivially.

So are you claiming that if I fully expand the commutator until I only have $$\left[ a , a^{\dagger} \right]$$ I get all the operators that I've pulled outside in the correct order to be able to write your final result?