Commuting creation and annihilation operators

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

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

I have managed to derive
[tex] \left[ a^m , (a^{\dagger})^n \right]= m a^{m-1} \left[ a , a^{\dagger} \right] [/tex]
(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?
 
161
0
Hello, I have the missfortune of having to calculate a commutator with some powers of the creation and the annihilation operators, something like:

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

I have managed to derive
[tex] \left[ a^m , a^{\dagger} \right]= m a^{m-1} \left[ a , a^{\dagger} \right] \quad\cdots (*)[/tex]
(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?
(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
[tex]\left[ a^m , (a^{\dagger})^n \right] = nma^{m-1}(a^\dagger)^{n-1}\left[a,a^\dagger\right][/tex].
What you only have to notice is, the commutator is a c-number.
Application of eq(*) to [tex]\left[ a^m , (a^{\dagger})^n \right][/tex] 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
 
6
0
Yeah, you're right it's supposed to be only [tex]a^{\dagger}[/tex] and no powers of n.
Still, I'm not entirely convinced, the [tex] \left[ a , a^{\dagger} \right] [/tex] is indeed a c-number, but the [tex] \left[ a^m , a^{\dagger} \right] [/tex] is an operator, and when I expand the full [tex] \left[ a^m , (a^{\dagger})^n \right] [/tex] 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 [tex] \left[ a , a^{\dagger} \right] [/tex] I get all the operators that I've pulled outside in the correct order to be able to write your final result?
 

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