Name of this relation, and struggle proving it.

[M. next]

[a, a[itex]^{+(n)}[/itex]] = na[itex]^{+(n-1)}[/itex]

1) What's the name of this relation if it has any?

2) I tried to prove this by induction, I started by saying that for n=1, this holds since [a, a[itex]^{+}[/itex]] = 1 (as we all know and as we can all prove)

then I assumed it true for (n-1), but I didn't go too far afterwards. Can someone give me a hint concerning its proof.

Thanks!

 

[Bill_K]

Is it homework?

 

[stevendaryl]

[a, a[itex]^{+(n)}[/itex]] = na[itex]^{+(n-1)}[/itex]

1) What's the name of this relation if it has any?

2) I tried to prove this by induction, I started by saying that for n=1, this holds since [a, a[itex]^{+}[/itex]] = 1 (as we all know and as we can all prove)

then I assumed it true for (n-1), but I didn't go too far afterwards. Can someone give me a hint concerning its proof.

Thanks!

Induction works just fine.

There's an identity for working with commutators that helps:

[itex][A, B C] = [A, B] C + B [A, C][/itex]

Apply to the case [itex]A = a[/itex], [itex]B = (a^\dagger)^{n-1}[/itex], [itex]C=a^\dagger[/itex].
Then it should be easy to prove it by induction.

 

[M. next]

Oh thank you! I am now convinced, I tried proving it several ways, and didn't use what you proposed, and the procedures kept turning me down! Thank you, again!

Bill_K,this wasn't a homework, I was trying to convince myself that its true. I asked my professor and he told me to prove it by induction and didn't give further hints.

Thanks guys.