Quantum Mechanics - Ladder Operators

[Tangent87]

I'm trying to show that N=a^\dagger a and K_r=\frac{a^\dagger^r a^r}{r!} commute. So basically I need to show [a^\dagger^r a^r,a^\dagger a]=0. I'm not quite sure what to do, I've tried using [a,a^\dagger] in a few places but so far haven't had much success.

 

[latentcorpse]

I'm trying to show that N=a^\dagger a and K_r=\frac{a^\dagger^r a^r}{r!} commute. So basically I need to show [a^\dagger^r a^r,a^\dagger a]=0. I'm not quite sure what to do, I've tried using [a,a^\dagger] in a few places but so far haven't had much success.

What's [AB,CD] equal to?

 

[Tangent87]

What's [AB,CD] equal to?

Ah, thanks latentcorpse!

[AB,CD]=A[B,CD]+[A,CD]B=A[B,C]D+AC[B,D]+[A,C]DB+C[A,D]B

Also, if I want to show \sum_{r=0}^\infty (-1)^r K_r=|0><0| is it sufficient to show \sum_{r=0}^\infty (-1)^r K_r|n>=|0><0|n>=0?

 

[sgd37]

redundant

 

[Tangent87]

Ah, thanks latentcorpse!

[AB,CD]=A[B,CD]+[A,CD]B=A[B,C]D+AC[B,D]+[A,C]DB+C[A,D]B

Also, if I want to show \sum_{r=0}^\infty (-1)^r K_r=|0><0| is it sufficient to show \sum_{r=0}^\infty (-1)^r K_r|n>=|0><0|n>=0?

No wait of course it's not, what am I thinking!