# Quantum Mechanics - Ladder Operators

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.

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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?

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 $$\sum_{r=0}^\infty (-1)^r K_r|n>=|0><0|n>=0$$?

redundant

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!