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 Quote by dimensionless I'm having trouble figuring out the following commutator relation problem: Suppose A and B commute with their commutator, i.e., $$[B,[A,B]]=[A,[A,B]]=0$$. Show that $$[A,B^{n}]=nB^{n-1}[A,B]$$ I have $$[A,B^{n}] = AB^{n} - B^{n}A$$ and also $$[A,B^{n}] = AB^{n} - B^{n}A = ABB^{n-1} - BB^{n-1}A$$ I don't know where to go from here. I'm not positive the above relation is correct either.
Now, use with $C= B^{n-1}$.
, that is use $[A,B^n] = B[A,B^{n-1}] + [A,B] B^{n-1}$.
Now, repeat this again on the first term using now $C= B^{n-2}$. You will get a recursion formula that will give you the proof easily.