- #1
ice109
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i've never really done a proof by induction but i would like to prove a statement about commutator relations so can you please check my proof:
claim: [itex][A,B^n]=nB^{n-1}[A,B][/itex] if [itex][A,B]=k\cdot I[/itex] where A,B are operators, I is the identity and k is any scalar.
proof: [tex] [A,B^2] = [A,B]B+B[A,B] = 2B^{2-1}[A,B] [/tex] where the B is brought out of the commutator by almost obvious identity and the last equality follows from the commutativity of the identity with any operator.
so for n=2 the equality is proven. we now assume that the equality holds for n-2 case and proceed to prove it holds for n case:
[tex][A,B^n]=[A,B^2B^{n-2}]=[A,B^{2}]B^{n-2}+B^2[A,B^{n-2}][/tex]
[tex] =2B[A,B]B^{n-2}+B^2(n-2)B^{n-3}[A,B] [/tex]
[tex] =2B^{n-1}[A,B]+(n-2)B^{n-1}[A,B][/tex]
[tex] =nB^{n-1}[A,B][/tex]
so how 'bout it? was successful?
claim: [itex][A,B^n]=nB^{n-1}[A,B][/itex] if [itex][A,B]=k\cdot I[/itex] where A,B are operators, I is the identity and k is any scalar.
proof: [tex] [A,B^2] = [A,B]B+B[A,B] = 2B^{2-1}[A,B] [/tex] where the B is brought out of the commutator by almost obvious identity and the last equality follows from the commutativity of the identity with any operator.
so for n=2 the equality is proven. we now assume that the equality holds for n-2 case and proceed to prove it holds for n case:
[tex][A,B^n]=[A,B^2B^{n-2}]=[A,B^{2}]B^{n-2}+B^2[A,B^{n-2}][/tex]
[tex] =2B[A,B]B^{n-2}+B^2(n-2)B^{n-3}[A,B] [/tex]
[tex] =2B^{n-1}[A,B]+(n-2)B^{n-1}[A,B][/tex]
[tex] =nB^{n-1}[A,B][/tex]
so how 'bout it? was successful?