Prove the following for operators A and B.
e^A B e^-A = B + [A,B] + (1/2!) * [A,[A,B]] + (1/3!) * [A,[A,[A,B]]] + ...
e^A = 1 + A + (1/2!)A^2 + (1/3!)A^3 + ...
The Attempt at a Solution
I have no clue how to start.
For the highly special case of [A,B] = constant and [A,B] commutes with A and B, we can prove that e^A B e^-A = B + [A,B]
through the following:
Take the given identity [A,F(B)] = [A,B] dF/dB and set F = e^B
[A,e^B] = Ae^B - e^B*A = e^B * [A,B]
multiply both sides by e^-B to left.
e^-B A e^B - A = [A,B]
e^-B A e^B = A + [A,B]
This is the first 2 terms of the series and if we take [A,B] = c, then the other terms are zero. How do I prove it for general [A,B] then?