Prove Commutator Identity: e^xA B e-xA = B + [A,B]x + ...

[kreil]

Homework Statement

Prove the following identity:

e^{x \hat A} \hat B e^{-x \hat A} = \hat B + [\hat A, \hat B]x + \frac{[\hat A, [\hat A, \hat B]]x^2}{2!}+\frac{[\hat A,[\hat A, [\hat A, \hat B]]]x^3}{3!}+...

where A and B are operators and x is some parameter.

Homework Equations

e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...
e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...

The Attempt at a Solution

\hat B e^{-x \hat A} = \hat B - [\hat B, \hat A]x + \frac{[[\hat B, \hat A], \hat A] x^2}{2!}+...

It seems after I rearrange the commutation orders, the signs all become positive and this is the required result, so I know I must be doing something wrong. I think it has to do with how I'm multiplying out B into the series..

i.e. \hat B (\hat A x)^2 = [\hat B, \hat A \hat A] x^2...??

or \hat B (\hat A x)^2 = [[\hat B, \hat A], \hat A] x^2...??

 

[gabbagabbahey]

It seems after I rearrange the commutation orders, the signs all become positive and this is the required result, so I know I must be doing something wrong. I think it has to do with how I'm multiplying out B into the series..

i.e. \hat B (\hat A x)^2 = [\hat B, \hat A \hat A] x^2...??

or \hat B (\hat A x)^2 = [[\hat B, \hat A], \hat A] x^2...??

Neither is correct.

\hat{B}(x\hat{A})^2=BA^2x^2[/itex]<br /> <br /> You won&#039;t have anything involving commutators until you multiply by both exponentials and collect terms in powers of the parameter x.

 

[kreil]

ahh I see it now I think..

e^{x \hat A} \hat B e^{-x \hat A} = \left ( 1+\hat A x + \frac{1}{2} \hat A^2 x^2 + \frac{1}{6} \hat A^3 x^3+... \right ) \left ( \hat B - \hat B \hat A x + \frac{1}{2} \hat B \hat A^2 x^2 - \frac{1}{6}\hat B \hat A^3 x^3+... \right )

So I multiply this out, collect terms in powers of x, and simplify to the commutator relations

Thanks