- #1

- 668

- 68

## Homework Statement

Prove the following identity:

[tex]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!}+...[/tex]

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

## Homework Equations

[tex] e^x = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...[/tex]

[tex] e^{-x} = 1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+...[/tex]

## The Attempt at a Solution

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

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

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