- 42

- 0

- Problem Statement
- Show that ##e^\left(-Â\right)*\hat B*e^Â = \hat B - \left[ Â, \hat B \right] + \frac {1} {2!} *\left[ \hat A, \left[ Â, \hat B \right] \right] - ... ##

- Relevant Equations
- ## e^x=\sum_{n=0}^\infty \frac {x^n} {n!} ##

That's my attempting: first I've wrote ##e## in terms of the power series, but then I don't how to get further than this $$ \sum_{n=0}^\infty (-1)^n \frac {Â^n} {n!} \hat B \sum_{n=0}^\infty \frac {Â^n} {n!} = \sum_{n=0}^\infty (-1)^n \frac {Â^2n} {\left( n! \right) ^2} $$. I've alread tried to expand it but it leads me to a weird sum of terms...