Show the formula which connects the adjoint representations

[Mutatis]

Homework 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...

 

[fresh_42]

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...

Where did you get this from? The proof is normally not done by direct computation; and it is usually not homework.

 

[Mutatis]

Well, this is one exercise from my quantum mechanics class...

 

[fresh_42]

Looks a bit troublesome to do it this way. The way I know is (details aside):

Given a Lie group $G$ and its Lie algebra $\mathfrak{g}$.

$G$ operates on itself via conjugation $x.y :=xyx^{-1}$ which gives rise to
$G$ operates on $\mathfrak{g}$ by $x.Y := \operatorname{Ad}(x)(Y) = x Y x^{-1}$ which gives rise to
$\mathfrak{g}$ operates on itself by $X.Y := \operatorname{ad}(X)(Y) = [X,Y]$

Now the two adjoint representations are related by $\operatorname{Ad}(\exp(x)) = \exp(\operatorname{ad}(X))$ since the exponential function is basically the integration from the tangent space, the Lie algebra $\mathfrak{g}$, back into the Lie group $G$.

What you have here is exactly this formula:
\begin{align*}
\operatorname{Ad}(\exp(-\hat A))(\hat B) &= \exp(-\hat A) \hat B \exp(\hat A) \\
&= \hat B - [\hat A, \hat B] +\frac{1}{2!} [\hat A,[\hat A,\hat B]] \mp \cdots \\
&= \left(1 + \operatorname{ad}(-\hat A) + \frac{1}{2!} (\operatorname{ad}(-\hat A))^2 \mp \cdots \right)(\hat B)\\
&= \exp(\operatorname{ad}(-\hat A)) (\hat B)
\end{align*}
If it is only an exercise in calculation with matrices, then you will probably have to use $\operatorname{ad}(-\hat A)(\hat B) = [-\hat A,\hat B] = -\hat A \hat B +\hat B \hat A$ and a bit of patience multiplying those sums.