Consider two linear operators $\hat{A}$ and $\hat{B}$. Show that $e^{\hat{A}}e^{\hat{B}}=e^{\hat{A}+\hat{B}}e^{[\hat{A},\hat{B}]/2}$ 
For operators $\hat{A}$ and $\hat{B}$:
   $[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A} \not=0$ but
  $[[\hat{A},\hat{B}],\hat{A}]=[[\hat{A},\hat{B}],\hat{B}]=0$ 
Show that:
$$e^{\hat{A}}e^{\hat{B}}=e^{\hat{A}+\hat{B}}e^{[\hat{A},\hat{B}]/2} \space \space \space \space \space \space \space \space \space \space \space \space (1)$$

Hints:


*

*Consider: $\hat{U}(t)=e^{t\hat{A}}e^{t\hat{B}}$ and show that $\frac{d}{dt}\hat{U}(t)=\hat{U}(t)e^{-t\hat{B}}\hat{A}e^{t\hat{B}}+\hat{U}(t)\hat{B}$

*Consider $\frac{d}{dt}e^{-t\hat{B}}\hat{A}e^{t\hat{B}}$ to show that $e^{-t\hat{B}}\hat{A}e^{t\hat{B}}=[A,B]t+\hat{A}$


I verified 1. by differentiating: $\frac{d}{dt}(e^{t\hat{A}}e^{t\hat{B}})=e^{t\hat{A}}\hat{A}e^{t \hat{B}}+e^{t\hat{A}}e^{t\hat{B}}\hat{B}=e^{t\hat{A}}e^{t\hat{B}}e^{-t \hat{B}} \hat{A}e^{t\hat{B}}+e^{t\hat{A}}e^{t \hat{B}}\\=(e^{t\hat{A}}e^{t \hat{B}})(e^{-t \hat{B}}\hat{A}e^{t \hat{B}}+\hat{B})=\hat{U}(t)e^{-t\hat{B}}\hat{A}e^{t \hat{B}}+\hat{U}(t)\hat{B} \space \checkmark$
How do I show 2. though? Shouldn't it read: $e^{-t\hat{B}}\hat{A}e^{t\hat{B}}=[\color{red}{\hat{A},\hat{B}}]t+\hat{A}$?
How can I use these two hints to show $(1)$? I don't really see the connection. Does this identity have a name?
 A: The result you are trying to prove is a special case of the Campbell-Baker-Hausdorff formula:
$$e^{A+B} = e^Ae^B e^{-\frac{[A,B]}{2!}}e^{\frac{2[B,[A,B]] + [A,[A,B]]}{3!}}\cdots$$

We start with a small fact that will be useful below:

If $f$ is an analytic function and $[X,Y] = 0$ then $Yf(X) = f(X)Y$ 

The proof is simple: By induction $[X,Y] = 0$ implies $[X^n,Y] = 0$ for any integer $n$. Next write $f$ as a power-series and use that $Y$ commutes with every term in the series
$$Yf(X) = Y(a_0 + a_1X + a_2 X^2 +\ldots) = (a_0 + a_1X + a_2 X^2 + \ldots)Y = f(X)Y$$

You show 2) by evaluating and simplifying the derivative and then integrating it up again. By the chain-rule we have
$$\frac{d}{dt}\left(e^{-Bt}Ae^{Bt}\right) = \left(\frac{d}{dt}e^{-Bt}\right)Ae^{Bt} + e^{-Bt}A\left(\frac{d}{dt}e^{Bt}\right)$$
We further have $\frac{d}{dt}e^{B} = Be^{B} = e^{B}B$ (since $B$ commutes with $B$) which gives us
$$\frac{d}{dt}\left(e^{-Bt}Ae^{Bt}\right) = e^{-Bt}(-BA)e^{Bt} + e^{-Bt}(AB)e^{Bt} = e^{-Bt}[A,B]e^{Bt}$$
Finally since $[A,B]$ commutes with $B$ ($[[A,B],B] = 0$) the result above gives us 
$$[A,B]e^{Bt} = e^{Bt}[A,B]$$ 
which leads to the desired result
$$\frac{d}{dt}\left(e^{-Bt}Ae^{Bt}\right) = e^{-Bt}[A,B]e^{Bt} = e^{-Bt}e^{Bt}[A,B] = [A,B]$$
since $e^{-tB}e^{tB} = e^{-tB+tB} = e^0 = 1$ (this fact the same as the result you are trying to show the simpler case when $[A,B] = 0$, i.e. $e^{A}e^{B} = e^{A+B}$). Integrating up we find
$$e^{-Bt}Ae^{Bt} = C + t[A,B]$$
where $C = A$ is determined by taking $t=0$.

Using the two hints we get
$$\frac{d}{dt}U(t) = U(t)(A + t[A,B]) + U(t)B = U(t)(A + B + t[A,B])$$
Integrating up and taking $U(0) = 1$ gives us
$$U(t) = e^{At}e^{Bt} = e^{At + Bt + \frac{t^2[A,B]}{2}}$$
which for $t=1$ is the result we are after. Another option to integrating is simply differentiating $e^{At + Bt + \frac{t^2[A,B]}{2}}$ and show that it satisfies the same ODE as $U(t)$ and has the same value $1$ at $t=0$.
