Product of exponential of operators in the context of non-relativistic quantum mechanics I want to show that, for any $A$ and $B$ operators
$$e^{A}e^{B}=e^{A+B} $$
if and only if
$$[A,B]=0$$
I remember my professor told use about looking for a differential equation  but I don't remember the details, and I want to be able to prove it. Brute force the series doesn't seem to be a good idea.
Any hint will be appreciated thanks.
 A: Define the function $f(u)=e^{uA}e^{uB}$. If $A$ and $B$ commute, you can take the derivatives of the $f(u)$ as follows:
$$\frac{df(u)}{du}=\frac{de^{uA}}{du}e^{uB}+e^{uA}\frac{de^{uB}}{du}=Ae^{uA} e^{uB}+e^{uA}Be^{uB}$$
$$\frac{d^2f(u)}{du^2}=A^2e^{uA} e^{uB}+2Ae^{uA}Be^{uB}+e^{uA}B^2e^{uB}$$
etc.
Now take the Maclaurin series expansion:
$$f(u)=1+(A+B)u+\frac{1}{2!}(A^2+2AB+B^2)u^2+\ldots=\\
=1+(A+B)u+\frac{1}{2!}(A+B)^2 u^2+\frac{1}{3!}(A+B)^3 u^3+\ldots=e^{(A+B)u}$$
For $u=1$ you get the desired result:
$$f(1)=e^{A}e^{B}=e^{A+B}$$
Also check this: 
Baker–Campbell–Hausdorff formula
Matrix exponential
A: I've never heard of trying to find a differential equation to prove this; I've only done by brute forcing the series expansion. That isn't as bad as it sounds. On the one side of the equation you get a single infinite series with terms of $(A + B)^n$. The other side of the equation gives you two infinite series multiplied together, with terms of $A^m$ and $B^l$. Don't make any assumptions about $[A,B]$ yet, and expand everything out until you have all the terms that contain $AB$ and $BA$. You should find that the constant terms and the terms of order $A$ and $B$ all cancel out, which will leave you with your condition for the commutator. 
A: There is no 'only if' because it is not true:
\begin{align}
e^{A+B} = e^A e^B
\end{align}
does not necessarily imply $[A,B] = 0$.
One can easily find an example of this using matrices. Here's one:
\begin{align}
A=
\begin{pmatrix}
0 & 0 \\
0 & 2\pi i
\end{pmatrix},
B=\begin{pmatrix}
0 & 1 \\
0 & 2 \pi i
\end{pmatrix}.
\end{align}
$[A,B] \neq 0$ but $e^{A+  B} = e^A e^B = I$.
Edit:
Let me help with the if part, using a differential equation as OP desires.
Compute
\begin{align}
\frac{d}{dt}(e^{t(A+B)}e^{-tA}e^{-tB}),
\end{align}
and show that it is $0$ if $[A,B] = 0$.
That implies that $e^{t(A+B)}e^{-tA}e^{-tB}$ is independent of $t$. In particular, plugging in $t = 0$ gives $e^{t(A+B)}e^{-tA}e^{-tB} = I$ for all $t$. Then plug in $t = 1$ to get $e^{(A+B)}e^{-A}e^{-B} = I$.
QED.
A: Remember that you are working with operators.
Since 
$$e^{A}e^B =(1+\frac{A}{2}+\frac{AA}{3!}+...)(1+\frac{B}{2}+\frac{BB}{3!}+...)$$
and 
$$e^{A+B} =(1+\frac{A+B}{2}+\frac{(A+B)^2}{3!}+...)$$.
Try matching the terms up using commutators.
A: The proof (due to Glauber, given in Messiah) is as follows: consider f(x)= exp(xA)+exp(xB)  
differentiate it ---> df/dx= Aexp(xA)exp(xB)+exp(xA)exp(xB)B
                  df/dx=f(x)[exp(-xB)Aexp(xB)+B] here we have to take care of the order of operators
                  df/dx=f(x)[exp(1-xB)A(1+xB) + B]  expand exponential

df/dx=f(x)[A+x[A,B]+B]     
It is easy to check that the solution to this first-order differential equation is one at x = 0 which is given by  
f(x)=exp(x[A+B]+1/2*x^2[A,B])
for x=1 and [A,B] =0; you can get the required answer
