Proving e^Ae^B=e^{A+B} for Commuting Matrices

[kreil]

Homework Statement

Show, by series expansion, that if A and B are two matrices which do not commute, then $e^{A+B} \ne e^Ae^B$, but if they do commute then the relation holds.

Homework Equations

$$e^A=1+A$$
$$e^B=1+B$$
$$e^{A+B}=1+(A+B)$$

The Attempt at a Solution

Assuming that the first 2 terms in the expansion is sufficient to use here (is it?), I got the following:

$$e^Ae^B=(1+A)(1+B)=1+A+B+AB$$

This would be equal if the AB term were not tacked on the end...does this term somehow become zero when the two matrices commute? If I'm on the wrong track please let me know.

Josh

 

[kreil]

I think I figured it out after a little thought...

$$e^Ae^B=(1+A)(1+B)=1+A+B+AB-BA=1+A+B+[A,B]$$

So if A and B commute, [A,B]=0 and the relation holds, else [A,B] does not equal zero and $e^{A+B} \ne e^Ae^B$