Proof of a commutator algebra exp(A)exp(B)=exp(B)exp(A)exp([A,B])

1. Sep 11, 2008

ismaili

I want to prove this formula
$$e^Ae^B = e^Be^Ae^{[A,B]}$$
The only method I can come up with is expand the LHS, and try to move all the B's to the left of all the A's, but it is so complicated in this way. i.e.
$$e^Ae^B=\frac{A^n}{n!}\frac{B^m}{m!} = \frac{1}{n!m!}\Big(A^{n-1}BAB^{m-1} + A^{n-1}[A,B]B^{m-1} \Big)= \cdots$$
Is there any good way of proof of this formula?

2. Sep 11, 2008

tiny-tim

Hi ismaili!

Yes … too complicated! …

so don't try to handle them all at once …

pick 'em off one at a time!

Hint: AeB = … ?

3. Sep 11, 2008

ismaili

Hi
Thanks!
By using $$[A,e^B] = [A,B]e^B$$ where I forgot to say that $$[A,B] = \text{c-number}$$ here, I can prove the identity right now.
Just back from the swimming pool, I will post my derivation later.

Cheers

4. Sep 11, 2008

jostpuur

If we set

$$A=\left(\begin{array}{cc} 0 & 1 \\ 0 & 0 \\ \end{array}\right),\quad\quad B=\left(\begin{array}{cc} 0 & 0 \\ 1 & 0 \\ \end{array}\right)$$

Then we get

$$[A,B] = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \\ \end{array}\right)$$

and

$$e^A = \left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array}\right ),\quad\quad e^B = \left(\begin{array}{cc} 1 & 0 \\ 1 & 1 \\ \end{array}\right),\quad\quad e^{[A,B]} = \left(\begin{array}{cc} e & 0 \\ 0 & e^{-1} \\ \end{array}\right)$$

$$e^A e^B = \left(\begin{array}{cc} 2 & 1 \\ 1 & 1 \\ \end{array}\right)$$

$$e^B e^A = \left(\begin{array}{cc} 1 & 1 \\ 1 & 2 \\ \end{array}\right)$$

$$\left(\begin{array}{cc} 1 & 1 \\ 1 & 2 \\ \end{array}\right)\left(\begin{array}{cc} e & 0 \\ 0 & e^{-1} \\ \end{array}\right) = \left(\begin{array}{cc} e & e^{-1} \\ e & 2e^{-1} \\ \end{array}\right)$$

So that formula doesn't seem to be correct. Where have you found it?

5. Sep 11, 2008

jostpuur

I just recalled a distant memory... according to which we are supposed to assume [A,[A,B]] = 0 = [B,[A,B]] with this formula? Could this be the case? Anyway, if you don't find mistake in my example, we will need to assume something for sure.

6. Sep 11, 2008

ismaili

Sorry that I forgot to say that formula holds when $$[A,B]$$ is a c-number.
I met this formula in Green and Witten's string theory book.
One of the derivation which I just found and benefit from Tim's hint is as follows.
We first calculate $$[A,e^B] = [A,B]e^B$$.
Then, we found
$$A^ie^B = e^BA^i + ie^BA^{i-1}[A,B] + C^i_2e^BA^{i-2}[A,B]^2 + \cdots$$
Hence, we can see that
$$e^Ae^B = e^Be^Ae^{[A,B]}$$

7. Sep 11, 2008

jostpuur

I see. Actually you mentioned that in the next post, but I wasn't reading anything else than the first post carefully.