# Non-commutative Binomial

1. Sep 12, 2010

### Oerg

1. The problem statement, all variables and given/known data
Well, I was trying to expand say for the third power of (A+B), where A and B are non-commutative.

3. The attempt at a solution

I get

$$(A+B)^3=(A^2+AB+BA+B^2)(A+B)=A^3+ABA+BA^2+B^2A+A^2B+AB^2+BAB+B^3$$

but from a few sources online, it should be

$$(A+B)^3=(A^3+3A^2B+3AB^2+B^3)-3(AB-BA)(A+B)$$

Now, these 2 results don't seem to be equivalent and I have checked this by expanding it myself. I have been thinking for quite some time about what is wrong with my answer but I seem to be getting nowhere. Any help would be greatly appreciated.

2. Sep 12, 2010

### vela

Staff Emeritus
Your answer is correct. I don't see you can get the second result without some extra condition on A and B.

3. Sep 12, 2010

### Oerg

the larger problem at hand is actually this:

http://www.voofie.com/content/103/proving-the-identity-eab-e-lambda2-ea-eb-if-ablambda/ [Broken]

and similarly

The question is to prove that

$$e^{\mu(A+B)}=e^{\mu A}e^{\mu B}e^{-\mu^2 [A,B]}$$-equation 1

, where $$\mu$$ is some complex number.

in the first post in the pf thread in the link above, the user got the following

and similarly the other user in the other link had the same results. Additionally, when I expanded the RHS in equation 1 with respect to $$\mu$$, I got the same results for power 3 that the 2 users got. But when I tried to expand the LHS in a power series, I did not get the RHS and I got results similar the my first post in this thread.

Last edited by a moderator: May 4, 2017
4. Sep 12, 2010

### vela

Staff Emeritus
Oh, OK. You need the assumption that A and B both commute with [A,B], which is indeed true if [A,B]=k where k is a complex number. In that case, you can show that

$$(A+B)^3=(A^3+3A^2B+3AB^2+B^3)-3[A,B](A+B)-[2A+B,[A,B]]=(A^3+3A^2B+3AB^2+B^3)-3[A,B](A+B)$$

Last edited: Sep 12, 2010
5. Sep 12, 2010

### Oerg

ahh I see many thanks