Is there a mistake in my non-commutative binomial expansion for (A+B)^3?

[Oerg]

Homework Statement

Well, I was trying to expand say for the third power of (A+B), where A and B are non-commutative.

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.

 

[vela]

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

 

[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/

and similarly

https://www.physicsforums.com/showthread.php?p=2796908

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

Notation I use: C^n = (A+B)^n respecting order, c^n = (a+b)^n with all A's before B's.
e.g.: (A+B)^2 = AA + AB +BA + BB, and (a+b)^2 = AA + 2AB + BB
(A+B)^0 = I
(A+B)^1 = A+B
(A+B)^2 = (a+b)^2 - k
(A+B)^3 = (a+b)^3 - 3k(A+B)
(A+B)^4 =(a+b)^4 - 6k(a+b)^2 + 3k^2
(A+B)^5 =(a+b)^5 - 10k(a+b)^3 + 15k^2(a+b)
(A+B)^6=(a+b)^6 - 15k(a+b)^4 + 45k^2(a+b)^2 - 15k^3

, where [A,B]=k

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.

 

[vela]

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

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

 

[Oerg]

ahh I see many thanks