# Commutator question. [A,B] =.lambda proof

1. Nov 14, 2013

### Jreyes613

1. The problem statement, all variables and given/known data
Hello!
I'm having troubles with this proof.
given two operators A &B , such that [A,B] = λ where λ is complex,and μ is also complex, show that

exp{μ(A+B)} = exp{μA}exp{μB}exp{(-μ^2λ)/2}

2. Relevant equations

[A,B] = λ.
[A,B] = AB-BA = λ

3. The attempt at a solution
I have tried expanding the right side of the equation into three different exponentials by subbing in λ=AB-BA.
I get exp{μ(2A+2B-μAB+μBA}
but i seem to have a factor of two extra...
I'm not even sure if i am on the right track, so if anyone could give me a starting point it would be greatly appreciated!

2. Nov 14, 2013

### MisterX

It seems like maybe you tried adding the exponents on the right hand side. That's not allowed. If $e^Ae^B$ generally is not equal to $e^{A+B}$. That would contradict what you are trying to show. Anyway, I'm not sure the best way to solve this problem, but what you are trying to show is.

$$\sum_{n=0}^\infty \frac{\mu^nA^n}{n!} \sum_{m=0}^\infty \frac{\mu^mB^m}{m!} \sum_{a=0}^\infty \frac{\left(-1\right)^a\mu^{2a}\lambda^a}{2^a a!}$$
$$=\sum_{b=0}^\infty \frac{\mu^b (A+B)^b}{b!}$$

You could try seeing if you can figure out where at the terms come from in the first expression.
$(A+B)^0= 1$
$(A+B)^1= A + B$
$(A+B)^2= A^2 + 2AB - [A,B]+ B^2 = A^2 + 2AB + B^2 - \lambda$
$(A+B)^3= (A+B)\left( A^2 + 2AB + B^2 - \lambda\right)$

However I'm really not sure exactly how to get this result.

3. Nov 14, 2013

### hilbert2

That's the Baker-Campbell-Hausdorff formula. Search with Google and you will probably find help in the proof.

4. Nov 15, 2013

### Jreyes613

Thank you guys.
Yes , my TA said that you can use the Baker -Campbell-Hausdorff formula.

I will be able to solve it now, thanks!!