Prove $\hat{A}+\hat{B}$ Commutator $\lambda$ Complex Number Relation

[kcirick]

Question:
If $\hat{A}$ and $\hat{B}$ are two operators such that $\left[\hat{A},\hat{B}\right] = \lambda$, where $\lambda$ is a complex number, and if $\mu$ is a second complex number, prove that:

$$e^{\mu\left(\hat{A}+\hat{B}\right)}=e^{\mu\hat{A}}e^{\mu\hat{B}}e^{-\mu^{2}\frac{\lambda}{2}}$$

What I have so far:
One can expand the exponential into power series:

$$e^{\mu\left(\hat{A}+\hat{B}\right)}=\sum_{n=0}^{\infty}\frac{\left(\mu\left(\hat{A}+\hat{B}\right)\right)^{n}}{n!}$$

$$e^{\mu\hat{A}}e^{\mu\hat{B}}e^{-\mu^{2}\frac{\lambda}{2}}=\sum_{p,q,r=0}^{\infty}\frac{\mu^{p}\hat{A}^{p}}{p!}\frac{\mu^{q}\hat{B}^{q}}{q!}\frac{\left(-\frac{\mu^2}{2}\right)^{r}\left(\hat{A}\hat{B}-\hat{B}\hat{A}\right)^{r}}{r!}$$

But I don't get any clues from expanding. So I went onto doing something different:

$$\left[\hat{A},\hat{B}\right]=\lambda \Rightarrow \left[\hat{A},\hat{B}\right]-\lambda=0$$
$$\hat{A}\hat{B}=\hat{B}\hat{A}+\lambda$$
$$\hat{B}\hat{A}=\hat{A}\hat{B}-\lambda$$

$$\left(\hat{A}+\hat{B}\right)^{2}=\hat{A}^2+\hat{A}\hat{B}+\hat{B}\hat{A}+\hat{B}^2=\hat{A}^2+\hat{A}\hat{B}+\hat{A}\hat{B}-\lambda+\hat{B}^2=\left(\hat{A}^2+2\hat{A}\hat{B}+\hat{B}^2\right)-\lambda$$

... then I proceeded to the power of 3, but I didn't get a pattern that I was hoping to get, so I don't know what I would get for power of n. Then I don't know what else to try. If any of you can help me out, It will get greatly appreciated!

Thank you!

 

[Tomsk]

Try writing out the first few terms for $\mu, \mu^2, \mu^3 etc$ for the sum

$$e^{\mu\hat{A}}e^{\mu\hat{B}}e^{-\mu^{2}\frac{\lambda}{2}}=\sum_{p,q,r=0}^{\infty}\frac{\mu^{p}\hat{A}^{p}}{p!}\frac{\mu^{q}\hat{B}^{q}}{q!}\frac{\left(-\frac{\mu^2}{2}\right)^{r}\left(\hat{A}\hat{B}-\hat{B}\hat{A}\right)^{r}}{r!}$$

and then rearrange. Remember that p, q, and r are all independant of each other, so you can have p=1, q=r=0 etc. I think that should work.

 

[dextercioby]

Just use the Campbel-Baker-Hausorff formula.

Daniel.