# Operator Identity

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!

Related Advanced Physics Homework Help News on Phys.org
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