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Operator Identity

  • Thread starter kcirick
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  • #1
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Question:
If [itex]\hat{A}[/itex] and [itex]\hat{B}[/itex] are two operators such that [itex]\left[\hat{A},\hat{B}\right] = \lambda[/itex], where [itex]\lambda[/itex] is a complex number, and if [itex]\mu[/itex] is a second complex number, prove that:

[tex] e^{\mu\left(\hat{A}+\hat{B}\right)}=e^{\mu\hat{A}}e^{\mu\hat{B}}e^{-\mu^{2}\frac{\lambda}{2}} [/tex]

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

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

[tex]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!} [/tex]

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

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

[tex]\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[/tex]

... 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!
 

Answers and Replies

  • #2
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Try writing out the first few terms for [itex]\mu, \mu^2, \mu^3 etc[/itex] for the sum

[tex]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!} [/tex]

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.
 
  • #3
dextercioby
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Just use the Campbel-Baker-Hausorff formula.

Daniel.
 

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