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Operator Identity

  1. Nov 6, 2006 #1
    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\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{B}\hat{A}=\hat{A}\hat{B}-\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!
  2. jcsd
  3. Nov 6, 2006 #2
    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.
  4. Nov 10, 2006 #3


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    Science Advisor
    Homework Helper

    Just use the Campbel-Baker-Hausorff formula.

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