1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expectation value in coherent state

  1. Oct 30, 2016 #1
    1. The problem statement, all variables and given/known data

    In a coherent state ##|\alpha\rangle##, letting ##P(n)## denote the probability of finding ##n^{\text{th}}## harmomic oscillator state. Show that

    $$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)=|\alpha|^{2}}$$

    2. Relevant equations

    3. The attempt at a solution

    A coherent state ##\alpha\rangle## is defined as ##D(\alpha)|\alpha\rangle = \exp(\alpha a^{\dagger}-\alpha^{*}a)|\alpha\rangle## so that

    $$\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)= \sum\limits_{n}n\ |\langle n|\alpha\rangle|^{2}}$$

    Therefore, it is prudent to evaluate ##|\alpha\rangle## in terms of ##|n\rangle## as follows:

    ##\displaystyle{|\alpha\rangle = \exp\ (\alpha a^{\dagger}-\alpha^{*}a)|0\rangle}##

    ##\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \exp\ (\alpha a^{\dagger})|0\rangle}##

    ##\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{n'!}(a^{\dagger})^{n'}|0\rangle}\bigg)##

    ##\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}|n'\rangle}\bigg)##

    ##\displaystyle{\langle n|\alpha\rangle = \langle n|\left(\exp (-|\alpha|^{2}/2) \exp\ (-\alpha^{*}a) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}\right) |n'\rangle}.##

    Am I correct so far?
     
  2. jcsd
  3. Oct 30, 2016 #2

    strangerep

    User Avatar
    Science Advisor

    OK, so you're using the Zassenhaus formula in this step, but note that the formula can be written in 2 ways. I.e., you could put ##\exp(-\alpha^*a)## on the right and change the sign of the ##|\alpha|^2## exponent. Then, what is $$\exp(-\alpha^*a)|0\rangle $$ ?
     
  4. Oct 30, 2016 #3
    ##\displaystyle{\exp(-\alpha^*a)|0\rangle=|0\rangle}##, but then, using Baker-Campbell-Hausdorff, I get

    ##\displaystyle{\exp(\alpha a^{\dagger}-\alpha^{*}a)}|0\rangle = \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) \exp\left(\frac{1}{2}[\alpha a^{\dagger},-\alpha^{*}a]\right)|0\rangle##

    ##= \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) \exp\left(-\frac{1}{2}|\alpha|^{2}[a^{\dagger},a]\right)|0\rangle##

    ##= \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) \exp\left(\frac{1}{2}|\alpha|^{2}\right)|0\rangle##

    ##= \exp\left(\frac{1}{2}|\alpha|^{2}\right) \exp(\alpha a^{\dagger}) \exp(-\alpha^{*}a) |0\rangle##

    ##= \exp\left(\frac{1}{2}|\alpha|^{2}\right) \exp(\alpha a^{\dagger}) |0\rangle,##

    when, in fact, I should have

    ##= \exp\left(-\frac{1}{2}|\alpha|^{2}\right) \exp(\alpha a^{\dagger}) |0\rangle.##

    What am I doing wrong here?
     
  5. Oct 30, 2016 #4

    strangerep

    User Avatar
    Science Advisor

    Check the Zassenhaus formula. Shouldn't the 3rd exponent have a minus sign?

    (BTW, you can shorten your latex a bit by using just double-dollar, instead of double-hash and \displaystyle.}
     
  6. Oct 30, 2016 #5
    Why use the Zassenhaus formula and not the BCH formula?
     
  7. Oct 30, 2016 #6

    strangerep

    User Avatar
    Science Advisor

    Look up the "Zassenhaus formula" on Wikipedia. It is simply a special case of BCH.
     
  8. Oct 30, 2016 #7
    But then the BCH formula says that

    $$\exp(A+B)=\exp(A)\exp(B)\exp\left(\frac{1}{2}[A,B]\right)$$

    and the Zassenhaus formula says that

    $$\exp(t(X+Y))=\exp(tX)\exp(tY)\exp\left(-\frac{t^{2}}{2}[X,Y]\right)$$.

    I don't see how the Zassenhaus formula is a special case of the BCH formula, if ##A=tX## and ##B=tY##.
     
  9. Oct 30, 2016 #8

    strangerep

    User Avatar
    Science Advisor

    Where are you getting your BCH formula from? There's another form of BCH which looks like this:
    $$e^A e^B ~=~ e^{A+B+[A,B]/2 ~+~ \dots} ~.$$
     
  10. Oct 30, 2016 #9
    Sorry, my bad!

    How do you prove the Zassenhaus formula from the BCH formula?
     
  11. Oct 30, 2016 #10
  12. Oct 30, 2016 #11

    strangerep

    User Avatar
    Science Advisor

    Actually, I think you do, at least for the current simple case. In fact that's trivial because we're only dealing here with the case where ##[A,B]## is a c-number (meaning it commutes with everything). So you can factor it out and transfer it across to the other side of the equation.

    The ##t## is a red herring. Just think in terms of A and B (i.e., put ##t=1## in those Wiki formulas). Then put ##A = \alpha a^\dagger##, etc.
     
  13. Oct 30, 2016 #12
    Ah, I see! So, I have

    ##\displaystyle{|\alpha\rangle = \exp\left(-\frac{1}{2}|\alpha|^{2}\right)\exp\left(\alpha a^{\dagger}\right)\exp\left(-\alpha^{*}a\right)|0\rangle}##

    ##\displaystyle{|\alpha\rangle = \exp\left(-\frac{1}{2}|\alpha|^{2}\right)\exp\left(\alpha a^{\dagger}\right)|0\rangle}##

    ##\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{n'!}(a^{\dagger})^{n'}|0\rangle}\bigg)##

    ##\displaystyle{|\alpha\rangle = \exp (-|\alpha|^{2}/2) \bigg(\sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}}|n'\rangle}\bigg)##

    ##\displaystyle{\langle n|\alpha\rangle = \langle n|\left(\exp (-|\alpha|^{2}/2) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}} |n'\rangle\right)}##

    ##\displaystyle{= \left(\exp (-|\alpha|^{2}/2) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}} \langle n|n'\rangle\right)}##

    ##\displaystyle{= \left(\exp (-|\alpha|^{2}/2) \sum\limits_{n'=0}^{\infty} \frac{\alpha^{n'}}{\sqrt{n'!}} \delta_{nn'}\right)}##

    ##\displaystyle{= \left(\exp (-|\alpha|^{2}/2) \frac{\alpha^{n}}{\sqrt{n!}}\right)},##

    which is a Poisson distribution with mean ##|\alpha|^{2}## and ##n## events in each interval so that

    ##\displaystyle{\langle\hat{n}\rangle \equiv \sum\limits_{n}n\ P(n)}##

    ##\displaystyle{= \sum\limits_{n}n \left(\exp (-|\alpha|^{2}/2) \frac{\alpha^{n}}{\sqrt{n!}}\right)\left(\exp(-|\alpha|^{2}/2)\frac{{\alpha^{*}}^{n}}{\sqrt{n!}}\right)}##

    ##\displaystyle{= \sum\limits_{n}n \exp (-|\alpha|^{2}) \frac{|\alpha|^{2n}}{n!}}##

    is the mean squared of the Poisson distribution and is equal to ##|\alpha|^{2}##.

    What do you think?
     
  14. Oct 31, 2016 #13

    strangerep

    User Avatar
    Science Advisor

    It's looking a lot better. Unfortunately, I have to dash out now. I'll try to check it more carefully later.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted