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Homework Help: Coherent States of Harmonic Oscillator

  1. Nov 29, 2014 #1
    1. The problem statement, all variables and given/known data
    Given the coherent state of the harmonic oscillator [tex]|z>=e^{-\frac{|z|^2}{2}}\sum_{n=0}^\infty\frac{z^{n}}{\sqrt{n!}}|n>[/tex]
    compute the probability for finding n quanta in the sate [itex]|z>[/itex] and the average excitation number [itex]<z|n|z>[/itex]

    2. Relevant equations


    Average excitation number = [itex]<z|n|z>[/itex]

    3. The attempt at a solution

    For the first part I did [itex]|<n|z>|^2=|e^{-\frac{|z|^2}{2}}\sum_{m=0}^\infty\frac{z^{m}}{\sqrt{m!}}<n|m>|^2[/itex]

    Which is correct but was I right in using m to specify a different energy eigenstate to n to obtain the kroncker delta?

    For the second part Im not sure how to even write out the problem. Should the eigenstates in [itex]|z>[/itex] and [itex]<z|[/itex] be [itex]|n>[/itex] or something different

    What I think is that it should be is

    [itex]<z|n|z>=e^{-\frac{|z|^2}{2}}\sum_{m=0}^\infty\frac{z^{\dagger m}}{\sqrt{m!}}e^{-\frac{|z|^2}{2}}\sum_{m'=0}^\infty\frac{z^{m'}}{\sqrt{m'!}}<m|n|m'>[/itex]

    But im not sure what <m|n|m'> is equal to, [itex]\delta_{nm}\delta_{nm'}[/itex] maybe?

    Im getting [itex]e^0[/itex] which isnt the right answer as per page 2 of this http://www.phas.ubc.ca/~joanna/phys501/coherent-states.pdf

    Any help would be great.
  2. jcsd
  3. Nov 29, 2014 #2


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    Gold Member

    The first part is correct.
    About your question regarding the second part, Pay attention that [itex] n=a^\dagger a [/itex] and [itex] a^\dagger |n>=\sqrt{n+1} |n+1 > [/itex] and [itex] a |n>=\sqrt{n}|n-1> [/itex].
  4. Nov 29, 2014 #3


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    Staff Emeritus
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    You pretty much have it, except that by definition [itex]n |m'\rangle = m' |m' \rangle[/itex]. So [itex]\langle m |n| m' \rangle = m' \langle m | m' \rangle[/itex].
  5. Nov 29, 2014 #4
    This gives me m' as the answer but the answer is [itex]|z|^2[/itex]?

    Nvm got it using the ladder operators
  6. Nov 29, 2014 #5


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    Staff Emeritus
    Science Advisor

    No, it doesn't give [itex]m'[/itex] as the answer, because [itex]m'[/itex] is summed over. Since [itex]\langle m | n |m' \rangle = m' \delta_{m, m'}[/itex], you have something like:

    [itex]e^{-z^2} \sum_{m=0}^\infty \sum_{m'=0}^\infty \frac{z^m z^{m'}}{\sqrt{m! m'!}} m' \delta_{m,m'}[/itex]

    Because of the [itex]\delta_{m, m'}[/itex] the second sum simplifies to just one term, the one where [itex]m'=m[/itex].
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