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!

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
    [itex]|z>=e^{-\frac{|z|^2}{2}}\sum_0^\infty\frac{z^{n}}{\sqrt{n!}}|n>[/itex]

    Probability=[itex]|<n|z>|^2[/itex]

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

    ShayanJ

    User Avatar
    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

    stevendaryl

    User Avatar
    Staff Emeritus
    Science Advisor

    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

    stevendaryl

    User Avatar
    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].
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted