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Fock states as integrals of coherent states

  1. Nov 11, 2017 #1
    Edit: I'm pretty sure I have answered my own question. I think I need to sandwich the integral between a bra and ket to pick out one term from the sum.

    1. The problem statement, all variables and given/known data


    Show that a Fock state ##|n\rangle## can be represented by the integral

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} e^{-\frac{r^2}{2}} \int_0^{2 \pi} d\theta e^{-in\theta} |re^{i\theta}\rangle$$

    where ##|r e^{i\theta}\rangle## is a coherent state ##|\alpha\rangle##

    2. Relevant equations


    3. The attempt at a solution

    The Fock state expansion is ##|\alpha\rangle = e^{-\frac{|\alpha|^2}{2}} \sum_{n = 0}^\infty \frac{\alpha^n}{\sqrt{n!}} |n\rangle## and I substitute this into the expression for ##|n\rangle## to get

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} e^{-\frac{r^2}{2}} \int_0^{2 \pi} d\theta e^{-in\theta} e^{-\frac{|\alpha|^2}{2}} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} |m\rangle$$

    Substituting ##\alpha = re^{i \theta}## allows for cancellation of the ##e^{r^2 /2}## term to give

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} |m\rangle$$

    I'm stuck here because i'm unsure of how to deal with the sum inside the integral. If m = n, can I bring all of the terms containing n inside the sum?

    That would give me the answer I want, but i'm not sure if i'm allowed to do this.

    Or is there something I can do to pick out one term from the sum? like acting with a bra and ket of n sandwiching the integral. I think this makes more sense

    Thank you
     
    Last edited: Nov 11, 2017
  2. jcsd
  3. Nov 13, 2017 #2

    Orodruin

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    So what happens when you do that? Note that you need to do it using a general ket ##\langle k|##. To reach the sought conclusion, this should be one when ##n=k## and zero otherwise.
     
  4. Nov 13, 2017 #3
    The bra can pass through everything on the RHS until it meets the ket, giving a Kronecker delta, which picks out the 'n' term from the sum. This leaves behind my new ket, everything cancels and i'm left with ##|n\rangle = |n\rangle##
     
  5. Nov 13, 2017 #4

    Orodruin

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    This seems to be doing exactly what I warned you about. Please write out your argumentation explicitly.
     
  6. Nov 13, 2017 #5
    Ok, starting with

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} |m\rangle$$

    I sandwich both sides between a pair of bra-kets.

    $$\langle k|n\rangle|k\rangle = \frac{\sqrt{n!}}{2 \pi r^n} \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} \langle k|m\rangle |k\rangle$$

    If I say that k = n, then due to orthonormality

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} \delta_{nm} |n\rangle$$

    this reduces the sum to a single term

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} \int_0^{2 \pi} d\theta e^{-in\theta} \frac{\alpha^n}{\sqrt{n!}} |n\rangle$$

    which after cancelling everything and evaluating the integral just leaves

    $$|n\rangle = |n\rangle$$
     
  7. Nov 13, 2017 #6

    Orodruin

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    With that you would also need to check all of the other choices of ##k## to make sure that they are also valid. It will be easier if you sum over ##k## since ##\sum_k |k\rangle\langle k | = 1## is the identity operator.
     
  8. Nov 13, 2017 #7
    Ah, I think I see what you are saying.

    So I would insert the identity $$\frac{\sqrt{n!}}{2 \pi r^n} \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} \sum_{k=0}^{\infty}| k\rangle\langle k |n\rangle$$

    which yields a ##\delta_{kn}##

    but how do I then deal with the sum over m? Do I do this in conjunction with sandwiching everything between the bra-kets?
     
    Last edited: Nov 13, 2017
  9. Nov 13, 2017 #8

    Orodruin

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    The ##|n\rangle## in the sum should be an ##|m\rangle##. I also suggest leaving the sum over ##k## for last so take it outside the integral and everything.
     
  10. Nov 13, 2017 #9
    Sorry, that was a typo.

    I'm not sure exactly how this works:

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} |m\rangle$$

    Do I sandwich everything between k, and then insert the identity as follows:

    $$\langle k|n\rangle |k\rangle = \frac{\sqrt{n!}}{2 \pi r^n} \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} \langle k|m\rangle |k\rangle$$

    $$\langle k|n\rangle |k\rangle = \frac{\sqrt{n!}}{2 \pi r^n} \sum_{k=0}^{\infty} |k\rangle \langle k| \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} \langle k|m\rangle |k\rangle$$

    which is zero everywhere except for when k = n

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} |n\rangle \langle n| \int_0^{2 \pi} d\theta e^{-in\theta} \sum_{m = 0}^\infty \frac{\alpha^m}{\sqrt{m!}} \delta_{nm}|m\rangle$$

    now with n=m, this pulls out one term from the sum,

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} |n\rangle \langle n| \int_0^{2 \pi} d\theta e^{-in\theta} \frac{\alpha^n}{\sqrt{n!}} |n\rangle$$

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} |n\rangle \langle n|n\rangle \int_0^{2 \pi} d\theta e^{-in\theta} \frac{\alpha^n}{\sqrt{n!}}$$

    $$|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} |n\rangle \int_0^{2 \pi} d\theta e^{-in\theta} \frac{\alpha^n}{\sqrt{n!}}$$

    then everything cancels leaving ##|n\rangle = |n\rangle##
     
  11. Nov 13, 2017 #10

    Orodruin

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    I would not start from this, it is what you want to prove. Instead, I suggest starting from the integral expression and multiplying it by the completeness relation. Since everything else is just numbers, you can move the ##|k\rangle\langle k|## around freely until it meets the ##|m\rangle## while still keeping the sum outside.
     
  12. Nov 13, 2017 #11
    That is just the integral expression with the fock state expansion applied. Surely I have to do that so the ket bra from the completeness relation can meet ##|m\rangle##?
     
  13. Nov 13, 2017 #12

    Orodruin

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    Surely you have to do what? Move the sum inside? No. Move the ket-bra expression inside, yes, but everything you have to move it past is numbers.
     
  14. Nov 13, 2017 #13
    Sorry, I think we're misunderstanding each other. I was a bit confused by what you meant about the starting point. Let me start again.

    I am told that ##|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} e^{r^2 / 2} \int^{\infty}_{0}d\theta e^{-in\theta}|re^{i\theta}\rangle##

    If I multiply by the completeness relation and bring the ket-bra inside, I get

    ##\sum_{k=0}^{\infty}|k\rangle \langle k||n\rangle = \sum_{k=0}^{\infty} \frac{\sqrt{n!}}{2 \pi r^n} e^{r^2 / 2} \int^{\infty}_{0}d\theta e^{-in\theta}|k\rangle \langle k||re^{i\theta}\rangle##

    I now apply the Fock state expansion of the coherent state to get

    ##|n\rangle = \sum_{k=0}^{\infty} \frac{\sqrt{n!}}{2 \pi r^n} e^{r^2 / 2} \int^{\infty}_{0}d\theta e^{-in\theta}|k\rangle \langle k|e^{-r^2/2} \sum_{m=0}^{\infty} \frac{(r e^{i\theta})^m}{\sqrt{m!}} |m\rangle##

    and I can push the ket-bra further inside until it meets the m

    ##|n\rangle = \sum_{k=0}^{\infty} \frac{\sqrt{n!}}{2 \pi r^n} e^{r^2 / 2} \int^{\infty}_{0}d\theta e^{-in\theta}e^{-r^2/2} \sum_{m=0}^{\infty} \frac{(r e^{i\theta})^m}{\sqrt{m!}}|k\rangle \langle k| |m\rangle##

    I now get the kronecker delta that picks out 1 term from the sum

    ##|n\rangle = \frac{\sqrt{n!}}{2 \pi r^n} e^{r^2 / 2} \int^{\infty}_{0}d\theta e^{-in\theta}e^{-r^2/2} \sum_{m=0}^{\infty} \frac{(r e^{i\theta})^m}{\sqrt{m!}}|n\rangle \delta_{nm}##

    which cancels down to just ##|n\rangle##
     
  15. Nov 13, 2017 #14

    Orodruin

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    You are not told that. You are asked to show that it is true so you should start from one of the sides and show that it is equal to the other.
     
  16. Nov 15, 2017 #15
    Right, I understand what you're saying now. Sorry, for the delay in responding, I got sidetracked by other work.

    Is this better?

    Starting with the RHS and applying the Fock state expansion:

    ##\frac{\sqrt{n!}}{2\pi r^n} e^{r^2/2} \int^{2\pi}_0 d\theta e^{-in\theta} |re^{i\theta}\rangle = \frac{\sqrt{n!}}{2\pi r^n} e^{r^2/2} \int^{2\pi}_0 d\theta e^{-in\theta} e^{-r^2/2} \sum_{m=0}^{\infty} \frac{(r e^{i\theta})^m}{\sqrt{m!}} |m\rangle##

    inserting the completeness relation

    ##RHS = \frac{\sqrt{n!}}{2\pi r^n} \sum^{\infty}_{k=0} \int^{2\pi}_0 d\theta e^{-in\theta} \sum_{m=0}^{\infty} \frac{(r e^{i\theta})^m}{\sqrt{m!}} |k\rangle \langle k|m\rangle##

    ##RHS = \frac{\sqrt{n!}}{2\pi r^n} \sum^{\infty}_{k=0} \int^{2\pi}_0 d\theta e^{-in\theta} \sum_{m=0}^{\infty} \frac{(r e^{i\theta})^m}{\sqrt{m!}} |k\rangle \langle k|m\rangle##

    from the ##\delta_{km}## I get

    ##RHS = \frac{\sqrt{n!}}{2\pi r^n} \sum^{\infty}_{k=0} \int^{2\pi}_0 d\theta e^{-in\theta} \frac{(r e^{i\theta})^k}{\sqrt{k!}} |k\rangle ##

    If I now sandwich the expression between ##\langle n|##...##|n\rangle## I get another Kronecker delta that kills the sum over k, and turns all of the k's to n's, reducing to ##|n\rangle##
     
  17. Nov 15, 2017 #16

    Orodruin

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    You cannot sandwich the expression between ##\langle n|## and ##|n\rangle##, it is a state vector! Instead, perform the integral over ##\theta##.
     
  18. Nov 15, 2017 #17
    $$\int^{2\pi}_0 e^{i\theta(k-n)} = -\frac{i(-1 + e^{2 i \pi (k - n)})}{k-n}$$

    $$RHS = \frac{\sqrt{n!}}{2 \pi r^n} \sum_{k=0}^{\infty} -\frac{i(-1 + e^{2 i \pi (k - n)})}{k-n} \frac{r^k}{\sqrt{k!}} |k\rangle$$

    when k = n this is zero...
     
  19. Nov 15, 2017 #18

    Orodruin

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    When ##k = n## you cannot do the integral like that. When ##k \neq n## it is zero.
     
  20. Nov 15, 2017 #19
    Ah, I see!

    Ok, I got there in the end. Thank you for your help and patience
     
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