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Quantum harmonic oscillator: average number of energy levels

  1. Feb 8, 2015 #1
    1. The problem statement, all variables and given/known data

    I must find the average number of energy levels of quantum harmonic oscillator at temperature T, and the answer is given as

    upload_2015-2-8_17-11-27.png

    I must use Boltzmann distribution and the sum of geometric progression. For finding the average value I must use the equation

    <F>=trace(F*rho)

    Where rho is the density matrix, given as

    upload_2015-2-8_17-15-39.png

    Where p is the probability and n is the energy eigenstate of oscillator. For F I must use b+b

    2. Relevant equations


    3. The attempt at a solution

    upload_2015-2-8_17-30-31.png
     

    Attached Files:

  2. jcsd
  3. Feb 8, 2015 #2

    TSny

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    Your expression for p(n) needs to be normalized so that the sum of all probabilities equals 1.

    If Amn are the matrix elements of a matrix A, how do you find the trace? If A is the matrix inside of your trace expression, can you find an explicit expression for Amn?
     
    Last edited: Feb 8, 2015
  4. Feb 9, 2015 #3
    It would be sum ∑Amm over m
     
  5. Feb 9, 2015 #4

    TSny

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    Yes,

    For an operator A, how would you express Amm in bra-ket notation?
     
  6. Feb 10, 2015 #5
    That would be <m|A|m> . So applying that I think I would get

    upload_2015-2-10_19-2-29.png

    But still, how to apply b+b to that sum?
     
  7. Feb 10, 2015 #6
    I got it wrong, <m|n>=kronecker's delta not unit matrix .... but that means I would lose the sum altogether!

    EDIT:
    Still wrong, the trace is also sum Σ<m|A|m> over m so in the end it should be the way I showed in the last post. But the question of b+b remains
     
    Last edited: Feb 10, 2015
  8. Feb 10, 2015 #7

    TSny

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    You have neglected the effect of the operator ##b^{\dagger}b##. You can bring ##b^{\dagger}b## inside your sum over n and let it act on |n>.
     
  9. Feb 10, 2015 #8
    That would bring n in front of |n> and I would get the sum:

    upload_2015-2-10_19-28-32.png

    But this is not geometric progression
     
  10. Feb 10, 2015 #9

    TSny

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    Remember, you did not normalize your probabilities. So, your result so far is off by an overall normalization factor ##Z##. The first exponential in your result does not depend on the index n, so you can pull it out of the sum. You are left with$$\sum_{n=0}^{\infty} ne^{-n \beta \hbar \omega}$$ where ##\beta =1/ kT##.

    The trick is to note that $$ne^{-n \beta \hbar \omega} = -\frac{1}{\hbar \omega} \frac{\partial }{\partial \beta}e^{-n \beta \hbar \omega}$$.
     
  11. Feb 10, 2015 #10
    Thank you. The sum of arithmetic-geometric series would yield the same result: http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence#Sum_to_infinite_terms

    I get

    upload_2015-2-10_20-50-37.png

    However, are you sure it is the right result? I can't see how it simplifies to upload_2015-2-10_20-51-40.png . Or do I just need some extra normalization?

    Edit:

    I think according to normalization I should just take

    upload_2015-2-10_20-59-55.png

    And I can see how it gives me the answer I require ;)
     
    Last edited: Feb 10, 2015
  12. Feb 10, 2015 #11

    TSny

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    Once you put in your normalization factor for the probabilities, you'll see that you'll get the desired result.
     
  13. Feb 10, 2015 #12

    TSny

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  14. Feb 10, 2015 #13
    I need to divide the result with Z?
     
  15. Feb 10, 2015 #14

    TSny

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    Yes. It's important to understand that your original expressions for the p(n)'s needed to be divided by ##Z##.
     
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