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Quantum harmonic oscillator: average number of energy levels

  • #1
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Homework Statement



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

Homework Equations




The Attempt at a Solution


[/B]
upload_2015-2-8_17-30-31.png
 

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Answers and Replies

  • #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:
  • #3
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It would be sum ∑Amm over m
 
  • #4
TSny
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It would be sum ∑Amm over m
Yes,

For an operator A, how would you express Amm in bra-ket notation?
 
  • #5
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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?
 
  • #6
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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:
  • #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>.
 
  • #8
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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
 
  • #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 $$ne^{-n \beta \hbar \omega} = -\frac{1}{\hbar \omega} \frac{\partial }{\partial \beta}e^{-n \beta \hbar \omega}$$.
 
  • #10
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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:
  • #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.
 
  • #12
TSny
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  • #13
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I need to divide the result with Z?
 
  • #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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