Quantum harmonic oscillator: average number of energy levels

[Earthland]

Homework Statement

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

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

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

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[TSny]

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

If A_mn 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 A_mn?

 

[Earthland]

It would be sum ∑A_mm over m

 

[TSny]

It would be sum ∑A_mm over m

Yes,

For an operator A, how would you express A_mm in bra-ket notation?

 

[Earthland]

That would be <m|A|m> . So applying that I think I would get

But still, how to apply b⁺b to that sum?

 

[Earthland]

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

 

[TSny]

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>.

 

[Earthland]

That would bring n in front of |n> and I would get the sum:

But this is not geometric progression

 

[TSny]

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 $$ne^{-n \beta \hbar \omega} = -\frac{1}{\hbar \omega} \frac{\partial }{\partial \beta}e^{-n \beta \hbar \omega}$$.

 

[Earthland]

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

However, are you sure it is the right result? I can't see how it simplifies to

. Or do I just need some extra normalization?

Edit:

I think according to normalization I should just take

And I can see how it gives me the answer I require ;)

 

[TSny]

Once you put in your normalization factor for the probabilities, you'll see that you'll get the desired result.

 

[TSny]

Thank you. The sum of arithmetic-geometric series would yield the same result: http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence#Sum_to_infinite_termsI think according to normalization I should just take

View attachment 78901

And I can see how it gives me the answer I require ;)

The expression does not equal 1. It equals $Z$ (the partition function). But $Z$ is what you need to use to normalize the probabilities.

 

[Earthland]

I need to divide the result with Z?

 

[TSny]

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