Harmonic Oscillator and Volume of Unit Cell in Phase Space

[RawrSpoon]

Long time no see, PhysicsForums. Nevertheless, I have gotten myself into a statistical mechanics class where the prof is pretty brutal and while I can usually manage, this problem finally has me stumped. I'd like to be nudged in the right direction, not outright given the answer if possible. I want to learn and keep the memory of how to solve this in my mind, not just get an A. Thanks in advance! Well, here goes.

1. Homework Statement
In class we postulated that integrations over that integrations over the phase space (x, p) of a classical particle are normalized by the Planck constant, h, so that the volume of a unit cell in phase space is equal to h. Here we prove this fact by considering the classical limit of a quantum system which can be calculated exactly.

a) The energy eigenvalues of a quantum harmonic oscillator with frequency $\omega$ are given by $E(n)= \hbar \omega (n+\frac{1}{2})$ with n=0,1,2,... and $\hbar = \frac{h}{2 \pi}$
Calculate the canonical partition function $Z_{qm}(T)$ of the quantum harmonic oscillator.

b)Find the leading term in the asymptotic expansion of $Z_{qm}(T)$ in the limit $\beta \hbar \omega \rightarrow 0$ corresponding to the classical limit $\frac {k_{B} T}{\hbar \omega} \rightarrow \infty$

c) The energy function of a classical harmonic oscillator with mass m, position x, and momentum p is given by $E(x,p) = \frac{p^{2}}{2m} + \frac {1}{2} m {\omega}^{2} x^{2}$

Calculate the canonical partition function $$Z_{cl} (T) = \frac {1}{v} \int_{-\infty}^{\infty} dx \int_{-\infty}^{\infty} dp\ e^{-\beta E(x,p)}$$

Here v is a constant with physical units length $\times$ momentum = energy $\times$ time required to make the partition function dimensionless. This constant will be determined in d) below.

d) Determine the constant v by comparing $Z_{cl} (T)$ found in c) with the classical limit of $Z_{qm} (T)$ found in b).

Homework Equations

$$Z_{qm}(T) = \sum_{n=0}^{\infty} e^{-\beta E(n)}$$ $$\sum_{n=0}^{\infty} x^{n} = \frac {1}{1-x}$$ $$\int_{-\infty}^{\infty} du \ e^{- \frac{1}{2} au^{2}} = \sqrt{\frac{2 \pi}{a}}$$

The Attempt at a Solution

[/B]
a)
$$Z_{qm}(T) = e^{-\frac{1}{2} \beta \hbar \omega} \sum_{n=0}^{\infty} e^{- \beta \hbar \omega n}$$ $$Z_{qm}(T) = e^{-\frac {1}{2} \beta \hbar \omega} \frac {1} {1-e^{- \beta \hbar \omega}}$$

However, I'm lost as for how to start b. I think I have to do a Laurent series, but I'm not quite sure how to go about finding the first term of the Laurent series. Mathematica says the first term is $\frac{1}{2x}$ but I'm not sure whether that's correct, or how to arrive at that answer manually if it IS, in fact, correct.

Again, many thanks in advance.

 

[kuruman]

However, I'm lost as for how to start b.

Start with $x=\beta \hbar \omega$ in which case
$$
Z_{qm}(T) = \frac {e^{-x/2}} {1-e^{- x}}=\frac{1}{e^{x/2}-e^{- x/2}}$$
You are looking for the asymptotic expression of the term on the far right as $x \rightarrow 0$. The denominator is $e^{x/2}-e^{- x/2}$. What does it become for small values of $x$?

 

[RawrSpoon]

Start with $x=\beta \hbar \omega$ in which case
$$
Z_{qm}(T) = \frac {e^{-x/2}} {1-e^{- x}}=\frac{1}{e^{x/2}-e^{- x/2}}$$
You are looking for the asymptotic expression of the term on the far right as $x \rightarrow 0$. The denominator is $e^{x/2}-e^{- x/2}$. What does it become for small values of $x$?

First off, thank you so much for replying!

I did some progress on this, I'm not sure if it's right. So I did something similar, albeit $x=\frac{\beta \hbar \omega}{2}$ This gave me $$Z_{qm}(T) = \frac {e^{-x}}{1-e^{-2x}} = \frac {1}{e^{x}-e^{-x}} = \frac {1}{2 sinh(x)} = \frac {csch(x)}{2}$$

I plugged in increasingly smaller values of x into the above equation, and realized that for every factor of 10 I reduced x, the output was changed by an inversely proportional factor of 10 as well. This made me believe that the first term of any corresponding series for csch(x) would be 1/x. Multiply this by the coefficient associated to csch(x) above, or $\frac{1}{2}$, and I got the first term of the series being $\frac{1}{2x}$, just as Mathematica mentioned the first term of the Laurent expansion would be. It bothers me a bit that I didn't obtain the result as cleanly, but it makes sense to me where this came from.

I then solved c) through the following:
$$Z_{cl}(T)=\frac {1}{v} \int_{-\infty}^{\infty} e^{-\frac{\beta m \hbar x^{2}}{2}}dx \int_{-\infty}^{\infty} e^{-\frac{\beta p^{2}}{2m}}dp$$

Which was easily solved as Gaussian integrals since
$$\int_{-\infty}^{\infty} e^{\frac {a \ u^{2}}{2}}du = \sqrt{\frac{2 \pi}{a}}$$

This gave me a result of
$$Z_{cl}(T) = \frac{2 \pi}{v \beta \omega}$$

So d) just asked to basically combine the two to find what v above is, which I obliged as so:
$$Z_{qm}(T) = Z_{cl}(T)$$
$$\frac{1}{2 \beta \hbar \omega} = \frac{2 \pi}{v \beta \omega}$$
$$\frac{2 \pi}{2 \beta h \omega} = \frac {2 \pi}{v \beta \omega}$$
This gives me an answer of $v=2h$. Is this correct? I think it looks fine, but I just need someone to look over my work I suppose. Again, thank you so much for replying!

 

[kuruman]

Your expression for $E(x,p)$ is incorrect. $E(x,p)=\frac{m \omega^2 x^2}{2}+\frac{p^2}{2m}$. That's not what you have in the $x$ integral. The $p$ integral is OK.

On edit: Also, your limiting expression for $Z_{qm}$ is a factor of 2 too large. I suggest that you find the limit the way I indicated in #2.

 

[RawrSpoon]

Your expression for $E(x,p)$ is incorrect. $E(x,p)=\frac{m \omega^2 x^2}{2}+\frac{p^2}{2m}$. That's not what you have in the $x$ integral. The $p$ integral is OK.

On edit: Also, your limiting expression for $Z_{qm}$ is a factor of 2 too large. I suggest that you find the limit the way I indicated in #2.

I seem to have just copied E incorrectly from my paper, but I have exactly what you mentioned on my paper, so sorry about the mixup!

Gotcha, so multiplying by 1/2 is incorrect. This makes sense, since upon reflection on your response, I've concluded that multiplying by 1/2 was a mistake.

Thank you so much for all your help! I've made tons of progress and I think I've got it down