# Interpretation verification: Partition function vs. number of states

1. May 3, 2014

### HJ Farnsworth

Greetings,

I have been studying stat mech lately, and while I have gotten good at using partition functions to solve problems, I wanted to check my interpretation of what a partition function is, and especially to contrast it with the number of states. So, I'm just looking for a yes or no to what I am saying, and if no, an explanation of why I am wrong and what the correct interpretation would be.

Let $\Omega (E)$ denote the number of states with energy less than or equal to $E$, let $\Omega_{\delta E} (E)$ denote the number of states with energy between $E-\delta E$ and $E$ (so essentially, the degeneracy of the energy level $E$), and let $\Omega$ denote the total number of states at all energy levels. (Note that we usually have $\Omega_{\delta E} (E) \approx \Omega (E)$ due to rapidly increasing density of states as a function of energy).

For a given system, $\Omega = \sum_{E_{n}} \Omega_{\delta E} (E_{n})$. For a system in contact with a heat bath, $Z=\sum_{n} e^{-\beta E_{n}} = \sum_{E_{n}} \Omega_{\delta E_{n}} (E) e^{-\beta E_{n}}$ (just using canonical, interpretation below proceeds similarly for grand canonical).

So, both $\Omega$ and $Z$ count all of the possible states, but while $\Omega$ does this directly by simply adding the degeneracy at each energy level over all energy levels, $Z$ weights the degeneracy at each energy level by the Boltzmann factor $e^{-\beta E_{n}}$, which I guess I would think of as a measure of the degree to which the heat bath would "allow" the system to be at that energy level.

Do people agree with this?

Thanks very much for any help that you can give.

-HJ Farnsworth

2. May 3, 2014

### WannabeNewton

Hi HJ! I would say that's a perfectly fine interpretation of the canonical partition function. The thing to keep in mind is that the number of microstates accessible to the heat bath is an extremely rapidly increasing function of its energy, particularly since by assumption the heat bath has far more degrees of freedom than the system in question, and so the probability of the system in question being in a microstate of very high energy (relative to its ground state energy) is very low since this would require the heat bath to have access to only a few microstates as per conservation of energy. What the partition function then does is count all accessible microstates of the system in question, weighting each with the Boltzmann factor in accordance with the above remark.

3. May 3, 2014

### HJ Farnsworth

Hi WannabeNewton,

Awesome, thank you very much for the response and elaboration!

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