# Molecular partition function

1. Apr 16, 2012

### luitzen

The probability of finding the system in microscopic state $i$ is:
$p_{i}=\dfrac{1}{Q}e^{-\beta E_{i}}$

Where $Q$ is the partition function.

Assumption: molecule $n$ occupies the $i_{n}$th molecular state (every molecule is a system).

The total energy becomes:
$E_{i_{1},i_{2},...,i_{N}}=\epsilon_{i_{1}}+ \epsilon_{i_{2}}+...+\epsilon_{i_{N}}$

$Q=\underset{i_{1},i_{2},...,i_{N}}{\sum}e^{-\beta\left(\epsilon_{i_{1}}+ \epsilon_{i_{2}}+...+\epsilon_{i_{N}}\right)}$

$=\underset{q}{\underbrace{\left(\underset{i_{1}} {\sum} e^{-\beta e_{i_{1}}}\right)}}\times\left(\underset{i_{2}} {\sum} e^{-\beta\epsilon_{i_{2}}}\right)\times...\times\left(\underset{i_{N}}{\sum}e^{-\beta\epsilon_{i_{N}}}\right)$

Where $q$ is the molecular or particle partition function.

The partition function becomes $Q=q^{N}$ . This is valid for distinguishable particles only (why?).

The probability of finding molecule $n$ in molecular state $i'_{n}$ is obtained by summing over all system-states subject to the condition that $n$ is in $i'_{n}$

$p_{i'_{n}}=\dfrac{1}{Q}\underset{i_{1},i_{2},...,i_{N}}{\sum}\delta_{i_{n},i'_{n}}e^{-\beta\left(\epsilon_{i_{1}}+ \epsilon_{i_{2}}+...+ \epsilon_{i_{N}}\right)}=\dfrac{1}{Q}e^{-\beta\epsilon_{i'_{n}}}q^{N-1}=\dfrac{1}{q}e^{-\beta\epsilon_{i'_{n}}}$

So why is $Q=q^{N}$ only true when the particles are distinguishable and what does it mean when it is stated that "the probability of finding molecule $n$ in molecular state $i'_{n}$ is obtained by summing over all system-states subject to the condition that $n$ is in $i'_{n}$"

Last edited: Apr 16, 2012
2. Apr 18, 2012

### Bill_K

Q = qN says that each particle occupies the available states independently of the others, thus the overall probability distribution is the product of the individual probability distributions. This is true for classical (distinguishable) particles but not for quantum (indistinguishable) particles.

For example consider how two particles 1, 2 would occupy two states A, B, and suppose that εA = εB so we may ignore the exponentials. For distinguishable particles the individual probabilities are 1/2, independently, so the probability of finding both particles in state A is 1/4. And likewise 1/4 for both in state B. But finding one in A and one in B can happen two ways, so the probability of that is 1/2. By contrast, boson states are counted by occupation number so there are three equally like cases: both in A, both in B or one in each. For fermions there is only one possibility: one in A and one in B.

3. Apr 23, 2012

### luitzen

Thanks for clarifying. I now fully understand it.