# Why does this formula for the partition function not include the multiplicity?

I am having problems understanding the formulas used for describing the partition functions and the probability distributions for canonical ensembles.

In the first case I have two formulas for the partition function: I can label each system microstate with $j$, associate it with an energy $E_j$, and state that:

$$p_j = e^{-\beta E_j/Z}\quad\text{with}\quad Z=\sum_je^{-\beta E_j/Z}$$

On the other hand, in my lecture notes, the canonical partition function and the probability distribution are given by:

\begin{align} Z &= \sum_j\Omega(N,Q)e^{-\beta E_j/Z} \\ P(Q) &=(1/Z)\Omega(N,Q)e^{-\beta Q} \end{align}

for a given system of $N$ weakly coupled oscillators and $Q$ quantas.

My question is: Why is the multiplicity $\Omega(N,Q)$ not taken into account in the first case?

• – Janus Boffin Apr 3 '16 at 11:11
• you linked my question @JanusBoffin – SkyTalentz Apr 3 '16 at 11:44
• my guess is that in the first case you are summing over states, whilst in the second you are summing over energies. – Quantum spaghettification Apr 3 '16 at 12:40
• @Quantumspaghettification According to the formula both sums are for the energies $E_j$ – SkyTalentz Apr 3 '16 at 13:04
• @SkyTalentz I have seen the notation you have given where $j$ denotes the state rather then purely an index for energy so $E_j$ would be the energy of the $j$th state and there is nothing stopping $E_j=E_i$ for two states $i$ and $j$. – Quantum spaghettification Apr 3 '16 at 13:08

## 1 Answer

(I am not sure this is an answer but it is to long to be a comment)

Let us create a simple example of a system of $3$ states, state $1$, state $2$ and state $3$. Let state $1$ and state $2$ both have an energy of $E$ and state $3$ have an energy of $E'\ne E$.

Your first summation is summing over individual states. I.e. it is saying 'let us call the energy of state $1$; $E_1$, the energy of state $2$; $E_2$ and the energy of state $3$; $E_3$. The sum then looks something like this: $$Z=\sum_{j=1}^3e^{-\beta E_j}$$ $$=e^{-\beta E_1}+e^{-\beta E_2}+e^{-\beta E_3}$$ $$=2e^{-\beta E}+e^{-\beta E'}$$

Whilst your second summation is summing over individual energies. I.e. it is saying 'let us call the energy $E$; $E_1$ and the energy $E'$; $E_2$. With $\Omega_1=2$ and $\Omega_2=1$ (i.e. the number of states with each energy). our sum now looks something like this: $$Z=\sum_{j=1}^2 \Omega_j e^{-\beta E_j}$$ $$=2e^{-\beta E}+e^{-\beta E'}$$ I hope this clears it up a bit, let me know if you have any further problems.

• Oh ok so these two sum should always give the same partition function for a given system, it's just a question of "perspective"? – SkyTalentz Apr 3 '16 at 16:34
• @SkyTalentz yep that's right, it simply depends on how you look at the problem. – Quantum spaghettification Apr 3 '16 at 16:40