# Relation between the $N$ particle partition function and probability?

For the 1 particle partition function the probability that the particle is in the state with energy $\varepsilon_i$ is given by: $$P_i =\frac{e^{-\varepsilon_i \beta}}{Z_1}$$ where $Z_2$ is the 1 particle partition function. In the case of an $N$ particle partition function, if the particles are distinguishable we have: $$Z=Z_1^N$$ and if they are not we have: $$Z=\frac{Z_1^N}{N!}$$ But do these relate in anyway to a probability(in the same way that $Z_1$ acts as a normalising factor) and if so the probability of what?

## 1 Answer

Yes, the Z partition function acts as a normalizing factor to compute the probability that the system of the N particles has an energy $\epsilon_i$. It is obviously more complex in the case of many particles system since a lot of configurations may lead to that energy level.

For example, consider a 2 particles system with 0 and $\epsilon$ as the allowable energy levels for each particles. Obviously, there is only 1 state (0,0) that leads to the total energy 0 and one state($\epsilon$,$\epsilon$) that leads to the energy level $2\epsilon$. However, there are 2 states leading to the total energy level $\epsilon$ which are (0,$\epsilon$) and ($\epsilon$,0). If both particles are identical we would consider that these 2 states are actually the same which accounts for the $N!$ in the expression you gave.

To conclude, the Z partition is a way to list all the allowable configurations leading to a macroscopic state. It is the most important quantity in statistical physics since all macroscopic quantities can be computed from its knowledge. Indeed, the entropy(average uncertainty of the system),average energy, Free energy, etc... can all be deduced from the partition function.

• I agree with the answer, but disagree with the characterization of the partition function as the most important quantity in statistical physics (even though it is often presented as such in textbooks). This central role should be reserved to the probability measure itself. Indeed, the latter gives you access to much more information (even about macroscopic quantities) than the partition function, which only gives you access to the statistical properties of the macroscopic quantities appearing in the Boltzmann weight: energy, magnetization, etc.). – Yvan Velenik Feb 1 '16 at 15:32
• In the example you have given would I therefore be right in saying that the probability the system is in the state with energy $2\varepsilon$ would be; $$P=\frac{2! e^{-2\varepsilon \beta}}{Z_1^2}$$ – Quantum spaghettification Feb 1 '16 at 16:32
• The partition function of the 2-particles system is : $Z=1+2e^{-\beta\epsilon}+e^{-2\beta\epsilon}$ which is the same as the square of the 1-particle system : $Z_{1}^{2}=(1+e^{-\beta\epsilon})^{2}$. Now the probability that a state with $2\epsilon$ arises is : $P(\varepsilon=2\epsilon)=\frac{e^{-2\beta\epsilon}}{Z}$ However, this must be divided by 2! if the 2 particles are supposed identical. Notice that this $2\epsilon$ state is less likely than the $0\epsilon$ state even though it has the same number of occurrence. – Ronan Tarik Drevon Feb 2 '16 at 15:15