# 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?

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.
• 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; $$P=\frac{2! e^{-2\varepsilon \beta}}{Z_1^2}$$ Feb 1, 2016 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. Feb 2, 2016 at 15:15