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Average number of particles in a certain energy level in the Canonical Ensemble

Physics Asked on March 10, 2021

A quantum system has $r$ discrete energy levels $varepsilon_1,varepsilon_2,varepsilon_3,…,varepsilon_r$ and $N$ particles distributed in these levels, with the number of particles at each level denoted by $n_1,n_2,n_3,…,n_r$. I’m trying to find the average number of particles in the $i$-th energy level, $leftlangle n_irightrangle$, and the fluctuation of this average, $leftlangle(Delta n_i)^{2}rightrangle$, using the Canonical Ensemble.

My attempt

The average energy of the system at the state $R$ determined by the occupation numbers $(n_1,n_2,n_3,…,n_r)_R$ can be computed by

$$
langle Erangle=leftlangle E_{R}rightrangle=sum_{R} P_{R} E_{R}
=frac{1}{Z}sum_{R} E_{R} e^{-beta E_{R}}
=-frac{1}{Z}bigg(frac{partial Z}{partial beta}bigg)_{N, V}
=-bigg(frac{partial ln Z }{partial beta}bigg)_{N, V}
$$

With a similar process, keeping in mind that $E_{R} = sum_{r} n_r varepsilon_{r}$, one gets that

$$langle n_irangle = sum_{R} P_{R} n_i
=frac{1}{Z}sum_{R} n_r e^{-beta sum_{r} n_i varepsilon_{r}}
=-frac{1}{beta}bigg(frac{partial ln Z}{partial varepsilon_i}bigg)_{N, V}
$$

Which is supposed to be the correct result. However, I am not sure that this $langle n_i rangle = sum_{R} P_{R} n_i$ is valid for this average since $P_r$ is the probability that the system is in the $R$-state, not that the $r$-th energy level has a certain number of particles…

Is the procedure I have performed in this correct?

One Answer

This isn't quite right. A microstate of your system is defined by the $r$-tuple $R=(n_1,n_2,ldots,n_r)$ which gives the occupation numbers of each energy level. Each $r$-tuple has a corresponding energy given by $E_R=sum_{i=1}^r n_{i,r} epsilon_i$ (where $n_{i,R}$ is the occupation number of the $i^{th}$ energy level in microstate $R$) and the probability that the system occupies each microstate is $P_R = e^{-beta E_R}/Z$, where $Z$ is the partition function.

It makes sense to compute the average energy of the system via this probability distribution: $$left<Eright> = sum_R P_R E_R = frac{sum_R E_R e^{-beta E_R}}{Z} = -frac{partial}{partial beta} log(Z)$$

It doesn't make sense to talk about $left<E_Rright>$, however. For each microstate $R$, $E_R$ is a fixed number.


The expected number of particles in energy level $i$ can be computed precisely the same way. We're averaging over all possible microstates, weighted by the probability of that microstate being inhabited by the system:

$$left<n_iright> = sum_R P_R n_{i,R}$$

Expanding this out more, $$left<n_iright> = frac{1}{Z}sum_R expleft[-beta sum_j n_{j,R} epsilon _jright]n_{i,R}= frac{1}{Z}sum_R -frac{1}{beta}frac{partial}{partial epsilon_i}expleft[-betasum_j n_{j,R} epsilon_jright]$$ $$= -frac{1}{beta}frac{partial}{partial epsilon_i} log(Z)$$

Correct answer by J. Murray on March 10, 2021

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