1
$\begingroup$

Suppose we have a system filled with $N$ particles. There are $k$ energy levels in this system, labeled by $\epsilon_i$, each with a degeneracy of $g_i$. Let us imagine $n_j$ particles out of these $N$ particles occupy the energy level $\epsilon_j$.

The single particle partition function $Z_{sp}$ is given by : $Z_{sp}=\sum g_ie^{-\beta\epsilon_i}$

Let $Z_N$ be the multi-particle partition function. Its expression is a little more complicated, and can be given by: $Z_N=\sum G_ie^{-\beta E_i}$.

Here $E_i$ is the total energy of all the $N$ particles and we are checking over all the combinations. Similarly, $G_i$ is the degeneracy of these total energies.

Here is my question :

What is the difference between asking how many particles are in the energy level $\epsilon_j$, and the probability of the system to have a total energy $E_m$?

The second question is simple, and the answer is given by : $$P(E_m)=\frac{G_me^{-\beta E_m}}{Z_N}$$

However, how do I find out the answer to the first question, i.e. find $n_j$, the number of particles in the $\epsilon_j$ energy level ?

According to Wikipedia, the number of particles in a particular energy level is nothing but the probability of a single particle to occupy that level, multiplied by the total number of particles. Since the probability of a single particle being in energy level $\epsilon_j$ is given using the single-particle partition function, our final answer comes to be :

$$n_j=\frac{g_j e^{-\beta \epsilon_j}}{Z_{sp}}N$$

Is this correct?

Like for a system with only one particle, the probability of that one particle to be in a particular energy level is the same as the probability of the entire system having that energy. However, for a system with $N$ particles, it seems that the probability of a particle being in a particular state is very different from the probability of the entire system having a certain total energy.

Are the statements all correct ?

$\endgroup$
2
  • $\begingroup$ Can you link to the wikipedia article? $\endgroup$
    – Andrew
    Nov 20, 2021 at 23:42
  • $\begingroup$ @Andrew here is the link $\endgroup$ Nov 21, 2021 at 0:39

1 Answer 1

3
$\begingroup$

It sounds like the main thing that would convince you is the check that I recommended in the last question :). I hedged my bets there by saying that the formula I gave for $N_j$ would agree with Wikipedia at large $N$. But it actually holds for general $N$.

The expected number of particles at level $j$ is found by summing $n_j$ over all microstates with their Boltzmann factor probabilities as weights. So again, that means \begin{equation} N_j = Z_N^{-1} \sum_{n_1 + \dots + n_k = N} n_j G(n_1, \dots, n_k) e^{-\beta (\epsilon_1 n_1 + \dots \epsilon_k n_k)}. \end{equation} But notice that this sum looks exactly like the the full partition function (which we know to be $Z_{\mathrm{sp}}^N$) differentiated with respect to $\epsilon_j$! Therefore, \begin{align} N_j &= -\beta^{-1} Z_N^{-1} \frac{\partial}{\partial \epsilon_j} Z_N \\ &= -\beta^{-1} Z_{\mathrm{sp}}^{-N} \frac{\partial}{\partial \epsilon_j} Z_{\mathrm{sp}}^N \\ &= -N\beta^{-1} Z_{\mathrm{sp}}^{-1} \frac{\partial}{\partial \epsilon_j} Z_{\mathrm{sp}} \\ &= N Z_{\mathrm{sp}}^{-1} g_j e^{-\beta \epsilon_j} \end{align} and everything checks out.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.