Multiplicity of particles on a lattice

[Arjani]

Homework Statement

A particle can exist in three microstates, with energies E0 < E1 < E2. Consider N >> 1 such particles, fixed on a lattice. There are now n0 particles with energy E0, n1 particles with energy E1 and n2 = N - n0 - n1 particles with energy E2. We have that n_j >> 1 for j = 0, 1, 2. What is the multiplicity?

The Attempt at a Solution

I've established that there are 2N + 1 possible macrostates for this system, but that doesn't really help me further. I've done this problem before, but then it was only two groundstates and the answer would be something like $\Omega = \frac{N!}{\frac{N + M}{2}!\frac{N - M}{2}!}$, so I was thinking it should be something like:

$$\Omega = \frac{N!}{\frac{N - n_1 - n_2}{3}! \frac{N - n_0 - n_2}{3}! \frac{N - n_0 - n_1}{3}!}$$

But somehow I'm guessing that's not going to be correct?

 

[anathema]

Thank you for your question. Your attempt at a solution is on the right track, but there are a few things that need to be clarified. Firstly, in order to determine the multiplicity of a system, we need to consider the number of ways that the particles can be distributed among the microstates. In this case, we have three different microstates, each with a different energy level. Therefore, the multiplicity can be calculated as follows:

\Omega = \frac{N!}{n_0!n_1!n_2!}

Where n_0, n_1, and n_2 represent the number of particles in each energy level. However, the given information tells us that n_j >> 1 for j = 0, 1, 2. This means that the number of particles in each energy level is much larger than 1, which allows us to use the Stirling's approximation for factorials:

n! \approx \sqrt{2\pi n}\left(\frac{n}{e}\right)^n

Using this approximation, we can rewrite the multiplicity as:

\Omega \approx \frac{\sqrt{2\pi N}\left(\frac{N}{e}\right)^N}{\sqrt{2\pi n_0}\left(\frac{n_0}{e}\right)^{n_0}\sqrt{2\pi n_1}\left(\frac{n_1}{e}\right)^{n_1}\sqrt{2\pi n_2}\left(\frac{n_2}{e}\right)^{n_2}}

Simplifying this expression, we get:

\Omega \approx \frac{N!}{n_0!n_1!n_2!}\left(\frac{e}{n_0}\right)^{n_0}\left(\frac{e}{n_1}\right)^{n_1}\left(\frac{e}{n_2}\right)^{n_2}

Now, since n_j >> 1, we can assume that n_j \gg 1, and thus, we can use the approximation \left(\frac{e}{n_j}\right)^{n_j} \approx 1. Therefore, the final expression for the multiplicity is:

\Omega \approx \frac{N!}{n_0!n_1!n_2!}