# Counting Microstates with Lateral Thinking

I’ve spent some time this morning stuck on the following undergraduate problem

Consider a quantum system with $N$ distinguishable particles, each of which can have energy $E=n\epsilon$. Show that the number of microstates consistent with a macrostate of energy $E$ is given by the binomial coefficient

$\displaystyle \left(\frac{E/\epsilon + N - 1}{N - 1}\right)$

This is a classic example of a problem which needs some lateral thinking! It’s pretty trivial when you get the right idea, but I think it’s not entirely obvious at first. I’ll share with you my (embarrassingly slow) reasoning – let me know whether you agree with my philosophy in the comments.

To start with, I tried some examples – attempting to arrange $3$ particles in $5$ energy levels for instance. My approach was to work out how I could partition $5$ as a sum of smaller numbers, then evaluate the number of possible configurations consistent with this.

More concretely, I had a configuration where the particles had energies $\{0,1,4\}$. There are $3 \times 2=6$ such arrangements, because the particles can be distinguished (think of them as different coloured balls, if you like).

That’s all very well, but generalizing this idea is tricky. The problem is that the total energy constraint $\sum E_i = E$ makes it hard to enumerate all possible configurations. So I sat, stumped, for a good few minutes.

But thankfully, my failure contained a vital clue. My difficulties lay with that irritating total energy constraint. What if I could remove it from the problem altogether?

To do this requires a bit of lateral thinking. We’ve been trying to fit particles into energy levels. But you can turn the problem around. Equivalently we can try to distribute the $E/\epsilon$ units of energy among the particles. This effectively trivializes the troublesome constraint.

We’re not quite out of the woods yet. We need to work out how to distribute the energy blocks into the particle buckets. Here a second piece of lateral thinking helps us out. Rather than throwing the energy into buckets, we can think of partitioning it into sections. It’s just like being at the supermarket till – different customers (particles) separate their shopping (energy) with plastic dividers.

So how many ways can we divvy up the shopping on the conveyor belt? Well, there are $N$ customers so we’ll need $N-1$ dividers. We’ve also got $E/\epsilon$ items ready to be bought. This means that you have

$\displaystyle (E/\epsilon + N - 1)!$

possible arrangements of the items and dividers. But hang on, every unit of energy looks exactly the same. It’s as if every customer has bought exactly the same product! And clearly it doesn’t matter if you exchange the dividers – the overall partition is unchanged. Taking this into account, the correct number of microstates is

$\displaystyle \frac{(E/\epsilon + N - 1)!}{(E/\epsilon)!(N - 1)!}$

This is exactly the binomial coefficient in the question above!

Although this problem was pretty simple, there are two important morals. First, always examine a problem from every angle. Second, never completely discard your failed attempts. Chances are they hold vital clues which will point you in the right direction!