# Thermal Statistics - Microstate Probabilities

## Homework Statement

For a system in equilibrium at temperature T, the probability of finding it in a microstate m is:

P(m) = ($1/Z$)exp($-E/kT$)
where Z is the partition function.

There are three accessible microstates, two with energy E$_{a}$ and one with energy E$_{b}$.

Two identical and indistinguishable bosons can occupy these microstates. What's the probability that both these bosons have energy E$_{b}$?

## Homework Equations

I have the values for E$_{a}$ and E$_{b}$.
It asked me to give a formula for Z which I put as Z = $\Sigma$exp($-E/kT$)
T = 300K
I calculated Z to be 1.08

## The Attempt at a Solution

What I've got at the moment is this:
Let the number of particles in states a, a, b be n$_{1}$, n$_{2}$, n$_{3}$:
n$_{1}$ n$_{2}$ n$_{3}$
1 1 0
1 0 1
0 1 1
2 0 0
0 2 0
0 0 2

So the probability of both bosons being in state n$_{3}$ aka b would be $1/6$, but this seems way too obvious for 4 marks. I also tried to use the Bose-Einstein distribution but I don't know the chemical potential.

Thanks

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