Thermal Statistics - Microstate Probabilities

[9202]

Homework Statement

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

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

There are three accessible microstates, two with energy E[itex]_{a}[/itex] and one with energy E[itex]_{b}[/itex].

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

Homework Equations

I have the values for E[itex]_{a}[/itex] and E[itex]_{b}[/itex].
It asked me to give a formula for Z which I put as Z = [itex]\Sigma[/itex]exp([itex]-E/kT[/itex])
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[itex]_{1}[/itex], n[itex]_{2}[/itex], n[itex]_{3}[/itex]:
n[itex]_{1}[/itex] n[itex]_{2}[/itex] n[itex]_{3}[/itex]
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[itex]_{3}[/itex] aka b would be [itex]1/6[/itex], 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

 

[mark726]

for your help!



Thank you for your post. It seems like you have a good understanding of the concept of partition function and the probabilities of microstates in equilibrium. Your approach of listing out the different combinations of particles in the three microstates is a good start. However, I would like to clarify a few things and provide some additional information to help you solve the problem.

Firstly, the formula for the partition function Z is correct. It is defined as the sum of the Boltzmann factors for all the microstates in the system. In this case, since there are three microstates with different energies, the partition function can be written as:

Z = exp(-E_a/kT) + exp(-E_a/kT) + exp(-E_b/kT)

= 2exp(-E_a/kT) + exp(-E_b/kT)

Now, for the probability of both bosons being in state b, we need to consider all the possible combinations of particles in the system. As you have correctly listed, there are six possible combinations. However, not all of them have the same probability. Let's consider each one separately:

1. Both bosons in state b:
This case has a probability of (1/Z) x (1/Z) = (1/Z)^2, since both bosons need to be in state b and the probabilities are independent.

2. One boson in state b and one in state a:
This case has a probability of (1/Z) x (1/Z) = (1/Z)^2, since the probabilities are independent.

3. One boson in state b and one in state a:
This case has a probability of (1/Z) x (1/Z) = (1/Z)^2, since the probabilities are independent.

4. Both bosons in state a:
This case has a probability of (1/Z) x (1/Z) = (1/Z)^2, since both bosons need to be in state a and the probabilities are independent.

5. Both bosons in state a:
This case has a probability of (1/Z) x (1/Z) = (1/Z)^2, since both bosons need to be in state a and the probabilities are independent.

6. Both bosons in state b:
This case has a probability of (1/Z) x (1/Z) = (1/Z)^2, since both bosons need to be