Canonical Partition function for 2 Fermions with two energy levels

[PhysicsRock]

Homework Statement

Consider two identical particles that follow Fermi-Dirac statistics in a thermodynamical system with two energy levels, $E_0$ and $E_1$, $E_1 > E_0$, that do no interact and are in contact with a heat reservoir at temperature $T$. Determine the canonical partition function.

Relevant Equations

$Z = \sum_\text{all configurations} e^{-\beta E}$

My solution is

$$
Z = e^{-2\beta E_0} + 4 e^{-\beta (E_0 + E_1)} + e^{-2\beta E_1},
$$

since Fermions have non-zero spin and there are four options for distributing spins (assuming only $\pm \frac{1}{2}$) among $E_0$ and $E_1$ such that $E_\text{tot} = E_0 + E_1$ and one option each that leads to a total energy of $2E_{0,1}$. However, I'm not sure whether or not the terms $e^{-2\beta E_{0,1}}$ should also be weighted with a degeneracy factor ($2$ in this case), because both up / down and down / up are possible. The quantum state remains identical under swapping the spins, but it also leads to an additional option to obtain the same total energy.

Help is appreciated, have a great day everyone.

 

[kuruman]

Just list the possible total energies and the particle configurations for each. You have
$(E_0,E_0)\rightarrow 2E_0$
$(E_1,E_0)\rightarrow E_0+E_1$
$(E_0,E_1)\rightarrow E_0+E_1$
$(E_1,E_1)\rightarrow 2E_1$
The degeneracy for the middle energy is 2 not 4.

 

[PhysicsRock]

Just list the possible total energies and the particle configurations for each. You have
$(E_0,E_0)\rightarrow 2E_0$
$(E_1,E_0)\rightarrow E_0+E_1$
$(E_0,E_1)\rightarrow E_0+E_1$
$(E_1,E_1)\rightarrow 2E_1$
The degeneracy for the middle energy is 2 not 4.

Alright, thank you!

 

[TSny]

Homework Statement: Consider two identical particles that follow Fermi-Dirac statistics in a thermodynamical system with two energy levels, $E_0$ and $E_1$, $E_1 > E_0$, that do no interact and are in contact with a heat reservoir at temperature $T$. Determine the canonical partition function.

The homework statement doesn't specify the spin of the particles. Also, it doesn't specify the degree of degeneracy of each of the one-particle energy levels ($E_0$ and $E_1$). So, to me, the statement is ambiguous.

To make the problem specific, you've taken the particle spin to be 1/2. Also, you've assumed that the only degeneracy of an energy level is that due to spin; that is, flipping the spin of a particle does not change the energy of the particle.

With these assumptions, there is only one two-particle state with total energy $2E_0$ and one two-particle state with total energy $2E_1$. There are four two-particle states with total energy $E_0 + E_1$.

So, I agree with your solution for $Z$ for the two-particle system:

My solution is

$$
Z = e^{-2\beta E_0} + 4 e^{-\beta (E_0 + E_1)} + e^{-2\beta E_1},
$$

However, @kuruman disagrees. So, I could be wrong. Or, maybe we have different interpretations of the problem statement.

 

[kuruman]

There are four two-particle states with total energy $E_0+E_1.$

How do you count 4? This is how I count 2.

We are told that two particles are at thermal equilibrium with a reservoir and that each has two energy states characterized by $E_0$ and $E_1.$ To me this means that they can absorb thermal energy $E_1-E_0$ from the reservoir and transit from $E_0$ to $E_1$ or that they can deposit thermal energy $E_1-E_0$ into the reservoir and transit from $E_1$ to $E_0.$

We are also told that they are non-interacting. To me this means that each particle does its own thing independently of the other. Now I count microstates of the two-particle ensemble (it's like tossing two coins and looking at the possibilities of heads and tails) and I get the results posted in #2:
$(E_0,E_0)\rightarrow 2E_0$
$(E_1,E_0)\rightarrow E_0+E_1$
$(E_0,E_1)\rightarrow E_0+E_1$
$(E_1,E_1)\rightarrow 2E_1$

The spin does not enter the picture. Even if it did, and we further assumed that particles interact and can be spin-coupled, we would still have a total of 4 microstates, one triplet ($S=1$) with a degeneracy of 3 and a non-degenerate singlet ($S=0$) separated by some energy proportional to the coupling constant.

 

[TSny]

We are told that two particles are at thermal equilibrium with a reservoir and that each has two energy states characterized by $E_0$ and $E_1.$

There are two energy levels, but we are not given any information about the number of single-particle states with energy $E_0$ and the number of states with energy $E_1$. If there is only one single-particle quantum state available for each energy level, we cannot put two particles into the same energy level. That would violate the Pauli exclusion principle for Fermi-Dirac statistics. So, $(E_0, E_0)$ and $(E_1, E_1)$ would not be allowed.

Consider the one-dimensional particle in a box with ground-state energy $E_0$ and first excited-state energy $E_1$. Ignore all higher energy levels. For a single spin 1/2 particle in the box, there are two states available for $E_0$. One state has "spin up" and the other "spin down". So, $E_0$ is doubly degenerate. Likewise for $E_1$.

If we put two identical, noninteracting, spin-1/2 particles in the box, there is one allowed two-particle state where both particles have energy $E_0$. In this state, the two spins are opposite and form a spin singlet. Likewise, there is one two-particle singlet state where both particles have energy $E_1$.

There are four two-particle states with one particle in the ground state and one in the excited state. One of these states is a spin singlet. The other three states correspond to a spin triplet.

 

[kuruman]

If there is only one single-particle quantum state available for each energy level, we cannot put two particles into the same energy level. That would violate the Pauli exclusion principle for Fermi-Dirac statistics. So, $(E_0, E_0)$ and $(E_1, E_1)$ would not be allowed.

I know what you are trying to say but it does not come out right. $(E_0,E_0)$ is allowed if one particle is spin-up and the other spin down. It is not allowed if both are spin up (or down).

That said, I confess that I interpreted rather loosely the "not interacting" part and disregarded the statement that the particles "follow Fermi-Dirac statistics." To redeem myself, I post a schematic of the six microstates below for indistinguishable fermions. The number of microstates having energy $E_0+E_1$ is indeed 4. My apologies for the confusion.

 

[TSny]

I know what you are trying to say but it does not come out right. $(E_0,E_0)$ is allowed if one particle is spin-up and the other spin down. It is not allowed if both are spin up (or down).

Agreed. For a spin 1/2 particle, there is a state with energy $E_0$ having spin up and another state with energy $E_0$ having spin down. So, there are two states of the particle with energy $E_0$ and, likewise, two states of the particle with energy $E_1$. So, we can have $(E_0, E_0)$ or $(E_1, E_1)$ without violating the exclusion principle.

My comment that $(E_0, E_0)$ would violate the exclusion principle was regarding your statement:

We are told that two particles are at thermal equilibrium with a reservoir and that each has two energy states characterized by $E_0$ and $E_1.$

(I've added emphasis to the word states.) I interpreted this to mean that each particle (not necessarily spin 1/2) has only one state characterized by $E_0$ (not two states) and one state characterized by energy $E_1$ (not two states). So, in this case, $(E_0, E_0)$ would imply both particles are in the same quantum state in violation of exclusion. I probably misinterpreted your statement. I think we're on the same page.

The problem statement is unclear to me because it does not provide any information for deciding how many single-particle states have energy $E_0$ or energy $E_1$. For spin 1/2 particles there are systems for which there are more than two single particle states associated with a particular energy level.

For example, consider the system to be a two-dimensional box with length in the x-direction equal to length in the y-direction. Let $n_x$ and $n_y$ denote the quantum numbers associated with the different spatial wavefunctions. The states $[n_x, n_y] = [1, 0]$ and $[n_x, n_y] = [0, 1]$ would then be different states with the same energy $E_1$. So, even without spin, there is two-fold degeneracy of level $E_1$. If the particles are spin 1/2, we have additional degeneracy due to spin. Thus, there would be a total of four distinct quantum states of energy $E_1$. For a cubical box (three dimensions), the degeneracy of $E_1$ would be 6 for spin 1/2 particles.

If we don't know the degeneracy of the energy levels, we can't derive the partition function $Z$.

 

[kuruman]

If we don't know the degeneracy of the energy levels, we can't derive the partition function $Z$.

Sure, but since the degeneracy of the levels (other than the implicit spin degeneracy) is not explicitly mentioned I would take the default assumption that $E_0$ and $E_1$ are non-degenerate.

 

[TSny]

Sure, but since the degeneracy of the levels (other than the implicit spin degeneracy) is not explicitly mentioned I would take the default assumption that $E_0$ and $E_1$ are non-degenerate.

Yes. That would be the natural assumption. It’s the assumption made by @PhysicsRock.