Canonical Partition function for 2 Fermions with two energy levels

PhysicsRock
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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.
 
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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.
 
kuruman said:
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!
 
PhysicsRock said:
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:
PhysicsRock said:
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.
 
TSny said:
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.
 
kuruman said:
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.
 
TSny said:
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.

Microstates.png
 
kuruman said:
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:
kuruman said:
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##.
 
TSny said:
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.
 
  • #10
kuruman said:
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.
 
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