How to Find the Canonical Partition Function for Two Quantum Particles?

[RawrSpoon]

Homework Statement

Consider a system of two quantum particles. Each particle has two quantum states, one with zero energy and one with energy ε>0. For each of the three cases, draw a table of the possible microstates α of the system, and find the canonical partition function Z(β).

a)The two particles are distinguishable

b) The two particles are indistinguishable bosons

c) The two particles are indistinguishable fermions

Homework Equations

$$Z(\beta,N)=\prod_{i=1}^{N}\sum_{E_{i}=0,\varepsilon} e^{-\beta E_{i}}=(1+e^{-\beta \varepsilon})^{N}$$

The Attempt at a Solution

a) The four states are pretty simple, they're {AB, 0}, {A, B}, {B, A}, {0, AB}. The partition function is also equally simple, being just $$Z(\beta, N) = (1+e^{-\beta \varepsilon})^{2}$$

b) This is where I get lost. The states are also pretty simple, being {AA, 0} {A, A}, {0, AA}. But the partition function is where I don't really understand how to proceed.

c) I get lost even worse here. The only possible state for this one is {A, A}. Again, the partition function is what gets me.

Thanks in advance for any possible help

 

[TSny]

$$Z(\beta,N)=\prod_{i=1}^{N}\sum_{E_{i}=0,\varepsilon} e^{-\beta E_{i}}=(1+e^{-\beta \varepsilon})^{N}$$

This way of expressing $Z$ is not valid for parts (b) and (c) where the particles are indistinguishable.

For parts (b) and (c), use the more fundamental expression $$Z=\sum_i e^{-\beta E_{i}}$$ where the sum is over all possible microstates of the two-particle system. Here, $i$ labels the microstates and $E_i$ is the energy of the $i^{\rm th}$ microstate.

 

[RawrSpoon]

This way of expressing $Z$ is not valid for parts (b) and (c) where the particles are indistinguishable.

For parts (b) and (c), use the more fundamental expression $$Z=\sum_i e^{-\beta E_{i}}$$ where the sum is over all possible microstates of the two-particle system. Here, $i$ labels the microstates and $E_i$ is the energy of the $i^{\rm th}$ microstate.

Ok so then $$Z(\beta , N) = \sum_i e^{-\beta E_{i}} = (e^{-\beta (0)})+(e^{-\beta \varepsilon})+(e^{-2 \beta \varepsilon})$$

I assume that $E_{i}$ is the total energy of each microstate, so given the microstates {AA,0}{A,A}{0,AA} I assume the energy is 0, $\varepsilon$, and $2 \varepsilon$

Is this headed in the right direction or am I just totally off-base? I appreciate the responses :)

 

[TSny]

Ok so then $$Z(\beta , N) = \sum_i e^{-\beta E_{i}} = (e^{-\beta (0)})+(e^{-\beta \varepsilon})+(e^{-2 \beta \varepsilon})$$

I assume that $E_{i}$ is the total energy of each microstate, so given the microstates {AA,0}{A,A}{0,AA} I assume the energy is 0, $\varepsilon$, and $2 \varepsilon$

Is this headed in the right direction or am I just totally off-base? I appreciate the responses :)

That looks right.

 

[RawrSpoon]

That looks right.

Ahhhh it makes sense now. The reason the distinguishable particles is $$Z(\beta , 2) = (1+e^{-\beta \varepsilon})^{2}$$ is because expanding the expression gives two different states where $E_{i}=\varepsilon$ right? That means that for fermions, $$Z=e^{-\beta \varepsilon}$$ since $E_{i}$ can ONLY be $\varepsilon$ due to the Pauli Exclusion Principle. I think I get it now, thanks!

 

[TSny]

Ahhhh it makes sense now. The reason the distinguishable particles is $$Z(\beta , 2) = (1+e^{-\beta \varepsilon})^{2}$$ is because expanding the expression gives two different states where $E_{i}=\varepsilon$ right?

Yes, that's correct.

That means that for fermions, $$Z=e^{-\beta \varepsilon}$$ since $E_{i}$ can ONLY be $\varepsilon$ due to the Pauli Exclusion Principle.

Right. For this case, the system can only be in one state. So, you already know everything about the system. The partition function doesn't serve any purpose as far as I can see. It's just an academic exercise.