1
$\begingroup$

So say we have $2$ indistinguishable electrons that can occupy $2$ energy states $E=0$ and $E=\Delta$. If we only had one fermion then are both states doubly degenerate since an electron can be spin up or spin down in each state? If that is true then the partition function for $1$ electron becomes:

$$Z_1=2+2e^{-\beta \Delta}$$

And then we have that:

$$Z_2=\frac{1}{N!}Z_1^2=\frac{1}{2}\left(2+2e^{-\beta\Delta}\right)=2+4e^{-\beta\Delta}+2e^{-2\beta\Delta}$$

But when I'm actually counting the states I find that there are $4$ different possibilities for the energy state $\Delta$ but only $1$ possible way for both electrons to occupy the ground the state or excited state since the electrons are indistinguishable. That would imply that:

$$Z_2=1+4e^{-\beta\Delta}+e^{-2\beta\Delta}$$

So which is it? What is going on here? Does the formula for $Z_2$ in terms of $Z_1$ for indistinguishable particles not hold for fermions? What about Bosons? Bosons don't have any spin so for $1$ boson the two level partition function should be:

$$Z_{1b}=1+e^{-\beta\Delta}$$

And then using the same formula for indistinguishable particles we have that:

$$Z_{2b}=\frac{1}{2}Z_{1b}^2=\frac{1}{2}\left(1+e^{-\beta\Delta}\right)=\frac{1}{2}+e^{-\beta\Delta}+\frac{1}{2}$$

But actually counting the states for $2$ bosons in a $2$ level system I get that there is $1$ possible way for both bosons to occupy the ground state or excited state, and $1$ way for one particle to occupy the ground state while one particle to occupies the excited state thus the partition function should be:

$$Z_2=1+e^{-\beta\Delta}+e^{-2\beta\Delta}$$

So I guess I'm ultimately asking why the formula for the partition function of $N$ non interacting indistinguishable particles isn't working for bosons and fermions. What am I missing?

$\endgroup$

1 Answer 1

2
$\begingroup$

So I guess I'm ultimately asking why the formula for the partition function of N non interacting indistinguishable particles isn't working for bosons and fermions. What am I missing?

You're not missing anything. $N!$ is a naive overcounting factor which is wrong for precisely the reasons you've worked out, both for bosons and for fermions. For fermions, $N!$ is too high because it allows for the possibility of more than one particle being in each state; for bosons, $N!$ is also too high because it overcounts multiply-occupied states.

If the temperature of the system is sufficiently high, then the most probable microstates will largely consist of singly-occupied energy levels, and for these microstates $N!$ is the correct overcounting factor. Put differently, if the average occupation number of each energy level is $\ll 1$, $N!$ is approximately the correct overcounting factor, but if the average occupation number a significant number of energy levels is of order $1$, you have to count more carefully.

This is ultimately a dividing line between classical and quantum statistics. If the temperature of the system is sufficiently high that $N!$ is a good approximation for the overcounting factor, you're essentially in the regime of classical physics; once multiply-occupied states become common, you need to start taking into account the bosonic or fermionic nature of the particles.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.