# Distinguishable, indistinguishable paramagnetic ideal gas [closed]

In the canonical ensemble, the partition function for an ideal gas is given by:

$$\frac{Z}{N!}$$

The factor $N!$ accounts for the indistinguishability of the particles of the ideal gas.

What happens if you consider a system of paramagnetic ideal gas particles, such that $N = N_\uparrow + N_\downarrow$? What does the required factor become?

I think it is:

$$\frac{Z}{N!} \frac{N!}{N_\uparrow!N_\downarrow} = \frac{Z}{N_\uparrow!N_\downarrow}$$

Can someone tell me if this is correct, and possibly explain why this factor is right in this case?

• Can you give us your definition of $Z$? Apr 11, 2014 at 10:40
• I suspect that $Z=z_1^N$ where $z_1$ is the number of states a single particle of ideal gas can have. Dec 19, 2014 at 22:29
• Define $Z$, otherwise this question is not well defined. Dec 7, 2017 at 2:22

Use the definition of partition function as a sum over all states. Then if you interchange two indistinguishable particles you get the same state. Now you have a standard ball into bins combinatorial problem.

• I understand this, but why do we multiply by this factor and not divide? For $N$ distinguishable particles, we have $N!$ distinguishable states. We divide by this number to get the partition function for indistinguishable particles. Now, if you have $N_\uparrow$ and $N_\downarrow$ distinguishable from each other, but indistinguishable amongst the two types, you have $\frac{N!}{N_\uparrow!N_\downarrow!}$ possible states. Why not divide by this factor? Nov 9, 2013 at 21:29
• Look at it in a different way. If you have only one group of indistinguishables, you divide by their possible combinations, which is N!. However now you have two, so you have to divide by their respective possible combinations as well. But since one group is distinguishable from the other, you get two sets of combinations, or two bins like the answer says, so you do the combinations separately (the two factorials) and then multiply them. Nov 9, 2013 at 22:06
• I suspect you're confusing two different problems - The fraction you wrote is the number of state a system could be in, if you had everything else (such as allowed positions in the system, momentum) set and only had to determine whether each particle was spin up or down. What we're looking for here is just to eliminate indistinguishable micro states and therefore only want the number of such "permutations" which you would over-count, had you ignored it while calculating $Z$ (If for example you did $z_1^N$ there where $z_1$ is the multiplicity for a single particle). Dec 19, 2014 at 22:37

That depends on how you define the partition function; specifically, how much you over-counted when computing it. If you were to compute the partition function using only physically distinguishable states then the $N!$ would not be necessary at all.

So, when you go to the case of spins, just how are you computing the partition function?

Why do you say for N particles you have N! distinguishable state? Lets just make sure you got it right.

Lets say you have N particles on a lattice having M sites. In the ideal case, no excluded volume therefore each particles can access M sites. The total number of states for distinguishable particles in M^N. Now, if you considere that the particles are indistinguishable, for each indistinguishable states, correspond N! distinguishable states (the number of permutations). So for each indistinguishable states, you overcounted N! state. Conclusion, you need to DIVIDE by N!.

Now if the spins can be up or down, using the same reasoning, for each indistinguishable state, correspond N1!N2! distinguishable states. Which is the number of permutation among N1 and N2. So you divide again by this new factor

As Nanite pointed, in the notation you choosed, Z is computed using distinguishability. Its not always the case, I am more used to see the N! inside the partition function Z

Btw you forgot a factorial sign

$$Z=\frac{z_{1}^{N}}{N!}$$
$$z_{1}=z_{\rm space}z_{\rm spin}$$