# Partition Function

1. Mar 12, 2006

### ak416

This is a question about thermal physics. There's this partition function Z = sum over all states s of the system ( exp(-E_s/T)). And its just used to calculate the probability of any state by taking the Boltzman factor exp(-E_s/T) of that state and dividing over the partition function. Theres one question that asks to show that the partition function for a combined system, Z(1and2) = Z(1)*Z(2). I understand the way its proved, you just take a double sum and say that E(1and2) =E(1)+E(2), so you can separate the sums. But by using a double sum arent you possibly overcounting some states? For example if E_s1 + E_s2 = 1 + 3 and E_s1 + E_s2 = 3 + 1, also 2+2... Shouldnt this just count as one state of the system, call it E_s = 4. Or would it be better to just keep it this way and then whenever you want to count the probability of observing an E = 4 of the double system, you would have to add all the possible boltzman factors corresponding to E = 4. I think i just answered my question..., but im just wondering whats the right way to think about it, because theres another part in the book about ideal gases talking about how when you have a system with distinct particles you can overcount, but when you have a system with identical particles, you have to multiply the partition function by 1/N! They also say at the end that in our argument we have assumed that all N occupied orbitals (i guess they mean energy levels) are always different orbitals. How does this change anything?

2. Mar 12, 2006

### Physics Monkey

The assumptions here are independence and distinguishability. It isn't true that that the partition function of a composite system is simply the product of the partition functions of its subsystems if the subsystems interact or if they are identical. The $$1/N!$$ in the classical partition function of an ideal gas is the classic example of this subtlety.

3. Mar 12, 2006

### Gokul43201

Staff Emeritus
ak : It sounds like your doubt is just another way of stating the Gibbs Paradox. You might want to look that up too.

4. Sep 25, 2009

### katewhitney

Do you suppose you could explain how you arrived at the answer in the first place?

Z = sum[exp(e/T)]

so Z(1+2) = Z(1)Z(2)

....

a double sum -- as in sum[ exp(e1/t) + exp(e2/t) ] = sum[exp(s1/t)*sum[sum[exp(s2/t)]
??

I understand the partition function idea - but I'm poor with sums :|