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partition function |
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| Apr11-08, 06:03 PM | #1 |
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partition function
1. The problem statement, all variables and given/known data
If we have a system of N independent particles and the partition function for one particle is Z_1, then is the partition function for the N particle system Z=(Z_1)^N? 2. Relevant equations 3. The attempt at a solution I'm pretty sure that this is true for a classical system, but I'm not sure if it's true for a quantum system. Does the Pauli exclusion principle spoil this somehow? |
| Apr11-08, 06:14 PM | #2 |
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Even without quantum considerations, you end up with over counting if the particles are identical. See Gibb's Paradox.
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| Apr11-08, 06:43 PM | #3 |
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Right. Sorry, I meant to write
[tex]Z={1\over {N!}}(Z_1)^N[/tex] Does that take care of over counting? What about the quantum case? |
| Apr12-08, 02:36 PM | #4 |
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partition function
It is not true for the quantum case. The quantum case is easier handled via grand-partition function.
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| Apr12-08, 04:27 PM | #5 |
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So in the quantum case, if we want to use the canonical ensemble, we have to calculate the whole partition function all in one shot?
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| Apr13-08, 05:06 AM | #6 |
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Yep. But like I said, usually, you calculate the grand-partition function (which factorises neatly into a function of single particle states).
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