- #1
- 8,943
- 2,955
There doesn't seem to be a forum that is specifically about statistical mechanics, so I'm posting this question here. I apologize for the long-winded introduction, but I think it's needed to provide context for my question:
If you have a discrete collection of single-particle energy levels \epsilon_i, then the grand canonical partition function (for noninteracting particles) is defined by:
\mathcal{Z}(\mu, \beta) = \sum_i \sum_{N_i} exp(N_i \beta(\mu - \epsilon_i))
= \sum_i \sum_{N_i} exp(\beta(\mu - \epsilon_i))^{N_i}
where N_i is the occupancy number: the number of particles in state i. To get Bose statistics, the allowable values for N_i are N_i = 0, 1, 2, ..., leading to
\mathcal{Z}(\mu, \beta) = \sum_i \dfrac{1}{1 - exp(\beta(\mu - \epsilon_i))}
(because 1+x+x^2 + ... = \frac{1}{1-x})
For Fermi statistics, the only possible values for N_i are N_i = 0, 1, leading to:
\mathcal{Z}(\mu, \beta) = \sum_i (1 + exp(\beta(\mu - \epsilon_i)))
Here's the question: Suppose that the energy levels for an electron are independent of spin direction. That means that for every single-particle state, there is a second state with the same energy and opposite spin state. Then it seems to me that there are two different ways to take this degeneracy into account:
(1) Replace \sum_i ... by \sum_i g_i ..., where g_i is the degeneracy of energy level i, and where the index i ranges only over states with distinct energies. In this case, g_i = 2, so the result is just to multiple \mathcal{Z} by 2.
(2) Modify the allowable occupancy number N_i to range from 0 to g_i, rather than just 0 or 1.
These two approaches give different answers, but I don't understand, physically, why. It seems that the only thing that should be important is how many electrons can have energy \epsilon_i.
If you have a discrete collection of single-particle energy levels \epsilon_i, then the grand canonical partition function (for noninteracting particles) is defined by:
\mathcal{Z}(\mu, \beta) = \sum_i \sum_{N_i} exp(N_i \beta(\mu - \epsilon_i))
= \sum_i \sum_{N_i} exp(\beta(\mu - \epsilon_i))^{N_i}
where N_i is the occupancy number: the number of particles in state i. To get Bose statistics, the allowable values for N_i are N_i = 0, 1, 2, ..., leading to
\mathcal{Z}(\mu, \beta) = \sum_i \dfrac{1}{1 - exp(\beta(\mu - \epsilon_i))}
(because 1+x+x^2 + ... = \frac{1}{1-x})
For Fermi statistics, the only possible values for N_i are N_i = 0, 1, leading to:
\mathcal{Z}(\mu, \beta) = \sum_i (1 + exp(\beta(\mu - \epsilon_i)))
Here's the question: Suppose that the energy levels for an electron are independent of spin direction. That means that for every single-particle state, there is a second state with the same energy and opposite spin state. Then it seems to me that there are two different ways to take this degeneracy into account:
(1) Replace \sum_i ... by \sum_i g_i ..., where g_i is the degeneracy of energy level i, and where the index i ranges only over states with distinct energies. In this case, g_i = 2, so the result is just to multiple \mathcal{Z} by 2.
(2) Modify the allowable occupancy number N_i to range from 0 to g_i, rather than just 0 or 1.
These two approaches give different answers, but I don't understand, physically, why. It seems that the only thing that should be important is how many electrons can have energy \epsilon_i.