Question

If I had `N` fermions, how could I determine the number of ways I can place these into `g` distinguishable containers? For example, how could I determine how many ways I could place `54` electrons into `54` `p` orbitals?

These indistinguishable particles can exchange. The `i`th particle can exchange with any of the other particles (assume nearly-degenerate orbitals), and this follows for each particle. Each particle has two orientations (up/down).

Up to two particles of opposite spin are allowed in each container, but zero particles in any given container is also allowed. That is, `0-2` particles are allowed per container, in integer increments.

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Another way to say this is, how many singlet configuration state functions would I need to describe all possible configurations of 54 electrons in 54 nearly-degenerate `p` orbitals?

Answer 1

Here is my take on this as far as I can understand. I use this idea to calculate the exchange energy for a particular arrangement of electrons. Exchange energy is a quantum - mechanical effect with no classical analogue. It depends on the fact that electrons with the same spin are indistinguishable from each other. It is a stabilising effect and is proportional to the numbers of pairs of electrons with parallel spins which can exchange with each other.

If you had 3 unpaired electrons of the same spin occupying 3 p orbitals then each electron can exchange with 2 others. This means 3 x 2 = 6 exhanges are possible. In general the number of exchanges possible is given by n(n-1) where n is the number of electrons with parallel spins. Because the exchange energy depends on the number of pairs of electrons we divide this by 2.

So in general the exchange energy is proportional to

= `n"n−12"/2`

. This means that the exchange energy for 3 electrons of the same spin for
`p^3` `= 3 "3 -1"/2 = 3`

What about `d^10` ? You have 5 electrons spin up and 5 spin down. For the spin up electrons you have 5 electrons that can make 4 exchanges which is a total of 5 x 4 = 20 exchanges. Divide this by 2 to get the exchange energy which is 20/2 = 10. The 5 spin up electrons will also have 10 exchange energy units by the same reasoning giving a total of 20. This is greater than any partially filled arrangement and goes some way to explaining why full sets of orbitals tend to be stable.

In your example of 54 electrons in 54 p orbitals (which is impossible) the number of possible exchanges will therefore be 54 x 53 = 2862.

Hope it helps