A quick question about the Fermi-Dirac distribution

[davon806]

Homework Statement

An electron has two spin states and a set of energy levels E1,E2,...
By the Fermi-dirac distrbution,the mean number of electrons in energy level Ek is
https://en.wikipedia.org/wiki/Fermi–Dirac_statistics#Distribution_of_particles_over_energy
Does it mean that,for an electron, the mean number is 2*F(E) ? But this sounds weird to me,because the mean number doubles just because of the degeneracy ? I am quite confused.

*Additional
If we take degeneracy into account,suppose there are N electrons(weakly-interacting).At some temperatures close to 0K,the N/2 states of lowest energy are occupied rather than N states?

Homework Equations

The Attempt at a Solution

Described in 1.

 

[mfb]

The probability for each state is given by the distribution. If you sum over two states (e. g. two spin states), the mean number of electrons will be twice that value, sure.

If we take degeneracy into account,suppose there are N electrons(weakly-interacting).At some temperatures close to 0K,the N/2 states of lowest energy are occupied rather than N states?

The lowest N states are occupied, N/2 of them will be spin up, N/2 will be spin down (neglecting couplings between spin and anything else).

 

[davon806]

The probability for each state is given by the distribution. If you sum over two states (e. g. two spin states), the mean number of electrons will be twice that value, sure.
The lowest N states are occupied, N/2 of them will be spin up, N/2 will be spin down (neglecting couplings between spin and anything else).

But for each state, can we accommodate 2 electrons? 1 with spin up and 1 with spin down,so that eventually the lowest N/2 states will be occupied?

 

[DrClaude]

But for each state, can we accommodate 2 electrons? 1 with spin up and 1 with spin down,so that eventually the lowest N/2 states will be occupied?

It depends how you define the state. @mfb is using state to mean unique single-particle states, including spin (so there is no degeneracy factor appearing). You seem to mean state being an energy state, a state defined only by the its energy, which can be degenerate due to spin.

Both approaches are valid, but you have to be clear what you are considering.

 

[davon806]

It depends how you define the state. @mfb is using state to mean unique single-particle states, including spin (so there is no degeneracy factor appearing). You seem to mean state being an energy state, a state defined only by the its energy, which can be degenerate due to spin.

Both approaches are valid, but you have to be clear what you are considering.

Could you explain what's meant by a single-particle state?This term comes up in my notes frequently but I have no idea what it is referring to?

 

[DrClaude]

Could you explain what's meant by a single-particle state?

It is a state in which a single-particle can fit

Let me take an atom as an example. If you solve the (non-relativistic) Schrödinger equation for a hydrogenic atom (1 electron, charge of the nucleus ≥ 1), you will find a series of eigenstates, which can be characterized by the quantum numbers $n,l,m$ and $m_s$. Each valid combination of these quantum numbers gives you one single-particle state.

The full state of an atom, while neglecting electron-electron interaction, would correspond to putting a series of electrons in these single-particle states, with at most one electron in each (Pauli exclusion principle).