# Energy per particle of polarized electron gas

1. Sep 26, 2010

### cscott

1. The problem statement, all variables and given/known data

Calculate the total internal energy per electron at zero temperature of a free noninteracting gas of electrons of density $n$, in the following two cases.

a) First assume that states with both spin directions are populated equally.

b) Now assume that the gas is fully polarized: only states corresponding to one spin direction, say "up", are populated.

3. The attempt at a solution

Know internal energy per particle is given by,

$$\frac{U}{N} = \frac{1}{N} \sum_{k\sigma} \epsilon_k n_F(\epsilon_k)$$

where $\sigma$ is summing spin states giving me a factor of one or two.

Eventually I get to,

$$\frac{U}{N} = \frac{1}{n} \int_{0}^{\inf} g(\epsilon})\epsilon n_F(\epsilon) d\epsilon$$

where $\sigma$ is wrapped into the density of states $g(\epsilon)$ and $n_F(\epsilon)$ is the Fermi-Dirac distribution.

So is it correct that the last difference between (a) and (b) (that isn't carried by $\sigma$) is the chemical potential $\mu = \epsilon_F$ in $n_F(\epsilon)$, where the polarized spins will have a Fermi radius twice as large as the non-polarized equally populated spins?

Trying to understand setting up the calculation right now more than performing it...

Last edited: Sep 26, 2010