Energy per particle of polarized electron gas

[cscott]

Homework Statement

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.

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...

 

[mark726]

Yes, your understanding is correct. In case (a), the Fermi level will be at the same energy for both spin directions, so the Fermi radius will be the same. But in case (b), the Fermi level will only correspond to one spin direction, so the Fermi radius will be twice as large. This will result in a difference in the total internal energy per electron between the two cases.