Contribution of one fermion to entropy of one-particle state

[Suske]

Homework Statement

Hello,

I'm studying for my final exam on statistical physics, and I found an exercise of which I think it is really easy but I'm unsure of how to do it! So now I wonder if I actually don't understand what I'm doing at all!

The question is as follows:

Calculate for fermionic particles the contribution to the entropy of a one-particle
state with energy ε when the particles chemical potential is μ , and the temperature
is T.

Homework Equations

call exp(β(ε - μ) = a

U = TS - PV + μN (1) (contributions to total)
βPV = log Z (2)
Z = Ʃ_sexp(-βE_s + βμN_s) (3)
U = ε/(a-1) (4)
N = 1/(a-1) (5)

The Attempt at a Solution

I rewrite (1) to get S = (U + PV - μN)/T
use (2), (4) and (5) to get S = (ε-μ)/(T*(a-1)) + (1/VTβ)*log Z
then rewrite Z as in (3) to ∏_s(1/((1-a^1)); <- not sure

so log Z would then be Ʃ_n -log(1 - a^1) ?
so then you would have:
contribution to S = (ε-μ)/(T*(a-1)) + (1/VTβ)*log Ʃ_n -log(1 - a^1) ?

but now i still have V in the equation! Help! and the sum isn't very pretty as well..
I honestly don't know how to get rid of those, and I hope someone will help me!

Thank you<,

Suske

 

[TSny]

For fermions, I think that the occupation number of a state s should be

n_s = 1/(a_s+1) where a_s = exp[β(ε_s-μ)].

Also, I think you didn't quite handle the PV/T term correctly.

PV = kTlogZ = kT\sum_s log(1+1/a_s)

We see that the contribution of one state, s, to PV is kT log(1+1/a_s).

So, there is no occurrence of V and there is no sum.

I think your handling of U and \muN is essentially ok, except for the sign change in the denominator of n_s and I would put a subscript s on the ε and the a (ε_s, a_s ) to denote the particular 1-particle state. So, the contribution of one state, s, to U-\muN is (ε_s -\mu)/(a_s+1).

A nice way to express the overall result is in terms of the occupation number n_s = 1/(a_s+1).

You should be able to show that the contribution to S of one state s: -k[(1-n_s)log(1-n_s)+n_slog(n_s)]