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Contribution of one fermion to entropy of one-particle state

  1. Jun 23, 2012 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    call exp(β(ε - μ) = a

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


    3. 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 Ʃn -log(1 - a^1) ?
    so then you would have:
    contribution to S = (ε-μ)/(T*(a-1)) + (1/VTβ)*log Ʃn -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
     
  2. jcsd
  3. Jun 24, 2012 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

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

    ns = 1/(as+1) where as = exp[β([itex]ε[/itex]s-μ)].

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

    PV = kTlogZ = kT[itex]\sum[/itex]s log(1+1/as)

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

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

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

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

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