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Derivation of Fermi-Dirac distribution

  1. Nov 5, 2015 #1
  2. jcsd
  3. Nov 6, 2015 #2
    I don't see the expression in the text. Where is it?

    you simply take the derivative of the expression in brackets in 2.5.12. Use the expression for lnW in 2.5.8.
     
  4. Nov 6, 2015 #3
    sorry, it should be eq(2.5.12), f/f(gifi).

    Also, can you tell me what variable do we take if we find a derivative of eq 2.5.12
     
  5. Nov 7, 2015 #4

    vanhees71

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    This is a very cumbersome derivation. It's much easier in terms of 2nd quantization. Let's take free non-relativistic fermions (ideal gas). The grand canonical operator for thermal equilibrium (as derived from the maximum-entropy principle under the constraint of given mean energy and particle number) is given by
    $$\hat{R}=\frac{1}{Z} \exp(-\beta \hat{H}+\alpha \hat{N}),$$
    where I choose convenient Lagrange multipliers. Their relation to the usual thermodynamic quantities follow later when we've evaluated the partition sum. It will turn out that ##\beta=1/(k_\text{B} T)## and ##\alpha=\mu/T## (where ##T## is the temperature and ##\mu## the chemical potential of the gas).

    We assume the particles to be in a cubic box of length ##L##. Assuming periodic boundary conditions, the possible single-particle momentum eigenvalues are ##\vec{p} \in \frac{\hbar}{2 \pi L} \mathbb{Z}^3##. In addition we have the ##2s+1## spin states with ##\sigma_z \in \{-s,-s+1,\ldots,s \}##, where ##s \in \{1/2,3/2,\ldots \}## is the spin of the fermions. The possible occupation numbers for each momentum-spin particle states are only ##0## or ##1## (because we've fermions, and the field operators thus obey canonical equal-time anti-commutation relations). Thus the partion sum is
    $$Z=\mathrm{Tr} \exp(-\beta \hat{H}+\alpha \hat{N})=\prod_{\vec{p},\sigma_z} \sum_{N(\vec{p},\sigma_z)=0}^{1} \exp[-N(\vec{p},\sigma_z) E(\vec{p})+\alpha N(\vec{p},\sigma_z)]= \prod_{\vec{p},\sigma_z} [1+\exp(-\beta E(\vec{p})+\alpha)].$$
    Thus we have
    $$\hat{R}=\frac{\exp(-\beta \hat{H}+\alpha \hat{N})}{\prod_{\vec{p},\sigma_z} [1+\exp(-\beta E(\vec{p})+\alpha)]}.$$
    Now the mean occupation number of each single-particle state is given by
    $$\mathrm{Tr}[\hat{R} \hat{N}(\vec{p},\sigma_z)]=\frac{\exp[-\beta E(\vec{p})+\alpha]}{1+\exp(-\beta E(\vec{p})+\alpha)} = \frac{1}{\exp[\beta E(\vec{p})-\alpha]+1},$$
    which is the Fermi-Dirac distribution.
     
  6. Nov 7, 2015 #5
    By f/f(gifi) you mean [itex] \partial / \partial (g_i f_i ) [/itex] , right ? Ok. I'm not familiar with the physics here. I haven't looked at this for a long time. Just following the math. The author says you need to find the maximum of the function in 2.5.11. This is with respect to the quantities [itex]g_i f_i[/itex] for each i , where you are treating each [itex]g_i f_i[/itex] as a single variable. Mentally replace the [itex]g_i f_i[/itex] with [itex]x_i [/itex] if it helps.

    To maximize (or minimize) a function with respect to a set of N variables [itex]x_i [/itex], you take the partial derivative with respect to each one and set it equal to zero, giving N equations in N unknowns. (Which may or may not be solvable.) In this case, the variables are [itex]g_i f_i[/itex] (treated as a single variable) for each i.
     
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