# 25.2: Fermi-Dirac Statistics and the Fermi-Dirac Distribution Function

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Let us consider the total probability sum for a system of particles that follows Fermi-Dirac statistics. As before, we let $${\epsilon }_1$$, $${\epsilon }_2$$,, $${\epsilon }_i$$,. be the energies of the successive energy levels. We let $$g_1$$, $$g_2$$,, $$g_i$$,. be the degeneracies of these levels. We let $$N_1$$, $$N_2$$,, $$N_i$$,. be the number of particles in all of the degenerate quantum states of a given energy level. The probability of finding a particle in a quantum state depends on the number of particles in the system; we have $$\rho \left(N_i,{\epsilon }_i\right)$$ rather than $$\rho \left({\epsilon }_i\right)$$. Consequently, we cannot generate the total probability sum by expanding an equation like

$1={\left(P_1+P_2+\dots +P_i+\dots \right)}^N. \nonumber$

However, we continue to assume:

1. A finite subset of the population sets available to the system accounts for nearly all of the probability when the system is held in a constant-temperature environment.
2. Essentially the same finite subset of population sets accounts for nearly all of the probability when the system is isolated.
3. All of the microstates that have a given energy have the same probability. We let this probability be $${\rho }^{FD}_{MS,N,E}$$.

As before, the total probability sum will be of the form $1=\sum_{\{N_i\}}{W^{FD}\left(N_i,{\epsilon }_i\right)}{\rho }^{FD}_{MS,N,E} \nonumber$

Each such term reflects the fact that there are $$W^{FD}\left(N_i,{\epsilon }_i\right)$$ ways to put $$N_1$$ particles in the $$g_1$$ quantum states of energy level $${\epsilon }_1$$, and $$N_2$$ particles in the $$g_2$$ quantum states of energy level $${\epsilon }_2$$, and, in general, $$N_i$$ particles in the $$g_i$$ quantum states of energy level $${\epsilon }_i$$. Unlike Boltzmann statistics, however, the probabilities are different for successive particles, so the coefficient $$W^{FD}$$ is different from the polynomial coefficient, or thermodynamic probability, $$W$$. Instead, we must discover the number of ways to put $$N_i$$ indistinguishable particles into the $$g_i$$-fold degenerate quantum states of energy $${\epsilon }_i$$ when a given quantum state can contain at most one particle.

These conditions can be satisfied only if $$g_i\ge N_i$$. If we put $$N_i$$ of the particles into quantum states of energy $${\epsilon }_i$$, there are

1. $$g_i$$ ways to place the first particle, but only
2. $$g_i-1$$ ways to place the second, and
3. $$g_i-2$$ ways to place the third, and
4. $$g_i-\left(N_i-1\right)$$ ways to place the last one of the $$N_i$$ particles.

This means that there are

$\left(g_i\right)\left(g_i-1\right)\left(g_i-2\right)\dots \left(g_i-\left(N_i+1\right)\right)= \nonumber$

$=\frac{\left(g_i\right)\left(g_i-1\right)\left(g_i-2\right)\dots \left(g_i-\left(N_i+1\right)\right)\left(g_i-N_i\right)\dots \left(1\right)}{\left(g_i-N_i\right)!}=\frac{g_i!}{\left(g_i-N_i\right)!} \nonumber$

ways to place the $$N_i$$ particles. Because the particles cannot be distinguished from one another, we must exclude assignments which differ only by the way that the $$N_i$$ particles are permuted. To do so, we must divide by $$N_i!$$. The number of ways to put $$N_i$$ indistinguishable particles into $$g_i$$ quantum states with no more than one particle in a quantum state is $\frac{g_i!}{\left(g_i-N_i\right)!N_i!} \nonumber$

The number of ways to put indistinguishable Fermi-Dirac particles of the population set $$\{N_1\mathrm{,\ }N_2\mathrm{,\dots ,\ }N_i\mathrm{,\dots }\}$$ into the available energy states is

$W^{FD}\left(N_i,g_i\right)=\left[\frac{g_1!}{\left(g_1-N_1\right)!N_1!}\right]\times \left[\frac{g_2!}{\left(g_2-N_2\right)!N_2!}\right]\times \dots \times \left[\frac{g_i!}{\left(g_i-N_i\right)!N_i!}\right]\times \dots =\prod^{\infty }_{i=1}{\left[\frac{g_i!}{\left(g_i-N_i\right)!N_i!}\right]} \nonumber$

so that the total probability sum for a Fermi-Dirac system becomes

$1=\sum_{\{N_j\}}{\prod^{\infty }_{i=1}{\left[\frac{g_i!}{\left(g_i-N_i\right)!N_i!}\right]}{\left[{\rho }^{FD}\left({\epsilon }_i\right)\right]}^{N_i}} \nonumber$

To find the Fermi-Dirac distribution function, we seek the population set $$\{N_1\mathrm{,\ }N_2\mathrm{,\dots ,\ }N_i\mathrm{,\dots }\}$$ for which $$W^{FD}$$ is a maximum, subject to the constraints

$N=\sum^{\infty }_{i=1}{N_i} \nonumber$ and $E=\sum^{\infty }_{i=1}{N_i}{\epsilon }_i \nonumber$

The mnemonic function becomes

$F^{FD}_{mn}=\sum^{\infty }_{i=1}{ \ln g_i!\ } -\sum^{\infty }_{i=1}{\left[\left(g_i-N_i\right){ \ln \left(g_i-N_i\right)\ }-\left(g_i-N_i\right)\right]}-\sum^{\infty }_{i=1}{\left[N_i{ \ln N_i-N_i\ }\right]+\alpha \left[N-\sum^{\infty }_{i=1}{N_i}\right]} +\ \beta \left[E-\sum^{\infty }_{i=1}{N_i}{\epsilon }_i\right] \nonumber$

We seek the $$N^{\textrm{⦁}}_i$$ for which $$F^{FD}_{mn}$$ is an extremum; that is, the $$N^{\textrm{⦁}}_i$$ satisfying

\begin{align*} 0&=\frac{\partial F^{FD}_{mn}}{\partial N_i}=\frac{g_i-N^{\textrm{⦁}}_i}{g_i-N^{\textrm{⦁}}_i}+{ \ln \left(g_i-N^{\textrm{⦁}}_i\right)\ }-1-\frac{N^{\textrm{⦁}}_i}{N^{\textrm{⦁}}_i}-{ \ln N^{\textrm{⦁}}_i\ }+1-\alpha -\beta {\epsilon }_i \\[4pt] &={ \ln \left(g_i-N^{\textrm{⦁}}_i\right)\ }-{ \ln N^{\textrm{⦁}}_i\ }-\alpha -\beta {\epsilon }_i \end{align*}

Solving for $$N^{\textrm{⦁}}_i$$, we find

$N^{\textrm{⦁}}_i=\frac{g_ie^{-\alpha }e^{-\beta {\epsilon }_i}}{1+e^{-\alpha }e^{-\beta {\epsilon }_i}} \nonumber$

or, equivalently,

$\frac{N^{\textrm{⦁}}_i}{g_i}=\frac{1}{1+e^{\alpha }e^{\beta {\epsilon }_i}} \nonumber$

If $$1\gg e^{-\alpha }e^{-\beta {\epsilon }_i}$$ (or $$1\ll e^{\alpha }e^{\beta {\epsilon }_i}$$), the Fermi-Dirac distribution function reduces to the Boltzmann distribution function. It is easy to see that this is the case. From

$N^{\textrm{⦁}}_i=\frac{g_ie^{-\alpha }e^{-\beta {\epsilon }_i}}{1+e^{-\alpha }e^{-\beta {\epsilon }_i}}\approx g_ie^{-\alpha }e^{-\beta {\epsilon }_i} \nonumber$

and $$N=\sum^{\infty }_{i=1}{N^{\textrm{⦁}}_i}$$, we have

$N=e^{-\alpha }\sum^{\infty }_{i=1}{g_i}e^{-\beta {\epsilon }_i}=e^{-\alpha }z \nonumber$

It follows that $$e^{\alpha }={z}/{N}$$. With $$\beta ={1}/{kT}$$, we recognize that $${N^{\textrm{⦁}}_i}/{N}$$ is the Boltzmann distribution. For occupied energy levels, $$e^{-\beta {\epsilon }_i}=e^{-\epsilon_i}/{kT}\approx 1$$; otherwise, $$e^{-\beta \epsilon_i}=e^{-\epsilon_i/kT}<1$$. This means that the Fermi-Dirac distribution simplifies to the Boltzmann distribution whenever $$1\gg e^{-\alpha }$$. We can illustrate that this is typically the case by considering the partition function for an ideal gas.

Using the translational partition function for one mole of a monatomic ideal gas from Section 24.3, we have

\begin{align*} e^{\alpha } &=\frac{z_t}{N}=\left[\frac{2\pi mkT}{h^2}\right]^{3/2} \frac{\overline{V}}{\overline{N}} \\[4pt] &=\left[\frac{2\pi mkT}{h^2}\right]^{3/2} \frac{kT}{P^0} \end{align*}

For an ideal gas of molecular weight $$40$$ at $$300$$ K and $$1$$ bar, we find $$e^{\alpha }=1.02\times {10}^7$$ and $$e^{-\alpha }=9.77\times {10}^{-8}$$. Clearly, the condition we assume in demonstrating that the Fermi-Dirac distribution simplifies to the Boltzmann distribution is satisfied by molecular gases at ordinary temperatures. The value of $$e^{\alpha }$$ decreases as the temperature and the molecular weight decrease. To find $$e^{\alpha }\approx 1$$ for a molecular gas, it is necessary to consider very low temperatures.

Nevertheless, the Fermi-Dirac distribution has important applications. The behavior of electrons in a conductor can be modeled on the assumption that the electrons behave as a Fermi-Dirac gas whose energy levels are described by a particle-in-a-box model.

This page titled 25.2: Fermi-Dirac Statistics and the Fermi-Dirac Distribution Function is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Paul Ellgen via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.