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Fermi Distribution Functions
Fermi Distribution Functions
Fermi-Dirac (F-D) Statistics Derivation
The following derivation covers the thermodynamic probability and the occupation index for fermions, which are indistinguishable particles that obey the Pauli exclusion principle.
1. Thermodynamic Probability (W)
Consider $n_i$ fermions to be distributed among $g_i$ energy cells. Due to the Pauli Exclusion Principle, each cell can hold at most one fermion.
- The 1st fermion can be accommodated in $g_i$ ways.
- The 2nd fermion can be accommodated in $(g_i - 1)$ ways
- The 3rd fermion can be accommodated in $(g_i - 2)$ ways.
- The $n_i$-th fermion can be accommodated in $(g_i - n_i + 1)$ ways.
The total number of ways for one energy level is given by:
W_i = g_i(g_i-1)(g_i-2)...(g_i - n_i + 1) = g_i!/ (g_i - n_i)!
Since fermions are indistinguishable, we divide by $n_i!$:
W_i = g_i! / [n_i! (g_i - n_i)!]
The total thermodynamic probability for the entire system is the product across all energy levels:
W = Π [g_i! / (n_i! (g_i - n_i)!)]
2. Occupation Index (n_i / g_i)
To find the distribution in equilibrium, we maximize $\ln W$ subject to the constraints of constant total number of particles ($\sum n_i = N$) and constant total energy ($\sum n_i E_i = U$)
Step A: Logarithm and Differentiation
Taking the natural logarithm and applying Stirling's Approximation
ln W = Σ [ln g_i! - ln n_i! - ln(g_i - n_i)!]
Differentiating for the maximum ($d \ln W = 0$) leads to
Σ ln[(g_i - n_i) / n_i] dn_i = 0
Step B: Lagrange Multipliers
Using multipliers $\alpha$ and $\beta$ for the constraints[cite: 421, 423]:
ln[(g_i - n_i) / n_i] = α + β E_i
Exponentiating both sides:
(g_i - n_i) / n_i = e^(α + β E_i)
(g_i / n_i) - 1 = e^(α + β E_i)
Final Result: The Occupation Index
The average number of particles per state (occupation index) is:
f(E_i) = n_i / g_i = 1 / [e^(α + β E_i) + 1]
\[\begin{align}
{{W}_{1}}&={{g}_{1}}\times \left( {{g}_{1}}-1 \right)\times \left( {{g}_{1}}-2 \right).....\left( {{g}_{1}}-\left( {{n}_{1}}-1 \right) \right) \\
& ={{g}_{1}}\left( {{g}_{1}}-1 \right)\left( {{g}_{1}}-2 \right).....\left( {{g}_{1}}-{{n}_{1}}+1 \right) \\
& =\frac{{{g}_{1}}\left( {{g}_{1}}-1 \right)\left( {{g}_{1}}-2 \right).....\left( {{g}_{1}}-{{n}_{1}}+1 \right)\left( {{g}_{1}}-{{n}_{1}} \right)\left( {{g}_{1}}-{{n}_{1}}-1 \right)....3\times 2\times 1}{\left( {{g}_{1}}-{{n}_{1}} \right)\left( {{g}_{1}}-{{n}_{1}}-1 \right)....3\times 2\times 1} \\
& =\frac{{{g}_{1}}!}{\left( {{g}_{1}}-{{n}_{1}} \right)!} \\
& \Rightarrow {{W}_{2}}=\frac{{{g}_{2}}!}{\left( {{g}_{2}}-{{n}_{2}} \right)!},{{W}_{3}}=\frac{{{g}_{3}}!}{\left( {{g}_{3}}-{{n}_{3}} \right)!}... \\
& \Rightarrow W={{W}_{1}}{{W}_{2}}{{W}_{3}}.... \\
& =\frac{{{g}_{1}}!}{\left( {{g}_{1}}-{{n}_{1}} \right)!}\frac{{{g}_{2}}!}{\left( {{g}_{2}}-{{n}_{2}} \right)!}\frac{{{g}_{3}}!}{\left( {{g}_{3}}-{{n}_{3}} \right)!}... \\
& =\prod\limits_{i=1}^{i=n}{\frac{{{g}_{i}}!}{\left( {{g}_{i}}-{{n}_{i}} \right)!}} \\
\end{align}\]
\begin{align}
\ln W&=\ln \prod\limits_{i=1}^{i=n}{\frac{{{g}_{i}}!}{{{n}_{i}}!\left( {{g}_{i}}-{{n}_{i}} \right)!}} \\
& =\sum{\ln {{g}_{i}}!}-\sum{\ln \left( {{n}_{i}}!\left( {{g}_{i}}-{{n}_{i}} \right)! \right)} \\
& =\sum{\ln {{g}_{i}}!}-\sum{\ln {{n}_{i}}!}-\sum{\ln \left( {{g}_{i}}-{{n}_{i}} \right)!} \\
& =\sum{\ln {{g}_{i}}!}-\sum{{{n}_{i}}\left( \ln {{n}_{i}}-1 \right)}-\sum{\left( {{g}_{i}}-{{n}_{i}} \right)\left( \ln \left( {{g}_{i}}-{{n}_{i}} \right)-1 \right)} \\
& =\sum{\ln {{g}_{i}}!}-\sum{{{n}_{i}}\left( \ln {{n}_{i}} \right)}+\sum{{{n}_{i}}}-\sum{\left( {{g}_{i}}-{{n}_{i}} \right)\ln \left( {{g}_{i}}-{{n}_{i}} \right)}+\sum{\left( {{g}_{i}}-{{n}_{i}} \right)} \\
& =\sum{\ln {{g}_{i}}!}-\sum{{{n}_{i}}\left( \ln {{n}_{i}} \right)}-\sum{\left( {{g}_{i}}-{{n}_{i}} \right)\ln \left( {{g}_{i}}-{{n}_{i}} \right)}+\sum{{{g}_{i}}} \\
& \Rightarrow d\left( \ln W \right)=0-\sum{d{{n}_{i}}\left( \ln {{n}_{i}} \right)-}\sum{{{n}_{i}}\left( \frac{1}{{{n}_{i}}} \right)}d{{n}_{i}}-\sum{\left( -d{{n}_{i}} \right)\ln \left( {{g}_{i}}-{{n}_{i}} \right)}-\sum{\left( {{g}_{i}}-{{n}_{i}} \right)\frac{1}{\left( {{g}_{i}}-{{n}_{i}} \right)}}\left( -d{{n}_{i}} \right)+0 \\
& =-\sum{d{{n}_{i}}\left( \ln {{n}_{i}} \right)-}\sum{d{{n}_{i}}}+\sum{\ln \left( {{g}_{i}}-{{n}_{i}} \right)d{{n}_{i}}}+\sum{d{{n}_{i}}} \\
& =\sum{\ln \left( {{g}_{i}}-{{n}_{i}} \right)d{{n}_{i}}}-\sum{\left( \ln {{n}_{i}} \right)d{{n}_{i}}} \\
& =\sum{\ln \frac{\left( {{g}_{i}}-{{n}_{i}} \right)}{{{n}_{i}}}d{{n}_{i}}} \\
& \Rightarrow \sum{\ln \frac{\left( {{g}_{i}}-{{n}_{i}} \right)}{{{n}_{i}}}d{{n}_{i}}}=0\left( \because d\left( \ln W \right)=0 \right) \\
& U=\sum{{{n}_{i}}{{E}_{i}}} \\
& N=\sum{{{n}_{i}}} \\
& \Rightarrow dU+dN=\beta \sum{d{{n}_{i}}{{E}_{i}}}+\alpha \sum{d{{n}_{i}}} \\
& \Rightarrow \beta \sum{d{{n}_{i}}{{E}_{i}}}+\alpha \sum{d{{n}_{i}}}=0 \\
& \Rightarrow \sum{\left( \beta {{E}_{i}}+\alpha \right)d{{n}_{i}}}=0 \\
& \Rightarrow \sum{\ln \frac{\left( {{g}_{i}}-{{n}_{i}} \right)}{{{n}_{i}}}d{{n}_{i}}}=\sum{\left( \beta {{E}_{i}}+\alpha \right)d{{n}_{i}}} \\
& \Rightarrow \ln \frac{\left( {{g}_{i}}-{{n}_{i}} \right)}{{{n}_{i}}}=\left( \beta {{E}_{i}}+\alpha \right) \\
& \Rightarrow \frac{\left( {{g}_{i}}-{{n}_{i}} \right)}{{{n}_{i}}}={{e}^{\left( \beta {{E}_{i}}+\alpha \right)}} \\
& \Rightarrow \frac{{{g}_{i}}}{{{n}_{i}}}-1={{e}^{\alpha +\beta {{E}_{i}}}} \\
& \Rightarrow \frac{{{g}_{i}}}{{{n}_{i}}}=1+{{e}^{\alpha +\beta {{E}_{i}}}} \\
& \Rightarrow \frac{{{n}_{i}}}{{{g}_{i}}}=\frac{1}{{{e}^{\alpha +\beta {{E}_{i}}}}+1} \\
& \Rightarrow f\left( {{E}_{i}} \right)=\frac{1}{{{e}^{\alpha +\beta {{E}_{i}}}}+1} \\
\end{align}
Fermi–Dirac Statistics – MCQ Quiz
