Visual walkthrough — Fermi-Dirac statistics — fermions, Fermi energy

We assume you know only two things from the parent: fermions are particles that refuse to share a state (the Pauli Exclusion Principle), and each allowed configuration carries a statistical "weight". Everything else — every symbol — we build here.

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Step 1 — One shelf slot, two possibilities

WHAT. Zoom all the way in. Forget the whole metal. Look at one single quantum state — think of it as one slot on a bookshelf. A slot has an energy label $\epsilon$ ("epsilon", just a name for how much energy a particle sitting here would carry).

WHY. Because fermions are antisocial, slots don't interfere with each other's counting. Each slot is its own tiny world that can be either empty ($n=0$) or full ($n=1$). Never $n=2$ — that would be two identical fermions in one state, which the Pauli Exclusion Principle forbids. This independence is the whole reason the hard problem breaks into easy one-slot problems.

PICTURE. Two slots side by side: one dark (empty), one glowing (occupied by a single electron). The number under each is its occupation $n$.

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Step 2 — Weigh each possibility (the two-term partition function)

WHAT. We give each possibility a statistical weight. In the Grand Canonical Ensemble — the bookkeeping where our slot can trade both energy and particles with a giant reservoir — a configuration holding $n$ particles gets the weight $w(n)=e^{-n(\epsilon -\mu )/k_{B}T}.$

Let me name every symbol right where it lives:

• $n$ — occupation (0 or 1), the thing we just defined.
• $\epsilon$ — the slot's energy: the "cost" of putting a particle here.
• $\mu$ — the chemical potential: the energy "budget" the reservoir is willing to spend to hand over one particle. Big $\mu$ = the reservoir is eager to fill slots.
• $k_B$ — Boltzmann's constant: the exchange rate between temperature and energy ($1.38\times 10^{-23}\text{J/K}$). It just converts $T$ into energy units.
• $T$ — temperature: how much thermal "jiggle" the reservoir supplies.
• $e^{-x}$ — the exponential. WHY this tool and not a plain fraction? Because independent random choices multiply their weights, and only the exponential turns adding energies into multiplying weights ($e^{-a}{e}^{-b}=e^{-(a+b)}$). Nature's weight for "energy $E$ above budget" is always $e^{-E/k_{B}T}$ — the Boltzmann factor.

WHY. To get a probability we need a total to divide by. Add the weights of all allowed configurations. There are exactly two: $\mathcal{Z}=\underset{=1}{\underbrace{w(0)}}+\underset{=e^{-(\epsilon -\mu )/k_{B}T}}{\underbrace{w(1)}}=1+e^{-(\epsilon -\mu )/k_{B}T}.$ $\mathcal{Z}$ (the grand partition function) is the normalizing total — the sum of all the ways this slot can be.

PICTURE. A balance scale: left pan holds the empty state with weight $1$; right pan holds the full state with weight $e^{-(\epsilon -\mu )/k_{B}T}$. Which pan is heavier decides whether the slot prefers to be full or empty.

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Step 3 — Average the occupation → the Fermi-Dirac distribution

WHAT. The average occupation $\bar{n}$ is each possible $n$ times its probability. Probability of a configuration = its weight ÷ $\mathcal{Z}$: $\bar{n}=\frac{0\cdot 1+1\cdot e^{-(\epsilon -\mu )/k_{B}T}}{1+e^{-(\epsilon -\mu )/k_{B}T}}.$ The $n=0$ term vanishes (multiplied by zero), leaving only the "full" weight on top.

WHY divide top and bottom by $e^{-(\epsilon -\mu )/k_{B}T}$? To clean the algebra into a memorable shape. Multiply numerator and denominator by $e^{+(\epsilon -\mu )/k_{B}T}$: $\bar{n}=\frac{1}{e^{+(\epsilon -\mu )/k_{B}T}+1}\equiv f(\epsilon ).$

PICTURE. The famous S-curve of $f$ against $\epsilon$, drawn for a warm temperature. It slides from $1$ (full, low energy) down through $\frac{1}{2}$ (right at $\epsilon =\mu$) to $0$ (empty, high energy).

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Step 4 — Read all three regimes off the curve

WHAT. One formula, three stories depending on the sign of $(\epsilon -\mu )$:

Regime	Exponent $(\epsilon -\mu )/k_{B}T$	$e^{(\epsilon -\mu )/k_{B}T}$	$f$
$\epsilon \ll \mu$	large negative	$\to 0$	$\to \frac{1}{0+1}=1$ (full)
$\epsilon =\mu$	exactly $0$	$=1$	$\frac{1}{1+1}=\frac{1}{2}$ (half)
$\epsilon \gg \mu$	large positive	$\to \infty$	$\to \frac{1}{\infty }=0$ (empty)

WHY it matters. The middle row is a gift: at $\epsilon =\mu$ the temperature cancels out entirely, so $f=\frac{1}{2}$ for any $T>0$. That is how experimentalists physically pin down $\mu$ — find the energy that's exactly half-occupied. (Contrast Maxwell-Boltzmann Statistics, whose $f=e^{-(\epsilon -\mu )/k_{B}T}$ has no such fixed crossing and can exceed 1 — nonsense for a fermion probability.)

PICTURE. The same S-curve, now with the three regions shaded and the half-filling point marked with a crosshair on the $\epsilon =\mu$ vertical line.

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Step 5 — Freeze to absolute zero: the curve becomes a cliff

WHAT. Send $T\to 0$. Look at the exponent $(\epsilon -\mu )/k_{B}T$: as $T\to 0$ the denominator $k_{B}T\to 0$, so any nonzero numerator blows up to $\pm \infty$.

• $\epsilon <\mu$: numerator negative → exponent $\to -\infty$ → $e^{-\infty }=0$ → $f=1$.
• $\epsilon >\mu$: numerator positive → exponent $\to +\infty$ → $e^{+\infty }=\infty$ → $f=0$.

WHY. We now define $E_F\equiv \mu (T=0)$, the Fermi energy — the chemical potential's value at absolute zero. The smooth S-curve collapses into a perfect step: every slot below $E_F$ is full, every slot above is empty.

PICTURE. Warm S-curves for shrinking $T$, converging onto the sharp vertical cliff at $E_F$. The filled block to the left is the Fermi sea.

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Step 6 — Fill k-space: the Fermi sphere and counting slots

WHAT. Now we ask how high is $E_F$. In a box of volume $V$, a free electron's state is labelled by a wavevector $\mathbf{k}$ (its momentum is $\hslash \mathbf{k}$). Allowed $\mathbf{k}$-points sit on a uniform grid; the volume per state in $\mathbf{k}$-space is $(2\pi {)}^3/V$. Energy grows with distance from the origin: $\epsilon ={\hslash }^2{k}^2/2m$, so filling from the bottom up means filling a solid sphere of radius $k_F$ centred at the origin.

WHY a sphere? Because $\epsilon$ depends only on $|\mathbf{k}|=k$, all states with the same energy sit at the same radius. "Fill lowest energies first" = "fill smallest radii first" = grow a ball. The surface of that ball, at radius $k_F$, is the Fermi surface.

Count the electrons = (states inside the sphere) × (spin factor 2, since each $\mathbf{k}$ holds one spin-up and one spin-down electron): $N=\underset{\text{spin}}{\underbrace{2}}\times \underset{\text{density of }\mathbf{k}\text{-points}}{\underbrace{\frac{V}{(2\pi {)}^3}}}\times \underset{\text{volume of Fermi sphere}}{\underbrace{\frac{4}{3}\pi k_F^3}}=\frac{V}{3{\pi }^2}{k}_F^3.$

Solve for the Fermi radius: $k_F={\left(3{\pi }^2\frac{N}{V}\right)}^{1/3}=(3{\pi }^{2}n{)}^{1/3},n\equiv \frac{N}{V}\text{(number density)}.$

PICTURE. A cutaway of $\mathbf{k}$-space: a grid of dots, a filled violet sphere of radius $k_F$, dots outside it empty. The label reads "fill lowest energy = fill smallest radius".

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Step 7 — Convert radius to energy → the Fermi energy formula

WHAT. Feed $k_F$ into $\epsilon ={\hslash }^2{k}^2/2m$ evaluated at the surface: $E_F=\frac{{\hslash }^2{k}_F^2}{2m}=\frac{{\hslash }^2}{2m}[(3{\pi }^{2}n{)}^{1/3}{]}^2.$

Term by term:

• $\hslash$ — reduced Planck constant, sets the quantum scale ($1.055\times 10^{-34}\text{J⋅s}$).
• $m$ — electron mass ($9.11\times 10^{-31}\text{kg}$); lighter particles → higher $E_F$.
• $(3{\pi }^{2}n{)}^{1/3}$ raised to the power $2$ gives the $2/3$ below.

PICTURE. A "ladder to energy" panel: the flat sea surface at $k_F$ mapped through the parabola $\epsilon ={\hslash }^2{k}^2/2m$ up to the height $E_F$.

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Step 8 — The edge case: a warm sea only ripples at the surface

WHAT. Reintroduce a small $T$ with $k_{B}T\ll E_F$ (true for every metal at room temperature). The cliff of Step 5 softens into a narrow ramp. From the parent's Example 3, $f$ falls from $0.9$ to $0.1$ over a width $\Delta \epsilon \approx 4.4k_{B}T,$ centred on $\mu$. Solve $f=0.9$: $e^x+1=1/0.9$ gives $x=\ln (1/9)\approx -2.20$; $f=0.1$ gives $x=\ln 9\approx +2.20$; the gap is $2\times 2.20=4.4$, where $x=(\epsilon -\mu )/k_{B}T$.

WHY it matters. Only electrons in this thin surface shell can absorb a nudge of thermal energy and move — the deep sea is frozen solid by exclusion (no empty slot nearby to jump into). This is why metals' heat capacity is tiny compared to the classical guess, and it's the launch pad for the Sommerfeld Expansion, which also reveals that $\mu (T)$ dips slightly below $E_F$ as $T$ rises — so $\mu =E_F$ only at $T=0$.

PICTURE. Zoom on the Fermi surface: the sharp cliff with a shaded "fuzzy shell" of width $\sim k_{B}T$ where $f$ ramps, arrows showing electrons hopping from just-below to just-above.

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The one-picture summary

The whole derivation on one canvas: one slot (0 or 1) → weigh the two states → average into the S-curve $f$ → freeze it to a step → stack the step into a filled sphere in $\mathbf{k}$-space → read its surface height as $E_F$ → warm it and only the surface shell ripples.

Recall Feynman retelling — say it back in plain words

Imagine a giant staircase of energy levels, and a crowd of electrons that flatly refuse to stand on the same step as anyone else. Cool everything to absolute zero and the crowd doesn't collapse to the bottom — they pack the staircase from the ground up until they run out of people. The last occupied step is the Fermi energy. Below it: shoulder-to-shoulder, every step full. Above it: empty. Now warm the crowd a little: the ones deep down are pinned — no free step nearby to move to — but the few standing right at the top can shuffle up or down over a band about $k_{B}T$ tall. That thin, jittery top layer is all the action in a metal. And where did the whole staircase picture come from? Just weighing "empty" against "full" for one step with a Boltzmann factor, dividing to get a probability, and stacking those probabilities across every step.

Recall

The two allowed occupations of a fermion slot ::: $n=0$ (empty) and $n=1$ (full) — the cap at 1 is Pauli exclusion. Why the exponential appears in the weight ::: independent energies add, so their weights must multiply, and $e^{-a}{e}^{-b}=e^{-(a+b)}$ is the only function that does this. Value of $f$ exactly at $\epsilon =\mu$, and why it's $T$-independent ::: $f=\frac{1}{2}$; the exponent is $0$ so $e^0=1$ regardless of $T$. What the Fermi sphere is ::: the ball of filled $\mathbf{k}$-states of radius $k_F=(3{\pi }^{2}n{)}^{1/3}$ at $T=0$. How high is the sea surface in energy ::: $E_F=\frac{{\hslash }^2}{2m}(3{\pi }^{2}n{)}^{2/3}$, set only by density $n$. Width of the thermal "fuzzy shell" ::: about $4.4k_{B}T$ around $\mu$ (between $f=0.9$ and $f=0.1$).