Fermi level and Fermi-Dirac distribution: Visual walkthrough

We are answering one question the whole way down:

"If there is an allowed energy state sitting at energy $E$ , what is the chance an electron is actually sitting in it right now?"

Call that chance $f (E)$ . The letter $f$ just stands for "fraction filled" — a number between $0$ (surely empty) and $1$ (surely full). Everything below earns that number.

Step 1 — Draw a single seat that can hold 0 or 1 electron

WHAT. Forget billions of electrons. Zoom all the way in to one allowed quantum state — think of it as one chair at some height $E$ on a ladder of energies. Because of the Pauli exclusion principle, that chair holds at most one electron.

WHY. If two identical electrons could share a state, we could use ordinary (classical) counting. They cannot. So the number of electrons in our chair, call it $n$ , has only two allowed values: $n \in {0, 1}$

$n = 0$ → the chair is empty.
$n = 1$ → the chair is occupied.

That "only two options" is the seed of the whole formula. Everything hard about Fermi–Dirac lives in this single rule.

PICTURE. One chair, two possible worlds side by side.

Figure — Fermi level and Fermi-Dirac distribution

Step 2 — Let the chair trade electrons with a huge reservoir

WHAT. Our chair is not alone. It touches a gigantic bath of electrons — the rest of the crystal — held at temperature $T$ . Electrons hop in and out of the chair from this bath.

WHY. We cannot predict a single hop, but statistics loves large baths. The bath sets two dials that never change:

$T$ — the temperature, how much random thermal jiggling there is.
$μ$ — the chemical potential: the energy "price" the bath charges to hand over one electron. (In our problem $μ$ ends up equal to the Fermi level $E_{F}$ , but let us keep the honest symbol $μ$ for now.)

PICTURE. The chair connected by a little door to a large tank of electrons labelled with $T$ and $μ$ .

Step 3 — Weigh the two possible worlds with the Gibbs factor

WHAT. Statistical mechanics gives a rule for "how likely is a configuration": a configuration whose net energy is $E$ carries a weight $weight = e^{- E / k_{B} T} .$

Let us decode every piece:

$e^{(\dots)}$ — the exponential function. WHY the exponential and not, say, a straight line? Because independent random costs multiply, and the only function turning multiplication of probabilities into addition of energies is the exponential. High energy → the minus sign makes the exponent very negative → weight near zero → "rarely happens". That is exactly the behaviour we want.
$k_{B}$ — Boltzmann's constant, the exchange rate between temperature and energy. At room temperature $k_{B} T \approx 0.0259$ eV.
$E$ — the net energy of the world we are weighing, using the $E - μ$ bookkeeping from Step 2.

Now apply it to our two worlds. Adding $n$ electrons to the chair costs net energy $E = (E - μ) n$ :

world	net energy $E$	weight $e^{- E / k_{B} T}$
$n = 0$ (empty)	$0$	$e^{0} = 1$
$n = 1$ (full)	$E - μ$	$e^{- (E - μ) / k_{B} T}$

WHY these two rows and nothing else. Because Step 1 said only $n = 0$ and $n = 1$ exist. A boson chair would have infinitely many rows; Pauli chops us down to exactly two. That truncation is why the answer will be so clean.

PICTURE. Two bars whose heights are the weights $1$ and $e^{- (E - μ) / k_{B} T}$ , drawn for a state above $μ$ (so the full-world bar is short).

Step 4 — Average the occupation: filled-weight over total-weight

WHAT. Probability = (weight of the outcome you want) ÷ (weight of all outcomes). We want the average number of electrons, which for a $0/1$ variable equals the probability of being filled:

f (E) = ⟨ n ⟩ = \frac{0 \cdot w _{0} + 1 \cdot w _{1}}{w _{0} + w _{1}} = \frac{e ^{- (E - μ) / k_{B} T}}{1 + e ^{- (E - μ) / k_{B} T}} .

WHY each piece.

Numerator $0 \cdot w_{0} + 1 \cdot w_{1}$ : multiply each occupation value $n$ by its weight and add. The $n = 0$ term vanishes (nothing there), so only $w_{1}$ survives on top.
Denominator $w_{0} + w_{1}$ : the sum of all weights, so the fractions of the two worlds add up to $1$ . This is called normalisation — it forces $f$ to be an honest probability between $0$ and $1$ .

PICTURE. The filled bar sitting on top of the total (both bars stacked), the ratio marked as a fraction of the full height.

Step 5 — Tidy the fraction into the famous form

WHAT. Divide top and bottom by the numerator $e^{- (E - μ) / k_{B} T}$ :

f (E) = \frac{e ^{- (E - μ) / k_{B} T}}{1 + e ^{- (E - μ) / k_{B} T}} \div e^{- (E - μ) / k_{B} T} \frac{1}{e ^{+ (E - μ) / k_{B} T} + 1} .

WHY this move. Nothing physical changes — we divided top and bottom by the same thing, which never alters a fraction's value. But look what happened: the numerator became a clean $1$ , and the exponent flipped sign (dividing by $e^{- x}$ is multiplying by $e^{+ x}$ ). Replacing $μ$ with the Fermi level $E_{F}$ (they coincide here) gives the boxed result:

PICTURE. The smooth S-shaped curve, with the three landmark points ( $f \to 1$ , $f = \frac{1}{2}$ , $f \to 0$ ) called out.

Step 6 — The $T = 0$ edge case: the curve snaps into a cliff

WHAT. Send temperature to absolute zero, $T \to 0$ . Then $k_{B} T \to 0$ , so the exponent $(E - E_{F}) / k_{B} T$ blows up in size:

If $E < E_{F}$ : exponent $= \frac{negative}{tiny} = - \infty$ , so $e^{- \infty} = 0$ and $f = \frac{1}{1 + 0} = 1$ .
If $E > E_{F}$ : exponent $= \frac{positive}{tiny} = + \infty$ , so $e^{+ \infty} = \infty$ and $f = \frac{1}{\infty} = 0$ .

WHY it matters. This is the "everyone sits as low as possible" world. Every chair below $E_{F}$ is filled, every chair above is empty, with a vertical cliff exactly at $E_{F}$ . It is the reason people wrongly think " $E_{F}$ = highest filled level" — true only here.

PICTURE. The smooth curve of Step 5 with an arrow showing it stiffening into a perfect step as $T \to 0$ .

Step 7 — Finite $T$ : the cliff melts into a ramp of width a few $k_{B} T$

WHAT. Turn heat back on. The cliff softens into a slope. How wide is the fuzzy region? Where the exponent is around $\pm 1$ , i.e. $∣ E - E_{F} ∣ \approx k_{B} T$ . So the transition spans a few $k_{B} T$ , centred on $E_{F}$ .

WHY. Thermal jiggling lets some electrons from just below $E_{F}$ jump to just above, leaving empty chairs (holes) behind. The curve records exactly this: a bit less than $1$ just below $E_{F}$ , a bit more than $0$ just above.

A beautiful symmetry falls out. Compare a state $Δ$ above with one $Δ$ below:

f (E_{F} + Δ) + f (E_{F} - Δ) = 1.

Read it aloud: "chance a state $Δ$ above is filled" = "chance a state $Δ$ below is empty." The curve is antisymmetric about the pivot point $(E_{F}, \frac{1}{2})$ .

PICTURE. The finite- $T$ curve with the width $\sim k_{B} T$ marked, and two mirrored points showing $f (E_{F} + Δ) + f (E_{F} - Δ) = 1$ .

Step 8 — The far-tail edge case: Fermi–Dirac becomes Boltzmann

WHAT. Go far above the Fermi level, $E - E_{F} ≫ k_{B} T$ (think the conduction band of a semiconductor). Then $e^{(E - E_{F}) / k_{B} T}$ is enormous, so the " $1$ " in the denominator is a rounding error:

f (E) = \frac{1}{1 + huge e ^{(E - E_{F}) / k_{B} T}} \approx \frac{1}{e ^{(E - E_{F}) / k_{B} T}} = e^{- (E - E_{F}) / k_{B} T} .

WHY. Once chairs are so high up that they are almost never occupied, electrons stop tripping over each other — Pauli's cap is irrelevant, and we recover the classical exponential decay. This single simplification is what makes carrier concentrations in ordinary (non-degenerate) semiconductors easy to compute.

PICTURE. The true S-curve and the Boltzmann exponential drawn on top; they hug each other in the far-right tail and split near $E_{F}$ .

The one-picture summary

Everything above, compressed: one chair → two weights → their ratio → the S-curve → its two edge cases (a cliff at $T = 0$ , a Boltzmann tail far above $E_{F}$ ).

Recall Feynman retelling of the whole walkthrough

Picture one chair on a tall staircase, and a strict rule: one bottom per chair (that's Pauli). The chair is next to a warm, crowded room full of people (the electron bath), kept at some warmth $T$ , and the room "charges" an energy price $μ$ to send someone over. Nature weighs "chair empty" and "chair full" using a fairness rule: costlier situations happen less often, and the cost enters through $e^{- cost / k_{B} T}$ . Empty costs nothing, so its weight is $1$ ; full costs $E - μ$ , so its weight is $e^{- (E - μ) / k_{B} T}$ . The chance the chair is taken is just "full weight ÷ (empty + full) weight." Clean that fraction up and you get $f = \frac{1}{1 + e ^{(E - E_{F}) / k_{B} T}}$ — the " $1$ " is the empty chair refusing to let more than one person sit. Freeze everything ( $T = 0$ ) and the staircase fills perfectly to a water line called the Fermi level: full below, empty above. Warm it back up and a few people near the line hop upstairs, leaving gaps below — the sharp line melts into a fuzzy ramp a few $k_{B} T$ wide. And way up high where chairs are almost always empty, people never bump into each other, so the fancy rule relaxes into the plain exponential Boltzmann guess.

Quick self-check

Where does the "

1

" in the denominator come from?

The weight of the empty state (

w_{0} = e^{0} = 1

); it is what caps

f

at 1 and encodes Pauli exclusion.

Why does the exponent flip sign between Step 4 and Step 5?

We divided top and bottom by

e^{- (E - E_{F}) / k_{B} T}

; dividing by

e^{- x}

multiplies by

e^{+ x}

T = 0

, why is

f

a step?

k_{B} T \to 0

makes the exponent

\pm \infty

, so

e

goes to

0

\infty

, giving

f = 1

below

E_{F}

and

f = 0

above.

When may you replace

f

e^{- (E - E_{F}) / k_{B} T}

When

E - E_{F} ≫ k_{B} T

, so the "

1

" is negligible (non-degenerate tail).

Connections

Density of states — this page gives $f (E)$ ; multiply by $g (E)$ and integrate for real electron counts.
Carrier concentration n and p — the Boltzmann tail from Step 8 is the workhorse.
Pauli exclusion principle — the $n \in {0, 1}$ rule of Step 1.
Chemical potential and equilibrium — the $μ$ dial from Step 2.
Band gap and conduction/valence bands — where "far above $E_{F}$ " physically lives.