Here is a puzzle that stumped physics for a quarter of a century. A copper wire conducts electricity superbly — its outer electrons roam almost freely through the metal, which is exactly why it carries current so well. Classical physics said those same free electrons, being a gas of little particles, should also soak up heat: each one ought to store about \tfrac{3}{2}k_BT of thermal energy, just like the atoms of an ideal gas. That would nearly double the measured heat capacity of the metal. But it doesn't. Measure copper's heat capacity at room temperature and the conduction electrons contribute almost nothing — perhaps one percent of what classical physics demands. How can electrons be free enough to carry current across kilometres of wire, yet refuse to store the heat that freedom seems to imply?

The answer is one of the great early triumphs of quantum mechanics, and it turns on a single idea: the Pauli exclusion principle. Electrons are fermions — no two can share a quantum state — so they cannot all crowd into the lowest energy. They are forced to stack, filling states from the bottom up into a deep "Fermi sea" of enormous energy. And once the sea is full, almost no electron can absorb a modest thermal kick, because the states just above it are already taken. Only the thin sliver at the very surface of the sea is free to move. This page develops that picture — the free-electron model of a metal — from Drude's classical guess to Sommerfeld's quantum fix, building on the physics of the ideal quantum gas and the broader story of condensed matter.

Drude's brilliant, broken start

In 1900 Paul Drude pictured a metal's conduction electrons as a classical gas of tiny charged balls, rattling around between the fixed positive ions like molecules in a box. It was a bold, productive idea. It gave a physical account of Ohm's law, predicted the roughly constant ratio of thermal to electrical conductivity (the Wiedemann–Franz law) — and got that ratio very nearly right, which felt like a triumph. For a while the classical electron gas looked like the answer.

But it carried a fatal flaw. A classical gas obeys equipartition: every electron should hold \tfrac{3}{2}k_BT of kinetic energy, contributing \tfrac{3}{2}k_B per electron to the heat capacity. Add that to the 3k_B per atom from the vibrating lattice and metals should have a much larger heat capacity than insulators. Experiment flatly refused: metals and insulators have almost the same heat capacity. Drude's electrons were off by a factor of around a hundred. Something was deeply wrong with treating electrons as an ordinary hot gas.

Sommerfeld's fix: electrons in a box, Pauli-stacked

Keep Drude's picture of non-interacting electrons in a box, but now treat them by quantum mechanics. A single electron confined to a cube of side L has quantised standing-wave states labelled by a wavevector \mathbf{k}, with energy

Each allowed \mathbf{k} is a grid point in "k-space", and each holds two electrons (spin up and spin down) by the Pauli principle. Now pour in the metal's N conduction electrons at absolute zero. They fill the lowest-energy states first, and because energy grows with |\mathbf{k}|, "lowest energy" means "closest to the origin of k-space". The electrons fill a solid sphere of radius k_F — the Fermi wavevector — and everything outside it is empty. The surface of that sphere is the Fermi surface.

Counting: the sphere's volume divided by the volume per state (2\pi/L)^3, times two for spin, must equal N. With electron density n = N/L^3 this pins down the Fermi wavevector purely from how densely the electrons are packed:

Notice what the Fermi energy is not: it has nothing to do with temperature. It is set entirely by how many electrons you have squeezed into the metal. Cram them tighter (raise n) and E_F rises — a purely quantum pressure born of Pauli exclusion.

The density of states: how many rooms at each energy

To do statistics we need to know how many electron states lie in each little slice of energy. Because states fill a sphere in k-space and E \propto k^2, counting the states in a shell of radius k and converting to energy gives the density of states g(E) — states per unit energy per unit volume. In three dimensions it grows as the square root of the energy:

The \sqrt{E} shape is characteristic of three dimensions (in two dimensions g is constant; in one dimension it falls as 1/\sqrt{E}). Higher up in energy there is simply more room in k-space, so more states per slice. Below, the pale curve is g(E); the shaded curve is the number of electrons actually occupying those states, g(E)\,f(E). Turn up the temperature and watch the sharp cliff at E_F soften into a gentle slope — but only right at the edge.

The Fermi–Dirac distribution and the thin active layer

At finite temperature the occupation of a state of energy E is given by the Fermi–Dirac distribution,

where \mu \approx E_F is the chemical potential. At T = 0 this is a perfect step: every state below E_F is full (f = 1), every state above is empty (f = 0). Warm it up and the step blurs — but only over an energy width of about k_BT around E_F. Explore it below: the chart is the same distribution as above, plotted on its own so you can watch the cliff at E_F sharpen as T \to 0.

This is the crux. Because k_BT at room temperature is roughly \tfrac{1}{40}\,\text{eV} while E_F for a metal is several electron-volts, the blurred layer is tiny — only a fraction \sim k_BT/E_F \approx 1\% of the electrons sit close enough to the surface to be nudged into an empty state. All the rest are locked in the depths, walled in by Pauli exclusion: there is nowhere for them to go. A deep electron that tried to absorb k_BT of energy would land in a state that is already occupied — forbidden. So it stays put, and stores nothing.

The heat-capacity puzzle, solved

Now the classical mystery evaporates. Only the fraction \sim k_BT/E_F of electrons near the Fermi surface can absorb heat, and each of those grabs about k_BT. So the electronic thermal energy scales as

Differentiate with respect to T and the electronic heat capacity comes out linear in temperature and suppressed by the huge factor T/T_F:

Two predictions fall straight out, and both are confirmed. First, the electronic heat capacity is small (the missing factor of a hundred is accounted for). Second, it is linear in T, so at very low temperature — where the lattice's phonon contribution dies as T^3 — the electron term \gamma T takes over and dominates. Plot C/T against T^2 for a cold metal and you get a straight line whose intercept is \gamma: a clean, direct measurement of the Fermi sea. Sommerfeld's one quantum swap turned Drude's embarrassment into a precision tool.

Worked examples

Example 1 — the Fermi wavevector of copper. Copper has about n = 8.5\times 10^{28} conduction electrons per cubic metre. Then

That is comparable to the inverse atomic spacing — the Fermi wavelength is a few ångströms, on the scale of the lattice itself.

Example 2 — the Fermi energy and temperature. Feeding k_F into E_F = \hbar^2 k_F^2/2m gives, for copper, E_F \approx 7.0\ \text{eV}. The corresponding Fermi temperature is

Eighty thousand kelvin! Room temperature (300\ \text{K}) is a tiny fraction, T/T_F \approx 0.004, of that. This is the quantitative meaning of "the Fermi sea is deep": the metal would have to be heated to tens of thousands of degrees before thermal energy rivalled the Fermi energy, which is why the electron gas is called degenerate at all ordinary temperatures.

Example 3 — the Fermi velocity. With k_F \approx 1.36\times 10^{10}\ \text{m}^{-1},

Over a million metres per second — and this is the speed of the fastest electrons at absolute zero. They are not going anywhere useful without an applied field (the sphere is symmetric, so all those velocities cancel), but they are most certainly not at rest.