Conduction and the Free-Electron Gas

Metals conduct because they contain electrons free to move through a lattice of fixed positive ions. Paul Drude proposed this picture around 1900, three years after the electron was discovered. It predicts Ohm's law and links electrical to thermal conduction, but it gets three quantities wrong, and each failure is repaired by the same two quantum corrections: the electrons obey the Fermi-Dirac distribution rather than the classical one, and they scatter as waves rather than as particles.

The Drude model

In the classical model the free electrons move at high thermal speed even with no applied field. At $T = 300 K$ ,

⟨ v ⟩ = \frac{8 k T}{π m _{e}} = 1.08 \times 1 0^{5} m/s .

An applied field $E$ superimposes a small drift velocity $v_{d}$ opposite to the field. If there are $n$ electrons per unit volume, the current through area $A$ is

I = n e A v_{d}, j = \frac{I}{A} = n e v_{d} .

The Drude picture: an electron moves at high thermal speed and scatters off lattice ions, with a small drift velocity v_d superimposed by the applied field; the charge in the shaded length passes area A per unit time.

For copper carrying $1 A$ in a wire of radius $0.815 mm$ , the free-electron density (one per atom) is $n = 8.47 \times 1 0^{28} m^{- 3}$ and the drift velocity is $v_{d} \approx 3.5 \times 1 0^{- 5} m/s$ — about $0.01 mm/s$ , ten orders of magnitude below the thermal speed. Charge moves slowly even though the signal travels near light speed.

The resistivity follows from the mean free path $λ = ⟨ v ⟩ τ$ , the average distance between collisions:

ρ = \frac{m _{e} ⟨ v ⟩}{n e ^{2} λ}, σ = \frac{1}{ρ} = \frac{n e ^{2} λ}{m _{e} ⟨ v ⟩} .

The model fails in three ways:

Magnitude. Using $⟨ v ⟩$ from the Maxwell-Boltzmann distribution, the predicted resistivity of copper is about 7 times the measured value.
Temperature dependence. Experiment gives $ρ \propto T$ , but the model gives $ρ \propto ⟨ v ⟩ \propto T^{1/2}$ .
Heat capacity. A classical electron gas should add $\frac{3}{2} R$ to the molar heat capacity, so metals would have $C_{v} = 3 R + \frac{3}{2} R = \frac{9}{2} R$ . The measured value is very nearly $3 R$ ; the electron contribution is only about $0.02 R$ .

All three come from the same mistake: electrons are not a classical gas.

The Fermi gas

Because electrons obey the exclusion principle, no more than two (opposite spins) share a level. At $T = 0$ they fill the lowest levels up to the Fermi energy $E_{F}$ . For a three-dimensional box of volume $V$ with $N$ electrons, filling states up to $E_{F}$ gives

E_{F} = \frac{h ^{2}}{2 m} (\frac{3 N}{8 π V})^{2/3},

and the average electron energy is

⟨ E ⟩ = \frac{3}{5} E_{F},

not the classical $\frac{3}{2} k T$ . For copper, $N / V = 8.47 \times 1 0^{28} m^{- 3}$ gives $E_{F} = 7.06 eV$ . Defining the Fermi temperature $T_{F} = E_{F} / k$ , copper has $T_{F} \approx 82, 000 K$ — far above any temperature at which it is solid.

The number of electrons per energy, n(E) ∝ √E, filled to the Fermi energy at T = 0 (solid edge); raising T smears only a slice of width ~kT near E_F (dashed), so most electrons cannot absorb thermal energy.

Only electrons within about $k T$ of $E_{F}$ have empty states to move into, so only they can absorb thermal energy. At $T = 300 K$ , $k T = 0.026 eV ≪ E_{F}$ , so only a tiny fraction $\sim k T / E_{F}$ participate. This immediately fixes the heat capacity: a full calculation (Sommerfeld) gives

C_{v} (electrons) = \frac{π ^{2}}{2} \frac{T}{T _{F}} R .

For copper at $300 K$ this is $\approx 0.018 R$ , matching experiment, because the huge $T_{F}$ suppresses it. The electron gas is nearly frozen out of the heat capacity.

The classical Dulong-Petit prediction adds a constant (3/2)R electron term at all temperatures; the quantum electron contribution is instead linear in T and tiny, so metals show only the (3R) lattice value near room temperature.

Quantum conduction

The other two failures are fixed by the wave nature of the electron. In an electric field the whole Fermi distribution shifts slightly in velocity, and the net current comes from the electrons near the Fermi surface. So the classical resistivity formula still holds if $⟨ v ⟩$ is replaced by the Fermi speed

u_{F} = \frac{2 E _{F}}{m _{e}},

about $1.6 \times 1 0^{6} m/s$ for copper — some 15 times the classical thermal speed. That would make the predicted resistivity worse, not better, unless the mean free path is also reconsidered.

In an electric field the entire Fermi distribution in velocity shifts by a small amount, so the net current is carried by the electrons near the Fermi speed u_F rather than by all of them; the shift is exaggerated here.

The key correction: a perfectly ordered lattice does not scatter an electron wave at all — the mean free path would be infinite. Scattering arises only from deviations from perfect order, chiefly the thermal vibrations of the ions. Treating the ions as points vibrating with mean-square displacement $⟨ r^{2} ⟩$ , the scattering cross section is $π ⟨ r^{2} ⟩$ , and the equipartition theorem gives $\frac{1}{2} M ω^{2} ⟨ r^{2} ⟩ = k T$ , so $⟨ r^{2} ⟩ \propto T$ . The mean free path becomes

λ = \frac{1}{n π ⟨ r ^{2} ⟩} \propto \frac{1}{T} .

Quantum scattering: the relevant cross section is not the ion's geometric area but the area π⟨r²⟩ swept by its thermal vibration, which grows with temperature and shrinks the mean free path as 1/T.

Two results fall out. Since $ρ \propto 1/ λ \propto T$ , the observed linear temperature dependence is recovered. And the numerical value comes out about 100 times larger for $λ$ than the ion-size estimate — for copper $λ \approx 39 nm$ — which cancels the extra factor from $u_{F}$ and brings the resistivity into agreement with experiment.

Impurities break the lattice order too, and their effect is temperature independent. The resistivity splits into two additive parts (Matthiessen's rule),

ρ = ρ_{t} + ρ_{I},

with $ρ_{t}$ from thermal vibrations and $ρ_{I}$ from impurities. As $T \to 0$ the thermal part vanishes and the resistivity approaches the constant residual $ρ_{I}$ — a purer sample has a lower floor.

The free-electron gas explains conduction, heat capacity, and thermal conductivity, but it treats every metal alike and says nothing about why some solids are insulators or semiconductors. The Sommerfeld model of the next lesson makes the counting quantitative, and the eventual distinction between metals, insulators, and semiconductors requires the effect of the periodic lattice on the electron energies themselves — the band theory that closes this half of the course.