University Physics IV · Molecular and Solid-State Physics · 12.5

The Free-Electron Fermi Gas

Sommerfeld's answer to why a metal stuffed with mobile electrons still obeys Dulong-Petit. Count the states, fill them by Fermi-Dirac rather than Boltzmann, and the electronic heat capacity collapses to a linear term you can only see below a few kelvin. Learn what that buys — and what deleting the lattice costs.

Open the lab ↓Practice & worked examples ↓

Build the model

Connect the measurement to the mechanism.

Drude had the right picture and the wrong statistics: a metal's valence electrons do roam free, but they are identical fermions, and at metallic densities Fermi-Dirac occupancy looks nothing like Maxwell-Boltzmann. Sommerfeld keeps the box and throws out the classical counting. Solve the free-particle Schrödinger equation in a cube with periodic boundaries and the allowed wavevectors form a grid in k-space, one state per (2π)³/V of volume; exclusion then makes filling the levels an exercise in geometry.

At T = 0 the electrons occupy a sphere of radius k_F = (3π²n)¹⁄³ whose surface sits at E_F = (ℏ²/2m)(3π²n)²⁄³ — several electronvolts in any real metal, which is tens of thousands of kelvin in temperature units. Everything the model explains follows from that one number dwarfing k_B T. Only electrons within a few k_B T of the Fermi surface have empty states to move into, so heating a metal stirs a fraction of order T/T_F of them rather than all of them, and the electronic heat capacity comes out linear in T and about a hundred times smaller than equipartition demands.

The cost is the lattice. With the periodic potential deleted, g(E) ∝ √E is finite at every positive energy, so this model can never open a gap, and every insulator and semiconductor falls outside it.

Simple definition: The free-electron Fermi gas is a model of a metal's valence electrons as non-interacting plane waves in a box, filling the lowest states two at a time up to the Fermi energy, with the lattice potential ignored entirely.
Example: Copper donates one electron per atom, n = 8.47 × 10²⁸ m⁻³, so E_F = (ℏ²/2m)(3π²n)²⁄³ = 7.0 eV and T_F = E_F/k_B = 8.2 × 10⁴ K — still degenerate at the melting point.

Fermi wavevectork_F = (3π² n)¹⁄³, with n = N/V

One number — the free electrons per cubic metre — fixes the radius of the filled sphere in k-space.

n in m⁻³, k_F in m⁻¹; the factor 2 for spin is already inside the 3π²

Fermi energy and Fermi temperatureE_F = ℏ²k_F²/2m = (ℏ²/2m)(3π² n)²⁄³, T_F = E_F/k_B

Sets every other scale in the model, and depends on density alone — not on temperature, chemistry or sample size.

Cu: n = 8.47 × 10²⁸ m⁻³ gives E_F = 7.0 eV and T_F = 8.2 × 10⁴ K

Density of states in three dimensionsg(E) = (V/2π²)(2m/ℏ²)³⁄² √E = (3N/2E_F) √(E/E_F)

√E is nonzero at every positive energy, so this model has no forbidden band and can describe metals only.

states per joule; g(E_F) = 3N/2E_F, about 0.21 eV⁻¹ per electron in copper

Fermi-Dirac occupancyf(E) = 1/[exp((E − μ)/k_B T) + 1], with μ → E_F as T → 0

The zero-temperature step feathers over a few k_B T — 0.026 eV at 300 K against a 7 eV cliff.

f(E_F) = 0.5 exactly; f = 0.27 at E_F + k_B T and 0.12 at E_F + 2k_B T

Electronic heat capacityCₑₗ = (π²/3) k_B² T g(E_F) = (π²/2) N k_B (T/T_F) = γT

Linear in T, and (π²/3)(T/T_F) = 1.2% of the classical (3/2)Nk_B for copper at 300 K.

Cu free-electron γ = 0.50 mJ mol⁻¹ K⁻², measured 0.695 mJ mol⁻¹ K⁻²

Ground-state energy and degeneracy pressureU/N = (3/5)E_F, P = (2/5) n E_F, B = (5/3)P = (2/3) n E_F

Exclusion alone, with no lattice and no Coulomb term, gets a metal's stiffness within a factor of two.

Cu at T = 0: 4.2 eV per electron, P = 38 GPa, B = 64 GPa against 137 GPa measured

Count states in k-space, then fill the sphere

Sommerfeld keeps Drude's box and replaces his statistics. Solve the free-particle Schrödinger equation in a cube of side L with periodic boundaries: the allowed wavevectors form a grid with one state per (2π/L)³ of k-space, a volume (2π)³/V that depends on the sample's size and not its shape. Each state holds two electrons, spin up and spin down. At T = 0 exclusion forces you to fill outward from the origin, so the occupied region is a sphere. Setting N = 2 × (4πk_F³/3) ÷ ((2π)³/V) gives N = V k_F³/3π², that is k_F = (3π²n)¹⁄³ with n = N/V. Copper's one free electron per atom gives n = 8.47 × 10²⁸ m⁻³, so k_F = 1.36 × 10¹⁰ m⁻¹ and E_F = ℏ²k_F²/2m = 7.0 eV. Notice what E_F does not depend on: not the chemistry of the metal, not the size of the crystal, not the temperature. Density alone.

The density of states goes as √E, and never vanishes

Count states below an energy instead. A sphere of radius k = √(2mE)/ℏ holds N(E) = V k³/3π² = (V/3π²)(2mE/ℏ²)³⁄², so g(E) = dN/dE = (V/2π²)(2m/ℏ²)³⁄² √E. That square root is the three-dimensional signature: in two dimensions g is constant, and in one it diverges as E⁻¹⁄². Dropping the constants gives a tidier form, g(E) = (3N/2E_F)√(E/E_F), so g(E_F) = 3N/2E_F — about 0.21 states per electronvolt per electron in copper. Now the structural point. √E is nonzero for every energy above the band bottom, so wherever you put the Fermi level there are empty states immediately above it. This model cannot produce a forbidden band, which means it describes metals and nothing else; silicon, diamond and every insulator lie outside it. Opening a gap needs the periodic lattice potential Sommerfeld deleted, which is the business of Bloch's theorem in the next topic.

Fermi-Dirac occupancy and the width of the blur

The occupancy of a state of energy E is f(E) = 1/[exp((E − μ)/k_B T) + 1], with the chemical potential μ tending to E_F as T tends to zero. At absolute zero it is a step: every state below E_F full, every state above it empty. Warm the metal and the step feathers, but only over a few k_B T — f = 0.73 at E_F − k_B T, exactly 0.5 at E_F, 0.27 at E_F + k_B T, and 0.12 at E_F + 2k_B T. Put numbers on it. At 300 K, k_B T = 0.0259 eV; against copper's E_F = 7.03 eV that is a feather 0.37% of the height of the cliff. Equivalently T_F = E_F/k_B = 8.2 × 10⁴ K, so across the whole range in which copper is a solid, T/T_F stays under 2%. That is what degenerate means: Fermi-Dirac and Maxwell-Boltzmann bear no resemblance to each other here, and the classical limit would need temperatures no metal survives.

Why heating a metal barely touches its electrons

The argument is one line long. An electron well below E_F cannot absorb k_B T of energy, because every state k_B T above it is already occupied — Pauli blocking, not any shortage of energy. Only those within a few k_B T of the Fermi surface have somewhere to go, and they are a fraction of order T/T_F of the whole. Each gains about k_B T, so the thermal energy is U ≈ N k_B T²/T_F and Cₑₗ = dU/dT is proportional to T. The Sommerfeld expansion supplies the coefficient: Cₑₗ = (π²/3)k_B²T g(E_F) = (π²/2)N k_B (T/T_F). Take two things from that. The heat capacity is linear in T, not constant, so it is a different function altogether from the classical (3/2)Nk_B. And it is small: the ratio is (π²/3)(T/T_F), which for copper at 300 K is 1.2%. That is why Dulong-Petit's 3R survived the discovery that every copper atom donates a mobile electron — those electrons contribute 0.15 J mol⁻¹ K⁻¹ against the lattice's 24.9.

Measuring γ from a plot of C/T against T²

Below the Debye temperature the lattice contributes Cₗₐₜ = (12π⁴/5)R(T/θ_D)³, so a measured molar heat capacity is C = γT + AT³. Divide through by T: C/T = γ + AT². Plot C/T against T² and you get a straight line whose intercept is the electronic coefficient γ and whose slope A gives θ_D. The two terms are equal where γT = AT³, that is at T = √(γ/A), which for sodium is 1.5 K — so the linear law is a liquid-helium measurement, never a room-temperature one. What comes back is close to, but not equal to, the free-electron prediction. Copper: 0.50 mJ mol⁻¹ K⁻² predicted against 0.695 measured. Potassium: 1.67 against 2.08. The usual bookkeeping is a thermal effective mass, m*/m = γ/γ_free, giving 1.38 and 1.25. That ratio is the receipt for what the model discarded — the lattice potential, and the interaction between the electrons themselves.

What degeneracy holds up, and where the model stops

The filled sphere carries real energy and real pressure. Averaging E over it gives U/N = (3/5)E_F, which is 4.2 eV per electron in copper at absolute zero, and the degeneracy pressure P = (2/5)nE_F is 38 GPa, with a bulk modulus B = (2/3)nE_F of 64 GPa against copper's measured 137 GPa. A factor of two, from a model with no lattice and no Coulomb interaction, is a real success. So is Wiedemann-Franz: the same Fermi-surface electrons carry charge and heat, so κ/σT = (π²/3)(k_B/e)² = 2.44 × 10⁻⁸ W Ω K⁻², a constant most metals obey near room temperature. The failures are all one failure. There is no gap, so the model predicts every solid is a metal and cannot say why diamond insulates. Aluminium and beryllium show positive Hall coefficients, as though their carriers were positive. And fitting copper's conductivity demands a mean free path of 39 nm, some 150 atomic spacings, which is absurd for a particle bouncing off a dense array of ions. Each of those needs the periodic potential back.

Change one variable at a time

Make the relationship visible.

Interactive model

Temperature T300 K

Fermi energy E_F7.00 eV

Leave T at 300 K: the step edge is razor sharp, and that alone is why Cₑₗ sits near 1% of the classical value. You must drag T to thousands of kelvin before the blur is even visible. Then reset T and pull E_F down to 3 eV — same temperature, more than twice the response.

FERMI TEMPERATURE T_F81228 K

ELECTRON DENSITY n8.41 ×10²⁸ m⁻³

kT / E_F0.0037

Cₑₗ ÷ (3/2)Nk_B0.0121

Live interpretationFERMI TEMPERATURE T_F: 81228 K. ELECTRON DENSITY n: 8.41 ×10²⁸ m⁻³. kT / E_F: 0.0037. Cₑₗ ÷ (3/2)Nk_B: 0.0121

Catch the common trap

Explain before calculating.

Misconception check

Reading "only a k_B T-wide slice is active" as "the electrons are sluggish, or nearly frozen". They are neither. At absolute zero copper's electrons average (3/5)E_F = 4.2 eV each, more than a hundred times the 0.039 eV that equipartition would hand them at 300 K; the ones at the Fermi surface travel at 1.6 × 10⁶ m s⁻¹; and the gas pushes outward at 38 GPa while the metal sits still. What is small is not the energy but the accessibility. An electron 1 eV below E_F cannot absorb a 0.026 eV phonon, because every state 0.026 eV above it is already occupied. Pauli blocking, not lethargy, is what makes Cₑₗ small. For the same reason T_F is not a temperature the metal is at — it is the temperature at which the gas would stop being degenerate, and no solid metal comes near it.

Copper's conduction electrons have E_F = 7.0 eV, so T_F = 8.2 × 10⁴ K. At 300 K, which statement correctly describes their contribution to the molar heat capacity?

Choose an answer to test the model.

Practice & worked examples

Reason from the model, then test the result.

EasySodium is monovalent, with a conduction-electron density n = 2.65 × 10²⁸ m⁻³. Find its Fermi wavevector, its Fermi energy in electronvolts, and its Fermi temperature. Take ℏ = 1.055 × 10⁻³⁴ J s, m = 9.11 × 10⁻³¹ kg, k_B = 1.381 × 10⁻²³ J K⁻¹ and 1 eV = 1.602 × 10⁻¹⁹ J.

The two spin states per k-point are already inside the constant, so k_F = (3π²n)¹⁄³. First the bracket: 3π² = 29.61, and 29.61 × 2.65 × 10²⁸ = 7.85 × 10²⁹ m⁻³.
Cube root: k_F = (7.85 × 10²⁹)¹⁄³ = 9.22 × 10⁹ m⁻¹. Sanity check — the Fermi wavelength 2π/k_F is 0.68 nm, a couple of atomic spacings, which is the right size for an electron shared across a lattice.
E_F = ℏ²k_F²/2m = (1.055 × 10⁻³⁴)² × (9.22 × 10⁹)² ÷ (2 × 9.11 × 10⁻³¹) = 9.46 × 10⁻⁴⁹ ÷ 1.822 × 10⁻³⁰ = 5.19 × 10⁻¹⁹ J.
Convert, then divide: E_F = 5.19 × 10⁻¹⁹ ÷ 1.602 × 10⁻¹⁹ = 3.24 eV, and T_F = E_F/k_B = 5.19 × 10⁻¹⁹ ÷ 1.381 × 10⁻²³ = 3.76 × 10⁴ K.

Answerk_F = 9.22 × 10⁹ m⁻¹, E_F = 3.24 eV, T_F = 3.76 × 10⁴ K. Sodium melts at 371 K, about one hundredth of T_F, so its electron gas is degenerate everywhere the metal exists.

MediumUsing T_F = 3.76 × 10⁴ K for sodium, find the molar electronic heat capacity at 300 K and compare it with the Dulong-Petit lattice value 3R. Then, given a Debye temperature θ_D = 158 K and a lattice term Cₗₐₜ = (12π⁴/5)R(T/θ_D)³, find the temperature below which the electrons dominate. Take R = 8.314 J mol⁻¹ K⁻¹.

Per mole N = N_A and N k_B = R, so Cₑₗ = (π²/2)R(T/T_F) = γT with γ = (π²/2)R/T_F = 4.935 × 8.314 ÷ 3.76 × 10⁴ = 41.03 ÷ 37600 = 1.09 × 10⁻³ J mol⁻¹ K⁻².
At 300 K: Cₑₗ = 1.09 × 10⁻³ × 300 = 0.327 J mol⁻¹ K⁻¹, against 3R = 24.9 J mol⁻¹ K⁻¹. The electrons carry 1.3% of the lattice's share — which is why Dulong-Petit outlived the discovery of the electron.
The two terms carry different powers of T. The phonon coefficient is A = (12π⁴/5)R/θ_D³ = 233.8 × 8.314 ÷ 158³ = 1944 ÷ 3.944 × 10⁶ = 4.93 × 10⁻⁴ J mol⁻¹ K⁻⁴.
They are equal when γT = AT³, so T = √(γ/A) = √(1.09 × 10⁻³ ÷ 4.93 × 10⁻⁴) = √2.21 = 1.49 K.

AnswerCₑₗ(300 K) = 0.33 J mol⁻¹ K⁻¹, about 1.3% of 3R; the electronic term overtakes the phonon term only below 1.5 K, which is why the linear law is a liquid-helium result.

HardLow-temperature heat-capacity data for potassium fit C = γT + AT³. Two points on the C/T against T² line are (T = 1.00 K, C/T = 4.66 mJ mol⁻¹ K⁻²) and (T = 1.50 K, C/T = 7.88 mJ mol⁻¹ K⁻²). Extract γ and A, get the Debye temperature from A, and compare γ with the free-electron prediction, given n = 1.40 × 10²⁸ m⁻³ and hence T_F = 2.46 × 10⁴ K.

Divide C = γT + AT³ through by T: C/T = γ + AT². Against T² that is a straight line, intercept γ, slope A — the standard way to separate electrons from phonons, because only the electronic term survives as T² tends to zero.
Slope: A = (7.88 − 4.66) ÷ (1.50² − 1.00²) = 3.22 ÷ 1.25 = 2.58 mJ mol⁻¹ K⁻⁴. Intercept: γ = 4.66 − 2.58 × 1.00 = 2.08 mJ mol⁻¹ K⁻².
Debye temperature from A = (12π⁴/5)R/θ_D³: θ_D³ = 233.8 × 8.314 ÷ 2.58 × 10⁻³ = 7.54 × 10⁵ K³, so θ_D = 91 K.
Free-electron prediction: γ_free = (π²/2)R/T_F = 41.03 ÷ 2.46 × 10⁴ = 1.67 mJ mol⁻¹ K⁻².
Ratio: m*/m = γ/γ_free = 2.08 ÷ 1.67 = 1.25. The heat capacity is linear in T exactly as Sommerfeld says, but the electrons respond as though 25% heavier — the lattice potential and the electron-electron interaction, reappearing as a mass.

Answerγ = 2.08 mJ mol⁻¹ K⁻², A = 2.58 mJ mol⁻¹ K⁻⁴, θ_D = 91 K, and a thermal effective mass m* = 1.25 m. The model gets the functional form exactly right and the coefficient 25% wrong.