research44 min studyLesson 64 in Physics

Fermi surfaces and Fermi liquids

Count a filled Fermi sea and explain how long-lived interacting quasiparticles organize low-energy metallic behavior.

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What you’ll understand

  • Calculate Fermi momentum and energy from a three-dimensional density.
  • Explain why low-energy rearrangements concentrate near a Fermi surface.
  • Distinguish a Landau quasiparticle from a bare electron and state where the description can fail.

Exclusion changes what cold electrons can do. Filling low-energy states creates a Fermi sea with a momentum-space boundary. Ordinary metals often respond through small rearrangements near that boundary. Interactions can preserve this organization by dressing excitations into long-lived quasiparticles. A Fermi liquid is this testable low-energy description, rather than a universal property of compressible quantum materials.

Count states before interpreting the surface

For a homogeneous, noninteracting three-dimensional gas, take two spin states with equal populations and dispersion ε(k)=ℏ2k2/(2m)\varepsilon(k)=\hbar^2k^2/(2m). At zero temperature, all states inside a sphere of radius kFk_F are occupied. Periodic boundary conditions put one momentum state per volume (2π)3/V(2\pi)^3/V per spin. Counting the sphere therefore relates number density to kFk_F. The Fermi energy is the band energy at that radius measured from the chosen band minimum. This counting uses a continuum gas; a crystal requires its actual bands and Brillouin zone.

The Fermi surface is a surface in momentum space, not an outer layer of electrons in real space. In a general band it is the set where the energy equals the chemical potential, and it can contain several sheets rather than one sphere. At nonzero temperature, the occupation step is smoothed over an energy range of order kBTk_BT. The chemical potential also adjusts to maintain the prescribed density. A sharp zero-temperature surface remains a useful reference when that thermal scale is small compared with the relevant electronic scales.

n=kF33π2,kF=(3π2n)1/3,EF=ℏ2kF22m,f(ε)=1e(ε−μ)/(kBT)+1n=\frac{k_F^3}{3\pi^2},\qquad k_F=(3\pi^2n)^{1/3},\qquad E_F=\frac{\hbar^2k_F^2}{2m},\qquad f(\varepsilon)=\frac1{e^{(\varepsilon-\mu)/(k_BT)}+1}
The density formula assumes three dimensions, two equally populated spin species, and a spherical noninteracting continuum Fermi sea.↔ Scroll wide equations horizontally.

Pauli blocking selects a thin active shell

An electron deep below the zero-temperature chemical potential cannot respond to an arbitrarily small added energy by moving to a nearby occupied state. The destination is already filled. Particle-hole excitations at low energy instead remove an electron just below the surface and place it just above. This phase-space restriction explains why the electronic heat capacity of an ordinary degenerate metal is proportional to temperature at sufficiently low temperature, rather than approaching the classical constant per electron. The proportionality coefficient depends on the density of available low-energy states.

For interacting systems, Landau’s construction associates low-energy excitations with those of a reference Fermi sea as the interaction is turned on without crossing a phase transition. A quasiparticle carries the electron’s conserved quantum numbers but includes a rearrangement of the surrounding medium. Its energy and effective mass can differ from bare values, and quasiparticles interact with one another. This association is a physical low-energy hypothesis, not an assertion that the exact many-body wavefunction is a product of independently dressed orbitals. Collective response requires interaction parameters as well as a dispersion.

A quasiparticle must remain recognizable

In a conventional Fermi liquid with regular short-range interactions, decay is restricted by energy conservation and Pauli blocking. Near the surface, a characteristic decay rate scales quadratically with excitation energy or temperature, up to dimension-dependent and interaction-dependent corrections. In three dimensions the familiar leading behavior is proportional to ξ2+(πkBT)2\xi^2+(\pi k_BT)^2, divided by an appropriate microscopic energy and by ℏ\hbar. As the excitation energy tends to zero, its width becomes small compared with that energy, so the excitation remains a useful propagating degree of freedom.

One-dimensional interacting fermions generically require Luttinger-liquid ideas; critical fluctuations, disorder, or pairing can also change the description. Resistance alone does not settle quasiparticle behavior: momentum relaxation and single-particle decay are different observables. The quadratic lab is only a one-dimensional energy-versus-coordinate analogy: with coefficient positive it shows a parabola and its slope. It does not populate a three-dimensional Fermi sphere, calculate occupations, or predict interacting lifetimes. Compare the local slope with velocity, then return to state counting.

vF=1ℏdε∗dk∣kF,τ−1∼C ξ2+(πkBT)2ℏE∗v_F=\left.\frac1\hbar\frac{d\varepsilon^*}{dk}\right|_{k_F},\qquad \tau^{-1}\sim C\,\frac{\xi^2+(\pi k_BT)^2}{\hbar E_*}
The rate expression is schematic conventional three-dimensional Fermi-liquid scaling, with model-dependent C and energy E*. It is not a universal numerical lifetime formula.↔ Scroll wide equations horizontally.

Density fixes the free-electron energy scale

A model metal has electron density n=1.0×1028 m−3n=1.0\times10^{28}\,\mathrm{m}^{-3}, two equally populated spin states, and the vacuum electron mass. Estimate kFk_F, EFE_F, and the degeneracy temperature TF=EF/kBT_F=E_F/k_B.

  1. Compute kF=(3π2n)1/3≈6.67×109 m−1k_F=(3\pi^2n)^{1/3}\approx6.67\times10^9\,\mathrm{m}^{-1}. Its units are inverse length because density enters under a cube root.

  2. Use ℏ=1.0546×10−34 J s\hbar=1.0546\times10^{-34}\,\mathrm{J\,s} and m=9.1094×10−31 kgm=9.1094\times10^{-31}\,\mathrm{kg}. Then EF≈2.71×10−19 J≈1.69 eVE_F\approx2.71\times10^{-19}\,\mathrm{J}\approx1.69\,\mathrm{eV}.

  3. Divide by kB=1.3806×10−23 J K−1k_B=1.3806\times10^{-23}\,\mathrm{J\,K}^{-1} to find TF≈1.96×104 KT_F\approx1.96\times10^4\,\mathrm{K}. At 300 K, T/TF≈0.015T/T_F\approx0.015, so this free-gas reference is strongly degenerate even though the material is not cryogenic.

The result

A density of 10²⁸ m⁻³ gives a free-electron Fermi energy around 1.69 eV and a Fermi temperature near 19,600 K; a real band can modify these estimates.

Explore the model

Interactive lab

Interactive lab · function

Energy curvature and slope

The lab plots only y=ax2y=ax^2 with its tangent and integral. Set positive a, interpret x as a dimensionless momentum coordinate, and compare the slope at opposite x. This illustrates an even quadratic dispersion with opposite velocities; it does not construct a Fermi sea, density of states, or interacting quasiparticle.

Energy curvature and slopeThis lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Current readings: y at chosen x: 0.608; Tangent slope · 2ax: 1.35; Signed area · ax³/3: 0.182. All parameters are adjustable with the labeled controls below.−2.4−1.201.22.4−9−4.504.59xy(x, ax²)
y = ax²TangentSigned integral
y at chosen x
0.608
Tangent slope · 2ax
1.35
Signed area · ax³/3
0.182
t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forAn energy slope determines a velocity only after the physical momentum scale and ℏ are restored.

Check your understanding

Why do low-energy particle-hole excitations of a cold ideal Fermi gas mainly involve states near its Fermi surface?

Try it yourself

Keeping the mass and spin degeneracy fixed in the same three-dimensional free-gas model, increase density by a factor of eight. How do kFk_F, EFE_F, and TFT_F change? Does this counting by itself prove that an interacting material is a Fermi liquid?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Fermi momentum scales as the cube root of density and the quadratic energy scales as its square.

Compare with the solution

The new Fermi momentum is twice the old value because 81/3=28^{1/3}=2. Both Fermi energy and Fermi temperature rise by a factor of four because they scale with kF2k_F^2. These relations are exact for the stated free-gas model. They do not prove the survival of long-lived quasiparticles in an interacting material; that requires a justified low-energy description and appropriate experimental or theoretical evidence.

Take it with you

  • The Fermi surface is a momentum-space boundary organizing low-energy excitations.
  • Pauli blocking makes the nearby available phase space central to cold fermion dynamics.
  • Fermi-liquid quasiparticles are long-lived dressed excitations within a specified regime, not universally valid bare electrons.

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