advanced27 min studyLesson 505 in Physics

Count electrons from occupied reciprocal volume

Separate Fermi-surface shape, spin channels, filled bands, and interacting assumptions.

On this page

What you’ll understand

  • Relate occupied Brillouin-zone fraction to electrons per primitive cell.
  • Distinguish the Fermi surface from the Fermi sea.
  • State the regime behind a conventional Luttinger-volume argument.

At zero temperature an independent-electron metal fills states up to a chemical potential. The boundary between occupied and unoccupied states is its Fermi surface; the occupied region is the Fermi sea. Their geometry strongly affects transport, but electron counting depends on enclosed reciprocal volume rather than the apparent length or area of a plotted line. Interactions can preserve a related counting rule under suitable conditions without preserving the original band shape or a step of unit height in occupation.

Count the sea, not just its boundary

For each spin channel, integrate the occupied part of every band using the reciprocal-space state measure. A primitive cell’s Brillouin-zone volume is (2π)ᵈ divided by its real-space volume, so a completely occupied band contributes one electron per cell per independent spin channel. In a simple spin-degenerate band, an occupied fraction f therefore contributes two f electrons per cell. Filled lower bands add integers even when they have no Fermi surface of their own.

Several electron and hole pockets require a consistent reference count. A nearly filled band may be described by its small unoccupied hole volume, but its electron contribution is full-band capacity minus that missing volume. Summing only visible small pocket volumes as if all were electron pockets can be badly wrong. Spin-orbit coupling or magnetic order can remove simple twofold degeneracy, and antiferromagnetic unit-cell enlargement changes the Brillouin zone used for counting.

ne=∑n,σ∫BZddk(2π)d Θ(μ−Enσ(k)),νe=nevcelln_e=\sum_{n,\sigma}\int_{\rm BZ}\frac{d^dk}{(2\pi)^d}\,\Theta(\mu-E_{n\sigma}(\mathbf k)),\qquad \nu_e=n_ev_{\rm cell}
ne is electron density per d-dimensional real-space measure, vcell is primitive-cell measure, and νe electrons per cell. Spin σ is summed explicitly, so no extra factor two is implicit.↔ Scroll wide equations horizontally.

Shape determines velocities and orbits

The Fermi surface is the constant-energy locus at the chemical potential, but its normal energy gradient sets the group velocity. Two surfaces with equal enclosed volume can have different curvatures, nesting features, and transport anisotropy. A large density of states reflects small energy gradients or extended energy-shell measure; it does not uniquely determine a surface shape. Measurements sensitive to orbital character may also weight different portions of the surface unequally.

In an ordinary translationally invariant Fermi liquid adiabatically connected to a noninteracting state, Luttinger counting relates particle density to the Fermi volume, with filled-band ambiguities handled consistently. The momentum distribution can have a discontinuity smaller than one, yet the quasiparticle residue does not multiply the enclosed-volume electron count. Symmetry breaking, topological order, and non-Fermi-liquid settings require examining the theorem’s hypotheses; the conventional statement is not a license to count every correlated metal by an arbitrary free-band picture.

νpartial=gsVFVBZ,vF=1ℏ∇kE∣E=μ\nu_{\rm partial}=g_s\frac{\mathcal V_F}{\mathcal V_{\rm BZ}},\qquad \mathbf v_F=\frac1\hbar\nabla_{\mathbf k}E\bigg|_{E=\mu}
VF is occupied reciprocal volume in one partially filled band and gs its actual independent degeneracy. The first formula is direct independent-band counting; its conventional interacting continuation needs Fermi-liquid and translation assumptions.↔ Scroll wide equations horizontally.

Filling alone is a first diagnostic

A nonmagnetic spin-degenerate one-orbital band at one electron per cell is half filled. In a simple independent-band picture it cannot be a conventional band insulator without changing the state count or splitting the relevant bands. Strong interactions can instead produce a Mott state, so the filling argument diagnoses the limits of that band picture rather than proving universal metallicity. Conversely, an even integer filling permits a band insulator but does not ensure one if bands overlap.

The probability lab illustrates an explicit degeneracy-independent count. Reset and set probability to 0.25 and trials to 8; predict mean occupation count 2. Reset and set probability to 0.75 and trials to 8; predict 6. This represents eight independent Bernoulli occupations with a common probability, as in a simplified thermal population model. It has no momentum labels or Fermi surface and does not determine zero-temperature pocket geometry or interacting Luttinger volume.

νe=2Nfull+2ffor spin-degenerate independent bands\nu_e=2N_{\rm full}+2f\qquad\text{for spin-degenerate independent bands}
Nfull is the integer number of fully occupied orbital bands and f the occupied fraction of one additional orbital band. The formula presumes precisely two independent spin states per orbital band and a chosen primitive cell.↔ Scroll wide equations horizontally.

Read a pocket fraction

A two-dimensional crystal has two completely filled spin-degenerate orbital bands and one additional spin-degenerate band with occupied reciprocal area 30% of its Brillouin zone. Find electrons per primitive cell and the hole count needed to fill the last band.

  1. The two full bands contribute 2×2=4 electrons per cell. They contain occupied states even though their energies do not cross the chemical potential and they contribute no boundary Fermi contour.

  2. The partly filled band contributes 2×0.30=0.60 electrons per cell. Adding the full bands gives 4.60 electrons per primitive cell, using the specified twofold degeneracy only once.

  3. The remaining empty fraction of the last band is 0.70, corresponding to 2×0.70=1.40 holes per cell relative to that band being full. These holes are a missing-state count, not an additional 1.40 occupied electrons.

The result

The filling is 4.60 electrons per cell, with 1.40 holes per cell needed to complete the partially filled band.

Explore the model

Interactive lab

Interactive lab · probability

Count independent occupation variables

Reset and set p=0.25, n=8; predict mean count 2. Reset and set p=0.75, n=8; predict mean 6.

Count independent occupation variablesThese are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. Current readings: Mean · np: 5; Variance · np(1−p): 2.5; Total probability: 1. All parameters are adjustable with the labeled controls below.03581000.070.150.220.3Successes kP(X = k)k = 0 · P = 9.77e-4
Exact probabilityHighlighted outcomeMean np
Mean · np
5
Variance · np(1−p)
2.5
Total probability
1
t = 0 s

These are exact binomial probabilities for independent trials with one fixed success probability. Playback tours the distribution; it does not resample outcomes. 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 forThe exact binomial model has independent equal-probability occupations. It contains no wavevectors, spin-split bands, or interacting Fermi-volume calculation.

Check your understanding

In a conventional Fermi-liquid volume count, does a quasiparticle residue Z<1 multiply the enclosed-volume carrier count?

Try it yourself

A single orbital band is fully spin polarized, so each wavevector has only one available channel. If it contains 0.40 electrons per cell, find occupied and empty Brillouin-zone fractions. Compare with a twofold spin-degenerate band at the same electron count.

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

Give me a hint

Write the band capacity per cell before dividing the electron count.

Compare with the solution

The polarized band has capacity one, so occupied fraction is 0.40 and empty fraction 0.60. With twofold degeneracy the same 0.40 electrons occupy fraction 0.20 of each equally filled spin channel, leaving fraction 0.80 empty. Applying a factor two without specifying spin would misidentify the Fermi volume.

Take it with you

  • Filled bands and partial pockets contribute differently to a complete count.
  • Spin and primitive-cell conventions must be stated before converting pocket volume into filling.
  • Conventional Luttinger volume is not multiplied by quasiparticle residue.

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