Fermi surface

For noninteracting fermions at zero temperature, the Fermi surface is the momentum-space boundary $ϵ(\mathbf{k})=\mu$ between occupied and unoccupied states. Its shape can be a circle, several pockets or a more complicated surface, depending on the dispersion and filling.

In an interacting Fermi liquid, occupation is not simply zero or one on either side. Instead the momentum distribution has a discontinuity with quasiparticle weight $Z$, and low-energy Landau quasiparticles define the Fermi surface. This distinguishes a sharp quasiparticle boundary from an ideal-gas step distribution. The geometry controls low-energy phase space, transport and collective response. Quantum oscillations constrain extremal orbit areas, while momentum-resolved spectral measurements can probe excitations near the chemical potential; the interpretation requires the relevant interaction and measurement model.

For a two-dimensional free electron gas with two spin states and $z$ electrons per square cell of side $a$, counting occupied wavevectors gives $z/a^2=2\pi k_F^2/(2\pi {)}^2$. The square first Brillouin zone has half-width $\pi /a$. A monovalent circle lies inside its faces; a divalent circle crosses its faces but not its corners. A real periodic potential can deform and reconnect the crossings, as in Fermi-surface reconstruction in a weak square potential; valence alone is insufficient to prove insulation.

For a free circle of radius $R$ and square reciprocal period $Q$ with $\sqrt{2}R<Q<2R$, the circle cuts the faces but not the corners of the first zone. Weak Bragg gaps reconnect the folded branches into two inequivalent edge-centred electron pockets and one corner-centred hole pocket. The zero-potential limit retains band crossings instead of avoided crossings.

For $s=Q/(2R)$, each face cap has area $C=R^2[\arccos s-s\sqrt{1-s^2}]$. The total second-band electron area is $4C$; the first-band hole area is $Q^2-\pi R^2+4C$. Multiplying either area by the spin degeneracy two and dividing by $(2\pi {)}^2$ gives the carrier densities. This is an infinitesimal-potential count, not an exact finite-gap formula.

A hole pocket surrounds unoccupied states near a band maximum. Its hole carrier density counts missing states in the nearly filled band, not all occupied electrons. In two dimensions the count is $g{A}_h/(2\pi {)}^2$. Four corner pieces of a square-zone plot can be one pocket on the periodically identified zone.

An electron pocket surrounds occupied states near a conduction-band minimum. In two dimensions its carrier density is $gA/(2\pi {)}^2$, where $A$ is its total reciprocal-space area and $g$ the unresolved degeneracy. Portions drawn at opposite zone faces can belong to the same pocket after periodic identification.

For an isotropic three-dimensional free electron gas at zero temperature, occupied wavevectors fill a sphere of radius $k_F$. Including two spin states, $n=k_F^3/(3{\pi }^2)$. The volume in reciprocal space counts occupied states; folding an extended-zone sphere into a first Brillouin zone does not remove states outside that first zone.

A Fermi point is a point of the zero-temperature Fermi surface in one-dimensional momentum space. For a spin-degenerate band with one occupied interval, its two boundaries are usually $-k_F$ and $+k_F$. Low-energy excitations near them form left- and right-moving branches. A perturbation at wavevector $2k_F$ can mix those branches, as in a Peierls distortion.

For a two-dimensional energy band, a Fermi contour is the set of wavevectors at the Fermi energy. It is the two-dimensional counterpart of a Fermi surface. At a regular value it consists locally of curves; at a band extremum or a crossing it can acquire isolated points or intersecting branches. At zero temperature, each band's occupied region satisfies $E_n(\mathbf{k})<E_F$, so the contours must be labelled by band when determining occupations. The Fermi contours of an isotropic Rashba band illustrate why two contours need not belong to two different bands.

The circular zero-temperature occupation boundary of an isotropic two-dimensional electron gas. The occupied disk has area $\pi k_F^2$; converting its area to particle number requires the reciprocal-space state density and a separately specified spin degeneracy.

Translating a Fermi circle of radius $p_F$ by $p_0\hat{\mathbf{x}}$ preserves its area. For $p_0<p_F$, its radial displacement from the original centre is the displayed expression. Its Taylor expansion is $\nu =p_0\cos \theta -p_0^2{\sin }^2\theta /(2p_F)+O(p_0^4/p_F^3)$. The second-order term cancels the apparent area increase from the first-order shape: ${\int }_0^{2\pi }(p_F\nu +{\nu }^2/2)d\theta =0$. This distinction matters when converting a grand-potential variation into a fixed-particle number energy change.