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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.

Let $\mathcal{A}$ be the total two-dimensional area. There are $\mathcal{A}/(2\pi {)}^2$ wavevector states per reciprocal-space area per spin. Including spin degeneracy two and filling a disk givesFor areal number density $n_e=N_e/\mathcal{A}$ this yieldsThe disk is centred at zero wavevector, and its boundary is a fermi circle. Omitting the two spin states would incorrectly double the energy inferred for a given number density.

For one electron per square cell, $n_e=1/a^2$, soNumerically, $\pi /a\simeq 2.17\times 10^{10}{\mathrm{m}}^{-1}$ exceeds $k_F\simeq 1.73\times 10^{10}{\mathrm{m}}^{-1}$, so the fermi circle lies entirely inside the first Brillouin zone.

The state-counting explanation is that a spin-degenerate band has two electron states per primitive cell. One electron per cell therefore occupies half of that band. In the circular free-electron approximation the occupied reciprocal-space area isA disk of that area fits inside this square: its radius is less than the distance to an edge. Half filling alone is not a shape-independent theorem that every possible Fermi surface must avoid every zone boundary; the square geometry and circular free-electron dispersion make the statement true here.

For an isotropic two-dimensional free-electron model, the occupied states form a disk centered at the reciprocal origin. Its Fermi circle first touches a zone side when its radius equals half the distance to a nearest reciprocal point:The contact points have coordinates $(\pm \pi /(2a),\pm \pi /(2a))$ with all four sign combinations. The distance to a zone vertex is $\pi /a$, which is larger; that distance would describe contact with a corner after the circle had already crossed a side.

For one electron per primitive unit cell, the areal number density is $n_s=1/(ab)$. The Fermi circle has radius $k_F=\sqrt{2\pi /(ab)}$. The nearer first-zone boundary is $k_x=\pm \pi /a$, because $a>b$. The circle protrudes precisely whenAt equality the Fermi surface only touches the boundary. Crossing an extended-zone boundary alone does not make a gap: the free electron gas remains metallic when the ionic periodic potential is neglected.

The shifted Fermi circle satisfies $|\mathbf{p}-p_0\hat{\mathbf{x}}|=p_F$. Its radial boundary at polar angle $\theta$ is the positive solution of $p^2-2p{p}_0\cos \theta +p_0^2=p_F^2$. Thus, for $p_0<p_F$,and its leading Taylor expansion isForward-pointing momenta gain occupied states and backward-pointing momenta lose them. With the paper's explicitly reversed Heaviside step function convention, ${\Theta }^{\prime }(x)=-\delta (x)$; consequentlywith a positive leading sign.

For the later quadratic energy calculation it is useful to distinguish the first-order shape from exact particle number conservation. The particle-conserving displacement of a Fermi circle hasThe second-order inward correction cancels the apparent area increase one would obtain by treating $\nu =p_0\cos \theta$ as exact.