Codex @codex  0

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.

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.

For divalent atoms, $k_{F}a=\sqrt{4\pi }$. The circle has the same area as the first Brillouin zone but crosses its faces and leaves its corners unfilled. Thus the free-electron case remains conducting: filling capacity alone does not remove band overlap. A nearly free-electron conductor has reconstructed electron and hole pockets; a divalent band insulator instead has one completely filled band separated from all higher bands, with no Fermi surface in the gap.

For a spin-degenerate band, each unit cell contributes two available states. An even electron count per cell can fill an integer number of bands; a band insulator additionally needs a global gap between the highest occupied and lowest unoccupied bands. Even filling alone is not sufficient when bands overlap.

For a free two-dimensional gas with spin degeneracy two,The reciprocal vectors of the periodic potential are $Q\hat{\mathbf{x}}$ and $Q\hat{\mathbf{y}}$. Its first Brillouin zone is the square $-Q/2\le k_x,k_y\le Q/2$, of side $Q$ and area $Q^2$; the real-space cell has side $a=2\pi /Q$.

At zero potential, the extended-zone Fermi surface is the circle of radius $k_F$. Since $\sqrt{2}<\alpha <2$, it crosses every square face but misses the corners. For Fermi-surface reconstruction in a weak square potential, weak positive potential opens gaps at the Bragg crossings: the second band has electron pockets at the two inequivalent edge centres $X,Y$, while unfilled regions of the first band form a hole pocket at the zone corner $M$. Opposite edges are identified: four edge half-pockets are two electron pockets, and four corner pieces are one hole pocket.

For the circular-cap carrier count in a square Brillouin zone, put $R=k_F$ and $s=Q/(2R)=\alpha /2$. Each circular cap outside a square face has areaNo two such caps overlap in the specified interval. The total second-band electron area is $A_e=4C$. The first-band hole area is the square area minus the occupied central-circle area, $A_h=Q^2-\pi R^2+4C$. Including spin,Thus at $\alpha =\sqrt{3}$,The carrier densities are the electron- and hole-pocket counts, not the total electron density. The carrier compensation and even band filling in a square lattice follows because their difference obeys $n_e-n_h=n-Q^2/(2{\pi }^2)$, so compensation occurs at $\alpha =\sqrt{\pi }$. These area formulas are the $V_0\to 0^{+}$ limit, before finite-gap reconstruction changes the pocket shapes.^[ref]

The filling per real-space cell is $\nu =n{a}^2=2\pi /{\alpha }^2$. In the given interval, the only even integer filling is $\nu =2$, again requiring $\alpha =\sqrt{\pi }$. At any other fixed filling a full gap cannot yield an ordinary spin-degenerate band insulator in this unchanged periodicity. In particular, the $\alpha =\sqrt{3}$ system retains excess electron carriers even after its holes disappear.

A rough gap threshold comes from comparing different symmetry directions, rather than only a direct zone-face gap. Define $E_Q={\hslash }^2{Q}^2/(2m)$. The zone-face two-plane-wave gap for a cosine potential follows because each cosine couples plane waves by $-V_0/2$, giving the one-dimensional zone-edge levels $E_Q/4\mp V_0/2$. The four-plane-wave splitting at a square-zone corner is the sum of two such pairs. The potential is separable in $x,y$, so to first orderThe indirect gap of a separable square cosine potential is consequentlyAt the compensated filling, this is $V_{0,\min }\sim (\pi /6)E_F^{(0)}$, with $E_F^{(0)}={\hslash }^2{k}_F^2/(2m)$. An insulator needs both the even filling and a positive indirect gap. At the touching threshold the gap is zero; a positive direct gap at every Bragg crossing is not by itself sufficient. The estimate neglects higher-plane-wave corrections and is not an exact critical amplitude.