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Consider two overlapping energy bands on a square lattice with spin degeneracy two and primitive reciprocal period $Q$. Let $n$ count the areal electron density in this pair, $n_e$ the occupied density in upper-band electron pockets, and $n_h$ the missing density in lower-band hole pockets. A completely filled lower band provides the reference densityCounting filled states gives $n=n_0-n_h+n_e$, hence $n_e-n_h=n-n_0$. Compensation, meaning equal electron pocket and hole pocket carrier densities, therefore occurs at $n=n_0$. For total pocket areas $A_e,A_h$, the same state counting in a Bloch band gives $n_{e,h}=2A_{e,h}/(2\pi {)}^2$; all equivalent pockets must be counted.

If the geometry is parameterized by $Q=\alpha \sqrt{2\pi n}$ with $\alpha >0$, compensation requires $\alpha =\sqrt{\pi }$. The real-space period is $a=2\pi /Q$, so the filling is $n{a}^2=2$ electrons per primitive unit cell. Compensation can coexist with both kinds of charge carriers; a global band gap is additionally required for a band insulator.

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 the electron-doped sample, sketch six closed ellipsoids centred on the $\Gamma$--X axes, just inside the six square faces of the face-centred-cubic first Brillouin zone, with long axes pointing radially toward X. In the parabolic approximation,The radial-to-transverse semiaxis ratio is $\sqrt{m_2/m_1}\simeq 2.2$. The readable conduction minimum near X and its crossing of $\mu$ support these electron Fermi surface pockets. The two $[001]$ cyclotron resonance fields arise from the two distinct orientations of the six ellipsoids, rather than two unrelated carrier species. The left panel of the original sketch includes axes, $\Gamma$, and the three positive-axis X points; the negative-axis X points are symmetry-equivalent.

The surviving valence maximum lies roughly $1.2\mathrm{eV}$ below the original $\mu$. Lowering $\mu$ by $1.5\mathrm{eV}$ therefore places it in the upper valence band in the conventional silicon band interpretation. The electron pockets disappear and one sketches a hole pocket centred at $\Gamma$, as in the middle panel. Detailed valence-band warping and the number of spin-orbit-resolved sheets cannot be extracted from the damaged figure, and the single drawn surface is schematic rather than a claim that all such sheets coincide.

A discriminating physical property is the Hall coefficient: electron-dominated transport gives a negative value, whereas hole-dominated transport gives a positive value in the ordinary single-dominant-carrier regime. One should not infer a numerical resistivity change without the carrier densities and mobilities.

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.