= Solution
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; \b[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,
$$
n=\frac{2\pi k_F^2}{(2\pi)^2},\qquad\boxed{k_F=\sqrt{2\pi n}.}
$$
The reciprocal vectors of the periodic potential are $Q\hat{\mathbf x}$ and $Q\hat{\mathbf y}$. Its <first Brillouin zone> is the square $-Q/2\leq k_x,k_y\leq Q/2$, 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 $\sqrt2<\alpha<2$, 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>.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2015/paper-1-fermi-pockets.png]
{title=A square Brillouin zone reconstructs the free Fermi circle into edge-centred electron pockets and a corner-centred hole pocket}
{source=paper-1-fermi-pockets.py}
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 area
$$
C=R^2[\arccos s-s\sqrt{1-s^2}].
$$
No 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,
$$
\boxed{\frac{n_e}{n}=\frac4\pi[\arccos(\alpha/2)-(\alpha/2)\sqrt{1-\alpha^2/4}],\qquad\frac{n_h}{n}=\frac{\alpha^2}\pi-1+\frac{n_e}{n}.}
$$
Thus at $\alpha=\sqrt3$,
$$
\boxed{n_e=\left(\frac23-\frac{\sqrt3}\pi\right)n\simeq0.11534n,\qquad n_h=\left(\frac{3-\sqrt3}\pi-\frac13\right)n\simeq0.07027n.}
$$
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 \b[compensation occurs at $\alpha=\sqrt\pi$]. These area formulas are the $V_0\to0^+$ limit, before finite-gap reconstruction changes the pocket shapes.<Solid State Physics by David Tong>[Chapters 2–3: band-state counting, zone folding, nearly free electrons and holes]{ref}
The filling per real-space cell is $\nu=na^2=2\pi/\alpha^2$. 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 $\alpha=\sqrt3$ 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 $E_Q=\hbar^2Q^2/(2m)$. 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 order
$$
E_{1,\max}(M)\simeq E_Q/2-V_0,\qquad E_{2,\min}(X,Y)\simeq E_Q/4+V_0/2.
$$
The <indirect gap of a separable square cosine potential> is consequently
$$
\boxed{\Delta_{\rm ind}\simeq\frac32V_0-\frac14E_Q,\qquad V_{0,\min}\sim\frac{E_Q}{6}=\frac{\hbar^2Q^2}{12m}.}
$$
At the compensated filling, this is $V_{0,\min}\sim(\pi/6)E_F^{(0)}$, with $E_F^{(0)}=\hbar^2k_F^2/(2m)$. \b[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.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2015/paper-1-indirect-band-gap.png]
{title=A local Bragg gap can coexist with a global band overlap; a stronger square potential separates the full first band from the second}
{source=paper-1-indirect-band-gap.py}
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