= Solution
Use two <plane waves>, $|\mathbf k\rangle$ and $|\mathbf k-\mathbf G\rangle$, as a <basis> near their Bragg degeneracy. The potential has <Fourier coefficients> $V_{\pm\mathbf G}$; its zero <Fourier coefficient> vanishes. All other <plane waves> are off <resonance> and are neglected in the leading <nearly-free electron model>. With $\varepsilon_{\mathbf k}=\hbar^2k^2/(2m_e)$, the <Schrödinger equation> reduces to the <Hermitian matrix> problem
$$
\begin{pmatrix}
\varepsilon_{\mathbf k}&V_G\\
V_G^*&\varepsilon_{\mathbf k-\mathbf G}
\end{pmatrix}
\begin{pmatrix}c_1\\c_2\end{pmatrix}
=E\begin{pmatrix}c_1\\c_2\end{pmatrix}.
$$
Setting the <characteristic polynomial> equal to zero gives
$$
(E-\varepsilon_{\mathbf k})(E-\varepsilon_{\mathbf k-\mathbf G})-|V_G|^2=0.
$$
Solving this <quadratic equation> gives
$$
E_\pm=\frac{\varepsilon_{\mathbf k}+\varepsilon_{\mathbf k-\mathbf G}}2
\pm\sqrt{\left(\frac{\varepsilon_{\mathbf k}-\varepsilon_{\mathbf k-\mathbf G}}2\right)^2+|V_G|^2}.
$$
For the real <cosine> potential its coefficient $V_G$ is real; the absolute-value form also allows an arbitrary choice of origin and corresponding Fourier phase.
At the Brillouin-zone boundary, $\mathbf k\cdot\mathbf G=G^2/2$, the free energies are equal. The two <standing waves> have different <probability densities> relative to the <lattice> potential and are split by
$$
\boxed{\Delta E=2|V_G|.}
$$
Repeating this <avoided crossing> at the <reciprocal lattice> boundaries joins smooth branches into <energy bands>. In one dimension a nonzero <Fourier coefficient> opens a gap at the corresponding crossing. In higher dimensions a direct gap at one <wavevector> need not give a common forbidden <energy> interval at every <wavevector>: different bands may still overlap in <energy>. The present two-dimensional example remains metallic even though the crossings are locally gapped.
For the rectangular <lattice>, choose real-space primitive vectors $a\widehat{\mathbf x}$ and $b\widehat{\mathbf y}$. The <reciprocal lattice> vectors are
$$
\mathbf G_x=\frac{2\pi}{a}\widehat{\mathbf x},\qquad
\mathbf G_y=\frac{2\pi}{b}\widehat{\mathbf y}.
$$
Define the half-widths of the first <Brillouin zone> by $K=\pi/a$ and $L=\pi/b$. Their values and the zone are
$$
\boxed{K=2.41661\,\mathrm{nm^{-1}},\quad L=3.92699\,\mathrm{nm^{-1}},\quad
-K\leq k_x\leq K,\quad -L\leq k_y\leq L.}
$$
A primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first <Brillouin zone> here. Its area is $4\pi^2/(ab)=37.9600\,\mathrm{nm^{-2}}$.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2012-rectangular-cells.png]
{title=Real-space primitive cell and reciprocal-space first <Brillouin zone>}
Including both <spin> states, counting the occupied free-electron states per unit area gives
$$
n_{2D}=2\frac{\pi k_F^2}{(2\pi)^2}=\frac{k_F^2}{2\pi}
=\frac2{ab}.
$$
Consequently
$$
\boxed{k_F=\sqrt{\frac{4\pi}{ab}}=3.47607\,\mathrm{nm^{-1}},\qquad
E_F^{(0)}=\frac{\hbar^2k_F^2}{2m_e}=0.46036\,\mathrm{eV}.}
$$
The occupied <disk> has precisely the area of the first <Brillouin zone>, but their shapes differ: $K<k_F<L$. It protrudes through both vertical zone boundaries and does not reach the horizontal ones. Thus two <Electrons> per cell do not imply a filled first band in the absence of a separating <band gap>.
Fold the protruding caps by $\pm\mathbf G_x$. The original <circle> and the translated <circle> intersect at $k_x=K$, $k_y=\pm\sqrt{k_F^2-K^2}$. The potential mixes these states and replaces each crossing by an <avoided crossing>. The lower branch reconnects into two open sheets running across successive zones in the $k_x$ direction; the upper branch forms a closed lens-shaped pocket about a vertical zone boundary. In a reduced-zone sketch its two halves sit at opposite edges, but periodic identification makes them \b[one closed sheet]. This is <rectangular-lattice Fermi-surface reconstruction>.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2012-fermi-reconstruction.png]
{title=Free and reconstructed <Fermi contours>: two open sheets and one closed pocket per reciprocal cell}
The largest pocket in this weak-potential construction is the <lens-shaped intersection of two disks> enclosed by the intersecting free circles. Each has radius $k_F$ and their centres are separated by $2K$. If $\theta=\arccos(K/k_F)$, two sectors minus their two triangles give the <lens bound on a reconstructed Fermi pocket>:
$$
A_{\max}=2k_F^2\theta-2K\sqrt{k_F^2-K^2},\qquad
\boxed{A_{\max}=7.30680\,\mathrm{nm^{-2}}.}
$$
Opening a gap raises the upper branch at fixed <chemical potential> and shrinks its pocket. The reported <quantum oscillation> area satisfies
$$
\boxed{0<A_k=5.74\,\mathrm{nm^{-2}}<A_{\max},\qquad
\frac{A_k}{A_{\max}}=0.786.}
$$
It is therefore consistent with this closed sheet. <Onsager quantization> gives $F=\hbar A_k/(2\pi e)\simeq601\,\mathrm T$. An orbit on this closed contour can return to itself under the <Lorentz force> and occupy quantized <Landau levels>. The open sheets do not enclose a finite orbit area for a perpendicular field in this two-dimensional model. The observation is consistent with the reconstructed pocket, without uniquely determining a completely arbitrary band structure.
For a quantitative rough coupling estimate, use the supplied fixed-chemical-potential approximation and retain the two waves coupled by $\mathbf G_x$. Write
$$
C=\frac{\hbar^2}{2m_e}=0.0380998\,\mathrm{eV\,nm^2},\qquad
v=|V_G|,\qquad k_x=K+q.
$$
Their upper <eigenvalue> is
$$
E_+(q,k_y)=C(K^2+q^2+k_y^2)+\sqrt{4C^2K^2q^2+v^2}.
$$
Setting it equal to $Ck_F^2$ gives the pocket boundary
$$
k_y^2=k_F^2-K^2-q^2-\sqrt{4K^2q^2+(v/C)^2}.
$$
Its intersection with $k_y=0$ occurs at $q=\pm q_m$, where
$$
q_m^2=k_F^2+K^2-\sqrt{4k_F^2K^2+(v/C)^2}.
$$
The <two-wave estimate of a Bragg-gap coupling from a pocket area> therefore uses
$$
A(v)=4\int_0^{q_m}
\sqrt{k_F^2-K^2-q^2-\sqrt{4K^2q^2+(v/C)^2}}\,dq.
$$
At $v=0$ this reproduces the <lens bound on a reconstructed Fermi pocket>; the pocket disappears at $v=C(k_F^2-K^2)=0.23786\,\mathrm{eV}$. <Numerical integration> gives $A(0.075\,\mathrm{eV})=6.22\,\mathrm{nm^{-2}}$ and $A(0.100\,\mathrm{eV})=5.54\,\mathrm{nm^{-2}}$. Matching the measured area gives $v\simeq0.0931\,\mathrm{eV}$, so the justified rough result is
$$
\boxed{|V_G|\sim0.1\,\mathrm{eV}\qquad(\Delta E\sim0.2\,\mathrm{eV}).}
$$
This estimate neglects remote plane-wave admixtures and other <Fourier coefficients>. It also holds the <chemical potential> fixed as requested; an exact fixed-density calculation would determine it self-consistently after reconstruction. The numerical root is a way to implement the approximate model, not a measurement of $V_G$ to four significant figures.
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