Rectangular-lattice Fermi-surface reconstruction

Consider a spin-degenerate two-dimensional free electron model with rectangular periods $a>b$ and two Electrons per unit cell. Write $K=\pi /a$, $L=\pi /b$, and $k_F^2=4\pi /(ab)$. If $K<k_F<L$, the circular Fermi contour crosses only the vertical boundaries of the first Brillouin zone. Folding translates its protruding caps back by a reciprocal lattice vector $(2K,0)$. A weak periodic potential mixes the intersecting plane waves, reconnecting the contours into two open first-band sheets and a closed second-band pocket. The pocket is centred on a zone boundary: its halves on opposite sides of a reduced-zone plot are identified periodically and belong to one closed sheet.

Put $C={\hslash }^2/(2m)$, $v=|V_G|$, $\mathbf{G}=(2K,0)$, and $k_x=K+q$. The upper eigenvalue of the two coupled plane waves isAt fixed Fermi energy $C{k}_F^2$, its closed Fermi contour hasThus $A(v)=4{\int }_0^{q_m}y(q)dq$ can be inverted numerically to estimate the Bragg scattering coupling. The pocket disappears at $v=C(k_F^2-K^2)$. The estimate assumes a fixed chemical potential and neglects coupling to other plane waves; a measured pocket area alone does not determine all Fourier coefficients of an arbitrary lattice potential.

In rectangular-lattice Fermi-surface reconstruction, the zero-coupling closed orbit is the overlap lens of two equal circles of radius $k_F$ with centres separated by $2K$. Two circular sectors minus their triangular portions giveAt fixed chemical potential, the upper hybridized energy band rises when the magnitude of the Fourier coefficient increases. Its occupied pocket therefore shrinks, so this lens is an upper bound in the two-wave model. It is not a universal bound for arbitrary changes of chemical potential or additional energy bands.