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For the Rashba effect with $k_R>0$, put $D=\sqrt{k_R^2+2m{E}_F/{\hslash }^2}$. If $-E_R<E_F<0$, both Fermi contours belong to the lower branch and bound the singly occupied annulus $k_R-D<k<k_R+D$; the central disk and exterior are empty. At $E_F=0$, the lower occupied disk has outer radius $2k_R$, while the origin is a degenerate point at the Fermi energy. For $E_F>0$, the lower Fermi contour has radius $k_{-}=D+k_R$ and the upper Fermi contour radius $k_{+}=D-k_R$. The inner disk has two occupied spin-split eigenstates at each momentum, the intervening annulus has one, and the exterior is empty. Consequently $k_{-}-k_{+}=2k_R$, whereas below the crossing the two lower-branch radii add to $2k_R$. At zero temperature, changing occupation values only on these Fermi contours affects a null set and leaves the integrated carrier density unchanged.

Use two plane waves, $|\mathbf{k}⟩$ and $|\mathbf{k}-\mathbf{G}⟩$, 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 ${\epsilon }_{\mathbf{k}}={\hslash }^2{k}^2/(2m_e)$, the Schrödinger equation reduces to the Hermitian matrix problemSetting the characteristic polynomial equal to zero givesSolving this quadratic equation givesFor 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 byRepeating 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 $a\hat{\mathbf{x}}$ and $b\hat{\mathbf{y}}$. The reciprocal lattice vectors areDefine the half-widths of the first Brillouin zone by $K=\pi /a$ and $L=\pi /b$. Their values and the zone areA primitive reciprocal cell can be drawn with a corner at the origin; its centred translate is the first Brillouin zone here. Its area is $4{\pi }^2/(ab)=37.9600\mathrm{n}{\mathrm{m}}^{-2}$.

Including both spin states, counting the occupied free-electron states per unit area givesConsequentlyThe 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 one closed sheet. This is rectangular-lattice Fermi-surface reconstruction.

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:Opening a gap raises the upper branch at fixed chemical potential and shrinks its pocket. The reported quantum oscillation area satisfiesIt is therefore consistent with this closed sheet. Onsager quantization gives $F=\hslash A_k/(2\pi e)\simeq 601\mathrm{T}$. 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 ${\mathbf{G}}_x$. WriteTheir upper eigenvalue isSetting it equal to $C{k}_F^2$ gives the pocket boundaryIts intersection with $k_y=0$ occurs at $q=\pm q_m$, whereThe two-wave estimate of a Bragg-gap coupling from a pocket area therefore usesAt $v=0$ 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 $A(0.075\mathrm{eV})=6.22\mathrm{n}{\mathrm{m}}^{-2}$ and $A(0.100\mathrm{eV})=5.54\mathrm{n}{\mathrm{m}}^{-2}$. Matching the measured area gives $v\simeq 0.0931\mathrm{eV}$, so the justified rough result isThis 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.

Diagonalization and conventions. The spin-dependent part is $\lambda (-k_y{\sigma }_x+k_x{\sigma }_y)$. For a Hermitian Hamiltonian $\lambda$ is real; the supplied lower-band spinor uses the convention $\lambda >0$, which will be adopted below. The squared spin matrix is ${\lambda }^2(k_x^2+k_y^2)I$. Hence the two ordered energies areIn the $k_z=0$ plane, let $k=\sqrt{k_x^2+k_y^2}$ and $q_R=2m\lambda /{\hslash }^2>0$. ThenIf $\lambda <0$, use $q_R=2m|\lambda |/{\hslash }^2$ to retain the energy ordering, and interchange the spin directions assigned below. This is an isotropic Rashba effect; $q_R$ has units of inverse length, whereas $\lambda$ has units of energy times length.

Band minima. Define $k_R=q_R/2$ and $E_R={\hslash }^2{q}_R^2/(8m)$. Completing the square givesAlong $k_y=k_z=0$, the radial coordinate is $|k_x|$, so the lower branch has two minima:In the full plane these are points on a circle of minima $k=k_R$. Both branches meet at $k=0$, $E=0$. Their ordered energy curves contain $|k_x|$: the lower curve descends from the crossing on either side, while the upper rises. Treating them as globally labeled smooth spin branches without switching labels across $k_x=0$ would obscure the lower-band minima.

Occupied regions and Fermi contours. At zero temperature each spin-split eigenstate with energy below $E_F$ is occupied. PutFor $-E_R<E_F<0$, $0<D<k_R$. The lower-band roots are $k_R-D$ and $k_R+D$, while the upper band is entirely empty. Thus the central disk $k<k_R-D$ is unoccupied, the annulus $k_R-D<k<k_R+D$ is singly occupied, and the exterior is unoccupied. Both Fermi circles belong to the lower band; there is no doubly occupied region.

At $E_F=0$, the lower band is occupied for $0<k<2k_R=q_R$, giving a singly occupied disk with outer Fermi circle at $q_R$. The two bands also meet the Fermi energy at the origin. That isolated degenerate point is not a finite doubly occupied disk, and occupation exactly at the Fermi energy is a boundary convention of zero area. Outside the outer circle the states are empty.

For $E_F>0$, $D>k_R$. The lower branch has outer radius $k_{-}=D+k_R$, while the upper branch has inner radius $k_{+}=D-k_R$. The disk $k<k_{+}$ is doubly occupied, with one state in each band at every momentum. The annulus $k_{+}<k<k_{-}$ is singly occupied, in the lower band only, and the exterior is unoccupied. These are the Fermi contours of an isotropic Rashba band; double occupancy here refers to two orthogonal spinor states at the same momentum, not two Electrons occupying one identical quantum state.

Inferring the splitting. Above the crossing, the two positive radii differ by $2k_R=q_R$:The minimum radius is therefore $k_R=0.05{\mathring{\mathrm{A}}}^{-1}$. It is the difference of the measured radii that is required for $E_F>0$; below the crossing the two lower-band radii instead add to $q_R$.

Spinors and spin direction. Write $a=(k_y+i{k}_x)/k$, which has $|a|=1$ for $k>0$. An orthogonal normalized partner to the supplied lower-band state isIndeed $u_{-}^{†}{u}_{+}=(-|a{|}^2+1)/2=0$. Direct multiplication by the spin matrix gives eigenvalues $-\lambda k$ and $+\lambda k$ for $u_{-}$ and $u_{+}$, respectively, so orthogonality has selected the other band rather than merely another arbitrary spin basis.

For a normalized spinor $(\alpha ,\beta {)}^T$, the Pauli expectation values are $(2\mathrm{Re}({\alpha }^{\ast }\beta ),2\mathrm{Im}({\alpha }^{\ast }\beta ),|\alpha {|}^2-|\beta {|}^2)$. Consequently the Rashba spin texture isThe spin lies in the plane and is perpendicular to momentum. On the outer circle $k_1$, which belongs to $E_{-}$, it points clockwise as viewed from $+z$; at $\mathbf{k}=(k_1,0)$ it points along $-y$. On the inner circle $k_2$, belonging to $E_{+}$, it points counterclockwise; at $\mathbf{k}=(k_2,0)$ it points along $+y$. Opposite momenta have opposite spins, consistent with time-reversal symmetry. At the origin the eigenstates are degenerate and no unique spin orientation is defined.

For $H_R=\lambda (k_x{\sigma }_y-k_y{\sigma }_x)$ with $\lambda >0$, the lower- and upper-band spinors of unit norm can be chosen asTheir spin expectation values are $⟨\mathbf{S}{⟩}_{\pm }=\pm (\hslash /2)(-k_y,k_x,0)/k$. They are tangential to circles of constant momentum magnitude, opposite between branches, and reversed at opposite momenta as required by time-reversal symmetry. When $E_F>0$, the outer contour belongs to the lower branch and the inner one to the upper branch. For $E_F<0$ the two Fermi contours both belong to the lower branch and have the same tangential spin orientation at a given angle in polar coordinates. At $k=0$ the Hamiltonian has quantum degeneracy in spin, and no unique spin texture direction is selected.

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.