Rashba effect ($H_R=\lambda (k_x{\sigma }_y-k_y{\sigma }_x)$)

The Rashba effect, named for Emmanuel Rashba, is a momentum-dependent spin-orbit interaction associated with broken spatial inversion symmetry. In an isotropic two-dimensional model, choose $\lambda >0$ and write the Hamiltonian $H={\hslash }^2{k}^{2}I/(2m)+\lambda (\hat{\mathbf{z}}\times \mathbf{k})\cdot \mathbf{\sigma }$, where $I$ is the identity matrix and $\mathbf{\sigma }$ is the vector of Pauli matrices. The eigenvalues of the spin term give $E_{\pm }(k)={\hslash }^2{k}^2/(2m)\pm \lambda k$. Define $k_R=m\lambda /{\hslash }^2$ and $E_R={\hslash }^2{k}_R^2/(2m)$. The lower branch is $E_{-}(k)={\hslash }^2(k-k_R{)}^2/(2m)-E_R$ and has a ring of minima at $k=k_R$. A negative $\lambda$ reverses the spin directions; ordering the two energies still uses $|\lambda |$. The alternative parameter $q_R=2k_R$ must not be mistaken for the minimum radius.

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.

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.