= Fermi contours of an isotropic Rashba band
{c}
For the <Rashba effect> with $k_R>0$, put $D=\sqrt{k_R^2+2mE_F/\hbar^2}$. 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.
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