= Solution
\b[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 are
$$
E_\pm(\mathbf k)=\frac{\hbar^2(k_x^2+k_y^2+k_z^2)}{2m}\pm|\lambda|\sqrt{k_x^2+k_y^2}.
$$
In the $k_z=0$ plane, let $k=\sqrt{k_x^2+k_y^2}$ and $q_R=2m\lambda/\hbar^2>0$. Then
$$
\boxed{E_\pm(k)=\frac{\hbar^2}{2m}(k^2\pm q_Rk),\qquad q_R=\frac{2m\lambda}{\hbar^2}.}
$$
If $\lambda<0$, use $q_R=2m|\lambda|/\hbar^2$ 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.
\b[Band minima.] Define $k_R=q_R/2$ and $E_R=\hbar^2q_R^2/(8m)$. Completing the square gives
$$
E_-(k)=\frac{\hbar^2}{2m}(k-k_R)^2-E_R.
$$
Along $k_y=k_z=0$, the radial coordinate is $|k_x|$, so the lower branch has two minima:
$$
\boxed{k_x=\pm\frac{q_R}{2},\qquad E_{-,\min}=-\frac{\hbar^2q_R^2}{8m}=-E_R.}
$$
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.
\b[Occupied regions and <Fermi contours>.] At zero <temperature> each spin-split <eigenstate> with <energy> below $E_F$ is occupied. Put
$$
D=\sqrt{k_R^2+\frac{2mE_F}{\hbar^2}}.
$$
For $-E_R<E_F<0$, $0<D<k_R$. 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 \b[unoccupied], the <annulus> $k_R-D<k<k_R+D$ is \b[singly occupied], and the exterior is \b[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 \b[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 \b[doubly occupied], with one state in each band at every <momentum>. The <annulus> $k_+<k<k_-$ is \b[singly occupied], in the lower band only, and the exterior is \b[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.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2014-rashba-occupations.png]
{title=Rashba dispersion and occupied momentum regions below, at and above the band crossing}
\b[Inferring the splitting.] Above the crossing, the two positive radii differ by $2k_R=q_R$:
$$
\boxed{q_R=k_1-k_2=(0.11-0.01)\,\mathrm{\mathring A^{-1}}=0.10\,\mathrm{\mathring A^{-1}}=1.0\times10^9\,\mathrm{m^{-1}}.}
$$
The minimum radius is therefore $k_R=0.05\,\mathrm{\mathring A^{-1}}$. 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 $q_R$.
\b[Spinors and <spin> direction.] Write $a=(k_y+ik_x)/k$, which has $|a|=1$ for $k>0$. An orthogonal normalized partner to the supplied lower-band state is
$$
\boxed{u_+(\mathbf k)=\frac1{\sqrt2}\begin{pmatrix}-a\\1\end{pmatrix}=\frac1{\sqrt2}\begin{pmatrix}-(k_y+ik_x)/k\\1\end{pmatrix}.}
$$
Indeed $u_-^\dagger u_+=(-|a|^2+1)/2=0$. 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 $(2\operatorname{Re}(\alpha^*\beta),2\operatorname{Im}(\alpha^*\beta),|\alpha|^2-|\beta|^2)$. Consequently the <Rashba spin texture> is
$$
\boxed{\langle\mathbf S\rangle_- =\frac\hbar2\left(\frac{k_y}{k},-\frac{k_x}{k},0\right),\qquad \langle\mathbf S\rangle_+ =\frac\hbar2\left(-\frac{k_y}{k},\frac{k_x}{k},0\right).}
$$
The <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.
\Image[/past-exam-of-the-quantum-condensed-matter-course-of-the-university-of-cambridge/2014-rashba-spin-texture.png]
{title=Opposite tangential spin textures on the outer lower-band and inner upper-band Fermi circles}
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