= Solution
Take a normalized <spin one-half> state in the $S_z$ <eigenbasis>. Since $\mathbf S=(\hbar/2)\boldsymbol\sigma$, direct multiplication of the <Pauli matrices> gives
$$
\langle S_x\rangle=\frac\hbar2(\alpha^*\beta+\beta^*\alpha),\quad \langle S_y\rangle=\frac\hbar2(-i\alpha^*\beta+i\beta^*\alpha),\quad \langle S_z\rangle=\frac\hbar2(|\alpha|^2-|\beta|^2).
$$
Substitution of the half-angle parametrization gives
$$
\boxed{\langle\mathbf S\rangle=\frac\hbar2(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta).}
$$
Thus $\theta$ and $\phi$ are precisely the polar and azimuthal angles of the mean <spin>. Two complex coefficients contain four real parameters; normalization removes one and an unobservable common <quantum phase> removes another. The resulting pure states form the <Bloch sphere>. One may choose $\alpha$ real and nonnegative away from the patch where $\alpha=0$. At the poles the azimuth is redundant. The length $|\langle\mathbf S\rangle|=\hbar/2$ should not be confused with $\sqrt{\langle S^2\rangle}=\sqrt3\hbar/2$.
For a spatial <plane wave>, $-i\hbar\boldsymbol\nabla$ becomes $\hbar\mathbf k$, so the <Hamiltonian operator> reduces to the <Weyl Hamiltonian> <matrix>
$$
H(\mathbf k)=\hbar v\,\boldsymbol\sigma\cdot\mathbf k.
$$
Define $\mathbf n=(\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta)$ and $\chi_+=(\cos(\theta/2),e^{i\phi}\sin(\theta/2))^{\mathsf T}$. The <matrix>
$$
\boldsymbol\sigma\cdot\mathbf n=\begin{pmatrix}\cos\theta&e^{-i\phi}\sin\theta\\e^{i\phi}\sin\theta&-\cos\theta\end{pmatrix}
$$
satisfies $(\boldsymbol\sigma\cdot\mathbf n)\chi_+=\chi_+$ by the angle-addition identities. Consequently the displayed <spinor> is an <energy eigenstate> for either $\mathbf k=+k\mathbf n$ or $\mathbf k=-k\mathbf n$, with respective energies $+\hbar vk$ and $-\hbar vk$, for $k\geq0$. Equivalently, at each fixed nonzero <wavevector> the two branches are
$$
\boxed{E_\pm(\mathbf k)=\pm\hbar v|\mathbf k|\quad(v>0).}
$$
If $v<0$, the two <energy> labels exchange; the ordered <eigenvalues> are $\pm\hbar|v||\mathbf k|$. Taking $v>0$ sets the convention in the sketches. An <orthogonal> <spinor> $\chi_-=(-e^{-i\phi}\sin(\theta/2),\cos(\theta/2))^{\mathsf T}$ obeys $(\boldsymbol\sigma\cdot\mathbf n)\chi_-=-\chi_-$ and supplies the other branch at $+k\mathbf n$.
The <Pauli matrix multiplication law> independently gives $H(\mathbf k)^2=(\hbar v)^2|\mathbf k|^2I$. The <dispersion relation> is therefore a pair of cones touching at zero <energy> and <momentum>; a signed straight-line cut has upper and lower branches $\pm\hbar v|k_x|$. In the $k_z=0$ plane, $\theta=\pi/2$, and the upper-branch mean <spin> points radially outwards, the lower-branch <spin> radially inwards. This is <spin-momentum locking>, not a tangential <spin> texture. At the touching point the two states are degenerate and the <spin> direction is not selected by the <Hamiltonian operator>.
\Image[/past-exam-of-the-advanced-quantum-physics-course-of-the-university-of-cambridge/2018-weyl-dispersion.png]
{title=Radial spin locking and a shifted Weyl cone, including a plane that misses its node}
Now let a spatially constant perturbation have <matrix> $V=\begin{pmatrix}u&w\\w^*&z\end{pmatrix}$, with $u,z$ real. The <Hermitian two-by-two matrix in the Pauli basis> decomposition follows explicitly from
$$
\begin{pmatrix}u&w\\w^*&z\end{pmatrix}=dI+a\sigma_x+b\sigma_y+c\sigma_z=\begin{pmatrix}d+c&a-ib\\a+ib&d-c\end{pmatrix},
$$
where $d=(u+z)/2$, $c=(u-z)/2$, $a=\operatorname{Re}w$, and $b=-\operatorname{Im}w$ are real. Equivalently, $d=\operatorname{tr}V/2$ and $(a,b,c)_i=\operatorname{tr}(V\sigma_i)/2$. The four <matrices> are <linearly independent> under the <trace> <inner product> and span the four-dimensional <real vector space> of Hermitian two-by-two <matrices>.
Spatial constancy means the perturbation does not mix different <wavevectors>, so each <momentum> sector can still be diagonalized independently. Write $\mathbf b=(a,b,c)$. The full <matrix> and its exact <eigenvalues> are
$$
H'(\mathbf k)=dI+(\hbar v\mathbf k+\mathbf b)\cdot\boldsymbol\sigma,\qquad \boxed{E_\pm(\mathbf k)=d\pm\sqrt{(\hbar vk_x+a)^2+(\hbar vk_y+b)^2+(\hbar vk_z+c)^2}.}
$$
This <constant matrix perturbation of a Weyl Hamiltonian> translates the cone to
$$
\boxed{\mathbf k_*=-\frac{(a,b,c)}{\hbar v},\qquad E_*=d.}
$$
In three-dimensional <momentum space> the <energy bands> still touch; the perturbation produces no full <spectral gap>. The new branch <spin> directions are parallel and antiparallel to $\hbar v\mathbf k+\mathbf b$, so their radial texture is centered on the shifted point.
A restricted cut must be distinguished from the full <dispersion relation>. In the fixed plane $k_z=0$, the energies are $d\pm\sqrt{(\hbar vk_x+a)^2+(\hbar vk_y+b)^2+c^2}$, with minimum branch separation $2|c|$. This plane misses the crossing if $c\ne0$. Along any line through $\mathbf k_*$ the <dispersion relation> still crosses linearly. Thus a sketch of a gapped fixed-plane cut alone would not describe the gapless three-dimensional <Hamiltonian operator>. The cone after a general perturbation is also no longer a function of $|\mathbf k|$ about the old origin alone.
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