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Adding a spatially constant Hermitian matrix $V=dI+\mathbf{b}\cdot \mathbf{\sigma }$ to an isotropic Weyl Hamiltonian preserves each momentum sector. By the Hermitian two-by-two matrix in the Pauli basis decomposition, the exact energies are $d\pm |\hslash v\mathbf{k}+\mathbf{b}|$. The touching point moves to ${\mathbf{k}}_{\ast }=-\mathbf{b}/(\hslash v)$ at energy $d$, without opening a gap in three-dimensional momentum space.

For a fixed $k_z=0$ slice, however,This slice has minimum branch separation $2|b_z|$ if $b_z\ne 0$, because it misses the shifted node. A one-dimensional cut can likewise look gapped while a crossing exists elsewhere. No nonzero two-by-two matrix anticommutes with all three Pauli matrices, which is another way to see why an independent constant mass matrix is unavailable for this isolated three-dimensional two-component model.

Take a normalized spin one-half state in the $S_z$ eigenbasis. Since $\mathbf{S}=(\hslash /2)\mathbf{\sigma }$, direct multiplication of the Pauli matrices givesSubstitution of the half-angle parametrization givesThus $\theta$ and $\varphi$ 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 $|⟨\mathbf{S}⟩|=\hslash /2$ should not be confused with $\sqrt{⟨S^2⟩}=\sqrt{3}\hslash /2$.

For a spatial plane wave, $-i\hslash \nabla$ becomes $\hslash \mathbf{k}$, so the Hamiltonian operator reduces to the Weyl Hamiltonian matrixDefine $\mathbf{n}=(\sin \theta \cos \varphi ,\sin \theta \sin \varphi ,\cos \theta )$ and ${\chi }_{+}=(\cos (\theta /2),e^{i\varphi }\sin (\theta /2){)}^{\mathsf{T}}$. The matrixsatisfies $(\mathbf{\sigma }\cdot \mathbf{n}){\chi }_{+}={\chi }_{+}$ 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 $+\hslash vk$ and $-\hslash vk$, for $k\ge 0$. Equivalently, at each fixed nonzero wavevector the two branches areIf $v<0$, the two energy labels exchange; the ordered eigenvalues are $\pm \hslash |v||\mathbf{k}|$. Taking $v>0$ sets the convention in the sketches. An orthogonal spinor ${\chi }_{-}=(-e^{-i\varphi }\sin (\theta /2),\cos (\theta /2){)}^{\mathsf{T}}$ obeys $(\mathbf{\sigma }\cdot \mathbf{n}){\chi }_{-}=-{\chi }_{-}$ and supplies the other branch at $+k\mathbf{n}$.

The Pauli matrix multiplication law independently gives $H(\mathbf{k}{)}^2=(\hslash v{)}^2|\mathbf{k}{|}^{2}I$. 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 $\pm \hslash v|k_x|$. 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.

Now let a spatially constant perturbation have matrix $V=\left(\begin{array}{cc}u& w\\ w^{\ast }& z\end{array}\right)$, with $u,z$ real. The Hermitian two-by-two matrix in the Pauli basis decomposition follows explicitly fromwhere $d=(u+z)/2$, $c=(u-z)/2$, $a=\mathrm{Re}w$, and $b=-\mathrm{Im}w$ are real. Equivalently, $d=\mathrm{tr}V/2$ and $(a,b,c{)}_i=\mathrm{tr}(V{\sigma }_i)/2$. 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 areThis constant matrix perturbation of a Weyl Hamiltonian translates the cone toIn 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 $\hslash v\mathbf{k}+\mathbf{b}$, 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 $d\pm \sqrt{(\hslash v{k}_x+a{)}^2+(\hslash v{k}_y+b{)}^2+c^2}$, with minimum branch separation $2|c|$. This plane misses the crossing if $c\ne 0$. Along any line through ${\mathbf{k}}_{\ast }$ 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.

Spin-momentum locking means that a state's spin direction is constrained by its momentum and its energy branch. In the isotropic Weyl Hamiltonian with $v>0$, the upper and lower branches have mean spin parallel and antiparallel to the momentum. Their textures in an equatorial momentum plane are radial outwards and inwards. Other Hamiltonians can lock spin transverse to momentum, so radial locking is an example rather than the general definition. In an effective model one must check whether its Pauli matrices represent physical spin or pseudospin before interpreting the texture as a measured spin.

The isotropic Weyl Hamiltonian acts on a two-component spinor as $H=v\mathbf{\sigma }\cdot \mathbf{p}$. For a plane wave of momentum $\hslash \mathbf{k}$, its two energy eigenvalues are $E_{\pm }=\pm \hslash |v||\mathbf{k}|$. For $v>0$ the corresponding Bloch vectors point along and opposite $\mathbf{k}$, respectively. If the Pauli matrices describe physical spin one-half, the mean spin is $\pm (\hslash /2)\hat{\mathbf{k}}$; in an effective two-band model the same algebra can describe an internal pseudospin instead. The bands meet linearly at the origin. ^[ref]