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The ordering of Weyl components depends on the supplied basis transformation. In this paper use $u_W=U{u}_D=(\xi ,\eta {)}^{\mathsf{T}}$. Applying the printed $U$ to the printed Dirac matrices givesThus $\xi$ is right chiral and $\eta$ left chiral in this ordering. The free massless Weyl equations areThey describe spin-one-half massless fields in the two inequivalent chiral spinor representations of the Lorentz group. Their rotation generators agree, but their boost generators have opposite signs; spatial parity interchanges the two. A positive-energy particle in the $\xi$ field has helicity $+1/2$, and one in the $\eta$ field has helicity $-1/2$. Their antiparticles have the opposite helicities, so a Weyl field is not a field containing only one helicity for both particles and antiparticles.

Neutrinos provide the observed ultrarelativistic approximation to a left-chiral particle and right-helicity antiparticle. However, neutrino oscillation establishes nonzero neutrino mass differences, so physical neutrinos are not all exactly massless Weyl particles. Chiral Weyl fields occur in the Standard Model, but an isolated exactly massless elementary Weyl particle is not established by neutrino observations. Whether neutrino mass eigenstates are Dirac or Majorana remains a separate question.^[ref]

Assume the scalar potential is real, so the Weyl equation has a Hermitian Hamiltonian $H=\mathbf{\sigma }\cdot \mathbf{p}+V$. Multiplying the equation and its adjoint by ${\psi }^{†}$ and $\psi$ respectively givesThus with $\rho ={\psi }^{†}\psi$ and $\mathbf{J}={\psi }^{†}\mathbf{\sigma }\psi$, the lower-index four-current $j_{\mu }=(\rho ,-\mathbf{J})$ obeys $\boxed{{\partial }^{\mu }{j}_{\mu }=0}$.

For $K=\mathbf{\sigma }\cdot \mathbf{p}$,Consequently $K$ is conserved for spatially constant $V$; a spatially uniform time-dependent scalar offset would also commute with it. For general nonuniform $V$ it is not conserved. At nonzero momentum, $(\mathbf{\sigma }\cdot \hat{\mathbf{p}}{)}^2=I$, so the helicity eigenvalues are $\boxed{+1,-1}$. In the chosen constant-potential Weyl Hamiltonian, the positive $E-V$ branch has helicity $+1$; the negative branch has $-1$. An expectation value in a superposition need not itself be an eigenvalue.

For a free massless Weyl spinor, yes: each component obeys the massless Klein-Gordon equation. In $c=\hslash =1$ units, take either chiral Weyl equation, for exampleApply its opposite-sign partner. The Pauli matrix multiplication identity makes the antisymmetric part vanish against the symmetric pair of spatial derivatives, soThus $\boxed{{\partial }_{\mu }{\partial }^{\mu }u=0}$ componentwise. This does not turn the field into a scalar: its components still mix under a Spinor representation of the Lorentz group. The second-order equation is a necessary consequence of the first-order equation, but does not by itself impose the chiral helicity relation.

There is no unconditional massless-wave conclusion from merely being a two-component spinor, or from the null bilinear of part d. For example, $u(t,\mathbf{x})=e^{-i\omega t}(1,0{)}^T$ has a null bilinear at every point, yet $({\partial }_t^2-{\nabla }^2)u=-{\omega }^{2}u\ne 0$ for $\omega \ne 0$. Massive free two-component chiral fields coupled through a mass term instead obey the massive Klein-Gordon equation; interacting fields can have additional terms. The affirmative result therefore uses the free, massless dynamics.

A two-component complex wave function describing the two spin states of a nonrelativistic spin-$1/2$ particle. Under a spatial rotation by angle $\theta$ about $\hat{\mathbf{n}}$ it transforms as $\psi \mapsto e^{-i\theta \hat{\mathbf{n}}\cdot \mathbf{\sigma }/2}\psi$, with Pauli matrices $\mathbf{\sigma }$. A Pauli spinor may describe a massive particle; two components alone do not imply the massless Weyl equation.