Past exam of the symmetry and relativity course of the University of Oxford / 2019 / 4 / e

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.