Weyl equation

Remarks

Since $(\boldsymbol{\sigma} \cdot \nabla)^2 = \Delta$, one has $(\partial_t - \boldsymbol{\sigma}\cdot\nabla)(\partial_t + \boldsymbol{\sigma}\cdot\nabla) = \partial_t^2 - \Delta$, so every component of a solution of either equation solves the wave equation.

Example: let $\mathbf{p} \in \R^3 \setminus \set{0}$ and let $u \in \C^2$ be an eigenvector of $\boldsymbol{\sigma} \cdot \mathbf{p}$ with eigenvalue $\abs{\mathbf{p}}$. Then $\psi = u\, e^{-i(\abs{\mathbf{p}} t - \mathbf{p}\cdot\mathbf{x})}$ satisfies $\partial_t \psi + \boldsymbol{\sigma}\cdot\nabla\psi = \left(-i\abs{\mathbf{p}} + i\, \boldsymbol{\sigma}\cdot\mathbf{p}\right)\psi = 0$. It is a positive-energy solution of the first equation whose spin points along $\mathbf{p}$.

Non-example: if $u$ is instead the eigenvector of $\boldsymbol{\sigma}\cdot\mathbf{p}$ with eigenvalue $-\abs{\mathbf{p}}$, then $\psi = u\, e^{-i(\abs{\mathbf{p}} t - \mathbf{p}\cdot\mathbf{x})}$ still solves the wave equation, but $\partial_t\psi + \boldsymbol{\sigma}\cdot\nabla\psi = -2i\abs{\mathbf{p}}\psi \ne 0$; it solves the second Weyl equation instead.

Which of the two equations is called right-handed and which left-handed depends on conventions. In the Weyl representation of the gamma matrices, the Dirac equation with mass $m = 0$ decouples into the two Weyl equations for the upper and lower pairs of components.