Massless Weyl spinor components

Question

Let's say there is a massless Weyl spinor moving in an arbitrary direction $n$. What are the analytical solutions for the two components of this spinor?

Answer 1

The equation of motion can be written down in the form $$ ( \partial_t \pm \vec \sigma \cdot \vec\nabla ) \psi (t, \vec x)=0,$$ where $\psi$ is two component column vector. Choice of $\pm$ sign corresponds to right or left handed Weyl neutrinos. By squaring this equation we see that $\psi$ satisfies also the wave equation, $$ (\partial_t^2 - \nabla^2) \psi (t, \vec x)= 0 .$$ Any reasonable function can be written down as a superposition of plane waves. If we do this, the wave equation tells simply that we only have components on mass-shell, i.e. plane waves $e^{ikx}$ where $k$ is a null vector. This is just Einstein's energy momentum relation $E^2 = m^2 + p^2$ for $m=0$. Therefore we can write $$ \psi(t, \vec x) = \int \frac{\mathrm d^3 k}{(2 \pi)^3 |\vec k|} \left( \psi^p(\vec k) e^{-i|\vec k|t +i \vec k \cdot \vec x} + \psi^n(\vec k) e^{i|\vec k|t -i \vec k \cdot \vec x} \right), $$ where $\psi^p$ and $\psi^n$ are so far undetermined spinor-valued functions. This integral is the most general solution to the wave equation, but what we wanted to solve in the first place was Weyl equation. Therefore we plug it in and the result is $$ (\sigma \cdot \hat k) \psi^{p/n}(\vec k)=\pm \psi^{p/n} (\vec k). $$ This eigenvalue equation has unique (up to normalization and phase) solution for every $\vec k$. For example for wave vectors in the $z$ direction this is $(1, 0)$ for right-handed spinors (spin in the direction of momentum) and $(0,1)$ for left-handed (spin in the direction opposite to momentum). Denote this solution as $\psi_0(\vec k)$. We now have the most general solution to the Weyl equation $$ \psi(t, \vec x) = \int \frac{\mathrm d^3 k}{(2 \pi)^3 |\vec k|} \psi_0(\vec k) \left( \alpha(\vec k) e^{-i|\vec k|t +i \vec k \cdot \vec x} + \beta(\vec k) e^{i|\vec k|t -i \vec k \cdot \vec x} \right), $$ where $\alpha(\vec k)$ and $\beta(\vec k)$ are now completely arbitrary complex functions.