= Solution
Continue with $m=0$. Along a characteristic $x=x_0+t$, the <chain rule> gives
$$
\frac{d}{dt}\psi_R(t,x_0+t)=(\partial_t+\partial_x)\psi_R=0.
$$
Thus the initial <wave packet> $\psi_R(0,x)=f(x)$ evolves as $\psi_R(t,x)=f(x-t)$. Similarly, along $x=x_0-t$, the left component is constant, giving
$$
\boxed{\psi_R(t,x)=f(x-t),\qquad \psi_L(t,x)=h(x+t).}
$$
A feature initially at $x_0$ therefore appears at $x_0+t$ for the right component and $x_0-t$ for the left component. The profiles translate without changing shape at the <speed of light>, which is 1 in these units. For positive-<energy> <plane waves>, the right component has $E=p>0$ and the left component has $E=-p>0$, another way to identify their propagation directions.
The conclusion concerns the <massless Dirac equation>. For $m\ne0$, the components couple as in <3/c>, and a positive-<energy> packet has <group velocity> $p/\sqrt{p^2+m^2}$, rather than a universal speed $\pm1$. For example, a massive positive-<energy> rest solution has $p=0$ and $\psi_R=\psi_L$, so neither component alone describes an independently moving light-speed packet.
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