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Continue with $m=0$. Along a characteristic $x=x_0+t$, the chain rule givesThus 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, givingA 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 $m\ne 0$, the components couple as in c, and a positive-energy packet has group velocity $p/\sqrt{p^2+m^2}$, rather than a universal speed $\pm 1$. 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.

A continuous Lorentz transformation in one spatial dimension must preserve $t^2-x^2$. In the component connected to the identity, take the image of the time axis to have velocity $v$, with $|v|<1$, and write $B_v(1,0{)}^{\mathsf{T}}=\gamma (1,v{)}^{\mathsf{T}}$. Preservation of the Minkowski norm gives ${\gamma }^2(1-v^2)=1$, so $\gamma =(1-v^2{)}^{-1/2}$. The spacelike column must be orthogonal to the timelike column and have squared Minkowski norm $-1$; continuity fixes it to $\gamma (v,1{)}^{\mathsf{T}}$. ThusHere $x=(t,x{)}^{\mathsf{T}}$ denotes the coordinate pair. This is the positive off-diagonal convention in the source; describing a passive frame moving at $+v$ instead reverses the sign of $v$.

Define the rapidity $\alpha =\mathrm{artanh}v$. The hyperbolic function identities giveIn light-cone coordinates, $t^{\prime }+x^{\prime }=e^{\alpha }(t+x)$ and $t^{\prime }-x^{\prime }=e^{-\alpha }(t-x)$. Therefore the derivatives transform oppositely:Use the field-transformation convention ${\psi }^{\prime }(x^{\prime })=S(\alpha )\psi (x)$. The component action in c is invariant underIndeed, the right-component kinetic term gains $e^{\alpha /2}{e}^{-\alpha }{e}^{\alpha /2}=1$, and the left-component kinetic term gains $e^{-\alpha /2}{e}^{\alpha }{e}^{-\alpha /2}=1$. The two factors in each Dirac mass term cancel as well. Finally $\det B=1$, so $d{t}^{\prime }d{x}^{\prime }=dtdx$ and the whole action is unchanged. These are the chiral boost weights in two spacetime dimensions.

In spinor notation, $S(\alpha )=\exp (\alpha {\gamma }^5/2)$ satisfies $S^{-1}{\gamma }^{\mu }S=B^{\mu }{}_{\nu }{\gamma }^{\nu }$ and $S^{†}{\gamma }^{0}S={\gamma }^0$, so ${\bar{\psi }}^{\prime }=\bar{\psi }{S}^{-1}$. These identities directly verify covariance of the Dirac equation and the Lorentz invariance of its action. The matrices $S$ are not unitary with respect to the ordinary component norm; the Dirac adjoint supplies the invariant bilinear.

The statement about there being only Lorentz boosts applies to the identity component of the One-dimensional Lorentz group. The full Lorentz group also has discrete transformations: the concrete spatial reflection $\mathrm{diag}(1,-1)$ preserves the Minkowski metric but has determinant $-1$, so it cannot equal any displayed $B(\alpha )$, whose determinant is $+1$.