Particular solution of Weyl representation of Dirac equation Okay, so I've been having some trouble studying by myself the solutions of the Dirac equation and I need help. Suppose $\psi$ is a solution of $(i\gamma^{\mu}\partial_\mu-m)\psi = 0$ and let $\psi = \binom{\psi_{L}}{\psi_{R}}$ be defined by:
$$\binom{\psi_{L}}{0} = \frac{1}{2}(I-\gamma^{5})\psi \quad \mbox{and} \quad \binom{0}{\psi_{R}} = \frac{1}{2}(I+\gamma^{5})\psi$$
I've already showed that $\psi_{L}$ and $\psi_{R}$ satisfy the coupled equations:
$$i\bar{\sigma}^{\mu}\partial_{\mu}\psi_{L}-m\psi_{R} = 0\quad \mbox{and}\quad i\sigma^{\mu}\partial_{\mu}\psi_{R}-m\psi_{L}=0 \tag{1}$$
where $\sigma^{\mu}$ are the Pauli matrices and $\bar{\sigma}^{\mu}$ is the $2\times 2$ identity if $\mu = 0$ and $-\sigma^{\mu}$ otherwise. If the particle described by $\psi$ is at rest at some reference frame $\mathcal{O}'$, the above equations become:
$$i\partial_{0}\psi_{L}' = m\psi_{R}'\quad \mbox{and} \quad i\partial_{0}\psi_{R}' = m\psi_{L}'$$
since the spatial derivatives must be zero.
Now consider an inertial frame $\mathcal{O}$ with respect to which $\mathcal{O}'$ and the particle described above are moving with velocity ${\bf{v}} = (0,0,v)$ in the $x^{3}$ direction and let:
$$\Lambda = \begin{pmatrix}
\cosh\theta & 0 & 0 & -\sinh\theta \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
-\sinh\theta & 0 & 0 & \cosh\theta
\end{pmatrix}
$$
which is a Lorentz boost taking $\mathcal{O}$ to $\mathcal{O}'$.
The exercise asks to prove:
(a) To show that $m\cosh\theta = m \gamma = E$ and $m\sinh\theta = mv\gamma = p$.
(b) To use item (a) and the solutions $\psi_{R}'$ and $\psi_{L}'$ above to show that:
$$\psi_{L} = e^{i(-Et+px^{3})}\binom{e^{-\frac{\theta}{2}}}{0} \quad \mbox{and} \quad \psi_{R}= e^{i(-Et+px^{3})}\binom{e^{\frac{\theta}{2}}}{0}$$
is the solution of the coupled equations (\ref{1}).
I have very little background with special/general relativity and since I've been studying these topics by myself, I don't have any help. But I don't know even how to start addressing (a) and (b). Any help is welcome!
 A: Hints:

*

*In the rest frame, the four-momentum is $(m,0,0,0)$. Writing this as a column matrix and multiplying by $\Lambda^{-1}(\theta)=\Lambda(-\theta)$ gives $(E,0,0,p)$ in the boosted frame. This is independent of the Dirac equation.


*For any coordinate transformation $x\to \bar x$, we have $d\bar x^a\bar \partial_a=d x^a\partial_a$. Use this to deduce that tf $\bar x^a=\Lambda^a_b x^b$ for any linear transformation $\Lambda$, then $\bar\partial_a=(\Lambda^{-1})_a^c\partial_c$, where $\Lambda^a_b(\Lambda^{-1})_a^c=\delta_b^c$. This is independent of the Dirac equation.


*Define
$$
 K\equiv \exp\left(\gamma^0\gamma^3\theta/2\right)
 =\cosh(\theta/2)+\gamma^0\gamma^3\sinh(\theta/2),
\tag{1}
$$
and notice that
$$
 K^{-1}\gamma^a\partial_a K = \gamma^a(\Lambda^{-1})_a^b\partial_b,
\tag{2}
$$
with $\Lambda$ defined as in the question. To derive (2), use
\begin{align}
 K^{-1}\left(\gamma^0\partial_0 + \gamma^3\partial_3\right)K
 &=
 \left(\gamma^0\partial_0 + \gamma^3\partial_3\right)K^2
\\
 K^{-1}\left(\gamma^1\partial_1 + \gamma^2\partial_2\right)K
 &=
 \gamma^1\partial_1 + \gamma^2\partial_2
\tag{3}
\end{align}
with
$$
 K^2 = \exp\left(\gamma^0\gamma^3\theta\right)
 =\cosh(\theta)+\gamma^0\gamma^3\sinh(\theta).
\tag{4}
$$


*Use 2 and 3 to see that if we are given one solution of the Dirac equation, then we can construct another solution by multiplying by the matrix $K$ and using $\Lambda$ to transform the spacetime variables as usual.
Don't split it into $\psi_{L/R}$ first, because that will only complicate things. Working with the original Dirac equation is easier. You can split it into $\psi_{L/R}$ after you're done with everything else, because $\gamma^5$ commutes with $K$.
