How to prove that Weyl spinors equations are Lorentz invariant? The Dirac equation is given by:
$[iγ^μ ∂_μ − m] ψ(x) = 0$ . 
We can prove that it's Lorentz invariant when: 
$ψ(x) \to S^{-1} \psi'(x')$ and $\partial_\mu \to \Lambda^\nu_\mu \partial'_\nu$, where 
$S(\Lambda) = 1 -\frac{i}{4} \sigma_{\mu\nu}\epsilon^{\mu\nu}$
which applying that for Dirac equation covariance that $\Lambda^\nu_\mu S \gamma^\mu S^{-1}=\gamma^\nu$ 
Now the Weyl spinors equations are given by:
$i \bar{\sigma}.\partial\psi_L=0,~~~~~~~(1)$
$i \sigma.\partial\psi_R=0, ~~~~~~~~~~(2)$
where $\sigma^\mu\equiv (1,\underline{\sigma})$ and $\bar{\sigma}^\mu\equiv (1,-\underline{\sigma})$
Look for instance Peskin's book "An Introduction To Quantum Field Theory"
where the Lorentz transformation of $\psi_L$ and $\psi_R$ are given by:
$\psi_L \to (1-i\theta.\frac{\sigma}{2}- \beta.\frac{\sigma}{2}) \psi_L$,
$\psi_R \to (1-i\theta.\frac{\sigma}{2}+ \beta.\frac{\sigma}{2}) \psi_R$,
How one can show that (1) and (2) are Lorentz invariant ? 
 A: The key thing is you are contracting $\partial_\mu$ into either $\sigma^\mu,\bar{\sigma}^\mu$. Consider contracting a general vector $p^\mu$
$$\sigma^\mu p_\mu = \left(\begin{array}{cc}
   p^0-p^3& -p^1+ip^2\\
   -p^1-ip^2 &p^0+p^3
   \end{array}\right)$$
Note that the determinant is just the norm of the vector: $(p^0)^2-(p^1)^2-(p^2)^2-(p^3)^2$.
If you sandwich this matrix between some $U$ and $U^{-1}$ the determinant is unchanged
$$\det\left(U\sigma^\mu p_\mu U^{-1}\right)=\det\left(\sigma^\mu p_\mu \right)$$
so it corresponds to some Lorentz transformation. 
So we can write
$$\sigma^\mu (\Lambda^{\mu'}_\mu p_{\mu'})=U(\Lambda)\sigma^\mu p_\mu U(\Lambda)^{-1}$$
for some 2 by 2 matrix $U(\Lambda)$.
It turns out that $U(\Lambda)$ is the very same matrix you are using to transform your right spinor, so that is why the Weyl equations are invariant.
For instance as an exercise, take a finite rotation about the z-axis: $$I-i\frac{\theta}{2}\sigma^3+\dots=\exp(-i\frac{\theta}{2}\sigma^3)=\left(\begin{array}{cc}
   \exp(-i\frac{\theta}{2})& 0\\
   0 &\exp(+i\frac{\theta}{2})
   \end{array}\right)$$
You can show that
$$\exp(-i\frac{\theta}{2}\sigma^3)\,\sigma^\mu p_\mu\,\exp(+i\frac{\theta}{2}\sigma^3)$$
does transform the $p^1,p^2$ components as a rotation by angle $\theta$.
