Explicit demonstration of the relativistic invariance of the Weyl equation It can be demonstrated explicitly that the Dirac equation is relativistically invariant. This is a proof (borrowed from Peskin & Schroder, see the unnumbered equation after the eqn. 3.31): https://dl.dropboxusercontent.com/u/6602265/dirac_inv.png
Now, two questions.
Question 1. Can you please show, in a similarly direct manner, that the Weyl equation is relativistically invariant? 
Starting from
$$
p^\mu \sigma_\mu \psi = 0
$$
and changing $\psi$ to:
$$
\psi (x) \to T \psi(\Lambda^{-1} x) \quad\mbox{where}\quad T = T(\Lambda) \in SL(2,C)
$$
we should arrive at the same equation.
Question 2.
Please show why the proof will fall apart if you introduce, by hand, a mass into the Weyl equation -- i.e., if we try to write the equation as
$$
(p^\mu \sigma_\mu - m) \psi = 0
$$
Why can we introduce a mass in the equation of Dirac but not in that of Weyl? Stated differently, what is the special property which the $\gamma$ matrices have but the $\sigma$ matrices lack, and which turns out to be critical here (so we cannot carry out for Weyl what Peskin & Schroeder did for Dirac) ?
My guess is that the combination $p^\mu \sigma_\mu$ should transform as a scalar, up to a multiplier, - then, indeed, the equation without a mass term will be Lorentz-invariant, while the one with massive term will not. Is my guess right?
 A: Well I can show you that the Weyl equations are relativistic invariant, at least. It relies on an identity I haven't found in any standard QFT books,  but it's easy to show.
First of all take our gamma matricies in the chiral representation 
$\gamma^\mu=
\left(\begin{array}{cc}
0 & \sigma^\mu \\
\bar{\sigma}^\mu & 0
\end{array}\right) $.
Under Lorentz transformations a Dirac spinor 
$\Psi(x)=\left(\begin{array}{c}
\psi_L  \\
\psi_R 
\end{array}\right)$ 
transforms to $\Psi'(x')=S\Psi(x)$. Here $\psi_L$ and $\psi_R$ are left- and right-handed Weyl spinors, respectively.
We can write $S$ as 
$S=
\left(\begin{array}{cc}
L & 0 \\
0 & R
\end{array}\right) $ such that 
$\psi_L\rightarrow L\psi_L$ and $\psi_R\rightarrow R\psi_R$. 
We also have that $S^{-1}\gamma^\mu S=\Lambda^\mu_{\text{ }\nu}\gamma^\nu$. We can then easily show that this is true if and only if the following hold
$L^{-1}\sigma^\mu R=\Lambda^\mu_{\text{ }\nu}\sigma^\nu$, 
$R^{-1}\bar{\sigma}^\mu L=\Lambda^\mu_{\text{ }\nu}\bar{\sigma}^\nu$. 
We can now show that the Weyl equations are invariant. I;ll do one of them. 
We have $i\sigma^\mu\partial_\mu\psi_R=0$. The Lorentz transformed version is 
$i\sigma^\mu\partial'_\mu\psi'_R=i\sigma^\mu (\Lambda^{-1})^\rho_{\text{ }\mu}\partial_\rho R\psi_R=iL(L^{-1}\sigma^\mu R)(\Lambda^{-1})^\rho_{\text{ }\mu}\partial_\rho \psi_R$
which gives us, after our identity, 
$iL\sigma^\nu \Lambda^\mu_{\text{ }\nu}(\Lambda^{-1})^\rho_{\text{ }\mu}\partial_\rho \psi_R$.
Then $\Lambda^\mu_{\text{ }\nu}(\Lambda^{-1})^\rho_{\text{ }\mu}=\delta^\rho_\nu$ giving us
$L(i\sigma^\nu \partial_\nu \psi_R)=0$.
The Weyl equations are derived from the massless Dirac equation, which can only happen since the left- and right-handed spinors decouple from each other. So adding a mass term to the Weyl equations in the first place doesn't really make sense since by definition they describe massless spin-1/2 particles.
But the proof does indeed fall apart with a mass term. If we (somehow) had 
$(i\sigma^\mu\partial_\mu-m)\psi_R=0$ 
the Lorentz transformed version of this would be 
$(iL\sigma^\mu\partial_\mu-mR)\psi_R$
which is obviously not equal to the original form.
