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?


1 Answer 1


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


the Lorentz transformed version of this would be


which is obviously not equal to the original form.


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