If you write the Majorana spinors as

$$\chi = \begin{pmatrix}\psi_L\\ i\sigma_2\psi_L^* \end{pmatrix} \tag1$$

It satisfies the Dirac equation that leads you to the Majorana equation $$i\bar{\sigma}^\mu\partial_\mu\psi_L = im\sigma_2\psi_L.$$

But if $\chi$ satisfies Dirac, is Dirac's Lagrangian the Lagrangian for $\chi$? My questions arises from one of my QFT classes where the professor said that Majorana fields like Eq. (1) don't have $U(1)$ symmetry. Nevertheless, if it is satisfying Dirac's equation, its Lagrangian should be Dirac's and therefore, it should have $U(1)$ symmetry.

Moreover, how do you know that Majorana's are self-conjugated? Do you impose it or it's a result from the Lagrangian or Eq. (1) or somewhere? I've been trying to understand this but I'm really stuck.


Perhaps a simpler example will help show the issue. Consider the complex scalar field Lagrangian, $$\mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - m^2 \phi^* \phi.$$ This Lagrangian has a $U(1)$ symmetry by phase rotations.

Now consider a real field $\varphi$. We already know how to deal with them, but out of laziness, we could choose to write the real field as a complex field $\phi$ that happens to be its own conjugate, $\phi^* = \phi$. This is useful because we can just use the very same complex scalar field Lagrangian, $$\mathcal{L} = (\partial_\mu \phi^*)(\partial^\mu \phi) - m^2 \phi^* \phi.$$ However, this Lagrangian does not have a $U(1)$ symmetry like the original one, even though it looks the same, because $\phi$ is actually real. You simply can't rotate its phase in the first place.

The same logic holds for the Majorana spinor field. We start with a Weyl spinor field $\psi_L$. If we only know about the Dirac Lagrangian, then we don't know how to write down a Lagrangian for this Weyl spinor alone. So we choose to write it as a full Dirac spinor $\chi$ which is constrained to be its own conjugate, and just use the Dirac Lagrangian for $\chi$. However, this Lagrangian does not have a $U(1)$ symmetry like the original one, even though it looks the same, because $\chi$ is self-conjugate. You simply can't rotate its phase in the first place.

  • $\begingroup$ I've made the explicit computation of the Lagrangian for $\chi$ , ${\cal L} = \bar{\chi}(i\gamma^\mu \partial_\mu - m)\chi$, and I get terms that goes with $\psi_L \psi_L, \psi_L^* \psi_L^*, \psi_L\psi_L^*$. The first 2 ones make the $U(1)$ symmetry impossible, true? Actually it can be seen that a rotation of $\chi$ would imply that $\psi_L$ transforms with the rotation and $\psi_L^*$ with the inverse of the rotation, which makes no sense. Is this what you want me to explain? $\endgroup$ – Vicky Nov 27 '18 at 12:08
  • $\begingroup$ And what about the self-conjugate thing? Is it something that we impose or do we deduce it somehow? $\endgroup$ – Vicky Nov 27 '18 at 12:10
  • $\begingroup$ @Vicky It depends on how you phrase it, which is why it's confusing. I think the most straightforward thing is to say, we want a theory with just a left-handed Weyl spinor. Given a field, you can always conjugate it to get another right, which is a right-handed Weyl spinor. If you stack these two objects together, you will get a 4-component object, which happens to be a Dirac spinor that is self-conjugate. $\endgroup$ – knzhou Nov 27 '18 at 13:06
  • $\begingroup$ @Vicky So the fundamental thing is that you start with one massive Weyl spinor. It just so happens that its degrees of freedom can be packaged into a self-conjugate Dirac spinor. However, other people never mention Weyl spinors and instead say a Majorana spinor is a Dirac spinor which is self-conjugate, which is kind of going in the opposite direction. At the end of the day the theory is the same anyway, so it doesn't really matter. $\endgroup$ – knzhou Nov 27 '18 at 13:07

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