My question concerns the plane wave solutions to the Majorana equation. First, recall the Dirac equation:

$$(i\gamma^\mu \partial_\mu-m)\psi=0$$ I suggest a solution in the form of a plane wave with $\vec{p}=0$ and $c=\hbar=1$: $$\psi=\psi_0 e^{\pm imt}$$ Plugging this into the Dirac equation, I find the equation

$$\gamma^0 \psi_0=\pm \psi_0$$

This leads to four real-valued terms for $\psi_0$, which correspond to the two spin solutions for positive and negative energies. I tried doing the same exact thing for the Majorana equation, but now I have

$$\widetilde{\gamma}^0\psi_0=\pm \psi_0$$

where $\widetilde{\gamma}^0$ is given in the Majorana basis as

$$\widetilde{\gamma}^0=\begin{pmatrix} 0 & 0 & 0 & -i\\ 0 & 0 & -i & 0 \\ 0 & i & 0 & 0\\ i & 0 & 0 & 0 \end{pmatrix}$$ When I solve the above for $\psi_0$, I then get a complex solution, which doesn't make sense. I thought that, under the Majorana basis, solutions would be of the form $\psi=\psi^*$? Also, a plane wave solution should still work for the Majorana equation, because, like its Dirac counterpart, it is still a factorization of the Klein-Gordon equation. Any explanation of what I'm doing wrong would be greatly appreciated.

  • 2
    $\begingroup$ A Majorana fermion is one in which $\psi^* = B \psi$ for some matrix $B$. $\psi$ itself does not have to be real (and typically is not). $\endgroup$
    – Prahar
    Apr 15, 2016 at 23:09
  • $\begingroup$ @Prahar but wouldn't the fact that the $\gamma$ matrix is purely imaginary ensure a real $\psi$? Also, I would have thought that the positive and negative energy solutions would have been the same in the Majorana basis--however, one can see from the above that is not what one gets when one solves for $\psi_0$. $\endgroup$ Apr 16, 2016 at 0:02
  • $\begingroup$ Can you explain that comment more? Why should a purely imaginary $\gamma$ imply a real $\psi$? $\endgroup$
    – Prahar
    Apr 16, 2016 at 17:47
  • $\begingroup$ @Prahar If $\gamma$ is imaginary, then I thought that the equation $(i\gamma\partial_\mu-m)\psi=0$ would yield a real solution $\psi$. $\endgroup$ Apr 16, 2016 at 17:55

1 Answer 1


In the Majorana representation, the Dirac equation is real and therefore the real and imaginary part of the solution separate. So far, so good. By plugging in the plane wave a complex exponential, the solution stay of course complex. Try $\sin(mt)$ and $\cos(mt)$


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