In Chiral representation, a Majorana spinor looks like:

$$\psi=\begin{pmatrix} \psi_L\\ -i\sigma^2\psi_L^*\end{pmatrix}$$

In this representation the Right handed field is the charge-conjugate of the left handed field. i.e., $(\psi_R)^c=\psi_L$, where $$\psi_R=\begin{pmatrix} 0\\ -i\sigma^2\psi_L^*\end{pmatrix}$$

and also $\psi^c=e^{i\phi}\psi$

  1. How does it look like in Majorana Representation, explicitly in the form of a column vector? What is the usefulness of Majorana representation?

  2. Can I use the condition $\psi^c=e^{i\phi}\psi$ to be the definition of a Majorana fermion?


2 Answers 2


What is the usefulness of the Majorana representation?

Majorana spinors are used frequently supersymmetric theories. In the Wess-Zumino model - the simplest SUSY model - a supermultiplet is constructed from a complex scalar, auxiliary pseudo-scalar field, and Majorana spinor precisely because it has two degrees of freedom unlike a Dirac spinor. The action of the theory is simply,

$$S \sim - \int d^4x \left( \frac{1}{2}\partial^\mu \phi^{\ast}\partial_\mu \phi + i \psi^{\dagger}\bar{\sigma}^\mu \partial_\mu \psi + |F|^2 \right)$$

where $F$ is the auxiliary field, whose equations of motion set $F=0$ but is necessary on grounds of consistency due to the degrees of freedom off-shell and on-shell.

Can I use the condition $\psi^{(c)}=\mathrm{e}^{i\phi}\psi$ to be the definition of a Majorana fermion?

Yes, Majorana fermions are fermions whose charge conjugate are equal to the original field; my lecture notes suggest this is the defining property. Upon canonical quantization, one finds that Majorana fermions have real Fourier coefficients/operators in their expansion.


I'm not sure if I understand your first question. Because, as far as I know what you wrote is the Majorana column vector already!! This spinor (this column vector is actually a spinor, since it does not transform as a vector under lorentz tranformations, but as a spinor) is useful to represent particles that are their own antiparticles!

The condition you wrote in question 2 is just the definition of ${\psi}_c$. The Majorana condition would actually be:

${\psi}_c = \psi$, without the operator. So the conjugated spinor is the spinor itself. That can only happen for a chargeless particle. From this definition you can see why it is useful to use this spinor for a particle that is its own anti-particle.

this lectures can probably help a lot in your question.

  • $\begingroup$ @user41847- The column vector I have written is the Majorana fermion written in chiral representation and I am asking how will it look in the Majorana representation? Because in Majorana representation $\psi=\psi^*$. I do not think that the condition written in question 2 is the definition of $\psi^c$. $\psi^c$ is defined as $\psi^c=C{\bar\psi}^T$. $\endgroup$
    – SRS
    Commented Mar 10, 2014 at 17:27

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