Question about Majorana fermion and Majorana representation 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 $\psi^c=e^{i\phi}\psi$


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

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

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*S. Willenbrock, Symmetries of the Standard Model, arXiv:hep-ph/0410370.

