We know that we can describe a spin $1/2$ massless particle using only a single Weyl field (lets say left-handed $\psi_{L}$). To introduce a mass term we have to use two spinor fields (one left-handed and one right-handed) and this gives the Dirac mass term.
The question is now that if we can describe a massive particle with a single Weyl field. Well yes, due to the fact that given a left-handed Weyl spinor, it is possible to construct a right-handed spinor $\psi_{R}=i\sigma^{2}\psi_{L}^{*}$. Thus, we can write the Dirac equation using $i\sigma^{2}\psi_{L}^{*}$

$$\hspace{43mm} \bar{\sigma}^{\mu}i\partial_{\mu}\psi_{L}=im\sigma^{2}\psi_{L}^{*} \hspace{30mm}(1)$$

The known algebraic methods performed for the Dirac equation to prove that it implies a massive Klein-Gordon equation can be performed here without any problems. Thus, the above equation implies $(\Box+m^{2})\psi_{L}=0$. Here we have constructed a mass term using only $\psi_{L}$ and this is known as Majorana mass. The similarity with the Dirac mass can be seen by writting $(1)$ in terms of the four component Majorana spinor $\psi_{m}$ in the chiral representation

$$\psi_{m}=\begin{pmatrix} \psi_{L}\\i\sigma^{2}\psi_{L}^{*} \end{pmatrix}$$

Now, equation $(1)$ becomes

$$(i\gamma_{\mu}\partial^{\mu}-m)\psi_{m}=0$$

The Majorana mass has a very important physical difference when compared to the Dirac mass. We know that the Dirac action with a mass term is invariant under a global $U(1)$ transformations of $\psi_{L}$ and $\psi_{R}$ (i.e. $\psi_{L}\rightarrow e^{i\alpha}\psi_{L},\hspace{2mm}\psi_{R}\rightarrow e^{i\alpha}\psi_{R}$).But for Majorana spinors, $\psi_{L}$ and $\psi_{R}$ are not independent, they are related by complex conjugation. So, if $\psi_{L}$ transforms as $\psi_{L}\rightarrow e^{i\alpha}\psi_{L}$then $\psi_{R}$ transforms like $\psi_{R}\rightarrow e^{-i\alpha}\psi_{R}$. The Majorana equation $(1)$ is not invariant under global $U(1)$ symmetries. This fact implies that a spin $1/2$ particle with a $U(1)$ conserved charge cannot have a Majorana mass. All spin $1/2$ particles with an electric charge cannot have a Majorana mass. Also leptons that have a Majorana mass violate the lepton number (because this is a $U(1)$ symmetry).

One possible particle that could have a Majorana mass is the neutrino. But this is yet to be determined. (I didn't answer your questions point by point but I hope this clarifies some of them).

The transformation rule for $\hat{C}$ or that of $\hat{C}\gamma_{0}^{T}$ is different from the usual one.

The whole charge conjugation operation is given by $VK$, where $V\equiv\hat{C}\gamma_{0}^{T}$ is unitary and $K$ is complex conjugation. (i.e., $VK$ is antiunitary.) Then, under a basis transformation,
$$
VK \ \rightarrow \ U^{\dagger}VK U = U^{\dagger}VU^{\ast} K.
$$
Hence $V$ transforms as $V \ \rightarrow \ U^{\dagger}VU^{\ast}$. My guess is that this is the piece that was missing in your derivation.

## Best Answer

Majorana fermions as defined in http://en.wikipedia.org/wiki/Majorana_fermion are really fermions, as its name indicates. So Majorana fermion really have Fermi statistics. It is not proper to say Majorana fermions obey non-abelian statistics, since fermion always obey Fermi statistics by definition.

The thing that people said to have non-Abelian statistics are defects (such as vortices's) that carry a zero-energy mode. Such a thing is not Majorana fermions. Calling a zero mode as a Majorana fermion is really confusing.