Are the two definitions of a Majorana fermion equivalent? There seems to be two definitions of a Majorana fermion and I can't work out how to show they are equivalent.


*

*The first definition, which is on Wikipedia, is that a Majorana fermion is described by an operator $\gamma$ such that $\gamma^\dagger = \gamma$. This seems to be the condensed matter way to define a Majorana.

*The second definition is that a Majorana fermion is a fermion for which $\psi^c = \psi$, where $\psi^c = C \psi^*$ and $C$ is the charge conjugation operator. This seems to be the high energy definition.
Are these definitions equivalent? If so, I would greatly appreciate a hint to help me show it for myself.
 A: No, they are not equivalent, and in fact the CM version is not actually a fermion at all, which makes the name extraordinarily confusing.
The HE definition concerns relativistic fields $\Psi$, which are genuine fermionic fields. As with any relativistic field, the solutions to the noninteracting Dirac equation can be expressed as linear combinations of both annihilation operators $b$ and creation operators $d^\dagger$ satisfying the usual canonical anticommutation relations $\{b_s, b_{s'}^\dagger\} = \{d_s, d_{s'}^\dagger\} \propto \delta_{ss'}$ with all other anticommutators zero. In the case of a Majorana fermion field, the "own-antiparticle property" $\Psi = \Psi^C$ holds at the level of the complete relativistic field, and requires that $d = b$; i.e. that the "particle part" and the "antiparticle part" of the field correspond to the same particle ladder operator.
In the CM version, there's no requirement for Lorentz invariance, so the concept of antiparticle is not always well-defined. We often consider the ladder operators to be the more fundamental objects, and work with (non-Lorentz invariant) fields that only have a creation operator without an accompanying annihilator operation. In this case, the (roughly speaking) "own-antiparticle property" applies at the level of the ladder operators: $\gamma = \gamma^\dagger$. This is not the case in the relativistic version, where we have the very different requirement that $b = d$. The canonical anticommutation relation $\{\gamma, \gamma^\dagger\} = \{\gamma, \gamma\} = 2\gamma^2 = 2(\gamma^\dagger)^2 \propto 1$ continues to hold, but the canonical anticommutation relation $\{\gamma, \gamma\} = 0$ does not hold. So there's no Pauli exclusion principle - you can indeed put two CM Majorana "fermions" into the same mode, but they just cancel each other out and are equivalent to the vacuum. This violation of the fermionic anticommutation relations is what gives them their anyonic statistics, and explains why they're not actually fermions (at least under the usual definition).
(Note that there's a culture clash between the communities as to which commutations relations are the "fundamental" ones. In HE, the relativistic-field commutation relations are postulated to be the fundamental ones, and the ladder-operator commutation relations are just derived results that only make sense in the context of free fields. In CM, the ladder-operator commutation relations are considered to be more fundamental.)
A: No they are not equivalent. One is a property of a single mode, one is the property of a  particle  being its own antiparticle. For 2d relativistic space-time  one can expand the field for a  chiral  right-going  particle  as 
$$
\psi(x,t)= \sum \hat a_k e^{ik(x-ct)},
$$
and the condition of it being the field of a Majorana particle will be $\hat a_k= \hat a_{-k}^\dagger$. There will be a Majorana mode in the CM sense if $k=0$ is an allowed value of the momentum so that $\hat a_0= \hat a_0^\dagger$.
The terminology is not deliberately confusing, but the two things are often conflated in news releases. 
