# Is there a vacuum for Majorana fermions?

I know this question sounds somewhat odd. But if $$\psi = \psi^\dagger$$ is the Majorana operator, what would $$\psi |0\rangle$$ be? Is it the zero state or a Majorana fermion? This is confusing, because $$\psi$$ is both the annihilation operator and the creation operator.

• For Dirac fermions $\psi(x)$ also contains both creation and annihilation operators and so $\psi(x)|0\rangle$ is a one-particle state – OON Jan 20 '20 at 23:05
• @OON Really? I always thought that for a Dirac fermion, $\psi$ is just the annihilation operator, and $\psi |0\rangle$ is just the zero of the Hilbert space. – Yantao Wu Jan 20 '20 at 23:17

If you mean a relativistic four-component Majorana fermion field $$\Psi$$ (which in the absence of interactions solves the Dirac equation), then it is $$\Psi^\dagger \neq \Psi$$. Instead the correct statement is that $$\Psi^C := \mathcal{C} \overline{\Psi}^T := \mathcal{C} (\Psi^\dagger \beta)^T = \Psi$$, where $$\mathcal{C}$$ is the charge conjugation matrix. In this case the plane-wave expansion of the relativistic four-component operator $$\Psi$$ contains both fermionic creation and annihilation operators, so $$\Psi|0\rangle$$ is a one-fermion state.
But I assume that you are actually referring to the nonrelativistic Majorana anyon operator which is often denoted by the letter $$\gamma$$. This operator is kind of like a fermionic ladder operator, but despite its name, it actually isn't a fermion at all, because it obeys the non-standard anticommutation relation $$\{\gamma, \gamma\} = 2 \neq 0$$. This is the operator that is analogous to the fermionic creation and annihilation operators $$a$$ and $$a^\dagger$$.
In this case, the answer to your question is that $$\psi|0\rangle = |1\rangle$$, which could be called the one-majorana fermion state. However, since $$\psi^2 = I$$, only the parity of the number of Majorana anyons is physically meaningful. So the Majorana state space has the algebraic structure of the monoid $$\mathbb{Z}_2$$ rather than the monoid $$\mathcal{N}$$, and it's probably better to call the two states $$|\text{even}\rangle$$ and $$|\text{odd}\rangle$$ rather than $$|0\rangle$$ and $$|1\rangle$$. The vacuum state $$|\text{even}\rangle$$ represents the complete absence of Majorana anyons, but it also represents any state with an even number of Majorana anyons. In this case, the Majorana anyon operator $$\psi$$ is best interpreted as the "switching" operator that exchanges the states $$|\text{even}\rangle$$ and $$|\text{odd}\rangle$$, rather than as a raising or lowering operator.