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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.

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    $\begingroup$ For Dirac fermions $\psi(x)$ also contains both creation and annihilation operators and so $\psi(x)|0\rangle$ is a one-particle state $\endgroup$ – OON Jan 20 '20 at 23:05
  • $\begingroup$ @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. $\endgroup$ – Yantao Wu Jan 20 '20 at 23:17
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The answer depends on which type of "Majorana operator" you are considering.

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.

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