I'm reading a book on topological quantum theory and one of the exercises says that Majorana fermions $\gamma_j$ are such that $\{\gamma_j,\gamma_i\}=\delta_{ij}$ and that $\gamma_j=\gamma_j^\dagger$, and asks to show that if the Hamiltonian is 0, then the ground state degeneracy of $2N$ Majorana fermions is $2^N$.
I have only a passing familiarity with the notation here, so my interpretation of this is: The $\gamma_j$ is not a Majorana fermion itself (whatever that would mean), but the creation/annihilation operator for one (and it would have to be both since it's self-adjoint). As I understand creation/annihilation operators, they represent the creation/annihilation of a particle in a particular state, i.e., at a particular point in space, or with a particular momentum. Thus, $\gamma_i$ must be a creation operator for the $i$th state that the particle could be in - perhaps different discrete locations.
Since the Hamiltonian is 0, then its ground state is the entire space, so the degeneracy is just the dimension of this space. Thus, the exercise is really asking how many configurations of these particles are possible. It seems like this would depend on the number of different states they could occupy: If there are $k$ possible states, then there should be $\binom{2N}{k}$ states, modulo the symmetries implied by $\{\gamma_i,\gamma_j\}$.
Is any of this correct? If so, then there are natural follow-up questions:
1) How would I determine the number of possible states? It seems like this is completely system dependent, and just saying ``$2N$ Majorana fermions'' isn't enough information to decide this.
Possibly I should conclude that there are $2N$ states, since there are $2N$ Majorana fermions, but then is there any sort of conservation law I should conclude about these states? That is, is the vacuum state still allowed, since it doesn't have $2N$ "particles"?
2) How exactly would the creation operator act? For example, take a state of $\gamma_i |n_1,\dots,n_k\rangle$. If they are really fermionic, then this should be 0 if $n_i=1$, and otherwise $c_i| n_1,\dots, n_{i-1},1,n_{i+1}\dots,n_k\rangle$ for some constant $c_i$. Is that right, and is $c_i$ something determined by the commutation relation, or should it be the same as that for a regular fermion?