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?

  • $\begingroup$ Just wondering: what book? $\endgroup$
    – d_b
    Mar 11, 2020 at 16:19
  • $\begingroup$ Actually on closer inspection they are course notes that want to be a book: www-thphys.physics.ox.ac.uk/people/SteveSimon/topological2019/… $\endgroup$
    – Sam Jaques
    Mar 12, 2020 at 9:55
  • $\begingroup$ Ah very nice. I didn't know Steve was writing such a book but it looks excellent. $\endgroup$
    – d_b
    Mar 12, 2020 at 18:30

1 Answer 1


The point with the Majorana fermions is exactly that they cannot be described with standard creation or annihilation operators where the state is either "occupied" or "empty". The fact that $\gamma^2 = 1$ means that we can act with the Majorana operator $\gamma$ as many times as you like on the state, without ever getting zero (such as with normal fermionic annihilation operator, for example, where $d^2=0$).

However, from each pair of Majorana fermions one can create a single fermionic operator (and vice versa). If one has $\gamma_1$ and $\gamma_2$ which maintain $\{\gamma_i,\gamma_j\} = 2\delta_{i,j}$ it is straight-forward to see that $$ d=\frac{\gamma_1+i \gamma_2}{2}, \; \; d^{\dagger}=\frac{\gamma_1-i \gamma_2}{2}$$ maintain standard fermionic anti-commutation relations. Now we can discuss the space using the familiar fermionic Fock space, where we can arbitrarily group into pairs the Majorana fermions (different pairings will just correspond to a change of basis) and if we start with $2N$ Majorana fermions we are left with $N$ normal fermionic states, each one of the can be occupied or unoccupied, leaving us with a degeneracy of $2^N$.

This can also give us an intuition into what acting with the Majorana operator do. By reversing the relations, we can have $\gamma_1 = d+d^{\dagger}$ (and this is sometimes just called the 'real' part of the fermion) and similarly $\gamma_2 = -i(d-d^{\dagger})$ (which is called the 'imaginary' part of the fermion). So acting with the Majorana changes the number of fermions in the system, but in a superposition of increasing the number and decreasing it. The quantity that definitely changes is the parity of the state - $P = (-)^N$ where $N$ the number of particles. So $\gamma$ is an eigenoperator of $P$ with eigenvalue $-1$.

This is why, in condensed matter at least, Majorana fermions are intimately linked with superconductivity. The ground-state of a superconductor doesn't have a definite number of particles, as it is a Bose-Einstein condensate of Cooper pairs of electrons. But the parity is often well-defined. If we can create a degeneracy in the ground state between the even and odd states, we can create a pair of Majorana fermions.

  • $\begingroup$ This is helpful but I still feel fundamentally confused. It seems like $\gamma_1=d+d^\dagger$ and $\gamma_2=-i(d-d^\dagger)$ should be the definition of a Majorana fermion, since it implies the space that $\gamma_i$ must act on, and the action it will have. Trying to define it without an underlying fermion (or a space on which it can act like a fermion) seems like you would need to do a lot more work to define it. $\endgroup$
    – Sam Jaques
    Mar 12, 2020 at 9:53
  • $\begingroup$ you can think of the fermionic operators $d_n = (\gamma_{2n-1}+i\gamma_{2n})/2$ as a way to define the space in which the Majorana live. It is just a formality, and we will not necessarily be interested in these operators specifically. Anyhow, any different choice of pairing will lead to the same space (described in a different basis). As with any quantum system, once you define the set of operators that span the space you essentially described the space. It doesn't matter if you have $2N$ Majorana operators or $N$ fermionic operators, they are exactly mapped one to the other $\endgroup$
    – user245141
    Mar 12, 2020 at 10:52
  • $\begingroup$ Was the definition given enough to define the operator, though? I suppose the definition defines an algebra, and that gives a Hilbert space it can act on, but will that be sufficiently unique? $\endgroup$
    – Sam Jaques
    Mar 13, 2020 at 11:39

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