In the language of second quantization, we describe fermions in terms of the annhilation \(c\) and creation operators \(c^\dagger\). Consider a system of \(N\) fermionic modes. Then, the anti-commutation relations are as follows:
\[\{ c_i,c_j^\dagger \} = \delta_{ij} , \{ c_i^\dagger,c_j^\dagger \} = 0, \{ c_i,c_j \} = 0\]The corresponding Majorana operators are defined as,
\[\gamma_i := (a_i + a_i^\dagger) , \bar{\gamma}_i := -\iota (a_i - a_i^\dagger)\]These objects are Hermitian, \(\gamma = \gamma^\dagger\) and \(\bar{\gamma} = \bar{\gamma}^\dagger\), and they satisfy the following anti-commutation relations,
\[\{ \gamma_i,\gamma_j \} = 2\delta_{ij}\]Another interesting object to define in terms of the Majorana operators are edge and vertex operators [1]. For each \(i,j \in [N]\),
\[E_{ij} := -\iota \gamma_i\gamma_j, V_j:= -\iota \gamma_j \bar{\gamma}_j\]These objects are Hermitian, traceless, and self-inverse. They obey a nice graph-theoretic notion of (anti-)commutativity. That is, if a vertex is shared among these objects, they anticommute. If they are disjoint, they commute.
\[\{ E_{jk},V_j \} = 0, \{ E_{ij},E_{jk} \} = 0\]and if \((i,j,k,l)\) are all distinct, then
\[[V_i,V_j] = 0, [E_{ij},V_k] = 0, [E_{ij},E_{kl}] = 0.\]The number operator for mode \(i\) is defined as \(n_i = c_i^\dagger c_i\), and the total number operator is \(n = \sum_i n_i\). With this, we define the parity operator as follows,
\[\mathcal{P} := (-1)^n.\]We say that a state is in a definite parity state if it is an eigenstate of the parity operator. Note that \(\mathcal{P}\) only has \(\pm 1\) eigenvalues. The parity superselection rule tells us the following: nature does not allow physical states to be in coherent superposition of states with different parity. Phrased another way, every physical state has to be in a definite parity sector. Furthermore, physical operations then must respect parity: they must commute with the parity operator. That is why one never sees a term like \(c + c^\dagger\) in a fermionic Hamiltonian, or why one can never measure such an operator. To obtain insight into this “rule,” let us consider the following gedankenexperimente [2,3]. Consider two parties, Alice and Bob. They share the following (non-normalized) state
\[|\psi\rangle = (1 + c_B^\dagger) |\Omega\rangle\]where \(\Omega\) is the vacuum. We see that this state is a superposition of different parities. Now suppose Alice wants to send one bit of classical information to Bob, with \(b \in \{ 0,1 \}\). Define
\[U_0 := I ,U_1 := \iota(c_A^\dagger - c_A)\]Corresponding to the value of the bit \(b\), Alice applies the unitary \(U_b\), and Bob measures the operator \(\mathcal{O}=(1 + c_B + c_B^\dagger)\). Then, we claim that this allows deterministic superluminal communication of the bit \(b\). Note that this is not a teleportation like claim, where we actually have to send some information to tell Bob the correct basis to measure in. The weird thing about this is that, the basis is determined already. We now see that this is true just because of the anticommutation relations, thereby showing (empirically, atleast) parity superselection. The result after applying the unitary is,
\[|\psi_0\rangle = (1 + c_B^\dagger) |\Omega\rangle\] \[|\psi_1\rangle = (1-c_B^\dagger)c_A^\dagger |\Omega\rangle\] \[\mathcal{O}|\psi_0\rangle= (1 + c_B + c_B^\dagger) (1 + c_B^\dagger) |\Omega\rangle =2(|\Omega\rangle + c_B^\dagger|\Omega\rangle) = 2|\psi_0\rangle\] \[\mathcal{O}|\psi_1\rangle = (1 + c_B + c_B^\dagger)(1-c_B^\dagger)c_A^\dagger |\Omega\rangle = c_A^\dagger |\Omega\rangle - c_B^\dagger c_A^\dagger |\Omega\rangle - c_A^\dagger|\Omega\rangle + c_B^\dagger c_A^\dagger |\Omega\rangle = 0\]Thus, both of these states are eigenstates of the operator \(\mathcal{O}\), viz.
\[\mathcal{O}|\psi_b\rangle = E_b|\psi_b\rangle\]and measuring this observable thus perfectly distinguishes them, and Alice manages to transmit the bit perfectly at superluminal speed. We reach a contradiction.