# Eigenvalues of Majorana fermions

I have a somewhat naive question about Majorana fermions. Typically, two Majorana fermion modes $$\gamma_{i,1}$$ and $$\gamma_{i,2}$$ are defined by writing a single ordinary fermion $$c_i$$ in terms of "real and imaginary parts": $$c_i = \frac{\gamma_{i,1} + i \gamma_{i,2}}{\sqrt{2}}, \quad c^{\dagger}_i = \frac{\gamma_{i,1} - i \gamma_{i,2}}{\sqrt{2}}$$ We therefore obtain $$2N$$ Majorana fermions from $$N$$ fermionic modes. These Majorana fermions satisfy the commutation relations $$\{ \gamma_i, \gamma_j \} = \delta_{ij}$$, but since they are Hermitian the naive occupation number one might write down is trivial: $$\gamma_i^{\dagger} \gamma_i = \gamma_i^2 = 1/2$$. Instead, typically pairs Majoranas into pairs and measures the operator $$i \gamma_i \gamma_j$$, which has eigenvalues $$\pm 1/2$$. Fair enough.

Now, here's the question: is there anything in particular which stops me from considering the eigenstates and eigenvalues of the single Majorana fermion $$\gamma_i$$? It's a Hermitian operator, so it can certainly be diagonalized. It's easy to see by inspection that $$\gamma_i$$ has eigenvalues $$\pm 1/\sqrt{2}$$. Additionally, if I think about my fermions in the sense of Jordan-Wigner, it appears that $$\gamma_{i,1}$$ plays the role of $$\sigma^x_i$$ in the spin language while $$\gamma_{i,2}$$ plays the role of $$\sigma^y_i$$, albeit with Jordan-Wigner strings attached.

One potential reason I can think of is that we clearly cannot assign each of $$2N$$ Majoranas different eigenvalues, since our Hilbert space is only $$2^N$$ dimensional. But this would not be much of a complaint in the corresponding spin language (nobody would be upset to hear that $$\sigma^x_i$$ and $$\sigma^y_i$$ are not independent); it's really just a reminder that the $$2N$$ Majoranas are not all independent. Is this the only reason not to consider the eigenvalues of single Majoranas alone, or have I missed something else? For example, is there a physical reason why $$\gamma_i$$ might not be a well-defined observable?

One reason such states are kinda boring is that they do not describe realizable states. An eigenvector of a Majorana operator does not have well-defined fermion parity: $$\gamma_i$$ does not commute with $$(-1)^F$$ and hence they are not simultaneously diagonalizable. Ergo, eigenvectors of $$\gamma_i$$ are neither bosons nor fermions, but a mixture instead. Such mixtures do not exist in nature (they violate the most fundamental super-selection rules: that enforced by $$(-1)^F$$). You cannot prepare eigenstates of $$\gamma_i$$ in a laboratory, not can you measure them in any experiment.
• I must admit, this answer does not seem extremely compelling to me. Just because a state cannot be prepared in a laboratory does not mean that the state is not physically interesting, or that it can't help us solve theoretical problems (ex: you certainly could not prepare a fermionic coherent state in the lab, it's not even a physical state in the Hilbert space). I'm also not entirely sure what you mean by saying the eigenvectors of $\gamma_i$ are a mixture of bosons and fermions; to me, it makes more sense to say that they are states with superpositions of different fermion particle number.