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?


1 Answer 1


There is nothing stopping us from considering eigenvalues/vectors of Majorana operators. It's just -- they do not have any particularly interesting use.

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.

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    $\begingroup$ 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. $\endgroup$
    – Zack
    Jun 25, 2021 at 16:17
  • $\begingroup$ @Zack 1) The answer to the OP is my first paragraph: we don't use these states very often because there aren't many situations where they show up. My second paragraph was just one possible reason they do not show up; if you do not find this reason convincing that's perfectly fine. That doesn't change my first paragraph. Even if we admit that the obstruction to being realizable is not important, that doesn't change the fact that, still, these states do not have any particularly interesting applications (at least as far as I'm aware). 2) By mixture I indeed mean superposition. $\endgroup$ Jun 25, 2021 at 18:07
  • $\begingroup$ As far as the question as-written is concerned (ie, "can we consider eigenvectors/eigenvalues of the Majorana operators?"), I suppose this is a reasonable answer: yes, we can consider them. But I'll a bit longer on awarding the answer in case anyone has additional insights as to why they are extremely uncommon in the literature. $\endgroup$
    – Zack
    Jun 25, 2021 at 20:33
  • $\begingroup$ @Zack Sure, don't feel the need to award any bounty, not now, not when it is over. If you don't find the answer convincing there is no need to accept it. I would definitely wait until the bounty is about to expire. If you don't get better answers by then, you can reconsider. Keep in mind that lack of answers may in fact confirm my claim that"these states are just boring". (But it need not) $\endgroup$ Jun 28, 2021 at 17:45

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