In a similar manner to how we can define Majorana fermionic operators $\gamma_j$ via $$ c_j \propto \gamma_{2j+1} + i \gamma_{2j}^\dagger, $$ where the $c$'s are fermionic creation/annahilation operators. These operators are super useful when dealing with fermionic systems. Im wondering if one can define and meaningfully use bosonic Majorana operators, i.e. $$ b_j \propto \tilde{\gamma}_{2j+1} + i \tilde{\gamma}_{2j}^\dagger, $$ where the $b$'s are bosonic creation/annahilation operators.

Is there a way to legalize these Majoranas?

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    $\begingroup$ The analog of a Majorana fermion for bosons are the excitations of a real scalar field $\phi$ as far as I know (where $\phi^\dagger = \phi$). The reason for this is that for both of these situations the particles are their own antiparticles (ie. the field expansion is $\phi \propto u_{k} \hat{a}_{k} + u^{\ast}_{k} \hat{a}^{\dagger}_{k}$, as opposed to the non-Majorana case, where you would have some other kind of creation operator $\hat{b}^{\dagger}_{k}$) $\endgroup$ Dec 16, 2021 at 21:21
  • $\begingroup$ I'm not familiar with condensed matter Physics, so what I say is in the context of QFT. The Majorana condition can be seem as a reality condition. Indeed, when representing the Clifford algebra gamma matrices $\Gamma^\mu$ one may choose the Majorana representation. In that representation a Majorana spinor has real components. In that setting, a Majorana spinor would be the fermionic analog of a real bosonic field. $\endgroup$
    – Gold
    Dec 16, 2021 at 21:44
  • $\begingroup$ Your expression for $\gamma_j$ is not hermitian, so it is not a Majorana operator. $\endgroup$
    – Meng Cheng
    Dec 17, 2021 at 2:13
  • $\begingroup$ @MengCheng you are right, there is a typo I meant $\gamma \leftrightarrow c$ and so on. $\endgroup$ Dec 17, 2021 at 10:32

1 Answer 1


The correspondingly defined objects for bosons are the position operator $x=(a+a^\dagger)/2$ and the momentum operator $p=(a-a^\dagger)/2i$, respectively.

  • $\begingroup$ Cant you also define those for fermions? Also, say you create a boson at position $x$, it's state then is defined by $a_x^\dagger|0\rangle$. Now with your definition, if we measure it's position $\langle x\rangle \stackrel{?}={} \langle (a_x+a_x^\dagger)/2\rangle =1/2$. I know Im butchering your answer a bit, but what does that mean? Shouldn't $\langle x\rangle=x$ for this state? $\endgroup$ Dec 19, 2021 at 23:47
  • $\begingroup$ @FriendlyLagrangian This is for a single harmonic oscillator, as presented in any QM textbook where creation/annihilation operators are introduced. $a_x$ is a completely different thing, in an entirely different space. In essence, $x$ and $p$ are the canonically conjugate operators. For a single light mode, these could e.g. correspond to the sine and cosine part of the wave. $\endgroup$ Dec 29, 2021 at 14:55

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