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Ok, i found the (silly) error: $$\bar\Phi\gamma^5\Phi=\Phi^\dagger\gamma^0\gamma^5\Phi$$ so under hermitian conjugation this becomes $$\Phi^\dagger\gamma^5\gamma^0\Phi=-\Phi^\dagger\gamma^0\gamma^5\Phi=-\bar\Phi\gamma^5\Phi$$ that imply $$\bar\Phi\gamma^5\Phi+h.c.=0$$ the same result that we found exploiting the two component structure.

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First, you should sort out if the components of the spinor are c-numbers or Grassmann numbers in your problem. If they are c-numbers, then the pseudoscalar built on the Majorana spinor vanishes, if I am not mistaken.

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Let $a$ be any spinor; then, by definition $\bar a\equiv a^\dagger \gamma^0$, where $\dagger$ stands for hermitian conjugation (transpose+complex conjugation: $a^\dagger=(a^T)^*$), and $\gamma^0$ is one of the Dirac matrices. With this in mind, the steps are as follows: first, we transpose the spinor:  u^T=\sqrt{E}\begin{pmatrix} c & s\mathrm ...

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The reason your logic fails is because $\psi$ is not simply a Grassmann variable; it is a four-component vector of complex Grassmann numbers (in four dimensions): $$\psi=\left(\begin{array}{c} \theta_1 \\ \theta_2 \\ \theta_3 \\ \theta_4 \end{array}\right)$$ With this knowledge, try computing $\bar{\psi}\psi\bar{\psi}\psi$ and ...

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The argument is false in four dimensional space. The error is the assumption that you get one Grassman number per spinor. In fact, you get one Grassman number per spinor component! In 4d, spinors have multiple components. (Both Weyl spinors have 2 components, and Dirac spinors have 4.) In 1d space, this is a correct argument. In 2d, it is correct for ...

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Are fermions non-local objects, in a sense in which gauge bosons are not? As far as I understand, the answer is definitely NO. Fermionic particles are local objects as bosonic ones. Based merely on the non-local form of the bosonized Jordan-Wigner fermions, one cannot conclude that fermions are non-local. Jordan-Wigner transformation, like any ...

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Okay, I think I have a semi-convincing picture of this in my head. Both of the other answers contain at least part of the story I wanted; I will put the whole thing here in hopes of feedback and that it is useful to someone else. As SM Kravec points out, fermionic parity is a non-local symmetry of a fermionic system. This suggests, as various people have ...

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Are fermions intrinsically non-local? Yes, definitely. It's quantum field theory, not quantum point-particle theory. An electron's field is what it is. And that field doesn't have a surface. From a great distance it will be swamped and undetectable, but there is no defineable place where that field stops. When one studies quantum mechanics of more ...

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I posted the question a few hours ago, and realized the answer lies in the fact that ${(\gamma^m)^{\alpha}}_{\beta}$ has two kinds of indices. It is indeed true that so does ${(\gamma^\mu)^\alpha}_{\beta}$. But the fact is that the flat space gamma matrices are invariant tensors of the Lorentz group $SO(d-1,1)$. The fact that ...

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Comments to the question (v1): There are three types of indices at play: (i) spinor indices, (ii) flat (vector) indices, and (iii) curved (vector) indices. The gamma matrices with flat indices are constants. They don't transform under local Lorentz transformations (LLTs). They can be viewed as intertwiners between spinor indices and flat indices. (LLTs ...

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So what people mean by 'non-local' varies from context to context and person to person. Wen has a very particular meaning to this. 1) In fermionization in $D=1+1$ the Jordan-Wigner fermions are, in the bosonic language, operators supported over many sites. The emergent (mutual)-fermions in the toric code are also supported at the ends of strings. 2) ...

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Prahar has already given a good answer. Here we will instead focus on the pertinent Lie group (as opposed to the Lie algebra and its generators). The Lie group $SL(2,\mathbb{C})$ is the double cover of the restricted Lorentz group $SO^+(3,1)$, cf. e.g. this Phys.SE post. The fundamental/defining representation $V\cong\mathbb{C}^2$ of the Lie group ...

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