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I think I might have misconceptions about the conceptual core of QFT. Let me explain where I am puzzled.

In QM, the measurement process is accounted by the postulate of collapse of the wave function: when a measurement is made on a state, the unit vector defining the state is projected on one of the eigenvectors of the basis defining the observable; now, (let us assume that the Hilbert space is finite-dimensional) two self-adjoint commute if and only if they have a common basis of eigenvectors. So, commuting self-adjoint operators model observables that can be simultaneously measured.

Let me now assume that the quantum fields, in quantum field theories, are to be thought as observables on spacetime, and that the measurement of an observable yields to a wave function collapse; I understand why one would ask that two fields with space-like separated supports should commute. But Wightman's axiom 3 also allows two such fields to be anti-commuting. What would happen if two space-like separated observers repeatedly made the same measurements (each one with respect to a field, and such that the two fields would anti-commute)? Would they randomly see their results switch sign, once in a while, thus telling them that the other has made his/her measurement? Wouldn't this imply super-luminal communication?

So, where am I wrong?

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The anticomutation at spacelike separation does not cause problems because observables in volving fermions are always bilinear in the Fermi fields, for example, $\bar\psi \gamma^\mu \psi$, and these commute at spacelike separation.

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  • $\begingroup$ I'm sorry, I don't get it at all. I might be mistaken, but I thought that the fields were themselves the observables. What are the observables, in QFT? $\endgroup$
    – Plop
    Dec 3, 2020 at 16:13
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    $\begingroup$ For scalar fields the fields are the observables, and measurements yield a numerical value. But if a Fermi field had a value, it would have to be an anticommuting Grassmann number and as far as we know there is no way to measure an anticommuting number. The Fermi bilinears like $\bar\psi \gamma^\mu \psi$ can have numerical values and so are observables. $\endgroup$
    – mike stone
    Dec 3, 2020 at 18:12
  • $\begingroup$ Again, I'm not getting it. As far as I understand, a QFT should be a Hilbert space $\mathcal{H}$, and a set of self-adjoint operators-valued distributions (the "operator" par meaning "operator on $\mathcal{H}$") on spacetime verifying the Wightman axioms. Trying to gain intuition from ordinary QM, I thought that the elements of $\mathcal{H}$ must still be considered as the states and the self-adjoint operator-valued distributions to be thought of observables. Am I correct? $\endgroup$
    – Plop
    Dec 3, 2020 at 19:40
  • $\begingroup$ If I get Wikipedia right, a Grassmann number is an element of the exterior algebra of something. How could it be an observable? $\endgroup$
    – Plop
    Dec 3, 2020 at 19:40
  • $\begingroup$ Exactly. If the linear operators $ \psi(x)$ and $ \psi(y) $ anticommute and have a common eignevector so that $\psi(x)|\alpha \rangle= \alpha(x) |\alpha \rangle$, $\psi(y)|\alpha \rangle =\alpha(y) |\alpha\rangle$ then $\psi(x)\psi(y)=- \psi(y) \psi(x)$ tells us that $\alpha(x)\alpha(y)=-\alpha(y)\alpha(x)$ so the eigenvalues anticommute and cannot be ordinary numbers. Anticommuting "numbers" are Grassmann objects. So the $\psi(x)$'s cannot be observables. $\endgroup$
    – mike stone
    Dec 3, 2020 at 19:55

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