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In QFT fields are Hermitian operators. And observables are represented by operators. I am confused are fields also observables?

Qmechanic
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physics
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Not all fields in QFT are Hermitian. Non-Hermitian fields include the complex scalar field (the Higgs boson is of this type) and the Dirac field (electrons and many other particles).

There are also Hermitian ones, including the real scalar field and the vector boson field. But fields do not need to be Hermitian.

In general, fields are not used in QFT in the same way that operator observables are in QM. By this I mean that we don't use their eigenvectors to assign probabilities to measuring particular values of observables. In a basic way you could just think of them as building blocks that we use to express the Lagrangian and Hamiltonian density.

However, it is generally possible to diagonalize fields and find a basis of their eigenvectors. This is an important step in the path integral. It's just that this basis is not typically used to assign probabilities as is done in QM.

doublefelix
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  • are fields also observables? – physics Apr 05 '23 at 09:02
  • People definitely don't think of them that way. Regardless of that, whether you could get some result by treating the hermitian fields as an observable, I'm not sure. For the non-hermitian fields it seems even less likely. – doublefelix Apr 05 '23 at 18:24