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Main issue: What are the legal and possible values of the quantum field can take?

Clarify by examples:

(1) For example, for the spin-0 Klein Gordon field $\phi$, we may choose it to be:

  • real $\mathbb{R}$.
  • complex $\mathbb{C}$.

(2) For the spin-1/2 fermion field $\psi$, we may choose it to be a spinor which needs to be

  • Grassman variable

but can also be

  • complex $\mathbb{C}$. (Dirac or Weyl spinor/fermion)
  • We can ask: Can it be in real $\mathbb{R}$? (Majorana or Majorana-Weyl spinor/fermion)

(3) For the spin-1/ boson field $A_\mu$, we may choose it to be a vector which needs to be

  • real $\mathbb{R}$ usually for photon field.

  • but can it be complex $\mathbb{C}$?

(3) How about the spin-3/2 fermion field $\psi_\mu$?

  • can it be in real $\mathbb{R}$, complex $\mathbb{C}$, quaternion $\mathbb{H}$, ..?
ann marie cœur
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    In some theories the field takes values in a Lie group. – G. Smith May 28 '19 at 00:52
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    Not sure why this is put on hold. It's a pretty good question. It boils down to, at least, as far as I can think, studying the stability of a theory resulting from such generalizations. 'Stability' here could mean unitarity, absence of negative energy states, etc. One example I know of a similar generalization is when you make the worldsheet coordinates in string theory non-Archimedean, p-adic valued, giving you p-adic string theory (e.g. https://www.sciencedirect.com/science/article/pii/0550321388902076) – Avantgarde May 28 '19 at 02:16
  • @AccidentalFourierTransform I was thinking of the fields in nonlinear sigma models, not gauge fields. These models were not part of my QFT course, so I may completely misunderstand them. I thought the fields in these models have values in Lie groups such as $O(3)$. Is that not the case? – G. Smith May 28 '19 at 02:42
  • @AccidentalFourierTransform The Scholarpedia article “Nonlinear sigma model” (http://www.scholarpedia.org/article/Nonlinear_Sigma_model) says that the fields map spacetime into a Riemann target manifold $M$. It then goes on to say, “All the isometries of the target space $M$ form the group $G$ representing the global symmetry of $M$. The most important case is given by the target space that is a Lie group $G$ itself.” So this seems to confirm my statement. Do you agree? – G. Smith May 28 '19 at 04:19
  • I re-ask a different and focused one here: https://physics.stackexchange.com/q/482858/ – ann marie cœur May 28 '19 at 04:31
  • @all, https://physics.stackexchange.com/q/482858/42982 https://physics.stackexchange.com/q/482860/42982 – ann marie cœur May 28 '19 at 04:36
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    An answer would depend on which model that we're talking about. And there are a lot of models to choose from... – Qmechanic May 28 '19 at 07:45
  • @G.Smith Ah, yes, you're absolutely right. I thought you were referring to gauge fields, thus my comment. But non-linear sigma models do take values in a generic homogeneous manifold, which may indeed be a Lie group. Although strictly speaking, the fields can be thought of as coordinates on the manifold, and so $\mathbb R$-valued. But anyway. – AccidentalFourierTransform May 28 '19 at 12:39

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