# Legal values of quantum field can take? $\mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$, ..?

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 $$\mathbb{H}$$, ..?
• In some theories the field takes values in a Lie group. – G. Smith May 28 '19 at 0:52
• 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. sciencedirect.com/science/article/pii/0550321388902076) – Avantgarde May 28 '19 at 2:16
• @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 2:42
• @AccidentalFourierTransform The Scholarpedia article “Nonlinear sigma model” (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 4:19
• 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 7:45