Do boson fields actually have values in QFT?

Question

In QFT a path integral involving fermion fields can be expanded out as a series giving the Green's function propagators. $G^{\alpha \beta}(x,y)$ and so on. The Fermion fields $\psi^\alpha(x)$ are seen as Grassman valued placeholders which don't take on actual values.

On the other hand boson fields such as the electromagnetic field are commuting fields. They can act as placeholders in an expansion to give the boson Green's functions $G^{\mu\nu}(x,y)$ and so on. But it seems like the fields $A_\mu(x)$ can actually be assigned values. (Because real numbers are commutative). And we could talk about the transition function from one value of the electromagnetic field at time $t$ to another at time $t'$ e.g. $G[A,A']$.

Indeed in quantum cosmology we talk of the transitions of one value of the gravitational field to another.

So my question is, is it legitimate to assign values to the boson fields. Or should they just be seen as placeholders like the fermion fields? And only the transition functions be used?

I suppose this is equivalent to the question of whether boson fields should always be thought of as collections of bosons.

If we can assign values to boson fields but not fermion fields, what does this tell us about what is special about boson fields?

On the other hand is there a way Fermion fields can be assigned values (not real numbers but maybe another kind of number?)

Answer 1

The distinction between bosonic and fermionic fields that the question seems to imply does not exist:

1. Classically, a boson field is a real/complex-valued function $M\to \mathbb{R}^n$ (or $\mathbb{C}^n$) and a fermion field is a Graßmann-valued function $M\to \Lambda \mathbb{R}^n$ (or $\Lambda\mathbb{C}^n$). Commonly this is also formalized with the concept of supernumbers, cf. this answer by Qmechanic.

2. Quantumly, what the fields become depends on your quantization procedure:

   • In the canonical operator formalism, the fields become operator-valued distributions. They have "definite values" in the sense that these operators do have a specific action on states, and there is no appreciable difference between bosonic or fermionic field operator except for their (anti-)commutation relations, obviously.

   • In the path integral formalism, the question "What value does the field have" does not really make sense. What does make sense is to ask for its expectation value, which then will be exactly the same type of value the classical field has. The difference between bosonic and fermionic fields here is that quantities with non-zero Graßmann degree not observable, but that does not make them "placeholder" or anything - to the formalism, they're just another kind of values.