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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?)

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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.

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  • $\begingroup$ So what you're saying is that fermion fields are not observable. Isn't this exactly the distinction that I put in my question? In a way the A fields are not observable since they are not gauge invariant. Only $F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu$ is. But also you could assign $A_3(4,2,5)=8$ for example. You can't assign a value to a fermion field. But can you assign values to combinations of fermion fields? say $\psi\gamma \overline{\psi}$ $\endgroup$
    – user84158
    Commented Aug 19, 2017 at 11:12
  • $\begingroup$ @zooby Of course I can assign a value for a fermion field: $\psi(x) = 3\theta_1$, where $\theta_1$ is a Graßmann generator. It's not a real value, but it is a value (maybe you have a non-standard definition of what the value of a function means). Being (un)observable is not special to fermion fields, you already pointed out bosonic gauge fields as a counter-example. $\endgroup$
    – ACuriousMind ♦
    Commented Aug 19, 2017 at 11:17
  • $\begingroup$ But that's not true is it? Because $\psi(x)\psi(y)=-\psi(y)\psi(x)$. So to assign values to each value in space you would need infinite Grassman numbers. $\psi(x)=\theta_x$. But then you're back where you started! Just changed the symbol $\psi$ to $\theta$. $\endgroup$
    – user84158
    Commented Aug 19, 2017 at 12:03
  • $\begingroup$ @zooby What? The Graßmann numbers themselves anticommute by definition. There are $n$ Graßmann generators (e.g. four for a typical spin-1/2 field) $\theta_i$ with $\theta_i \theta_j = - \theta_j \theta_i$ - this property is the whole reason why we say that fermionic fields are Graßmann-valued. I can perfectly well take a constant Graßmann-valued field $\psi(x) = \theta_1$, because it fulfills $\theta_1 \theta_1 = 0 = - \theta_1 \theta_1$ by definition of the $\theta_i$. $\endgroup$
    – ACuriousMind ♦
    Commented Aug 19, 2017 at 12:10
  • $\begingroup$ Yeah but with two grass man numbers $\psi(3)=\theta_1$, $\psi(8.32)=\theta_2$. Now you've run out of Grassman numbers. What will you assign to $\psi(25.6)$ ? $\psi(x)$ IS a continuum of Grassman numbers. Otherwise like you say you would just get $\psi(x)^2=0$ $\endgroup$
    – user84158
    Commented Aug 19, 2017 at 13:55

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