In what sense is a scalar field observable in QFT?

Question

Consider a QFT consisting of a single, hermitian scalar field $\Phi$ on spacetime (say $\mathbb R^{3,1}$ for simplicity). At each point $x$ in spacetime, $\Phi(x)$ is an observable in the sense that it is a hermitian operator (operator-valued distribution) on the Hilbert space of the theory, but is each such operator observable in a stronger, more physical sense? Is there an experiment one could hypothetically perform to measure the value of such a field at a given spacetime point?

This is one of those questions I glossed over while learning QFT, but now it's bugging me. In particular, I think this point is central in preventing me from understanding certain basic assumptions in QFT such as microcausality which I also never really think about anymore.

Answer 1

Every observable in the technical or mathematical sense (linear Hermitian operator on the Hilbert space) is, in principle, observable in the physical operational sense, too. That's why it's called this way.

Magnetic fields may be measured, for example, by compasses. Analogous methods exist for electric fields, scalar fields, or any other fields. For example, if you want to measure the Higgs field, you may, in principle, place a top quark (or an even heavier particle if there is one) at that point and measure its induced inertial mass.

Let me mention that a true observable must be gauge-invariant. So if a complex field carries a charge $Q$, it is not gauge-invariant. One has to combine it to expressions such as $\phi^\dagger \phi$ to get gauge-invariant objects. These are true observables. This extra requirement doesn't contradict the original definition because gauge-non-invariant operators are not well-defined linear operators acting on the physical Hilbert space (because physical states are equivalence classes and the action of a gauge-non-invariant operator would depend on the representative of the class). Yes, by the Hilbert space, I always meant the physical ones, after all identifications that should be made are made and unphysical states such as longitudinal photons are removed.

Also, fermionic fields may be called observables but they can't have nonzero eigenvalues. Only products that are Grassmann-even – contain an even number of fermionic factors – are measurable due to the existence of superselection sectors that divide bosonic and fermionic states according to the eigenvalue of $(-1)^F$. But formally speaking, we could imagine states in the Hilbert space with Grassmann-odd coefficients and the "fermionic coherent states" would be eigenvalues of fermionic operators. However, Grassmann-odd probability amplitudes aren't physical so such a construction is purely formal.

Answer 2

One cannot observe, even in principle, $\Phi(x)$, as it does no qualify for an ''observable''.

The reason is that observations must happen in space and time, and this is inevitably associated with smearing the field. Indeed, it is well-known from algebraic quantum field theory that $\Phi(x)$ is not a Hermitian operator, but just a label for the (nonexisting) value of an operator-valued distribution $\Phi$.

In principle, observable are at best the smeared operators $\int dx f(x)\Phi(x)$ with sufficiently regular test functions $f$ that have a support that covers the region of spacetime in which the whole observation is made. (The latter aspect was swept under the carpet in Lubos Motl's answer and in the subsequent discussion there. He alludes to the standard discussions of quantum measurements, but these assume unlimited repeatability. Since repeating something changes its spacetime position, these arguments work only for processes that are either periodic, or essentialy stationary at the scale of repetition.)

However, from a practical point of view, what is observable are only smeared field expectations $\langle\int dx f(x)\Phi(x)\rangle$ and (Fourier convolutions of) smeared field correlations $\langle\int dxdy f(x,y)\Phi(x)\Phi(y)\rangle$. This is sufficient for the applications of QFT to high energy experiments, nuclear fuels, quantum optics, semiconductors, and the early universe (and probably everything else).

Answer 3

A QFT is a formalism of occupation numbers. The latter are observable. Normally we speak of plane waves (free particles) and study their occupation number evolutions. The field $\Phi(x)$ is an auxiliary tool for making calculations. Its "physical" properties are dictated by the properties of free particles whose "superposition" gives $\Phi(x)$.

EDIT: Seeing so many comments, I would like to underline again: properties of $\Phi (x)$ (including microcausality) follow from properties of $a_p$ and $a^+_p$ and from the way how $\Phi$ is constructed.