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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.

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    $\begingroup$ @JerrySchirmer: You're quite wrong on this. (Note that the model joshphysics asks about has no gauge symmetry.) Consider the quantum mechanics of a particle moving in a rotationally symmetric potential on the plane. There is a symmetry which rotates the $x$ and $y$ axes into each other. But we do not claim the $x$ and $y$ positions are unobservable. $\endgroup$
    – user1504
    Feb 21, 2013 at 17:02
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    $\begingroup$ @JerrySchirmer: But the field $\phi$ is not a wave function. It's directly analogous to the position observables in quantum mechanics. $\endgroup$
    – user1504
    Feb 21, 2013 at 17:09
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    $\begingroup$ @JerrySchirmer: The question wasn't about the standard model. It was about scalar field theory. Pure scalar field theory does not have gauge symmetries. Lubos is correct that gauge symmetries must be accounted for in determining the observables, but his comments aren't really relevant here, because the group of gauge symmetries of the pure scalar field theory is the trivial group. $\endgroup$
    – user1504
    Feb 21, 2013 at 18:17
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    $\begingroup$ @JerrySchirmer Yes, the complex scalar field has a U(1) symmetry. But this symmetry is not a gauge symmetry! It is not modded out by. $\endgroup$
    – user1504
    Feb 21, 2013 at 18:28
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    $\begingroup$ @JerrySchirmer One is not required to gauge any symmetries unless there are massless higher spin particles around! The purpose of gauge invariance is to remove unphysical polarizations from the non-trivial representations of the Lorentz group. This issue doesn't crop up at all in scalar field theories. Just because something looks like a U(1) gauge symmetry (which is really a redundancy) doesn't mean it is. It could be a real global symmetry - something that connects physically different states instead of different representations of the same state. $\endgroup$
    – Michael
    Feb 22, 2013 at 9:54

3 Answers 3

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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.

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    $\begingroup$ I'd add a question Lubos, so $\phi$ is Hermitean and gauge invariant thus dubbed observable, but, excluding decoherence, would you see that it is possible for an observer, in the sense someone doing an experiment, to measure the value of $\phi$ at some point $(\vec{x_0},t_0)$. I am puzzled by the fact that we can perform a redefinition of the field, wave function renormalization, so one should maybe add a specific energy scale to this measurement. $\endgroup$ Feb 21, 2013 at 11:30
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    $\begingroup$ Ok. Although you've stated that every hermitian operator is, in principle, observable in the sense of measurement, that doesn't completely answer the question. Could you be more specific about how one would justify such an assertion. What sort of measurement would one perform in this case? $\endgroup$ Feb 21, 2013 at 15:52
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    $\begingroup$ To clarify: what does it mean to measure the value of the field itself at a given spacetime point? $\endgroup$ Feb 21, 2013 at 17:11
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    $\begingroup$ @joshphysics: fields are "operator distributions", not just "operators", in the same sense as delta(x) is a distribution and not proper function. This is a detail. To get totally ordinary operators, one has to e.g. average the operators over an (arbitrarily small) region. In practice, only sufficiently "diluted" operators will be easy to measure, but that's just in practice. In principle, one may measure any (gauge-invariant) function of the operators. $\endgroup$ Feb 22, 2013 at 9:19
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    $\begingroup$ Dear joshphysics, the procedure of the measurement may be hard but it may be done in principle, whatever the operator is. You may construct each operator as a linear superposition of $|i\rangle\langle i|$ objects. Those are projection operators. So every measurement is reconstructible (among other things) by a sequence of measurements yes/no whether $|\psi\rangle$ is in a particular state. And any state may be unitarily rotated to one that you may measure by a standardized procedure (particle is there/not). A sketch how to measure any observable. $\endgroup$ Feb 25, 2013 at 5:47
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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).

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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.

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  • $\begingroup$ (NB: I didn't downvote you.) Let me see if I understand you right. By this exact logic the effective quantum field theory of phonons is a theory of phonon occupation numbers, and the actual displacements of atoms from their lattice sites is an unphysical auxiliary tool. Pardon me if I say that seems bass-ackwards. By what criterion am I meant to judge ahead of time whether a given qft is a theory of occupation numbers only or whether the fields themselves are physical? $\endgroup$
    – Michael
    Feb 21, 2013 at 10:36
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    $\begingroup$ @MichaelBrown: I speak of QFT, not of classical field theory. In a QFT the field $\Phi(x)$ is an operator constructed from other operators including operators of anti-particles. So the interpretation of $\Phi(x)$ is somewhat more sophisticated than that of a classical field. For example, a static electric field determines the force in the charge equation and therefore can be measured by a dynamometer. $\endgroup$ Feb 21, 2013 at 10:59
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    $\begingroup$ @MichaelBrown: Any function of an observable is an observable too, but in case of QFT we deal with operators, not regular functions of eigenvalues. $\endgroup$ Feb 21, 2013 at 11:01
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    $\begingroup$ @user10001 The downvotes are because it is not a correct answer. QFT is not just the occupation number formalism (unless you think conformal field theories are not field theories). And the field operators are not just auxiliary tools (unless you think the electric and magnetic fields are not observable). $\endgroup$
    – user1504
    Feb 21, 2013 at 13:13
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    $\begingroup$ @VladimirKalitvianski: Lubos' answer is correct. Why waste effort on duplication? $\endgroup$
    – user1504
    Feb 21, 2013 at 13:17

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