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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.

joshphysics
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  • Good question. Not a complete answer but maybe a start (I would like to learn more about this myself): a state with a definite value of the field is a coherent state, involving superpositions of arbitrary numbers of field quanta. So you are essentially looking at a "big" classical field configuration where, as my supervisor puts it, "the field doesn't notice if you add or remove a particle." So... depending on the interactions you have your scalar field might be observable in the same sense that a classical electromagnetic field is observable. – Michael Feb 21 '13 at 09:34
  • Well, at minimum, only the modulus of $\Phi(x)$ is even in principle observable, since $\Phi \rightarrow e^{ia}\Phi$ with $a$ constant is a symmetry of, certainly the free field, and also of most physical matter fields. – Zo the Relativist Feb 21 '13 at 09:37
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    The obvious actual physical example I can think of is the Higgs vev, which we measure classically every time a lepton comes to rest. Maybe the chiral QCD condensate is another example - though I really don't know as much about that as I'd like. – Michael Feb 21 '13 at 09:38
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    @JerrySchirmer: That's not correct. You're assuming that $\Phi \mapsto e^{ia}\Phi$ is a gauge symmetry, which it is isn't necessarily. – user1504 Feb 21 '13 at 10:25
  • @user1504: I didn't say that $a$ is a function of spacetime. Almost every physical field theory has the leas restrictive "global gauge invariance" – Zo the Relativist Feb 21 '13 at 14:07
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    @JerrySchirmer: The existence of a symmetry that acts on a scalar field does not imply that the value of the field is unobservable. That's the mistake. In joshphysics' gedankenuniverse, where there's a single real scalar field, you can actually observe the value of that field, not just its modulus. – user1504 Feb 21 '13 at 15:31
  • @user1504: only after fixing the global gauge. If a symmetry makes the numerical value of the field arbitrary, then that quantity is not observable. – Zo the Relativist Feb 21 '13 at 16:57
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    @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. – user1504 Feb 21 '13 at 17:02
  • @user1504: sure, but the value of the wave function is unobservable. internal symmetries of the description are not the same thing as physical symmetries of the system. – Zo the Relativist Feb 21 '13 at 17:06
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    @JerrySchirmer: But the field $\phi$ is not a wave function. It's directly analogous to the position observables in quantum mechanics. – user1504 Feb 21 '13 at 17:09
  • @user1504: but it's value isn't physical! It can be set arbitrarily at a point by a gauge choice. Lubos says the identical thing below. – Zo the Relativist Feb 21 '13 at 17:16
  • And irrespective to this, name me a field in the standard model, (as in "most physical matter fields" in my original comment) that doesn't have a local gauge symmetry, modulo spontaneous symmetry breaking. – Zo the Relativist Feb 21 '13 at 17:20
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    @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. – user1504 Feb 21 '13 at 18:17
  • @user1504: No. you can shift the entire field by a constant phase. If there are no spatial and time derivatives to the phase, then you don't get the bit that has to be cancelled by the E&M coupling, and the scalar field action is invariant. – Zo the Relativist Feb 21 '13 at 18:20
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    @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. – user1504 Feb 21 '13 at 18:28
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    @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. – Michael Feb 22 '13 at 09:54
  • @MichaelBrown: I think this conversation is beginning to be circular, but all I"m saying is that the actual value of the field is meaningless unless you choose a basis for the scalar field, which means fixing the global U(1). – Zo the Relativist Feb 22 '13 at 14:56

3 Answers3

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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.

Luboš Motl
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    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. – Learning is a mess Feb 21 '13 at 11:30
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    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? – joshphysics Feb 21 '13 at 15:52
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    To clarify: what does it mean to measure the value of the field itself at a given spacetime point? – joshphysics Feb 21 '13 at 17:11
  • @joshphysics: To clarify: do you speak of a quantized or a classical field $\Phi$? – Vladimir Kalitvianski Feb 21 '13 at 19:09
  • @VladimirKalitvianski Quantized. – joshphysics Feb 22 '13 at 01:08
  • Dear @Learningisamess, well, if you make a field redefinition or switch to another renormalization scheme etc., the same symbols will represent different operators/observables - they'll be functions of the previous ones - but they will still be observables, right? Forget QFT. Just take ordinary QM. I may still redefine what I mean by x,y,z of an electron in a hydrogen atom, right? That doesn't mean that this freedom to redefine things will invalidate the fact that x,y,z are observables whatever convention I pick. – Luboš Motl Feb 22 '13 at 09:17
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    @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. – Luboš Motl Feb 22 '13 at 09:19
  • @LubošMotl Yeah, I'm aware of the operator-valued distributional nature of quantum fields. I must be missing something here since this measurability seems somewhat obvious and/or intuitive to you and others. Perhaps I'll think more and try to better understand my confusion...thanks all the same. – joshphysics Feb 22 '13 at 09:24
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    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. – Luboš Motl Feb 25 '13 at 05:47
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    @LubošMotl That's reasonable thanks. Also thanks for the publicity :) – joshphysics Feb 28 '13 at 15:12
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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).

Arnold Neumaier
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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.

Vladimir Kalitvianski
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  • (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? – Michael Feb 21 '13 at 10:36
  • If you say (the only thing I could guess might make sense): "when you can take the lattice spacing $a\rightarrow 0$ then you have a theory of occupation numbers only", well... nobody really cares if their qft's have a genuine continuum limit anymore - they are all effective theories with a cutoff. I prefer to withhold judgement on ontologies that are sensitive beyond the Planck scale until we actually have experiments at the Planck scale. :) – Michael Feb 21 '13 at 10:39
  • @MichaelBrown: Yor wrote: "...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." That's nearly right. I simply did no say "unphysical". I meant it depends. In your particular case the atoms displace often together from their lattice sites ;-) – Vladimir Kalitvianski Feb 21 '13 at 10:44
  • It seems to me that saying the field "is an auxiliary tool for making calculations," then saying "physical" in scare quotes and saying that the physical properties are those of free particles strongly implies something very different than what you apparently mean... – Michael Feb 21 '13 at 10:50
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    @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. – Vladimir Kalitvianski Feb 21 '13 at 10:59
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    @MichaelBrown: Any function of an observable is an observable too, but in case of QFT we deal with operators, not regular functions of eigenvalues. – Vladimir Kalitvianski Feb 21 '13 at 11:01
  • Number states (eigenstates of your free Klein-Gordon Hamiltonian) are usually not robust to decoherence, (assuming an interacting Hamiltonian system bath of the form $ba^{\dagger}+b^{\dagger}a$, RWA approximation b and a being the annihilation operators of the bath and the scalar field respectively, and so the pointer states would be the eigenstates of $a$ or the coherent states). – Learning is a mess Feb 21 '13 at 11:06
  • @MichaelBrown: Concerning an "effective QFT", in case of phonons it is apparently meaningless to consider phonons with the wavelength much shorter than the lattice constant, but in a correctly formulated theory such phonons are so energetic that they are under threshold and their occupation numbers are negligible (not harmful), so no cutoff is really necessary. Contributions of virtual shortwave phonons in calculations depends on correctness of the interaction terms and is also negligible in a correct formulation, in my humble opinion. – Vladimir Kalitvianski Feb 21 '13 at 11:29
  • This i think is the correct answer +1. Why downvotes ? – user10001 Feb 21 '13 at 12:18
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    @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). – user1504 Feb 21 '13 at 13:13
  • @user1504: So give us your reply to the questions. Why are you shy? – Vladimir Kalitvianski Feb 21 '13 at 13:16
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    @VladimirKalitvianski: Lubos' answer is correct. Why waste effort on duplication? – user1504 Feb 21 '13 at 13:17
  • @user1504: Indeed. Measuring a magnetic field with a compass is the right response to the QFT in question. – Vladimir Kalitvianski Feb 21 '13 at 13:28
  • @VladimirKalitvianski: I think we can safely interpret 'compass' as an informal way of saying 'something with a magnetic moment'. – user1504 Feb 21 '13 at 15:34
  • @user1504: And how about the quantized scalar field? – Vladimir Kalitvianski Feb 21 '13 at 16:18
  • @user1504 I think from QM point of view its not electric and magnetic fields which are physical but the states in Hilbert space of U(1) YM theory. Quantum mechanically there can be no such thing as a "continuous (observable) field", we can only have states with discrete particle content (or their linear combinations). – user10001 Feb 21 '13 at 17:17
  • @user10001: The Maxwell field lacks a mass gap, so its Hilbert space is not spanned by states with discrete particle content. Think about it a bit, and you'll see that (in flat space) the Hilbert space is spanned by eigenstates of the (smeared) electric and magnetic field operators. – user1504 Feb 21 '13 at 18:25
  • @user1504 I always had the viewpoint that observables have to do with symmetry of the theory. They form a representation of symmetry group on space of states. We can work in terms of fields or creation - annihilation operators or may be something else. The physical content of the theory is not in fields but in the space of states and the representation of the symmetry group on it. – user10001 Feb 21 '13 at 18:55
  • @user1504: Nobody speaks of Maxwell field here, but of the scalar Hermitian field $\Phi (x)$. QED in cavity has a kind of "mass gap" - the photon spectrum starts at $\omega > 0$, so the particle interpretation holds. – Vladimir Kalitvianski Feb 21 '13 at 19:07
  • @VladimirKalitvianski: Actually, user10001 brought up the Maxwell field, AKA, U(1) YM theory. – user1504 Feb 21 '13 at 19:27
  • @user1504: I meant joshphysics and his quantized scalar field. – Vladimir Kalitvianski Feb 21 '13 at 19:37
  • If you are planning on continuing this conversation, could you please move it to chat? – Manishearth Feb 22 '13 at 05:52