Are observables in QFT actually observable?

Question

Consider some interacting QFT on a lattice (just to avoid infinitely large momentums). The size of the lattice is assumed to be much smaller than the size of the emergent particles (like in our world). Lets assume there are scientists living in such system and they try to measure the field (which is an "observable" by the standards of QM) at a lattice site. Would they be able to do this? Wouldn't the fact that all particles are much larger than the size of the lattice restrict probing the field at a single site? In other words would the observables in such system actually be observable (at least in theory)?

Edit. There have been two excellent answers by Chiral Anomaly and ACuriousMind, but I feel they are addressing a bit different question from what I intended. I would like to be more specific. In the world I described above, what are the restrictions on the resolution of the details of the fields the scientists can measure? For instance, if they smashed electrons faster and faster, would they get a better value of electric field near the center of an electron? Or, if they sent an electron through a non-uniform electric field would they be able to extract the value of the field in an arbitrarily small region?

Answer 1

A quantum mechanical secret kept very well in plain sight is that the word "observable" when defined as "self-adjoint operator on a Hilbert space" does not actually mean "you can construct an apparatus that can measure this like the Born rule promises".

1. If you want to be formal about it, you have to consider quantum fields in the continuum as operator-valued distributions $\phi(x)$ that only yield actual operators when smeared with a test function $f(x)$ as $\phi(f) := \int f(x)\phi(x)\mathrm{d}x$.

   The algebra of observables we need associate to a QFT is not the distributions $\phi(x)$, but the Haag-Kastler net of observables where on each region of spacetime $R\subset \mathbb{R}^{1,3}$ we have the algebra of operators $$ \mathcal{A}(R) = \left\{\phi(f) \mid \mathrm{supp}(f)\subset R, \phi \text{ is a quantum field}\right\},$$ where I'll wave my hands a bit and say that some expressions in the usual "fields" like their derivatives $\partial_\mu \phi$ also count as a field here. The self-adjoint operators in this algebra are the observables. (Don't try to figure out how to rigorously construct this - we do not have rigorous constructions of most quantum field theories)

   For instance in QED, you can smear the electric field $F^{0i}(x)$ with a test function to get an observable $F^{0i}(f)$ that corresponds to the electric field in the region $\mathrm{supp}(f)$ weighted by the value of $f$. Whether or not this is something you can "in theory" measure depends on what sort of measurement apparati your "theory" here can construct. In the end, when the $\mathrm{supp}(f)$ gets very narrow, you'll have to concede that there probably isn't any realizable apparatus that could detect it. This, however, is not a phenomenon unique to quantum fields!

2. Consider an operator with unbounded continuous spectrum in ordinary quantum mechanics, such as position: You will have to admit that there is no realizable apparatus that could distinguish positions $x$ and $x+\epsilon$ for arbitrarily small $\epsilon$. But this doesn't mean we "can't measure position", it means we have to broaden our conception of measurement: What we might be really measuring is not some sharp projection onto an eigen"state" of position with eigenvalue $x_0$ (these states rigorously don't exist as states just like the QFT $\phi(x)$ is not an operator), but the projection onto some sharply but not infinitely localized state with wavefunction $\psi(x)$ centered at $x_0$ and falling off quickly far from it (how quickly depending on the details - i.e. "accuracy" - of the measurement apparatus), i.e. the projection of the original wavefunction onto some subset $X\subset\mathbb{R}^3$, but where $X$ is not a point. Note that this is morally very similar to "smearing" the position operator, just like we had to smear the quantum fields.

   This was a hand-wavy description, if you are interested in formalizations of this, a more general theory of measurement considers positive operator-valued measures, and for a treatment of how to construct measurement processes for continuous observables see Ozawa's "Quantum measuring processes of continuous observables ". Notably, Ozawa proves that a measurement process for continuous observables can never be considered just as resulting in projections onto eigenstates, and does not lead to the usual property of the repeated measurement yielding exactly the same state that we are used to from discrete observables.

Answer 2

As emphasized in the excellent answer by ACuriousMind, some of the operators that we usually call "observables" might not be feasibly measurable through any physical process that the theory can actually describe.

This makes sense intuitively. A theory might formally allow us to consider doing something like a double-slit interference experiment with the whole earth, but our intuition tells us that we cannot actually perform such a measurement. This intuition can be substantiated by quantifying the physical resources that such a measurement would require and comparing it to the totality of all resources available in the known universe. See Omnès (1994), The Interpretation of Quantum Mechanics, pages 308-309.

I'm posting this supplementary answer for two reasons: (1) to explain how that general conclusion can be deduced easily, using only a few basic principles, and (2) to apply it to the case described in the question.

How to deduce that some "observables" cannot be measured

Measurement is a physical process, not metaphysical magic. Measurement involves physical equipment sitting in a physical environment, and those things are all made of molecules (etc). Molecules can be described using quantum theory, so a sufficiently comprehensive quantum model should be able to describe the physical process of measurement just like it describes any other physical process, even if the calculations are too difficult for us to do.

Of course, to use the theory itself to describe the process of measuring one of its own observables, we must use a sufficiently comprehensive model — one that is rich enough to include physical equipment sitting in a physical environment, all made of jillions of molecules. Single-particle models are obviously inadequate for this purpose, but something like QED+QCD should be sufficient, so consider QED+QCD. This is consistent with the original question, because we know exactly how to formulate QED+QCD on a lattice.

I'll use this notation:

• Let $\Gamma$ be the set of all self-adjoint operators whose spectra are purely discrete and whose number of eigenspaces is finite. (Focusing on $\Gamma$ eliminates distracting technical complications with things like unbounded operators and continuous spectra.)

• Let $\Omega$ be the set of all observables that could feasibly be measured.

• For any operator $A\in\Gamma$, let $A_k$ denote the projection operator onto the $k$-th eigenspace of $A$.

• Let $|\psi\rangle$ be a state-vector that accounts for everything we know about how the physical system was prepared, including the measurement equipment and the environment.

I will explain why $\Omega$ cannot include all of $\Gamma$.

Suppose $A\in\Gamma$ and $A\in\Omega$. How would we use the theory itself to recognize whether or not $A$ has been measured? Any physical process that qualifies as a "measurement of $A$" should at least have this property: after the measurement, the projections $A_k|\psi\rangle$ all differ from each other in a practically irreversible way.

"Practically irreversible" is too vague. For one thing, if the Hilbert space is finite-dimensional (say $10^{100000}$-dimensional), then a loose translation of a general theorem about unitary time-evolution says that history always eventually repeats itself. So requiring permanent practical irreversibility might be too strong. The following argument only assumes that the measurement is temporarily practically irreversible.

To make this more precise, let $I$ be a finite time-interval in the not-too-distant future of the alleged measurement event, and let $\Omega_I$ be the subset of $\Omega$ consisting of observables localized in $I$. (I'm using the Heisenberg picture. Also, $I$ should be large compared to the lattice spacing.) The projections $A_k|\psi\rangle$ differ from each other in a "temporarily practically irreversible way" if they cannot be significantly mixed with each other by any operators in $\Omega_I$. More explicitly, the off-diagonal matrix elements $\langle\psi|A_j BA_k|\psi\rangle$ ($j\neq k$) should be close to zero compared to $\langle\psi|B|\psi\rangle$, for all $B\in\Omega_I$.

Because of the loose words "significantly" and "close to," that criterion is still not precise, but we can make it precise by choosing an artificial threshold. As long as the threshold is constraining enough to make the criterion useful (I mean useful in principle), we can deduce that $\Omega$ cannot contain all of $\Gamma$. The reason is simple. The time-slice principle, one of the basic principles of quantum theory, requires that all of the theory's observables can be generated (as operators) by the observables associated with any nonzero time-interval. In particular, $\Omega_I$ must generate all of $\Omega$. More precisely: the von Neumann algebra generated by $\Omega_I$ contains all of $\Omega$. If $\Omega$ contained all of $\Gamma$, then the measurement-diagnosis condition introduced above would never be satisfied, because the projections $A_k|\psi\rangle$ can always be significantly mixed with each other by some operator in $\Gamma$. Therefore, $\Omega$ cannot include all of $\Gamma$.

This explains why the set of feasibly-measurable observables cannot include all self-adjoint operators, not even all self-adjoint operators whose spectra are purely discrete with a finite number of eigenspaces.

Application to single-site observables

The argument shown above is general, but it's non-constructive: it doesn't tell us which self-adjoint operators do or don't represent observables that could feasibly be measured. That would probably be too much to expect from such a simple argument.

However, for the case described in the question, the argument is sufficient, at least if we interpret the question with a little bit of poetic license. The question asks about the feasibility of measuring an observable localized at a single lattice site. Using a little poetic license, and motivated by symmetry, I'll interpret this to mean that $\Omega$ is assumed to contain all observables associated with individual lattice sites, links (nearest-neighbor pairs), and plaquettes (smallest squares). Including plaquette observables is important because the gauge fields (the EM field and the gluon field) don't have any observables associated with individual sites, but they do have observables associated with individual plaquettes. Including link observables is important so that the argument works. Hopefully that's still consistent with the spirit of the question.

But if $\Omega$ includes all observables associated with individual sites, links, and plaquettes, then the same kind of reasoning used above implies that no observable localized in any finite region of spacetime can ever be measured! That's because all such observables in QED+QCD are generated by the site-, link-, and plaquette-observables.

To avoid that absurd situation, we must exclude at least some site-, link-, and/or plaquette-observables from $\Omega$, and then our love of symmetry compels us to exclude all of them. We can still take $\Omega$ to include local observables that are smeared over regions that are large compared to the lattice spacing, as explained in ACuriousMind's answer. Such smeared observables are sufficient for all practical experimental purposes, if we take the lattice spacing to be small enough.

Comments

• The measurability criterion shown above cannot tell us what $\Omega$ should be. It can only test whether or not a given ansatz for $\Omega$ is self-consistent. The same thing is true for other general principles like the time-slice axiom and microcausality. They are conditions that the set of observables should satisfy, but they don't tell us what the set of observables should be.

• I described an $\Omega$ that doesn't satisfy the measurability condition, but I didn't describe one that does. I didn't even prove that one exists. I don't think that's been done before, maybe because it requires choosing an arbitrary threshold. Choosing a threshold amounts to choosing a definition of measurement. Measurement is a physical phenomenon, so its definition is inherently ambiguous, just like the definition of "river" is inherently ambiguous. Choosing an arbitrary threshold doesn't make the ambiguity go away.

• A good comment by the OP challenged my assumption that $\Omega$ should satisfy the time-slice property. In physical terms, the time-slice property says that if we use observables at one time to characterize the state, then the state's characterization in terms of observables at any other time should be uniquely determined. If that becomes untrue when we limit it to measurable observables, then physics would be even less predictable than quantum theory's probablity rules indicate. That seems unlikely, which is why I felt comfortable applying the time-slice property to $\Omega$, but beware that this hasn't been studied in any depth as far as I know.