Can we ever "measure" a quantum field at a given point?

Question

In quantum field theory, all particles are "excitations" of their corresponding fields. Is it possible to somehow "measure" the "value" of such quantum fields at any point in the space (like what is possible for an electrical field), or the only thing we can observe is the excitations of the fields (which are particles)?

Answer 1

For linear bosonic fields it might be possible -- in some sense -- to evaluate the field at a given event of the spacetime, otherwise there should not be a classical corresponding satisfying the same equations (think of the Maxwell equations). Fermionic fields are not directly observable since they violate causality (associated currents do not, however).

I henceforth restrict attention to a real scalar bosonic (Klein-Gordon) field $\phi$.

As amatter of fact $\varphi(x) =\langle \Psi| \hat{\phi}(x) \Psi\rangle$ makes sense if $\Psi$ is a coherent state. In concrete cases coherent states show properties expected by macroscopic systems. In general however $\hat{\phi}$ has to be understood as a operator valued distribution and the theoretical object which enters the theoretical arguments of QFT is $$\hat{\phi}(f) = "\int_M \hat{\phi}(x) f(x) d^4x"$$ for a smearing (smooth) function $f: M \to \mathbb{R}$ (or $\mathbb{C}$ depending on the used formalism).

If $\Psi$ is coherent $\langle \Psi| \hat{\phi}(f_n) \Psi\rangle$ converges to a number if $f_n(x)\to \delta(x,x_0)$ in weak sense, and thus the natural intepretation of this limit is $\langle \Psi| \hat{\phi}(x_0) \Psi\rangle$. This idea is corroborated from the fact, for instance, that $M \ni x \mapsto \langle \Psi| \hat{\phi}(x) \Psi\rangle$, defined as above, satisfies the classical equation of motion (Klein-Gordon for us).

Personally, after many years I deal with this stuff (and I am supposed to be an expert), I still find unsatisfactory this general viewpoint when I have to compare the theoretical approach with practicical procedures in laboratories. Barring the case of coherent states and $f$ replaced by $\delta$, how can I choose, in practice, the smearing function $f$? What happens if I change $f$ to $g$? I mean, how can I pratically check if the function is $f$ instead of $g$? Roughly speaking one may say something, for instance, it seems that the supports of the functions play a relevant role, e.g., dealing with causality issues. But the details of a given $f$ seem not important. There is a too large redundancy apparently in the formalism. And, finally, how to relate $f$ with, for instance the measurement instruments? E.g., Is the support of $f$ the region occupied by the laboratory?

There is an alternative intepretation which uses a smearing procedure with solutions of the motion equation $\psi$:

$$\hat{\psi}[\psi] := ''\int_{\Sigma} \left( \hat{\phi}(x)\partial_n \psi(x) - \psi(x)\partial_n \hat{\phi}\right)\: d\Sigma(x)''$$
where $\Sigma$ is a spacelike (Cauchy) surface with normal unit vector $n$.

There is a relation between classical solutions $\psi$ and associated one-particle states $|\psi\rangle$ of the field. If $\Psi_0$ is a Gaussian vacuum state $\hat{\phi}[\psi]\Psi_0 = |\psi\rangle$ and so on by applying many times the smeared field operator to the vacuum state, according to the Fock space structure. This use of the field operator plays a central role in various aspects of the formalism like in the LSZ formulas.

This notion is affected by a smaller number of intepretative issues since, after all, we know the meaning of $\psi$. The two intepretations are realated

$$\hat{\phi}[Ef]= \hat{\phi}(f)\:,$$

where $E: C_c^\infty(M) \to S$ -- $S$ being the space of solutions of the field equations -- is the causal propagator.

The map $E$ is not injective, so an answer to my previous issues can be searched along the idea that actually $f$ and $g$ are equivalent if they give rise to the same solution $Ef=Eg$. This point of view is consistent even if $f$ and $g$ may have disjoint supports. Unfortunately the smearing procedure with solutions is not allowable (for technical reasons) when the theory is (self-) interacting, i.e., when it is physically interesting!

My impression is that the intepretation of $\hat{\phi}(f)$ where, roughly speaking, $f$ accounts for the spacetime region where we perform experiments is too naive, though popular. $\hat{\phi}(f)$ essentially is a theoretical object, useful to implement various notions like locality and causality. However, its relation with experimental practice is not yet completely understood.

Answer 2

Yes, you can measure a quantum field at a point, in the following sense. We know that quantum fields are actually operator-valued distributions, so a quantum field operator comes as $Q(f)$ where $f$ is a test function on spacetime. Formally, this can be read as an integral of an operator valued function $Q(x)$ depending on single spacetime coordinate points $x$ against $f(x)$ with respect to $x$. $Q(x)$ is formally the "quantum field at a point" that one wants to measure. One can show under very general circumstances that $Q(x)$ is not an operator. However, there is a (in a suitable sense, dense) set of state vectors $|v \rangle$ so that the expectation value $\langle v|Q(x)|v \rangle$ has a well-defined, finite value which arises as limit of $\langle v|Q(f)|v\rangle$ when the test function $f$ tends to the Dirac-delta distribution concentrated at $x$. (One may want to assume that the quantum field is an "observable field" and invariant under gauge transformations etc.) Now here comes the catch. I mentioned above that $Q(x)$ is not an operator. This means that typically (and again, this can be proved under very general assumptions) the expectation value of the "squared field" $ \langle v| Q(x) Q(x) |v \rangle$ does not exist (is infinite). Thus, one can measure the expectation value of a "quantum field at a point", but that may not be of much use (yet that depends somewhat on the context, or what further conclusions one wants to draw) since the statistical variance, which is defined using $\langle v| Q(x) Q(x) |v\rangle$, is infinite. In other words, on making repeated (statistical) measurements of the quantum field at a point $x$ (if that were practically feasible) to determine $\langle v|Q(x)|v\rangle$, a distribution of values with an infinite spread is obtained - which however has a well-defined, finite mean value.

Answer 3

One must be careful not to confuse formalism for physics. What do we mean when we talk about a field? In quantum field theory (which is a formalism) a field is often represented by an operator. What does that operator represent in the physical world? The field operator is in fact used to define the observables (correlation functions and all that). These observables are then used to observe states (which are the things that we have in the physical world). In other words, the measurement would look like $\langle\psi| \hat{\phi}(x) \hat{\phi}(y) |\psi\rangle$, where $\hat{\phi}(x)$ is the field operator and $|\psi\rangle$ is the state.

So what would it then mean to "measure a field"? Another way to look at it is to consider the path integral formulation of quantum field theory where the field becomes an integration variable for the path integral. In that sense, the field is any solution of the equations of motion. In certain cases, one can have a state associated with a specific solution. In that case the solution becomes a parameter function for the state. When one makes measurements of that state, one can in fact observe (aspects of) its parameter function, which may then in some sense correspond to the idea that one "measures the field."

Does that make sense?

Answer 4

The comment by flippiefanus is curious. First, that contributor criticizes that the answer given is - if I understand correctly - too formalistic, to then go on and refer to the path integral formulation using hypothetical paths which are even more formalistic. Especially if one takes into account that the paths (or here: field configurations) which contribute significantly to the (the measure of the) path integral are extremely "rough" and are not smooth or continuous. In my view, that is far more formalistic than the idea that one could measure a field at a point. Yes, that is clearly an idealization. But pointlike idealizations appear elsewhere in physics, like point particles in classical meachnics. This idealization means the dimensions of an object or, in this case, the process of measurement, are very small compared to other reference items used. As the formalism of quantum field theory then shows (and therein lies its strength), if that can be assumed for a quantity like "field strength" (by its effect on test objects) in the sense of expectation values, then it cannot be assumed for the uncertainties. Remember that quantum field theory is a statistical theory so a single experiment of "field strength measurement in a small extension of space and time" by action on a test object must be repeated multiple times to have a reliable statistics to determine mean values and variances and such. The divergence of the variance for measurements for the true "pointlike" measurement localization limit is then the expression of the quantum mechanical uncertainly relations. This is how the idealization is to be read. Seen in this way, the formal elements of quantum field theory have a clear relation to operational measurement procedures.

Answer 5

Quantum Fields can't be physical, you can see this from the Equivalence Theorem which states that if I have a quantum field $\Phi(x)$, I can perform a field redefinition in my action $\Phi(x)\rightarrow \Phi'(x) = f(\Phi(x))$, so that as long as $f(\Phi(x))$ satisfies some simple properties, all S-matrix elements (basically everything we can measure) are invariant. The value of the field can't possibly be an observable.

Answer 6

In QFT, it's not possible to measure the value of quantum fields at any point in space. This is because quantum fields are not in spacetime (per the Copenhagen Interpretation, Transactional Interpretation, and others which include the concept of wave function collapse). They are calculated entities which we infer from the behavior of particles (which are in spacetime). When a measurement is made, a particle appears. For example, when a photon hits a photographic plate, it is a real thing and we can measure it's position. But the underlying electromagnetic quantum field can only be calculated.

The wave function calculates the quantum field at any point. But the calculation does not tell us the strength of the field. It tells us the probability of detecting (measuring) a particle.

For example, let's say that we're talking about the electromagnetic field and the calculation is for predicting the likelihood of detecting a photon in a particular position. Here's an image of the relationship of the quantum field to the particle:

The red grid graphs the wave function's calculation of the probabilities that a particle will be detected. The green film shows where a photon has actually been measured in spacetime. In this sequence, its path has been measured in 4 positions.

This image is a still from an excellent 5-minute film by Fermilab on QFT 3. It addresses your question. Also, see this article on measurement in quantum mechanics in an encyclopedia for laypeople which addresses this issue in straightforward terms.