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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)?

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    $\begingroup$ Does the electromagnetic field count? $\endgroup$ – probably_someone Dec 11 '18 at 0:17
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    $\begingroup$ Notice that technically you can't even measure classical fields (what you measure are the forces and the equation of motion). $\endgroup$ – gented Mar 17 at 3:30
  • $\begingroup$ @gented Ok but then I'm going to argue that you can't measure force either. You just measure the position of the needle on your force-o-meter. $\endgroup$ – DanielSank Mar 28 at 23:59
  • $\begingroup$ I think you can't measure them since fields in QFT are a superposition of creation and annhilation operators, so they "live" in Fock space. Actually, most of them (except real scalar fields, i.e., self-adjoint) are not self-adjoint so accordingly to QM postulates they can't be related to observables quantities. QED's potential (phtoton field) is related to "measurable" quantity (magnetic and electric potentials) because each componente is a real scalar field. I use quoting mark in <<measurable>> due to the right comment given by @gented (cont.) $\endgroup$ – Vicky Mar 29 at 0:13
  • $\begingroup$ (cont.) You determine them up to a value given by gauge invariance $\endgroup$ – Vicky Mar 29 at 0:13
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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: Red grid represents quantum field; Green film represents particles in spacetime.

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.

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    $\begingroup$ Unfortunately, this answer appears to try address a different question than the one that is being asked. $\endgroup$ – Peter Kravchuk Mar 17 at 4:13
  • $\begingroup$ I have added a few sentences at the beginning to more directly answer the question. $\endgroup$ – AlexH Mar 28 at 23:35

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