I'm watching some lectures on Relativistic Quantum Information and in one of them - available here - the instructor is talking about measuring quantum fields.

He then says basically that in real experiments, we don't measure quantum fields directly with projective measurements as in non-relativistic quantum mechanics, but that we rather couple the system to a non-relativistic particle detector and then measure the detector.

If I'm getting it this will ultimately lead to things like the Unruh-deWitt detector.

Now in fact in the QFT course I've taken measurements on quantum fields weren't really covered, so I'm wondering about what he said.

My question here is the following: in practice, always in real world relativistic QFT experiments (like in LHC and other particle accelerators), is what he said true?

Does a measurement of a quantum field always occurs by coupling to a first-quantized system and measuring the first-quantized system?

To be honest, I find this quite weird, because if that's the case, then there would be a limitation imposed by the detector on what we can actually measure. This seems to never be mentioned in non-relativistic QM.

So if this is true, why it must be done that way? Why can't one do a projective measurement on a quantum field directly?

  • $\begingroup$ It's true. Think of how we measure electric fields, classically, for example. $\endgroup$ – Gabriel Golfetti Mar 29 '19 at 23:45
  • $\begingroup$ From what I know, in real world QFT experiments, the field (aka a function of position) isn't actually measured. The predictions at the LHC come from projective measurements in momentum space, with the same formalism as in QM albeit with a different momentum operator etc. $\endgroup$ – doublefelix Mar 30 '19 at 0:30
  • $\begingroup$ If single particles are the smallest quantized units of excitation of the field, why would you expect to be able to measure something other than the smallest unit of excitation? Put another way: what kind of measurement do you think could return something other than a “particle” signature? $\endgroup$ – JPattarini Mar 30 '19 at 3:40

Non relativistic quantum mechanics has solutions of two body interactions and can predict spectra of atoms. When many bodies are involved then one can use a quantum mechanical field theory .

A qm field theory assumes for each particle/body involved in the interactions a field, taking the plane wave solution as representing all over space the particle, and quantum mechanical creation and annihilation operators generate the many body systems in scattering and decay situations, which can be calculated using Feynman diagrams. Quantum field theory is a useful tool in many branches of physics, for example here is a publication about a field theory for atomic nuclei.

In particle physics the assumption is that all elementary particles in the table are fields covering all space time , the fields represented by plane wave solutions of the corresponding equations (klein gordon, dirac,....), on which fields, creation and annihilation operators propagate the particles, and the calculation of measurable quantities happens through Feynman diagrams , which is where they were first applied, to calculate two to many crossections and decays.

My question here is the following: in practice, always in real world relativistic QFT experiments (like in LHC and other particle accelerators), is what he said true?

From the above definition of a field, of course you cannot measure it as you cannot measure the wavefunction, but only the probability of $Ψ*Ψ$ can predict the data to be seen. The data itself is secondary interactions in the detectors which will measure the energy momentum four vectors, at the LHC , or the space-time footprints in the double slit experiment one electron at a time.

It is not a matter of first and second quantization, unless a very specific experiment is designed.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.