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Do we observe dots on a position detector like in the double slit experiment? I doubt that because the double slit experiment is non-relativistic QM, and so a "position measurement" is defined. However, there's no "particle position" operator in Quantum Field Theory.

In ordinary QM, I visualised particle collisions by thinking of the colliding particles' wave functions in the position space. I thought of two wave functions, each with a sharp enough mean position, heading towards each other.

However, in QFT, the wavefunction is a wave-functional. There's no position operator anymore. There's no mean position. Are the wavefunctionals heading towards each other somehow?

What exactly are we colliding in particle accelerators and why do we call them particles? What kind of entity/information do we detect in particle accelerators? Are we detecting a small quantum of energy? But we must be detecting it at some position, right? Wherever the energy measurement device is located. So does it mean that particles are a quantum of energy but located at a position?

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Do we observe dots on a position detector like in the double slit experiment?

Of course we do. They are classical, relativistic BB pellets, for all intents and purposes, as they don't interfere with each other. The position reflects the angle of their momentum vector, and we measure their momentum and energy: we basically work in momentum space--this is where the infinite coupled oscillators in QFT decouple, after all!

However, there's no "particle position" operator in Quantum Field Theory.

It is true there is no crisp position operator used in HEP, but that's only because it is basically useless: at the level of detection there is no particle interference, as quantified in the linked question.

However, in QFT, the wavefunction is a wave-functional. There's no position operator anymore. There's no mean position. Are the wavefunctionals heading towards each other somehow?

The wave functional formulation is a terrible, terrible, formulation of QFT for the purposes of visualization! (It only helps some in formal correspondence procedures...) The conventional canonical formulation relies on quantum field operators, quite narrowly localized, in practice! These are instrumental in describing the scattering process in the minuscule interaction region: a fireball of the size of fermis!

The proper visualization might entail two classical BB pellets headed at each other, colliding on occasion, and acting quantum in the region of collision; then, once out of it, the products being many classical pellets, gears, cherries, gnats, leaving the interaction region as classical objects, and hitting detectors, detected as basically classical objects: lumps of momentum and energy. QFT was crucial in calculating what happened in the interaction region, but hardly operative outside it.

So does it mean that particles are a quantum of energy but located at a position?

Of course: given the huge momenta and energies involved, most of the time, the uncertainty principle that you imagine is relevant hardly enters at all! Again, outside the interaction region, particles are quite classical.

NB. Caveat for fussbudgets: To be sure, there are QM interference phenomena along the way in special types of experiments, involving neutrino or flavor oscillations, but these are outliers and merit separate discussions, where paradoxes of the UP do enter. But that is a very different question... A further quantum effect is HBTwiss-like interference patterns at the detector, magnifying quantum interference phenomena in the interaction region, not outside it.

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  • $\begingroup$ The QM interference in "special types of experiments" is seen statistically from the detection of multiple particles. Experimentally, the only waves you can directly detect are classical, where QM models tell you that you are in the Correspondence Principle limit of high quantum number. $\endgroup$
    – John Doty
    May 27 at 15:20
  • $\begingroup$ Don't all formulations have to deal with a wavefunctional? The interaction picture is half Heisenberg half Schrodinger, so there must be a wavefunctional. The path integral formulation also describes a wavefunctional $\endgroup$
    – Ryder Rude
    May 27 at 15:23
  • $\begingroup$ So, outside of the tiny interaction region, QFT can be approximated by high speed billiard balls. $\endgroup$
    – Ryder Rude
    May 27 at 15:35
  • $\begingroup$ @ Ryder No, wavefunctionals are a technically sweet, but definitely off-mainstream formulation of QFT! Normally one computes correlation functions of quantum fields, often perturbatively. The path integral is an alternate technique to compute those, but without wavefunctionals. Yes, in empty space, particles are ultra relativistic billiard balls. (I chose BB-shot to connect to speed: 10 ppl off the speed of light for LHC protons, these days...) $\endgroup$ May 27 at 16:27
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    $\begingroup$ @RyderRude Before you study formulations (mathematics), you should study the experiments (physics). $\endgroup$
    – John Doty
    May 27 at 17:31

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