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I have a train of thought which leads me to believe that we should be able to observe a fraction of a particle. Pelase help me:

  1. In Quantum Field Theory, we model particles as total energy eigenstates of the field energy. This accurately models the probabilities of particle creation/annihilation observed in the lab.

  2. The total energy is $\int H (x) d^3x$. This is a non-local quantity. The only way we'd have been able to measure this is if we were an omnipresent being performing a measurement on the entire $R^3$ space at once.

  3. We should only be able to measure local quantities like the energy density $H(x) $. In spite of this, point no. 1 above accurately models scattering experiments.

  4. So it must be that the total energy measurement serves as a good approximation for the local energy density measurement.

  5. One we to reconcile point 1. and 3. that I can think of is : Let's say we take a total energy eigenstate $|p\rangle$. And then we plot on $R^3$, the expected value of the energy density $f(x) =\langle p|H(x)|p\rangle$. (Since $|p\rangle$ is not a physical state, take it to be a Gaussian). If $f(x) $ is sharply localised near some $x$, and $0$ everywhere else, then an energy density measurement at that point would serve as a good approximation to the total energy measurement. So, if we happen to have an energy measurement device located at that point, we'd end up measuring the total energy as a whole particle (instead of a fraction).

Is everything correct upto this point? Moving on:

  1. Point no. 5. should only approximately apply to high momentum particles, that we deal with in scattering experiments. This is because point no. 5 approximately simultaneously assigns a position and a momentum to a particle. For high momentum particles, we can approximately bypass the uncertainty principle by having a Gaussian with a sharp mean position and momentum.

  2. So, as we move to the $v\ll c$ regime, the the plot of $f(x)$ should be more spread out on $R^3$. This means that, if we attempted to measure the energy density $H(x) $ at a point, we'd end up measuring a fraction of the total energy.

  3. However, from non-relativistic QM, we know point no. 7 is not true. Whenever we attempt a measurement, either the entire particle gets summoned to the point with probability $|\psi (x) |^2$, or we observe no particle. There is no such thing as detecting a fraction of the particle.

I really want to understand where I'm going wrong. If my reasoning upto point 5. was correct, I want to know how to derive point no.8 from it in the non-relativistic regime, i.e. how do we derive the experimentally observed fact that the entire particle gets summoned with probability $|\psi (x) |^2$?

J.G.
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Ryder Rude
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    Your question seems to boil down to: "how can we have localized particles in quantum mechanics, if particles are momentum eigenstates?" If you agree with that simplification (maybe I oversimplifed it!), then my response would be that to understand the outcome of a position measurement you are better off working with states in the position basis. In relativistic QFT this gets hairy, but in non-relativistic quantum mechanics there is no problem. You can work with the subspace of single particle (momentum) states in the full Fock space, and superpositions of these are well localized. – Andrew Jan 12 '23 at 05:33
  • Section 2.8.1 of Tong's notes might be relevant: https://www.damtp.cam.ac.uk/user/tong/qft.html – Andrew Jan 12 '23 at 05:33
  • @Andrew Thanks for the link. Is it possible to define the position operator from local field operators? In my point 5, it seems like there is a correspondence established between a position measurement and an energy density measurement. But this correspondence does not hold in the non-relativistic limit, as in the limit, the energy density measurement would only measure a fraction of the particle, while the position measurement summons the entire particle – Ryder Rude Jan 12 '23 at 05:50
  • @Andrew I want to be able to derive the "particle summoning" feature of non-relativistic QM from the energy density measurements (or some other field density measurement) of Quantum Field Theory. It seems to me like we artificially define a position operator and just postulate that it summons the particle, instead of defining it in terms of field density operators and deriving that it summons the particle. – Ryder Rude Jan 12 '23 at 05:52
  • @Andrew Since the entire theory is based on the correspondence between classical field densities and quantum field operators, it feels a bit artificial to define a position operator directly in terms of its eigenstates and postulate its interpretation as "probability of observing a particle at position $x$ in physical space". This operator doesn't have a direct counterpart in classical field theory, either in relativistic fields or the Schrodinger field. But in special cases like geometric optics, we can define a position observable in the classical theory using the energy observable. – Ryder Rude Jan 12 '23 at 06:08
  • One of the potential issues here is that the field operator and the number operator are incompatible observables -- they don't commute. So there is a tension between looking at observables defined in terms of particle eigenstates--like the position of a particle--and observables like energy density that are defined in terms of field operators. – Andrew Jan 12 '23 at 14:01
  • The "why" is that we don't. That was a great surprise when discovered, at least for photons. The theory can't possibly explain this, since the theory was crafted to match the phenomena. – John Doty Jan 12 '23 at 22:44
  • #5: I'm stuck on: how does one measure energy density? I may have completely missed the point, but it seems that your argument hinges on that question. – garyp Jan 13 '23 at 02:57
  • @garyp Point 3 was that we should only be able to measure the energy content of the local region in which the measurement device is located, and not the total energy of the entire $R^3$ space. Inspite of this, we are able to make accurate predictions by pretending that we're measuring the total energy operator. In point 5, I calculated the expected value of the energy density operator on the wavefunction. I argued that if its plot is concentrated near some $x$, then we should be able to approximately measure the total energy by having our measurement device located at $x$ – Ryder Rude Jan 13 '23 at 03:44
  • @garyp also think about the classical theory. One will only be able to measure the local value of the electric fields by conducting experiments. – Ryder Rude Jan 13 '23 at 04:07
  • I guess I wasn't clear. What device do you use to measure energy density of a quantum field? – garyp Jan 13 '23 at 17:16
  • @garyp I mean that the energy measurement devices that we have don't actually measure the total energy of the $R^3$ space, but only of the local region. That is, they measure the integral of the energy density over a small region. I was sloppy with my wording (but the argument in the post is still fine).We don't measure the density. We measure the energy over a small region. – Ryder Rude Jan 13 '23 at 17:23

2 Answers2

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There are many important things to think about in your question but I don't agree with point 5. If $\left | p \right >$ were truly an energy eigenstate of a translation invariant theory, then $f_p(x) = \left < p | H(x) | p \right >$ would be independent of $x$. For $f_p(x)$ to make sense as a collider observable, it would need to be somewhat localized in position space. I.e. the state $\left | p \right >$ needs to be somewhat smeared in momentum space.

The most sensible thing to write is \begin{align} \left | p \right > = \int_{\mathbb{R}^3} \rho_p(k) a^\dagger_k \left | 0 \right > \end{align} where $\rho_p(k)$ attains its peak value at $k = p$. But I don't see how the particle being relativistic or not interferes with our ability to prepare states like this. As long as $\rho_p(k)$ is not so narrow that we lose all information about position, experiments can be built which give it any width. Whether $|p| \approx m$ or $p \gg m$. So that's my objection to the idea that the non-relativistic limit requires $f_p(x)$ to be more and more spread out.

Of course you can have states which are too spread out to be interpreted as wavepackets. Solitons are the ones that typically attract the most interest. But I don't see any barrier in principle towards designing an experiment which detects a fraction of a particle in the sense you're describing.

Connor Behan
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    I agree that you can create Gaussians even in non-relativistic QM. But in non-relativistic QM, we also learn about states that are almost plane waves. Even for these states, the entire particle gets summoned at a point with probability $|\psi (x)|^2$. We never talk about observing a fraction of the particle. – Ryder Rude Jan 12 '23 at 04:35
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You ask in the title:

Why can't we observe a fraction of a particle?

Define "observe" and define "particle".

Observations in our everyday world culminated, after the time of Newton, into measurements of velocities etc, in classical mechanics, starting with "classical particles" defined as tiny parts of solid matter. Particle of dust for example.

The theoretical models of classical mechanics electrodynamics etc, when brought to describe particles approaching in mass or dimension zero, break down, because 1/r singularities appear.

Quantum mechanics theory has solved this problem, and the current mainstream physics defines elementary particles as point particles, i.e. zero dimensions, which means they cannot be fragmented.

This axiomatic proposition in the standard model of elementary particles fits the data very well and is inherent in the quantum field theory models.

There exist other quantum field theories where what the creation and annihilation operators create are composites of elementary particles. There the mathematics you are contemplating might fit that data, but not the elementary particles of the standard model of particle physics.

It may be that future observations and experiments dispose of this axiomatic assumption by finding discrepancies with the standard model, but this is the status at the moment. Elementary particles are zero point particles and cannot be fragmented.

anna v
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    anna, the OP defines a particle as an eigenstate of a Hamiltonian and defines observe as calculating a matrix element of that Hamiltonian. You may disagree with their definitions, however it is fine at least as a catchy title for a SE question. Your answer potentially answers a different question, but certainly not this one. – ɪdɪət strəʊlə Jan 12 '23 at 17:39
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    If you notice, I am answering the title, which is what comes up in searches on the net. There can be a huge number of hamiltonians that do not connect with physical observations, and this is a physics site. Physics uses mathematics for modeling, itis not mathematics. That is why I added the other QFTs for physical observation. Back in 1963 I sat through a course that had a QFT for nuclear physics. – anna v Jan 12 '23 at 18:04
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    @ɪdɪətstrəʊlə You can define anything you want in mathematics. You cannot do that in physics: the phenomena are in charge. – John Doty Jan 12 '23 at 22:48