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

Question

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:

6. 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.

7. 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.

8. 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 $|\psi (x) |^2$?

Answer 1

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.

Answer 2

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.