PeterDonis said:
No, they don't. Each particle is a degree of freedom of the system, and the wave function describes probabilities for the system. There is no general sense in which the individual particles "contribute" to the wave function.
Particles are not degrees of freedom of the system. The particle is the system. If it's prepared in a pure state, in non-relativistic QM the particle can be described by a wave function ##\psi(t,\vec{x})##, which obeys the Schrödinger equation,
$$\mathrm{i} \hbar \partial_t \psi(t,\vec{x})=\hat{H} \psi(t,\vec{x}),$$
where ##\hat{H}## is the Hamilton operator (in position representation). For a particle in an external potential it reads
$$\hat{H}=-\frac{1}{2m} \hat{\vec{p}}^2 + V(\hat{\vec{x}})=-\frac{\hbar^2}{2m} \Delta + V(\vec{x}).$$
The physical meaning is that
$$P(t,\vec{x})=|\psi(t,\vec{x})|^2$$
is the probability density for observing the particle at ##\vec{x}## at time, ##t##.
As demonstrated by the nice Youtube movie (though it uses photons rather than particles, where the description with a wave function is impossible and QED must be used, but the qualitative picture is of course the same, i.e., you can for sure find movies of this kind, where massive particles, electrons in the non-relativistic regime are used), a single particle does not appear as some "smeared cloud" but manifests itself as making one dot when registered by the detector (here obviously a scintillation screen or something similar, unfortunately the Youtuber doesn't give a scientific reference/paper, where to find the details of the experiment). Only when accumulating a lot of equally prepared particles going through the slits the predicted probability distribution forms. It clearly shows and interference pattern as the Schrödinger equation predicts.
Of course, also the wave-particle duality is resolved since 1926 with Born's probability interpretation, but that's another story.
PeterDonis said:
No, it doesn't.But the quantum wave function is not a wave with a source. It's not radiation. It's a different thing.
Indeed it's a representations of a single-particle pure quantum quantum state with the above given probabilistic meaning.
PeterDonis said:
No. In ordinary non-relativistic QM you can't add particles to the system. You have to include all the particles at the start.
Here you simply have one particle at a time going through a double slit and being registered at the scintillation screen (and it's position stored).
PeterDonis said:
In relativistic QFT, particles can be created or destroyed, but "particles" aren't fundamental anyway; quantum fields are. "Adding a particle" is just a particular quantum field operation. Viewing it as "changing the probability amplitudes in the system" is not a useful way to view what is going on.
Indeed, and photons cannot be in any sense interpreted as particles. They don't even admit the definition of a position observable in the strict sense. Nevertheless, what also QED provides in this case of the double-slit experiment is the probability for detecting a photon at a certain place on the screen (it's given by the energy density of the electromagnetic field), and this forms an interference pattern when many single photons are used. It's similar as for particles, although there's no "wave function" for photons.
PeterDonis said:
Your whole conceptual scheme here seems to be wrong. This question isn't even answerable.
I think the standard formulation of QT perfectly answers the somewhat vague question in #1.
