My problem appears to be with the initial state of the system, which I have written as
$$\left| \Psi\right\rangle=\left|\psi_1\right\rangle\left|\psi_2\right\rangle+\left|\psi_2\right\rangle\left|\psi_1\right\rangle,$$
where $\left|\psi_1\right\rangle$ is packet from one source, and $\left|\psi_2\right\rangle$ is packet from another one.
This state says that the system is in a superposition of states, in each of which one of the particles comes from one source, and another necessarily from another source. I.e. the system is highly entangled. Such system could be created e.g. by some generator of pairs of particles with opposite momenta.
But two independent sources are clearly not such a source of entangled pairs. As the particles are indistinguishable, and there's no symmetry which would allow us to determine that the particles come from different sources, we can't say which source the particle has come from. If we watch single particle emitting from the sources, they might come one after another from different sources, or they could repeatedly come from single source, then several times from another one. I.e. there's no rule that if one particle is from source A, next detected one is from source B. So, the initial state must be in the following form:
$$\left|\Psi\right\rangle=\left(\left|\psi_1\right\rangle+e^{i\phi}\left|\psi_2\right\rangle\right)\otimes\left(e^{i\psi}\left|\psi_1\right\rangle+e^{i\chi}\left|\psi_2\right\rangle\right)+\\ +\left(e^{i\psi}\left|\psi_1\right\rangle+e^{i\chi}\left|\psi_2\right\rangle\right)\otimes\left(\left|\psi_1\right\rangle+e^{i\phi}\left|\psi_2\right\rangle\right),$$
where $\phi,\psi,\chi$ are constants, which depend on the experimental setup.
Now from the form of the initial state it's obvious that the interference pattern will be present, and it's confirmed by numerical simulation.
Thus, it appears that even in this multiparticle experiment particles interfere with themselves, rather than with each other, to produce visible pattern on the screen.
When a quantum of light arrives at a double slit, it passes through both slits as a wave and arrives upon a second screen with the interference pattern of a single wave that has been split into two waves, that have then interfered with each other.
This is not correct. The photons arrive one at a time whole, not split in space. In any case, in quantum mechanics what is waving is the probability of detecting the particle not the particle itself.
Here is the double slit experiment displaying one photon (quantum of light) at a time, and what happens when many photons are accumulated.
Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.
At the frame on the far left the footprints of the individual photons are seen. The photons do not leave a signal all over the place, they hit at a specific (x,y)at a distance z, according to the probability of the solution for the setup "photons scattering off two slits with specific width and distance". This probability is given by the $Ψ*Ψ$ of the specific wavefunction and it looks random in the first frame on the left.
The accumulation of photons shows the classical interference pattern, which for the quantum level means the probability distribution $Ψ*Ψ$.
A detector after one of the slits intercepting the photon, changes the boundary conditions to a different system, and thus a different $Ψ*Ψ$. It is no longer the same experimental setup. It should be obvious that if the detecting instrument after the slit , absorbs the photon like the screen does, only the untouched slit will give a signal on the far screen, which could not interfere with itself .( A sophisticated experiment with electrons which tries to minimally show the effect came to the conclusion that the detecting level acts as a point source for the electrons going through it, i.e. a different $Ψ*Ψ$ for the electron which is no longer coherent so as to show the interference pattern.)
Therefore can one assume that detection has 'caused' the collapse of the wave portion of the duality?
Detection at the screen has picked ("collapsed ")an instance of (x,y,z) of the original wavefunction and removed that photon from the final screen. In general after the detection of "which slit" the photons are in a different wave function with new boundary conditions.
How has detection precisely influenced the duality? Can anyone clarify?
The duality is not affected by detection, the mathematical model that describes the probabilities , $Ψ*Ψ$, has a different Ψ because the boundary conditions have changed and the coherence necessary to display interference is lost.(coherence in the phases describing the photons in spacetime). Again, the term wave particle duality has to do with the mathematics of the quantum mechanical probabilities. The probability is a wave, (a solution of a quantum mechanical system) the particle manifests as a point in (x,y,z,t) when interacting in a measurement, in accumulation of many particles with the same boundary conditions, the probability distribution is built up.(It is the same as throwing dice. The probability distribution versus the numbers 1-6 is seen in the accumulation of many throws).
Best Answer
When we are talking of elementary particles we are talking of quantum mechanics.
The wave nature of quantum mechanics comes because the equations are wave equations and the solutions of these wave equations squared have been defined , Born rule, as the probability of observing the particle at an (x,y,z,t). Thus interference in a quantum mechanical setup means: interference patterns in a probability density distribution, not in energy or mass .
The photons, as elementary particles, due to the peculiarity of their masslessness and the Maxwell equations have the same frequency in the single photon double slit interference patterns ( probability distributions) as the frequency displayed by the electromagnetic wave that may emerge from a huge number of photons. (The classical EM wave does display interference patterns in its energy distribution, hence the confusion between classical and quantum interferences).
Now two single particles quantum mechanically will also have a single solution in quantum mechanics that will be defined by the boundary conditions. These solutions will be different than if they are far apart and can be considered independent. Thus the probability of their manifesting in an (x1,y1,z1) (x2,y2,z2) at time t will be different and thus they may be considered to interfere with each other.
Consider an electron and a proton, many boundary conditions could exist:
a) a bound state governed by their potential
b) a resonance if the relative energy is higher than the hydrogen bound state
c) an elastic scattering both changing directions
d) inelastic scattering emitting a photon in each other's field
e) if the energy is high enough a generation of new particles due to the scatter
Different boundary conditions will show different dependances, but yes, they will interfere/change the probabilities for each other.