How does QFT treat Young’s Double-Slit Experiment?

[Quant]

TL;DR Summary

The QM wavefunction/probability distribition in view of QFT modes (in cavity)

How does QFT treat the Young’s DSE? Is there a wave function (wave packet) attached (and created at the moment of launching of the photon) or the modes of the EM quantum field are pre-existing due to experimental configuration (including the screen) and do they play the role the wave function is supposed to do in QM?

 

[Nugatory]

The interference demonstrated by Young's double-slit experiment is a classical phenomenon explained by the classical wave solutions of Maxwell's equations of classical electrodynamics - there is no quantum mechanics involved. Quantum mechanical double-slit experiments are a different thing, and it is unfortunate that the analogy between these and Young's experiment is so often used in popularizations.

The essential characteristic of the quantum-mechanical experiment is that the interference pattern builds up one detection at a time, a behavior that is cannot be explained by classical physics. At a very rough level, the quantum field theory explanation works by writing the electromagnetic field as a sum of oscillatory modes (in momentum space, not position space); in this form we can calculate the probability of a particular mode absorbing or releasing one quantum of energy at a given point in spacetime.
If you are not already comfortable with this approach to quantum physics, you could do worse than working through Lancaster and Blundell's book "Quantum Field Theory for the Gifted Amateur".