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@PeterDonis another equivalent formulations is the operator Heisenberg picture, right? If it's equivalent, and we are discretising (ie removing all the theory and leaving pure numbers and artihmetic), why does the picture used matter?
If you are referring to the equivalence in non-relativistic QM between the Heisenberg and Schrodinger pictures, no, that does not exist in QFT either.another equivalent formulations is the operator Heisenberg picture, right?
That's not a good description of what one is doing when discretizing a continuum theory.discretising (ie removing all the theory and leaving pure numbers and artihmetic)
It's obvious, why one needs to explain a toy model when looking at your struggeling with it: It's to teach the next generation of physicists about the methodology how to tackle real-world QFT, and indeed QCD is a pretty delicate subject. So it's good to first study simple toy models first. As already Platon knew, there's no king's way to the wisdom. You have to go the whole way from the beginning to the end. There's no shortcut, and indeed you should learn physics from good text books and then from physics papers rather than from youtube videos!@PeterDonis the model is a toy model. I do not see how 100 years of QFT is needed to explain a toy model.
This is true, and the answer to the question I asked you in post #24 will help to clarify this. (Hint: "adjacent times" are "adjacent cells", in the time direction on the lattice.)the time evolution is always defined in the end in terms of how ##\Phi## interacts between adjacent cells or adjacent times.