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I have heard it said that Richard Feynman was a proponent of a particle approach to QFT while Julian Schwinger preferred a local fields description. What is meant by “local fields”? Surely when one uses Feynman diagrams to calculate Green’s functions, for example en route to calculating scattering amplitudes with the LSZ reduction formula, they are evaluating time correlation functions of field operators through a sum of Feynman diagrams. In what ways are Feynman diagrams a “particle approach” and what are “local fields”?

Qmechanic
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Ian
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1 Answers1

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Local fields refers to Quantum Field Theory, which, for elementary particle physics, assumes that for all particles in the table (and their antiparticles) in every space-time point there exists a field represented mathematically by the plane wave solution of the appropriate quantum mechanics wave equation, covering all spacetime. On these fields creation and annihilation operators create and annihilate the particles measured in the laboratory.

Feynman diagrams show input real time four vectors of interacting particles, interacting through virtual exchanges with the appropriate forces, leading to real time fourvectors as the output.

virtual

Internal lines are place holders, carrying the correct quantum numbers for the particle indicated except it is off mass shell, because it is under the integration for the specific calculation, in the case above electron electron repulsion.

The two approaches are mathematically connected. I remember back in 1963 , after spending a semester solving field theory problems for Bogoliubov's book, going to a CERN school and learning what the whole fuss was about during the lectures by Veltman! The whole fuss after all is for calculating crossections, decay rates etc.

Please note that quantum field theory is a mathematical tool that can be used for other quantum systems than elementary particles.

anna v
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  • But aren’t Feynman diagrams simply a pictorial representation of the terms in the perturbative expansion of time correlation functions of field operators? – Ian Sep 12 '19 at 06:32
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    Yes, except they give a very easy recipe to calculate physical results, which is not easily discovered when fighting with creation and annihilation operators, the way we had to, in our course when I was a graduate student. – anna v Sep 12 '19 at 06:40