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Feynman path integrals and Schrodinger wave functions are calculationally equivalent. Any calculation of a physical measurable quantity will have to match in both cases if done right. This does not answer the underlying ontological status of both formalisms, though. Which formalism has more ontological reality and why?

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Maybe this should be asked in philosophy instead? – FrankH Oct 14 '11 at 0:43
Yeah, this question is the sort of thing that physics doesn't address. Check the Philosophy site. – David Zaslavsky Oct 14 '11 at 0:56
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The two formalisms are also "ontically" equivalent, since the wavefunction is the physical object in both. The path integral is just a different tool for getting the evolution of the wavefunction. For a genuinely (slightly) ontically inequivalent formulations, you can consider wavefunctions vs. density matrices. – Ron Maimon Oct 14 '11 at 1:10

closed as off topic by David Zaslavsky Oct 14 '11 at 0:56

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1 Answer

Ontology is a philosophy question about the "true nature of reality" and is not a question that physics tries to answer at all. As you note, both the Feynman path integral and the Schrodinger wave equation approach give identical answers if done correctly, so from a physics point of view, both are correct and there is no reason to think that one or the other is a better answer to the philosophical ontological question of the "true nature of reality". So physics gives no answer to this question. We just let the philosophers argue about this question and we just get on with the business of creating theories that give better and better fits to all the experimental data that we generate.

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