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The concept requires all possible path's to be mapped out, and any remaining paths not cancelled out represent the most probable path of the object.

Considering this:

i) If "infinite" paths are mapped out (since there are infinite paths for every movement) , isn't it guaranteed for every path to be canceled out? Every path has an opposite, and if infinite paths are mapped, the opposites are guaranteed to be present.

ii) How do you map out infinite paths? Nothing infinite can be completely mapped, on the basis that it is infinite.

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Sorry, what are exactly "infinite paths" and what sort of cancellation are you talking about? Your question makes no sense to me. Feynman's path integral is an infinite-dimensional version of $\int d^n x \exp(iS[x_j])$ where $S$ is complicated enough function of the variables $x$. There's no cancellation that is both exact and trivial here. It's a complicated integral which produces a complicated result. We sometimes say that if $S$ varies quickly, most of the terms tend to cancel, so only places of stationary $S$ contribute in some classical expansion, but that's just an approximation. – Luboš Motl Apr 20 '12 at 16:18
For a physical understanding of why the contribution from certain paths "adds up", i.e., an answer to your first question, see Chap 19, Vol 2 of the Feynman lectures. – Vijay Murthy Apr 20 '12 at 16:21

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It's your comment:

Every path has an opposite

that's the problem. We're integrating the action along the path and it's not true that for every path there is a path with an equal but opposite action integral.

Re the infinite number of paths, summing an infinite number of quantities is exactly what integration does. The only problem is showing that the integral doesn't give an infinite result.

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This answer is ok, but actually, it is "true" in a matching sense that every path has a path with opposite sign, but as you correctly say, this doesn't make any difference for an integral. The way to make sense of the path integral is to give an infinitesimal imaginary part to the time, so that large action paths are exponentially suppressed, not just phase-cancelling, and this is glossed over by Feynman, but not by others. Feynman understood this intuitively, since it is how he makes sense of oscillatory integrals in practice--- he puts an exponential cutoff on them and relaxes the regulator. – Ron Maimon May 1 '12 at 21:43

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