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The paths of the Feynman path integral are not actually taken. The phrase "takes every possible path" is a mangled statement of the mathematical instruction to take the integral of $\exp(-\mathrm{i}S)$ over all possible paths for the action $S$ to get the probability amplitude of something happening. It is a fact of quantum mechanics that this integral ...


5

I would like to add a few things to ACuriousMind's answer. What Greene most certainly intends to say is every path(even faster than light ones i.e. those which are not time-like everywhere) contributes to the propagation amplitude. In fact,since every path in space-time contributes with equal weight, there are also paths which go "back in time" and "come ...


2

The way to see to this a la Peskin and Schroeder is by staring at the renormalized propagator in equation 7.75, \begin{equation} P _{ \mu \nu } \equiv \frac{ - i g _{ \mu \nu } }{ q ^2 ( 1 - \Pi ( q ^2 ) ) } \end{equation} In $ 4 $ dimensions, $ \Pi ( q ^2 ) $ doesn't contain a pole (as expected) and so, \begin{equation} P _{ \mu \nu } = \frac{ - i g _{...


1

I'm afraid that linear polarization is not as interesting an example as you may have hoped. First, the answer: in quantum optics, whether or not a quantum state exhibits linear polarization is independent of the photon count for that state. A single-photon state can be linearly polarized. Now, the explanation: in quantum electrodynamics (QED) it is ...



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