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Let's assume we are in a perfect vacuum and let a photon travel from a point $A$ to a point $B$. As far as my understanding of quantum physics goes, the photon takes all possible paths between $A$ and $B$, as shown in this picture:

enter image description here

Here, I have drawn 3 possible paths $a$, $b$ and $c$. Path $a$ goes directly from $A$ to $B$ and its length is thus equal to the distance $AB$. it takes light $\frac{d(A,B)}{c}$ seconds to get from $A$ to $B$ on this path.

As you can see, the paths $b$ and $c$ are not very straight, so their length is greater than $AB$. If the photon were to travel along one of these paths, the time it would take to get from $A$ to $B$ would be equal to the length of the path $l(b)$ or $l(c)$ divided by the speed of light $c$. As for both $l(b)$ and $l(c)$ is $l \gt d(A,B)$, the photon has to travel longer along the paths $b$ and $c$ than on $a$.

However, if I understand correctly, in quantum physics, a particle such as our photon travels along all paths at once. This means that it arrives at $B$ after $\frac{d(A,B)}{c}$ seconds, but somehow also travels paths which are longer than $d(A,B)$.

How is this possible without the particle travelling faster than light (which is not possible, according to special relativity)? I know that the photon does not "split up" or something like this (which maybe was implied by my question) and I am not very familiar with quantum physics, so I am convinced that this is just a misconception of mine.

As I am still in school, I would, if possible, prefer a non-mathematical answer; possibly an answer with both the concept and the math, so that it is also useful to more advanced users.

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To be more precise, it sounds like what you're describing is the path integral formulation of quantum mechanics. Within this framework, every possible path between $A$ and $B$ contributes a complex number to an infinite sum, which is used to calculate the "actual" path the particle takes.

It is tempting at first to ask how a particle could take a path that in theory requires it to travel faster than light, and these paths are considered in this approach and do contribute to the sum, however the classical path the particle takes never requires it to travel faster than light. The paths further from the classical path effectively cancel each other out in the infinite sum. For a nice visual representation of this see here.

Saying that the particle does traverse all paths is doing something that "pop-physics" outlets often do: assume that exactly what happens in a mathematical model also happens in real life. This is analogous to "the fabric of spacetime is bent by mass and energy", it certainly allows people to visualise what's going on but it is an analogy that can only be taken so far.

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    $\begingroup$ Why do you think a quantum particle like a photon has a “true path” like a classical particle does? If it did, it would have a well-defined position and momentum at each instant, violating the Uncertainty Principle. $\endgroup$
    – G. Smith
    Aug 5 '20 at 20:36
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    $\begingroup$ Also, in QED the photon propagator is nonzero outside the light cone. $\endgroup$
    – G. Smith
    Aug 5 '20 at 20:39
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    $\begingroup$ There is nothing “invalid” about the other paths. They contribute to the quantum amplitude in exactly the same way that the paths you consider “valid” do. $\endgroup$
    – G. Smith
    Aug 5 '20 at 20:42
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    $\begingroup$ I am not trying to claim that a photon “actually traverses” all paths. I don’t think it “actually traverses” any of them, because that is classical language. $\endgroup$
    – G. Smith
    Aug 5 '20 at 20:53
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    $\begingroup$ @G.Smith Also just referring to your last comment, the phrase "path" is still valid in this context is it not? Even if we're calculating the propagator not a literal path through space we are still evaluating the action along worldlines which are paths, no? There are better answers than the one I've given that more experienced people could give, but the OP doesn't seem to be asking for a brutally formal response. $\endgroup$
    – Charlie
    Aug 5 '20 at 21:08
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A particle cannot travel faster than the speed of light in the sense that you cannot measure the particle to be in position $A$ and then subsequently measure the particle to be in position $B$ unless enough time has passed (so that $\frac{|B-A|}{\Delta t}\leq c$). This is actually not true in normal quantum mechanics, but it is true in Quantum Field Theory. All paths contribute to the Feynman path integral. However, the question of whether the particle is actually travelling faster than light along some of these paths is not well defined. We can measure the particle at one point, and then at another point. The paths that you get by connecting these observation points will never be faster than light. What happens between the observation points is not something you can see or measure.

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Yes, within the Feynman integral approach the photon (and any other particle treated with this machinery) may travel with a velocity grater than $c$. (This is the intuitive corresponding of a virtual particle one uses in quantum field theory.)

However the path integral method cannot be taken too literally. Not everything one may imagine by looking at this representation from an intuitive point of view reflects the observable physical reality. It is a powerful mathematical procedure to compute probabilities of observable physical events.

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  • $\begingroup$ I think it's interesting to draw an analogy with a seismic wave. A seismic wave might travel 100km from A to B at 3km/s but it isn't only the houses sitting on top of the AB line that shake. $\endgroup$ Aug 8 '20 at 14:32

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