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Feynman say's that a photon takes every path while reflecting off a mirror when going form A to B, but we only see the middle one(where incident angle = reflected angle) because all the others are cancelled out as they have longer routes and while following them their phase arrows cancel out each other.

If you can, please explain it in general terms without mathematics, as I am studying it the same way.

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    $\begingroup$ A phase arrow is a complex number with unit magnitude. Like, something of the form $e^{i\theta}$. $\endgroup$ – knzhou Aug 20 '16 at 3:01
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    $\begingroup$ As Larry Gopnik rightly said, trying to understand quantum mechanics without the mathematics is a doomed endeavour, and you will only get a very partial view of the subject. Past a certain (very early) point, you need the mathematics. $\endgroup$ – Emilio Pisanty Aug 20 '16 at 17:45
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This question is based, as pretty much everyone knows, on the book "QED, the strange theory of light and matter", by Feynman.

My answer is based on my limited understanding of the book.

Phase Arrows has a good description of Feynman's analogy.

As well the above source, if you visit Feynman Lecture Using Phase Arrows , you can watch RPF's description between times 29:41 and 36:27 of part He draws the "arrows" diagram on the chalkboard.

Similarly, if you visit Feynman Lecture On Photons, Feynman talks about this subject between times 59:33 and 60:32.

One of the differences between classical mechanics and quantum mechanics is that classically, objects have a definite path,or trajectory, so if you kick a soccer ball, you can be pretty certain where it will end up, and the path it will take.

Whereas in q.m.you can't say that a photon will definitely either follow a path or be found at a particular place. You can only assign a probability that a photon will be found at a particular place.

So Feynman sort of took this idea to the extreme, in a mathematical sense. Mathwise,you can say the photon took every available path, but all the paths that don't confirm to classical notions cancel each other out, by destructive interference. So you are left with just the paths that take the least time.

This is sometimes known as the sum over histories approach to quantum mechanics.

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    $\begingroup$ Although I didn't use feynman's book, your resources were good. $\endgroup$ – John_Nash Sep 17 '16 at 22:11
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Feynman's understanding of this effect probably came from optics where it is known as Fermat's principle. The mechanism behind this principle is based on the notion of stationary phase, which is the phenomenon that light will go in that direction where the phase coming from different paths varies slowly.

I'll try to explain this without mathematics. The scenario is one where there is some field that propagates linearly through space or through some linear system, as in the case where we have linear optical components. At the output, one would then find locations where the field has a large amplitude and other locations where the amplitude is small or even zero.

If one looks at a particular output point, the field amplitude at that point is the sum of all the bits of field that arrived there via different paths. The phase of these bits is determine by the length of the path that they followed to get to the output point. If this phase value changes rapidly as a function of the varying path, then the different values will tend to cancel each other. You can think of it as a bunch of arrows pointing in all directions and you are adding these arrows to find the combined arrow. For the rapidly varying case the amplitudes (lengths of the arrows) will vary slow compared to the phase (directions of the arrows). Arrows that point in opposite directions will cancel each other. So the net result is a very small amplitude.

If, on the other hand, the bits of the fields arriving at the output point, have a slow varying phase, then the arrows would all point in more or less the same direction. So they would add up to give a larger amplitude (longer arrow).

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  • $\begingroup$ But if I shoot a photon at a mirror, I can be 100% sure of where it will end up and which path it will take(other paths will be cancelled,but still), how is the probability of that photon reaching that point after hitting the mirror be 100% $\endgroup$ – John_Nash Sep 17 '16 at 22:14
  • $\begingroup$ The catch is, you cannot shoot a photon. You need to direct a beam of light and light beams always have a finite size. Furthermore, visible light has a very small wavelength. This makes the effect seems 100% precise. If one were able to see light with much longer wavelength the behaviour would appear less precise. $\endgroup$ – flippiefanus Sep 18 '16 at 4:05

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