# Feynman's QED lectures: quickest path and reflection

theorem4.5.9
Feynman's first topic in his second lecture on QED is the nature by which light reflects off of a mirror. We work in ##\mathbb{R}^2##. Suppose we have a light source sitting at ##(-1,1)## and a photomultiplier sitting at ##(1,1)##, with a mirror along the x-axis. We also place a block between the source and photomultiplier. Let's say it extends from ##(0,+\infty)## to ##(0,1/2)##.

Suppose the following event occurs: the photomultiplier clicks. We want to find the expected path of the photon (I assume we'd want the path of minimal mean square). Feynman's heuristic argument, roughly stated, is that if a path allows many nearby paths of drastic time difference, these wave functions will cancel out. Since the path of minimal length (or minimal time elapse) will have nearby paths of similar length, the wave functions will constructively add. This heuristic leads us to conclude that the expected path will be approximately the path of shortest length.

In this example, Feynman only considers straight line paths that reflect upon the mirror. Later, he says that light doesn't have to follow straight paths, that we could apply the above over some set of reasonable paths. So if we revisit the example, with the same event (photomultiplier clicks) but with more (reasonable) sample paths, we should arrive at the same conclusion. However, the path connecting the source to the bottom of the block at ##(0,1/2)##, and then to the photomultiplier is clearly the path of shortest distance. Furthermore, the heuristic of nearby paths having similar distance is even more valid than the reflected path.

Following the same outline as Feynman, I'm left to conclude that QED predicts that the "classical" path of light would be to travel to the tip of the block, and then to the photomultiplier, and not to be reflected by the mirror.

So, something is wrong here, and I'm not sure what.

Homework Helper
The famous QED lecture series is intended to give a math-light sense of how QED works, not to be a rigorous course in the methods. Thus it is incomplete - as you have noticed.

The "heuristic" is only a rule of thumb. The part to stress is how the calculation is carried out.
Indeed some light may follow a trajectory that does not strike the mirror - in Feynman's example, since he does not place any barrier in the setup, this would most likely be the the straight-line path from the source to the detector. That's the path of least time. The mean of all possible paths in that setup does not tell us anything useful.  Which is why Feynman restricts his discussion to the light which is reflected off the mirror only ... just like the secondary school lessons on mirrors only consider the reflected light.

The sumation of the phasor from each path being considered is the important part - did you have a go working out the resulting probability distribution for different y positions of a detector at x=1 using that method? But, really, if you want to learn about how the QED calculations are done IRL, you need a more mathematical introduction.

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• bhobba and dlgoff
theorem4.5.9
Thanks for replying. A proper text on the subject is probably the way to go, but I don't have the time to invest in that right now, but I think I can get a lot for very cheap with QED.

Correct me if I'm wrong, but I think the heuristics are much more than a rule of thumb. I'm under the impression that the summations involved in the path integral formulation were a case of functional integration where the underlying space could not be shown to be a measure space. Contrast this with the Wiener measure, which is a form of justified functional integration. In this way, explanation to the calculations is no more justified than passing explanation to heuristic reasoning.

Even when I try to sidestep functional integration by introducing finitely-many-parameter family of paths, there's still but there are too many unstated assumptions in QED for calculation to be possible. It's easy enough find the direction of one of these vectors, as the angle is just a linear function of the length of path for fixed frequency, but I've nowhere to guess the length of the vector. Supposing the length is known, then there's the question of which paths should be allowable. Piecewise continuous paths of two straight lines should be an OK approximation, but the bigger trouble I see is how to sum over them. I could parameterize these paths by specifying where they intersect, say at ##(\alpha, \beta)##, and then sum by integrating over these two parameters, but this is not canonical. Indeed, one can parameterize these paths by saying they instead intersect at ##(f(\alpha),g(\beta))## for any reasonable ##f,g## (e.g. diffeomorphic functions). The different parameterizations lead to different measures on the sample space of paths. So even in the simplified case, the integration isn't justified without some heuristic.

But this seems to be getting away at the question at hand. Feynman says in his introduction that he will remain true to the physics, that you won't have to unlearn anything after reading the book. So I want to push his reasoning and see where it goes. In his example, he actually does place a barrier between the source and the detector. It's just that he assumes light can only travel in straight lines and be reflected. Doing away with that notion, how can one predict, heuristically or not, that light will prefer the longer path of reflection over the shorter, curved path around the barrier?

Feynman's heuristic argument, roughly stated, is that if a path allows many nearby paths of drastic time difference, these wave functions will cancel out. Since the path of minimal length (or minimal time elapse) will have nearby paths of similar length, the wave functions will constructively add. This heuristic leads us to conclude that the expected path will be approximately the path of shortest length.

No. This is not the general conclusion one can draw from that. The (rough) conclusion is that any path, which does not allow many nearby paths of drastic time difference - resulting in destructive interference - will have a non-vanishing probability. You cannot conclude that this will always leave you only with the shortest path just because this is the only sensible path in the first scenario.

• sanpkl
Homework Helper
I think the heuristics are much more than a rule of thumb
The rule of thumb is that the classical path is always the path of least time.
You have discovered that this is not always the case.
Follow the math rather than the hand waving.