# Feynman's QED Lectures: Questions about partial reflection

1. Jul 12, 2008

### STS816

In Feynman's lectures he describes partial reflection as occuring because the "little arrows" cancel out when the glass is certain thickness. I realize that there isn't an actual model of how QED works but what exactly does he mean by "cancel out"?

Also when do you shrink and turn the little arrows and when do you "add" them together, head to tail? Do you shrink and turn arrows until you get a final arrow and then proceed to "add" several final arrows together? If so, does that mean that shrink and turning arrows gets you one step(i.e. whether a photon is reflected or transmitted) in an event(i.e. whether a photon will pass through a pane of glass)?

Thanks for the help. And FYI, I'll more than likely be back with more questions as I read more :).

EDIT:4 minutes after my original post and I already have another question lol. Feynman states that an electron "orbits" a proton in a hydrogen atom because the two are continuously exchanging photons. In his diagram, however, I can't find any path of a photon that would "push" the electron towards the proton. All of his pathes seem like they should "push" the electron away, not toward, the proton. I have a sneeking suspicion that this has something to do with positrons traveling backward in time but I'm not sure. Once again,thanks for the help.

Last edited: Jul 12, 2008
2. Jul 12, 2008

### Fredrik

Staff Emeritus
It just means that the arrow associated with the shorter path (reflection off the front surface) is in the opposite direction of the arrow associated with the longer path (reflection off the back surface), so they add up to zero.

You add the amplitudes of different paths from event A to event B to get the amplitude for detection at B given emission at A. I'm not sure what shrinking and turning you're referring to.

I don't know if there's any way to see that an interaction that can be described as an exchange of virtual particles can be attractive, other than doing the QFT calculations explictly, but maybe someone else has a better answer.

3. Jul 12, 2008

### gremezd

I don't understand one thing. Why the two arrows, one corresponding to front reflection and the other to back reflection, are subtracted and not added?
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If the angle difference between the stopwatch hands is phi, then from what says Feynman, the percentage of the reflected light would be:

P=0.16*cos( (pi-phi)/2 )^2 .

But from the theory of interference we have that intensity of the sum of two plane waves with phase difference delta is:

I=2*I_0( 1 + cos(delta) )

Where I_0 is the intensity of each of the two reflected waves. It should be I_0=0.04, since 4 percent of the light is reflected.

If we want that P=I, then the following relation should be valid:

phi+delta = pi.

But this would mean that the angle phi between the two hands is not the phase difference delta.

Somewhere in these considerations I'm definetely wrong. Where do I do a mistake? Or at least how to show that Feynman's stopwatch approach is equivalent to simple interference? For example, how to show with classical interference methods that there is almost no reflection from a very very thin film?

Last edited: Jul 12, 2008
4. Jul 14, 2008

### gremezd

Ok. I guess, I found the answer myself. Everything is well, when we account for the phase change by pi after reflection from optically densier medium :)

5. Jul 14, 2008

### neu

Well it is the actual model of quantum mechanics. Think of the spinning arrow as tracing out the oscillation of a wave. This wave is the probability amplitude i.e. the wavefunction

6. Jul 14, 2008

### limey

I am not sure that I understand what you mean by "tracing out".
Could the doubleslit experiment, i.e. the interference pattern on the back wall, be modelled by a figure of Feynman arrows? I suppose the areas of destructive inteference should have a wide array of arrows changing fast and cancelling each other out, and the constructive areas more slowly changing arrows that (mostly) add up to a length larger than zero?