Feynman couldn't explain how the photon becomes instantaneously aware of the glass thickness. Do we have a better understanding now?

Question

I remember reading in Richard Feynman's QED about this unknown physics mechanism which possibly involves information propagating instantly and it blew my mind:

The probability of photon to reflect or refract on a slide of glass depends on the thickness of the slide. Feynman said that we don't know how the photon is "aware" of the thickness of the slide when it interacts with its side. The information about the thickness of the slide would have to "travel" from the other side of the glass to the side the photon interacts with and this appears to be instantaneous. (*)

Other quantum processes like quantum tunneling have been shown to not actually break the speed of light so I was wondering if we have a better understanding of this physics process now.

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*) I read QED a while ago and I don't currently have a copy of it, so I hope I remember correctly. I know the electron is modeled as a wave until it is measured, so I am pretty sude I have misused some terminology like "photon interacting with the side of the slide". Please correct me. But this doesn't change the fact that Feynman couldn't explain the aparent instantaneous information travel, as far as I remember from his book.

Answer 1

Actually, the photon doesn't have to know the thickness. Moreover, if we speak of a wave with a well-defined "beginning", like e.g. $\psi(x,t)=\sin(\omega t-kx)\theta(\omega t-kx)$ (with $\theta$ being Heaviside function), incident on the glass, part of this wave will reflect as if the glass were semi-infinite. But then the reflection from the far side of the glass will come back to the near side and, after being transmitted through the near side, it will start interfering with the initial reflection from this side. After some traveling time, the secondary reflections will add up to the outgoing wave, and only in the long term would you get the final steady state with the reflectance being defined, as Feynman says, by the thickness of the glass.

By that time, the initial part of the reflected wave will have already travelled away. So, even if the reflectance, as calculated from the glass thickness, is exactly zero, you'll still get a pulse of light reflected before the process reaches steady state of no reflection.

Answer 2

In Feynman's QED The Strange Theory of light and Matter, he devotes a large amount of time explaining how one goes about calculating the probability of reflection/transmission of a photon from a thin layer of glass. Feynman's account of thin film interference can be found on pages $69$-$72$.

Briefly speaking, he uses the idea of a clock hand as a revolving arrow attached to every little photon. The arrow serves as an analogy to the theoretical wavefunction.

He then states the key principle that to calculate the probabilities of events, one adds the arrows of all the independent ways the event can occur (and squares its length etc).

So in the context of explaining the dependence of reflection (which rotate and/or shrink the arrows) on the thickness of the glass layer,

1. the photon doesn't need to receive any information about the thickness of the glass upon its arrival at the top layer - instantaneously or not.

2. the photon isn't a center of computation which when provided with the thickness of glass, would make the split second decision to reflect or not.

So how does the photon sense the thickness of the glass?

1. One considers all paths the photon could possibly have taken. Turns out paths suggested by classical wave theory ray diagrams contribute the maximum to the probabilities.

2. While adding up the probability for alternate paths, some paths involve passing through the thickness of the glass.

3. The probability contributions from these paths enable the photon to be able to "sense" the glass thickness.

There is no need to stipulate (nor any way to know really) that a single individual photon senses the layer's thickness and then decides at the reflection point. All one can say is that for a bunch of them, the correct odds for observed behaviour are obtained by including all paths - some of which get affected by the thickness.

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A1. In the language of wavefunctions, the reflection probability is determined by the superposition of all wavefunctions including those "reflected" from bottom surface - thus being affected by the thickness in their phase.

A2. In pages $16$-$23$ Feynman expresses the conundrum faced by Newton and others in trying to model a photon's behaviour during reflection from a glass layer. On page $24$ he states how physics has given up and settled for probabilities. During this discussion he never states instantaneous information travel.

Answer 3

I can only comment on the basic problem: photon scattering off definite field conditions.

As an example here is a lowest order scattering of a photon with an electric field, represented by virtual photons,

The on mass shell photon enters on the top left and leaves on the top right. The diagram will give the probability of scattering when calculated and the electric field value is used.

When hitting the electric field of a lattice, whether transparent or not, the field is a boundary condition for the scattering of the photon.

All boundary conditions in this sense are instantaneous, otherwise one could not do calculations. There is no velocity of light involved , except if the field is changing, when the field can only change within the limits of velocity of light.

If your memory of Feynman's statement is correct, the answer is that the information of the thickness of the lattice, on which the photon impinges, is already embedded in the topology of the electric and magnetic fields of the lattice on with which the photon is interacting. If there is some change in the lattice, that information has to travel with the velocity of light for changes to embed in the field.

Edit in order to make clear with this simple experiment the difference between the probability nature of the photon wave function, and the energy in space electromagnetic wave functions . It also demonstrates the existence of the fringe fields of matter on which the photon scatters.

Here is an experiment one photon at a time:

Single-photon camera recording of photons from a double slit illuminated by very weak laser light. Left to right: single frame, superposition of 200, 1’000, and 500’000 frames.

The experiment is: single photon at a time , of given energy, scattering.

The boundary conditions are: two slits of a given width, a tiny distance apart

On the left each photon footprint shows up as a dot in the (x,y) of the screen, (the (z is the distance of the slits to the screen), time is not recorded. There is nothing wavy of the footprint, it looks like a classical particle footprint hitting a plane.

The photons directions look random.

As one progresses from left to right and more photons are accumulated an interference pattern slowly appears and on the far right we see the expected and well defined mathematically classical interference pattern of double slit interference.

The experiment shows the probability wave nature of same energy and same boundary condition photons, and also of how the classical electromagnetic wave emerges from the seemingly random quantum behavior.

This is because the quantum wavefunction is modeled by a quantized Maxwell's equation. How the classical fields develop from the quantum substrate is seen here .

The boundary conditions for the photon solution are given by the electric field around the two slits, on/through which the photon scatters as it goes through. These conditions are embedded, the way the thickness of a lattice is embedded in the field the photon scatters off. The information that it is two slits of given width and distance is there, whether there are photons impinging or not.

The same is true for a scattering from a lattice, the information of its thickness is embedded in the fringe field topology off which the photon scatters in the question above.

Answer 4

The sinusoidal pattern of the reflection rate as a function of thickness is due to constructive/destructive interference between reflections off the near and far surfaces of the glass, and it's correctly explained by Maxwell's electromagnetism without quantum mechanics. As Ruslan said, the "detection" of the thickness happens at the speed of light: wave fronts reflect off the far surface, travel back, and interfere with later wave fronts that have just arrived at the near surface.

The wave-like behavior of light is not mysterious by itself. To explain the combination of particle-like and wave-like behavior exhibited by light, you need quantum mechanics. In the book QED, Feynman flat-out states that light is definitely made of particles, not waves, and he presents all of the wave-like behavior of those particles as mysterious. There may have been a time after the discovery of Feynman perturbation theory (which is the method of "adding arrows" described in the book) that it was widely believed to be more fundamental than field/wave theory, but by the time QED was published, it was clear that fields are more fundamental, and Feynman diagrams, and even the whole concept of particles, are only special-case approximations. I like QED and it was my introduction to quantum field theory, but you shouldn't take what he says about the particle nature of light too seriously.

Answer 5

The photon does not reflect off the surface. What is a surface anyway? It is two-dimensional, infinitely thin, a mathematical construct. It does not exist.

The things that exist are the atoms and electrons of the glass. The photon is interacting with all that, and it is of course not possible to do a full treatment of this in quantum electrodynamics.

It can be approximated as scattering by the atoms, as with x-rays in the Ewald theory of optical reflection. Each atom contributes to the scattered amplitude. At some thicknesses, the scattered amplitudes interfere constructively, at other thicknesses destructively.

Answer 6

No faster than light communication or quantum mechanics needed to describe the interference that occurs when light reflects off of a piece of glass.

The glass is a Fabry-Perot etalon. An etalon is an optical component which has two glass surfaces parallel to each other. Suppose the etalon has thickness $L$ and speed of light $v = \frac{c}{n}$ where $n$ is the index of refraction. It will take light time $\tau = \frac{v}{L}$ to traverse from one side of the glass to the other.

The physics proceeds as follow. Suppose for time $t < 0$ there is no light shining on the etalon. Then, a little prior to $t=0$ a monochromatic beam of light is turned on such that at $t=0$ that beam of light arrives at the etalon. What will happen? Suppose the beam carries a constant power (energy per second) of 1 mW flowing towards the etalon.

When the beam arrives at the first reflective surface of the etalon 4% or the light will reflect, no more, no less. This will continue to be the case for time $t<\tau$. that is, during this time period the reflected light level would be 40 $\mu W$. Note at this time the power that has transmitted through the etalon is 0% since it hasn't gotten to that side yet.

Then, at time $\tau$, the light will arrive at the second reflective surface. At this time 96% of the light will transmit through the etalon and another 4% of the light will reflect off of the back reflective surface. There is now 4% reflected directly off the front surface, 96% travelling forwards through the etalon, 96% of the first 96% transmitting through the whole device and 4% of the first 96% travelling backwards through the etalon towards the first surface.

After another time $\tau$ the reflection from the second surface will finally reach the first surface. This is when the interference will begin to occur. 96% of this light will transmit out adding to the first reflected either constructively or destructively.

The light will continue to bounce back and forth in this way and after every time $2\tau$ another round trip will be completed and the total reflected power will be slightly modified due to interference with the next bit of circulated power.

Eventually the reflected and transmitted power will exponentially, asymptotically approach a steady-state level which is determined by the length of the etalon modulo the wavelength of light. For a thin piece of glass I guess this timescale would be just a small factor larger than $\tau$. If the thickness of the piece of glass is 1 cm then $\tau = 33 ps$.

So we see that the interference doesn't build up instantaneously as suggested in the original question here, rather, it takes less than a ns to build up to its final value. While this is very fast it is still 100% consistent with relativistic causality.

Note: I believe the questioner is referring to the first chapter of "QED the Strange Theory of Light and Matter" By Feynman in which this experiment is described. I don't see any claims of faster than light information like the OP suggests. There is some rhetorical puzzling about how this works but the rest of the chapter/book seems set up to answer this and other questions.

Answer 7

It is very important to understand that this experiment was done using a light ray consisting of many photons, not just shooting a single photon at a glass slide.

Actually when they did this experiment, they did only check the part of the wave that was refracted by way of checking whether that part of the ray exited the glass slide on the other (far) side. They did not check whether the light ray was actually traveling inside the glass slide (without exiting on the far side).

the way any transparent sheet of glass partly reflects any light shining on it.

https://en.wikipedia.org/wiki/QED:_The_Strange_Theory_of_Light_and_Matter

Now when the initial ray enters the glass slide, some of the photons that make up the light ray will reflect (elastically scattered) and some of the will be refracted, and some will be absorbed (cease to exist as photons). It is just the ratio of these that is different.

For glass, most of the photons will be reflected or refracted, very little will be absorbed.

Now as some part of the photons travel through the glass (refract), their intensity will decrease, because some of them will be absorbed (heat up the glass), and at the far side of the glass slide, even some of these photons will be reflected back, and only the rest of the photons will be exiting the glass slide on the far side.

If the glass slide is thick enough, a higher ratio of them will be absorbed and the ones that are not absorbed, will be reflected back from the far side or will exit the glass slide on the far side.

Now as the photons travel through the glass, some of them will actually be elastically scattered in directions other then the direction of the wavefront, including in the opposite direction.

The ones that reflect back from the far side will interfere with the initial wave that enters, causing the refracted part of the ray to decrease even more.

So as the glass gets thicker, the intensity of the refracting part of the light ray will decrease because:

1. some of the photons will be absorbed (cease to exist as photons)

2. some of the photons will be scattered in the opposite direction of the wavefront, causing interference

As the glass reaches a certain thickness, the number of photons that actually reach the far side and exit diverges to 0. We will only see in the part of the light ray that is reflected back from the glass slide. This is when they say in the experiment "the light ray has reflected back from the glass slide, and no part of it has refracted".

Just to clarify, if we try to do this experiment with a single photon at a time, we would see that the single photon has a decreasing probability of refracting and exiting the far side as the thickness increases, and this is caused by the fact that as the thickness increases, the photon has to travel through a glass consisting of an increasing number of atoms, thus the probability of the photon being absorbed (and not make through the glass and not exit the far side at all) increases.

So the answer to your question is that no, nothing is instantaneous, the photons travel and interact through the glass slide at the speed of light (please note that in this case it is even less then c, the vacuum speed), and interfere with the original light ray, causing at a certain thickness the phenomenon of an only appearance of a part of the light ray that reflects (and none that refracts).

Answer 8

There is no mistery here: the solution of Maxwell equations include all the space $\square A = j$. All points of space take part in forming a wave.

The problem arises when we think of photon as of a localized point particle which is obviously not always a fruitful approximation.