If I understood Feynman correctly, a single photon reflects with a chance determined by the added amplitude arrows of each surface it encounters. I'd still like to know how that is determined if the photon is already on its way back before it would reach the last surface, as he asked at 30:41.
$\begingroup$
$\endgroup$
5
-
1$\begingroup$ A single photon is either absorbed or it's not. If not it will continue on its way. If it is absorbed it will be re-emitted in a Random directions. It takes millions of photons to average what we call a reflection. $\endgroup$– Bill AlseptCommented Jul 5, 2017 at 17:10
-
$\begingroup$ Feynman was talking about a single photon, though, with an adjustable angle of reflection. $\endgroup$– Cees TimmermanCommented Jul 5, 2017 at 17:35
-
$\begingroup$ Single photon does not have an adjustable reflection angle. It can be emitted randomly in any direction. $\endgroup$– Bill AlseptCommented Jul 5, 2017 at 18:05
-
$\begingroup$ @BillAlsept In a cavity, (the Feynman talk with the thin glass is a very low Q cavity), only certain electromagnetic modes are allowed. This changes the spontaneous emission rate(Purcell effect) and also limits the directions that are allowed. If you want to look into it search Cavity Quantum Electrodynamics so in that sense the direction of emission doesn't have to be random. $\endgroup$– UVphotonCommented Jun 4, 2022 at 14:59
-
$\begingroup$ Reflection is not absorption followed by reemission. $\endgroup$– my2ctsCommented Dec 8, 2023 at 13:34
Add a comment
|
3 Answers
$\begingroup$
$\endgroup$
2
I didn't take the time to watch that movie. However, a critical thing in qm is the measurement of a wave function. Once we measure a photon, the wave function collapses. Hence, the probability amplitude of all "points" which are not consistent with the measurement become zero.
-
$\begingroup$ So it can only reach the last surface when not already measured, but the last surface does affect the chance of measuring it. FTL effect? $\endgroup$ Commented Jul 5, 2017 at 10:54
-
$\begingroup$ Wave function collapse is purely interpretation and has no counterpart in the formalism of quantum theory. $\endgroup$– my2ctsCommented Dec 8, 2023 at 13:36
$\begingroup$
$\endgroup$
2
I didn't go though the whole video, but looking at your comments, a couple of things that might help clear things up.
Feynamm was tilting the back surface of the glass so it wouldn't bounce directly back and fourth between the two parallel surfaces. He was just talking about that some so he could talk about only having a front surface refection going back to the photomultiplier near the light source.
In the case where the the two surfaces are parallel, you form a cavity. I don't think Feynman's vide talks about how a cavity can only support certain electromagnetic modes. The cavity formed by the two surfaces is not very good, but it is still a cavity...
The single photon when it encounters the cavity is only allowed to occupy the allowed modes. In this case the photon is weakly coupled to the cavity, but people work pretty hard to make very high finesse cavities where the surfaces are very reflective where the photon could be strongly coupled to the cavity and the effects can be very strong.
There is a branch of photonics called Cavity Quantum Electrodynamics where people study light matter interactions in cavities.
Purcell effect - changes the spontaneous emission rate of emitters in cavities
Rabi oscillations - changes in how many states in energy levels.
Cavities don't have to be mirrors, but can also be photonic crystals or Bragg reflectors, or grating etc.
But in any case, the main point is that you don't need to invoke FTL, or track the photon as it travels through the glass to reach the back surface if you think about it from a cavity QED point of view.
If the cavity is there then only certain modes are allowed, and the photon will interact with the allowed modes depending on the nature of the cavity. If the surfaces are moved apart or the reflectivity of the surfaces, then the allowed modes are changed. If the surface are very far apart compared to the wavelength of the photon then there is not enough coherence for a good cavity.
For a lot of cavity problems like making lasers you don't need the full CQED description and you can look at the Fabry-Perot cavity from an interference of waves.
More recently with the interest in quantum computing and quantum sensors there is a lot more interest in understanding the details.
-
$\begingroup$ A photon interfering with itself proves it's a wave, but how can it when the measurement is done before the final surface affects it? That would explain the double slit experiment. $\endgroup$ Commented Jun 5, 2022 at 15:04
-
$\begingroup$ I think part of the issue is that the photon is not just by itself, but it coupled to a larger system, and the math to describe that interaction is not easily worked out. So it is not obvious how it all fits together when you are just looking at the photon by itself in terms of the timing. Most lectures and textbooks don't try to work out the double slit or these type of problems in ways where the coupling of the photon to the rest of the system. On the other hand for a lot of real world applications it dosen't matter that much. $\endgroup$– UVphotonCommented Jun 5, 2022 at 17:19
$\begingroup$
$\endgroup$
Special relativity says spacetime is flat (1D) at light speed in vacuum, compressing the future (explaining how a photon reflects before it reaches the final surface) and expanding the past.