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As I know, when photons hit the surface of a mirror, atoms on the surface of the mirror are excited and when they back to some lower states, they emit photons. I have two questions about this process. First, why atoms back to a lower state while photons are hitting with the surface of the mirror continuously? The second question is that why emitted photons are emitted in a way that their angles with the mirror are the same as the angle of the photons before hitting the mirror? In another word, why the image in the mirror is not disordered?

Qmechanic
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user201949
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    Possible duplicate of Explain reflection laws at the atomic level – Stéphane Rollandin Jul 19 '18 at 21:52
  • Regarding your last line: https://en.wikipedia.org/wiki/Fermat%27s_principle –  Jul 19 '18 at 21:55
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    "atoms on the surface of the mirror are excited and when they back to some lower states, they emit photons" This is not what is happening in a mirror. – my2cts Jul 19 '18 at 22:29
  • The image is not distorted, because the wave function of the photons does not collapse at the mirror. Why doesn't the wave function collapse? Apparently no one on this site knows, because no one has answered it correctly so far: https://physics.stackexchange.com/questions/368333/when-light-reflects-off-a-mirror-does-the-wave-function-collapse – safesphere Jul 19 '18 at 22:32
  • Also, there is no consensus on this site on how the reflection happens. Some say photons are reemitted; others say photons are "scattered". – safesphere Jul 19 '18 at 22:34
  • A correction regarding your question. In a metallic mirror, it is not atoms that absorb photons and get excited through very much discrete levels. Instead, photons are absorbed by free electrons whose distribution of levels (e.g. due to the quantized momentum) is virtually continuous. In a non-metallic full internal reflection, such as in a glass prism, - I wish the experts here would explain what exactly absorbs and reemits (or even "scatters") light. Overall, it would be a miracle if you get a correct answer and a miracle if you know, which answer is actually correct. – safesphere Jul 19 '18 at 22:39
  • I have considered that photons carry the energy (of excited electrons) away in random directions and not the other way, where photons are so called emitted as energy levels fall. Photons travel in random directions and do not match the angle of incidence. Feynman shows you how to derive reflections of many photons on a mirror from random direction across the whole surface. – Bill Alsept Jul 19 '18 at 22:46
  • @safesphere At the formal level of quantum field theories all non-trivial photon scattering is represented by the combination of a destruction and creation operator, meaning that all "scattering" is absorption and re-emission. These are decidedly not different explanations. If you attack the problem in the classical theory atoms or conduction electrons act as damped driven oscillators where the incidence field in the driver and the damping comes from radiating the absorbed energy back with the result that "absorption and re-emission" is a reasonable explanation in those terms as well. – dmckee --- ex-moderator kitten Jul 20 '18 at 20:57

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Very good questions. The comments are right that there is no consensus on this site about what happens when photons hit a mirror, so I will give you the version that seem the most logical and will answer your questions.

You figured out by yourself, that when you try to describe how photons are creating a mirror image with absorption, re-emission, it will not give you the correct image.

Why? Because there are three things that can happen when a photon hits a mirror and interacts with the atoms:

  1. elastic scattering, the photon keeps its energy and phase and changes angle

  2. inelastic scattering, the photon will give part of its energy to the atom, and will change angle

  3. absorption, the photon will give all its energy to the atom, and the valence electron will move to a higher energy level as per QM

Now in the case of a mirror, it cannot be absorption, because of what you say:

  1. the photons would change angle, decoherently

The emission can also change angle decoherently, so the original photon's angle will not be the opposite of the re-emitted angle, and all the photons will be re-emitted in different angles.

  1. the photons would not keep their energies, and phases

The problem with absorption is that when the atom re-emits the photon, it can re-emit in multiple steps. So the absorbed photon's energy will be divided into more then one photon and the re-emitted photons will have energies that will sum up the original photon. So the re-emitting electron will come down as per QM to a lower energy level in more then one step. This way a mirror image cannot be built.

so a mirror image will not be built with absorption-re-emission.

The only way to keep the photon's energy and phase is elastic scattering, that is Rayleigh scattering, when the photon's wavelengths' are big compared to the size of atoms.

With mirrors, some photons will be elastically scattered, some will be inelstically scattered, and some will be absorbed. In the case of a mirror, most of them will be elastically scattered, and that will build up the image.

Now mirror images are not only built by metals. If you look at glass, some can give you a really good mirror image. So why is that?

It is because it is elastic scattering, and it is almost the same thing in the case of mirror as refraction.

What is the difference in the case of glass between reflection (mirror image in glass) and refraction (image traveling in glass)?

The only difference is the angle. Reflected photons build up an image that is reflected by an angle almost opposite to the original angle. Refracted photons build up an image that is refracted by an angle that is almost the same as the original one.

It is just the angle, and that with refraction, photons travel inside the new medium, glass. With reflection, photos travel back in air. Both cases it is elastic scattering. It is called specular reflection.

Árpád Szendrei
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  • This answer is, to say the least, arguable on so may levels. What is called "photon scattering" is in fact a photon absorption and reemission mediated by a virtual electron. Furthermore, photons are particles upon measurement, but waves of probability in between. If photons behave as particles, the wave function is collapsed, the phase information is lost, and the image cannot persist from this point on. For the image to be undistorted, you need to treat photons as excitations of the quantum field generated by moving charges. This question requires an expert answer of the Quantum Field Theory. – safesphere Jul 20 '18 at 02:34
  • Thanks for your explanations. What I wrote is what I had found on many websites! Now I can see that the answer is much more complicated ( and more interesting) than I thought. I was thinking that is it possible to say that from each photon, exactly one photon is emitted or reflected from a mirror. Because if it is right, according to their momentum, we can say that the emitted or reflected photon must be opposite to the first photon. – user201949 Jul 20 '18 at 06:59
  • @safesphere it is best to forget the wave. Any light phenomena is best described or derived with individual photons. Billions of coherent photons only resemble a wave. – Bill Alsept Jul 20 '18 at 14:29
  • @BillAlsept Your statement, "Billions of coherent photons only resemble a wave", shows that you are referring to the classical wave. Please note that my comment instead referred to the quantum "waves of probability". You could try forgetting the classical wave, but I was not talking about it. However, you cannot forget about the quantum wave, because photons are not "little balls". Furthermore, you can forget about the classical wave only if you have a full QFT solution. Otherwise a classical wave still gives a better description than the corpuscular theory. This case is a perfect example. – safesphere Jul 20 '18 at 14:59
  • FYI - Despite my criticism, the downvote is not mine. – safesphere Jul 20 '18 at 15:01
  • I am interested in what’s really happening and not probabilities. This can only be explained with individual photons and not waves that cannot be physically described. Like a Galton board, you can calculate the probabilities but you can also physically describe how a ball bounces its way down to the bottom. – Bill Alsept Jul 20 '18 at 15:47
  • @user201949 when a photon is re-emitted, it can be re-emitted in multiple steps or just one step. If in one step, it might be possible that the energy and phase is the same, and the direction the opposite, but that has little probability, and that all the other photons will do that too, will have even lesser probability. That is why re-emission would give a decoherent image. If you like it please upvote. – Árpád Szendrei Jul 20 '18 at 16:27
  • @Bill Alsept yes I know that you are interested in exactly what is happening with the photon when it changes angle, and how it changes angle. My explanation on that was with the lattice and the wavelengths, just like the slits, and the constructive and deconstructive interference. It is the interference that changes the angle. – Árpád Szendrei Jul 20 '18 at 16:31
  • What you need to understand, is that the photon itself is only a photon when (in your classical view) when it interacts with particles. Now when it is traveling, it is a wave (QM), but no the classical wave. So when it travels, the wave itself can interact with itself, and because of that interference, it will change angle. The wave (QM) itself will change angle, that is the wavefunction, because of the interference. That is how it is possible for the photon to show up on the screen in the different parts of the interference pattern. – Árpád Szendrei Jul 20 '18 at 16:35
  • The photon as a particle does not change angle, it is not like a particle traveling and then interacting with the slits or the lattice and change angle like a classical object. It is traveling as a wave (QM), and here safesphere is right, because you need to look at the traveling photon as a wave (QM), and understand that it can interfere with itself (when traveling through both slits or through many spaces in the lattice) and so changing angle because of the interference. – Árpád Szendrei Jul 20 '18 at 16:37
  • @ÁrpádSzendrei We will just agree to disagree. And if you say wave you are just kicking the can down the road. Nobody can physically describe what a light wave is or how big it is. And I’m not sure way everyone is against a photon being a particle but it’s the only thing that makes sense. It’s the only way to derive every phenomena in light. The way you would explain one single photon going from earth to the moon is far far more complicated than it needs to be. – Bill Alsept Jul 20 '18 at 16:47
  • @Bill Alsept I think we do not disagree. I think we are talking about the same thing. I think that the fabric of spacetime (whic we do not yet know how it looks because it is much smaller then the Planck scale) will show how stress-energy can bend spacetime and what it bends. When we will know that, we will know too what is a wave that travels. I think it is the wave in the fabric of spacetime, just like stress energy bends spacetime, and GWs are a wave that travels in the fabric of spacetime. – Árpád Szendrei Jul 20 '18 at 16:55
  • Those will be the QFT fields, which we do not know how they look since those should be smaller then the Planck scale too. Those fields are everywhere and the particle that you are talking about is the excitation of those fields. We do not know what that excitation looks like as it travels, because we do not see how the fabric of spacetime and the fields look like at the Planck scale. – Árpád Szendrei Jul 20 '18 at 16:57
  • My question would be, could you explain the double slit experiment with the particle as a classical object? – Árpád Szendrei Jul 20 '18 at 16:58
  • @BillAlsept "Any light phenomena is best described or derived with individual photons." Utter nonsense. Models are only as good as your ability to obtain usable results from them. You don't get closer to understanding physics by choosing models that obscure cause and effect behind of blizzard of intractable math. Real physics is done using appropriate models and not by trying to out-fundamental the next scientist. – dmckee --- ex-moderator kitten Jul 20 '18 at 21:10
  • @dmckee How does committing to a real and physical model that describes cause-and-effect, obscure cause-and-effect? Wave functions and probabilities don’t describe cause-and-effect. They are not accountable for anything. My model can be derived with math as well. It just happens to be a model that can physically describe what’s going on and for some reason that bothers some. – Bill Alsept Jul 20 '18 at 21:29
  • It’s utter nonsense to ignore photons. There’s no basis for that belief. You may have been taught that and your teacher was taught that, but what is the basis of your belief. Of course there are individual photons. Wave functions and probabilities are far more of obscure than basic cause-and-effect. – Bill Alsept Jul 20 '18 at 21:40