The electron absorbs the energy of photon(with specific frequency)and re-emits the photon.The photon can be emitted in any direction. So why do they get re-emitted in a specific direction after reflection? On hitting normal to surface the photons follow path in reverse normal direction and hitting at an angle it follows V path.

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    $\begingroup$ Neither electrons, photons nor atoms are little balls of stuff that reflect off each other like billiard balls. That just isn't how the world works, and if you want to study physics, you need to come to terms with it. There are approximations that work in certain cases, models that give the right answer to some of the questions, but you must not stretch the model beyond where it's defined. Classical theories of light don't have photons or atoms - and tell you light reflects in a specific direction. QED has photons and atoms - and photons do not reflect off electrons. $\endgroup$ – Luaan Mar 31 '20 at 6:59

Although a single photon can only be absorbed and emitted by a single electron, it leaves that electron in exactly its original state. There is no record, and no way of knowing, which electron absorbed and emitted the photon. According to quantum theory, to calculate the result when any electron could have absorbed and emitted the photon, we must form a superposition of all the processes which could have taken place. The calculation in quantum mechanics takes the form of wave mechanics -- this does not have to mean that there is actually a wave, only that the mathematical theory behaves as though there was a wave. Since a wave would reflect at a particular angle, that is also what happens for photons.

The reasons for this strange quantum behaviour are deep and subtle. Quantum mechanics is actually a theory of probabilities, not a theory of physical waves. The mathematics tells us that the probability of a particular angle of reflection is actually a certainty; we can even show that this mathematical behaviour is necessary to a consistent probabilistic interpretation of quantum theory. It is much harder, probably impossible, to conceptualise exactly what is physically taking place at the level of elementary particles leading to these kinds of results.

  • $\begingroup$ Wait, what? Every other sentence either contradicts itself or the preceding one. $\endgroup$ – user257090 Apr 13 '20 at 7:59

The actual working-out on the level of single photons is quite tricky.

I can only give you this hint: There is a coherent interaction of the photon with a lot of electrons in the silver layer. Silver or aluminium coatings work, because these materials have free electrons (which is also why the are excellent conductors)

If there was (mainly) absorption+emission by single atoms and their electrons, it would not be a mirror but just an ordinary surface.

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    $\begingroup$ That's interesting. So a mirror is a mirror because the (all free?) electrons behave together as a collective? Do they behave similarly or how can this be seen? I'm trying to understand this on an atomic scale while avoiding QM in parallel :D $\endgroup$ – Ben Mar 31 '20 at 9:38
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    $\begingroup$ Electrons of the world, unite! $\endgroup$ – Vikki Mar 31 '20 at 18:10
  • $\begingroup$ It is not only silver or aluminium used for mirrors, nickel or chromium may be used too, even quicksilver was used. Not so excellent conductors do work too. $\endgroup$ – Uwe Apr 1 '20 at 18:18
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    $\begingroup$ @Ben : They're sort of free -- in a polished metal surface, this is an interaction of a photon with the electrons in the conduction band of the metal -- i.e. with thousands of electrons in delocalized states. It is worth noting that visible wavelengths are much bigger than atomic bond lengths, so even if you model a photon as a compact disturbance in the classical E-M field, that disturbance is wide enough to include many atoms and a horde of conduction band electrons. $\endgroup$ – Eric Towers Apr 1 '20 at 22:09
  • $\begingroup$ DoctorNuu, question specially to you: How you describe the interaction between photons and the electrons on edges (edge, slit, double slit)? $\endgroup$ – HolgerFiedler Apr 4 '20 at 6:03

You are confused because you try to imagine mirror reflection as absorption and reemission. Though on this site, this might be helpful for certain reasons, in this case you need to understand that reflection is elastic scattering.

Elastic scattering is a form of particle scattering in scattering theory, nuclear physics and particle physics. In this process, the kinetic energy of a particle is conserved in the center-of-mass frame, but its direction of propagation is modified (by interaction with other particles and/or potentials). Furthermore, while the particle's kinetic energy in the center-of-mass frame is constant, its energy in the lab frame is not. Generally, elastic scattering describes a process in which the total kinetic energy of the system is conserved.


Now what is very important to understand is that you are asking two questions:

  1. why is the relative angle of the photons kept in mirror reflection

  2. why is the angle of the photons the same (mirrored) as the incident angle

The answer to 1. is elastic scattering, which is specular reflection. This is the only way to keep the relative energy levels and relative angle of the photons, and build a mirror image.

Specular reflection, also known as regular reflection, is the mirror-like reflection of waves, such as light, from a surface. In this process, each incident ray is reflected at the same angle to the surface normal as the incident ray, but on the opposing side of the surface normal in the plane formed by incident and reflected rays. The result is that an image reflected by the surface is reproduced in mirror-like (specular) fashion.


enter image description here

Now the answer to 2. is the law of reflection.

The law of reflection states that for each incident ray the angle of incidence equals the angle of reflection, and the incident, normal, and reflected directions are coplanar.

Now the opposite is diffuse reflection, like a non-shiny surface, where the photons are re-emitted or reflected in random directions.

  • $\begingroup$ I am glad to see that you finally changed your point of view on this. $\endgroup$ – my2cts Mar 30 '20 at 22:28
  • $\begingroup$ So is it so that for mirror photon undergoes elastic scattering and for other objects it undergoes scattering which is electron emitting photon in some random direction? $\endgroup$ – Karan Bhatia Mar 31 '20 at 8:04
  • $\begingroup$ @KaranBhatia you are asking what materials surface atoms absorb photons, and then re-emit it in random directions, like a wall? $\endgroup$ – Árpád Szendrei Mar 31 '20 at 16:19

It's the same reason that light follows a straight path in a constant medium and follows Snell's laws when changing medium: there's destructive interference between all the other paths. You can actually get light to reflect at another angle if you remove parts of the mirror. If you do it right, the parts of the wavefunction going in the new direction remain, and the ones interfering with them disappear. Which parts of the mirror need to be removed depend on the frequency of the light (you can also do this with transmission: if you cut slits in a sheet strategically, you can get light going through it to bend by having only the diffraction in the desired direction survive).

Here's a Feynman lecture on it:


And here's an article on it:


  • $\begingroup$ It is perhaps noteworthy that the wave-like probability distribution with its pattern of destructive/constructive interference is not fundamentally different from any other wave which allows superposition. E.g., any section of raised water principally creates a wave in all directions as well; that water waves are directed, regular waves, and that they are reflected on a wall, is owed to the same interference as photon propagation and reflection on a mirror. $\endgroup$ – Peter - Reinstate Monica Apr 2 '20 at 14:33

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