Rule I learned:

only photons with quanta of energy match matching the exact difference between energy levels can be absorbed/ reemitted.

But to me optical density works opposite to this rule. Photons of any frequency (not just the resonance frequency, and frequency represents the energy it carries) can pass through a transparent material, and more importantly be slowed down to some extent. That the light (not just visible light) gets slow down is an evidence that atoms of any material is able to absorb the photon and release back into the inter -atomic voids.

Maybe I have a misconception. Perhaps electrons don't have to change their energy level in order to reemit the photons/ energy they have absorbed....

  • $\begingroup$ I think the first answer to this question also answers your question. Hopefully that helps :) P.s. I would've put this in the comments but I'm not quite there with the rep yet! $\endgroup$
    – Sean
    Aug 6, 2015 at 19:25
  • $\begingroup$ The author of that answer doesn't tell you where that "delay" comes from? He just said you will get it by doing some "math".... $\endgroup$ Aug 6, 2015 at 20:40
  • $\begingroup$ However, he does say that the electrons are driven back and forth by the photon energy. It doesn't answer my question entirely. What if a single photon from my flashlight carries less energy than is needed for electron to make even a single-gap jump? Does that mean the photon will not be absorbed, and so the light will not pass through? $\endgroup$ Aug 6, 2015 at 20:42
  • $\begingroup$ If I understand correctly, there is no electron transitions occurring within the atoms in this situation. Instead, the electrons are being driven by the electric field of the photon (i.e. being "pushed" from one side of the atom to the other, treating the nucleus as fixed). Therefore even a low energy photon should be able to pass through the medium (avoiding resonant modes of the atoms). $\endgroup$
    – Sean
    Aug 6, 2015 at 21:16
  • $\begingroup$ If the energy it takes for electron to jump from 1 to 3 is 15 eV, the photon has 15.2eV. Being certain that the extra .2 eV won't make full transition, what happen to this extra 0.2 that is different than 15eV used to complete the transition? $\endgroup$ Aug 6, 2015 at 22:06

1 Answer 1


The following is a solid: The atoms with the electrons in specific energy levels and a lattice where the atoms are held, with vibrational and rotational degrees of freedom. The energy levels for these degrees of freedom can be very small.

A photon impinging on the lattice will interact mainly with the electric field of the lattice. If it has an energy that can raise an electron to a higher level, it is absorbed, and then reemitted in a random direction. If this happens with high probability the lattice is not transparent, but opaque.

For optical frequencies the lattice of a piece of glass is transparent,there are no energy levels that can absorb the optical photons. This means that the photons scatter with the field of the lattice elastically, not loosing momentum and very little energy to the field of the lattice, because the exchanges are very small. (energy loss would be seen in a change in frequency, but it is too small to be observable).

A light beam emerges from a confluence of photons. Photons are quantum mechanical entities, and the superposition of their wavefunctions creates the classical electromagnetic wave.

The slowing down of the light beam through a transparent material comes from the random change in direction by the elastic scatters of the individual photons, which are traveling with speed c. This gives a larger path length to the individual photon, thus a different effective velocity to the light beam which emerges from these photons.


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