The excerpt below, about Rayleigh scattering, is taken from the following page: https://en.m.wikipedia.org/wiki/Rayleigh_scattering

Rayleigh scattering results from the electric polarizability of the particles. The oscillating electric field of a light wave acts on the charges within a particle, causing them to move at the same frequency. The particle therefore becomes a small radiating dipole whose radiation we see as scattered light.

Is Wikipedia trying to say that the light wave is causing the electrons, themselves, to move at the same frequency as the incoming light (i.e. visible light)? If so, how can this be? From my understanding, an electron, when in its orbital, doesn't vibrate in a classical manner, and it's not even known where the electron is when in an orbital.

If Wikipedia is not referring to the electrons in their orbitals, then what might they be referencing?

  • $\begingroup$ no it doesn't move at exactly the same frequency.. you can model the electron as a forced damped (or undamped) harmonic oscillator to get an idea of the frequency of oscillation (as well as the amplitude).... $\endgroup$
    – Bruce Lee
    Commented Feb 22, 2016 at 14:46
  • $\begingroup$ @BruceLee No, it does oscillate at exactly the same frequency. Same goes for the damped harmonic oscillator that you use as a model. $\endgroup$
    – garyp
    Commented Feb 22, 2016 at 14:52
  • $\begingroup$ @BruceLee Yes, it does. Using your model system, it moves at the same frequency because it is a linear and driven oscillator. $\endgroup$ Commented Feb 22, 2016 at 14:52
  • $\begingroup$ @garyp but the frequency should be changed if the system is damped... $\endgroup$
    – Bruce Lee
    Commented Feb 22, 2016 at 14:55
  • $\begingroup$ @BruceLee No. The resonant frequency changes, but not the frequency of oscillation when driven. It oscillates at the driving frequency. $\endgroup$
    – garyp
    Commented Feb 22, 2016 at 14:57

1 Answer 1


An electron in an atom is in a specific orbital only if it is completely isolated. The incoming radiation perturbs the system, so that it can no longer be described as being in a specific orbital. The electron will be in a superposition of all the orbitals of the isolated atom. In this state, the charge distribution in the atom is not fixed; it does oscillate at the driving frequency.

  • $\begingroup$ that helps make tremendous sense out of this issue. So that would go for any or all electrons within an atom... $\endgroup$
    – adam3033
    Commented Feb 22, 2016 at 15:12
  • $\begingroup$ @adam3033 Yes, any and all, although the contribution from inner-orbital electrons will be smaller. $\endgroup$
    – garyp
    Commented Feb 22, 2016 at 15:33
  • $\begingroup$ I wanted to resend my comment in case you'd missed it. I just needed some clarification. The process for Rayleigh, the electron(s) being in a superposition of all the orbitals, is not referring to electron(s) jumping levels? $\endgroup$
    – adam3033
    Commented Feb 23, 2016 at 11:52
  • $\begingroup$ ^^continuing.. something technically different than electronic transitions? $\endgroup$
    – adam3033
    Commented Feb 23, 2016 at 11:57
  • $\begingroup$ @adam3033 Yes, that's right. Related, but not the same. Maybe the simplest way to think about it: imagine an atom in a DC field electric field. The electron wave function will be a static combination of all orbitals, and the result will not be spherically symmetric. Now allow that DC field to oscillate... $\endgroup$
    – garyp
    Commented Feb 23, 2016 at 20:45

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