Light Traveling in Transparent Medium
Most recent answer: 05/29/2012
Q:
When visible light travels thru a tranparent medium does it always cause electrons in the medium's atoms to jump to an elevated subshell and then the electrons drop back while emitting a photon of the original frequency? Or does light just travel thru the 'gaps' between molecules without affecting atoms and electons?
- Jim Wawrzyniak (age 72)
Mentor, OH, USA
- Jim Wawrzyniak (age 72)
Mentor, OH, USA
A:
Since light travels more slowly through transparent materials than through a vacuum (that's why it bends as it enters or leaves glass), it must somehow be interacting with the materials. It's definitely not "traveling through gaps". There really aren't any gaps in the electronic waves in a solid or liquid. Even if there were gaps between the nuclei of the atoms, they'd be so much smaller than the wavelength of the light that the light couldn't propagate through them.
Your other idea is a little closer, but not quite accurate. In the presence of the electromagnetic field the electrons go into states that are combinations of the states that were present in the absence of the field. These include some of the original lowest-energy "ground" state and some of the original higher energy states. As the wave leaves from a region the electron states smoothly settle back to be the ones that were present originally in the absence of the field.
Your description of processes in which electrons "jump to an elevated subshell and then the electrons drop back while emitting a photon of the original frequency" sounds similar but without the possibility of the states being superpositions of the original definite-energy states. It's closer to a description of fluorescence, although in fluorescence the range of excited states allows the emitted photon to be a little lower energy than the absorbed photon. Processes with real absorption-emission cause materials not to be fully transparent.
Mike W.
Your other idea is a little closer, but not quite accurate. In the presence of the electromagnetic field the electrons go into states that are combinations of the states that were present in the absence of the field. These include some of the original lowest-energy "ground" state and some of the original higher energy states. As the wave leaves from a region the electron states smoothly settle back to be the ones that were present originally in the absence of the field.
Your description of processes in which electrons "jump to an elevated subshell and then the electrons drop back while emitting a photon of the original frequency" sounds similar but without the possibility of the states being superpositions of the original definite-energy states. It's closer to a description of fluorescence, although in fluorescence the range of excited states allows the emitted photon to be a little lower energy than the absorbed photon. Processes with real absorption-emission cause materials not to be fully transparent.
Mike W.
(published on 05/29/2012)
Follow-Up #1: light-atom interaction
Q:
Thank you for your reply. You provided insight to light/matter interaction when you said the light wave continues after exciting a nearby atom (molecule).
I have a series of questions:
1. At the instant the light wave excites the atom, is the electric field of the wave diminished (magnetic field too?) and then after passing the atom the wave field(s) return to original values?
2. To keep it simple, assume visible light from a perfect beam source producing n photons/second which have equal frequency and are in phase. Is it fair to sum the waves into a single wave with an electric (and magnetic) field equal to n times that of a single photon? Then how does the wave/atom behavior change when this wave with a larger electric field passes?
Jim W
- Jim W (age 72)
Mentor, OH USA
- Jim W (age 72)
Mentor, OH USA
A:
Let me just deal with question 2 for now.
Yes, it's fair to say that the classical E and B for n photons are n times those for one, because all those values are exactly zero. E and B only acquire non-zero expectation values for states which do not have a precise number of photons. For more typical states, with indefinite photon numbers and non-zero expectation of the fields, the field magnitude goes as the square-root of the average photon number. This makes sense because the energy goes as the square of the field and also as the photon number.
On question 1, it's easier for me to think in terms of classical electric and magnetic susceptibilities of a solid, i.e. a large collection of atoms. These are frequency-dependent complex numbers, with the imaginary part of the complex number showing that the response of the material is not quite instantaneous. For a wave pulse with a narrow frequency width, and hence a broad width in time, the oscillations from the material slightly lag those of the incoming wave. For pulses with a broader frequency range, this lag is different for the different frequency components, so the pulse shape is distorted.
Mike W.
Yes, it's fair to say that the classical E and B for n photons are n times those for one, because all those values are exactly zero. E and B only acquire non-zero expectation values for states which do not have a precise number of photons. For more typical states, with indefinite photon numbers and non-zero expectation of the fields, the field magnitude goes as the square-root of the average photon number. This makes sense because the energy goes as the square of the field and also as the photon number.
On question 1, it's easier for me to think in terms of classical electric and magnetic susceptibilities of a solid, i.e. a large collection of atoms. These are frequency-dependent complex numbers, with the imaginary part of the complex number showing that the response of the material is not quite instantaneous. For a wave pulse with a narrow frequency width, and hence a broad width in time, the oscillations from the material slightly lag those of the incoming wave. For pulses with a broader frequency range, this lag is different for the different frequency components, so the pulse shape is distorted.
Mike W.
(published on 05/30/2012)