What actually happens when light meets a surface(QED or QM)?

1. Aug 4, 2015

Christian Grey

I want to know what actually happens when light meets a surface like water or wood.

Quantum mechanics says that objects are neither "transparent" nor "opaque". Rather a system as a whole can accept "energy packets" in certain ranges of energy and has no states available to accept other ranges of energy. If something is "transparent" it means its energy level structure (for crystal systems this is called the "band structure") is such that it can't accept energy packets between ~1.7eV to 3.0eV which corresponds to light in the visible spectrum with wavelengths of 700nm (red) to 400nm (violet). There is of course nothing special about that particular energy range except that it happens to be the range that our human eyes have evolved to see. If I instead, for example, looked in the infrared range, like in this image:

http://www.jwst.nasa.gov/exhibit/ir2.jpg

I'd see that the man's eyeglasses are in fact opaque in that range (meaning that the glass does accept energy packets corresponding to the infrared range, even if it didn't in the visual range) and what he's holding there is a "black" garbage bag, which is in fact transparent to that range of light. Here's the same picture in the visual range:

http://www.jwst.nasa.gov/exhibit/ir1.jpg

While quantum electrodynamics says it works on by adding (magnitude only not direction) vectors (from the strange theory of light and matter) you start the stopwatch when photon leaves the light source and you stop the watch when the photon reaches the photo-detector, and then add the two vectors or more to get the resultant vector, which tells whether reflection or refraction will take place.

I am talking about what Feynman discusses about, in the book The strange theory light and matter,when he says that out of 100 photons 4 are partially reflected and 96 are transmitted,and how many photons are refracted or transmitted depends on the thickness of the glass, which he then demonstrates by using stopwatch,he places two photodetectors one above the glass and one below the glass, and starts the stopwatch when the photon moves from the light source and stops it when it has reached either of the photodetectors.

My question is which one is right QM(energy band) or QED(using stopwatch) ?
What actually happens when light meets surface like water, glass or wood? Which is right: QM or QED? Am I missing something in QED that I don't know (because it's called most complete theory of light) or it is best explained by condensed matter physics?

2. Aug 4, 2015

Staff: Mentor

Feynman is about as correct as you can get at the lay level - but you have to take with a bit of a grain of salt when he says you wont have to unlearn anything later.

Beyond that - it gets very very complicated - see the following for what actually happens with absorption and inside the medium

Like I said - it isn't correct to say you wont need to unlearn anything from Feynman - but you need to be quite advanced for it to be an issue.

Thanks
Bill

Last edited by a moderator: May 7, 2017
3. Aug 4, 2015

BvU

As you can expect, both are right. But you want to keep things separate: your first half is about absorption, but then Feynman's thingy is about transmission/reflection: zero absorption (all photons either get transmitted or reflected). His dealing is that there is a probability for each of the two that depends on -- among other things -- the thickness of the glass. Has to do with interference between reflected and transmitted waves. (hence the prominence of the wavelength in the quantitative expressions). For the question: how a single photon can be subject to these interference expressions when it either reflects or transmits ? you are back in quantum mechanics (but simple wave mechanics and even simpler optics come up with good descriptions as well).

Just like with Feynmans ecellent expose about the double slit experiment: an individual photon, electron (, baseball ) interferes with itself !