Seeing colors: photons vs waves

Question

As an atmospheric physics major I am familiar with electromagnetic radiation in the atmosphere and what dictates what wavelength objects will emit at. When observing radiation in the atmosphere it is always thought of as a wave, whether it be longwave or shortwave. Recently though I have been introduced to the quantum world and I am having trouble translating between the wave model and the particle model of light.

I understand that the sun, that has a temp of 6000k, has its peak emission in the visible spectrum. When that light makes it to Earth, and hits an object, that light will be absorbed and what is not absorbed will be reflected back and we will see that color. I am having trouble understanding how this works when thinking of photons. How do we see colors of an object when thinking about photons?

Recently I watched video on why glass is transparent. It said that the electrons in glass were arranged in such a way that when they encountered photons there was not enough energy for them to reach a higher energy level. This made me think of why absorbing that photon and reaching a higher energy level is necessary to not being transparent. It made it seem like that in order to not be transparent photons had to be absorbed so electrons could emit a photon of that color.

When thinking about photons, do we see colors because photons are being reflected back like a wave does? Or do we see electrons emitting a certain color of photon as it lowers energy levels? For instance, I have a green wall with a window. Are photons passing right through the glass but hitting my wall, exciting the electrons up a level then as they go to a lower level a green photon is emitted or do the photons hit my wall and jsut bounce back?

Answer 1

Thus is a very common confusion, and it occurs because light is neither a wave nor a particle but instead it's (currently best described as) a quantum field. The wave and particle descriptions are approximations that apply under some circumstances. In particular the photon model is a good way to describe how the electromagnetic field exchanges energy with it's environment. When the light transfers energy to something else the energy transferred is an integral number of photon energies.

So, in your example, glass is clear because for visible light there aren't any energy levels spaced one photon energy apart. Since light can only interact with the glass by exchanging energy in photon sized lumps the interaction can't occur. Glass does absorb in the ultraviolet because photon energy is proportional to the light frequency and at uv frequencies the photon energy is big enough to excite electrons in the glass.

The green pigment in the paint on your wall has been chosen to have electronic excitations that correspond to the photon energy of red and blue light, but none with an energy matching green light. This means that red and blue light falling on the wall is absorbed but green light is reflected. In general with solids, when a light excites an electronic transition and is absorbed, the energy of the excited electrons is dissipated as lattice vibrations. Only in some circumstances is it re-emitted as light, in which case you get fluorescence or phosphorescence. So it isn't the case that light is absorbed and re-emitted as green light. The green light is reflected and stays green, while the other colours are absorbed and their energy ends up heating the wall.

The green light reaching your retina has a photon energy that matches the optical pigments in the M-cones. Thus the light is absorbed (in photon sized chunks) by exciting electrons in the optical pigments.

Answer 2

John's answer is clear for the ensemble of photons that make up the electromagnetic wave

If you are really asking how individual photons end up making the classical electromagnetic wave, whether reflected or not, you have to dig in further in to quantum electrodynamics. Lubos Motl has in his blog an entry of how classical waves emerge from a large ensemble of photons.

Classical equations emerge from the quantum mechanical ones in a coherent fashion but electromagnetism has the extra elegance to have Maxwell's equations both for the classical waves and the quantum mechanical equation that gives a wave function for a photon. The same four dimensional potential enters, and thus the continuity of classical and quantum in the frequency for the photon energy , E=h*nu, and the frequency expressed in space by a large ensemble of photons of frequency nu.

Answer 3

The energy of a photon gives rise to its frequency:

E = h*f

where f is the frequency and h is the Planck constant.

Frequency is inversaly proportional to the wavelength according to:

lambda*f=c

where lambda is the wavelength and c is the speed of light.

Therefore you can state that the photons with higher energy level (particle nature of the light) have higher frequencies (wave nature of the light) and lower wavelenghts.