# How does quantum mechanics explain the color of macroscopic objects?

I am familiar with two explanations for the phenomenon of visible color.

The first is that, for any given object, there are some wavelengths of light it reflects and some it absorbs. The reflected rays reach our eyes and thus determine the color of the object from a human perspective. So if an object appears red under white light, it's because it absorbed light from the green-blue-violet end of the visible spectrum and reflected light from the red end.

However, my understanding of light from the quantum perspective seems to contradict this picture. In quantum mechanics (as I understand it) "reflection" is the process of a photon being absorbed and then re-emitted by an atom. In order to be absorbed by an atom, the photon must have energy corresponding precisely to the difference between two possible electron energy states, meaning only certain discrete wavelengths can be absorbed, and these wavelengths should be exactly the same as those emitted.

How is it that macroscopic objects absorb certain wavelengths and re-emit others given what's happening at the atomic level? What am I missing or misunderstanding here?

• Are you talking about the color that you see or the color as measured by an instrument such as a spectrograph? Mar 20, 2023 at 22:49
• Reflection (mirror) and scattering (more common) are both process where the incoming photon changes direction, i.e. is not absorbed. There is interesting discussion about whether this photon is the original or not ..... but in physics in general it is not proven or disproven ... but it may not matter. For your color question the process is scattering not truly reflection. Mar 21, 2023 at 0:56
• Mar 21, 2023 at 1:57
• It's worth noting that the perceived colour of a macroscopic object is not uniquely determined by its emission spectrum, since illusions can cause objects with the same emission spectra to appear different. Mar 21, 2023 at 2:41
• Side note: The macroscopic description is much older than our knowledge about and understanding of atomic spectra (which strictly speaking means only in gases; fluids, solids, plasma are different species in this respect). So conclusions from the first can be wrong or incomplete in several aspects, while the second provide more details and a surprise every now and then. Mar 21, 2023 at 5:43

## 2 Answers

There are two important details. One is that in a solid, the atoms interact with each other, which broadens the lines into a continuum.* The second is that when an atom/material absorbs a photon and goes into an excited state, it can also decay to a different state than the one it started in. A common example of this is fluorescence, where an atom absorbs a high energy photon, decays to an intermediate excited state (emitting a photon or phonon in the process), and then decays again to emit a lower energy photon. This process splits up the original photon's energy. A material that absorbs perfectly at some color would convert all the energy absorbed at that wavelength to phonons (lattice vibrations).

*Even an atom in free space does not have infinitely narrow lines. It interacts with the electromagnetic field around it, which broadens the lines. This interaction is precisely the interaction that leads to spontaneous emission (and in fact the linewidth is inversely proportional to the lifetime of the excited state).

A macroscopic substance is composed of countless fundamental particles in continuous mutual interaction with every other fundamental particle nearby. Every interaction has associated, in-principle-quantized energy states.

A picture of the absorption or emission spectrum of such a substance would then look like a near continuum of infinitesimally thin lines.

However, everything in the substance is also moving, vibrating, oscillating, etc, in random directions and magnitudes. Insofar as the interaction with light goes, this constant movement corresponds to continuously changing levels of doppler shift. Since any incident photon has many nearby particles to potentially interact with, this blurs the near continuum of infinitesimally thin lines into a continuous spectrum of overlapping finite bandwidth bands.