# Why can I see a diamond?

So a diamond has a energy band gap of $\approx 5$eV. If that is too much for visible light $\approx 1.6$eV to be absorbed and then it travels straight through a diamond.

Although, I still see a diamond. I know that when light goes through, there is a change in index of refraction. If visible light does not interact with the diamond then how does it reflect off the surface?

• Possible Duplicate: physics.stackexchange.com/q/43361 – user108787 Nov 28 '16 at 23:44
• Duplicate physics.stackexchange.com/q/7437 where it explains that the light does interact with a transparent material. – Farcher Nov 28 '16 at 23:47
• Mind you, this is a brilliant - a cut specifically designed to cause many refractions that add up to a nice shiny gemstone. If you made a plane of perfect diamond (good luck), it would be more "invisible" than glass. Note that similar cuts are also used on glass, with similar (though still very different) effects. – Luaan Nov 29 '16 at 8:59
• Through? Yes. Straight? No. – candied_orange Nov 29 '16 at 12:20
• What your picture seems to show is a collage of images of things outside the diamond, and you infer the presence of the diamond from this complicated collage. The various images result from refractions and total internal reflections on the surfaces of the diamond with the pattern depending on the angles of these surfaces. – Henry Nov 30 '16 at 13:16

Light does not have to make outer shell electrons leap the full band gap to interact with them. Electrons can be excited to virtual states, whence photons of the same energy and momentum are emitted. So although it is true that the absorption loss for pure diamond is very small as you've rightly inferred, a phase delay arises from this interaction, as I discuss further in this answer here.

In diamond, this phase delay is big: diamond has a refractive index of about 2.4 for visible light. So you see all the effects of the strong difference between the diamond's refractive index and that of the air around it: you see a diamond plate shift transmitted light sideways relative to the background, you see a strong specular reflexion from surfaces (the power reflexion ratio is about 17% for diamond) and, for white light, you see strong dispersion into colors for glancing reflexions and transmissions.

• this argument sounds a bit circular to me; after all the reason why the refractive index is 2.4 and not 1 is that there is interaction. – hyportnex Nov 29 '16 at 1:49
• @hyportnex Isn't then that the answer - that the phase-delay inducing interaction is why you see the diamond, and the refractive index is merely the quatifying of the effect's extent. I'm saying this, or at least, that's what I mean to say. – Selene Routley Nov 29 '16 at 2:03
• @WetSavannaAnimalakaRodVance You are giving the classical expalnation. At a photon level, electromagnetic vertices would mean a change in frequency and loss of phases if there really exists absorption and reemssion , unless it is elastic scattering . A transparent lattice is a macroscopic manifestation of a quantum mechanical state and individual photons interact with the whole lattice , imo. One could say "electrons can be..." but it gives the impression of interactions at an individual level, which has to be small for a lattice to be transparent. – anna v Nov 30 '16 at 4:57
• @annav "individual photons interact with the whole lattice , imo"; indeed they do, as I state in the latter parts of the answer I refer to. Although the explanation is often oversimplified to imply that the individual electrons act sequentially.. It still implies a phase delay. – Selene Routley Nov 30 '16 at 5:14

My take:

Photons have a wave function that carries the electric and magnetic field information which is a solution of the quantized Maxwell's equation. Thus there exist phases between photons and the superposition is again a wavefunction which macroscopically builds the electric and magnetic fields. Lubos Motl has a blog entry on how this happens at a QFT level.

When a photon hits a boundary condition , three things can happen: a) it can scatter elastically, which means it retains its frequency but changes angle, b)it can scatter inelastically, which means it changes frequency, or c) it can be absorbed raising the energy level of an electron ( in a lattice, in a molecule, in an atom) and a different photon is emitted and phases are lost.

For a reflective surface where images are retained a) happens: all phases in the emergent ensemble of photons are intact.

For an opaque surface c) happens

For a transparent lattice it is still a). The photon interacts elastically with the lattice, phases through the ensemble are kept coherent and thus we see through glass. It is a "photon + lattice" scattering at an individual level, but for a medium to be transparent the emergent ensemble of photons must retain coherence. Phases change coherently in the quantum mechanical solution, otherwise there would be no transparency

In a diamond there are reflective surfaces that back scatter part of the light , images are distorted but still phase information is coherent.

b) is the case where colors change if the scattering is with the whole lattice and phase coherence can be kept.

The Kramers-Kronig relations connect the real part of the permittivity $\epsilon_r(\omega)$ to the imaginary part $\epsilon_i(\omega)$ [or real and imaginary refractive index, $n_r(\omega)$ and $n_i(\omega)$], i.e., the phase velocity of light in a crystal at a given frequency to the absorption properties over the whole frequency range. Thus the high refractive index of diamond in the visible region is related to the absorption bands in the UV region related to the band gap.

• I like that take on things - highly original and correct! – Selene Routley Nov 29 '16 at 5:27
• I don't buy this argument. The Kramers-Kronig relations tell you that you cannot have dispersion, i.e. changes in $\epsilon_r(\omega)$ / $n_r(\omega)$, as a function of frequency, without having a nonzero absorption, and vice versa. (and, moreover, their $1/\Delta\omega$ integrand tells you that the two can't be very far from each other, either.) However, a constant $n=\mathrm{Re}(n)=n_0>1$ is also an analytical function of $\omega$, and it is also consistent with the Kramers-Kronig relations. – Emilio Pisanty Nov 30 '16 at 0:59
• Kramers-Kronig analysis is a common tool in relating measured index of refraction to extinction coefficient (or related connected variables) or vice versa in the investigation of optical properties of solids. Here you'll find specifically some of the many investigations using the Kramers-Kronig relations for the analysis of the optical properties specifically of diamond: Phillip HR and Taft EA, Phys. Rev. 127, p. 15 (1962) – freecharly Nov 30 '16 at 5:33
• @Emilio Pisanty - Continued: Sobolev et al. Optics and Spectroscopy 88,2,217 (2000) The Kramers-Kronig relations depend only on the assumption of causality and linearity and are generally valid. And they relate connected quantities also at different frequencies. I wonder what extinction coefficient frequency dependence you will get from your assumed (unphysical) constant n. Maybe you will be surprised. – freecharly Nov 30 '16 at 6:15
• @WetSavannaAnimalakaRodVance Yeah, except keep in mind that "dispersion" doesn't mean "$n>1$", it means $\frac{dn}{d\omega}\neq 0$. – Emilio Pisanty Nov 30 '16 at 22:44

The electrical field of the EM wave interacts strongly with the electrons. Best way to get an intuition is to regard them as Lorentz oscillators. In diamond these have resonant frequencies above 5 eV, so they will respond out of phase for visible light. This gives rise to scattered waves, that add to the incident wave, giving a phase velocity that is lower than $c$.