@JohnCuster's answer to Why is gold golden? is based on the band structure of the delocalized conduction electrons in gold, whereas other answers there tend to support the "Gold is gold due to relativistic shifts in atomic orbitals" comparing silver to gold for example:

The absorption you see is a $\mathrm{5d \rightarrow 6s}$ transition. For the silver $\mathrm{4d \rightarrow 5s}$ transition, the absorption is in the UV region, but the contraction gives gold a blue absorption (i.e. less blue is reflected). Our eyes thus see a yellow color reflected.

as does this answer to Why do “relativistic effects” come into play, when dealing with superheavy atoms?, and my answer to How does the thin gold film in the glass of spacesuit helmets block thermal IR but transmit visible? What's the property? also parrots the absorption line theory (citing an ACS "communities post") and lists specific gold atomic lines in the blue.


  1. Can the color of gold metal be reproduced by modeling only the conduction electron band structure, or is it necessary to invoke atomic absorption lines?
  2. Since the conduction bands depend on the details of the localized atomic structure, does it turn out that a relativistic effect affects those, and so indirectly affects the conduction bands and their optical properties?

metal reflectivities from near IR to visible to near UV; gold, silver and aluminum

Source, found in Wikipedia's Relativistic quantum chemistry.

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    $\begingroup$ I gave my +1 awhile back. I worry a bit that asking both these questions in the same post might discourage some people who would only feel comfortable answering just one of them. I'd recommend asking two separate questions and providing a link to the other one in each of them! $\endgroup$ Commented Jun 19, 2021 at 4:52
  • $\begingroup$ @NikeDattani I had a hunch that the two items were so intertwined that a single good answer could cover both. I'm peeling wzkchem5's answer like an onion and learning a lot in the process. I have a hunch that it is covering all of my question nicely thought, doesn't it? $\endgroup$
    – uhoh
    Commented Jun 19, 2021 at 21:22

2 Answers 2


A band can either be expressed as a linear combination of plane waves, or as a linear combination of atomic orbitals. By asserting that the color of gold can be explained by relativistic shifts of atomic orbitals, one have implicitly expanded the conduction band wavefunctions into a linear combination of atomic orbitals. Then it follows easily that if an atomic orbital contributes significantly to a band, and the energy of the atomic orbital changes, then the energy of the band will tend to change in the same direction. One can as well explain the changes of the band structure starting from a plane wave expansion, only that the wordings will become clumsier. For example, the usual conclusion that relativistic effects usually lower the energy of s orbitals and raise the energy of d orbitals, will have to be rephrased as: "relativistic effects tend to lower the energy of those linear combinations of plane waves that have no angular node near a nucleus, but raise the energy of those linear combinations that have two angular nodes near a nucleus." The latter phrasing is much less intuitive, but is valid.

If you are interested in a reproduction of the color of gold without recourse to atomic energy levels, just imagine that you perform an all-electron, relativistic band structure calculation of gold metal, with bare, "non-pseudized" nuclear attraction potentials, and using a huge plane-wave basis set that is capable of expanding even the 1s orbitals of gold. This reproduces the color of gold as long as the density functional, quasiparticle theory, or whatever theory you are using to calculate the band structure, is accurate enough. Of course this is probably not doable in practice since you need an insanely high cutoff energy to describe the 1s orbital of gold. You can circumvent this by using a pseudopotential, but then your calculation will involve atomic energy levels implicitly, since pseudopotentials are usually parameterized by atomic energy levels and/or atomic wavefunctions.

  • $\begingroup$ Thank you for your insightful and very instructive answer! I just wrote this comment. $\endgroup$
    – uhoh
    Commented Jun 19, 2021 at 21:23

Lets start with some data from Optical Transmission in Thin Metal Films where they deposited a range of metals on glass and measured their transmission. Sticking with their data on the noble metals, they see pretty much what one would expect - some changes in shape as the film thickness goes from 0.5 to 4nm of gold, coupled with the clear emergence of the plasmon absorption feature. There are absolutely no atomic absorption lines in the spectrum (completely expected since any intra-gold transitions are at much much higher energies than the visible range of a few eV. (Also note the comparison of Cu, Ag, and Au, clearly seeing the plasmon absorption features in each of them that are close together, but close in the visible range and thus we are highly sensitive to small differences).

Recall that plasmons are a feature of bulk (not atomic) systems with free carriers. In the noble metals the conduction bands are made up of the outer electrons that are donated to the cause of crystal binding. The plasmon frequency depends on the density of free carriers and their effective mass - that is pretty much all. All of the noble metals have highly similar band structure and Fermi surfaces (both bulk, not atomic, features). A peek at a table of electron binding energies (such as at LBL) shows you that the outermost electron of Au has a binding energy of 57.2eV. The next one is 74.2eV. For comparison, Ag has 58.3 and 63.7 eV, and Cu 75.1 and 73.3 eV, respectively. So, really, the outer electrons of Au are not particularly different from the outer electrons of the other noble metals as one would expect from their band structures noted in your first link.

But, remember that the nice picture we keep in our heads of hydrogen-like atomic orbitals is, well, wrong (except for one-electron atoms, sort of). Bethe and Salpeter wrote an entire book (Quantum Mechanics of One- and Two-Electron Atoms, and I count myself extremely lucky to have taken graduate quantum mechanics from Ed Salpeter) on the difficulties of adding that second electron. With more and more electrons being added, the difficulties mount rapidly because of screening and other factors. Sometimes I sit in awe and contemplate just how well the hydrogen-like orbitals survive all of that.

So, where do 'relativistic effects' come in to play? Looking back at the electron binding energy tables, one sees that the Au 1s electron has a binding energy of 80.725keV, about 15% of the rest energy of an electron. What this means is that, should you want to do a full-up multi-electron calculation of all 79 electron states, you are going to have to deal with the relativistic effects of those core orbitals which will have an influence on the outer states because they have a screening effect different from the simple hydrogen levels. All that means, is that the last few outer electrons of Au see a slightly different potential than they would have if there were no relativity. In the end, given the binding energies for Au compared with Cu and Ag there really is not much difference. The color differences of them are all done to the plasmon absorption in the visible, a bulk feature, with slight different plasmon energies for Cu, Ag, and Au. So, look to electron density and effective mass for your answer.

  • $\begingroup$ Thanks for your answer, I'm glad you stopped by :-) I will have to eventually redo this answer based on what I learn reading through answers here over and over, I'm learning a lot. I did mention two strong Au I transitions at 460.75 and 479.26 nm there, so I'm not sure what "...no atomic absorption lines in the spectrum (completely expected since any intra-gold transitions are at much much higher energies than the visible range of a few eV." means exactly, is it possible to clarify? $\endgroup$
    – uhoh
    Commented Jun 22, 2021 at 1:26
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    $\begingroup$ @uhoh - well, a gas is not a solid. Strong absorption lines would show up in the data in the paper. $\endgroup$
    – Jon Custer
    Commented Jun 22, 2021 at 2:05
  • $\begingroup$ Ah, yes I see what you mean. The world would look very different if we regularly saw atomic transitions from solids (absorption and emission from hot ones). Thanks! $\endgroup$
    – uhoh
    Commented Jun 22, 2021 at 2:07
  • $\begingroup$ @uhoh - just to clarify, the point is that an electron in an excited atomic state in a dilute gas has few options to lose the energy beyond emission (and only the atomic absoprtion lines to get the electron excited). By contrast, in a solid, there are many available states above the Fermi surface (essentially a continuum considering all wave vectors $k$), and once an electron is put up there it rapidly thermalizes back down through electron-electron and electron-phonon interactions (order picoseconds or less). The physics is just very different. $\endgroup$
    – Jon Custer
    Commented Jun 22, 2021 at 14:49
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    $\begingroup$ Well, that thermalizes quite readily as well, popping up to the Fermi surface and disappearing into the empty part of the conduction band. The density of things to thermalize with results in very fast non-radiative energy loss, much faster than any radiative lifetime. $\endgroup$
    – Jon Custer
    Commented Jun 22, 2021 at 15:05

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