Why Gold Is Gold

The question

Two metals with the same outer shell, two different colours

Silver and gold sit in the same column of the periodic table. Each has ten d electrons in a filled shell and a single s electron on the outside, and both crystallise in the same face-centred cubic lattice with nearly the same spacing. Silver reflects the whole visible spectrum and looks white. Gold absorbs blue and violet light and looks yellow. Silver does start absorbing too, but only in the ultraviolet, near 3.9 eV. Gold's absorption begins about 1.5 eV lower, inside the visible range.

The difference comes from the 79 protons in gold's nucleus. They accelerate its inner electrons to more than half the speed of light. Once that happens, Schrödinger's equation is no longer accurate enough and Dirac's is required. The rest of this page follows that effect step by step, from the nucleus to the colour on your screen:

Dirac equationG, F, κ

Atomself-consistent LDA

Metalrelativistic LMTO bands

Excitationstransition state

Lightε(ω), R(λ)

ColourCIE → sRGB

The code uses only three facts about gold from outside: the atomic number Z = 79, the lattice constant 4.078 Å, and the room-temperature resistivity. The resistivity sets the tiny Drude damping and barely affects the colour. The reference data further down (NIST atomic tables, published band energies) is used only to check the results. It is never fed into the calculation.

Step 1 · the equation

The Dirac equation, reduced to two radial functions

Dirac's relativistic wave equation for an electron in a potential \(V\) is a matrix equation for a four-component spinor:

\[ \big[\, c\,\boldsymbol{\alpha}\cdot\mathbf{p} + \beta\, m c^{2} + V(r) \,\big]\,\psi = E\,\psi \]

In a spherical atom the angular part separates exactly. What remains are two coupled radial functions: a large component \(G = r g\) and a small component \(F = r f\). In atomic units (\(\hbar = m = e = 1\)), with energy measured from the rest mass:

\[ \frac{dG}{dr} = -\frac{\kappa}{r}G + \frac{E - V + 2c^{2}}{c}\,F, \qquad \frac{dF}{dr} = \frac{\kappa}{r}F - \frac{E - V}{c}\,G \]

Solved numerically on a logarithmic grid (Adams–Moulton, 4th order), with eigenvalues found by node counting and matching at the classical turning point.

The quantum number \(\kappa\) combines orbital and spin angular momentum. Each \(l > 0\) splits into two channels, \(j = l - \tfrac12\) (\(\kappa = l\)) and \(j = l + \tfrac12\) (\(\kappa = -l-1\)). That split is spin–orbit coupling, and nobody puts it in by hand; it falls out of the equation. The same is true of the mass–velocity and Darwin corrections. Near the nucleus the solution behaves as \(G \propto r^{\gamma}\) with \(\gamma = \sqrt{\kappa^{2} - (Z/c)^{2}}\). For gold's 1s electron that gives \(\gamma = 0.817\) instead of Schrödinger's 1.

This page uses one solver for both worlds. Setting \(c \to \infty\) (numerically \(c \times 10^{4}\)) turns the Dirac equation into Schrödinger's. So when the colour changes between the two, the speed of light is the only thing that changed.

Check that the solver is right. For a bare point nucleus, Dirac's equation has an exact solution, \(E = c^{2}\big[(1 + (Z/c)^{2}/(n - |\kappa| + \gamma)^{2})^{-1/2} - 1\big]\). The table compares that formula with this page's numerical solver for any Z you choose. The last column is the Schrödinger energy \(-Z^{2}/2n^{2}\), so you can see how much relativity alone moves each level.

Step 2 · the atom

Seventy-nine electrons, solved self-consistently

Each electron moves in the field of the nucleus plus the average field of the other 78. That field depends on where the electrons are, and where they are depends on the field. The loop is run until the two agree (Kohn–Sham density-functional theory, with local-density exchange and correlation):

\[ V(r) = -\frac{Z}{r} + \int \frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|}\,d^{3}r' + V_{xc}[\rho],\qquad 4\pi r^{2}\rho(r) = \sum_{n\kappa} \text{occ}_{n\kappa}\,\big(G_{n\kappa}^{2} + F_{n\kappa}^{2}\big) \]

Each ring is one subshell, drawn at its computed mean radius ⟨r⟩ on a square-root scale so the core stays visible. Dots show occupancy. Toggle the two worlds and watch the 6s ring (gold) shrink under Dirac while the 5d shells (violet) swell.

6s Dirac6s Schrödinger5d_5/2 Dirac5d Schrödinger

Radial probability densities \(G^{2} + F^{2}\) from the self-consistent atom. The dotted line marks the Wigner–Seitz radius of the metal, where neighbouring atoms begin.

Orbital energies of the valence levels. Relativity lowers 6s by about 1.7 eV, because s electrons reach the nucleus and gain relativistic mass. It raises the 5d levels, because the contracted inner s and p shells screen the nucleus better, and it splits 5d into j = 3/2 and 5/2. The 5d–6s gap shrinks from both ends.

Validation only. The same atom from the NIST reference tables (Kotochigova et al., VWN correlation). The small remaining differences come from the correlation functional and the point nucleus. NIST Au data

Step 3 · the metal

From one atom to a crystal: relativistic LMTO bands

In the solid, each atom sits in a Wigner–Seitz sphere of radius \(S\) (3.01 bohr for gold). Inside the sphere, the Dirac equation is solved at every energy, and the result is summarised by how the wave leaves the sphere, the logarithmic derivative \(D_\kappa(E) = S\,g'(S)/g(S)\). Andersen's linear muffin-tin orbital (LMTO) method turns that into a potential function, and a crystal state exists wherever the potential function matches the lattice's canonical structure constants:

\[ P_\kappa(E) = 2(2l+1)\,\frac{D_\kappa(E) + l + 1}{D_\kappa(E) - l}, \qquad \det\!\big[\,P(E) - S(\mathbf k)\,\big] = 0 \]

Fitting \(P_\kappa(E) \approx (E - C_\kappa)/(\Delta_\kappa + \gamma_\kappa(E - C_\kappa))\) gives the tight-binding Hamiltonian \(H = C + \sqrt{\Delta}\,S^{\gamma}(\mathbf k)\,\sqrt{\Delta}\), a 32 × 32 complex matrix (s, p, d, f × 2 spins) with separate Dirac parameters for every κ.

The structure constants \(S(\mathbf k)\) depend only on the geometry of the fcc lattice. All of the chemistry, and all of the relativity, enters through \(C_\kappa\), \(\Delta_\kappa\) and \(\gamma_\kappa\). The valence charge from the filled bands is fed back into the sphere potential and the cycle repeated until it converges.

d characters–p characterd band top → Fermi level: the absorption threshold

Energy bands along the standard path through the fcc Brillouin zone, with the Fermi level at 0. Line colour shows how much d character each state has. Right: density of states (total, and the d part shaded).

Potential parameters and self-consistency log

Largest potential change per self-consistency cycle (log scale).

Step 4 · excitations

Band energies are not excitation energies

Kohn–Sham eigenvalues describe the ground state. When light knocks a d electron up to the Fermi level, it leaves a hole in a fairly localised d shell, and the surrounding electrons relax around it. Local-density eigenvalues miss that relaxation. They put the gold d bands too close to the Fermi level, a known shortfall of about 1 eV in the 5d–6sp gap (Rangel et al., 2012). With Kohn–Sham bands alone, the computed gold comes out grey.

The fix used here is also computed from scratch. It is Slater's transition-state idea: the excitation energy equals the eigenvalue difference taken halfway through the transition. In a metal the excited electron stays nearby and screens the hole, so the halfway atom is neutral, \(5d^{9.5}6s^{1.5}\). The difference between its potential and the ground-state potential is a self-energy potential, trimmed to the ion core. It is added to the crystal potential:

\[ V_{S}(r) = V\big[d^{9.5}s^{1.5}\big] - V\big[d^{10}s^{1}\big], \qquad V \to V + \Theta(r)\,V_{S}(r), \quad \Theta = \Big[1 - (r/r_{c})^{8}\Big]^{3} \]

The cut-off \(r_c\) is not fitted to any colour. It is fixed variationally, as the radius that maximises the d → Fermi-level gap (after Ferreira, Marques & Teles' DFT-½).

The variational scan: d → Fermi-level gap against cut-off radius.

Validation only. Band energies of gold at Γ and L (eV, Fermi level = 0), against a full-potential PBE + spin–orbit calculation and photoemission experiments, as tabulated by Rangel et al., PRB 86, 125125 (2012). The correction lowers the d bands toward experiment as intended, but it also lifts the bottom of the s band (Γ₆⁺) by about 0.5 eV. That is a side effect of adding a core-region potential after self-consistency, and it barely matters for the colour, which depends on states within a few eV of the Fermi level.

Step 5 · light

How the metal answers light: ε(ω) and reflectance

Light at frequency ω can be absorbed when it lifts an electron from a filled band to an empty one. Fermi's golden rule adds up every such transition across the Brillouin zone. The momentum matrix elements come from the same Hamiltonian, as \(\partial H/\partial\mathbf k\) plus on-site dipoles from the radial Dirac functions. The free electrons at the Fermi surface add a Drude term. Its plasma frequency is also computed from the bands, as an average of the squared Fermi velocity.

\[ \varepsilon_2(\omega) = \frac{4\pi^{2}}{\Omega\,\omega^{2}} \sum_{\mathbf k,n,m} \big|\langle m\mathbf k|\hat{\mathbf e}\cdot\mathbf p|n\mathbf k\rangle\big|^{2}\,\delta(E_{m\mathbf k} - E_{n\mathbf k} - \omega), \qquad \varepsilon_1(\omega) = 1 + \frac{2}{\pi}\,\mathcal P\!\!\int_0^\infty \frac{\omega'\varepsilon_2(\omega')}{\omega'^{2} - \omega^{2}}\,d\omega' \]

Then \(\tilde n = \sqrt{\varepsilon}\), and the normal-incidence reflectance is \(R = |(\tilde n - 1)/(\tilde n + 1)|^{2}\).

Interband absorption, the imaginary part ε₂ without the Drude term. The band marks the visible range. Gold's absorption begins inside it; silver's and non-relativistic gold's begin beyond violet.

Computed reflectance across the visible spectrum. Where the curve dips, that colour is absorbed instead of reflected.

Step 6 · colour

From a spectrum to a colour you can see

The eye reduces any spectrum to three numbers. The reflectance is multiplied by a daylight illuminant (a 6504 K black body) and weighted with the CIE 1931 colour-matching functions to give tristimulus values X, Y, Z. Those are converted to sRGB and white-balanced, so that a perfect mirror comes out exactly white.

\[ X = \int R(\lambda)\,I(\lambda)\,\bar x(\lambda)\,d\lambda, \quad Y = \int R\,I\,\bar y\,d\lambda, \quad Z = \int R\,I\,\bar z\,d\lambda \;\;\longrightarrow\;\; \text{sRGB} \]

The CIE colour-matching functions (lines, from Wyman, Sloan and Shirley's analytic fit) against computed gold reflectance (shaded). The reflectance falls away under the z̄ (blue) lobe, so the Z value drops and the colour turns yellow.

One bounce is what a flat polished surface returns. After two bounces, as in a crease, a carved letter or the inside of a ring, the reflectance is squared and the colour gets richer. That is part of why gold objects look more golden than a single reflection would suggest.

What was assumed

The approximations, and where they show

Ingredient	Treatment here	Effect
Exchange–correlation	Local density (Slater exchange + Perdew–Zunger)	Atomic levels agree with NIST to about 0.03 eV
Crystal potential	Atomic-sphere approximation, frozen core, point nucleus	sp band near X runs a few tenths of an eV low
Excitation energies	Slater transition state, cut-off radius chosen variationally	Opens the d gap by about 0.35 eV; the measured value is still about 0.3 eV further away
Lifetime broadening	Gaussian, set with the slider (default 0.30 eV)	Changes the shade, not the answer
Drude damping	From the 2.2 µΩ·cm room-temperature resistivity	Negligible in the visible

Because the computed absorption edge still sits a little too low in energy, the computed gold absorbs slightly into the green, which gives it a warm, rose-gold cast. Real gold's reflectance stays near 90% down to about 550 nm before it falls. The conclusion does not depend on any of these approximations. With the speed of light at its real value, gold's d → Fermi-level gap lands inside the visible spectrum and the metal is yellow. Push c toward infinity with everything else unchanged, and the gap moves into the ultraviolet and gold turns as white as silver. Try it with the speed-of-light control.