Part 1 · ~1 fs
Absorption: Retinal as a Particle in a Box
Retinal is bound to the protein through a lysine, forming a protonated Schiff base. Its π electrons are delocalised along a chain of alternating single and double bonds from carbon 5 to the nitrogen, which makes it a natural one-dimensional box.
In the free-electron model, each π electron sits in a level of a box of length L:
\[ E_n = \frac{n^2 h^2}{8 m_e L^2} \]
With N π electrons filling levels in pairs, the highest occupied level is n = N/2 and the lowest empty one is N/2 + 1. The lowest-energy absorption promotes one electron across that gap:
\[ \Delta E = \frac{(N+1)\,h^2}{8 m_e L^2}, \qquad \lambda = \frac{8 m_e c L^2}{(N+1)\,h} \]
For the retinal Schiff base, take N = 12 (six C=C bonds plus C=N) and L ≈ 11 bonds × 1.40 Å ≈ 15.4 Å. This gives λ ≈ 600 nm, the right order of magnitude for a crude model.
Full derivation
Step 1 — Schrödinger equation in the box
Inside the chain (0 < x < L) the potential is taken as zero; outside it is infinite, so the wavefunction must vanish at both ends:
\[ -\frac{\hbar^2}{2m_e}\frac{d^2\psi}{dx^2} = E\,\psi, \qquad \psi(0) = \psi(L) = 0 \]
Step 2 — Solve
The general solution is ψ = A sin kx + B cos kx with k² = 2meE/ℏ². The condition ψ(0) = 0 sets B = 0, and ψ(L) = 0 requires sin kL = 0:
\[ k_n = \frac{n\pi}{L}, \qquad \psi_n(x) = \sqrt{\frac{2}{L}}\,\sin\frac{n\pi x}{L}, \qquad n = 1, 2, 3, \dots \]
Step 3 — Energies
Substituting kn back, with ℏ = h/2π:
\[ E_n = \frac{\hbar^2 k_n^2}{2m_e} = \frac{\hbar^2\pi^2 n^2}{2m_e L^2} = \frac{n^2 h^2}{8 m_e L^2} \]
Step 4 — Fill the levels
By the Pauli principle each level holds two electrons of opposite spin, so N π electrons fill levels 1 to N/2. The lowest excitation lifts one electron from n = N/2 (HOMO) to n = N/2 + 1 (LUMO):
\[ \Delta E = \frac{h^2}{8 m_e L^2}\left[\left(\tfrac{N}{2}+1\right)^2 - \left(\tfrac{N}{2}\right)^2\right] = \frac{(N+1)\,h^2}{8 m_e L^2} \]
Step 5 — Wavelength
Setting ΔE = hc/λ:
\[ \lambda = \frac{hc}{\Delta E} = \frac{8 m_e c\, L^2}{(N+1)\,h} \]
Step 6 — Numbers
With L = 1.54 × 10⁻⁹ m, the energy unit is h²/(8meL²) ≈ 2.54 × 10⁻²⁰ J ≈ 0.159 eV. For N = 12:
\[ \Delta E \approx 13 \times 0.159\ \text{eV} \approx 2.07\ \text{eV}, \qquad \lambda \approx \frac{1240\ \text{eV}\!\cdot\!\text{nm}}{2.07\ \text{eV}} \approx 600\ \text{nm} \]
The simulation on the simulations page recomputes this and prints ΔE = 2.06 eV, λ = 602 nm.
Why a conjugated chain behaves like a box
Each carbon in the chain is sp² hybridised: three of its orbitals form the σ bonds of the backbone, and the fourth, a p orbital, sticks out perpendicular to the molecular plane. Neighbouring p orbitals overlap side by side, so the π electrons are not tied to one bond but spread along the whole conjugated stretch.
To an electron, the chain is a one-dimensional corridor: roughly flat potential along it, and a steep rise at the ends where conjugation stops. That is exactly the infinite square well, with three simplifications worth naming:
- The walls are not infinite. Real wavefunctions leak slightly beyond the end atoms, which is why L is often taken one bond length longer than the atom-to-atom distance.
- The floor is not flat. Each nucleus is a dip in the potential, and in neutral polyenes single and double bonds alternate in length. This opens an extra gap, so the free-electron model predicts colours that are too red for long neutral chains.
- Electrons interact. The model ignores electron–electron repulsion entirely. Better treatments (Hückel theory, then configuration-interaction calculations) fix this step by step.
Despite all three, the model captures the key trend: longer conjugation means smaller gaps and redder absorption. It explains why carrots (β-carotene, 11 conjugated double bonds) look orange while short polyenes are colourless.
Check your understanding
- Why does the HOMO–LUMO gap shrink as L grows, even though more electrons are added?
Answer: ΔE ∝ (N + 1)/L², and N grows roughly in proportion to L, so ΔE ∝ 1/L overall. - What would happen to λ if retinal lost its proton, so the positive charge disappeared?
Answer: bond alternation increases and the absorption shifts strongly to the blue, near 360 nm in the deprotonated Schiff base. The protein exploits this switch during its photocycle. - The wavefunction ψ6 has how many nodes inside the box?
Answer: n − 1 = 5.
Why the protein matters
The same chromophore absorbs near 440 nm in methanol, around 470 nm in channelrhodopsin-2 and near 570 nm in bacteriorhodopsin. This opsin shift comes from the protein's charges: the negative counter-ion near the nitrogen and polar residues along the chain stabilise the ground or excited state differently, widening or narrowing the gap. Engineers exploit this to build red-shifted variants that reach deeper into tissue.
Discussion point. The free-electron model works better for the protonated Schiff base than for neutral polyenes, because the positive charge reduces bond-length alternation and makes the box more uniform.
What this model leaves out
A single length L and a flat floor cannot capture bond alternation, electron correlation, or the protein field — which together are the whole reason the model lands at 602 nm while ChR2 measures 470 nm. The gap is not an error to be hidden; it is the size of the effect Part 2 and the opsin shift are about.