Past exam of the chemistry course of the year 3 of the natural sciences course of the University of Cambridge / 2022 / Paper 2A / Section A / 20 / b

Write $H=\alpha I+K$ with the same carbon diagonal parameter throughout, and define $P$ to have entries $+1$ on one sublattice and $-1$ on the other. Bipartite connectivity impliesThus $Kc=ϵc$ implies $K(Pc)=-ϵ(Pc)$. This proves energy pairing about $\alpha$, $E=\alpha \pm ϵ$, and equal site-probability magnitudes in each paired pair, since $|(Pc{)}_i{|}^2=|c_i{|}^2$. This is the Coulson-Rushbrooke pairing property.^[ref]

For the usual neutral half-filled closed-shell occupation, the lower member of each pair is doubly occupied. Combining paired coefficients with completeness of the orthonormal basis gives one pi electron per carbon, so the model predicts no net pi charge separation. The same argument makes off-diagonal pi bond orders between sites on the same sublattice zero. These statements assume a neutral half-filled density and do not apply unchanged to the charged six-electron four-site ion considered separately.

An imbalance of sublattice sizes also guarantees at least $|N_A-N_B|$ nonbonding orbitals at energy $\alpha$, because the rectangular coupling block has insufficient rank to remove all its null space. None of the spectral pairing, equal paired densities, neutral charge uniformity or sublattice bond-order restrictions is required for a non-alternant graph. Long-range same-sublattice couplings or unequal diagonal parameters can likewise destroy the simple proof.