= Solution
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 implies
$$
PKP=-K.
$$
Thus $Kc=\epsilon c$ implies $K(Pc)=-\epsilon(Pc)$. This proves \b[energy pairing about $\alpha$], $E=\alpha\pm\epsilon$, and \b[equal site-probability magnitudes in each paired pair], since $|(Pc)_i|^2=|c_i|^2$. This is the Coulson-Rushbrooke pairing property.<Note on the method of molecular orbitals by Charles Alfred Coulson and G. S. Rushbrooke>[1940, pp. 193–200, alternant-system pairing and density properties]{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 \b[one pi <electron> per <carbon>], so the model predicts no net pi charge separation. The same argument makes \b[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 \b[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.
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