A filled shell is not an empty statement

A trio of orbitals in which two have exactly no overlap with each other, and no resonance integral between them, still has a bond order between those two. At four electrons it is −0.954 — very nearly a full antibond between two functions that do not interact — because the third orbital decides which combinations are occupied, and the occupied set is what a bond order is computed from.

It closed with a prediction. Everything there was at four electrons, where the antisymmetric combination is occupied and supplies the −1. At two it is empty and the bond order should be positive; at six every level is full, the sum over occupied orbitals is a sum over a complete set, and the bond order must be exactly zero by completeness.

Two of those three are right.

Three counts, three behaviours #

With no electrons the bond order is identically zero at every third-orbital energy, which is the control: whatever this quantity is, it is a property of the occupation and not of the basis alone.

With two it is positive throughout, rising from 0.0459 with the third orbital deep below the pair to 1.0306 with it at zero — a bond order above one between two orbitals with no interaction at all.

With four it is negative throughout most of the range, running −0.954, −0.894, −0.770, −0.631, −0.431, −0.217, −0.047 and then crossing to +0.031. The −0.954 is the left-hand end of that curve.

With six it is a horizontal line.

The two open-shell limits are the two integers the argument predicts, and they arrive cleanly. At eighty electron volts down the two-electron order is $1.0\times10^{-4}$ and the four-electron order is −0.999899. In that limit the occupied set is the third orbital alone, or the third plus the antisymmetric combination of the pair — and the second of those is, on its own, a full antibond between two functions that have no interaction to be antibonded by.

The filled shell is not zero #

The six-electron line is flat to better than $10^{-9}$ over twenty-five electron volts of sweep. That much is what was expected: with everything occupied, the energies stop mattering, because there is no longer any question of which orbitals are filled.

What it settles at is not zero. It is 0.142857 — one seventh.

The argument for zero runs: the sum over all occupied orbitals of $c_i c_j$ is a sum over a complete set, a complete set resolves the identity, and the identity has zeroes off the diagonal. Every step of that is correct in an orthonormal basis and the second step is false in any other.

In a non-orthogonal basis a complete set of eigenvectors of the generalised problem satisfies $\sum_k \mathbf{c}_k \mathbf{c}_k^{\mathsf T} = \mathbf{S}^{-1}$, not the identity. So a filled shell’s density matrix is the inverse overlap matrix, and the question becomes whether the inverse of a matrix with a zero in it has a zero there.

It does not. The overlap matrix here is one on the diagonal, zero between the two orbitals that do not overlap, and a quarter between each of them and the third. Its determinant is 0.875, and the off-diagonal entry the pair acquires in the inverse is $0.0625/0.875 = 1/14$. Twice that — two electrons per orbital — is the one seventh.

The check is that the two routes share nothing. One diagonalises a three-by-three generalised eigenvalue problem, fills its levels and sums $c_i c_j$ over the occupied set; the other inverts a three-by-three matrix by cofactors and never computes an eigenvector. They agree to nine figures, which is the arithmetic saying the identification is right rather than the story being plausible.

What that means for the pair #

The interesting part is not the arithmetic; it is what the number is a property of.

At six electrons the bond order between the two orbitals is a function of the overlaps alone. It does not know the resonance integrals, the site energies, the third orbital’s position or anything else about the Hamiltonian. And the overlaps are the part of a calculation a basis choice fixes.

So the filled-shell bond order has traded one arbitrary dependence for another. The projector is unique and the basis is not is the general form of that trade, and this is the case where the quantity lands entirely on the wrong side of it. At two and four electrons it depends on where the third orbital sits, which is at least a physical quantity, in the way an overlap is not an interaction but is still something a geometry decides; at six it depends only on how the functions were chosen to describe the system. A quantity that is independent of the Hamiltonian and dependent on the basis is not a property of the molecule at all, and that is a sharper statement than the four-electron case supports, because there the number at least moves when the physics does.

There is a second reading worth having. The A–C bond order at six electrons is $-4/7 = -0.571$, and A and C do overlap, at a quarter. So the pair that overlaps gets a negative filled-shell bond order and the pair that does not gets a positive one — the sign is inverted relative to the overlap. That is the inverse matrix again: a positive off-diagonal entry in $\mathbf{S}$ produces a negative one in $\mathbf{S}^{-1}$ at leading order, and a zero produces whatever the second-order term supplies.

There is a way of stating the whole finding that makes it less surprising and more useful, and it is worth putting down because it is what a reader should take away. A bond order computed by summing over occupied orbitals is a projection, and a projection has two ingredients: which subspace is occupied, and what the metric is. At two and four electrons the subspace is doing the work, and the answer moves as the third orbital moves because the subspace does. At six the subspace is everything, so it stops carrying information, and what is left is the metric alone. The number is not an artefact appearing from nowhere; it is the second ingredient, which was there the whole time and was masked by the first.

That also says which of the four rows is the anomaly. It is not the six-electron row. It is the belief that the six-electron row would be empty, which is the belief that a metric is an identity — and a whole argument about what a basis lends rests on that not being true.

Everything, at one energy #

With the third orbital at the pair’s own energy, the four rows give: a bond order of 0.00000, +0.36940, −0.63060 and +0.14286; an A–C bond order of 0, +0.52241, +0.52241 and −0.57143; a binding of 0, −5.33, −5.33 and +5.83 electron volts.

The geometry is identical in all four rows. So is the Hamiltonian. So are the overlaps, the resonance integrals and the site energies. Only the number of electrons changes, and every quantity anybody would quote about this pair changes with it — including its sign, twice.

One column does not move. The pair bond — what removing the direct A–B interaction costs the total energy — is exactly zero in every row, because there is no direct interaction to remove. That is the preferred measure for the four-electron case, and it is the only one of the four that gives the same answer at every count, which is a point in its favour and is also the reason it is useless: a measure that is identically zero on this system cannot distinguish the four rows either.

A coincidence in that table is worth naming rather than leaving to be noticed. The two- and four-electron bindings are the same number, −5.3286 eV. It is not a deep fact: with the third orbital at the pair’s energy the three levels are symmetric about it, the middle one sits exactly at $\alpha$, and adding the third and fourth electrons to it adds exactly what removing them from the isolated reference takes away.

Across the sweep the binding is negative — bound — at two and four electrons everywhere, and positive at six over most of the range. So the same three orbitals in the same positions make a bound system or an unbound one depending on how many electrons are in them, which is the ordinary behaviour of a filled antibonding level and is not surprising. What is surprising is only that none of it belongs to the pair.

What is left of the four-electron argument #

Almost all of it, and it is worth saying so plainly, because the correction above is to one sentence rather than to a finding.

The four-electron claim is that a bond order between two orbitals with no interaction can be large, that it is set by which combinations the third orbital causes to be occupied, and that a reader taking it as a statement about the pair is taking it as a statement about something else. Every one of those survives, and the sweep strengthens the last of them: the same pair on the same Hamiltonian now returns four different verdicts rather than one surprising one.

What went wrong was a limiting case reasoned about rather than computed, using an argument that is correct in the setting it is usually made in. That is a good failure to have on record: the prediction was specific, and checking it takes one parameter sweep. A prediction vague enough to survive would have taught nothing.

What was computed, and how #

The trio is three orbitals with a generalised eigenvalue problem: site energies on the diagonal, resonance integrals proportional to the overlaps in the usual Wolfsberg–Helmholz form, and an overlap matrix in which $S_{AB}$ is set to exactly zero while $S_{AC} = S_{BC} = 0.25$. That is not an approximation to a small overlap; it is zero, so a claim about what happens when two orbitals do not interact is a claim about this system rather than about a limit of it.

The bond order is $\sum_k n_k c_{ki} c_{kj}$ over the occupied orbitals, with the coefficients from the generalised problem and the occupations from filling the levels with the stated number of electrons. That is the same definition every population analysis of this kind uses.

The binding is the trio’s total energy minus the energy of three isolated orbitals holding the same electrons, and that reference has to count the electrons actually present. Written as $2\alpha + 2\min(\alpha,\alpha_C)$, which is the four-electron case written out, it is right at four electrons and wrong the moment the count moves: it counts four electrons into a shell that has none and reports an empty system as bound by 54.4 electron volts. The general form and the four-electron one agree exactly at four electrons.

Four checks hold the reading up and one of them is a refusal. The empty shell must give exactly zero at every geometry. The two-electron order must be positive everywhere and the four-electron order must reach nearly a full antibond. The six-electron order must not move at all as the third orbital is swept — a spread below $10^{-9}$, which is the statement that it is independent of the energies rather than merely insensitive to them. And it must equal twice the inverse overlap, by the route with no diagonalisation in it.

Where the model stops #

This is a one-electron model with overlap kept and no repulsion. A real filled shell has electron–electron terms that make the density matrix something other than $\mathbf{S}^{-1}$, and the identification above is exact only for a single Slater determinant built from the eigenvectors of a one-electron Hamiltonian. What survives in a correlated calculation is the general point rather than the number: the density matrix of a complete occupation is a metric object, and a metric object has entries where the interaction does not.

Three orbitals is three orbitals, and the one-seventh is a property of this overlap matrix. A different $S_{AC}$ gives a different filled-shell order — the general form is $2(\mathbf{S}^{-1})_{AB} = 2S_{AC}^2/(1 - 2S_{AC}^2)$ for this geometry, which is second order in the overlap and therefore small when the overlaps are small. That is why nobody trips over it in an ordinary molecule: the effect is real and it is a few per cent, not a seventh, unless the shared overlaps are large.

And the bond order used here is the Mulliken-style product of coefficients rather than one of the overlap-weighted variants. A different convention gives a different filled-shell value, and some of them do give zero. That is not a rescue: it means the answer to is the filled-shell bond order zero is a question about which of several conventions is in use — the same shape as a partial charge that four scales disagree about, which is what this collection means when it says a basis is not a thing.

The generalisation #

Completeness is a statement about a resolution of the identity, and in a non-orthogonal basis the thing resolved is the metric. Every argument of the form the sum over all states is trivial is doing that step, and every one of them is exposed to the same failure. The same shape appears in a density that is identical between two descriptions, where the invariance holds because the transformation is orthogonal and would not otherwise, and the difference between the two cases is exactly whether an overlap was kept.

And a quantity whose value depends on the electron count is not a property of a pair. Four counts on one Hamiltonian give a bond, an antibond, a bond and nothing, between two orbitals whose relationship to each other never changes. Anything computed from the occupied set inherits the occupation, and calling it a bond order encourages the reading that it describes the two things it is indexed by. The bond order computed from the eigenvectors is where the definition is made; this is the third case in a row to find it reporting something other than what its two indices name.

Who found it, and when #

That $\sum_k \mathbf{c}_k\mathbf{c}_k^{\mathsf T} = \mathbf{S}^{-1}$ for a complete set in a non-orthogonal basis is standard linear algebra and appears in every careful treatment of population analysis — it is why Löwdin’s symmetric orthogonalisation exists, and why Mulliken populations are basis-dependent in a way that has been complained about since 1955. The filled-shell case is the one where the dependence is total rather than partial, and it is not usually pointed at because a filled shell of a whole molecule is not something people compute bond orders in.

What this does is take a prediction made in good faith from an argument that is correct in the orthogonal case, and find the one seventh. The value of the exercise is not the number; it is that the prediction was made explicitly enough to be checked, and that checking it needs one sweep of a single parameter.

Still open: separating physics from metric in one sweep #

The obvious continuation is the overlap’s size. The filled-shell order is $2S_{AC}^2/(1-2S_{AC}^2)$ on this geometry — second order in the shared overlap, and therefore something that grows fast once the overlaps are large. Sweeping $S_{AC}$ from a tenth to a half would say where the effect stops being a curiosity and starts being the size of a real bond order, and the closed form is available to be checked against the diagonalisation at every point rather than at one.

The nearer question is the asymmetry, now a sharper question than it was. Detuning one of the two outer orbitals breaks the symmetry that puts one level exactly at the free-atom energy, and the four-electron bond order should move first order in the detuning while the six-electron one cannot move at all — since it does not know the energies. A single sweep would therefore separate the part of the bond order that is physics from the part that is metric, on one system, which is a cleaner separation than comparing counts.