Two changes that do not add

Three orbitals in a row — two ends, A and B, each overlapping a middle one, C — have a level that stays at the free-atom energy however one end’s overlap with the middle is changed. No symmetry protects it; a count does. At three electrons that level holds the unpaired electron, and its spin is set by the ratio of the two overlaps: from half on each end to 0.236 and 0.764 as A’s overlap goes from 0.25 to 0.45, exactly the squared ratio, at every energy of the middle orbital.

Two changes were compared and they separated cleanly. The overlap asymmetry moves the spin and not the level. A coupling between the ends — a small A–B overlap, with the Wolfsberg–Helmholz interaction that goes with it — moves the level, by more than an electronvolt at an overlap of a tenth, and leaves the spin at exactly a half, because it keeps the two ends equivalent. Energy and composition, it seemed, were answering to different things.

Each result was established with the other change switched off. A real radical does not choose. A bent three-centre radical with one end nearer the middle than the other has both an asymmetry and a through-space coupling of its ends, and the question the separation leaves is whether the two effects then add or interfere.

At the origin they add #

The first answer is the one the separation predicts. At the symmetric, uncoupled trio, the mixed second derivatives — how the level’s rate of change with one change depends on the other — vanish: 3 × 10⁻⁴ eV per unit overlap squared for the level, where its rate against the coupling alone is ten electronvolts per unit, and 10⁻⁹ for the spin. Neither change alters the other’s first effect. For small enough changes, the separation holds.

It holds because each clean result was a symmetry or a count at the origin, and both survive a first-order perturbation by the other. But a separation that survives at first order can fail at second, and the interesting question is how far out it goes.

Beyond it, the level feels the asymmetry #

Switch the coupling on and sweep the asymmetry, and the level no longer ignores it. At an end-to-end overlap of 0.01 the asymmetry lowers the level by 0.016 eV by the time A’s overlap reaches 0.45; at 0.05 by 0.086; at 0.1 by 0.198, a sixth of the coupling’s own shift of 1.133. The dashed lines — what the two separate effects add to — are flat, because the asymmetry alone never moved the level. The solid lines fall away from them.

The count that held the level fixed needed the rows of the Hamiltonian belonging to A and B to be proportional: each touching only C, each with the same energy. A coupling puts an entry between A and B, and the asymmetry makes their entries in C’s column different; together they break the proportionality that each alone preserved. The interference grows with both — nearly in proportion to the coupling, and faster than in proportion to the asymmetry — which is why it is invisible at the origin and a fifth of an electronvolt at the corner of the sweep.

And the spin feels the coupling #

The spin behaves the same way from the other side. The coupling alone never moved it — the ends stay equivalent — so the dashed lines are the asymmetry’s own curve. With the coupling on, the solid lines run below it: at the largest asymmetry the spin on A is 0.2184 against 0.2358, pushed further from a half than the asymmetry alone pushed it. The coupling amplifies the asymmetry’s effect on the spin, where on its own it had none.

The electron reaches the middle #

With either change alone the singly occupied orbital has a node on C — no amplitude there at all, to the last digit — and the spin on C is zero. With both on, it is not. At an asymmetry of 0.2 and a coupling of 0.1 the middle orbital carries 0.0116 of the electron. At fixed asymmetry the amount grows as the square of the coupling: doubling the coupling from 0.01 to 0.02 multiplies it by 4.02.

That is the clearest sign that the separation has gone. The node on C was what made the level and the spin separable — it kept the orbital living on the two ends, where a ratio of overlaps could set the split and the middle orbital’s energy could not enter. Once the orbital reaches C, it can.

The middle orbital’s energy decides the spin #

It does. Uncoupled, the spin on A at the largest asymmetry is 0.2358 with the middle orbital at −18, −13.6, −9 and −5 eV, as the formula with no energy in it said it would be. Coupled, it is 0.177, 0.218, 0.331 and 0.456. With C well below the ends the coupling pushes the spin further onto B; with C well above them it pulls it almost back to an even split. The quantity that was a pure ratio of overlaps has become a function of every energy in the problem.

The level tells the same story at smaller scale: coupled, it moves from −12.69 to −12.50 eV across the same four energies of C, where uncoupled it sits at −13.6 whatever C does.

The spin becomes a partition #

The earlier result made a second claim that depends on the node. A spin “on an atom” is a partition of an orbital spread over overlapping functions, and different partitions usually disagree; for the uncoupled radical, the Mulliken partition and the Löwdin one agreed to every digit, and the reason given was that the orbital has no amplitude on C and the ends do not overlap each other, so no overlap term has anything to share.

With both changes on, both parts of that reason fail, and the partitions part. With C at −5 eV they disagree by 0.011 of the electron at the largest coupling; at −9 eV by 0.005; at −18 by 0.002. The spin distribution of a bent, coupled radical is a convention, as most atomic populations are.

But one curve does not rise. With the middle orbital at the ends’ own energy, −13.6 eV, the two partitions agree to 10⁻¹⁵ at every coupling and every asymmetry — including the corner where C carries a hundredth of the electron and A and B overlap by a tenth. The reason given for their agreement cannot be the reason here, because neither of its conditions holds.

A Hamiltonian that is a function of its overlap #

The reason is the coupling rule. The Wolfsberg–Helmholz rule sets each off-diagonal Hamiltonian element to K times the overlap times the average of the two site energies. When all three site energies are equal to α, every off-diagonal element is Kα times the overlap and every diagonal element is α, so the Hamiltonian is Kα times the overlap matrix plus (1 − K)α times the identity. It is a function of the overlap matrix, and a matrix commutes with any function of itself.

When the Hamiltonian and the overlap commute, the generalised eigenvalue problem’s eigenvectors are eigenvectors of the overlap matrix. For such a vector the Mulliken population on each site and the Löwdin population are both the eigenvalue times the squared coefficient, and after normalisation they are the same numbers. The commutator is computed rather than argued: it vanishes to 3.6 × 10⁻¹⁵ with the middle orbital at the ends’ energy, and is 2.1, 2.2 and 4.1 with it at −18, −9 and −5 eV.

So the agreement had two independent reasons, and the uncoupled radical of the earlier essay enjoyed both at once at its reference point. Away from equal site energies the node keeps the partitions together; away from the node, equal site energies do. Only a radical with neither — coupled, asymmetric, with a middle orbital at its own energy — makes its spin a matter of convention. That also corrects the earlier essay’s closing caution, that a small A–B overlap would part the partitions: at equal site energies it cannot.

What a hyperfine measurement would see #

Spin densities are not measured directly; hyperfine couplings are, and each nucleus’s coupling reports the spin at that nucleus. The uncoupled radical gave a clean prediction: the ratio of the two ends’ couplings is the squared ratio of their overlaps, and the middle nucleus couples not at all. That made the ends’ hyperfine ratio a reading of the overlap asymmetry and nothing else, and a middle-nucleus coupling of zero a test that the model applied.

Both changes on, each part of that prediction changes. The ends’ ratio now depends on the coupling and on the middle orbital’s energy, so it can no longer be read as an overlap ratio without knowing both. And the middle nucleus acquires a positive coupling, as the square of the coupling between the ends. A non-zero central hyperfine would be the fingerprint of a radical in which both effects act — a signature the uncoupled model cannot produce at all.

That fingerprint has a competitor, and it is worth naming because it has the opposite sign. In the allyl radical, the textbook three-centre case, the central carbon’s measured spin is negative: spin polarisation of the paired electrons, a correlation effect outside any single determinant, puts spin of the wrong sign on the middle atom. The coupling computed here puts spin of the right sign there. A real bent radical would carry both, and their relative size is what the second-determinant question below would settle; the sign of the central coupling would say which dominates.

How much of the earlier picture survives #

It is worth being exact about what the separation still gives, because it is not nothing. A bond order can appear between atoms that do not interact, and here, with A and B genuinely coupled, the level’s shift is still dominated by the coupling: at the corner the level has moved 0.935 eV, of which the coupling alone accounts for 1.133 and the interference takes back 0.198 — a correction of a sixth. Likewise the spin’s departure from a half is still dominated by the asymmetry: 0.264 of it at the asymmetry alone, and 0.017 more from the interference. The first-order picture is the right picture to start from, and the wrong one to finish with.

What does not survive in any form is the stronger claim that the level and the spin answer to different things. They answer to the same Hamiltonian, and they separated because two particular changes happened each to leave one of them fixed. Combine the changes and every quantity depends on every parameter, the partition included — the situation a weight that depends on how it is weighed found for ionic weights, reached here from a radical that had looked like an exception to it.

What was computed and how #

The trio is the same three-orbital model: A and B at −13.6 eV, B–C overlap 0.25, A–C overlap 0.25 + σ, an optional A–B overlap, Wolfsberg–Helmholz constant 1.75, the generalised eigenvalue problem solved with the overlap kept, three electrons. The spin is the singly occupied orbital’s population by the Mulliken and the Löwdin partitions. The grid is five asymmetries from 0 to 0.2 and five couplings from 0 to 0.1, repeated at four energies of the middle orbital; mixed derivatives at the origin are central differences with a step of 10⁻³.

The claims are stated where they can fail: that the mixed derivatives of the level and the spin vanish at the origin; that at the corner the level falls short of the sum by more than a tenth of an electronvolt, the spin overshoots it by more than a hundredth, and C carries more than half a per cent; that the spin on C quadruples when the coupling doubles; that C’s energy leaves the uncoupled spin fixed and moves the coupled one by more than a fifth of an electron; that Mulliken and Löwdin agree to 10⁻¹² at equal site energies everywhere, disagree by more than 10⁻⁴ otherwise with both changes on, and agree when the coupling is off. The refusal is the commutator itself, which must vanish at equal site energies and nowhere else, or the reason given for the agreement is the wrong one.

Where the model stops #

One determinant, three orbitals. The spin here is the singly occupied orbital’s population. A second determinant — promoting a paired electron into the singly occupied level — can put negative spin on C, and the correlated spin density of a real radical includes it. Whether that spin polarisation interferes with the coupling’s effect is the question the three-orbital problem can still answer exactly.

The Wolfsberg–Helmholz rule. The equal-energy agreement is a property of a coupling proportional to overlap. A rule with a different energy dependence — a harmonic mean, or a fixed resonance integral — would not make the Hamiltonian a function of the overlap matrix, and the partitions would part at equal energies too.

And the changes are of one geometry. Asymmetry and coupling are varied independently here. In a real bent radical they are tied together by the bond angle, which moves along a single line through this plane.

The ends are held at one energy. Every statement about the node, the proportional rows and the equal-energy agreement assumes A and B are the same kind of atom. A heteroatom at one end is the site-energy change of the earlier essay, which moves both the level and the spin on its own, and combining it with the two changes here would give a three-parameter family in which the clean results can be expected to fail at first order, since the equal end energies they rest on are gone.

Separable at the origin is not separable #

The general point is about what a clean separation licenses. A filled shell and a regime that belongs to the neighbours both ended in statements true in one limit and qualified away from it, and this is the same shape one level down. A level no symmetry was protecting and a symmetry that holds or does not both established exact zeros — derivatives that vanish identically — and a zero at the origin says nothing about the second order. Two effects that each vanish for the other’s reason can interfere as soon as both are finite, and here they do in every quantity measured.

The second point is about reasons. Two partitions agreed, and a reason was given that was true and sufficient. It was not the only one, and at the one point where the sweep sat most of the time, a different reason was doing the work. An agreement can have two causes, and finding one does not show the other is absent; it shows only that removing the one found will not necessarily end the agreement.

Still open: the path a real bent radical takes, and the second determinant #

The obvious open question is the line through this plane that a real molecule follows. Bending a three-centre radical changes the A–C and B–C distances and the A–B distance together, so asymmetry and coupling are functions of one angle, and the interference computed here would appear as a curvature in how the spin moves with the angle. Tracing that line for a stated geometry would turn two independent parameters into one prediction.

The nearer question is the second determinant. The lowest excitation that promotes a paired electron into the singly occupied level puts spin of the opposite sign on the middle orbital, and the three-orbital problem can include it exactly. Whether that negative spin grows with the coupling, as the positive spin on C does here, or ignores it as the level ignored the asymmetry, would say whether spin polarisation and the coupling’s leak onto C reinforce or cancel.