A symmetry holds or it does not

Two essays are built on a trio of orbitals in which two — call them A and B — have exactly no overlap with each other and no resonance integral between them, and both couple to a third. The first finds a bond order of −0.954 between the two that do not interact. The second finds that at a filled shell the bond order becomes one seventh and stops depending on any energy in the problem.

Both rest on something neither measured. One of the three levels sits at the free-atom energy exactly, at every third-orbital energy, and it does so because the two outer orbitals are equivalent: the group has an antisymmetric representation, the combination $(\text{A}-\text{B})/\sqrt2$ is the only function belonging to it, and a function alone in its representation has nothing to mix with.

The question is how fast that goes when the two are made inequivalent, and whether the exactness is protected to first order or lost at once.

It goes at once #

Raise A’s site energy by $\delta$ and leave B’s alone. At $\delta = 0.05$ eV the level has moved by 0.025106 eV: 0.502120 of the detuning.

At 0.1 it has moved by 0.050419, a ratio of 0.504190. At 0.2, 0.508176. At 0.5, 0.518882.

The ratio approaches one half as the detuning falls. That is not a small number and it is not going to zero: the exactness has no protection at all, and the level takes half of whatever is done to one of the two orbitals.

The half is not an empirical constant. First-order perturbation theory says a level’s displacement is the expectation of the perturbation in the unperturbed state, and the unperturbed state here is $(\text{A}-\text{B})/\sqrt2$, which puts half its weight on A. So the level moves to the mean of the two site energies, and half is what a half-and-half combination gives.

What is left after that is second order. The distance from the mean is $1.06\times10^{-4}$ eV at a detuning of 0.05, $4.19\times10^{-4}$ at 0.1 and $1.64\times10^{-3}$ at 0.2 — quadrupling as the detuning doubles, and the slope on logarithms over the smallest three points is 1.974. So first order is not merely a good description at small detuning; it is the whole of the behaviour, with the exact diagonalisation and the formula agreeing to four figures at a twentieth of an electron volt.

What protection would have looked like #

The question has a definite shape, and it is worth drawing.

If the level were protected to first order — if the symmetry that placed it were broken only at second order by this perturbation — the displacement would go as $\delta^2$, and on logarithms it would fall away from the linear line at every decade. It does not. The measurement lies on the linear reference at every point across two decades of detuning.

That is the answer, and it generalises past this system. A symmetry either holds or it does not. There is no sense in which a nearly-equivalent pair has nearly the exact level: the exactness is a statement about a representation, representations are counted rather than measured, and a perturbation that removes one removes it entirely. What survives is not a fraction of the result; it is a different result — the level’s new position, which first-order theory names in advance.

What replaces the exactness #

The pessimistic reading of the figure above is that the exact level was fragile and is therefore worth little. That reading is wrong, and separating it from the true one is most of what this essay is for.

An exactness produced by a symmetry is a statement of the form this level is here, whatever the parameters are. When the symmetry goes, that statement is gone completely — there is no weakened version of whatever the parameters are. But the level does not scatter. It goes to the mean of the two site energies, and it goes there to four figures at a detuning of a twentieth of an electron volt.

So what replaces a symmetry statement is a formula, and the formula is cheaper than the diagonalisation it replaces. A reader who knows the two site energies knows where the level is without solving anything, over the whole range where the second-order residue is negligible — which here is a detuning up to about a fifth of an electron volt, or roughly a twentieth of the resonance integral.

That is the useful form of the answer, and it is worth contrasting with the case where nothing replaces the symmetry. The trans influence rests on two ligands sharing an orbital; break the symmetry there and the two are not related by any formula, because the quantity being compared is a difference between two competing overlaps rather than a mean of two numbers. A symmetry that places a level and one that relates two quantities fail differently, and only the first leaves an expression behind.

Where the second-order term comes from #

The quadratic residue is small and it is not noise, and it has an identifiable source that is worth naming because it says what the first-order picture leaves out.

At zero detuning the antisymmetric combination is an exact eigenvector. At non-zero detuning it is not: the perturbation has a matrix element between it and the symmetric combination, of size $\delta/2$, and the two are separated by the splitting the third orbital produces. Second-order perturbation theory then gives a shift of $(\delta/2)^2$ divided by that separation — quadratic in the detuning, with a coefficient set by how far apart the two combinations already are.

That predicts something checkable, and it is not checked here: the residue should be larger when the third orbital is far from the pair, because the splitting between the symmetric and antisymmetric combinations is then smaller. Every number above is at one third-orbital energy, and the residue’s coefficient is the quantity a second sweep would measure.

What the source does say immediately is that the residue is a mixing between two states of the pair, not a leakage to the third orbital. The third orbital’s role is to have created the splitting that the mixing is divided by — which is the same role it plays in every result about the trio: it is never in the answer and it decides the answer.

Everything else moves too #

Breaking the symmetry does not only move the level.

The two bond orders A–C and B–C were both 0.522408 at zero detuning, equal because the two orbitals are equivalent. At $\delta = 0.05$ they are 0.525795 and 0.518996 — separated first order, symmetrically about their common value, by about 0.068 per electron volt each way.

And the pair’s own bond order, the one two essays have been about, moves as well: −0.630602 becomes −0.632329, deepening the antibond. It is worth noting the direction. Raising one of the two orbitals makes the pair more antibonded, and the pair still has no interaction — so the quantity is responding to a change in an orbital it is not connected to, through an orbital it is not indexed by.

The table’s first row is the undetuned system, and the level there is at the free-atom energy to $-8.9\times10^{-15}$ eV — machine zero rather than a small number. That is the control the whole argument needs: if the exactness at $\delta = 0$ were a coincidence of the arithmetic rather than a symmetry, there would be nothing to break.

What a chemist would do with this #

The construction is abstract and the situation it models is not. Two ligands related by a symmetry operation of a complex have exactly equal contributions to every property the operation preserves; make them slightly different — a substituent on one, a crystal field, a solvent molecule — and every one of those equalities goes.

The measurement above says what to expect when that happens, and the answer is more useful than “the equalities are approximate”. The two ligand quantities separate symmetrically about their common value, first order in the difference, so their mean is preserved to second order even though neither of them is. That is why a mean over symmetry-related positions is a robust quantity and an individual one is not — and it is the same arithmetic that makes an averaged coefficient better behaved than its parts in a quite different setting.

The size is easy to underestimate. A detuning of 0.05 eV is small by any chemical standard — smaller than a solvent shift — and it separates the two bond orders by 0.0068, which is a per cent and a third of their value. A calculation reporting bond orders to two decimal places would show them as equal; one reporting three would not, and the difference between the two reports is entirely the last digit.

What was computed, and how #

The trio is a generalised eigenvalue problem: three site energies on the diagonal, resonance integrals proportional to the overlaps, and an overlap matrix in which $S_{AB}$ is exactly zero while $S_{AC} = S_{BC} = 0.25$. The detuning is added to A’s site energy and to nothing else — not to its overlaps, not to its resonance integrals — so the perturbation is one number in one place and the comparison with first-order theory is a comparison with a formula that has that number in it and nothing else.

Every level is from an exact diagonalisation. The “displacement” is the distance from the free-atom energy of the level nearest it; the “distance from the mean” is the distance from $\alpha + \delta/2$, which is where first-order theory puts it. Those are two different reference points and the whole reading is the difference between them: the first is linear in the detuning and the second is quadratic.

Four things are checked. At zero detuning the level must be at the free-atom energy to machine precision. At small detuning the ratio must be one half to within two per cent. The distance from the mean must be a per cent or less of the detuning, and it must grow faster than linearly, so that a version of this in which first-order theory happened to be wrong would fail. And the bond orders must move first order as well, which is what says the symmetry breaking reaches every quantity rather than only the spectrum.

The bond orders use the same definition as the two trio essays, so the numbers are comparable directly, and the definition itself is the standard one.

Where the model stops #

This is a one-electron model with overlap kept and no repulsion, and the detuning is a change to a site energy rather than to a geometry. In a real molecule making two ligands inequivalent changes their distances and their overlaps as well, and those enter the resonance integrals — so the clean separation here, where one number moves and everything else is held, is a construction rather than a chemistry.

The perturbation is small in the right sense over the range where the first-order statement is checked, and the range is only two decades. Nothing here says what happens when the detuning is comparable with the resonance integral, except that the ratio has drifted to 0.53 by one electron volt and the second-order term is visible. At four electron volts the ratio has come back down to 0.517, which is not a return to first-order behaviour but the third and fourth orders arriving with opposite signs — a reminder that a ratio drifting back towards its limit is not evidence of anything.

The second-order residue is measured over three points and its slope is 1.974 rather than 2. Whether that is the fit’s window or a genuine third-order contamination is not separated here; two is what the theory says and 1.974 is within what three points at this spacing can distinguish.

And the exactness being broken is a non-degenerate one. A level alone in its representation is protected from mixing; two levels sharing a representation are a degenerate pair, and degenerate perturbation theory behaves differently — the splitting is first order in the perturbation but the eigenvectors rearrange discontinuously. That case is a degeneracy a group predicts, and the two should not be run together.

And the symmetry was sufficient, not necessary. The level is attributed above to the equivalence of the two outer orbitals, and that attribution is too strong. Changing the overlap of one of them with the third orbital breaks the equivalence completely and moves the level by nothing at all, because what holds it at the free-atom energy is that the pair sits at one energy with no coupling of its own and is seen by only one other orbital — a level no symmetry was protecting measures that. The site-energy result here stands; the reading of it as a symmetry statement does not.

The generalisation #

An exactness that comes from a symmetry is a discrete fact, and discrete facts do not degrade. They are replaced. When the symmetry goes, the quantity does not become approximately what it was; it becomes something else, and the useful question is what the something else is rather than how much of the old answer survives. Here the answer is unusually good — the level moves to the mean of two site energies, which is a formula — and that is worth separating from the pessimistic reading, because a symmetry breaking that leaves a first-order formula behind is a symmetry breaking that can be computed with.

This collection reached the same conclusion on a gap that only a tetrahedron closes, where an exactness from an absence was lost immediately along a distortion, and the two together make a rule for reading any exactly-zero quantity in this collection: ask which symmetry produces it, and then ask what the smallest realistic thing that breaks that symmetry is. If the answer to the second is anything at all, the zero is a statement about the model rather than about the molecule.

And a quantity indexed by two objects can respond to a third. The pair’s bond order moves when A’s site energy is raised, and A is one of the two it is indexed by — but it moves through C, since A and B have no interaction to transmit anything. Two essays have now been about that, and the detuning makes it visible in a derivative rather than in a value: $\partial P_{AB}/\partial\alpha_A$ is not zero for a pair with no coupling. That derivative is a cleaner statement of the whole finding than any of the values have been, because a derivative cannot be dismissed as a convention: whatever the bond order is measuring, it is measuring something that responds to an orbital the pair does not touch.

Who found it, and when #

Non-degenerate first-order perturbation theory is Rayleigh’s and Schrödinger’s, and the statement that a level’s displacement is the expectation of the perturbation in the unperturbed state is the first line of it. That a function alone in its irreducible representation cannot mix is Wigner’s, and the two together predict everything measured above before it is measured.

What the measurement adds is the arithmetic that a reader is entitled to want and rarely gets: the ratio at four detunings, the quadratic residue, and the machine-zero control at $\delta = 0$. The theory could have been quoted instead; the better habit is that a claim gets a computation it could fail, and the failure available here was that the residue might have been linear, which would have meant the first-order identification was wrong.

Still open: detuning an overlap rather than an energy #

The obvious open question is a detuning that is not a site energy. Making the two outer orbitals inequivalent by moving one of them changes $S_{AC}$ rather than $\alpha_A$, and that enters both the overlap matrix and the resonance integral — so first-order theory has two terms in it rather than one, and whether they add or cancel is a question this construction can answer with one more sweep. The interesting possibility is that they nearly cancel, which would leave the level looking protected without any symmetry protecting it.

The nearer question is the filled shell. The six-electron bond order is a function of the overlaps alone, independent of every energy — so a detuning of a site energy should move it by exactly nothing, while it moves every other quantity first order. That is a prediction with a sharp failure mode and it costs one sweep at a different electron count, and it would separate the part of a bond order that is physics from the part that is metric on one system rather than by comparing counts.