A ratio of exactly one is a tie

The intensity contrast of a removal spectrum — the weakest of the strongest few lines divided by the strongest of the rest — can be taken to the limit of infinite on-site repulsion, on eight small systems at two electrons each. The approach is first order in the reciprocal repulsion, so two points a factor of four apart give the limit to six figures, and for a ring the limit comes out as $(1 + 2\cos(\pi/n))^2$.

Every one of those eight systems held two electrons. Its closing paragraph asked for the filling: take the same limit at half filling, where the contrast is already known to cross the factor of two the distinguishability test needs, and find out whether it settles above two or below.

It settles below. It settles on exactly one.

Exactly one is a suspicious answer, and it is suspicious in a way already known. One is the floor of a ratio between two quantities that have been sorted, and a sorted ratio reaches its floor when the two things being compared are the same thing.

What the cut is cutting #

The contrast is built by ranking every line that carries weight, calling the strongest $n_\uparrow$ of them fundamentals because that is how many electrons there are to remove, calling the rest satellites, and dividing the weakest fundamental by the strongest satellite. The rank at which that cut falls is fixed by the electron count. What sits on either side of it is not.

At half filling on a ring of six the third and fourth strongest lines have weights $0.205414079$ and $0.205414079$ — agreeing to $6 \times 10^{-16}$ — and energies agreeing to $8 \times 10^{-15}$. They are two members of one degenerate level of the five-electron ion. The cut at rank three falls between them, and the contrast is one of them divided by the other.

That is not a statement about satellites. It is not a statement about intensity at all. It is a statement that a rank-based cut has been drawn through a degenerate multiplet, which any rank-based cut through a spectrum built of multiplets will eventually do.

The spectrum is becoming multiplets #

Why the cut lands inside a pair, and why it does so more reliably as the repulsion grows, is visible directly.

At a repulsion of eight, one hundred and twenty-four lines carry weight. At a repulsion of a thousand and twenty-four, twenty-nine do, and ten adjacent pairs among them are exactly degenerate. Lines are not disappearing; they are merging, as the large-repulsion limit organises the ion’s states into multiplets whose members become exactly equivalent.

A spectrum with a hundred and twenty-four distinct lines has many places to put a cut without hitting a tie. A spectrum with twenty-nine lines, most of them paired, has few. A hundred lines and no way to sort them established that the line count itself is not the problem; what this adds is that the structure of the surviving lines makes the sorting rule ill-posed rather than merely unhelpful.

Which degeneracy, and the guard that tests the other one #

The same mechanism appears at two electrons, stated plainly: on a ring of four at half filling, the two lines the cut falls between are members of one degenerate shell, and a ratio taken across them is a ratio of a quantity to itself. That reading is right and nothing here disturbs it — only its counting, which named the fourth line and the fifth where the pair is in fact the second and the third, both carrying a weight of $0.394940250$ at an energy of $-1.004320$. Two electrons put the cut after the second line, so that is where it has to be.

What it did next was build a guard around it. The ring of four has a degenerate level at the Fermi energy — a half-filled shell whose two orbitals are exactly equal in energy — and the guard was to require that every system in the table have a non-degenerate one.

That guard does not catch this case. A ring of six at half filling has a Fermi-level gap of $2.000$ against the ring of four’s $3 \times 10^{-17}$: it is as far from a degenerate shell as any system in the table. It ties anyway, at five of the eight repulsions measured.

The two degeneracies are different objects. The shell degeneracy is a property of the one-electron problem for the neutral molecule. The tie is a degeneracy of the ion — the system with one electron removed — and it is produced by the ion’s own symmetry and its own many-electron structure. Nothing about a gap in the neutral’s orbital diagram determines whether the ion’s states come in exact pairs.

It is not the ground state either #

There is a second explanation available and it is also wrong here, which is worth establishing because it is the one that would let the number be dismissed as numerical.

If the neutral’s ground state were degenerate, an eigensolver would be free to return any combination within the subspace and the weights would be arbitrary. The gap here runs from $0.349$ down to $2.67 \times 10^{-3}$ and never reaches zero, so the ground state is determined at every repulsion and the weights are measurements. The tie is exact — to $6 \times 10^{-16}$ in weight and $8 \times 10^{-15}$ in energy — where an arbitrary combination would give an arbitrary ratio near one rather than one to the last bit.

The same test overturns a smaller claim. The ring of four reads exactly one from $U = 0.5$ to $16$ and stops at $64$ and $256$, and the stopping was attributed to a nearly degenerate ground state confusing the solver. But that ring’s many-electron gap is $0.332$ at $U = 8$ and $0.014$ at $U = 0.5$, and the contrast is exactly one at both — at its widest and at its narrowest alike. Where the contrast is not one, at $U = 64$ and $256$, the gap is $0.062$ and $0.016$, in between. The gap and the tie are unrelated, so it cannot be the gap that decides.

What decides is the weight ordering. The tie flag flips between $16$ and $64$ because two lines exchange places in the ranking and the cut moves out of a pair — a discrete structural event, fully determined, needing no appeal to the solver. The earlier reading had the ordering in it: what does not change is their ordering by weight at the boundary. It changes at sixty-four.

There is a fair reading of the earlier result, and it is worth stating. Its ring of four was a deliberate refusal — a system the extrapolation should decline — and as a refusal it worked: the number came back as one and the system was excluded. What is disputed here is only the reason recorded for the exclusion, and the reason matters because it became the rule for admitting everything else.

The filling in between #

Between two electrons and six there is four, and it is the case that shows what the tie does to an extrapolation.

At four electrons the cut falls inside a pair at some repulsions and not at others, in no order — one at $8$, $16$ and $32$, a real ratio of $1.224$, $1.285$ and $1.295$ at $64$, $128$ and $256$, back to $1.060$ and $1.000$, then $1.292$ and $1.170$. The sequence is not converging because it is not one sequence: it is two interleaved quantities, a tie and a genuine ratio, sampled in an order decided by where the weight ranking happens to fall.

The extrapolation that worked at two electrons returns $1.1296$ for it. That number is not the limit of anything.

It is worth being precise about why this is worse than a straightforward divergence. A quantity that runs away is obviously not converging and nobody fits a line to it. A quantity that sits between $1.00$ and $1.30$ across a factor of two thousand in the repulsion looks like a converging quantity with some noise on it, and $1.1296$ looks like a reasonable summary of it. Everything about the presentation is plausible; the only thing wrong is that two different quantities are being read off one axis.

That is the same confusion more bands than there are orbitals had to untangle in the spectrum itself, where lines from different physical origins arrive interleaved and the count alone cannot separate them. Here it is not the lines that are interleaved but two definitions of one number.

The guard that licenses the extrapolation #

The extrapolation at two electrons anticipated exactly this with a test: the departure from the fitted limit multiplied by the repulsion must be one number across the tail, since a first-order approach makes that product constant. It is a good test and at two electrons it works — the product is $50.8$ across a factor of sixty-four in the repulsion, varying by under one per cent.

Applied at the two lower fillings it refuses both, and this is where it fails to help. At four electrons it refuses because the residuals run to $6.6 \times 10^{2}$ and will not settle, which is correct and informative. At half filling it refuses because the residuals are $9.9 \times 10^{-10}$ — machine noise around a quantity that is exactly constant — and their relative spread is therefore enormous.

The test is a relative spread, and its denominator is the departure it is measuring. A departure of zero is the best possible outcome for a first-order fit and the worst possible denominator for a test of one. So the instrument returns the same verdict for a sequence that has converged perfectly and one that is not converging at all, and the two are told apart instantly by the one number the test divides away.

What the answer to the question actually is #

Stripped of the instruments, the question has an answer and it is worth stating plainly, because the rest of this essay is about why the instruments could not deliver it.

At half filling on a ring of six the contrast falls monotonically — $1.513$, $1.448$, $1.178$ — and crosses the factor of two well before the first repulsion measured. It then reaches the floor of its own definition and stays there. So the contrast does not recover above two, and satellites at half filling cannot be told from fundamentals by intensity at any repulsion. That is the answer, and it does not depend on anything argued above.

What the argument above establishes is that the last figure in that sequence — the exact one — carries no information, so the answer rests on the three values before it. Three points is a weaker basis than ten and it is what there is. Koopmans’ theorem is exact for nothing is the essay that set the standard here: an answer is worth what its weakest step is worth, and saying so is part of giving it.

What was computed, and how #

Three fillings on a ring of six, at eight on-site repulsions from eight to a thousand and twenty-four, extended to sixteen thousand for the extrapolations. Each point is an exact diagonalisation of the neutral system and of the ion, followed by the overlap of every ion eigenstate with the neutral ground state with one electron removed — which is what a removal weight is, and is the calculation a satellite that never loses its place built.

A tie is defined as two lines agreeing in both weight and energy to $10^{-10}$. Requiring both is what stops an accidental coincidence between two unrelated lines from being counted as a degeneracy, and it is not a formality: at four electrons and a repulsion of sixty-four there are fourteen exact pairs among sixty lines, so coincidences in one quantity alone would be common.

The convention that the strongest $n_\uparrow$ lines are the fundamentals is inherited rather than derived, and what a photoelectron spectrum measures is where it comes from: a one-electron picture has exactly that many lines and each carries unit weight, so ranking by weight and cutting at the electron count reproduces it exactly when the repulsion is switched off. Every difficulty in this essay comes from carrying that rule into a regime where the object it was defined on no longer exists.

The refusal is a repulsion of zero. There the spectrum is the one-electron one — as many lines as there are electrons to remove, every one of unit weight, no satellite at all — and the right report is that there is nothing to be a ratio of, rather than a tie.

Where the model stops #

A ring of six is one system and the tie structure is a property of its symmetry. A chain of six has no rotational symmetry and so has fewer exact degeneracies in its ion, and whether its half-filled contrast also pins at one is not answered here. That is the first thing to check, and it is eight diagonalisations.

The tie is also exact only because the model is. A real molecule’s ion states are split by everything a Hubbard model leaves out — different orbitals on different atoms, spin-orbit coupling, vibrations — so the two lines would not be equal, they would be close. The contrast would then be near one rather than one, which is worse rather than better: near one looks like a measurement, and one looks like what it is.

And the contrast is one distinguishability test of two. The boundary belongs to the gap is about the other one — whether satellites fall inside the range the fundamentals span — and that test does not depend on a rank cut at all. It fails here too, at every repulsion above eight, which is why the honest summary of half filling is that satellites cannot be identified rather than that they are equally bright.

The generalisation #

Any quantity defined as the k-th largest divided by the (k+1)-th largest has this failure mode, and the failure mode is invisible in the value. Ratios of ranked quantities appear all over: a spectral gap taken between ordered eigenvalues, a participation ratio, a condition number formed from ordered singular values, a selectivity taken between the best and second-best score.

Each of them returns its floor when the two ranks it names fall inside a degenerate group, and each of them returns that floor as a number in the same units and the same range as a real answer. Nothing about the value says which happened.

The check costs nothing and is worth building in: alongside the ratio, report whether the two quantities at the cut are equal to within the arithmetic’s precision. A single boolean, computed from two numbers already in hand, separates these two things are equally large from these two things are the same thing. That is the column added here to the table, and it is the column that turns a limit of one from a result into a diagnosis.

The same lesson applies to the convergence test. A relative residual is the right instrument when there is something to converge and the wrong one when there may not be, and the repair is equally cheap: print the absolute residual beside it. Ten to the minus ten and six hundred are not hard to tell apart, and the ratio that hides the difference was constructed to be scale-free.

Who found it, and when #

That removal spectra develop multiplet structure at large on-site repulsion is standard many-body physics, following from the strong-coupling limit of the Hubbard model worked out in the 1960s and 1970s; the degeneracies are the ion’s spin and momentum multiplets and nobody discovered them here.

What is new here is narrower and is about instruments rather than about spectra: that a rank-based contrast reaches its floor by a tie rather than by a convergence, that a guard written in terms of the neutral’s shell degeneracy does not catch a degeneracy of the ion, and that a relative residual test refuses an exact answer and a divergent one in identical language. All three were found by taking the question literally and computing the six diagonalisations it costs.

Still open: the chain of six at half filling #

The obvious open question is the chain of six at half filling. It has the same electron count and much less symmetry, so it should have far fewer exact pairs in its ion, and if its contrast settles on a real number below two then the answer to the question is below two, and genuinely so on a system without the symmetry that produced the tie. That is the reading wanted, and it is eight diagonalisations away.

The nearer question is the tie’s own boundary. At four electrons the cut moves in and out of a pair as the repulsion changes, and the repulsions at which it moves are already recorded. What is not known is whether those crossings are level crossings in the ion — two multiplets exchanging their order in weight — or whether one line’s weight is passing another’s smoothly. The ion’s levels and weights are computed at every point already, so plotting the two lines at the cut against the repulsion would say which, and would turn the flickering into something with a mechanism rather than a list of repulsions where it happens.