The limit was two chains at once

Sorting a photoelectron spectrum by height rests on a contrast: the weakest of the lines that belong to orbitals over the strongest of the rest. With N↑ electrons of one spin to remove, the fundamentals are the N↑ strongest lines and the contrast is the N↑-th over the next; below two, height stops sorting a spectrum. For half-filled Hubbard chains that contrast was followed to very large repulsion, and its limit turned out to be a parity. At large repulsion the lines pair up at ±E with weights that agree as $1/U$, so the contrast goes to one when the cut after the N↑-th line falls inside a pair — on the chains of two and six, with one and three up electrons — and to something else when it falls between pairs. On the chain of four, with two, the limit came out at 1.3125.

The exact many-electron calculation behind those numbers diagonalises every configuration of the chain, and two things were left. That number looks like a fraction, 21/16, and nothing said whether it was one. And the rule itself was established on three chains: the chain of eight was predicted to miss one and the chain of ten to reach it, neither could be diagonalised exactly at the cost used, and the prediction was an argument from the levels of a single particle rather than a count.

Both are answered by building the limit instead of approaching it.

A hole in a chain carries two problems with it #

At very large repulsion the half-filled chain holds one electron on every site — the regime in which the molecular-orbital description has dissociated — and nothing moves: a hop would put two electrons on one site at a cost of $U$. What remains of the electrons’ freedom is their spins, and they are coupled only through virtual hops that go and come back, at an energy of order $t^2/U$ — the superexchange that couples two spins through an empty or doubly occupied intermediate. The ground state is therefore a spin chain’s ground state, the singlet of a Heisenberg chain of n spins, with every site singly occupied.

Removing an electron leaves a hole, and the hole can move at no cost in repulsion: an electron next to it hops in. On an open chain that motion has a property that makes everything after it simple. As the hole passes along the chain the electrons it passes shift by one site each, but their order along the chain never changes. Read left to right, skipping the hole, the spins form a string of n − 1 that the hole’s motion leaves alone. So a final state is a product: a wave for the hole on the n sites, times a state of the n − 1 spins in their order.

The hole’s part is one particle hopping on a chain of n sites, and its standing waves are the ones every chemist has met as Hückel orbitals of a linear polyene, with levels at $2\cos(k\pi/(n+1))$. That is why the lines at large repulsion sat at those energies to within a hundredth. Each level is a cluster rather than a line, because every state of the n − 1 spins has the same hole energy; the spins split the cluster only at order $t^2/U$.

The weight of a line follows from the product directly. Removing an up electron from site j leaves the hole at j and a string of n − 1 spins whose amplitude is the ground state’s amplitude for that string with an up spin at j. Call that set of amplitudes the spin factor for site j: it belongs to the spin chain alone and knows nothing about the hole. A line is then a hole level k and a spin state χ, and its weight is

$$W(k,\chi) = \sum_{j} \varphi_k(j)^2 \,\bigl(\chi \cdot A_j\bigr)^2 ,$$

the hole’s standing wave squared at each site times the square of the spin state’s overlap with what removal at that site left behind.

One consequence needs no calculation. Summed over every spin state in a level, the second factor becomes the probability that site j holds an up electron in the ground state, which is one half on every site of a singlet. The hole’s wave squared sums to one. Every hole level carries exactly half an electron, on every chain, whatever the spins do. The total of n/2 up electrons is shared out equally among the n levels, and the whole of the spin chain’s influence on the spectrum is in how each level’s half is divided among its lines.

Twenty-one sixteenths, nearly #

On the chain of four the division can be done by hand. Three spins are left behind, and with two down and one up they hold three states: a quartet component that removal from a singlet cannot reach, and two doublets. Reflection of the chain sends the string’s first spin to its last, and the two doublets have opposite parity under it — one even and one odd. The hole’s waves are each even or odd too, so whatever operator splits the cluster is symmetric under reflection and cannot mix an even doublet with an odd one. On the chain of four, symmetry alone decides which spin states the lines are.

Then the two factors are numbers. The four-site singlet, whose energy is $-\tfrac{3+2\sqrt3}{4}J$, has more of its weight on the end bonds than the middle one, and its spin factor for the even doublet is $\tfrac{2+\sqrt3}{8} \approx 0.467$ at an end site and exactly $\tfrac14$ at a middle one. An end spin has one neighbour to form a singlet with and does so more completely, so removing it leaves the remaining three closer to the doublet that matches. The hole’s waves on four sites are built from $\sin^2(\pi/5) = \tfrac{5-\sqrt5}{8}$ and $\sin^2(2\pi/5) = \tfrac{5+\sqrt5}{8}$, the golden-ratio numbers of a pentagon: the inner levels at ±0.618 put more of the hole on the end sites, and the outer levels at ±1.618 more on the middle ones.

The strongest line is where they agree. The inner levels’ hole sits where the singlet’s spin factor is largest, and their even line weighs

$$W_{\text{inner}} = \tfrac14 + \tfrac{\sqrt3}{16} + \tfrac{\sqrt{15}}{80} = 0.406665, \qquad W_{\text{outer}} = \tfrac14 + \tfrac{\sqrt3}{16} - \tfrac{\sqrt{15}}{80} = 0.309841 .$$

Each odd line is a half less its partner, as the level’s sum requires. With two up electrons the cut falls between the inner pair and the outer pair, and the limit of the contrast is their ratio:

$$\frac{20 + 5\sqrt3 + \sqrt{15}}{20 + 5\sqrt3 - \sqrt{15}} = 1.3124978 .$$

That is not 21/16. It misses it by 0.0000022, in the sixth figure, which is why four decimal places could not tell them apart. The exact spectra extrapolated from U = 4096 and 16384 give 1.3124972, six ten-millionths from the closed form and nearly five times as far from the fraction. The term that separates the strongest line from the next carries both irrational numbers at once: $\sqrt{15}$ is $\sqrt3$ from the spin chain times $\sqrt5$ from the hole, and an intensity ratio in a photoelectron spectrum turns out to be a product of a Heisenberg singlet’s end-weighting and a pentagon’s geometry. The first is a property of electrons that cannot move, the second of one hole that can, and in this limit they do not interact at all.

The construction is the limit the exact spectra are heading for, and not a model that happens to sit near them. For the chains of four and six every strong line of the exact spectrum was matched to its built counterpart from U = 64 to 16384, and the largest miss falls along a line of slope −1.000 on logarithmic axes on both chains: 0.028 at U = 64, $1.1 \times 10^{-4}$ at U = 16384. What remains at any finite repulsion is a correction in $t/U$, which is the size of the hopping the infinite-repulsion picture has frozen.

The energy that vanishes still decides #

The chain of four was easy because symmetry chose the spin states. On longer chains it does not. Five spins left behind on the chain of six hold five doublets, seven on the chain of eight hold fourteen, nine on the chain of ten hold forty-two, and reflection sorts them only into two groups. Which combinations are the lines is decided by whatever splits the cluster, and that is the order-$t^2/U$ part of the Hamiltonian acting inside the level.

That has an odd consequence. At infinite repulsion the splitting is zero, and every combination of spin states within a level has the same energy; a level’s half-electron could be shared among them any way at all. But the limit is the limit of a sequence, and at every finite repulsion the lines are the eigenstates of the splitting operator, whose eigenvectors do not depend on how small it is. So the intensities at infinite repulsion are set by an operator whose size has gone to zero, and getting them right means getting that operator right, not merely its order of magnitude.

The operator has two parts. One is exchange between neighbouring spins across bonds the hole is not part of, which is the superexchange of the ground state, weighted by the chance that the hole is not sitting between the two spins. The other is a hop of the hole across an occupied site: an electron next to the hole hops in, the site it came from briefly holds two electrons, and one of them hops out to the far side. The hole moves two sites at once and the spin it crossed may be flipped. This three-site process is the same order as exchange, and in the reading of the strong-coupling limit as spins plus a hole it is often dropped.

It cannot be dropped here. On the chain of eight, with exchange only, the strongest line at ±1.00 weighs 0.2581 and the one at ±1.53 weighs 0.2402, and the limit would be 1.0746. With the three-site hop the first falls to 0.2508 and the second barely moves, to 0.2398, and the limit is 1.0458. The two lines at the cut are shifted by amounts differing by a factor of about nineteen, and the contrast — a ratio of exactly those two — loses two fifths of its excess over one. The same happens on the chain of twelve, from 1.0364 to 1.0195. On the chain of four the two operators give identical lines, which is the symmetry argument above made visible: when there is nothing to choose between, how the choice is made cannot matter.

Which of the two is right is a question the exact spectrum can answer, and it was asked separately: the chain of eight diagonalised in full at U = 1,000 and 10,000, a configuration space of 4,900 states for the ground state and 3,920 for the hole. Its contrasts are 1.0361 and 1.0448, and extrapolated to infinite repulsion as $1/U$ they give 1.04577 — against 1.045772 from the construction with the three-site hop, and nowhere near 1.0746 without it. At U = 10,000 the two members of the pair at ±1.00 weigh 0.25091 and 0.25070, which average to the built 0.2508.

Every level holds a half, and the cut is between the second pair and the third #

Drawn level by level, the chain of eight shows the sum rule and the parity rule together. Every column is exactly one half, split among fourteen lines. The strongest line of each level is the bottom of its column, and those strongest lines rank by distance from the band centre: 0.316 at ±0.35, 0.251 at ±1.00, 0.240 at ±1.53 and 0.229 at ±1.88. Each appears twice, once at +E and once at −E, with weights equal not to some number of figures but identically.

That identity is worth stating precisely, because it is what turns the parity rule from an argument into a theorem in this limit. The hole’s waves at levels k and n + 1 − k differ only by a sign on alternate sites, so they have the same square everywhere. And every process in the splitting operator moves the hole by zero sites or by two, never by one, so alternating signs cancel in pairs and the operator inside level k is the same operator as inside level n + 1 − k. Same squares, same operator, same lines. Computed separately for every level on every chain up to ten, the ±E pairs agree to $10^{-9}$.

With four up electrons the cut after the fourth strongest line therefore falls after two complete pairs, between the pair at ±1.00 and the pair at ±1.53. The contrast is their ratio, 0.2508 ÷ 0.2398, and the limit is 1.0458. With an odd number of up electrons the cut must fall between the two members of a pair and the ratio is a number divided by itself.

Odd gives one, even gives less each time #

So the parity rule is now a count on six chains. The chains of two, six and ten have one, three and five up electrons and a limit of exactly one; the chain of ten, which had only been predicted, is built and checked. The chains of four, eight and twelve have two, four and six and limits of 1.3125, 1.0458 and 1.0195. Every even chain stays above one, as the rule required, but by a shrinking margin: the excess falls by a factor of 6.8 from four to eight and 2.3 from eight to twelve, and over the last step that is close to the inverse square of the length. Two steps do not establish a law, and the excess may be heading to zero or to a floor; what can be said is that the cut on an even chain falls between two pairs whose weights draw together as the levels crowd, so a long even chain’s limit is one in practice even though it is never one exactly.

That reframes what the tie on the ring of six and the number the tie got right were measuring. A contrast of one at strong repulsion is not evidence that fundamentals and the satellites a correlated spectrum adds have become indistinguishable. It is evidence that the rank cut has fallen between two lines the physics has made identical, and on a chain that happens exactly when the number of electrons of one spin is odd. A contrast of 1.02, read on a chain of twelve at the same repulsion, would be the same physics with the cut one line further along.

How the limit was built #

The construction is second-order degenerate perturbation theory taken from the hopping itself, not from a derived spin Hamiltonian. Every configuration with one electron on every site — or, after removal, with one empty site — is a state of the unperturbed problem. The operator at order $t^2/U$ is $-PTQTP/U$, where $T$ is the hopping, $Q$ projects onto configurations with a doubly occupied site and $P$ back onto the rest: one hop up into the costly intermediate, one hop back down, divided by the cost. Written that way the exchange and the three-site hop both appear without either being derived, and dropping the three-site hop means keeping only the terms that return the hole to the site it left.

The ground state is the lowest eigenvector of that operator on the half-filled configurations — 924 of them on the chain of twelve — found by a Lanczos iteration. For each hole level the operator is projected onto that level’s states, the hole’s standing wave times every string of n − 1 spins — 3, 10, 35, 126 and 462 strings on the chains of four to twelve — and diagonalised; degenerate eigenvalues are pooled into one line, since a spectrum cannot tell their members apart. The weights then follow from the formula above, with the fermion sign of removing an up electron from site j: minus one for each up electron to its left.

The checks, run wherever these numbers are drawn. On every chain the weights sum to the number of up electrons and every hole level to one half, to $10^{-8}$. The ±E levels are computed separately up to the chain of ten and agree to $10^{-9}$; on the chain of twelve the upper half is copied from the lower, which is what the operator’s even moves guarantee and what the smaller chains confirm. The chain of four’s eight lines equal their closed forms to $10^{-9}$ with or without the three-site hop, and its closed form differs from 21/16 by between one and ten millionths. Odd chains reach one to $10^{-9}$ and even ones exceed it; the three-site hop moves the chain of eight’s limit by more than 0.02. The exact spectra of the chains of four and six land on the built lines with a miss below $3 \times 10^{-4}$ at U = 16384 falling as $1/U$, and their extrapolated limits agree with the construction. The refusal is the fermion sign: built without it, the chain of four’s limit comes out at 1.0871, and the comparison with the exact spectrum rejects it — so the agreement is a statement about the construction and not about a check that would have passed anything.

What the product does not reach #

Infinite repulsion, and the first order beyond it. Everything here is the limit and the splitting that selects its lines. The approach on the long chains is not followed: the chain of six’s contrast reached its limit through exact ties and changes of satellite, and the chain of eight, at 1.036 for U = 1,000, is approaching from below, but whether it passes through ties on the way is not known from two points.

Open chains. The product rests on the hole leaving the spins’ order alone. On a ring a hole that goes all the way round shifts every spin by one place, so the string is not left alone and the factorisation needs the spin chain’s own momentum. The closed-form limits on rings below half filling were a different construction with the same flavour.

Twelve sites. The chain of fourteen needs blocks of 1,716 spin strings per level, which is within reach of a better eigensolver and not of the one used. The decline of the even chains’ excess is read off three points.

One kind of interaction. The model has only on-site repulsion, and its spin chain is the Heisenberg chain that model produces. A neighbour repulsion or a frustrating second-neighbour exchange would change both factors, and a real conjugated chain has both; what survives is the structure — a hole’s wave against a spin state — not the numbers.

And the removal channel. At half filling the addition spectrum is this one reflected, so nothing is lost; away from half filling there is no singlet of one electron per site to start from, and none of this applies.

Who found the product, and what it says about intensities #

That the infinite-repulsion Hubbard chain’s eigenstates are a product of spinless charges and a squeezed spin chain was shown by Masao Ogata and Hiroyuki Shiba in 1990, from the Bethe-ansatz solution; it is the cleanest statement of spin–charge separation in one dimension. The expansion in $t/U$ that produces both the exchange and the three-site hop goes back to the large-repulsion treatment of narrow bands by Harris and Lange in 1967. Neither was aimed at the question here, which is a spectroscopist’s: when lines are sorted by height, what decides the height at the place where the sorting is done.

The answer generalises. At strong repulsion a removal intensity is not a property of an orbital, as Koopmans’ picture would have it, nor of a many-electron state in some irreducible way. It is the overlap of two separate problems — where the hole is, and what the spins it left behind are doing — and either can be computed alone. That is why the chain of four has a closed form and the longer chains have numbers: four sites leave a spin problem small enough that symmetry solves it, while eight leave fourteen doublets that symmetry sorts only into two groups, and the lines are whatever eigenvectors the splitting operator has. And it is why a calculation that gets the hole right and the spins slightly wrong — exchange without the three-site hop — reads the chain of eight’s limit more than 60 per cent too far from one, while being right about every line’s position.

Still open: the ring’s twist, and the approach on eight #

The obvious open question is the ring. On a ring of n sites a hole that circles once shifts the spin string cyclically, so the hole’s wave and the spin state are tied together by a boundary condition: the hole moves with a momentum that includes the spin string’s own. The product survives with that twist, and the ring of six’s exact one — which came through a degeneracy, not a parity — should appear in it as two spin momenta whose hole levels coincide. Building rings of four, six and eight at infinite repulsion the same way would say whether their contrasts obey a rule about the number of up electrons, about momentum, or about both.

The nearer question is how the chain of eight gets to its limit. The chain of six’s approach had a tie at the cut and two changes of satellite on the way; the chain of eight sits at 1.036 at U = 1,000 and 1.045 at U = 10,000, below its limit and rising. Whether its approach is monotone, or whether its lines also exchange ranks on the way, needs its spectrum at a few dozen repulsions — a few minutes each at the size measured here, and the same following-by-continuity that took the chain of six apart.