Two more sites move the root, not the pole

The give-back is the fraction of the on-site correlation saving that an interaction reaching one bond hands back — forty-four per cent of it, when the interaction was only weighted rather than solved. Solved exactly on a ring of six, it changes sign near a neighbour repulsion of half the on-site one, and the essay that tried to locate that sign change found it changes sign twice: once at a root, where the pair count at one bond is back at its free value, and once at a pole, where the on-site count is and the fraction’s denominator vanishes. At on-site repulsions of 6, 8, 10 and 12 the root sat a little below U/2 and the pole a little above, and the window between them closed with the repulsion. Multiplied by $U$ it was 2.03, 2.11, 2.06 and 1.94 — a coefficient near two on a separation shrinking as $1/U$.

That essay said plainly what four points on one ring could not settle: whether the two was a property of the model or of six sites. It named the ring of eight as the test. The ring of eight has now been solved, and so has the ring of ten, and the answer arrived in an unexpected order: first a reason the obvious ring of eight cannot answer, then an answer that separates the two crossings completely.

The ring of eight that cannot be used #

A half-filled ring of eight with ordinary, periodic boundary conditions puts eight electrons into orbitals $-2\cos k$ at $k = 0, \pm\pi/4, \pm\pi/2, \ldots$, and the fourth electron of each spin lands in a degenerate pair at $k = \pm\pi/2$. So the free ring’s ground state is not one state but four, and “the free pair counts” every change is measured against are undefined until one of them is chosen. The essay that gave the interaction a 1/r tail saw this coming and proposed the obvious choice: the member the interaction itself selects, by diagonalising it inside the four. It also warned that at first order the selection might not happen. An open shell has defeated a reference here before: on the ring of four the correlation energy came out with the wrong power of the repulsion because its reference was never a single state, and the ring of eight is the same arithmetic at twice the size.

It does not happen. Projected into the four free states the interaction leaves its two lowest choices degenerate to within $10^{-12}$ at every repulsion scanned, so the member returned is whichever the eigensolver happens to produce. Its on-site count wanders from 1.875 to 2.125 pairs across the scan — with no change at all in the state it is supposed to stand for. A free value that moves when nothing physical does is a convention, and the give-back’s root and pole are both defined as returns to a free value.

There is a second problem and it is worse.

At $U = 8$ the interacting ground state of the periodic ring changes character at $V/U = 0.5115$: two states of different symmetry cross, the energy is continuous and the pair counts are not. Both changes jump there, and both change sign there. A bisection for the root lands on the jump; so does a bisection for the pole; they agree to every digit, which is what one discontinuity found twice looks like. On this ring there is no root and no pole in the sense the ring of six had them — there is a level crossing, reported twice in the same words. That is the defect the six-site essay was named for, one level down: a sign change that is not a zero, in a quantity whose denominator was never asked about.

Half a flux quantum, and a ring of ten #

Both problems have the same cure. Thread the ring with half a flux quantum — reverse the sign of the hopping across one bond — and the allowed wavevectors move to $k = \pm\pi/8, \pm 3\pi/8, \ldots$. The ring of eight becomes closed-shell, its free ground state unique, and its interacting ground state smooth through the whole region: the pair counts change steeply near U/2 but continuously: halve the steepest step of any scan and the change across it halves too, which is what a steep stretch does and a jump does not.

A ring of ten needs no flux. With ordinary boundary conditions it is closed-shell for the same reason six is, and it is 63,504 configurations against six’s 400 — a few seconds a point by Lanczos iteration rather than milliseconds.

So the comparison is between three closed shells: six periodic, eight antiperiodic, ten periodic. The boundary condition alternates along that sequence while the size grows steadily, and that is useful rather than awkward: anything that changes steadily from six to eight to ten is following the size, since the boundary condition went one way and then back.

The pole stays where six put it #

The poles are the same on all three rings. At $U = 6$ they are 0.5304, 0.5302 and 0.5297 of the on-site repulsion; at $U = 8$, 0.5199, 0.5195 and 0.5190; at $U = 10$, 0.5140, 0.5135 and 0.5132. At every repulsion the three rings share they agree to within a thousandth and a half of $U$. Whatever sets the pole, two and four more sites do not change it.

The roots do not agree. At $U = 6$ they are 0.4741, 0.4865 and 0.4941; at $U = 8$, 0.4869, 0.4975 and 0.5035; at $U = 10$, 0.4934, 0.5017 and 0.5059. Each two sites add a hundredth of $U$ or so, and the additions are shrinking — twelve thousandths from six to eight at $U = 6$, seven and a half from eight to ten — but they are all upward. The window between root and pole closes from one side only.

Why one moves and the other does not #

The difference between the two crossings is in what each is measured against, and it shows in the counts themselves.

The pole compares the on-site count with its free value, and at half filling that is a quarter per site on any ring — each site’s two spins are independently half occupied when nothing correlates them. The root compares the one-bond count with its free value, which is $1 - 2G^2$ per site, where $G$ is the free hopping amplitude per spin across one bond. That $-2G^2$ is the exchange hole: the correlation two electrons of the same spin carry in a determinant with no repulsion in it at all, and it is the only part of the free pair distribution that knows how many sites the ring has. That depends on which wavevectors are occupied, so it depends on the ring: 7/9 = 0.7778 on six, 0.7866 on the antiperiodic eight, 0.7906 on ten, rising toward $1 - 2/\pi^2 = 0.7974$ on an infinite ring.

Across the crossover both interacting counts get steeper with every two sites — at its steepest, per unit of $V/U$, the on-site count rises at 3.6, 5.7 and 8.3 pairs per site on six, eight and ten, and the one-bond count falls at 7.7, 12.2 and 17.2 — and they steepen about one point near $V/U = 0.52$, where the curves for all three rings nearly coincide. The on-site count’s free value of a quarter happens to be crossed right there — at 0.5199, 0.5195 and 0.5190 of $U$ — so steepening the curve about that point barely moves the crossing. The one-bond count’s free value is crossed out on the flank, where the curves differ most, and a steeper curve through the same pivot meets a given level closer to the pivot. Its target is also climbing, which pulls the crossing the same way.

So the root moves because it is a crossing on the side of a sharpening step, and the pole stays because it is a crossing at the step’s pivot. Neither is a statement about U/2.

The pivot itself is the ring’s charge-ordering crossover — the place the six-site essay already found the pole sitting in, on the steepest part of the rise of the alternating structure factor. That is a property of the interaction’s balance between the on-site and the one-bond terms, which is local and converges quickly with the ring’s size, so the crossover’s location barely moves while its width shrinks. A quantity tied to the location is size-independent; a quantity tied to the width, or to a free value that is itself a finite-ring number, is not, and the root is tied to both.

The coefficient was two on six sites only #

The quantity the six-site essay read its scaling from was (pole − root)·U, which is constant if the separation closes as $1/U$.

On the ring of six it is 2.03, 2.11, 2.06 and 1.94 from $U = 6$ to 12 — the published numbers, reproduced to the last figure — and then it keeps falling: 1.68 at 16, 1.45 at 20, 1.26 at 24 and 0.99 at 32. The range the six-site essay measured happens to sit across a shallow maximum, which is exactly where a coefficient looks most constant. On the antiperiodic ring of eight it is 1.57 at $U = 6$ and falls from there, to 0.66 at 16 and 0.20 at 32. On the ring of ten it is 1.28 at 6, 1.00 at 8, 0.73 at 10, 0.53 at 12 and 0.28 at 16.

So the separation does not close as $1/U$ on any ring. Read as a local power of $U$, it falls as roughly the first power on six up to 12 and steepens beyond; on eight and ten it is steeper from the start — between 16 and 32 the separation on the ring of eight falls by a factor of about six for a doubling of $U$, a local power near −2.7. The two was a property of six sites, read over the stretch of repulsion where six sites made it look like a constant.

U/2 is between them only for a while #

The six-site essay’s sharpest statement was that the naive location $V = U/2$ falls strictly between the root and the pole at every repulsion, the limit of both and neither of them. On the repulsions it measured that is true. It is not true for long.

On every ring the root rises through U/2. On six it does so near $U = 17.8$ — beyond the last repulsion the essay ran, which was 12 — and settles at about 0.5007 of $U$ through 24 and 32. On eight it crosses near 9.2 and on ten near 7.3, so on the ring of ten U/2 is already outside the window at $U = 8$, the repulsion most of this argument’s scans were run at. Above the crossing the root sits a few thousandths of $U$ above the half and the pole a few thousandths above the root, and U/2 is below both.

That turns the six-site picture around. The root was not approaching U/2 from below and stopping short; it was passing through on its way to the pole. Whether it reaches the pole is a question about rings larger than ten, and the two things already measured point in the same direction: the separation falls faster than $1/U$, and it falls faster on each larger ring.

How the rings were solved #

Every ground state is exact for its ring, in the sense the smallest many-electron calculation established: the full configuration space of half-filled rings of six, eight and ten sites with on-site repulsion $U$ and a repulsion $V$ between nearest neighbours, the lowest state found by Lanczos iteration with full reorthogonalisation from a fixed start vector, and the residual $\lVert H\psi - E\psi \rVert$ checked below $10^{-7}$ of $U$ at every point — the loosest is the ring of ten at $U = 16$, at $9 \times 10^{-8}$, which moves a pair count by about a millionth. The flux is a sign on the hopping across one bond. Pair counts at each separation are read off the ground state as the pair distribution was first built, and the free values from the Slater determinant of the occupied orbitals; on the closed rings the latter agree with $1 - 2G^2$ and with a quarter to $10^{-12}$.

Each root and pole is bracketed on a scan of $V/U$ from 0.44 to 0.57 in steps of 0.005 and then refined by regula falsi to $10^{-9}$ of $U$. The scan doubles as a check on smoothness, and the obvious check fails in an instructive way. Flagging any step more than six times the median step works at small repulsions and flags every ring from $U = 10$ or 16 upward, because the crossover sharpens with $U$ and its steepest step outgrows the flat ones without being a discontinuity. So the steepest interval of each scan is halved four times instead, keeping the half with the larger change: across a smooth stretch the larger half carries a little over half the change, and across a jump it carries all of it. Every closed ring at every repulsion passes as smooth by that test; the periodic ring of eight fails it. On the periodic ring of eight the jump is found and refined by bisection on the discontinuity itself, and the root and pole bisections are required to land on it.

What the comparison rests on: the ring of six reproduces its published root and pole at all four of its published repulsions to within a ten-thousandth of $U$ — the published pole of 0.5200 at $U = 8$ is the same crossing, $V = 4.1595$, rounded up from 0.51994; the three poles agree within 1.5 thousandths of $U$ wherever the rings overlap; the roots rise and the coefficients fall with size at every shared repulsion; every ring’s coefficient falls by more than a third across its range; and every root passes U/2, earlier on larger rings. The refusal is the periodic ring of eight: its reference must be four degenerate states with a selection gap below $10^{-10}$, its counts must jump by more than one where its energy does not, and its root and pole must coincide with the jump — if they came out as two distinct crossings, the smoothness test would be calling a level crossing a pair of zeros.

What three rings cannot say #

Three sizes are three points. The roots move by twelve thousandths and then seven and a half from six to eight to ten at $U = 6$; a geometric reading of that shrinkage would bring the root to the pole on an infinite ring, and a slower one would not. Ten sites are the largest this solver reaches in seconds, and the ring of twelve — 853,776 configurations — is where the question would be asked next.

The infinite ring has a transition, not a crossover. A finite ring’s charge ordering is a smooth crossover whose steepest point the pole sits near; the sharpening with size is the crossover on its way to becoming a transition. What kind of transition the infinite extended Hubbard ring has at these repulsions is a question this argument does not answer, and the answer matters: if the counts jump there, root and pole would meet in the jump — which is what the periodic ring of eight already shows, as an accident of a level crossing on eight sites.

One-bond interaction only. The numerator of the give-back has one term because the repulsion reaches one bond. With a second term the root stopped being a return of any one count to its free value and became a cancellation between two, and on rings of other sizes each of those counts would carry its own free value’s size dependence. With a 1/r tail the root depended on where the interaction’s zero was put, and nothing here tests whether the size dependence survives that.

Half filling, and the twist is a choice. Threading eight sites by half a flux quantum is the cure for its open shell, not a property of a real ring; the comparison treats the boundary condition as a nuisance and shows that the trend does not follow it, which is as far as three rings can take that.

A number read from one size #

The habit worth keeping is to treat any coefficient read off a single finite system as provisional until a second size has been tried — and to ask, before trusting a scaling law, which side of the computation it actually lives on.

The two that six sites gave was a careful measurement. It was reproduced here to the last figure. What it was not was a property of the model, because the root it was measured to is a crossing against a free value that changes with the ring and lies on the flank of a step that sharpens with the ring. The pole, measured the same way, is a crossing against a free value that does not change and lies at the step’s pivot, and it barely moved. The same arithmetic, the same solver and the same definition gave one size-independent number and one size-dependent one, and only a second and third size could say which was which.

Still open: the ring of twelve, and a reference for the periodic ring #

The obvious open question is the ring of twelve, which is closed-shell with antiperiodic boundary conditions and 853,776 configurations — beyond a dense solver, inside a sparse one given minutes a point. At $U = 8$ the root has climbed 0.0106 and then 0.0060 of $U$ from six to eight to ten; a third step would say whether the shrinkage is geometric, which would put the root and the pole together on an infinite ring, or slower, which would leave a finite window.

The nearer question is the periodic ring of eight itself. Its free reference is undetermined at first order, and the obvious next step is second order: diagonalise the interaction’s second-order effective operator inside the four free states and see whether it chooses. If it does, the periodic ring has a defined root and pole away from its level crossing, and a fourth point at size eight — with the other boundary condition — would test whether the trend in size survives a change of twist at fixed size.