A root that moves when nothing else does

A second-neighbour term turned the give-back’s root into a cancellation. With an interaction that reached one bond, the numerator was V times the change in the pair count at one bond, and its root was the repulsion at which that count returned to its free value. With a second term at two bonds the numerator became a sum of two contributions of opposite sign, and the root became a balance between an enhancement at one separation and a depletion at the next — at a repulsion where neither was at its free value.

The question that essay left was a genuine tail. A ring of six has three separations off-site: one bond, two bonds and the antipode. An interaction that falls off as one over the distance reaches all three, so the numerator has three terms, and three terms whose signs are not fixed in advance could vanish together at more than one repulsion. Whether they do decides whether the root of the give-back names one number in a model with a realistic interaction or several.

The short answer is one. The longer answer is that the one number turns out to depend on something nobody had thought of as a choice.

Three terms, and where they meet #

The interaction used first is $1/r$ with $r$ measured round the ring: V at one bond, V/2 at two, V/3 at the antipode. The on-site repulsion is U = 8 in units of the hopping, as throughout this argument, and the ground state is solved exactly at every V — the same exact treatment of a small ring with one more number in its diagonal. The numerator of the give-back is then

$$N = V\,d(1) + \tfrac{V}{2}\,d(2) + \tfrac{V}{3}\,d(3),$$

with d® the change in the number of electron pairs at separation r against the free ring. The denominator is the on-site saving, −U·d(0), exactly as before.

The three terms and their sum are drawn from V = 2 to V = 7. At weak coupling the one-bond term carries almost everything: electrons pushed off each other’s sites go to the neighbouring site, and the two longer separations barely move. As V rises the picture changes in a way the two-bond case never showed. The two-bond term turns positive and grows, the one-bond and antipodal terms turn negative, and the sum crosses zero at V = 5.9198 with the one-bond term at −1.526, the two-bond term at +1.741 and the antipodal term at −0.215.

So the cancellation at the root is a single separation paying for the two on either side of it. It is two against one, as the earlier essay’s two terms were one against one, and the one is the middle separation.

One root, found by looking everywhere #

A bracket and a bisection cannot answer how many roots. They find one, wherever the bracket happens to put it, and a pair of roots close together inside a coarse grid interval is invisible to them. The search that separated the root from the pole learned that the hard way: a bisection reports a root and a pole in identical words.

So the numerator was evaluated at every V from 0.02 to 12 in steps of 0.02 — six hundred ground states per interaction — and every sign change was then bisected to forty halvings.

All three numerators change sign exactly once. The two-bond case reproduces the earlier essay’s 5.5151 to every printed digit, which is the check that this calculation is that calculation with one term added. The 1/r tail measured round the ring gives 5.9198, and the same tail measured through the plane of a regular hexagon — where the second neighbour is $\sqrt{3}$ bond lengths away and the antipode two, so the ratios are 0.577 and a half — gives 5.9360.

And there is no near miss. Away from the root and above V = 1 the smallest the numerator gets is 0.2, against values of order one to ten elsewhere. Below V = 1 it goes to zero with V itself, which is the trivial root every give-back has and not a crossing. A pair of roots closer together than the scan’s step, both inside one interval of 0.02, would have to produce a curve that dips through zero and back within that interval while staying further than 0.2 from zero everywhere else, and the curve is smooth on a scale a hundred times coarser than that.

So the lead’s worst case does not happen on this ring. Three terms of mixed sign could cancel three times, and they cancel once. The root of the give-back names one number with a realistic tail, as it did with a truncated one.

Every separation turns over #

What changes is which separations are doing the cancelling.

With the two-bond reach of a half and nothing at the antipode, the root sits where one bond is still enhanced (+0.185), two bonds depleted (−0.370) and the antipode — which costs nothing in that model — heavily enhanced (+0.527). The antipode was where the electrons pushed out of the penalised separations went, because it was free.

Put a price on the antipode and that refuge closes. With the 1/r tail round the ring the same three changes at the root are −0.258, +0.588 and −0.109. Every one has turned over. Across the hexagon, where the antipode costs half of V rather than a third, they are −0.310, +0.704 and −0.194: the same pattern, more strongly.

The reversal is not an effect of the antipodal term’s size. At the root the antipodal contribution to the numerator is the smallest of the three, −0.215 against +1.741 for two bonds. What it changes is the arrangement the electrons have to settle into, and the numerator is a weighted sum over that arrangement. A small term that redirects where the displaced pairs go reweights every other term.

The pair counts are not free to do anything, either. Their changes add to zero exactly — to 1e-13 at every point scanned — because six electrons make fifteen pairs, counting a doubly occupied site as one, whatever the state. On the free ring those fifteen sit as 1.5 on-site, 4.667 at one bond, 6 at two and 2.833 at the antipode — already not the 1.5, 6, 6 and 3 that independent densities would give, which add to sixteen and a half because they let an electron pair with itself, and not even uniform in their departure from them, because same-spin electrons keep apart with no repulsion at all. A correlated state can move them between separations and cannot create or destroy any. With the antipode free the displaced pairs could all go there; with the antipode charged they go to the middle.

This is also where the earlier essay’s reading of its own root lands. It found the root a balance between one bond and two, with the one-bond count still enhanced at a reach of a half. With a genuine tail the one-bond count is already depleted at the root: the root comes 0.280 later than the repulsion at which one bond returns to its free value (5.640), not 0.355 earlier.

Turning the antipode up by hand #

The two tails sit at particular antipodal strengths, a third and a half of V. Holding the two-bond term at half of V and moving the antipodal one from nothing to a half by twelfths shows how the arrangement gets from one pattern to the other.

It does not get there in one step. At no antipodal term and at a twelfth, two bonds is the depleted separation against two enhanced ones. At a sixth the antipode is the enhanced one against two depleted. At a quarter one bond is the depleted one. From a third onward two bonds is the enhanced one, and it stays so to a half. So the partner the cancellation is made with is handed on three times between a reach with nothing at the antipode and the 1/r tail, and each hand-over happens somewhere inside a twelfth of V.

The root does not move monotonically either. It rises from 5.515 to 6.139 at a sixth and 6.158 at a quarter, and then falls, to 5.304 at a half. The pole follows the same shape a little above it, so the two are never further apart than at no antipodal term: 0.706 there, 0.354 on the 1/r ring, 0.285 at a half. The window between the two crossings that the second-neighbour term widened is narrowed again by the third.

There is a reading of the non-monotone root. Half filling on a bipartite ring keeps the ground state a singlet throughout — Lieb’s theorem rather than a computation — so nothing here is a change of spin state. A modest antipodal price competes with the two-bond price for the same displaced electrons, so a larger V is needed before the nearest-neighbour term wins by enough to change sign — the competition that pushed the root later at two bonds, pushed further. A large antipodal price makes the charge-ordered arrangement favourable sooner: in the alternating arrangement on a ring of six the doubly occupied sites are two bonds apart and never antipodal, so penalising the antipode costs it nothing. Past a quarter the second effect wins.

Lowering every interaction at once #

The pair changes adding to zero has a consequence that is easy to state and that undermines the root more thoroughly than anything above.

Add the same constant c to the interaction at every separation, on-site included. The Hamiltonian gains c times the number of pairs, which is fifteen in every configuration, so it gains the constant 15c and nothing else. Every state is unchanged — the ground state at (U, V, V/2, V/3) and at the same four numbers less 1.5 have pair changes that agree to 1e-9 at every separation. The energy is shifted and nothing physical is. It is the same kind of observation as the factor of 259 between two correlation energies of one state: a number that changes when nothing physical does is reporting a convention.

The give-back is not shifted by zero. Its numerator becomes N − c·d(0) and its denominator −(U + c)·d(0), and neither of those is N or −U·d(0).

The cleanest case is an interaction that is flat off-site — the same V at one bond, two bonds and the antipode. Lowering everything by V leaves the Hubbard model with on-site repulsion U − V and no off-site interaction at all, and the solver confirms that the two have identical pair changes. So the flat model’s state is a Hubbard state, its numerator is V times d(1) + d(2) + d(3), which is −V·d(0), and its give-back is exactly V/U — at every repulsion, whatever the state looks like, to 1e-8 on the scan.

On the 1/r ring at V = 2 the give-back as written is 0.228 of the on-site saving. Lower the interaction by V/3 everywhere, so that the antipode costs nothing and the tail is a two-bond interaction of V − V/3 and V/2 − V/3, and the same state’s give-back is 0.158. Nearly a third of the number disappears without one electron moving.

None of this is peculiar to the give-back’s denominator. The numerator alone is the energy the off-site interaction returns, and it changes too, by c times the change in the double occupancy. It is a statement about a part of the interaction energy, and a part of an energy has no zero of its own unless the model supplies one.

The pole stays, the root moves #

The two crossings are not affected equally, and the difference is the most useful thing on the page.

The pole is where d(0) vanishes — the double occupancy back at its free value — and d(0) is a property of the state, the quantity a correlation hole is measured by at contact. Shifting the interaction changes the denominator from −U·d(0) to −(U + c)·d(0), which vanishes at the same repulsion. So the pole stays at 6.274.

The root is where N − c·d(0) vanishes, and that depends on c. With the antipode made free it moves from 5.920 to 5.777, a seventh of a unit of V for a re-zeroing that changes no state at all. For comparison, moving the reach from nothing to a half moved the root by 1.62, and that was a change to the physics.

Nor is the antipode a privileged place to put the zero. Lower by c = κV for any κ, and the root sits wherever N/(V·d(0)) equals κ.

That ratio rises monotonically from −0.918 at weak coupling to infinity at the pole, so every choice of zero gives exactly one root, and the choices are not close together. Zero at the antipode puts it at 5.777. Zero at two bonds puts it at 5.646. Anything lower than 0.918 V puts it nowhere — with the interaction zeroed at one bond, so that the longer separations are attractive, the numerator never vanishes below the pole at all, and the give-back has no root to report.

The curve’s shape is the reason the root is so sensitive near κ = −0.9. Up to V = 3 the ratio stays between −0.918 and −0.903, because at weak coupling the pairs displaced from contact go almost entirely to one bond in a fixed proportion. A lowering anywhere between 0.918 V and 0.880 V puts the root anywhere from zero to four.

What the model does and does not supply #

So the question is whether the model fixes the zero, and on this ring it does not.

A Coulomb interaction in an extended system fixes its own zero at infinite separation. It goes to zero as the electrons move apart, and every electron in a neutral system is surrounded by a compensating charge, so the uniform part of the interaction is cancelled before any partitioning starts. That is the reason electron-gas theory works with a neutralising background, and the reason an exchange–correlation hole that integrates to the right number of electrons is worth more than one with the right shape. The sum rule here is that statement on six sites.

A ring of six has no infinity. The farthest any two electrons can be is three bonds, where the 1/r tail is still a third of V. Setting the interaction to zero there is a choice, setting it to V/3 is the 1/r form taken literally, and nothing in a ring of six says which is right. On a ring of n the antipodal value of a 1/r tail is 2V/n, so the ambiguity shrinks as the ring grows — but it shrinks as one over the size, which is slow, and every quantity on this ring is measured at the size where it is largest.

The Hubbard and two-bond models never had to face it. With the interaction zero beyond one bond or two bonds, the zero is set by the model’s own truncation, and the earlier essays’ roots inherited it without saying so. The truncation was a choice too; it just did not look like one.

Exactly what was solved #

The ring of six at half filling: three electrons of each spin, 400 configurations, hopping t = 1 round the ring, and a diagonal interaction given as one number per separation — on-site, one bond, two bonds, antipode. Three electrons of each spin means a hop across the bond that closes the ring passes two others of its spin, so every fermion sign is +1 and the ring is periodic without any extra care; that is a property of this filling and not of rings in general.

The ground state by Lanczos iteration with full reorthogonalisation, stopped when the bound on the lowest Ritz value’s residual falls below 1e-11, and then checked directly: ‖Hψ − Eψ‖ is below 1e-7 at every one of the points scanned. With no antipodal term the pair changes at contact, one bond and two bonds agree with the dense solver every earlier essay used to 1e-8 at three interactions.

The pair changes are counts, not ratios. d® is the expected number of electron pairs at separation r in the correlated state less the number in the free ring’s, counting a doubly occupied site as one pair. Their sum is checked to be zero at every point, and an antipodal list that counted each of its three pairs twice breaks it — a check that catches an interaction of the wrong strength, which is how such an error presents.

Lowering everything by a constant is checked to leave the state alone at one interaction to 1e-9, and the flat interaction is checked to be the Hubbard model at U − V separation by separation, against the solver rather than the algebra.

Where the ring stops #

Six sites is three separations. A real tail reaches as far as the system does, and on a ring of six the farthest separation is close enough for the tail to be a third of its nearest value. That is the size at which the zero of the interaction matters most and also the only size the earlier essays could solve. The Lanczos solver here is not limited to it: a ring of eight at half filling is 4,900 configurations, sparse, and costs a tenth of a second a point rather than the dense solver’s minute.

One repulsion, U = 8. Every comparison is made there, in the range where a mean field’s error has long since outgrown the answer and only the exact state is worth differencing. The non-monotone root along the antipodal sweep and the narrowing of the window between root and pole are both read at that single value, and at a different U the hand-over points along the sweep will move.

The tails are chosen. Round the ring and through the plane of a hexagon are the two obvious meanings of distance on a ring of six, and neither is screened. A screened form such as Ohno’s would lower the tail’s long end and change the antipodal ratio, and on the evidence of the sweep, changing that ratio by a twelfth can change which separation carries the cancellation.

A number that comes with a convention #

The habit this leaves: before reading a quantity’s zero, find out whether the quantity is invariant under every change that leaves the physics alone. The give-back is a share of an interaction energy, and an interaction energy can be split into parts only after its zero is fixed. On a ring where nothing fixes it, the share — and the repulsion at which it changes sign — is partly a statement about the convention.

The pole is the quantity that survives. It is where the double occupancy returns to its free value, a property of the ground state alone, and it moved by 0.04 between the two 1/r tails while the root of a single tail moved by 0.27 between two choices of zero. A report of either crossing in a model with a long-ranged interaction is worth more if it says which one it is, and for the root, where the interaction was taken to vanish.

Who computed what #

The extended Hubbard model with a nearest-neighbour term dates from the 1970s, and long-ranged interactions have been standard in the Pariser–Parr–Pople treatment of π systems since 1953, with Ohno’s screened form from 1964. That a uniform off-site interaction acts only through U − V follows from counting pairs and is not new. The give-back, its root and pole, and the two-bond cancellation are this argument’s own, from the essays before it.

What is computed here is the antipodal term, the full scan for roots, the reversal of every pair change at the root, the hand-overs along the antipodal sweep, and the dependence of the root, but not the pole, on where the interaction’s zero is set.

Still open: the ring of eight, and a zero the model chooses #

The size question that the dense solver refused twice is open now that the ground state of the larger ring costs a tenth of a second. A ring of eight would say whether the root–pole separation’s 1/U scaling has the coefficient near two that six sites gave — with a complication worth having: a ring of eight electrons at half filling is open-shell when free, its highest occupied level a degenerate pair, so the free ground state is four states rather than one and the free pair counts every change is measured against are not defined until one is chosen. The obvious choice — the member the interaction itself selects — fails at first order: projected into the four, the interaction leaves its lowest level still degenerate. Either second order chooses, or the reference becomes a second convention alongside the zero of the interaction. An antiperiodic ring of eight, threaded by half a flux quantum, is closed-shell and has no such problem, which makes it the cleaner system and a different one.

The nearer question is whether any model fixes the zero by itself. A ring whose tail is taken from the Ewald sum of a periodic lattice, with a neutralising background, has a zero set by the physics rather than by the reader, and the give-back’s root on that ring would be a property of the state in a way none of the roots here are.