When the atoms are not all the same

Three rungs of this ladder search for homometric pairs and all three search for the same object: a set of points. Four atoms on eight sites, the sumset construction that explains them, the same search moved onto a torus — in every one of them an atom is a position and every position counts once.

A crystal is not like that. A Patterson peak at a vector v has height

$$P(v) \;=\; \sum_{a-b\,=\,v} w_a\, w_b,$$

where w is the atom’s scattering power, so a sulfur contributes about sixteen times what a carbon does to the same vector. The difference multiset the searches compare is the special case in which every w is one.

The question that leaves is whether the ambiguity is an artefact of that special case. Two arrangements with the same vectors need not have the same weighted autocorrelation, and the natural expectation is that unequal atoms break most homometric pairs — that homometry is a degeneracy which distinguishing the atoms lifts, in the same way a perturbation lifts a degenerate energy level.

That expectation is wrong in an interesting direction. Distinguishing the atoms does break some pairs. It also creates more than it breaks.

The arithmetic, and why it is still exact #

The search is the previous rungs’ search with one thing added. A configuration is k occupied sites on a cycle of n, each carrying a weight from a small alphabet — here one or two, which is a light atom and a heavier one. Its weighted autocorrelation is the sum above, computed at every vector.

Everything remains an integer. The weights are whole numbers, the autocorrelation is a sum of products of them, and two configurations either agree in every entry or differ by at least one. There is no tolerance in the comparison and no question of how close counts as equal, which is the property that makes an exhaustive search a decision rather than a survey.

Two configurations are compared up to translation and reflection, with the weights carried along. Translation changes no vector; reflection reverses every vector and leaves the multiset alone, which is Friedel’s law in its simplest form. Quotienting by both is what the unweighted search does and it is what stops every configuration being trivially paired with its own copies.

The weights travelling with their atoms is the part that is new and it is the whole difficulty. Two configurations occupying the same sites with the weights permuted are genuinely different structures — one has the heavy atom here and the other has it there — so the canonical form is over the sequence of position-and-weight pairs, not over the positions with a weight histogram beside them. Comparing positions alone would identify structures a chemist would never confuse.

Ambiguity where there was none #

The sharpest case is a ring of nine sites with four atoms, and it is sharp because the unweighted answer is nothing at all.

No arrangement of four identical atoms on nine sites shares its vectors with another. That is a decision from an exhaustive search over every class, not a failure to find one. On that ring, in that setting, the phase problem has no homometric component: recovering the structure from the intensities is hard, and it is not ambiguous.

Allow two kinds of atom and there are six homometric groups. The structures in them differ in position and in which atom is where, and their weighted vectors agree exactly. So on the ring where identical atoms are perfectly determined, distinguishing the atoms introduces arrangements that no measurement of intensities separates.

That is worth stating carefully because it inverts a comfortable intuition. Adding information to the structure has added ambiguity to the measurement. The two are not in tension: what the experiment measures is the autocorrelation, and a richer set of structures maps into the same space of autocorrelations, so more structures collide. The extra information is in the object and not in the data.

What the shared function looks like #

The pair’s common autocorrelation is what an experiment would measure, and drawing it makes the loss concrete.

The origin peak is worth a sentence because weighting changes what it is. In the unweighted case it is the number of atoms, which a chemist knows already. Weighted, it is Σ w², which is the quantity Wilson statistics call the total scattering power of the cell — genuinely useful, and still nothing about the arrangement. The uninformative peak stays uninformative and merely becomes a different uninformative number.

Which weightings the construction survives #

Patterson’s construction explains why homometric pairs exist rather than merely exhibiting them: a set whose points are the sums b + c over two smaller sets has a partner obtained by reversing one factor, and the vectors do not notice. The question here is what weights do to it.

The construction survives exactly one family of weightings, and it is a small one. Give the atom at b + c the weight u(b)·v(c) — a product of a weight belonging to the first factor and one belonging to the second — and the weighted autocorrelation factors again, so the reversal is invisible again and the pair persists.

That is a strong condition. A structure’s atoms are whatever elements the chemistry supplies, and there is no reason at all for their scattering powers to factor along a decomposition of the positions. So most weightings break most instances of the construction — which is the expectation the essay opened with, and it is right about this family and wrong about the total.

Half of the classic pair #

Applying the same question to the pair every treatment of this subject opens with gives the cleanest number in the essay.

Eight of sixteen. Not none, which would say weighting resolves homometry; not sixteen, which would say weighting is irrelevant to it. Half, which says that whether an ambiguity survives depends on where the heavy atoms landed rather than on how heavy they are.

That is the practically important form of the result. A crystallographer facing two candidate structures with matching intensities cannot reason that the presence of a heavy atom settles the matter, because for half the placements it does not. What settles it is a specific check on the specific structures, and the check is cheap — compute both weighted autocorrelations and compare — which is a better answer than a rule of thumb in either direction.

Why the collisions multiply, counted #

The result stops being surprising once the two sides of the map are counted, and the count predicts the table row by row.

What the measurement supplies. The autocorrelation on a ring of n positions has n entries. Its value at the origin is the total scattering power, which is known from the composition; and P(v) = P(−v) for every v, because reversing a vector reverses the pair that made it. So the independent numbers a measurement delivers are the entries at v = 1 up to v = ⌊n/2⌋ — four of them on a ring of eight or nine, six on a ring of twelve.

What the structure needs. An arrangement of k identical atoms is k numbers, less one for the translation that changes nothing. An arrangement of k atoms each of which may be one of two kinds is k positions and k weights: twice as many, and the second half is not reduced by the translation at all.

Setting the two side by side explains every row. Four identical atoms on a ring of nine are four numbers against four measured ones, and the search finds no ambiguity — the map is injective, just. Four weighted atoms are eight numbers against the same four, and there is no way for eight parameters to be recovered from four measurements without collisions somewhere. The pairs the search returns are those collisions, found rather than predicted, and the parameter count says where to look for them.

That argument is a heuristic and it must not be read as a proof. Parameters and measurements here are integers under strong constraints, not real coordinates in a smooth problem, and an excess of parameters over data does not by itself force two particular configurations to coincide — the arithmetic could conspire to keep them apart, and on a ring of nine with identical atoms it very nearly does. What the count gives is the right expectation and the right direction: the weighted problem is more under-determined than the unweighted one, by a factor of about two in the parameters with no change at all in the data.

Read that way, the whole essay’s result is not a curiosity about small rings. Distinguishing the atoms doubles what a structure must specify and adds nothing to what a diffraction pattern returns, so the imbalance that homometry is a symptom of gets worse. The information the weights carry is in the crystal and not in the intensities, which is the same sentence the phase problem is about, said about a different missing quantity.

What this says about the heavy-atom method, and what it does not #

The heavy-atom method works, and nothing here threatens it, so the boundary is worth drawing precisely.

The method does not work by removing homometry. It works because a single dominant scatterer makes the phases of the structure factors approximately those of the heavy atom alone, which is a statement about approximating a sum, not about a degeneracy being lifted. The Patterson’s Harker sections then locate that atom because the vectors between symmetry-related copies of it are confined to a plane.

Every step of that is compatible with the pair persisting. Two homometric structures with the same weights have the same Patterson, hence the same Harker sections, hence the same heavy-atom position; the phases derived from it are the same; and the maps that come out differ in exactly the places the two structures differ. The method returns an answer and the answer is one of the two.

In practice the ambiguity is broken elsewhere, and it is worth being explicit about where, because none of it is diffraction. A homometric partner usually has impossible bond lengths, or a density no compound has, or an arrangement of the heavy atoms no chemistry supports. The intensities do not choose and the chemistry does — which is the same resolution the first rung reaches, arriving at it now with the atoms distinguished and finding it unchanged.

Where the exactness stops #

Computed here: for each ring size, every configuration of k sites with each atom given a weight from the alphabet; the canonical form of each under translation and reflection with the weights carried; the exact integer autocorrelation of each; the groups sharing one; the split of those groups into ones whose structures are uniformly weighted and ones that are not; the sixteen weightings of each half of the classic pair and how many find a partner; and Patterson’s construction with a product weighting, checked entry by entry.

Two weights, not ninety-two. The alphabet is {1, 2}, which is a light atom and one scattering twice as strongly. Real scattering factors are not small integers, they depend on the scattering angle, and a sulfur against a carbon is nearer sixteen to six than two to one. What the small alphabet buys is exactness — every comparison is between integers and no pair is reported because two floating-point numbers happened to agree. What it costs is that these are model structures, and the counts here are counts for this alphabet. A larger alphabet gives more configurations and, on the evidence of going from one weight to two, more collisions rather than fewer.

A ring, not a crystal. One dimension, a handful of sites, no thermal motion, no scattering-angle dependence, no absorption. The ring is a crystal with n sites in a one-dimensional cell, which is enough for the question — homometry is about whether the map from structures to autocorrelations is injective, and that is a question a small exact case answers honestly.

And the counts are counts of classes, not of structures. Two arrangements related by a translation or a reflection are one entry throughout, on both sides of every comparison. Reporting raw configurations would inflate every column by a factor of about 2n and would say nothing extra.

The sixth of those is the one that matters most, and it is the one that would have caught a wrong canonical form. Running the weighted search with an alphabet of one weight must return precisely the unweighted search’s answer — same count, same groups — because it is the same search. A canonicalisation that treated the weights inconsistently would give a different number there, and the test costs nothing.

What the count is a count of #

There is a reading of the census table that would be wrong and it is the obvious one, so it is worth blocking.

The fourth column is not “how much worse the problem gets”. It counts homometric groups among a set of configurations that is itself much larger — sixteen times larger at four atoms, since each may be either kind. The right comparison is a rate rather than a total, and by that measure the picture is less dramatic: eight groups among eighty-seven configurations against one among eight.

What survives that correction is the rows where the unweighted count is zero. A rate cannot explain six pairs where there were none; no normalisation turns nothing into something. So the claim this essay makes is the narrow one and it is the one the arithmetic supports: weighting does not resolve homometry, and on some rings it introduces it. How the density of pairs behaves as the alphabet grows is a different question, it is answerable by the same search at more cost, and it is not answered here.

Who asked this, and when #

Patterson raised the weighted case himself and did not settle it, which is worth knowing before treating it as a modern refinement.

His papers of 1939 and 1944 pose the problem for point sets because that is where the arithmetic is clean, and the 1944 construction is stated for indicator polynomials with coefficients of one. The generalisation to arbitrary positive coefficients is immediate — the factorisation argument never uses that the coefficients are one — and he says so. What he does not do is search, because searching was not available.

The exhaustive question is a computational one and it stayed open on that account rather than on any other. The number of weighted configurations on a ring of twelve with four atoms and two kinds is a few hundred after quotienting, which is nothing now and was a research project then; and the interesting sizes for a real structure are far beyond any search, then or now, which is why the subject moved to constructions and bounds rather than censuses.

The consequence is that a small exact case is still worth computing, and that its job is narrow. It cannot say how common homometry is in real structures — nobody knows, and the honest position is that a homometric partner is rarely looked for because it is rarely suspected. What it can do is refuse a plausible general claim, and the claim it refuses here is that distinguishing the atoms makes the ambiguity go away.

The rung above this one on the ladder: homometry under a space group #

The searches on this ladder have varied the dimension, the construction and now the atoms, and each time the answer has been that the coincidence is not an artefact of the last simplification removed. What has not been varied is the symmetry: every structure searched here sits on a bare ring with no operations imposed, and a real crystal sits in a space group.

That is the natural next question and it cuts both ways. A space group forces atoms into orbits, which reduces the configurations enormously and might eliminate the pairs; it also makes the Patterson more symmetric than the structure, which is a loss of information and might create them. Which effect wins is not obvious from here, it is decidable by the same exhaustive method with the orbits enumerated instead of the subsets, and the answer would say whether homometry is a hazard of the general position or of crystals as such.