Direct Methods and the Sign Relation

How often the relation holds

α-quartz, SiO₂, P3₂21 — 324 triplets from 36 non-zero coefficients, carrying detail to 0.12 Å.

Each marker is a band of triplets of similar strength, and the curve is Cochran’s prediction for this structure — not a fit to the points. The line at 0.5 is what guessing would give.

2 of the 5 bands here sit above the curve and 3 below it. This projection is close to what Cochran’s derivation assumes — atoms at random — so the prediction here is an estimate and the measurement should fall on either side of it. The test of that assumption is ⟨|E² − 1|⟩, which is 0.913 here against the ideal 0.968 for a centrosymmetric structure of randomly placed atoms.

What that says

The relation holds for 193 of 324 triplets here, 60 per cent against the 50 you would get by guessing. Restrict it to the 7 whose predicted probability is above 0.9 and it holds for every one of them. That is the regime a program works in: it does not use the weak relations, it starts from the strong ones.

The strength of the relation depends on how many scatterers the projection has — this page works in projection, so atoms sharing an x merge into one column, and it is those columns that count, not the atoms in the cell — and that is why direct methods are spoken of as a small-molecule technique. For N equal point atoms Cochran’s prefactor reduces to 1/$\sqrt{N}$, so a triplet of a given normalised magnitude is less likely to hold in a big structure than in a small one — not because the physics changes but because each atom contributes a smaller share of every structure factor, and the sum of many random contributions is what the relation has to beat. Doubling the number of scatterers costs a factor $\sqrt{2}$ in every triple product on this page. That is the whole reason the method solved small molecules in the 1960s and left proteins to isomorphous replacement.

This does not solve anything. A direct-methods program fixes an origin, gives a few strong reflections symbolic signs, propagates them through relations like these, and ranks the resulting sign sets by a figure of merit — a search whose answer nothing here could check. What this page does instead is measure the relation the search rests on: the signs in the table are computed from the published coordinates, so every prediction can be marked right or wrong, which is the one thing a solver’s output would not allow.

The strongest relations

Strongest first, which is the order a program uses them in: it starts from the relations it can trust and never reaches the weak ones. Showing 20 of 324; the counts above are over all of them.

One step of symbolic addition

The projection has a centre of symmetry at x = 0 and another at x = ½, and either will do: they describe the same crystal. Moving between them multiplies F(h) by (−1)^h, so the sign of F(15 0 0) is a fact about that choice and not about the structure. So choose it. Calling it + fixes the origin, costs nothing, and is the one sign on this page that cannot be wrong.

The signs in the two columns are in the origin fixed at the top — a solved structure is always described from an origin somebody picked, and comparing against a different one would mark half the answer wrong for no reason. The chosen and named rows have nothing to be right or wrong about: one is a choice and the other is still a letter.

Using only the 7 relations the theory calls near-certain, 6 signs follow, giving 10 of the 36 present reflections a sign. 3 of them are letters rather than values, so this is not one answer but 8 of them, and choosing between them is what a figure of merit is for. Give each letter the value the published structure has — the best any ranking could do — and every one of the derived signs is right. Drop the floor and propagate through every relation instead: 34 signs are derived rather than 6, and 10 of them are wrong. That is the trade the method is built around: a weak relation determines more and is worth less, and one wrong sign taken early is carried into everything after it.

Every relation the answer implies also holds: of the 10 pairs of determined reflections whose sum is determined too, all 10 satisfy s(h) s(k) = s(h+k) — including the ones the procedure never used. A relation it did use would prove nothing, since the third sign was derived from the other two and was made to fit.

This is symbolic addition, which Karle and Karle first applied in 1963 and set out in full in 1966, working from the sign relation Sayre published in 1952. It is three rules: choose the sign the origin makes free, give a letter to the strongest reflection you cannot derive, and take the third sign of a relation from the other two. A modern program does all of it with more reflections, in three dimensions, and ranks the sign sets by a figure of merit instead of asking anyone — but the arithmetic is this arithmetic.

Why a product of three

A single sign is not a fact about the crystal. This projection has a centre of symmetry at x = 0 and another at x = ½, and either will do as an origin; move it to the second and F(h) is multiplied by (−1)^h, so the sign of every odd reflection reverses — all 18 of them here. The triple product survives it, because (−1)^h(−1)^k(−1)^h+k = +1 whatever h and k are. That is what makes it a structure invariant, and it is the reason the amplitudes can predict it: an amplitude does not move when the origin does, so no function of amplitudes could ever fix a quantity that does.

The reflections

The E values are normalised in two steps, the same two the Wilson plot uses: divide |F|² by Σg² at each reflection’s own resolution, which removes the form-factor decay exactly, then by the mean over the reflections that are present. The second step makes ⟨E²⟩ equal to 1 by construction rather than by fit, and it is doing real work: without it copper comes out at exactly 2, because half its h00 are systematically absent and the survivors carry their share.

What the prediction was built from

The equivalent scatterer count is the number that decides everything: the relation reads 1/$\sqrt{N}$, so a structure with one heavy atom among light ones behaves like a much smaller one and gives up its signs easily. That is why the heavy-atom structures fell first, and why a page of equal light atoms is where these methods run out.