Crystallography & Symmetry
patterns derived from their groups
Series

Direct methods — the series

10 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. ⟨cos Φ⟩ against κ, 379 triplets. The mean cosine of the triplet, binned by the concentration κ = 2|E₁E₂E₃|/√N, for the 379 triplets of a structure of 24 atoms whose reflections all exceed |E| = 1.2. The curve is Cochran's I₁(κ)/I₀(κ), computed from the distribution and not fitted to anything; the points are measured, with the number of triplets in each bin printed above. They agree to 0.1 root-mean-square. The measured points sit slightly above the curve throughout, which is the finite structure showing: Cochran's derivation assumes atoms placed at random and there are only 24 of them.

    Three phases that do not move when the origin does

    A phase is a property of the description, not of the crystal: shift the origin and every one of them changes. A sum of three phases whose indices add to zero does not change, because the shifts cancel. That sum is the smallest thing about a structure that a diffraction experiment could in principle know, and it is not distributed at random.

    part 1 · diffraction
  2. the most consistent answer has 100 per cent of the signs. 60 runs of the sign procedure from 60 different random starts, each plotted at its self-consistency — a figure computed without any knowledge of the answer — against the fraction of its signs that are in fact right. Throwing out the 1 run that reached the uniform solution — every sign the same, perfectly consistent and physically a single peak — the highest consistency belongs to a run with 100 per cent of the signs right. The ranking works here, and the reason it works is that the cell is small. Nothing in the plot's horizontal axis knows the answer, which is the only reason a procedure of this kind is a procedure at all.

    The formula that has the answer already

    The tangent formula rebuilds each phase from all the others, and the true phase set is very nearly a fixed point of it — hand it the answer and it hands the answer back. Start it anywhere else and it does not arrive. Having a fixed point and finding it are different problems, and the second is where the subject spent twenty years.

    part 2 · diffraction
  3. A map from amplitudes alone, 0.59 grid steps out. The density after 150 cycles of flipping, with the atoms that produced the data drawn as rings — moved into the origin and the handedness the solution chose, because a phase set does not fix either and comparing without allowing for them measures the arbitrariness of the description. Every peak of the map is an atom and every atom has a peak. Nothing about the arrangement went into the calculation: the input was a list of amplitudes and a random set of phases.

    The solver that knows no symmetry

    Compute a map from amplitudes and random phases, reverse the sign of everything below a small threshold, transform back and keep the phases. Repeat. The structure appears — and so does its space group, which was never supplied.

    part 3 · diffraction
  4. The sign turns when the cross terms go weak. Every quartet among the strong reflections of a small structure, sorted by the mean of its three cross terms, with the mean cosine of the phase sum in each bin. Where the cross terms are strong the quartet behaves like a triplet and the cosine is near one. Where they are weakest it is -0.698 — negative — and 93 per cent of those quartets have a cosine below nought. The four reflections of the quartet are equally strong in every bin; what changes is three reflections that are not in the sum at all.

    The relation that can say no

    Every phase relation before this one pushes a sum towards zero, so none of them can contradict another — a phase set satisfying all of them badly is still satisfying them in the same direction. A quartet can be estimated at π instead, and it is when its three cross terms are weak, so the information arrives from the reflections nobody would have thought worth measuring.

    part 4 · diffraction
  5. An equality, with a known factor in it. Sayre's identity says the structure factor is proportional to the convolution of the structure factors with themselves — an equality, not a probability — with a factor that depends on the atoms and the resolution and on nothing else. Computed for a structure of Gaussian atoms, the ratio of the two sides falls with resolution exactly as predicted: the logarithm of the ratio is a straight line in the squared index, its slope is minus half the width in the atomic factor, and the ratio is very nearly real and positive, so the identity relates the phases and not only the amplitudes.

    The relation that is an equality

    Every relation of this kind so far is a probability — right nine times in ten, useless applied once and decisive applied ten thousand times. One is not. For a structure of equal, resolved atoms the squared density has peaks in the same places, so every structure factor is exactly a convolution of all the others, with a factor that depends only on the atom. It is exact, and on its own it is useless.

    part 5 · diffraction
  6. Heavier atoms spoil the line, and then mend it. Above, the scatter of Sayre's ratio about its fitted line, averaged over four random cells, as a quarter of the atoms is made heavier, for cells of eight and of twelve atoms. Below, one minus the square of ρ, the correlation between the density's transform and its square's, computed from the composition alone. The spread of atomic numbers grows without limit along this axis, but ρ does not keep falling: it is lowest near a weight of three and climbs back towards one as the heavy atoms come to dominate, because a cell ruled by a few equal heavy atoms is nearly a cell of equal atoms. The scatter rises and falls with it. At equal weight what scatter remains comes from atoms that overlap, more of them in the larger cell.

    Unequal atoms break the equality, and very unequal ones mend it

    Sayre's identity is exact for a cell of equal atoms. Give a quarter of them more weight and the line it predicts scatters, reflection by reflection, by exactly the difference between weighting atoms by Z and by Z squared. Make them heavier still and the scatter shrinks again, because a cell ruled by a few heavy atoms is nearly a cell of equal ones. The scatter measures the cell, but only if the phases are already known.

    part 6 · diffraction
  7. Which cells the flipping solver finds. For four cells of twenty atoms — all carbon, a quarter sulphur, one sulphur, one lead — the share of random starting phase sets that charge flipping takes to the right structure within 250 cycles, against the threshold below which density has its sign reversed. Each point pools several random arrangements of atoms with several starts each, and the atoms fall off with angle as real atoms do. The cells with sulphur are solved more often than carbon alone, the quarter-sulphur cell most of all, each at a threshold near one standard deviation; the lead cell is never solved at any threshold.

    The heavy atom the flipping solver cannot see past

    Sayre's identity degrades as a cell's atoms become unequal, most of all for a quarter of heavy atoms three times the weight of the rest, and the obvious prediction is that a solver built on the same sparseness should struggle there too. It does not. Charge flipping solves a cell of carbon with sulphur in it more often than carbon alone, and never solves a cell with one lead atom — because a threshold stated in standard deviations belongs to whichever atom owns the map's variance.

    part 7 · diffraction
  8. Four flipping rules on four cells. The best success rate of the flipping solver, pooled over five arrangements and eight starts, on four cells: twenty carbons, a quarter sulphur, one sulphur among twenty and one lead among twenty. Four rules for what is flipped: below a number of standard deviations of the map, below a height set by the lightest atom's weight, the same height above the map's mean, and the original rule with the known heavy atom's density removed first. On the lead cell only the last solves anything, 55% against nothing; on both sulphur cells it raises the rate to near certainty; on carbon alone it is the original rule.

    The lead taken out solves what no threshold could

    Charge flipping never solved a cell of nineteen carbons and one lead, and two repairs were proposed: a threshold set by the composition rather than by the map's spread, and the lead's own density removed before each flip. The threshold does nothing for the lead cell at any setting, measured from zero or from the map's mean. Removing the lead takes it from nought to fifty-five per cent, and lifts both cells with sulphur in them to near certainty. The trouble was never where the threshold sat.

    part 8 · diffraction
  9. Success of the flipping solver against how far off the subtracted lead is. The share of forty starts on the cell of nineteen carbons and one lead that the flipping solver solves, with the lead's density subtracted at a position displaced from the true one by up to sixteen grid steps, half the cell. The rate does not fall with the offset: the points scatter without a trend about the pooled rate of 40%, the band marking two binomial standard deviations, and the original solver, with nothing subtracted, solves none. Refining the position from the map does no better than holding it.

    The lead is wherever it is taken out

    Charge flipping solves a cell of nineteen carbons and one lead once the lead's own density is taken out before each flip and put back after. The obvious worry was precision: a lead subtracted a fraction of a grid step from where it really is leaves a residual larger than any carbon. The residual is there, larger than a carbon at a tenth of a step. It does not matter. Subtracted two steps away, or half the cell away, the lead still lets the solver succeed at the same rate, because the solver builds the whole structure round wherever it is put.

    part 9 · diffraction
  10. Success of the flipping solver against the error in the vector between two subtracted leads. The share of 80 starts on a cell of eighteen carbons and two leads that the flipping solver solves, with both leads' density subtracted, the first at its true place and the second displaced from its true place by an error in a random direction. The rate holds to a quarter of a grid step (21, 19, 18, 11, 0, 0, 0 successes at errors of 0, 0.1, 0.25, 0.5, 1, 2, 4 steps), falls by half at half a step, and is zero from one step on. With one lead the place did not matter at all; with two, the vector between them does.

    Two leads need their vector to half a step

    With one heavy atom in a cell with no symmetry, charge flipping did not care where the atom was subtracted: it rebuilt the structure round whatever place it was given. With two heavy atoms that freedom runs out, as predicted, because no translation can put both under wrong references at once. Subtract one of them a whole grid step from where it belongs and the solver never succeeds. But the tolerance is five times looser than the estimate it was predicted from: up to a quarter of a step the rate does not move, at half a step it halves, and the residual the solver absorbs is three carbons' worth, not one.

    part 10 · diffraction

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