A second wavelength completes the map

A map of the atoms that break the law built a Patterson synthesis from squared Bijvoet differences, the differences between the two halves of each Friedel pair. The map it produced holds the vectors between the anomalous scatterers and almost nothing of the hundred and twenty light atoms around them. Every coefficient of that map is the anomalous substructure’s intensity $|F_A|^2$ multiplied by the square of a sine, the sine of the phase difference $\Delta\varphi$ between the whole structure and the substructure. The sine squared changes from reflection to reflection, and that essay traced two of the map’s limits to it: a number of reflections below which the vectors did not top the map, and a number of sites beyond which they sank into background.

It ended by naming the obvious next measurement. Near an absorption edge the real part of the anomalous scattering, $f'$, changes with wavelength as well as the imaginary part $f''$. Two wavelengths therefore give a second difference, between datasets rather than within one. The two most anomalous wavelengths are the wrong pair showed that this dispersive difference carries the cosine of the phase where the Bijvoet difference carries the sine. A map built from both should have no sine squared left in it. That is computed here, and it turns out to be true and to matter much less than expected. The dispersive half earns its place differently, and it brings with it a way of failing that the Bijvoet map is immune to.

A sine from one dataset, a cosine from two #

The structure is the one the Bijvoet map was tested on: a hundred and twenty light atoms placed pseudo-randomly in a plane cell, and two anomalous scatterers at $(0.20, 0.20)$ and $(0.65, 0.45)$ carrying under five per cent of the scattering. The two wavelengths are the inflection and the high remote of the selenium-like edge that the wavelength-choice essay modelled, with $f'$ and $f''$ scaled onto this structure’s anomalous atom. At the inflection $f' = -0.66$ and $f'' = 0.27$. At the high remote $f' = -0.16$ and $f'' = 0.33$. The change in $f'$ between them, half a unit, is what the dispersive difference lives on.

To first order in the anomalous scattering, the two differences are

$$|F(h)| - |F(-h)| \approx \frac{2f''}{f_0}\,|F_A|\sin\Delta\varphi, \qquad \langle|F|\rangle_{\lambda_1} - \langle|F|\rangle_{\lambda_2} \approx \frac{f'_1 - f'_2}{f_0}\,|F_A|\cos\Delta\varphi ,$$

where $f_0$ is the anomalous atom’s ordinary scattering and the angle brackets average the two halves of a Friedel pair. Squared, the first gives $|F_A|^2\sin^2\Delta\varphi$ and the second $|F_A|^2\cos^2\Delta\varphi$. Each is a Patterson of the substructure in expectation, and each fluctuates from reflection to reflection by exactly what the other lacks. Divide each by its own coefficient and add them, and the fluctuation cancels:

$$\Bigl(\frac{f_0\,\Delta F_{\text{bij}}}{2f''}\Bigr)^2 + \Bigl(\frac{f_0\,\Delta F_{\text{disp}}}{\Delta f'}\Bigr)^2 \approx |F_A|^2 .$$

The identity is only first order, so it was checked reflection by reflection rather than trusted. Over 625 reflections the combined coefficient departs from the substructure’s own intensity, computed from the two atoms’ positions, by 2.2 per cent on average.

The sum is the substructure’s own map #

A Patterson synthesis of the combined coefficients should then be the Patterson of the two anomalous atoms, as if the light atoms were not there and the phases had never been lost. The picture at the head of this essay sets it beside the two maps made from one difference each and beside that exact map.

The combined map and the exact map are, for practical purposes, the same map. Their correlation over the whole cell is 0.999. The Bijvoet map correlates 0.78 with the exact one and the dispersive map 0.84. The missing fifth or so is the fluctuating sine or cosine, spread over the cell as a background of small peaks. All three maps put the two substructure vectors first after the origin, at about half the origin’s height, which is where two equal atoms put them in any Patterson.

This is the result the Bijvoet essay’s reasoning predicts, and it is exact to the order the linearisation holds.

Two backgrounds that are mirror images #

Why the sum works can be seen in the halves rather than in the sum. A sine squared is a half minus half the cosine of twice the angle, and a cosine squared is a half plus the same term:

$$\sin^2\Delta\varphi = \tfrac12 - \tfrac12\cos 2\Delta\varphi, \qquad \cos^2\Delta\varphi = \tfrac12 + \tfrac12\cos 2\Delta\varphi .$$

The constant half gives each map the substructure’s Patterson at half strength. The term in $\cos 2\Delta\varphi$ gives each its background, and the two backgrounds are the same pattern with opposite signs. The computed maps bear this out closely. Normalise each map to its own origin, subtract the exact substructure map, and correlate what is left over in the Bijvoet map with what is left over in the dispersive one. The correlation is $-0.995$. Where the Bijvoet map has a spurious peak, the dispersive map has a hole of the same shape, and adding the two fills every hole with its peak.

So the dispersive map is not a second, independent estimate of the substructure’s Patterson, to be averaged with the first for a gain of $\sqrt2$. It is the first map’s complement, and the gain from combining them is complete cancellation rather than averaging. The same complementarity one experiment gives the cosine, the other the sine found for a single reflection’s phase appears here in a whole map.

How exactly the weights must be known #

The cancellation needs each squared difference divided by its own coefficient, $2f''$ for the Bijvoet difference and $\Delta f'$ for the dispersive one, and an experimenter knows neither exactly. $f''$ at a remote wavelength is well determined by an absorption measurement. $f'$ at the inflection is the steepest point of a steep curve, and a few electronvolts of error in the beam’s energy moves it a good deal. The question is how much a misjudged $\Delta f'$ costs.

Remarkably little. With $\Delta f'$ taken fifteen per cent too small or twenty per cent too large, the combined map still correlates 0.99 with the exact one. At forty per cent too large, or thirty per cent too small, it correlates 0.97, which is better than either half alone. Only a factor of two, taken in either direction, brings it down to 0.91 or 0.93, where the dispersive half alone already stands at 0.84. A wrong weight leaves part of one background uncancelled, and the part left over grows faster than the mismatch does, close to its square for small errors. A careless estimate of $f'$ therefore costs a small share of what the second wavelength bought.

Something unexpected follows from the cancellation all the same.

What the threshold was #

If the sine squared caused the Bijvoet map’s limits, the combined map should need fewer reflections than the Bijvoet map to put the vectors on top. It needs exactly as many.

The combined map correlates 0.999 with the exact one at every data size, from 81 reflections to 1,681. It is the substructure’s Patterson at 81 reflections just as much as at 1,681. It still fails to put the two vectors on top below 289 reflections, and so does the exact map, the Patterson of the two atoms with no light atoms, no error and no phases lost. At 169 reflections and below, its strongest peaks after the origin are the origin’s own shoulders, two grid steps from the centre. At that resolution the origin peak is broad enough that its flanks outrank a vector at half its height, and a search for separated peaks counts them. It is the same truncation that as sharp as the sphere is wide measures in an electron-density map, a peak whose width is set by the highest index in the sum. From 289 reflections the origin is narrow enough and the vectors come first.

So the threshold of 289 reflections, on this structure, is a resolution limit and not an averaging limit. The Bijvoet essay read it as the point where the sine squared had averaged out, and on this structure the reading does not survive: a map with no sine squared in it has the same threshold. That essay has been corrected. Its other measurement, that a structure with only ten, twenty or forty light atoms needs more reflections, is a genuine statistical effect: the Bijvoet map then fails at 289 reflections where the exact map passes. Recomputed, those structures need 529 reflections, not the 841 once reported.

The dispersive map alone passed the test at 169 reflections, where the exact map failed. That is the fluctuation happening to suppress the origin’s shoulders once, in one structure. The same map correlates only 0.84 with the substructure’s Patterson at that size, and a single fortunate pass is not a reason to trust it.

Where the fluctuation does cost #

If not reflections, the sine squared should cost something, and it does. It costs sites. Each vector’s peak falls as the substructure grows, to about one part in $n$ of the origin for $n$ equal atoms, while the fluctuation’s background does not fall with it.

At this signal strength all three maps find every vector up to five sites, twenty vectors. At six sites, thirty vectors, the Bijvoet map has lost four of them into its background and the dispersive and combined maps have found all thirty. The combined map’s peaks stand at 0.18 of the origin against 0.12 for the Bijvoet map’s.

The Bijvoet essay found only seven of twelve vectors at four sites. That structure’s anomalous signal was four times stronger, with $f''$ at 40 per cent of $f_0$ where this edge gives 11. The linearisation behind every formula here holds less well at that strength, and the difference is a reminder that the site count is a property of the signal as well as of the map. Here the fluctuation’s cost is real and small: a few vectors at the edge of what either map can hold, bought back by the second wavelength.

Random error, and why the smaller signal tolerates more #

The mean Bijvoet difference here is 4.75 per cent of $|F|$ and the mean dispersive difference 4.16. The obvious expectation is that the larger signal tolerates more measurement error. It does not.

With independent error on every magnitude, the Bijvoet map loses a vector at an error of 2.15 times its mean difference. That falls inside the range of two to three and a half that the earlier essay found. The dispersive map lasts to 3.41 times its own, a ratio of 1.59 between them. The reason is in how each difference is formed. A Bijvoet difference subtracts two measurements, one for each half of the pair. A dispersive difference subtracts two Friedel means, four measurements in all, and the averaging halves each mean’s variance. It is the same arithmetic by which every reflection, several times over buys a smaller random error with redundancy, here built into the definition of the difference. For the same error on each magnitude, a dispersive difference carries $\sqrt2$ less noise. The rest of the ratio, above $\sqrt2$, is the cosine squared distributing the signal among reflections slightly differently from the sine squared. In absolute terms the Bijvoet map is lost at a ten per cent error on each magnitude and the dispersive map at fourteen.

This advantage is an accounting of independent errors only. Every magnitude above was perturbed independently, which is the right model for counting statistics and the wrong one for everything that differs between two datasets.

An error only one of them can see #

The two halves of a Friedel pair are measured in the same dataset, on the same crystal, usually close together in time and at the same wavelength. The two terms of a dispersive difference come from different datasets. Whatever differs between the datasets enters the dispersive difference and not the Bijvoet one. The simplest such difference is a scale factor.

A scale error multiplies every magnitude of the second dataset by $1 + \varepsilon$. The Bijvoet map’s contrast is unchanged to the last digit, because both halves of every pair are scaled together and a Patterson of scaled coefficients is a scaled Patterson. The dispersive map changes completely. Each dispersive difference gains $\varepsilon|F|$, and $|F|$ is the whole structure. The Patterson of that error is the ordinary Patterson of all hundred and twenty-two atoms, the map that needs no phases with its fourteen thousand vectors, added to the substructure’s two. The figure measures exactly that. The dispersive map’s correlation with the whole structure’s Patterson is 0.54 with no scale error, 0.76 at four per cent and 0.98 at thirty-two. At that point the map is the whole structure’s map with the substructure’s vectors faintly visible on it.

They remain visible for longer than the correlation suggests, and the reason is particular to this model. A hundred and twenty equal light atoms have a Patterson with no strong peaks: fourteen thousand small vectors spread over the cell like a haze. The substructure’s vectors stay the highest peaks until the scale error passes fifty per cent, because nothing in the returning map is a peak. That holds for an error that makes the second dataset too strong, the only sign tried here. An error of the other sign erases the substructure at under four per cent, whatever the crystal contains, for a reason a scale error between two wavelengths has a sign works out. A structure with a few moderately heavy atoms that do not scatter anomalously, a metal site or sulphur far from its own edge, would bring back strong ordinary vectors instead of a haze, and the loss would come far sooner. The contrast measures the right thing. A two per cent scale error costs the dispersive map about as much contrast as a random error of two per cent on every single measurement.

Scaling, and what it takes with it #

The remedy is to scale the second dataset onto the first before taking differences, by the factor that best matches their Friedel means. It is a relative version of the absolute scaling the average that knows the atoms derives from Wilson’s statistics, and it has the same blind spot: it can only see what is proportional to $|F|$. It removes a constant scale error exactly. With the datasets scaled, the dispersive map is the same map to nine decimal places whatever the error was.

It is not the same map as the one with no error and no scaling. Scaling the datasets onto each other removes whatever part of the dispersive difference is proportional to $|F|$, and part of the true dispersive signal is. When $f'$ changes between wavelengths, the whole structure’s scattering power changes with it, and on average the inflection dataset is weaker than the remote one for a physical reason. The least-squares scale cannot tell that from an error. With scaling, the substructure peaks fall from 0.52 of the origin to 0.37 and the contrast from 5.2 to 3.3. The correction takes real signal along with the error, and there is no scale factor that removes one without the other.

Real datasets differ by more than a constant. The commonest difference is a relative temperature factor, a fall-off with resolution that differs between the two, from radiation damage accumulated between them or from a change in the crystal. A single scale factor cannot remove an error that varies with resolution. Unscaled, the dispersive map loses a vector once the second dataset falls five per cent further than the first at the edge of the data. One scale factor rescues it at five per cent and fails by ten. Scaling in ten resolution shells holds to ten per cent and fails by eighteen. Each finer correction removes more of the error and costs a little more of the signal. Even with nothing to remove, the shell-scaled map’s contrast is 3.38 against 5.24 unscaled.

What the maps cannot say #

The structure is two-dimensional and has no symmetry. In a space group, vectors between symmetry-related anomalous atoms fall on the sections symmetry stacks them on, and that changes where a search looks, not the arithmetic of the differences.

The edge and the scale of $f'$ and $f''$ are a stated model, the same selenium-like edge the wavelength-choice essay used, scaled onto an atom of ordinary scattering three times a light atom’s. The numbers of reflections and sites are measurements on this structure. The ratio of 2.15 to 3.41 is the one quantity expected to carry over, because it comes from counting measurements.

The errors are of three kinds only. Independent random error, a constant scale and a resolution-dependent scale. Radiation damage also changes the structure between datasets, which no scale factor models, and absorption differs between the two halves of a Friedel pair measured at different orientations, which would give the Bijvoet map an error of its own. Neither is modelled here.

A map is not a solution. A difference Patterson gives the vectors between anomalous atoms. Turning them into positions is solving from the vector set, and turning positions into phases is the problem the phase problem names. Nothing here computes a phase.

Who worked it out #

Jerome Karle’s analysis of 1980 wrote the multi-wavelength intensities in the form used above. Wayne Hendrickson and his colleagues made multi-wavelength anomalous diffraction routine around 1990, with selenium brought into proteins as selenomethionine. The practice of forming dispersive differences only between carefully scaled datasets, and of measuring the wavelengths close together in time, is the working form of the scale argument above. Scaling in resolution shells, and the local scaling that refines it further, are standard steps in data reduction. The sine-and-cosine complementarity is the same one one experiment gives the cosine, the other the sine found between isomorphous replacement and anomalous scattering, here inside a single experiment.

What the calculation has to refuse #

Both refused claims were predictions this essay began with, and both are the kind that sound too obvious to test. The first fails because what matters is the error in each difference, and a dispersive difference averages twice as many measurements. The second fails because the threshold it would lower was set by the width of the origin peak, which no combination of differences can change. The test that makes the rest trustworthy is the exact map. Without it, the Bijvoet map’s threshold would still be attributed to an average it does not depend on.

Still open: a structure whose own Patterson has peaks #

The scale error was survivable here because the structure’s own Patterson is a haze. The case that matters in practice is a structure with a few heavier atoms that do not scatter anomalously at the chosen edge: a metal in a protein studied at the selenium edge, or sulphur, whose own edge is far away. Their vectors are the strongest features of the ordinary Patterson. A scale error between datasets writes them into the dispersive map at a height that grows as the square of the error, where they compete directly with the substructure’s. The scale error at which a given set of such atoms takes over the dispersive map, and whether scaling in shells removes them before it removes the substructure, is the next measurement. It needs a second species of atom in the model and nothing else.