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The Patterson Function

Transforming the structure factors needs their phases. Transform the intensities instead and you get a map of every interatomic vector in the cell — for nothing, from what a diffractometer actually records. A heavy atom stands out of it and can be read straight off.

You supply
One of the named structures, and how far the series should run. Everything else — the intensities, the vectors, the electron counts — is computed from the atoms.
Reading it
Summing over h alone gives the projected Patterson, so vectors sharing a u pile up — including on the origin, whose weight is then more than ΣZ2. And not every vector is its own maximum: a truncated series merges what it cannot resolve.

Worked examples: zirconia — the zirconium read straight off · aragonite, calcium at a quarter · quartz at 6 terms — the vectors merge · quartz at 30 terms — and separate · rock salt — where there is nothing to read

Input

up to h =

How far the series runs. Blank means 12. Fewer terms is lower resolution — the vectors broaden and neighbouring ones merge.

Everything summed here is an intensity. No phase appears anywhere on this page, which is the whole reason the function exists.

The Patterson function

0.000.00 Å0.251.29 Å0.502.58 Å0.753.86 Å1.005.15 Å040,33880,677121,015u = 0.0000 (0.000 A), weight 13,804, 24 pairs: Zr–Zr, O–Ou = 0.4484 (2.309 A), weight 6,390, 4 pairs: Zr–Zr; Harker vector, so x = u/2u = 0.5516 (2.841 A), weight 6,390, 4 pairs: Zr–Zr; Harker vector, so x = u/2u = 0.1665 (0.858 A), weight 2,558, 8 pairs: Zr–O, O–Zru = 0.2055 (1.058 A), weight 2,558, 8 pairs: Zr–O, O–Zru = 0.2819 (1.452 A), weight 2,558, 8 pairs: Zr–O, O–Zru = 0.3461 (1.783 A), weight 2,558, 8 pairs: Zr–O, O–Zru = 0.6539 (3.368 A), weight 2,558, 8 pairs: Zr–O, O–Zru = 0.7181 (3.699 A), weight 2,558, 8 pairs: Zr–O, O–Zru = 0.7945 (4.092 A), weight 2,558, 8 pairs: Zr–O, O–Zru = 0.8335 (4.293 A), weight 2,558, 8 pairs: Zr–O, O–Zru = 0.3720 (1.916 A), weight 512, 8 pairs: O–Ou = 0.4874 (2.510 A), weight 512, 8 pairs: O–Ou = 0.5126 (2.640 A), weight 512, 8 pairs: O–Ou = 0.6280 (3.235 A), weight 512, 8 pairs: O–Ou = 0.1154 (0.594 A), weight 256, 4 pairs: O–Ou = 0.1406 (0.724 A), weight 256, 4 pairs: O–Ou = 0.8594 (4.426 A), weight 256, 4 pairs: O–Ou = 0.8846 (4.556 A), weight 256, 4 pairs: O–Oall 19 interatomic vectorsu, in fractions of a and in ÅP(u)

baddeleyite, monoclinic zirconia, P21/c — 12 intensities out of 12, carrying detail to 0.42 Å. 12 atoms in the cell give 144 interatomic vectors, falling on 19 distinct values of u.

Every atom has a vector of length zero to itself, so the origin always carries the largest peak — and it says nothing about the structure. In three dimensions its weight would be ΣZ2 = 6,902, one term per atom. Here it is 13,804, because a projection also piles up every pair of different atoms that happens to share an x. The difference is exactly the information the projection threw away.

What went in

|F(100)|^2 = 597, d = 5.084 A|F(200)|^2 = 7,751, d = 2.542 A|F(300)|^2 = 2,244, d = 1.695 A|F(400)|^2 = 5,616, d = 1.271 A|F(500)|^2 = 4,158, d = 1.017 A|F(600)|^2 = 2,921, d = 0.847 A|F(700)|^2 = 3,648, d = 0.726 A|F(800)|^2 = 20, d = 0.635 A|F(900)|^2 = 2,716, d = 0.565 A|F(1000)|^2 = 17, d = 0.508 A|F(1100)|^2 = 2,161, d = 0.462 A|F(1200)|^2 = 174, d = 0.424 A123456789101112h, in |F(h00)|²|F|²

Every bar is above the line, because an intensity is a squared amplitude. The same reflections on the Fourier synthesis page are drawn with their signs, and the difference between the two pictures is exactly what a measurement loses.

The direct beam is not a reflection, so I(000) is never measured — but F(000) is the electron count of the cell, 224 here, which anybody who knows the formula can write down without measuring anything. It is the one coefficient that is free rather than missing, and it is why the mean of the curve comes out at F(000)2.

Gaps in the row of bars are systematic absences: a centred lattice or a glide plane makes whole classes of h00 vanish, so the series has fewer terms than its length suggests and the projection repeats more often than the cell does.

Reading a coordinate out of it

The heaviest atom is Zr, 40 electrons, and it dominates the map: a peak between two of them weighs as the product of their electron counts. Because the projection is centrosymmetric, an atom at x has density at −x, so the two are separated by 2x — the Harker vector. Find that peak, halve it, and you have the atom, with no phase used anywhere.

Harker vector u = 2x Weight there u/2 u/2 + ½
0.5516 6,390 0.2758 — Zr 0.7758
0.4484 6,390 0.2242 0.7242 — Zr

2x = u fixes x only to within a half, so u/2 and u/2 + ½ both satisfy it and the map cannot choose between them. The atom list below settles which one this crystal uses; a real structure solution tries both and keeps whichever gives sensible chemistry — and in a space group with a second inversion centre half a cell along a, both are right, because they are the same structure with the origin moved.

The vectors themselves

uÅ PairsHow many Weight
0.0000 0.000 Zr–Zr, O–O 24 13,804
0.4484 2.309 Zr–Zr 4 6,390
0.5516 2.841 Zr–Zr 4 6,390
0.1665 0.858 Zr–O, O–Zr 8 2,558
0.2055 1.058 Zr–O, O–Zr 8 2,558
0.2819 1.452 Zr–O, O–Zr 8 2,558
0.3461 1.783 Zr–O, O–Zr 8 2,558
0.6539 3.368 Zr–O, O–Zr 8 2,558
0.7181 3.699 Zr–O, O–Zr 8 2,558
0.7945 4.092 Zr–O, O–Zr 8 2,558
0.8335 4.293 Zr–O, O–Zr 8 2,558
0.3720 1.916 O–O 8 512
0.4874 2.510 O–O 8 512
0.5126 2.640 O–O 8 512
0.6280 3.235 O–O 8 512
0.1154 0.594 O–O 4 256
0.1406 0.724 O–O 4 256
0.8594 4.426 O–O 4 256
0.8846 4.556 O–O 4 256

Derived from the atom coordinates, not from the curve above — this is the answer the map is trying to give you. Every vector appears in both directions, which is why the function is centrosymmetric however the crystal is built.

What was summed

Cell edge a 5.1505 Å
F(000) 223.87 electrons
Σ|F|2 over the series 32,024
P(0) 114,165 — F(000)2 + 2Σ|F|2, and the largest value anywhere
Mean of the curve 50,117 — which is F(000)2 = 50,117, and also the total weight of every interatomic vector
Origin peak weight 13,804
ΣZ2, one term per atom 6,902 — smaller, because this projection puts more than one atom on the same x

Try it

zirconia — the zirconium read straight off · aragonite, calcium at a quarter · quartz at 6 terms — the vectors merge · quartz at 30 terms — and separate · rock salt — where there is nothing to read