xraytools.

The Patterson Function

https://xraytools.com/patterson?structure=albite

Single-crystal data

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.

Before this
This map is built from intensities, not from structure factors, which is exactly why it needs no phases. The transform that does need them is next door.
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 · albite — a Patterson with no line to read

Earlier on the path: Fourier Synthesis and the Phase Problem Next on the path: Direct Methods and the Sign Relation On From intensities to a structure, step 3 of 6

Notation here: I — what each one means here

Terms here: zone

See also: Fourier Synthesis and the Phase Problem · Direct Methods and the Sign Relation

What each input changes
Terms
How many intensities the sum runs over. The map is of vectors, so adding terms sharpens the vector peaks, not the atoms — there are no atoms in it.
Teaching with this page
Objective
After this page a learner can say what a Patterson map contains, and why it needs no phases.
Start from
this worked example
Ask first
A Patterson map has a strong maximum away from the origin. Is there an atom at those coordinates?
Watch for
“Yes — a maximum in a map is where the density is”
Then
Direct Methods and the Sign Relation
Check yourself: A Patterson map has a strong maximum away from the origin. Is there an atom at those coordinates?

No — a Patterson maximum is a vector between two atoms Yes — a maximum in a map is where the density is

The Patterson function is the autocorrelation of the electron density, so what it holds is interatomic vectors rather than positions. Its largest maximum is always at the origin, which is every atom paired with itself. That is why it needs no phases and why reading it as a structure is the first mistake everybody makes with it.

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 Å0.252.03 Å0.504.07 Å0.756.10 Å1.008.14 Å0124,584249,168373,753u = 0.0000 (0.000 Å), weight 5,558, 52 pairs: Na–Na, Al–Al, Si1–Si1, Si2–Si2, Si3–Si3, O1–O1, O2–O2, O3–O3, O4–O4, O5–O5, O6–O6, O7–O7, O8–O8u = 0.0000 (0.000 Å), weight 5,558, 52 pairs: Na–Na, Al–Al, Si1–Si1, Si2–Si2, Si3–Si3, O1–O1, O2–O2, O3–O3, O4–O4, O5–O5, O6–O6, O7–O7, O8–O8u = 0.5000 (4.069 Å), weight 5,558, 52 pairs: Na–Na, Al–Al, Si1–Si1, Si2–Si2, Si3–Si3, O1–O1, O2–O2, O3–O3, O4–O4, O5–O5, O6–O6, O7–O7, O8–O8u = 0.5000 (4.069 Å), weight 5,558, 52 pairs: Na–Na, Al–Al, Si1–Si1, Si2–Si2, Si3–Si3, O1–O1, O2–O2, O3–O3, O4–O4, O5–O5, O6–O6, O7–O7, O8–O8u = 0.2007 (1.633 Å), weight 983, 8 pairs: Al–Si2, Si2–Al, O3–O5, O5–O3u = 0.2007 (1.633 Å), weight 983, 8 pairs: Al–Si2, Si2–Al, O3–O5, O5–O3u = 0.2993 (2.435 Å), weight 983, 8 pairs: Al–Si2, Si2–Al, O3–O5, O5–O3u = 0.2993 (2.435 Å), weight 983, 8 pairs: Al–Si2, Si2–Al, O3–O5, O5–O3u = 0.7007 (5.702 Å), weight 983, 8 pairs: Al–Si2, Si2–Al, O3–O5, O5–O3u = 0.7007 (5.702 Å), weight 983, 8 pairs: Al–Si2, Si2–Al, O3–O5, O5–O3u = 0.7993 (6.504 Å), weight 983, 8 pairs: Al–Si2, Si2–Al, O3–O5, O5–O3u = 0.7993 (6.504 Å), weight 983, 8 pairs: Al–Si2, Si2–Al, O3–O5, O5–O3u = 0.0873 (0.710 Å), weight 871, 8 pairs: Na–Si3, Si3–Na, O1–O2, O2–O1u = 0.0873 (0.710 Å), weight 871, 8 pairs: Na–Si3, Si3–Na, O1–O2, O2–O1u = 0.4127 (3.358 Å), weight 871, 8 pairs: Na–Si3, Si3–Na, O1–O2, O2–O1u = 0.4127 (3.358 Å), weight 871, 8 pairs: Na–Si3, Si3–Na, O1–O2, O2–O1u = 0.5873 (4.779 Å), weight 871, 8 pairs: Na–Si3, Si3–Na, O1–O2, O2–O1u = 0.5873 (4.779 Å), weight 871, 8 pairs: Na–Si3, Si3–Na, O1–O2, O2–O1u = 0.9127 (7.427 Å), weight 871, 8 pairs: Na–Si3, Si3–Na, O1–O2, O2–O1u = 0.9127 (7.427 Å), weight 871, 8 pairs: Na–Si3, Si3–Na, O1–O2, O2–O1u = 0.0044 (0.036 Å), weight 864, 8 pairs: Al–O5, Si2–O3, O3–Si2, O5–Alu = 0.0044 (0.036 Å), weight 864, 8 pairs: Al–O5, Si2–O3, O3–Si2, O5–Alu = 0.1787 (1.454 Å), weight 864, 8 pairs: Al–O3, Si2–O5, O3–Al, O5–Si2u = 0.1787 (1.454 Å), weight 864, 8 pairs: Al–O3, Si2–O5, O3–Al, O5–Si2u = 0.3213 (2.614 Å), weight 864, 8 pairs: Al–O3, Si2–O5, O3–Al, O5–Si2u = 0.3213 (2.614 Å), weight 864, 8 pairs: Al–O3, Si2–O5, O3–Al, O5–Si2u = 0.4956 (4.033 Å), weight 864, 8 pairs: Al–O5, Si2–O3, O3–Si2, O5–Alu = 0.4956 (4.033 Å), weight 864, 8 pairs: Al–O5, Si2–O3, O3–Si2, O5–Alu = 0.5044 (4.104 Å), weight 864, 8 pairs: Al–O5, Si2–O3, O3–Si2, O5–Alu = 0.5044 (4.104 Å), weight 864, 8 pairs: Al–O5, Si2–O3, O3–Si2, O5–Alu = 0.6787 (5.523 Å), weight 864, 8 pairs: Al–O3, Si2–O5, O3–Al, O5–Si2u = 0.6787 (5.523 Å), weight 864, 8 pairs: Al–O3, Si2–O5, O3–Al, O5–Si2u = 0.8213 (6.683 Å), weight 864, 8 pairs: Al–O3, Si2–O5, O3–Al, O5–Si2u = 0.8213 (6.683 Å), weight 864, 8 pairs: Al–O3, Si2–O5, O3–Al, O5–Si2u = 0.9956 (8.101 Å), weight 864, 8 pairs: Al–O5, Si2–O3, O3–Si2, O5–Alu = 0.9956 (8.101 Å), weight 864, 8 pairs: Al–O5, Si2–O3, O3–Si2, O5–Alu = 0.2270 (1.847 Å), weight 800, 8 pairs: Na–O1, Si3–O2, O1–Na, O2–Si3u = 0.2270 (1.847 Å), weight 800, 8 pairs: Na–O1, Si3–O2, O1–Na, O2–Si3u = 0.2730 (2.221 Å), weight 800, 8 pairs: Na–O1, Si3–O2, O1–Na, O2–Si3u = 0.2730 (2.221 Å), weight 800, 8 pairs: Na–O1, Si3–O2, O1–Na, O2–Si3the 20 strongest of 646 interatomic vectorsu, in fractions of a and in ÅP(u)

Point at a mark along the foot to see which pair of atoms that vector is, and which maximum it belongs to. Click it to keep it; click it again, click empty space, or press Escape to let go.

low albite, NaAlSi3O8, C1 — 6 intensities out of 12, carrying detail to 0.61 Å. 52 atoms in the cell give 2704 interatomic vectors, falling on 646 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. Its weight here is Σf0(0)2 = 5,558, one term per atom, because this projection keeps every atom at an x of its own. That is the textbook ΣZ² with the scattering factor in place of the electron count: f0(0) is the electron count to the precision of the published fit, so the two differ in the last figure and neither is a rounding error.

What went in

|F(200)|² = 0, which is what a diffractometer records. The amplitude behind it is |F| = 0.3 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 3.638 Å. In the map above, this reflection contributes one cosine of period a/2 = 4.069 Å, weighted by that intensity.|F(200)|² = 0, which is what a diffractometer records. The amplitude behind it is |F| = 0.3 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 3.638 Å. In the map above, this reflection contributes one cosine of period a/2 = 4.069 Å, weighted by that intensity.|F(400)|² = 17,057, which is what a diffractometer records. The amplitude behind it is |F| = 130.6 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 1.819 Å. In the map above, this reflection contributes one cosine of period a/4 = 2.034 Å, weighted by that intensity.|F(400)|² = 17,057, which is what a diffractometer records. The amplitude behind it is |F| = 130.6 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 1.819 Å. In the map above, this reflection contributes one cosine of period a/4 = 2.034 Å, weighted by that intensity.|F(600)|² = 15,349, which is what a diffractometer records. The amplitude behind it is |F| = 123.9 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 1.213 Å. In the map above, this reflection contributes one cosine of period a/6 = 1.356 Å, weighted by that intensity.|F(600)|² = 15,349, which is what a diffractometer records. The amplitude behind it is |F| = 123.9 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 1.213 Å. In the map above, this reflection contributes one cosine of period a/6 = 1.356 Å, weighted by that intensity.|F(800)|² = 35, which is what a diffractometer records. The amplitude behind it is |F| = 5.9 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.909 Å. In the map above, this reflection contributes one cosine of period a/8 = 1.017 Å, weighted by that intensity.|F(800)|² = 35, which is what a diffractometer records. The amplitude behind it is |F| = 5.9 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.909 Å. In the map above, this reflection contributes one cosine of period a/8 = 1.017 Å, weighted by that intensity.|F(1000)|² = 7,207, which is what a diffractometer records. The amplitude behind it is |F| = 84.9 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.728 Å. In the map above, this reflection contributes one cosine of period a/10 = 0.814 Å, weighted by that intensity.|F(1000)|² = 7,207, which is what a diffractometer records. The amplitude behind it is |F| = 84.9 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.728 Å. In the map above, this reflection contributes one cosine of period a/10 = 0.814 Å, weighted by that intensity.|F(1200)|² = 1,505, which is what a diffractometer records. The amplitude behind it is |F| = 38.8 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.606 Å. In the map above, this reflection contributes one cosine of period a/12 = 0.678 Å, weighted by that intensity.|F(1200)|² = 1,505, which is what a diffractometer records. The amplitude behind it is |F| = 38.8 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.606 Å. In the map above, this reflection contributes one cosine of period a/12 = 0.678 Å, weighted by that intensity.123456789101112h, in |F(h00)|²|F|²

Point at any bar for the intensity, the amplitude behind it, and the cosine it contributes. Click it to keep it; click it again, click empty space, or press Escape to let go.

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, 519.90 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. The reason it is not a whole number: the scattering factors used here are a fitted expansion, and at s = 0 the fit reproduces Z to a few hundredths of an electron rather than exactly. The electron count itself is of course an integer.

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 Si1, 14 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
0.0074 392 0.0037 — Si1 0.5037 — Si1
0.9926 392 0.4963 — Si1 0.9963 — Si1
0.0074 392 0.0037 — Si1 0.5037 — Si1
0.9926 392 0.4963 — Si1 0.9963 — Si1

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 same intensities in two dimensions

Summing the same intensities over the hk0 zone instead of over h00 gives P(u, v) — the same expression with one more index, and still no phase in it. What the plane has that the axis cannot have is a Harker line. This structure has none, and that is a result rather than a gap: every operation that acts on this projection has (W − I) invertible, so its vectors run over the whole plane and pin nothing down. A centre of symmetry is the plainest case: it sends r to −r, so the vector is −2r and moves wherever the atom does. That is why a triclinic structure gives a Patterson with no line to read, and why triclinic is the hard case for this method.

uv

P(u, v) over one cell, 312 intensities to h, k = 12, sampled on 32 × 32. The origin peak is left out of the shading: it is several times anything else here, and scaling on it would leave the rest of the map blank.

With no Harker line there is nothing for the vectors to collect on, and they do not: every one of the 677 distinct vectors away from the origin is somewhere general. That is the control for the claim, on the one structure that fails it.

The projection down c keeps every atom apart: all 52 land on their own (x, y), so no vector here is an accident of the projection.

The vectors themselves

uÅ PairsHow many Weight
0.0000 0.000 Na–Na, Al–Al, Si1–Si1, Si2–Si2, Si3–Si3, O1–O1, O2–O2, O3–O3, O4–O4, O5–O5, O6–O6, O7–O7, O8–O8 52 5,558
0.5000 4.069 Na–Na, Al–Al, Si1–Si1, Si2–Si2, Si3–Si3, O1–O1, O2–O2, O3–O3, O4–O4, O5–O5, O6–O6, O7–O7, O8–O8 52 5,558
0.2007 1.633 Al–Si2, Si2–Al, O3–O5, O5–O3 8 983
0.2993 2.435 Al–Si2, Si2–Al, O3–O5, O5–O3 8 983
0.7007 5.702 Al–Si2, Si2–Al, O3–O5, O5–O3 8 983
0.7993 6.504 Al–Si2, Si2–Al, O3–O5, O5–O3 8 983
0.0873 0.710 Na–Si3, Si3–Na, O1–O2, O2–O1 8 871
0.4127 3.358 Na–Si3, Si3–Na, O1–O2, O2–O1 8 871
0.5873 4.779 Na–Si3, Si3–Na, O1–O2, O2–O1 8 871
0.9127 7.427 Na–Si3, Si3–Na, O1–O2, O2–O1 8 871
0.0044 0.036 Al–O5, Si2–O3, O3–Si2, O5–Al 8 864
0.1787 1.454 Al–O3, Si2–O5, O3–Al, O5–Si2 8 864
0.3213 2.614 Al–O3, Si2–O5, O3–Al, O5–Si2 8 864
0.4956 4.033 Al–O5, Si2–O3, O3–Si2, O5–Al 8 864
0.5044 4.104 Al–O5, Si2–O3, O3–Si2, O5–Al 8 864
0.6787 5.523 Al–O3, Si2–O5, O3–Al, O5–Si2 8 864
0.8213 6.683 Al–O3, Si2–O5, O3–Al, O5–Si2 8 864
0.9956 8.101 Al–O5, Si2–O3, O3–Si2, O5–Al 8 864
0.2270 1.847 Na–O1, Si3–O2, O1–Na, O2–Si3 8 800
0.2730 2.221 Na–O1, Si3–O2, O1–Na, O2–Si3 8 800

Derived from the atom coordinates, not from the curve above — this is the answer the map is trying to give you. The 20 strongest are listed; 626 weaker ones are not. Every vector appears in both directions, which is why the function is centrosymmetric however the crystal is built.

What was summed

Cell edge a 8.1372 Å
F(000) 519.90 electrons
Σ|F|2 over the series 41,152
P(0) 352,597 — F(000)2 + 2Σ|F|2, and the largest value anywhere
Mean of the curve 270,292 — which is F(000)2 = 270,292, and also the total weight of every interatomic vector
Origin peak weight 5,558
ΣZ2, one term per atom 5,558

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 · albite — a Patterson with no line to read

Where this comes from