xraytools.

The Patterson Function

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

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 corrected, scaled measurement gives. 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 Å0.251.41 Å0.502.82 Å0.754.23 Å1.005.64 Å015,10630,21145,317u = 0.0000 (0.000 Å), weight 6,269, 32 pairs: Na–Na, Na–Cl, Cl–Na, Cl–Clu = 0.0000 (0.000 Å), weight 6,269, 32 pairs: Na–Na, Na–Cl, Cl–Na, Cl–Clu = 0.5000 (2.820 Å), weight 6,269, 32 pairs: Na–Na, Na–Cl, Cl–Na, Cl–Clu = 0.5000 (2.820 Å), weight 6,269, 32 pairs: Na–Na, Na–Cl, Cl–Na, Cl–Clall 2 interatomic vectorsu, in fractions of a and in ÅP(u), electrons² per unit 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.

halite (rock salt), Fm3m — 6 intensities out of 12, carrying detail to 0.47 Å. 8 atoms in the cell give 64 interatomic vectors, falling on 2 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 Σf0(0)2 = 1,639 — the textbook ΣZ², with the scattering factor in place of the electron count, which is why it is not quite a whole number — one term per atom. Here it is 6,269, 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(200)|² = 7,291, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 85.4 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 2.820 Å. In the map above, this reflection contributes one cosine of period a/2 = 2.820 Å, weighted by that intensity.|F(200)|² = 7,291, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 85.4 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 2.820 Å. In the map above, this reflection contributes one cosine of period a/2 = 2.820 Å, weighted by that intensity.|F(400)|² = 3,507, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 59.2 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 1.410 Å. In the map above, this reflection contributes one cosine of period a/4 = 1.410 Å, weighted by that intensity.|F(400)|² = 3,507, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 59.2 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 1.410 Å. In the map above, this reflection contributes one cosine of period a/4 = 1.410 Å, weighted by that intensity.|F(600)|² = 1,954, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 44.2 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.940 Å. In the map above, this reflection contributes one cosine of period a/6 = 0.940 Å, weighted by that intensity.|F(600)|² = 1,954, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 44.2 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.940 Å. In the map above, this reflection contributes one cosine of period a/6 = 0.940 Å, weighted by that intensity.|F(800)|² = 1,171, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 34.2 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.705 Å. In the map above, this reflection contributes one cosine of period a/8 = 0.705 Å, weighted by that intensity.|F(800)|² = 1,171, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 34.2 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.705 Å. In the map above, this reflection contributes one cosine of period a/8 = 0.705 Å, weighted by that intensity.|F(1000)|² = 722, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 26.9 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.564 Å. In the map above, this reflection contributes one cosine of period a/10 = 0.564 Å, weighted by that intensity.|F(1000)|² = 722, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 26.9 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.564 Å. In the map above, this reflection contributes one cosine of period a/10 = 0.564 Å, weighted by that intensity.|F(1200)|² = 462, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 21.5 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.470 Å. In the map above, this reflection contributes one cosine of period a/12 = 0.470 Å, weighted by that intensity.|F(1200)|² = 462, which is what a corrected, scaled measurement gives. The amplitude behind it is |F| = 21.5 electrons, and its phase is not measured: that is the half of the measurement this page does without. d = 0.470 Å. In the map above, this reflection contributes one cosine of period a/12 = 0.470 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, 111.97 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.

Nothing to read out of this one

There is nothing to read out of this map. The heaviest atom here is Cl, and it sits at x = 0 or ½ — a special position. Its Harker vector 2x therefore falls on the origin, buried under every self-vector in the cell, so the map confirms the atom is on a special position and tells you nothing further. Try zirconia, whose zirconium is at a general x.

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. An operation takes an atom at r to Wr + w, so the vector between the two is (W − I)r + w, and everything follows from whether that matrix can be inverted. A centre of symmetry gives −2I, which can, so its vectors run over the whole plane and say nothing. A mirror or a glide line cannot, and its vectors collapse onto a line — 8 of them here.

Harker line u = 0 - it fixes 2yHarker line u = 0 - it fixes 2yHarker line u = 1/2 - it fixes 2yHarker line v = 0 - it fixes 2xHarker line v = 0 - it fixes 2xHarker line v = 1/2 - it fixes 2xHarker line u − v = 0 - it fixes x + yHarker line u − v = 1/2 - it fixes x + yHarker line u − v = 1/2 - it fixes x + yHarker line u + v = 0 - it fixes x − yHarker line u + v = 1/2 - it fixes x − yHarker line u + v = 1/2 - it fixes x − yCl against its image at 0.000, 0.000Cl against its image at 0.500, 0.000Cl against its image at 0.000, 0.000Cl against its image at 0.000, 0.500Cl against its image at 0.000, 0.000Cl against its image at 0.000, 0.500Cl against its image at 0.000, 0.000Cl against its image at 0.000, 0.500uv

P(u, v) over one cell, 168 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. The dashed lines are this space group’s Harker lines, drawn from the symmetry rather than found in the map. The rings mark where Cl lands on each of them.

Those lines cover no area at all — a line has none — and yet 100 per cent of the Patterson’s weight lies on them, counting the all 3 distinct vectors away from the origin. The origin peak is left out of both figures: it sits on every line through it, it is the largest peak in any Patterson, and it is the one peak that says nothing about the symmetry. The share is computed from the coordinates rather than read off the picture, so no grid or tolerance enters it.

The projection down c puts more than one atom on the same (x, y), so some of these vectors are shorter than the atoms really are apart and some pairs have collapsed onto the origin. That is the same loss the axis above suffers, one dimension less of it — and it is why the origin peak weighs more than the sum of the atoms’ own squares.

What each line fixes

A peak on a Harker line is worth a coordinate. The vector along the line is (m · r) times a fixed direction, so reading its position gives that one combination of the atom’s coordinates — one combination, not one coordinate, which is the part a diagram of a Harker section tends to leave unsaid. Where the combination is 2x the answer comes in two, half a cell apart: that is the same origin ambiguity the axis above runs into, and it is resolved the same way — by whether the rest of the structure then makes sense. Some of these vectors land on the origin, where every self-vector in the cell already is. Those are buried and tell a reader nothing, which is a property of where this atom sits rather than a failure of the method.

Harker line fixes peak at read off it giving
u = 0 2y the origin 0.0000 y = 0.0000 or y = 0.5000
u = 1/2 2y 0.500, 0.000 0.0000 y = 0.0000 or y = 0.5000
v = 0 2x the origin 0.0000 x = 0.0000 or x = 0.5000
v = 1/2 2x 0.000, 0.500 0.0000 x = 0.0000 or x = 0.5000
u − v = 0 x + y the origin 0.0000 x + y = 0.0000
u − v = 1/2 x + y 0.000, 0.500 0.0000 x + y = 0.0000
u + v = 0 x − y the origin 0.0000 x − y = 0.0000
u + v = 1/2 x − y 0.000, 0.500 0.0000 x − y = 0.0000

Cl is at x = 0.5000, y = 0.5000 in the published structure, which is what those readings have to agree with — and they are computed from the symmetry and the coordinates by two different routes, so the agreement is a check rather than a restatement.

The vectors themselves

uÅ PairsHow many Weight
0.0000 0.000 Na–Na, Na–Cl, Cl–Na, Cl–Cl 32 6,269
0.5000 2.820 Na–Na, Na–Cl, Cl–Na, Cl–Cl 32 6,269

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 5.6400 Å
F(000) 111.97 electrons
Σ|F|2 over the series 15,107
P(0) 42,752 electrons2 per unit u — F(000)2 + 2Σ|F|2, and the largest value anywhere
Mean of the curve 12,538 electrons2 per unit u — which is F(000)2 = 12,538, and also the total weight of every interatomic vector
Origin peak weight 6,269
ΣZ2, one term per atom 1,639 — 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 · albite — a Patterson with no line to read

Where this comes from