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.
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 atomsYes — 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
Everything summed here is an intensity. No phase appears anywhere on this page,
which is the whole reason the function exists.
The Patterson function
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.
α-quartz, SiO2,
P3221 —
30 intensities out of
30, carrying detail to
0.14 Å.
9 atoms in the cell give
81 interatomic vectors, falling on
35 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 = 972, 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
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, 89.99 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 Si, 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
0.9394
196
0.4697 — Si
0.9697
0.0606
196
0.0303
0.5303 — Si
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. 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 — 3 of them here.
P(u, v) over one cell, 960 intensities to h, k = 15, sampled on
32 × 32.
The curve above uses all
30; a transform on
32 points cannot carry more than
15, so the map stops there. 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
Si lands on each of them.
Those lines cover no area at all — a line has none — and yet 33 per cent of the Patterson’s weight lies on them, counting the 24 of 72 distinct vectors away from the origin that do. 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 keeps every atom apart: all 9 land on their own (x, y), so no vector here is an accident of the projection.
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. 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 + v = 0
x − y
0.530, 0.470
0.4697
x − y =
0.4697
2u − v = 0
y
the origin
0.0000
y =
0.0000
u − 2v = 0
x
0.061, 0.530
0.4697
x =
0.4697
Si is at x =
0.4697, y =
0.0000 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
u
Å
Pairs
How many
Weight
0.0000
0.000
Si–Si, O–O
9
972
0.4697
2.308
Si–Si
2
392
0.5303
2.606
Si–Si
2
392
0.1466
0.720
Si–O, O–O, O–Si
4
352
0.2669
1.311
Si–O, O–O, O–Si
4
352
0.4135
2.032
Si–O, O–Si, O–O
4
352
0.5865
2.882
Si–O, O–O, O–Si
4
352
0.7331
3.602
Si–O, O–Si, O–O
4
352
0.8534
4.193
Si–O, O–Si, O–O
4
352
0.0562
0.276
Si–O, O–Si
2
224
0.1168
0.574
Si–O, O–Si
2
224
0.2028
0.996
Si–O, O–Si
2
224
0.2634
1.294
Si–O, O–Si
2
224
0.3231
1.588
Si–O, O–Si
2
224
0.3837
1.885
Si–O, O–Si
2
224
0.6163
3.028
Si–O, O–Si
2
224
0.6769
3.326
Si–O, O–Si
2
224
0.7366
3.619
Si–O, O–Si
2
224
0.7972
3.917
Si–O, O–Si
2
224
0.8832
4.340
Si–O, O–Si
2
224
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; 15 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
4.9134 Å
F(000)
89.99 electrons
Σ|F|2 over the series
1,134
P(0)
10,365 electrons2 per unit u —
F(000)2 + 2Σ|F|2, and the largest value
anywhere
Mean of the curve
8,098 electrons2 per unit u — which is
F(000)2 = 8,098,
and also the total weight of every interatomic vector