Fourier Transform and Phase Determination

1
Fourier Transform and Phase Determination

• Fourier Series and Synthesis
• Direct Methods
• Patterson Methods

2
Fourier Series

• Eulers formula
• for any real number f,
• proved by Taylor
• Series Expansions

Fig.1 Eulers formula Image from Wikepedia
3
Fourier Series

• Definition
• Any periodic function, if it is
• piecewise continuous
• square-integrable in one period,
• it can be decomposed into a sum of sinusoidal and
• cosinoidal component functions---Fourier Series

4
Fourier Series

• e.g, f(t) is periodic with period T,
• Here,
• It is the nth harmonic (angular frequency) of the
  function f

5
Fourier Series

• In exponential form
• where, the fourier coefficients are
• Relationship between an,bn and cn

6
Fourier Series
Fig.2 Fourier Series for Identity
function Animation from Wikipedia
No cosine terms? Because this is an odd function
7
Fourier Series

• In exponential form
• where, the fourier coefficients are
• Relationship between an,bn and cn

Transform pair
8
Family of Fourier Transform
9
Fourier Series in XRD

• Fourier series
• Continuous, Periodic ? Discrete, Aperiodic
• Election Density Distribution ? Structure Factor
• ?(x,y,z) ? F(hkl)
• Real Space (R3) ? Reciprocal Space (Z3)

10
The Nature of F Transform

• Re-express a function in one domain into another
  domain (conjugate domain)
• e.g. Time domain ? Frequency domain
• Real space ? Reciprocal space
• Now, to know a function, we have two ways
• (1) From the original domain
• directly measure (x,f(x)), get enough sample
  points and do regression. (can we measure the
  electron density distribution as we measure water
  density in a pond?)
• (2) From the conjugate domain
• get Cn, and transform to original domain

11
About Cn

• The meaning of C0
• ?0 0,
• Integrated mean
• think that F(000) the amount of total electrons
  in a cell

12
About Cn
Assume f is real, thus an and bn are both
real When f is odd, an 0 When f is even, bn
0, so Cn is real, which means the phase angle of
Cn is kp Center of symmetry in real space ??
F(hkl) is real a-n an, b-n -bn, so C-n
Cn, Friedels law (symmetry in reciprocal
space) ?? ?(x,y,z) is real, which is definitely
right Generally speaking, reality in one domain
implies symmetry in the conjugate domain.
13
Reexamine Structure Factor
Coordinate Form Vector Form Where h h,
k, l, rj xj, yj, zj fj is the atomic
scattering factor, and it is relevant to the
modulus of reflection index
14
Reexamine Structure Factor

• F(hkl) is considered as the resultant of adding
  the waves scattered in the direction of the hkl
  reflection from all the atoms in the unit cell.
• Assumption the scattering power of the electron
  cloud surrounding each atom could be equated to
  that of a proper number of electrons (fj)
  concentrated at the atomic center.
• What is the scattering factor fj ?
• It is a continuous Fourier transform of electron
  distribution around an atom. Since we use sphere
  model of atoms, the scattering factor is also
  isotropic, only relevant to h.

15
Reexamine Structure Factor

• Another view
• Consider all electrons in one unit cell as a
  whole. In this case, F(hkl) is the sum of the
  wavelets scattered from all the infinitesimal
  portions of the unit cell. It is the Fourier
  Transform of electron density distribution in the
  cell.

The integration domain V is the space of one unit
cell. Also, no sphere model assumption is made in
this formula.
16
Fourier Transform pair

• In Crystallography

Note the exponential term, one has a minus sign,
the other hasnt the period (V) is involved in
one equation.
17
Reexamine Structure Factor

• If the center of symmetry exists

So F is real when center of symmetry exists. In
the same way, we can explore the effects of other
symmetry elements and systematic absences.
18
Wilson Plot

• Before knowing how phases are obtained, a general
  idea of how the magnitude is got from initial
  intensity data is useful.
• First, a series of corrections, like absorption
  correction, LP correctionafter these, we get
  Irel
• Then, we need to put these real intensities on an
  approximately absolute basis, that is what the
  Wilson plot deals with.

19
Wilson Plot
Theoretical average intensity It merely
depends on what is in the cell, not on where it
is. Ideally, the ratio of the two intensities
should be the scaling factor. To calculate, it is
not simple. First, fj varies with sin?/?. So the
averaging should be carried out in thin
concentric shells, not the whole range.
20
Wilson Plot
Second, f contains thermal motion effects, which
could be expressed as In which, fo stands for
the scattering factor of a static atom, also B is
not known. If we assume same B value for all
atoms, the exponential term is the same for all
foj,
21
Wilson Plot
Now if C is the coefficient that brings Irel to
an absolute basis, it is a constant. Now,
substitude Iobs and the equation could be
transformed to In which, B and C are what we
assume constant for all reflections. If we plot
this equation, the intercept is lnC, the slope is
-2B.
22
Wilson Plot
Fig.3 an illustration of Wilson plot Image from
http//www.ocms.ox.ac.uk/mirrored/xplor/manual/
23
Wilson Plot

• We get C by extrapolation, then we get
• Note, the structure factor in this lecture, also
  the term structure factor we generally use, is
  the Fobs, the scaled F.

24
Us and Es

• U(hkl) unitary structure factor
• Definition U(hkl) F(hkl)point/F(000)
• Fpoint is the structure factor when we replace
  the real atoms with point atoms, whose scattering
  power is concentrated at the nucleus, and f is no
  longer a function of (sin?)/?, but a constant
  equal to the atomic number Z.
• The right is a common
• approximation formula
• of Fpoint

25
Us and Es

• Note F(000) is equal to total atomic number, thus

Here, fj includes the effects of atomic
vibration. Ult,1, and the phase is same as
that of F(hkl)
26
Us and Es

• E(hkl) normalized structure factor

is the theoretical average intensity. And F(hkl)
is the scaled structure factor.
e is usually 1, but may assume other values for
special sets of reflections where symmetry causes
intensities to be abnormally high (e.g. m-b
causes (h0l) average intensity is twice that of
general (hkl), so e 2 for E(h0l).)
27
Us and Es

• E shows which reflections have above or below
  average intensities. It is easy to see that the
  expectation of E2 is one.
• Also, E value can help us see if center of
  symmetry exists
• ltEgt ltE2-1gt
• Centrosymmetric 0.798 0.968
• Non- 0.886 0.736

28
The Phase Problem

• As we have seen, we need F(hkl) to do the Fourier
  transform, but experiments can only give us F,
  so we need to know the phase by some other means.
• The Direct Method and Patterson Method are two
  ways to give us phase information.

29
Direct Methods

• Basis
• Some phase info is hided in the Magnitude of
  structure factors and their relationships.
• Inequality
• Harker and Kasper applied Cauchy inequality to
  unitary structure factor equation and got
• U2(hkl)lt,1/2 1/2 U(2h,2k,2l)
• Here, centrosymmetry is assumed.

30
Direct Methods

• U2(hkl)lt,1/2 1/2U(2h,2k,2l)
• If U2(hkl) 0.6, U(2h,2k,2l) 0.1,
• u must be positive for the inequality to hold.
• If U2(hkl) 0.3, U(2h,2k,2l) 0.4,
• Could be either, but to guess u is positive
  could probably be right
• Weakness
• rare reflections are qualified, since F
  declines with sin? rapidly.

31
Direct Methods

• Karle-Hauptman determinant

32
Direct Methods

• Iet us examine a third order determinant

If it is centrosymmetric If Us at the
right side are large, the left must be
positive s(F) is the symbol fuction, equal to
-1 when F is negative
33
Direct Methods

• If it is non-centrosymmetric,
• If Us are all large,

34
Direct Methods
Sayres Equation (1) This equation can
be derived from Fourier Series as
follows (2) Rewrite the equation by
setting that hLL, kL (3)
35
Direct Methods

• Since squared density is also periodic, so write
  it into its
• Fourier series form
• (4)
• Here G(h) is the structure factor of the squared
  cell.
• Compare (3) and (4), it follows that
• (5)
• The structure factor G(h) is
• (6)

36
Direct Methods
In equation (6), gj is the scattering factor of
the squared cell. If we assume equal atom model,
(3) reduces to (7) The normal structure
for equal atoms is (8) Thus we can
obtain (9)
37
Derect Methods

• Finally, from (9) and (5), we get Sayres
  equation
• (10)
• Dont care the fraction coefficient outside the
  summation.
• The summation contains a large number of terms
  however, in general it will be dominated by a
  smaller number of large F(k)F(h-k). Considering
  a reflection with large F(h), it can therefore
  be assumed that the terms with large F(h)F(h-k)
  have their angular part approximately equal to
  the angular part of F(h) itself

38
Direct Methods
So we get, Or If it is centrosymmetric, which
means phase could only be 0 or p, Same result
as deduced from Karle-Hauptman determinant Note,
these relationships are independent of origin
location.
39
Direct Methods

• Does the Magnitude of F(hkl) change with the
  shift of origin?
• No.
• Does the phase of F(hkl) change with the shift of
  origin?
• Yes.
• However, certain linear combinations of phases
  dont Change regardless of the arbitrary
  assignment of cell origins. These are called
  structure invariants.

40
Direct Methods

• Structure Invariants
• Assume atom js position is rj. Now the origin
  shifts through a vector r, the new position of
  this atom is rj. The beam, which contributes to
  reflection h and is scattered by this atom j now
  has a phase change equal to 2p(hr), independent
  of where the atom is.
• Thus, the phase change of F(h) is 2p(hr).
• If we want the sum of several reflection phases
  to be constant, which means their total phase
  changes should be zero, the condition is easily
  derived,
• h1h2h3hn 0
• This kind of combination is a structure invariant.

41
Direct Methods

• The choice of origin is not arbitrary other than
  P1 space group, the symmetry elements should be
  considered and the permissible origins are much
  less, so there are many more this kind linear
  combinations, which are called structure
  seminvariants.
• e.g. In space group P(-1), the origin must be
  located at one of the eight centers of symmetry
  in the cell, so the shift vector r can take eight
  values only,
• r (s1, s2, s3), s1,2,3 0 or ½
• Think about the phase of F(222), does it change?

42
Direct Methods

• Table 1 relative sign relationships for
  possible origins

43
Direct Methods

• Now, we have Inequality, Sayres equation, which
  use Intensity data to guess phase
• we have triple structure invariants and
  seminvariants, which show what remains unchanged
  when the origin is shifted.

44
Direct Methods

• General Process
• Pick out all stronger reflections (like Egt2.0)
  as a set, find all triple relationships among
  them (indices sum to zero), compute probabilities
  of each triple.
• Select three reflections for origin
  determination. These reflections are the most
  often and most reliably interconnected. The
  phases of these three are set freely, but do
  check the table on last page to avoid conflicts.
  After the initial set of phases, origin is fixed.
• Phase propagation (phase variables may be
  involved)

45
Direct Methods

• General Process
• Get enough phases, do Fourier Transform, a
  preliminary electron density model is obtained.
  Usually, we can get positions of several heavier
  atoms.
• Calculate phases based on this model, and connect
  the calculated phases with observed F(hkl), and
  transform again, the result is usually better, so
  more atom positions could be located, so our cell
  model is improved.
• Repeat (5), and also gradually lower the E
  cutoff, include more reflections in the
  calculation, until all the atoms are located.

46
Direct Methods

• Lets see a phase propagation process.
• E.g. a crystal with space group P21/c
• Assign phases to F(3 1 -17), F(3 4 11), F(5 0
  14), thus the origin is set
• Three starting reflections combine to imply two
  new ones,
• s(6 5 -6) s(3 1 -17) s(3 4 11)
• s(8 1 -3) s(3 1 -17) s(5 0 14)
• This example is from George Stout, Lyle Jensen,
  X-ray Structure Determination---A Practical
  Guide, second edition, page 271-273

47
Direct Methods

• (3) Based on the symmetry, in the case of P21/c,
• if kl is even, F(h k l) F(h -k l) F(-h k
  -l)
• if kl is odd, F(h k l) -F(h -k l) -F(-h
  k -l)
• so, s(-3 4 -11) -s(3 4 11) -
• s(8 -1 -3) s(8 1 -3)
• combine,
• s(5 3 -14) s(-3 4 -11) s(8 -1 -3) -
• At this point, lets assume we have used up all
  relationships we can find, thus to continue this
  process, we need to assign phase variable to
  another useful reflection.

48
Direct Methods

• Assign s(6 5 -12) a (a could be or -)
• s(2 -6 9) s(6 5 -12) s(8 -1 -3)
• s(2 -6 9) a
• s(2 -6 9) a
• s(3 6 5) s(-3 1 17) s(6 5 -12) a
• Now we have a checking relationship, all three
  reflections
• phases are known,
• s(2 -6 9) s(3 6 5) s(5 0 14)
• a a
• which is correct and lends confidence to the work
  so far.

49
Direct Methods

• Also, s(9 1 -1) s(3 -4 11) s(6 5 -12) -a
• In this relationship
• s(2 -6 9) s(7 7 -10) s(9 1 -1)
• two are known in terms of the variable
• a s(7 7 -10) -a
• But s(7 7 -10) is negative absolutely.
• In this way, more and more structure factor
  phases are determined or assumed, and Fourier
  transform could be carried out when phase
  information is enough.

50
Patterson Methods

• In 1935, A. L. Patterson pointed out that if we
  do Fourier Synthesis with F2 instead of F as
  coefficients, the result shows info on all of the
  interatomic vectors.
• In Patterson map, if (u, v, w) is a peak, it
  indicates there are atoms exist in the cell at
  (x1, y1, z1,) and (x2, y2, z2) such that ux1-x2,
  vy1-y2, wz1-z2.

51
Patterson Methods

• Formula
• The above is actually the electron density
  function convoluted with its inverse. This may
  give a better understanding why it shows the
  vectors between atoms.

52
Patterson Methods
Unit Cell Patterson Cell
Fig.4 Compare Patterson cell with unit
cell Figure from http//www-structmed.cimr.cam.ac.
uk/Course/
It illustrates that you can think of a Patterson
as being a sum of images of the molecule, with
each atom placed in turn on the origin. Also note
the patterson cell and the crystal unit cell has
the same size.
53
Patterson Methods

• How many peaks are there in one Patterson cell?
• If we have N atoms in one cell, there are N2
  interatomic vectors, in which N vectors are 0
  (just one atom to itself), so we have N2-N1
  peaks in a Patterson cell if there are no more
  overlaps.
• The huge origin peak, is usually useless and may
  cover other peaks near it. It is possible to
  remove it by subtracting the average value of
  F2 from each term. If we use E2, it will be
  easier. (why easier?)

54
Patterson Methods

• Two other properties of Patterson Peaks
• The peak spread is the sum of the two
  corresponding Fourier density peak and also peak
  amount is nearly squared, but cell size is the
  same. These two lead to a great overlap. But we
  can create a sharpened Patterson map, if we use
  Fpoint2.
• The peak height in Patterson map is approximately
  the multiple of two Fourier density peaks which
  generate it. Thus the vectors connecting heavy
  atoms will stand out sharply,

55
Patterson Methods

• Patterson Symmetry
• The Patterson space group is derived by original
  space group by replacing all translational
  elements by the correspondent nontranslational
  elements (axes, mirrors) and by adding a center
  of symmetry if it is not already present. Their
  lattice types dont change.

56
Patterson Methods

• Though translational elements are replaced,
  traces are left.
• Harker lines and planes
• ---unusually high average intensity of certain
  lines or planes
• Think about a m vertical to b axis in real space.
  In Patterson map, a dense line is along b axis.
• If above is not a mirror, but a c glide, can we
  tell the difference based on the patterson map?
• In the same way, think about axes and screw axes.

57
Patterson Methods
Fig.5 Harker section This image is
from http//journals.iucr.org/d/issues/2001/07/00/
gr2131/gr2131fig2.html Which symmetry element
generate the above harker section?
58
Patterson Methods

• On a Patterson Map, we can see strongest peaks,
  which are related to heavy atoms we can also see
  Harker lines and planes, which are related to
  symmetry elements. Therefore, we can probably
  tell heavy atoms positions relative to symmetry
  elements in real space. Since the origin is also
  set or restricted by the symmetry elements in
  each space group, we can tell where these heavy
  atoms are.

59
Patterson Methods

• Now, lets look at an example of one heavy atom
  in P21/c
• List out all equivalent positions (4 in this
  case)
• Caculate vectors between these positions (16
  vectors, 4 of which contribute to origin peak,
  other overlaps lead to stronger peaks in special
  positions)
• Compare with observed harker section, and get
  heavy atom coordinate

60
Patterson Methods
Fig.6 equivalent positions of P21/c from
International Table
61
Patterson Methods
Table 2 vectors between equivalent positions
also the peak positions in Patterson Maps
Same colors indicate overlap positions, but four
black coordinates are just four different
positions
62
Patterson Methods

• Now, collect redundant peaks
• peak position relative weight
• 0 0 0 4
• 0 ½2y ½ 2 along harker line
• 0 ½-2y ½ 2 along harker line
• 2x ½ ½2z 2 in harker plane
• -2x ½ ½-2z 2 in harker plane
• 2x 2y 2z 1
• -2x -2y -2z 1
• 2x -2y 2z 1
• -2x 2y -2z 1

63
Patterson Methods

• So, on the Harker line, which is parallel to b
  axis, equal to shift b axis up ½ unit along c, we
  may find two strong peaks, for example, the
  coordinates are (0,0.3,1/2) and (0,0.7,1/2), thus
  we can get y0.1 by solving a simple equation. In
  the same way, we can get x and z from Harker
  plane. So, this heavy atom is found.

64
Patterson Method

• After this, we get a simple unit cell model with
  only heavy atoms, which usually dominate the
  reflection phases.
• Following is just routine repetition until all
  atoms are solved.