Why does diffraction give
a Fourier transform?
|ki| = |ks| = 2π/λ
ki
ks
r
l1 = (λ/2π) × ki·r
l2 = −(λ/2π) × ks·r
Total additional pathlength is
(λ/2π) × (ki − ks)·r
Total phase shift is
(ki − ks)·r
ki
ks
Q = ki − ks
−ks
Total phase shift is
Q·r
sensitive to the actual wave-vectors ki and ks , but only to their difference. Because of this,
we can
formulate
our scattering equations
terms of
single
wave-vector pattern
Q = kiis−not
ks .
One
interesting
property
by this ·in
equation
is athat
the interference
ψ1 (x) +highlighted
ψ2 (x) = A exp(ik
(21)
s x) × {1 + exp[i(ki − ks ) · r]}.
This is precisely
the same
Q we have
already
in to
BL1
and
BL2. Rewriting
equation
sensitive
to the actual
wave-vectors
ki and
ks , butmet
only
their
difference.
Because
of this,
(21) in this
light, we
can
say that equations
the scatteredterms
beamof
has
been modified
by a Q
factor
we
formulate
our
scattering
single
= ki is
− not
ks .
Onecan
interesting
property
highlighted by thisinequation
isathat
the wave-vector
interference pattern
This
is precisely
the same
Q we have
already
met
in to
BL1
and
BL2. Rewriting
equation
sensitive
to the actual
wave-vectors
ki and
ks , but
only
their
difference.
Because
of this,
(21)
in this
light, weour
can
say that the
scattered
beam of
hasa been
modified
by a factor
we can
formulate
scattering
equations
in exp(iQ
terms
wave-vector
Q = ki −(22)
ks .
F (Q)
= 1 + exp(iQ
· r) =
· 0) +single
exp(iQ
· r).
This is precisely the same Q we have already met in BL1 and BL2. Rewriting equation
(21)
in this light, we F
can
say
the
scattered
beam ·has
been
modified
a factor
It
is straightforward
to
extend
this
idea
of +
arbitrarily
point
particles
at
(Q)
= 1that
+
exp(iQ
·tor)a=collection
exp(iQ
0)
exp(iQ
· many
r). by
(22)
positions r1 , r2 , r3 , . . . to give the general “scattering factor”:
It is straightforward to
extend
idea to
collection
at
F (Q)
= 1 this
+ exp(iQ
· r)a =
exp(iQ · of
0) arbitrarily
+ exp(iQ ·many
r). point particles
(22)
positions r1 , r2 , r3 , . . . to give the general “scattering factor”:
F (Q)
= exp(iQ
· r1 ) + exp(iQ · r2 ) + exp(iQ · r3 ) + . . .
(23)
It is straightforward
to extend
�this idea to a collection of arbitrarily many point particles at
positions r1 , r2 , r3 , . . . to =
give theexp(iQ
general
factor”:
· rj“scattering
).
(24)
F (Q) = exp(iQ
· r1 ) + exp(iQ · r2 ) + exp(iQ · r3 ) + . . .
(23)
j
�
=
exp(iQ · rj ).
(24)
IB Mineral Sciences
Indeed, Module
by converting
over· rall
space we ·allow
generalisaj to an
F (Q)this
= sum
exp(iQ
·and
rintegral
) + exp(iQ
r319
) + a. .further
.
(23)
B: Reciprocal
Space,
Symmetry
1Crystallography
2 ) + exp(iQBH
�
tion to arbitrary particle distributions (i.e. allowing for non-point particles):
exp(iQ
(24)
j ).colours in the Fourier images, but not
we are saying is that we =
measure the
intensity ·ofrthe
IB Mineral Sciences
Indeed, the
by colours
converting
this
sum
an integral
over alla space
allow a further generalisathemselves.
This
makes
difficult to reconstruct
real-spacewe
description
j toit very
�
Module B: Reciprocal Space, Symmetry and
Crystallography
BH
19
of structureparticle
from what isdistributions
an incomplete “colour-blind”
reciprocal-space
description.particles):
Known
tion to arbitrary
(i.e. allowing
for non-point
lySpace, Symmetry
and
Crystallography
as the
“phase
problem”, this issue
forms =
the subject
the final ·lecture
F (Q)
ρ(r)ofexp(iQ
r) dr.in our course,
(25)
arevery
saying
is that we
measure
intensity
of the
colours
Fourier
images, but
One
important
in solving
this
problem
isover
an
understanding
of symmetry
in not
Indeed, BL9.
by we
converting
thishelp
sum
to the
an
integral
allin the
space
we
allow
a further generalisa�
the colours
themselves.
makes
very learning
difficult to
reconstruct
a real-space
description
crystalline
materials.
We will This
spend
someit time
about
symmetry
and its effect
on
tion to arbitrary
particle
distributions
(i.e.
allowing
for non-point
particles):
of structure
from
is an incomplete
reciprocal-space
description.
Known
diffraction
patterns
— what
the subject
lectures
BL5 through
to
BL8. · r) dr.
Fof(Q)
=“colour-blind”
exp(iQ
(25)
l in
Space,
Symmetry
and
Crystallography
This
equation
should
be
entirely
familiar.
What
have
shown
here
as the
“phase
problem”,
this issue
subject we
of the
final lecture
in
our
course,
swe
the
Fourier
but
measure
theimages,
intensity
ofnot
the
colours
informs
thetheρ(r)
Fourier
images,
but
notis quite remarkable:
the
time
being,
will concern
ourselves
a nothing
little
more with
of of
diffraction
BL9.
One
verywe
important
help
in
solvingisthis
problem
is but
anthe
understanding
symmetry
in
namely, For
that
the
scattering
process
ageometry
Fourier
transformation
itself. In a
�
experiments,
and
the relationship
between
the time
various
quantities
The
first
struct
a real-space
description
es.
This
makes it very
difficult
to We
reconstruct
a real-space
description
crystalline
materials.
will spend
some
learning
aboutinvolved.
symmetry
and
itsthing
effect on
diffraction
experiment,
the
scattered
wave
characterised
by kswetells
aboutremarkable:
the Fourier
we will
doshould
is topatterns
relate
scattering
vector
Q to BL5
the
various
diffraction
concepts
first us
This
equation
beour
familiar.
What
we
shown
here
is quite
F (Q)
ρ(r)
exp(iQ
· r) dr.
(25)
diffraction
—entirely
the
subject
of lectures
through
to have
BL8.
we
the intensity
of
colours
ink=the
Fourier
images,
butlaw.
not waves for different
ocal-space
description.
Known
t is measure
an incomplete
“colour-blind”
reciprocal-space
description.
Known
encountered
in
IA:the
namely,
the Bragg
d-spacing
and the
the Bragg
component
atthe
the
wave-vector
Q =diffraction
− nothing
kangle
. little
By2θ,more
measuring
scattered
iis
sa
namely,
that
scattering
process
butwith
atheFourier
transformation
itself. In a
For
the
time
being,
we will
concern
ourselves
geometry
diffraction
Reminding
ourselves
of the
scattering
geometry
illustrated
in the
diagram
on pageof
BH
16,
es. final
This
makes
it
very
difficult
to
reconstruct
a
real-space
description
experiments,
and
the
relationship
between
the
various
quantities
involved.
The
first
thing
he
in
our
course,
em”,
thislecture
issue
forms
the
subject
of
the
final
lecture
in
our
course,
wave-vectors
k
(and,
effectively,
for
different
Q),
we
can
actually
piece
together
a
picture
of
s the relationship
diffraction
experiment,
the scattered
wave characterised
by ks tells
us
the Fourier
we begin
with
This equation
should
familiar.
we have
shown
here
is about
quite remarkable:
we will
do is to be
relateentirely
our scattering
vector QWhat
to the various
diffraction
concepts
we first
reciprocal
space.
Diffraction
isan
anBragg
experimental
technique
that
allows
us to view reciprocal
t is an
incomplete
“colour-blind”
reciprocal-space
description.
Known
at the
the
wave-vector
Q
=
ki diffraction
− knothing
measuring
the
scattered
different
understanding
of symmetry
rtant
help
incomponent
solving
this
problem
isprocess
understanding
of
symmetry
s . By
encountered
inin
IA: namely,
the
angle
2θ, d-spacing
and the
Braggin
law. waves for
namely,
that
scattering
is
but
a Fourier
transformation
itself.
In a
Reminding
ourselves
of the scattering
geometry
illustrated
incan
the diagram
on page
BH 16,
space
directly.
wave-vectors
ks (and,
effectively,
piece
together
picture
of
2 the
2 Q),
em”,
this spend
issue
forms
the
subject
ofabout
final
in actually
our
course,
Q2 =scattered
|Q|
=for|kdifferent
ks |2 = kilecture
+ ks2 weand
(26)
out
and
itsexperiment,
on
Wesymmetry
will
some
time
learning
symmetry
its
onus
diffraction
the
wave
characterised
byeffect
k
about a
the
Fourier
i−
s tells
weeffect
begin with
the relationship
2
space.
Diffraction
an
experimental
technique that
allows
to view reciprocal
==
ki2k+ k−
(27) us waves
s cos 2θ
s −
component
atThere
the
wave-vector
Q
k2k.ikBy
the Namely,
scattered
different
inreciprocal
solving
this
problem
isisan
understanding
of
symmetry
in
Well,
almost.
isthrough
(as always)
one
fundamental
problem.
that whenforwe
detect
�measuring
�i �
BL8.
—rtant
the help
subject
of lectures
BL5
to
BL8.
2 s �
2
2π
2π
space
directly.k (and, effectively,=for2 different
− 2are
cos
(28) wave
wave-vectors
Q),measuring
we2θcan actually
together
picture
of
s
a scattered
such
as FQ(Q),
what|kiwe
is not piece
the
itself,a but
some
2 about
We will spend
some wave
time
learning
= |Q|2 = λsymmetry
− ks |2 =λki2 + ks2 and its effect on
(26)
2
2
reciprocal
space.
Diffraction
is an
experimental
technique
that
allows
us
view
reciprocal
16π
2 2 2square
with
the
geometry
of
diffraction
we
will
concern
ourselves
a
little
more
with
the
geometry
diffraction
intensity
I(Q)
that
is (as
proportional
to
the
ofproblem.
itsof
modulus:
I(Q)
∝when
|F (Q)|
.detect
This
=
k
+
k
−
2k
(27)
Well,
almost.
There
always)
one
fundamental
Namely,
thatto
we
i ks cos 2θ
i
= BL8.
sin
θ,s
(29)
— the subjectspace
of lectures
BL5
through
to
� �2
λ 2 � �2
directly.
2π
2π are
is afirst
real
positive
value,
whereas
the
amplitude
Fisfirst
(Q)
a wave
complex
number.
By
aintensity
scattered
wave
such
as F
(Q),
what
not is
the
itself,
but some
involved.
The
entities
relationship
between
thething
various
quantities
thing
− 2 measuring
cosThe
2θ
(28)
= 2 weinvolved.
λ 2
λ
2
where
we that
have used
the identity
costo
2θ =
2 2sinsquare
θ.
magnitude Q
of information
measuring
intensity
rather
than1 −with
amplitude
weConsequently,
have
surrendered
allI(Q)
about
the
intensity
I(Q)
isis
proportional
the
ofproblem.
itsthe
modulus:
∝when
|F (Q)|
. detect
This
we
will
concern
ourselves
a
little
more
the
geometry
of
diffraction
16π
us
diffraction
concepts
we
first
Well,
almost.
There
(as
always)
one
fundamental
Namely,
that
we
e our
scattering
vector
Q
to
the
various
diffraction
concepts
we
first
2
scattering vector is related to the Bragg
θ byθ,the equation
= angle
sin
(29)
2
phase ofthe
the
amplitude.
In terms of
our
Fourier transform
illustrations
of BL1, what
λcat/duck
intensity
is
awave
real positive
whereas
the
amplitude
Fis(Q)
a complex
number.
By
a scattered
as value,
F quantities
(Q),
what
weinvolved.
are and
measuring
notisthing
the
wave itself,
but some
eamely,
relationship
between
the such
various
TheBragg
first
d-spacing
thediffraction
Bragg
law.
the and
Bragg
angle
2θ,amplitude
d-spacing
the
law.
4π
measuring
intensity
rather
than
we
the
2
2 have surrendered all information about
intensity
I(Q)
isthe
proportional
of its modulus:
I(Q)
=costhe
sin=θ.square
(30)
wherethat
we have
used
the identity 1Q−to
2θ
2 sin θ. Consequently,
the magnitude
Q of ∝ |F (Q)| . This
enofour
scattering
vector
Q
to
various
diffraction
concepts
we
first
λ
the
diagram
on
page
BH
16,
the scattering
inour
the
diagram
on page
BH 16,
scatteringillustrated
vector
related toof
Bragg
angle θ by Fourier
the equation
phase
ofgeometry
the
Inisterms
cat/duck
transform
BL1, what
intensity
is the
aamplitude.
real positive
value, the
whereas
the amplitude
F (Q) isillustrations
a complex of
number.
By
BH 19
BH 19
BH 19
Note
that because the
experimental
limit
of θ is 90 (i.e.and
2θ
=
180 )Bragg
then there is
a
amely, the Bragg
diffraction
angle
2θ,amplitude
d-spacing
the
law.
ationship
measuring
intensity
rather
we
surrendered all information about the
4π
maximum
measurable
value than
of Q: namely,
Q
= 4π/λ.
kθ.s haveon
Q=
sin
(30)
of the scattering
illustrated
inseeour
the
diagram
page
BH
16,
λ
phase ofgeometry
the amplitude.
of
cat/duck
Fourier
transform
illustrations
of BL1, what
Bragg’s
law told us thatIn
we
expect
to
a
diffraction
peak
corresponding
to
the
Bragg
kterms
i
angleNote
θ whenever
there were
a set of atomiclimit
planes
the distance
λ/2there
sin θ. is a
that because
the experimental
of θseparated
is 90 (i.e.by 2θ
= 180 d) =
then
ationship
2θ
◦
◦
max
max
◦
Recasting
this in
terms of Q,value
we have
maximum
measurable
of Q: namely, Qmax = 4π/λ.
max
◦
2told us that
2
to see a diffraction peak corresponding to the
Bragg
Q2 = |Q|2 = |kBragg’s
= ki2 +wekexpect
(26)
i − klaw
s |(26)
2π
sQ atomic
=
.planes separated by the distance d = (31)
angle θ whenever there were a set ofQ
λ/2 sin θ.
d
2Recasting
2 this in terms of Q, we have
θ
(27)
= ki + ks −
2ki ks cos 2θ
(27)
2 complete
2 sense in terms of our picture of the reciprocal lattice.
this
�
Q2 = |Q|2 = Hopefully
|k�
k�should
|22 =make
k�
(26)
2
i −know,
ithat+wekexpect
2π
We now
components in reciprocal space when2π s only toQsee
2π s from BL2,
=
.
(31)
ever2Q is a reciprocal
lattice vector Q = ha +kb +lc
. The corresponding magnitude of Q
2 (28)
cos 2θ
−
2
costhe2θlatticed (and
(28)
=
2
= iskquite
k
−
2k
k
cos
2θ
(27)
i
s
easy
to
calculate
in
the
case
where
reciprocal
lattice)
is
orthogonal:
i +
s
λ
λ
Hopefully
this
should
make
complete
sense
in
terms
of
our
picture
of
the
reciprocal
lattice.
�
�
�
�
by Pythagoras we
2 have
2
We2π
that we expect only to see components in reciprocal space when2now know, from BL2,2π
16π
Q is a 2
reciprocal
Q = ha
of Q
−
2 lattice vectorcos
2θ+kb +lc . The corresponding magnitude
(28)
=
(29)
= 2 ever
sin
θ,
(29)
is2quite
easy
to
calculate
in
the
case
where the lattice (and reciprocal lattice) is orthogonal:
λ
λ
λby Pythagoras
we have
16π 2
=
sin2 θ, 2
(29)
sequently,
the1 −
magnitude
of θ. Consequently, the magnitude Q of
the identity
cos λ2θ2 = 2Qsin
quation
is
related to the Bragg angle θ by the equation
the identity 1 − cos 2θ = 2 sin2 θ. Consequently, the magnitude Q of
is related to the Bragg 4π
angle θ by the equation
(30)
Q=
sin
θ.
(30)
λ
4π
Q=
2θmax
= 180◦ ) then
there
is θ.
a
he
experimental
limit
ofλ θsin
90◦ (i.e. 2θmax = 180◦ ) then there (30)
is a
e value of Q: namely, Qmax = 4π/λ.
he experimental limit of θ is 90◦ (i.e. 2θmax = 180◦ ) then there is a
ak
corresponding
the
hat
we expect
to to
see
a Bragg
diffraction peak corresponding to the Bragg
e value
of Q: namely,
Q
max = 4π/λ.
d by
theadistance
d = λ/2
sin θ.separated by the distance d = λ/2 sin θ.
re
were
set of atomic
planes
hatofwe
to see a diffraction peak corresponding to the Bragg
ms
Q, expect
we have
re were a set of atomic planes separated by the distance d = λ/2 sin θ.
ms of Q, we have
2π
Q = (31)
.
(31)
d
2π
Q =inlattice.
.
(31)
picture
of the reciprocal
make complete
sense
dterms of our picture of the reciprocal lattice.
entsthat
in reciprocal
whenBL2,
we expectspace
only to
see components in reciprocal space when∗ our picture of the reciprocal lattice.
make
sense
in terms
corresponding
of∗ +lc
Q of
latticecomplete
vector Qmagnitude
=
ha∗ +kb
. The corresponding magnitude of Q
BL2,
that
expect
onlythe
to lattice
see components
in reciprocal
wheneciprocal
lattice)
is orthogonal:
late in
thewe
case
where
(and reciprocal
lattice) isspace
orthogonal:
lattice
QSciences
= ha∗ +kb∗ +lc∗ . The corresponding magnitude of Q
ve
IBvector
Mineral
late inModule
the case
the lattice
reciprocal
is orthogonal:
B: where
Reciprocal
Space,(and
Symmetry
and lattice)
Crystallography
ve
∗
∗
∗
∗
∗
∗
� ∗ 2
�1/2
(ha ) + (kb∗ )2 + (lc∗ )2
� 2 2
�1/2
4π h
4π 2 k 2 4π 2 l2
=
+
+
,
a2
b2
c2
Q =
BH 20
(32)
(33)
which, when substituted into equation (31) gives us the (hopefully) familiar expression for
d-spacing in an orthogonal crystal:
1
h2 k 2
l2
= 2 + 2 + 2.
d2
a
b
c
(34)
Qmax = 4π / λrad
10-1
10
Radio
10-3
10-5
Microwaves
People
Eye
10-7
Infra-red
Needle
10-9
Ultraviolet
Cell
10-11
10-13
X-Rays
Molecule Atom
Microscope
Synchrotron
Nucleus
Particle Colliders
x-rays (λ ~ 1 Å)
neutrons (λ ~ 1 Å)
electrons (λ ~ 0.02 Å)
Kα
Kβ
Lα
K L M
Kα
Kβ
λmin
lab x-rays
10-15
Gamma Rays
Intensity
103
Wavelength
(metre)
Δrmin = 2π / Qmax = λ / 2
Wavelength
produce electromagnetic radiation. This mechanism gives rise to a broad distribution of
x-ray energies commensurate with the energies of the incident electrons. The second
(and typically more useful) mechanism of x-ray production occurs when an electron strikes
one of the metal atoms and causes an inner-shell electron to be ejected. An outer shell
electron then relaxes into the now-partially-vacant inner shell, and emits an x-ray photon of
a characteristic energy. For us, the important case is that of electrons falling from the L to
the K shells (i.e. from the second to the first shell), and this transition is labelled Kα. This
transition will have different energies (and hence produce x-rays of different wavelengths)
for different metal targets:
˚
Transition Wavelength (A)
TiKα1
2.47851
FeKα1
1.93597
CuKα
1.54184
AgKα
0.561603
WKα2
0.20901
˚
Transition Wavelength (A)
CrKα1
2.28962
CoKα1
1.78896
MoKα
0.711445
TaKα1
0.215947
AuKα1
0.180195
The traditional method of recording and measuring x-ray diffraction patterns is to use photographic film that has been optimised for x-rays. This is ideal for obtaining a wide sweep
of the diffraction pattern in one measurement. The “greyness” of each reflection can be
measured by optical methods and converted into a relative intensity. Several diffraction
photographs may be required in order to record the range from very strong to very weak,
since the strong reflections can easily saturate the film. The alternative method of measuring x-ray intensities is to use a detector (or “counter”). This has the advantage over films of
giving the intensity directly and more accurately. The disadvantage is that a single detector
will only record a single point in reciprocal space at any one time. This problem can be
overcome by using area detectors that contain a large number of pixels, each of which is
capable of recording its own signal. These are now the most widely used form of detector.
One fundamental limitation on the use of x-ray tubes is that the maximum intensity is limited
by the need to prevent overheating of the target. X-ray production is not very efficient, and
most of the energy of the incident electron beam is lost as heat. This can be overcome in
part by using rotating targets, so that the electron beam does not strike a single area of the
target. However, there is still a mechanical limitation on the maximum intensity produced.
Moreover, the wavelength of x-rays produced by a given x-ray tube is determined by the
nature of its metal target, and so is not variable.
nucleus
electron cloud
∗
∗
anes
tice planes
can becan
evaluated
be evaluated
via thevia
dotthe
product
dot product
d∗h1 k1 l1 d
· d∗h1∗hk21kl21l2· .dAlso,
= 1/|d
d=
| 1/|d∗ |
h2 k2 l2 . dAlso,
ween
ng between
(hkl) planes,
(hkl) planes,
and is and
given
is explicitly
given explicitly
by theby
formula
the formula
2 ∗ 2 2 2∗ 2∗ 2 2 ∗ 2 ∗ ∗
∗ ∗
∗
h+(ak∗2)(b
+) k+(bl (c
) +
) l+(c2klb
) +c 2klb
cos ∗αc∗∗ +
cos2lhc
α∗ +
a 2lhc
cos∗βa∗∗ +
cos2hka
β ∗ +∗ b2hka
cos ∗γb∗∗, cos γ ∗ ,
2 2
∗
∗
re
, γthe
are
internal
the internal
anglesangles
of the of
reciprocal
the reciprocal
lattice lattice
unit cell.
unit This
cell. equation
This equation
re
hefamiliar
more familiar
expression
expression
Z
f (Q)
AH 25
“form factor”
or
2 scattering
2 2 factor”2
“atomic
1 IB MINERAL
h1 kSCIENCES
h l k2
l2
=
+
=Solids:++
Module A: Degrees
of Order
in
The Role+
of Diffraction
d2
ad22 ba2 2 c2b2
c2
0
AH 25
(1)
(1)
Q
AH 25
IB MINERAL SCIENCES
AH 25
between lattice planes can be evaluated via the dot product d∗h1 k1 l1 · d∗h2 k2 l2 . Also, d = 1/|d∗ |
Module A: Degrees of Order in Solids: The Role of Diffraction
nal
tems.
is thesystems.
spacing between (hkl) planes, and is given explicitly by the formula
∗ 2
2 ∗be
∗ ∗ dot product
∗ ∗∗
∗ ∗ d = ∗1/|d∗ |
between
evaluated
· d∗∗ 2 k22hka
(d∗ )2lattice
= h2 (aplanes
) + kcan
(b )2 +
l2 (c∗ )2 + via
2klbthe
c cos α∗ + 2lhcd
ah1 kcos
b cos γ ,
l2 . Also,
1 l1 β h+
is the spacing between (hkl) planes, and is given explicitly by the formula
where α∗ , β ∗ , γ ∗ are the internal angles of the reciprocal lattice unit cell. This equation
∗ 2
(d∗ )2to=the
h2 (a
) +familiar
k 2 (b∗ )2 expression
+ l2 (c∗ )2 + 2klb∗ c∗ cos α∗ + 2lhc∗ a∗ cos β ∗ + 2hka∗ b∗ cos γ ∗ ,
reduces
more
nsform
of the of
unit
thecell
unit cell
∗
2
where
, β ∗ , γ ∗aare
internal
angles
lattice
unit
cell.
equation
1 of
h2 the
kreciprocal
l2atom
lins
contains
aαgroup
group
oftheatoms,
of atoms,
where
where
each
each
isatom
defined
is This
defined
by
a position
by a position
and has
and
a has a
=
+
+
(1)
2
2
2
reduces to the more familiar expression
d
a
b
c2
AH 25
AHIf
25 the position
IB MINERAL
MINERAL
SCIENCES
SCIENCES
AH 25by
AH 25
ctron
of electron
density.
density.
If= the position
ofIB2 each
ofatom
eachisatom
represented
is represented
a Dirac
by a delta
Dirac delta
2 of Order
2Solids:
for orthogonal systems. Module
Module
A: Degrees
Degrees
inl⊗
in Solids:
The Role
TheofRole
Diffraction
of Diffraction
1 A: h
k of Order
IB Mineral Sciences
= 2+
+ the
(1) as being
ne
think
can of
think
the
of
electron
theSpace,
electron
density
density
of
whole
the BH
whole
group
group
as being
described
described
by
by
2
d2 and
a Crystallography
bof
c2
Module
Reciprocal Sciences
Symmetry
37
IBB:Mineral
between
between
lattice
lattice
planes
planes
can becan
evaluated
be evaluated
via thevia
dotthe
product
dot product
d
d
·d
· .dAlso,. dAlso,
= 1/|d
d =| 1/|d |
Fourier transform
of
the
unit
cell
Module
B:between
Reciprocal
Space,
Symmetry
and
Crystallography
37
orthogonal
ffor
tion
the
of
delta
the
functions
delta
functions
with
the
with
functions
the
functions
describing
the BH
electron
density
density
issystems.
the
spacing
is the spacing
between
(hkl) planes,
(hkl)
planes,
and
is and
given
is explicitly
given
explicitly
by
theby
formula
thedescribing
formula the electron
�
�
�
unit cell
contains
a=group
of
atoms,
where
atom
is
has
a γ , about
(dThe
(dhF)Fourier
(a=)transform
h+(a
k=)(b+) k+
(bl −
(c
) (ha
+
) each
lof
+
2klb
)the
c 2klb
cosof
αcFdefined
cos
+the
2lhc
α(Q)+aby
2lhc
cosadensity
βaposition
+
cos2hka
β +and
b2hka
cosabout
γb ,cos
atom,
ρThe
The
ρjof(r).
Fourier
transform
electron
electron
density
a single
a single
atom atom
×
(48)
(Q)
δ[Q
+(ckb
++lc
)]
j (r). transform
Fourier
of) the
unit
cell
distribution
electron
density.
If the position of each atom is represented by a Dirac delta
�
�
F , β ,(Q
=
hkl)
=are
F the (hkl)
(49) cell.
where
where
α
α
γ
,
β
are
,
γ
the
internal
internal
angles
angles
of
the
of
reciprocal
the
reciprocal
lattice
lattice
unit
unit
This
cell.
equation
This
equation
ering
cfunction,
scattering
factor
factor
freduces
(Q)
fhkl)
(Q)
which
weexpression
derived
derived
earlier.
one can think
of thewhich
electron
density
of we
the�
whole
group asearlier.
being described by
jFto
�= the
= 0 familiar
The unit cell contains
a group
ofjfamiliar
atoms,
where
each atom
is defined by a(50)
position and has a
reduces
the(Qto
more
more
expression
∗
h1 k1 l 1
∗ 2
∗2 2 ∗ 2 2
∗2 2 ∗ 2 2
crystal
∗
∗
crystal
∗∗
∗
2∗ 2 ∗ 2 2
∗
∗ 2 ∗ ∗
∗
∗ ∗
∗
unitcell
∗ ∗ ∗
∗∗
h1hk21kl21l2
∗ ∗∗
∗
h2 k2 l 2
∗ ∗ ∗
∗
∗ ∗∗
∗
∗
hkl
∗
unitcell
× Funitcell (Q)
Fcrystal
(Q)the=functionsδ[Q
− (ha∗ +the
kb∗electron
+ lc∗ )] density
the convolution of the delta functions
with
describing
distribution of electron density. If the
position of each
atom
is
by a Dirac delta
2 hkl
22
2 2 represented
2
�
�
h1 electron
kh l k density
l
about each atom,
ρj (r).� The Fourier�
transform1 of the
about a single atom
�
� (1) (1)
�
= 2lc2 +
++ 2 2group
+
∗=whole
function, onej can thinkjFcrystal
of the
electron
density
of
as being
described
2×2 F(hkl)
(hkl)
=
(51)
δ[Q
−hkl)
(ha∗ d+2 kb
)] ba
unitcell
j by
j
dFunitcell
cb (Q) c2
Ffcrystal
=
=∗ +athe
is the atomic scattering factor
which
we
derived
earlier.
j (Q) (Q
hkl
the� convolution
the delta functions
with the functions describing the electron density
� forof
=systems.
Funitcell
(52)
orthogonal
for orthogonal
systems.
Fcrystal
(Q (hkl)
�= hkl) = 0
�
jatom
j position
about
each
ρj (r). The
transform
of delta
the electron
density
a single
atom
An
at atom,
the
r�j isFourier
represented
by the
function
δ(r − rabout
has Fourier
j ), and
�
�
is the atomic
scattering
factor
f
(Q)
which
we
derived
earlier.
transform
exp(iQ
· rFourier
).
The
electron
density
associated
with
the
atom
at
position
r
can
be
Fourier
transform
transform
of
the
of
unit
the
cell
unit
cell
j
j
j
j
j
j
j
�
2
2
represented by
the
convolution
ρ
(r)
⊗
δ(r
−
r
),
and
the
resultant
Fourier
transform
by
the
j
j
�
�+ lzj )] exp(−Bj sin θ/λ )
(hkl) = by the
f (Qdelta
+ ky
� position
�unit
hkl ) exp[2πi(hx
jδ(r
An atom at the
rcontains
is Frepresented
function
−isjrdefined
has
Fourier
unit
Thecell
contains
a group
a group
of atoms,
of atoms,
where
where
each
atom
eachisatom
defined
byand
a position
by
a position
and has
and
a has a
j� cell
j ),
product fj (Q)j The
× exp(iQ
·
r
).
j
j
j
�
transform exp(iQ
· rdistribution
electron
density
associated
the
atom
atisposition
cana delta
be delta
distribution
of electron
of electron
density.
density.
If the
position
If the position
ofwith
eachof
atom
each
isatom
represented
represented
by ar�jDirac
by
Dirac
j ). The
�
2 described
2by
function,
can
one
can
of
think
the
of
electron
the
electron
density
density
of
the
of
whole
the
whole
group
group
as
being
as
being
described
by
Fthink
(hkl)
=
b
exp[2πi(hx
+
ky
+
lz
)]
exp(−B
sin
θ/λ
)
represented
byfunction,
the
ρof
(r)
⊗
δ(r
−
r
),
and
the
resultant
Fourier
transform
by
the
The actual unit
cellconvolution
willone
consist
several
atoms,
the
contributions
from
which
are
simply
j
j
j
j
j
jj
� of the of
the
convolution
the
convolution
delta
the
functions
delta
functions
with
the
with
functions
the
functions
describing
describing
the
electron
the
electron
density
density
j
product
fj (Q) × The
exp(iQ
· rresult
).
added together.
end
for
the
Fourier
transform
of
the
unit
cell
is
the
“structure
j
IB Mineral
Sciences
about about
each
atom,
each atom,
ρj (r). The
ρj (r).Fourier
The Fourier
transform
transform
of the of
electron
the electron
density
density
about about
a single
a single
atom atom
factor”
Module B: Reciprocal Space, Symmetry
and Crystallography
�
2 earlier.
2BH 37
is
the
atomic
is
the
atomic
scattering
factor
factor
fj (Q)
which
fj (Q)
which
wecontributions
derived
we
derived
earlier.
The actual unit cell will consistscattering
of several
atoms,
the
which are simply
B
�
j = 8π �ufrom
j
F (Q) =
fj (Q) exp(iQ · rj ).
added together.
end
result
transform
of
unit
cell−isrδ(r
�
� the “structure
AnThe
atom
An
at
atom
the
at
position
thefor
position
rthe
is Fourier
r�j is represented
by the�
by
delta
thethe
function
delta
function
δ(r
and
r�j ),has
and
Fourier
has Fourier
jrepresented
j ), −
�j
�
�
�
factor”
�
transform
transform
exp(iQexp(iQ
· rj ). The
· rj ).electron
The electron
density
associated
associated
with the
with
atom
the at
atom
position
at position
r�j can rbe
�
j can be
∗
∗density
∗
× replicated
Funitcell (Q)
(48)
Fcrystal (Q)of=atoms
δ[Q
(ha +unit
kb +cell
lc )] is
In forming a crystal,
the group
in− one
atFourier
eachFourier
lattice
point,
represented
represented
by theF
by
convolution
the convolution
⊗ρδ(r
(r)−⊗rδ(r
−rjr).
the
and
resultant
the resultant
transform
transform
by theby the
(Q)
=hkl jρfj j(r)
(Q)
jexp(iQ
j ), ·and
j ),
j
j
j
�
and mathematically
this
operation
corresponds
to the convolution of the(49)
unit cell with the
Fcrystal
(Q×
hkl) ×=exp(iQ
product
product
fj (Q)
f=j (Q)
exp(iQ
·Frunitcell
· r�j ).
j ). j(hkl)
lattice. Thus the Fourier
transform
is the product of the (50)
Fourier transform
Fcrystal (Q
�= hkl) = j0of the crystal
j
In forming a crystal,
theactual
group
ofcell
atoms
in
cellatoms,
is
at from
eachwhich
lattice
point,
The actual
The
unit
cell
unit
will
consist
will consist
of one
several
ofunit
several
atoms,
thereplicated
contributions
the contributions
from
which
are
simply
are simply
of the unit cell (derived here)�and that of the lattice (i.e. the reciprocal lattice). Since the
addedadded
together.
together.
The end
The
result
end result
for thefor
Fourier
the Fourier
transform
transform
of
the
of
unit
the
cell
unit
is
cell
the
“structure
is
the
“structure
and mathematically
this
operation
corresponds
to
the
convolution
of
the
unit
cell
with
the
F (hkl) =
f for
(Qhkl )all
exp[2πi(hx
+ lzj )] exp(−B
θ/λ2 )
(51)
j + kyj when
reciprocal lattice
is zero-valued
Q except
Q isj sin
a 2reciprocal
lattice vector, we
factor”
factor”transform
j
lattice. Thus the
Fourier
of
the
crystal
is
the
product
of
the
Fourier
transform
�
�
∗
∗
∗
need only consider the values�Q = 2π(ha
+=
kb
+f=jlc
The
be expressed in terms
F
(Q)
F lz
(Q)
(Q)).jexp(iQ
2
2r·jrjcan
F (hkl)here)
=
bj exp[2πi(hx
kyj +
θ/λexp(iQ
) ). · rj ). (52)
j + the
j )] exp(−B
of the unit cell (derived
and
that of
lattice
(i.e.fsinj (Q)
the
reciprocal
lattice). Since the
j
jlattice vectors to give
of their components xj , yj , zj parallel
to the real-space
j
reciprocal lattice is zero-valued for all Q except when Q is a reciprocal lattice vector, we
In forming
In forming
a crystal,
a crystal,
the group
the group
of
of atoms
in one
unit
onecell
unit
is cell
replicated
is replicated
at eachatlattice
each lattice
point, point,
∗ ∗atoms
∗ in
∗
need only consider
the
values
= operation
2π(ha
+corresponds
kblc∗ ∗+
rjthe
be expressed
in
terms
Q ·mathematically
rj = Q
2π(ha
+ kb
+
) ·lc
(xj).a
+
b
+can
zconvolution
j c)
and mathematically
and
this
this
operation
corresponds
toThe
theyjto
convolution
of the of
unit
thecell
unit
with
cellthe
with the
of their components
xThus
zj=parallel
to+the
real-space
lattice
vectors
toofgive
j , yj ,the
lattice.lattice.
Thus
Fourier
the
Fourier
transform
transform
crystal
the
crystal
is the
is
product
the product
the of
Fourier
the Fourier
transform
transform
2π(hx
ky
+oflzthe
).of
crystal
(48)
osition
the position
r is represented
r is represented
by theby
delta
the function
delta function
δ(r − rδ(r
), −
and
r ),has
and
Fourier
(49)has Fourier
� (50)
�
xp(iQ
r ). The
· r ).electron
The electron
density
density
associated
associated
with the
with
atom
theat
atom
position
at position
rj can rbe
j can be
e by
convolution
the convolution
ρ (r) ⊗ρδ(r
(r)−⊗rδ(r
), and
− r ),
the
and
resultant
the resultant
Fourier
Fourier
transform
transform
by the by the
(51)
Q)
xp(iQ
× exp(iQ
· r ). · r ).
(52)
unit
l will
cell
consist
will consist
of several
of several
atoms,atoms,
the contributions
the contributions
from which
from which
are simply
are simply
he
her.end
Theresult
end result
for thefor
Fourier
the Fourier
transform
transform
of the of
unit
thecell
unit
is cell
the “structure
is the
(53) “structure
� �
F (Q) =
F (Q) f=(Q) exp(iQ
f (Q) exp(iQ
· r ). · r ).
aal,crystal,
the group
the group
of atoms
of atoms
in onein
unit
onecell
unit
is cell
replicated
is replicated
at eachatlattice
each lattice
point, point,
matically
ly this operation
this operation
corresponds
corresponds
to the convolution
to the convolution
of the unit
of thecell
unit
with
cellthe
with the
Fourier
s the Fourier
transform
transform
of the of
crystal
the crystal
is the product
is the product
of the of
Fourier
the Fourier
transform
transform
cell
erived
(derived
here) and
here)that
andofjthat
thej of
lattice
(i.e. the
(i.e.
reciprocal
the reciprocal
lattice).lattice).
Since the
Since the
j the lattice
of the of
unit
thecell
unit
(derived
cell (derived
here)
and
here)
that
andof
that
the of
lattice
the lattice
(i.e. the
(i.e.
reciprocal
the reciprocal
lattice).
lattice).
Since the
Since the
∗
∗
∗
Q for
·lattice
rj =
2π(ha
+
kb
+
)except
·factor,
(x
+we
ywhen
jb
reciprocal
reciprocal
lattice
is zero-valued
is zero-valued
forQ
alllc
for
Q
all
Qj a
except
when
when
Q+is
isQ
ageneral
reciprocal
lattice
vector,vector,
we we lattice
attice
sSubstituting
zero-valued
is zero-valued
all
for
Q
except
all
except
when
Q
iszajQc)reciprocal
athe
reciprocal
islattice
aform
reciprocal
lattice
vector,vector,
we we
this
expression
back
into
the
structure
obtain
need only
needconsider
only=consider
the
values
the
values
Q
=
2π(ha
Q
=
2π(ha
+
kb
+
kb
lc
).
+
The
lc
).r The
expressed
be expressed
in terms
in terms
2π(hx
� j ∗+ kyj + lz∗j ).
∗∗
∗ can rbecan
onsider
r the values
the
values
Q
=components
2π(ha
kb
lcreal-space
).lattice
+The
lclattice
).rjvectors
The
be
expressed
be expressed
in terms
in terms
of their
of
components
their
x ,=
yf , (Q
zx 2π(ha
,parallel
y+
, z kb
parallel
to the+
to
real-space
the
vectors
tocan
givetor
give
j can
F
(hkl)
=Q
j
hkl ) exp[2πi(hxj + kyj + lzy )].
Substituting this expression backj into the structure factor, we obtain the general form
Q · the
r Q=· rreal-space
2π(ha
= the
2π(ha
+ kbreal-space
+ kb
lc ) lattice
+
· (x
lc a) +
· (x
y ba lattice
+
zy c)
b + z c)vectors
ponents
ts xj , yj , zxjj ,parallel
yj , zj parallel
to
to
vectors
to giveto give
�
∗
j
j
j
jj
j
j
∗∗
∗∗
∗
j
j
j
∗
∗
∗∗
∗
j
jj
jj
j
= are
2π(hx
=spherical,
+ kyjj +
+ lz
ky
+ lz+
j2π(hx
jky
j ). j (Q)
Here, we have assumed
that=the atoms
that
F (hkl)
fj (Q
+j ).
hkl ) exp[2πi(hx
j so
j flz
y )]. depends only on the
j
magnitude of Q,
and
not
on
its
direction.
∗ expression
∗ into
∗structure
∗the structure
∗ factor,
Substituting
Substituting
this expression
this
back
backthe
into
factor,
we obtain
we obtain
the general
the general
form form
Q · rj Q=· rj2π(ha
= 2π(ha
+ kb + kb
lc ) +
· (xlcj a) +
· (x
yjjba + zyjjc)
b + zj c)
� �
Here, we have assumed that the Fatoms
are
spherical,
so thatj +fjky
(Q)
depends
(hkl)F=(hkl)
f=j (Qhklf)j (Q
exp[2πi(hx
+ lz
kyy j)].+ lzy )].only on the
hkl ) exp[2πi(hx
jj +
magnitude of Q, and not on itsjdirection.jj j
j jj
j
= 2π(hx
= 2π(hx
+ ky +
+ lz
ky).+ lz ).
Here, Here,
we have
we assumed
have assumed
that the
that
atoms
the atoms
are spherical,
are spherical,
so thatsofjthat
(Q) fdepends
only on
only
theon the
j (Q) depends
magnitude
magnitude
of Q, and
of Q,
not
and
onnot
its direction.
on its direction.
gxpression
this expression
back into
back
the
into
structure
the structure
factor, factor,
we obtain
we obtain
the general
the general
form form
� �
F (hkl)F=(hkl) f=j (Qhklf)j exp[2πi(hx
(Qhkl ) exp[2πi(hx
+ lz
kyy )].
j + kyjj +
j + lzy )].
j
j
ave
sumed
assumed
that the
that
atoms
the atoms
are spherical,
are spherical,
so thatsofjthat
(Q) fdepends
only on
only
theon the
j (Q) depends