Beginning Chapter 5
Reciprocal Space
The reciprocal lattice is based on the Bravais lattice (not crystal lattice) in direct space.
The collection of all wave vectors K that yield plane waves with the periodicity of a
given Bravais lattice (eiKir = e iKi(R+r) ) is known as its reciprocal lattice.
K belongs to the reciprocal lattice of a Bravais lattice of points R, provided that the
relation eiKiR = 1 holds for all R.
Let a1, a2, and a3 be a set of primitive vectors for the direct lattice. The reciprocal
lattice can be generated by the three primitive vectors
r r
r r
r r
r
r
r
a ×a
a ×a
a ×a
b1 = 2π r 2r 3r ; b 2 = 2π r 3r 1 r ; b3 = 2π r 1r 2 r
a1 ⋅ ( a 2 × a 3 )
a1 ⋅ ( a 2 × a 3 )
a1 ⋅ ( a 2 × a 3 )
as can be verified by
y
r r
ai ⋅ b j = 2πδ ij
The reciprocal lattice is itself a Bravais lattice.
Any vector in real or reciprocal space can be expressed in terms of the respective
r
r
r
primitive vectors: r
k = k b +k b +k b
r r
k ⋅ R = 2π (k1 n1 + k 2 n2 + k 3 n3 )
1 1
2 2
3 3
r
r
r
r
R = n1 a1 + n2 a 2 + n3 a3
Reciprocal of the Reciprocal Lattice
The reciprocal lattice of any reciprocal lattice (K) is the original direct Bravais
lattice (R), since all vectors R for which the relation eiKiR = 1 holds for all K
belong to the original direct lattice,
lattice and all points of the direct lattice satisfy the
relation eiKiR = 1 for all K.
r
r
r
r r r
r r r
Specifically, using vector identity A × ( B × C ) = B ( A ⋅ C ) − C ( A ⋅ B ) , it can be shown
r r
that
b ×b
r
a1 = 2π r 2r 3 r
and so on.
b1 ⋅ (b2 × b3 )
r
r r 4π 2 (ar3 × ar1 ) × (ar1 × ar2 )
4π 2 a
b2 × b3 =
= r r 1r
r r r 2
| a1 ⋅ (a 2 × a3 ) |
[a1 ⋅ (a 2 × a3 )]
The volume of a unit cell in the reciprocal lattice is
r r r
8π 3
8π 3
Ω cell =| b1 ⋅ (b2 × b3 ) | = r r r
=
| a1 ⋅ (a 2 × a3 ) | Vcell
The larger the unit cell in direct space, the smaller is the unit cell in reciprocal
space.
1
Examples
Simple Cubic
r
r
r
a1 = a xˆ ; a 2 = a yˆ ; a3 = a zˆ
r 2π
r 2π
r 2π
b1 =
xˆ ; b2 =
yˆ ; b3 =
zˆ
a
a
a
Face-Centered Cubic
r a
a1 = ( yˆ + zˆ ) ;
2
r 2π
b1 =
( − xˆ + yˆ + zˆ ) ;
a
r a
r a
a 2 = ( zˆ + xˆ ) ; a3 = ( xˆ + yˆ )
2
2
r 2π
r 2π
b2 =
( xˆ − yˆ + zˆ ) ; b3 =
( xˆ + yˆ − zˆ )
a
a
Body-Centered Cubic
r a
r a
r a
a1 = (− xˆ + yˆ + zˆ ) ; a 2 = ( xˆ − yˆ + zˆ ) ; a 3 = ( xˆ + yˆ − zˆ )
2
2
2
r 2π
r 2π
r 2π
b1 =
( yˆ + zˆ ) ; b2 =
( zˆ + xˆ ) ; b3 =
( xˆ + yˆ )
a
a
a
Other Examples
Monoclinic
90o rotation
angle and ratio conserved
Hexagonal: rules similar to monoclinic’s
Orthorhombic: bi = 2π/ai
Centered Structures: In general, reciprocal cell dimension doubles in all
directions affected by the additional points. Reciprocal lattice also becomes
centered.
Body centered Q Face-centered
Base-Centered Q Base-centered
2
Brillouin Zones
The Wigner-Seitz primitive cell of the reciprocal lattice is the first Brillouin Zone.
The “origin” of the reciprocal
lattice is known as the Γ-point.
The nth Brilloun zone is the region in
reciprocal space reachable from the origin
(Γ point) by crossing over (n-1) Bragg
planes that is unreachable by crossing over
only (n-2) Bragg planes.
6th
Alternative
A
i definition:
fi i i
For a point
i k in
i
reciprocal space, draw spheres of radius |k|
about every reciprocal lattice point. If k is
in the interior of n-1 spheres and on the
surface of one, then it lies in the interior of
the nth Brillouin zone.
What if a point lies on the surface of more
than one spheres?
3rd
Brillouin Zones
3
Lattice Planes
Vectors in reciprocal lattice are related to lattice planes in direct space.
A lattice plane is any plane containing at least three noncollinear Bravais lattice
points. Any such plane actually contains an infinite number of points.
Family of lattice planes: set of parallel, equally spaced lattice planes, which
together contain all the points of the three-dimensional Bravais lattice.
For any family of lattice planes separated by a distance d, there are reciprocal
lattice vectors perpendicular to the planes, the shortest of which have a length of
2π/d. Conversely, for any reciprocal lattice vector K, there is a family of lattice
planes normal to K and separated by a distance d, where 2π/d is the length of the
shortest reciprocal lattice vector parallel to K.
Miller Indices of Lattice Planes
The Miller indices (h, k, l) of a set of lattice planes, with an inter-planar spacing of
d, are the coordinates of the reciprocal lattice vector normal to that plane, with a
magnitude equaling 2π/d, i.e. k=hb1+ kb2+ lb3. This is a reciprocal space
definition.
fi i i
The Miller indices also have a definition in real space. With one lattice plane
assumed to cut through the origin,
of the adjacent plane on the
r the
r intersections
r
three axes of the direct axes x1 a1 , x 2 a 2 , x3 a3 are used to define (h, k, l) as (note
that xi may be infinity) :
h=
1
1
1
,k= ,l=
x1
x2
x3
0 used for infinite intercept
h , k , l is used for negative h, k, l
4
Cubic Lattice Planes
common cubic planes
Distance between adjacent crystal planes:
CUBIC
d hkl =
a
h + k2 + l2
2
Non-Cubic Systems Miller Indices
2-D Lattice
(Monoclinic)
h2
k2
l2
1
2hk cos γ
= 2
+ 2
−
+ 2
2
2
2
2
d hkl a1 sin γ a 2 sin γ a 2 a1 sin γ a3
orthorhombic
tetragonal
h2 k 2 l 2
1
= 2 + 2 + 2
2
d hkl a
b
c
2
2
l2
1 h +k
=
+
2
d hkl
a2
c2
Areal Density of Lattice Points on Plane = n * dhkl
5
Hexagonal Systems
4 indices
(h1 h2 h3 l)
sum of first three indices
always vanishes
Beginning Chapter 6
X-ray Diffraction
1. X-ray well suited for determination of crystal
structures. 1 angstrom wavelength Q 12.3 keV
hω =
hc
λ
2. Bragg found discrete intense peaks (Bragg peaks) of
scattered radiation from crystalline materials for certain
energy and incident directions.
3. Bragg accounted for sharp scattered peaks as
radiation reflected off of ordered crystallographic
planes specularly and constructively.
Bragg Condition
nλ = 2d sinθ
6
Why does a plane reflect specularly?
Suppose a plane wave with wave vector k impinges upon a collection of
atoms on a single plane. Assuming that atoms scatter x-ray elastically
(k=k’) and spherically symmetric, what is the amplitude of wave at a
point r far away from the sample along the direction k’ ?
r
A(r ) ∝ ∑ e
j
r r
ik ⋅ R j
r r r
ik ′⋅( r − R j )
r r
e
r r
| r − Rj |
Constructive interference if
eik ′⋅r
r
large r
r r
r
(k − k ′) ⊥ R j
∑e
r r r
i ( k − k ′ )⋅ R j
j
independent of in-plane
arrangement of atoms
von Laue’s Formulation of XRD
Von Laue did not assume specular
reflection from crystal planes, but general
scattering
g from identical units.
First consider just two scattering centers
displaced by d. The optical path difference
is
r
d cos θ + d cos θ ′ = d ⋅ (nˆ − nˆ ′)
k, k’, d are not coplanar, in general
which, for constructive interference, gives
r
d ⋅ ( nˆ − nˆ ′) = mλ , m = ± 1, ± 2, ...
r
r 2π
| k |=| k ′ |=
λ
r r r
d ⋅ (k − k ′) = 2πm
For the entire crystal to contribute constructively, the displacement between any two
scattering centers also satisfies
r r r
r r r
R ⋅ (k − k ′) = 2πm
e iR⋅( k − k ′) = 1
for all R,
which simply states that the vector K = k-k’ belongs to the reciprocal lattice.
7
Bragg Plane
A Bragg plane is any plane that is the perpendicular
bisector of the line joining the origin of k-space (Γ-point)
to a reciprocal lattice point. Von Laue’s condition
specifies
ifi constructive
i diffraction
iff
i whenever the change in
i
wave vector is a reciprocal vector, K= k-k’. Since |k|=|k’|,
the three vectors k, k’, and K form a triangle with two
sides equal. If the k vector is drawn from the Γ-point, its
tip falls on the plane bisecting Γ and K.
r
r
k ⋅ Kˆ = 12 | K |
π/2-θ
Note: K does not need to be a “primitive”
reciprocal lattice vector. What happens when
K is not a primitive reciprocal vector?
Equivalence
q
of the Bragg
gg and von Laue conditions
Bragg Condition
von Laue Condition
nλ = 2d sinθ
nK = 2k sinθ
multiply both sides by 2π/(kK)
In real space, what is the orientation of the
planes that reflect the X-rays?
Ewald Construction
EWALD CONSTRUCTION
G ve thee incident
Given
c de wave
w ve vec
vector
o k,, a sphere
sp e e of
o
radius k is drawn about the point k. Any
reciprocal lattice vector falling on the surface
of this Ewald sphere leads to Bragg
reflection. Note that this is a condition which
is not easily met for a randomly oriented
single crystal with a monochromatic x-ray
beam.
Laue Method of crystal structure
determination.
A “white” x-ray source (wide range of
wavelengths or energies)
8
X-Ray Diffraction (XRD)
Rotating Crystal Method
Monochromatic X rays, with crystal rotated.
The Ewald sphere is
i fixed
fi
in
i space.
Debye-Scherrer Method (Powder)
Monochromatic X rays,
with powder sample.
r
| K | = 2k sin 12 φ
Theta - 2 Theta Method
Detector scans at twice the
angular speed of the sample.
Theta - 2 theta scan is used to
identify phases of the sample
(powder diffraction pattern). It is
also used to measure the lattice
spacing perpendicular to the surface.
9
X-Ray Rocking Curve
Also known as double-crystal
diffraction, x-ray rocking curve is
used
d to
t measure the
th “mosaic
“
i
spread” of microcrystals or the
overall quality of a crystal (thin
film).
X-Ray Diffraction Structure Factor
Geometrical Structure Factor: lattice with a basis (of identical atoms).
n
SK = ∑ e
r r
iK ⋅ d j
j =1
Intensity of Bragg peak proportional
to |SK|2 in additional to other angular
dependencies.
Example: b.c.c. considered as s.c. with basis
r
S K =1 + exp iK ⋅ 12 a( xˆ + yˆ + zˆ)
[
]
r 2π
K=
(n1 xˆ + n2 yˆ + n3 zˆ )
a
⎧2, n1 + n2 + n3 even⎫
S K =1 + (−1) n1 + n2 + n3 = ⎨
⎬
⎩0, n1 + n2 + n3 odd ⎭
same as f.c.c. lattice with
double the cell linear
dimension
10
X-Ray Diffraction Structure Factor
Face Centered Cubic
S K = 1 + (−1) n1 + n2 + (−1) n3 + n2 + (−1) n1 + n3
⎧ 4, n1 , n2 , n3 all even or all odd ⎫
SK = ⎨
⎬
⎩0, n1 , n2 , n3 mixed even and odd ⎭
not applicable to (2,1,1)
Monatomic Diamond Lattice
[
r
S K =1 + exp iK ⋅ 14 a( xˆ + yˆ + zˆ )
]
r 2π
K=
(n1 xˆ + n2 yˆ + n3 zˆ )
a
⎧ 2, n1 + n2 + n3 is divisible by 4 ⎫
⎪
⎪
1 ± i, n1 + n2 + n3 is odd
SK = ⎨
⎬
⎪0, n + n + n = odd number × 2⎪
1
2
3
⎭
⎩
not applicable to (2,2,1)
Hexagonal Close-Packed, Centered Orthorhombic, Centered
Tetragonal, etc.
Atomic Form Factor
n
SK = ∑
j =1
r iKr ⋅dr j
f j ( K )e
r
r
− 1 r iKr ⋅rr
f j (K ) =
dr e ρ j ( r )
∫
e
Variation in the intensity of Bragg peaks can help distinguish crystal
structures.
11
Sample Shape Q Reciprocal Point Shape
Suppose a plane wave with wave vector k impinges upon a crystal of finite size.
r
r
r
v
Rn1 ,n2 ,n3 = n1a1 + n2 a2 + n3 a3
− N1 ≤ n1 ≤ N1 ; − N 2 ≤ n2 ≤ N 2 ; − N 3 ≤ n3 ≤ N 3
Assuming that atoms scatter x-ray elastically and spherically symmetric,
r
A( r ) ∝
∑e
r r
ik ⋅ Rn1 ,n2 ,n3
n1 , n2 , n3
r r r
ik ′⋅( r − R
If the sample is thin,
Express
∑e
n1 , n2 , n3
r r r
i ( k − k ′ )⋅ Rn1 ,n2 ,n3
r r
)
n1 ,n2 ,n3
e
r r
| r − Rn1 ,n2 ,n3 |
N1 ~ 10 ;
large r
e ik ′⋅r
r
∑e
r r r
i ( k − k ′ )⋅ Rn1 ,n2 ,n3
n1 , n2 , n3
N 2 = N3 = ∞
r r
r
r
r
(k − k ′) = x1b1 + x2b2 + x3b3
∝
N1
∑e
n1 = − N1
n1 ( 2πix1 )
∞
∑e
n2 = −∞
n2 ( 2πix2 )
∞
∑e
n3 ( 2πix3 )
n3 = −∞
Delta functions
Sample Shape Q Reciprocal Point Shape
Reciprocal lattice consists of lattice “points” because the
crystal is assumed to be infinite in size, i.e. reciprocal lattice is
th F
the
Fourier
i transform
t
f
off a reall infinite
i fi it lattice,
l tti and
d vice
i versa.
If the sample under study has a finite size (microscopic size)
along one or more of its dimension, the reciprocal lattice also
takes on finite width along those directions.
Examples: thin film Q rods, thin wire Q discs, small particles
Q blobs, etc. Ewald sphere drawn as usual, but chances of
cuttingg across some reciprocal
p
lattice “points”
p
are ggreatly
y
enhanced.
12
Surface Diffraction
a layer of regularly arranged atoms ……
….. is transformed into an array of rods
in reciprocal space.
The array of rods in
reciprocal space is then used
in an Ewald construction of
the diffraction pattern.
13