Solid State Physics
Lecture 4 – Reciprocal lattices
Professor Stephen Sweeney
Advanced Technology Institute and Department of Physics
University of Surrey, Guildford, GU2 7XH, UK
[email protected]
Solid State Physics - Lecture 4
Recap from Lecture 3
• The structure of crystals can be probed with X-rays using
diffraction techniques
• There is a relationship between the lattice spacing and the Miller
indices (hkl) which is straightforward for cubic structures
• There are various techniques for X-ray diffraction: single crystal,
powder method, Laue (energy dependent) technique
• Applications go well beyond condensed matter physics, e.g.
archaeology etc.
Solid State Physics - Lecture 4
More on diffraction
• Diffraction from a crystal has two
contributions:
• Diffraction due to the underlying
lattice
• Diffraction due to the basis
Solid State Physics - Lecture 4
Bragg revisited
2D Case
n  2d sin 
Solid State Physics - Lecture 4
Revision of Waves

A
A exp ik r  t 
Definition of wave:
(complex form)
amplitude
wave-vector
k
time dependence
(often ignore)
observation point
2

Solid State Physics - Lecture 4
Bragg revisited
General 3D Case
Path difference = AB+BC:
AB  d cos 
BC  d cos '
A
AB  BC  d cos '  d cos 
B
 n  d cos '  d cos 
C
k is wave-vector of incident wave
k’ is wave-vector of diffracted wave
d is the distance between two lattice points
(d is the displacement vector)
Solid State Physics - Lecture 4
Laue condition
Since d is the displacement vector
between the scattered waves:
General 3D Case
k d 
k
A
2
d cos 

2
k 'd  
d cos '

Combine with our previous result (previous slide):
d
B
k’
C
k is wave-vector of incident wave
k’ is wave-vector of diffracted wave
d is the displacement vector

k  k ' d
n  d cos    d cos ' 
2
K  k  k'
 2n  K  d
is the scattering vector
or
exp iK d  1
Laue (diffraction) condition
Solid State Physics - Lecture 4
The Reciprocal lattice
•
The diffraction pattern that is generated in X-ray
diffraction is a representation of a “reciprocal
lattice”
•
The reciprocal lattice is of fundamental
importance when considering periodic structures
and processes in a crystal lattice where
momentum is transferred (e.g. diffraction)
•
It is a set of imaginary points in which the direction of a vector from one point to
another corresponds to a direction normal to a plane in a real lattice
•
G
The separation of the reciprocal lattice
points (magnitude of the vector) is
proportional to the reciprocal of the
real separation between planes, i.e.
2
G 
d hkl
Solid State Physics - Lecture 4
The Reciprocal lattice
Things to remember:
•
Convention: the reciprocal lattice vector is 2 times the reciprocal of the
interplanar distance (NB: crystallographers often don’t bother with the 2 but we
always use it in Solid-State physics)
•
Real lattice points correspond to the location of real objects (e.g. atoms, ions etc.)
and have dimensions [L]
•
Reciprocal lattice points are rather more abstract (e.g. magnitude/direction of
momentum) with dimensions [L-1]
•
We can relate the real lattice to the reciprocal lattice using Fourier Transforms
(but we won’t in this course)
Solid State Physics - Lecture 4
2D Reciprocal lattice
•
The basis set of the reciprocal lattice vectors are defined from:
real lattice
vector
Therefore:
reciprocal lattice
vector
(where i,j are the directions)
b1 must be perpendicular to a2 and
a1 must be perpendicular to b2
Solid State Physics - Lecture 4
2D Reciprocal lattice
Real Lattice
Reciprocal Lattice
Solid State Physics - Lecture 4
2D Reciprocal lattice
General 2D lattice:
Real Lattice
Reciprocal Lattice
General 2D reciprocal lattice vector
Solid State Physics - Lecture 4
3D Reciprocal lattice
General Case (3D) lattice:
area of plane in unit cell
x vector perpendicular to plane
a1
a2
a3
volume of real unit cell
Thus, b1 is perpendicular to a2 and a3 with magnitude 2/a1
Solid State Physics - Lecture 4
3D Reciprocal lattice
General Case (3D) lattice:
a1  b 2  0
as before
General 3D reciprocal lattice vector
Solid State Physics - Lecture 4
Cubic Lattices
For simple cubic Bravais Lattice
a
a
a
Q. What are the reciprocal lattice vectors?

a2  a3 
b1  2
a1  a 2  a 3 
z
y
x
x y z
2
a 2  a 3  0 a 0  a 2 x  b1 
x
a
0 0 a
Solid State Physics - Lecture 4
Cubic Lattices – bcc structure
bcc cell primitive vectors:
Q. What are reciprocal lattice
primitive vectors?
Solid State Physics - Lecture 4
Cubic Lattices – bcc structure
Primitive reciprocal lattice vectors for bcc (real space):
The reciprocal lattice is a fcc structure with cube unit cell length of 2/a
(similarly a bcc real space lattice has a reciprocal lattice fcc structure)
Solid State Physics - Lecture 4
Laue condition (again)
From before:
d is the real space vector
separating the lattice points of
interest, i.e.
K is the scattering vector which
depends on the wavelength and
the geometry of the scattering
d  n1a1  n2a 2  n3a3
 2n  K  d
or
exp iK d  1
Laue (diffraction) condition
We will see that if K=Ghkl (where Ghkl is a reciprocal lattice vector)
that the Laue condition is satisfied
Solid State Physics - Lecture 4
Laue condition (again)
We may write that:
G hkl  hb1  kb 2  lb3
If
k’
k
K=k’-k
=Ghkl
(where h,k & l are the Miller Indices…)
d  a1
G hkl  d  hb1  kb 2  lb3  a1

a2  a3 
But b1  2
a1  a 2  a 3 
and
b2  a1  b3  a1  0

a2  a3 
 G hkl  d  2h
 a1  2h
a1  a 2  a 3 
Satisfies Laue condition
Solid State Physics - Lecture 4
Reciprocal lattice and Miller Indices
Since the scattering vector K for each diffracted beam corresponds to a point
on the reciprocal lattice, we can use (hkl) to label that beam
G hkl  hb1  kb 2  lb3
is orthogonal to the planes with indices (hkl)
G hkl  h  k  l
is inversely proportional to the spacing (dhkl)
of the (hkl) planes, as can be proven…
2
2
2
Solid State Physics - Lecture 4
Reciprocal lattice and Miller Indices
Real Space
Reciprocal Space
The vector normal to lattice plane (hkl) is parallel to
Solid State Physics - Lecture 4
Reciprocal lattice and Miller Indices
The vector normal to lattice plane (hkl) is parallel to
Remember:
 a 2 a1   a3 a1 
     
h
k h  l
 a 2  a3   a1  a 2   a3  a1 




 kl   kh   hl 
1
ha 2  a3  la1  a 2  ka3  a1 

hkl
1 volume
hb1  kb 2  lb3   volume G hkl

hkl 2
2
hkl
Solid State Physics - Lecture 4
volume
Reciprocal lattice and Miller Indices
(since ai.bj=2ij)
Solid State Physics - Lecture 4
Ewald Construction
An elegant construction to show the condition for diffraction
Procedure (2D):
A

k’
k
B
E G
O
1. Construct the reciprocal lattice
2. Draw a vector AO (k) of length 2/
ending on a lattice point
3. Draw a circle of radius 2/
4. Draw a vector AB (k’) to a point of
intersection with another lattice point
5. Draw vector OB (G) joining the
intersecting lattice points
6. Draw a line perpendicular to OB back to
A ( is the angle between the incident or
scattered wave and the plane AE (i.e. it
is the scattering angle)
7. Repeat for all points of intersection
Solid State Physics - Lecture 4
Ewald Construction
An elegant construction to show the condition for diffraction
Vector OB (G) joins two lattice points, is
normal to a set of real planes and has
length 2/dhkl
OE has length
A
2
OB has length

k’
2
2

sin 
2
k
B
E G

sin 
2
 2
sin  
   2d hkl sin 

d hkl
Braggs Law
O
Solid State Physics - Lecture 4
Ewald Construction
An elegant construction to show the condition for diffraction
Other points of note:
If length of AO<½ 2/a then diffraction is
not possible (i.e. if >2a)
Shorter wavelength (larger circle) leads to
more intersections (higher probability of
diffraction)
A

k’
In 3D this is a sphere (Ewald sphere)
k
B
E G
O
Solid State Physics - Lecture 4