Handout 5
The Reciprocal Lattice
In this lecture you will learn:
• Fourier transforms of lattices
• The reciprocal lattice
• Brillouin Zones
• X-ray diffraction
• Fourier transforms of lattice periodic functions
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Fourier Transform (FT) of a 1D Lattice
Consider a 1D Bravais lattice:

a1  a xˆ
Now consider a function consisting of a “lattice” of delta functions – in which a delta
function is placed at each lattice point:
f x 

f x     x  n a 

a1  a xˆ
x
n  
The FT of this function is (as you found in your homework):



n  
f k x    dx    x  n a  e  i k x
x

  ei kx
n  
na

2
a


m  
  k x  m

2 

a 
The FT of a train of delta functions is also a train of delta functions in k-space
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
1
Reciprocal Lattice as FT of a 1D Lattice
f x 
f k x 
FT is:
1

a1  a xˆ
x
2
a

2
b1 
xˆ
a
kx
The reciprocal lattice is defined by the position of the delta-functions in the FT of
the actual lattice (also called the direct lattice)
Direct lattice (or the actual lattice):

a1  a xˆ
x

2
b1 
xˆ
a
kx
Reciprocal lattice:
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Reciprocal Lattice of a 1D Lattice
For the 1D Bravais lattice,

a1  a xˆ



The position vector Rn of any lattice point is given by: Rn  n a1
f x 
 
f  x     x  Rn

n  


The FT of this function is:

a1  a xˆ
x


 
f k x    dx   x  Rn e i k x

n  


x

 
  e i k .Rn
n  
The reciprocal lattice in k-space is defined by the set of all points for which the kvector satisfies,
 
e i k . Rn  1

for ALL Rn of the direct lattice

 
For the points in k-space belonging to the reciprocal lattice the summation  e i k .Rn
n  
becomes very large!
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
2
Reciprocal Lattice of a 1D Lattice
For the 1D Bravais lattice,

a1  a xˆ



The position vector Rn of any lattice point is given by: Rn  n a1
The reciprocal lattice in k-space is defined by the set of all points for which the kvector satisfies,
 
e i k . Rn  1

for ALL Rn of the direct lattice
 


i k . Rn
For k to satisfy e
 1, it must be that for all Rn:
 
k . Rn  2   integer 
 k x na  2   integer
2
 kx  m
a

where m is any integer
Therefore, the reciprocal lattice is:
kx

2
b1 
xˆ
a
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Reciprocal Lattice of a 2D Lattice
Consider the 2D rectangular Bravais lattice:
If we place a 2D delta function at each lattice
point we get the function:

f x, y   
y

a2  c yˆ

   x  n a   y  m c 
n   m  

a1  a xˆ
x
The above notation is too cumbersome, so we write it in a simpler way as:

 
f r     2 r  R j
j


The summation over “j ” is over all the lattice points





A 2D delta function has the property:  d 2 r  2 r  ro  g r   g ro 
and it is just a product of two 1D delta functions corresponding to the x and y
components of the vectors in its arguments:  2 r  r     x  r . xˆ  y  r .yˆ 
o
o
o

Now we Fourier transform the function f r  :
 
 




 
f k   d 2 r f r  e  i k . r   d 2 r   2 r  R j e i k . r

j
 e
j
 
i k . R j

2 2



2  
2 

  kx  n
   ky  m

ac n   m   
a  
c 


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3
Reciprocal Lattice of a 2D Lattice
 

2 2     k  n 2    k  m 2 
i k . R j

f k  e


 x
  y

ac n   m   
a  
c 
j
ky
y

2
b2 
yˆ
Reciprocal lattice
c

a2  c yˆ

kx

2
x
a1  a xˆ
b1 
xˆ
a

Direct lattice
• Note also that the reciprocal lattice in k-space is defined by the set of all points for
which the k-vector satisfies,
 
e

for all R j of the direct lattice
i k . Rj
1
• Reciprocal lattice
lattice or as set of all points in k-space
 as
 the FT of the direct

for which exp i k . R j  1 for all R j , are equivalent statements


ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Reciprocal Lattice of a 2D Lattice
ky
y

a2  c yˆ

2
b2 
yˆ
c

a1  a xˆ
Reciprocal lattice
kx

2
b1 
xˆ
a
x
Direct lattice
• The reciprocal lattice of a Bravais lattice
a Bravais lattice and has its own

 is always
primitive lattice vectors, for example, b1 and b2 in the above figure

• The position vector G of any point in the reciprocal lattice can be expressed in
terms of the primitive lattice vectors:



G  n b1  m b2
For m and n integers
So we can write the FT in a better way as:
 2 2 
 

2  
2  2 2

2
f k 

  kx  n
 k  Gj
   ky  m

2 j
ac n   m   
a  
c 



where 2 = ac is the area of the direct lattice primitive cell
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
4
Reciprocal Lattice of a 3D Lattice
d
Consider a orthorhombic direct lattice:




R  n a1  m a2  p a3
c
Then the corresponding delta-function lattice is:

 
f r     3 r  R j

j

a2
where n, m, and p are integers

a3

A 3D delta function has the property:
3
 


 d r  r  ro  g r   g ro 

a1
a
3
The reciprocal lattice ink-space
is defined by the set of all points for which the k
vector satisfies:
exp i k

 . R j  1 for all R j of the direct lattice. The above relation
will hold if k equals G :






G  n b1  m b2  p b3

2
b1 
xˆ
a
and

2
b2 
yˆ
c

2
b3 
zˆ
d
Finally, the FT of the direct lattice is:
 
 




 
f k   d 3 r f r  e  i k . r   d 3 r   3 r  R j e  i k . r


j
 e
 
i k . R j
j



b2 

b1
b3
2 3  3 k  G   2 3  3 k  G 


j
j
acd
3
j
j
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Direct Lattice Vectors and Reciprocal Lattice Vectors
ky
y

a2  c yˆ

2
b2 
yˆ
c

a1  a xˆ
Reciprocal lattice
kx

2
b1 
xˆ
a
x
Direct lattice



R  n a1  m a2



G  n b1  m b2
Remember that the reciprocal lattice in k-space is defined by the set of all points for
which the k-vector satisfies,
 
e

for all R of the direct lattice
i k .R
1


So for all direct lattice vectors R and all reciprocal lattice vectors G we must have:


ei G . R  1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
5
Reciprocal Lattice of General Lattices in 1D, 2D, 3D



More often that not, the direct lattice primitive vectors, a1 , a2 , and a3 , are not
orthogonal
Question: How does one find the reciprocal lattice vectors in the general case?
ID lattice:

If the direct lattice primitive vector is: a1  a xˆ
and length of primitive cell is:  = a

2
Then the reciprocal lattice primitive vector is: b1 
xˆ

 
f r     r  R j

j
a
 2
 
f k 
  k  Gj
1 j



Note:
 
a1 . b1  2

e


i G p . Rm
1
2D lattice:


If the direct lattice is in the x-y plane and the primitive vectors are: a1 and a2
 

and area of primitive cell is:  2  a1  a2

aˆ  zˆ 
zˆ  a1
Then the reciprocal lattice primitive vectors are: b1  2 2
b2  2

 
f r     2 r  R j

j





2
2



2 2
2
f k 
  k Gj
2 j
Note: a j . bk  2  jk
and
e



i G p . Rm
1
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Reciprocal Lattice of General Lattices in 1D, 2D, 3D
3D lattice:



If the direct lattice primitive vectors are: a1 , a2 , and a3

 
and volume of primitive cell is:  3  a1 . a2  a3 
Note:
Then the reciprocal lattice primitive vectors are:


aˆ  a
b1  2 2 3
3



a a
b2  2 3 1
3

 
f r     3 r  R j

j
 
a j . bk  2  jk
 

a a
b3  2 1 2
3

 
2 3
3
f k 
  k  Gj
3 j




Example 2D lattice:
b

a2

a1


i G p . Rm
1

2
 xˆ  yˆ 
b1 
b

4
b2 
yˆ
b

a1  b xˆ

b
b
a2   xˆ  yˆ
2
2
 
b2
 2  a1  a2 
2
e

b2

b1
4 b
4 b
b
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
6
The Brillouin Zone
The Wigner-Seitz primitive cell of the reciprocal lattice centered at the origin is
called the Brillouin zone (or the first Brillouin zone or FBZ)
Wigner-Seitz primitive cell
1D direct lattice:

a1  a xˆ
Reciprocal lattice:
x
First Brillouin zone
kx

2
b1 
xˆ
a
2D lattice:

a2  c yˆ
ky
Wigner-Seitz
primitive cell
y
Reciprocal lattice

2
b2 
yˆ
c

a1  a xˆ
First Brillouin zone
x

2
b1 
xˆ
a
kx
Direct lattice
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
2D lattice:
Wigner-Seitz
primitive cell
First Brillouin zone

b
b
a1  xˆ  yˆ
2
2

b
b
a2  xˆ  yˆ
2
2
 
b2
 2  a1  a2 
2

a1

a2
b
The Brillouin Zone

2
 xˆ  yˆ 
b1 
b

2
 xˆ  yˆ 
b2 
b

b2

b1
4 b
4 b
b
Direct lattice
Reciprocal lattice
Volume/Area/Length of the first Brillouin zone:
The volume (3D), area (2D), length (1D) of the first Brillouin zone is given in the
same way as the corresponding expressions for the primitive cell of a direct lattice:
1D
2D
3D

 1  b1
 
 2  b1  b2

 
 3  b1 . b2  b3


Note that in all dimensions (d) the following
relationship holds between the volumes, areas,
lengths of the direct and reciprocal lattice
primitive cells:
d
d 
2 
d
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
7
Direct Lattice Planes and Reciprocal Lattice Vectors
There is an intimate relationship between reciprocal lattice vectors and planes of
points in the direct lattice captured by this theorem and its converse
Theorem:
If there is a family of parallel lattice
planes separated by distance “d ” and n̂
is a unit vector normal to the planes
then the vector given by,
3D lattice

G
 2
G
nˆ
d
is a reciprocal lattice vector and so is:
m
2
nˆ
d
 m  integer
d
d
Converse:

If G1 is
 any reciprocal lattice vector,
and G is the reciprocal lattice vector

of the smallest magnitude parallel to G1 ,
then there exist a family
of lattice planes

perpendicular to G1 and G , and
separated by distance “d ” where:
2D lattice
d

G
2
d 
G
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Example: Direct Lattice Planes and Reciprocal Lattice Vectors
ky
y

a2  c yˆ

2
b2 
yˆ
c

a1  a xˆ
Reciprocal lattice
kx

2
b1 
xˆ
a
x
Direct lattice
Consider:
 

 xˆ yˆ 
G  b1  b2  2   
a c

There must be a family of lattice planes normal to G and separated by:
Now consider:



 2 xˆ yˆ 
G  2b1  b2  2 
 
 a c

2
ac
 
2
G
a  c2
2
G
There must be a family of lattice planes normal to G and separated by:  
ac
2
a  4c 2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
8
The BCC Direct Lattice
4 a
a
y
a
x
4 a
z
4 a
a
Direct lattice: BCC
Reciprocal lattice: FCC
The direct and the reciprocal lattices are not necessarily always the same!
a
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The FCC Direct Lattice
4 a
a
y
a
z
x
4 a
4 a
a
Direct lattice: FCC
Reciprocal lattice: BCC
First Brillouin zone of
the BCC reciprocal
lattice for an FCC direct
lattice
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
9
The Reciprocal Lattice and FTs of Periodic Functions
The relationship between delta-functions on a “d ” dimensional lattice and its Fourier
transform is:
d


 
f r     d r  R j
j
 2 
 
d
f k 
  k  Gj
d j





Supper W r  is a periodic function with the periodicity of the direct lattice then
by definition:

 
W r  R j  W r 

for all R j of the direct lattice


One can always write a periodic function as a convolution of its value in the
primitive cell and a lattice of delta functions, as shown for 1D below:
W x 
a
W  x 
a 2
a 2
x
a2
a
a2
x
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Reciprocal Lattice and FTs of Periodic Functions
W x 
a
W  x 
a 2
a 2
x
a2
a
a2
x
Mathematically:

W  x   W  x      x  n a 
n  

And more generally in “d ” dimensions for a lattice periodic function W r  we have:



 
W r   W r     d r  R j
j
Value of the function
in one primitive cell

Lattice of delta
functions
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
10
The Reciprocal Lattice and FTs of Periodic Functions
For a periodic function we have:



 
W r   W r     d r  R j
j

Its FT is now easy given that we know the FT of a lattice of delta functions:


 
f r     d r  R j
j

 
2 d
d
f k 
  k  Gj
d j




We get:


 2 d
 
 
2 d
d
d
W k  W k 
 k  Gj 
  k  G j W G j
d j
d j






 
The FT looks like reciprocal
lattice of delta-functions with
unequal weights
If we now take the inverse FT we get:


 i k . r

ddk
ddk

W r   
W
k
e

2 d
2 d

W G j i G j . r

e
d
j

2 d
 

 
d
i k .r
  k  G j W G j e
d
 

j

 
A lattice periodic function can always
be written as a Fourier series that only
has wavevectors belonging to the
reciprocal lattice
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
The Reciprocal Lattice and X-Ray Diffraction
X-ray diffraction is the most commonly used method to study crystal structures

In this scheme, X-rays of wavevector k are sent into a
 crystal, and the scattered
X-rays in the direction of a different wavevector, say k ' , are measured

k'

k
If the position
 dependent dielectric constant of the medium is
given by  r  then the diffraction theory tells us thatthe
amplitude of the scattered X-rays in the direction of k ' is
proportional to the integral:
 
 




S k  k '   d 3r e  i k ' . r  r  e i k . r


For X-ray frequencies, the dielectric constant is a periodic
function with the periodicity of the lattice. Therefore, one can
 
write:

 r     G j  e

i Gj . r
j
Plug this into the integral above to get:


 X-rays will scatter in only those directions for which:
  
k'  k  G
  
Or: k '  k  G
  2 3 k  G j  k '



S k  k'    G j
j




where G is some reciprocal lattice vector

Because  G
 is also a reciprocal vector
whenever G is a reciprocal vector
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
11

k'
The Reciprocal Lattice and X-Ray Diffraction
 X-rays will scatter in only those directions for which:
  
k'  k  G

k
(1)
Also, the frequency of the incident and diffracted X-rays is the
same so:
'  


 k'c  k c


 k'  k
 2 2  2
 
k'  k  G  2 k . G
(1) gives:
 2 2  2
 
 k'  k  G  2 k . G
2 2  2
 
 k  k  G 2 k .G
2
  G
  k .G 
2
Condition for X-ray diffraction
ECE 407 – Spring 2009 – Farhan Rana – Cornell University

k'
The Reciprocal Lattice and X-Ray Diffraction
2
 
G
k .G  
2
The condition,

k
is called the Bragg condition for diffraction
Incident X-rays will diffract efficiently provided the incident
wavevector satisfies the Bragg
condition for some

reciprocal lattice vector G
A graphical way to see the Bragg condition is that the incident wavevector lies on a
plane in k-space (called the
 Bragg plane) that is the perpendicular bisector of some
reciprocal lattice vector G

k

k
  
k'  k  G

G
  
k'  k  G

G
Bragg plane
k-space
Bragg plane
k-space
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
12

k'
The Reciprocal Lattice and X-Ray Diffraction
The condition,
2
 
G
k .G  
2

k
can also be interpreted the following way:
Incident X-rays will diffract efficiently when the reflected
waves from successive atomic planes add in phase
**Recall that there are always a family of lattice planes in
real space perpendicular to any reciprocal lattice vector
Condition for in-phase reflection from
successive lattice planes:



2d cos   m 

G
2  2 
1  2 
m
 cos    m

  d 
2 d 
2
  G
 k .G 
2

d

k
Real space

2
Gm
nˆ
d
2
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Bragg Planes
2D square
reciprocal lattice
Corresponding to every reciprocal
lattice vector there is a Bragg plane in
k-space that is a perpendicular
bisector of that reciprocal lattice
vector
Lets draw few of the Bragg planes for
the square 2D reciprocal lattice
corresponding to the reciprocal lattice
vectors of the smallest magnitude
1D square reciprocal lattice
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
13
Bragg Planes and Higher Order Brillouin Zones
2D square
reciprocal lattice
Bragg planes are shown for the
square 2D reciprocal lattice
corresponding to the reciprocal lattice
vectors of the smallest magnitude
3
3
2
3
3
1
2
2
Higher Order Brillouin Zones
3
3
The nth BZ can be defined as the region
in k-space that is not in the (n-1)th BZ
and can be reached from the origin by
crossing at the minimum (n-1) Bragg
planes
3
The length (1D), area (2D), volume (3D)
of BZ of any order is the same
2
3
1st BZ
3
2
2nd BZ
1
2
3rd BZ
3
1D square reciprocal lattice
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Proof of the General Lattice FT Relation in 3D
This appendix gives proof of the FT relation:

 
f r     3 r  R j

j


 
2 3
3
f k 
  k Gj
3 j



for the general case when the direct lattice primitive vectors are not orthogonal




Let: R  n1 a1  n2 a2  n3 a3
Define the reciprocal lattice primitive vectors as:





aˆ  a
a a
b1  2 2 3 b2  2 3 1
3
3


Note:
a j . bk  2  jk
 

a a
b3  2 1 2
3
Now we take FT:
 




 
f k   d 3 r f r  e  i k . r   d 3 r   3 r  R j

 e
 
i k . R j
j

 
 e i k . r
j
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
14
Appendix: Proof

One can expand k in any suitable basis. Instead of choosing the usual basis:

k  k x xˆ  k y yˆ  k z zˆ
I choose the basis defined by the reciprocal lattice primitive vectors:




k  k1 b1  k2 b2  k3 b3


Given that: a j . bk  2  jk
I get:
 

i k . R j
f k  e


j




 i k . n1a1  n2 a2  n3 a3 
 e
n1 n2 n3
  k1  m1   k2  m2   k3  m3 

m1 m2 m3
Now:
 k1  m1   k2  m2   k3  m3    3 k  G 






where: G  m1 b1  m2 b2  m3 b3



But we don’t know the exact weight of the delta function  3 k  G

ECE 407 – Spring 2009 – Farhan Rana – Cornell University
Appendix: Proof
Since:

k  k x xˆ  k y yˆ  k z zˆ
This implies:
k x   b1x
k   b
 y   2x
 k z  b3 x




k  k1 b1  k2 b2  k3 b3
and
b2 x
b2 y
b3 y
Any integral over k-space in the form:
b3 x   k1 

b2 z  k2 
 
b3 z  k3 










can be converted into an integral in the form:  dk x
by the Jacobian of the transformation:


 dk x  dk y  dk z





 dk1  dk2  dk3


(1)

 k x , ky , kz

 k1 , k2 , k3 
Therefore:
 k1  m1   k2  m2   k3  m3  

 dk y  dk z






 dk1  dk2  dk3
 k x , ky , kz
  3 k  G 
 k1, k2 , k3 
ECE 407 – Spring 2009 – Farhan Rana – Cornell University
15
Appendix: Proof
From (1) on previous slide:

  b . b  b     2 3
1
2
3
3


 k x , k y , kz
 k1, k2 , k3
3
Therefore:
 

i k . R j

f k  e

j

2 3
3

j
3
  k1  m1   k2  m2   k3  m3 
m1 m2 m3
k  G j 


ECE 407 – Spring 2009 – Farhan Rana – Cornell University
16