VI. Reciprocal lattice
6-1. Definition of reciprocal
lattice from a lattice

 
with periodicities a, b , c in real space
 




*
*
*
b c
c a
a b
a     ;b     ;c    
a  (b  c )
b  (c  a )
c  (a  b )
 




* b  c * c  a * a  b
or a 
;b 
;c 
V
V
V
Remind what we have learned in chapter 5
Pattern  Fourier transform  diffraction
Pattern of the original pattern!
3-D: the Fourier transform of a function f(x,y,z)
F ( x, y , z )  



f ( x, y, z )e2i ( ux vy wz )dxdydz

r  xxˆ  yyˆ  zzˆ

u  uuˆ  vvˆ  wwˆ
Note that
ux+vy+wz: can be considered as a scalar
 
product of r  u
if the following conditions are met!
xˆ  uˆ  1; xˆ  vˆ  0; xˆ  wˆ  0
 
yˆ  uˆ  0; yˆ  vˆ  1; yˆ  wˆ  0  r  u  ux  vy  wz
zˆ  uˆ  0; zˆ  vˆ  0; zˆ  wˆ  1


What is r ? Then what is u ?
Consider the requirements for the basic
translation vectors of the “reciprocal lattice”
*
Say a  
 
 
a  a*  1; b  a*  0; c  a*  0
*  * 
i.e. a  b; a  c
 
*
*  
In other words, a || b  c  a  k (b  c )
 
 *
 * 
 a  a  1  a  a  a  k (b  c )  1
1
1
k     
a  (b  c ) V
 
 
*
b c
b c
a     
V
a  (b  c )
 
 
*
c a
c a
Similarly, b     
V
b  (c  a )
 
 
*
a b
a b
c    
V
c  (a  b )
＊ A translation vector in reciprocal lattice is
*
called
reciprocal
lattice
vector
G
hkl
* *
*
*
Ghkl  ha  kb  lc
* orthogonality; orthornormal set
 *
a  a  1;
 *
a  b  0;
 *
a  c  0;
 *
b  a  0;
 *
b  b  1;
*
b  c  0;
 *
c a  0
 *
c b  0
 *
c c 1
* In orthorhombic, tetragonal and cubic
systems,
* 1 1
a   
a a
* 1 1
b   
b b
* 1 1
c   
c c
*
* Ghkl
is perpendicular to the plane (h, k, l) in
c
real space
c
 
b a
AB  
k h
 
c a
AC  
/ h
The reciprocal lattice vector
* *
*
*
Ghkl  ha  kb  lc
l

a
a
h
A
C
b
k

B b
 
* *  b a 
*
*
Ghkl  AB  (ha  kb  lc )    
 k h 

*
  *  a    * b 
Ghkl  AB   ha 
   kb    0
h  
k


Similarly,

c
c
l
a
h
C
A
a
 

*
*
*  c a 
*
Ghkl  AC  ( ha  kb  lc )    
 l h


*
 *  a   * c 
Ghkl  AC   ha 
   lc    0
h  
l

*
*
Therefore, Ghkl  AB ; Ghkl  AC
*
Ghkl is perpendicular to the plane (h, k, l)
b
k

B b
*
1
Moreover, Ghkl 
interplanar spacing
d hkl
of the plane (h, k, l)

 *
 *
* *
c
b Ghkl b ha  kb  lc
c
d hkl    *  
*
C
b
l
k Ghkl k
G
hkl
k 
 *
a
b kb
1
B b
h
  *  *
A

k Ghkl
Ghkl
a
* *
 *
 *
a ha  kb  lc
1
or d hkl  a  G hkl
 
 *
*
*
h Ghkl h
Ghkl
Ghkl
* *
 *
 *
c Ghkl c ha  kb  lc
1
 *
*
or d hkl    *  
l Ghkl
l
Ghkl
Ghkl
Graphical view of reciprocal lattice!
 How to construct a reciprocal lattice from a crystal
Pick a set of planes in a crystal and using a direction
and a magnitude to represent the plane
parallel
Plane
set 2
d2
d1
Plane set 1
d
d 2*
d
Plane
set 3
d3
d1* : d 2*  (1 / d1 ) : (1 / d 2 )
k
k
*
*
d1  ; d 2 
d1
d2
*
1
d 2*
*
1
d 3*
Does it really form a
lattice?
Draw it to convince
yourself!
Example: a monoclinic crystal
Reciprocal lattice (a* and c*) on the plane
containing a and c vectors.
(b is out of the plane)
c
b
a
(-100)
(100)
c
(102)
(001)
c
(001)
(002)

(002)
a
O
a
(00-2)
c
(00-2)
(002)
(001) (101) c
(002)

O
a
O
a
d
d
*
d 001
*
d100
O
*
d10
1
*
d 00
1
O
c*
c
(002)
(00-1)
*
d 002
*
102
*
101
(10-1)
(00-2)
2D form a 3-D reciprocal lattice

*
a*
a
*
*
a *  d100
; c *  d 001
Lattice point in reciprocal space
*
*
*
a*  d100
; b*  d010
; c*  d001
*
*
*
*
*
*
*
*
Ghkl  dhkl  hd100  kd010  ld001  ha  kb  lc
Integer
Lattice points in real space
ruvw  ua  vb  wc
Relationships between a, b, c and a*, b*, c*:
Monoclinic: plane  y-axis (b)
 c *  a and c *  b  c *  a  0 and c *  b  0
c // a  b
*
Similarly,
a  b  0 and a  c  0; a // b  c
*
*
*
c
*
b  a  0 and b  c  0; b // c  a
*
: c  c*.
*
 cc* = |c*|ccos, |c*| = 1/d001  ccos = d001
 cc* = 1
b

c*
d001
a
Similarly, aa* = 1 and bb* =1.
 c* //ab,
Define c* = k (ab), k : a constant.

cc* = 1  ck(ab) = 1  k = 1/[c(ab)]=1/V.
V: volume of the unit cell
ab
bc
ca
*
*
*
Similarly, one gets a 
; b 
c 
V
V
V
6-2. Reciprocal lattices corresponding to crystal
systems in real space
(i) Orthorhombic ,tetragonal ,cubic
*
c

c

b

a
*
a

c
(ii) Monoclinic
*
b
*
c


a
*
a
*

* b
b
(iii) Hexagonal

a

c
*
c

o
b
30

60o
*
b
o
30
*
a
*
c
c 120o

a
We deal with reciprocal lattice
Transformation in Miller indices.
* 
a  b;
* 
b  a;
* 
a c
* 
b c
60o
*
a

b
*
b
 
b c
bc sin( / 2)aˆ
aˆ
*
a    

a  (b  c ) a  bc sin( / 2)aˆ a  aˆ
aˆ
2aˆ


o
a cos
30
3a


*
2bˆ
c a
ca sin( / 2)bˆ
bˆ

b    

3b
b  (c  a ) b  ca sin( / 2)bˆ b  bˆ
*
aˆ : unit vector of a
 
*
ab
bˆ : unit vector of b
*
a b *  60o
 
a b
ab sin( 2 / 3)cˆ
cˆ
*
c    

c  (a  b ) c  ab sin( 2 / 3)cˆ c  cˆ
cˆ
cˆ


o
c cos 0
c
6-3. Interplanar spacing
*
* *
1
1
Ghkl 
 Ghkl  Ghkl  2
d hkl
d hkl
* *
* *
1
*
*
 ( ha  kb  lc )  ( ha  kb  lc )
2
d hkl
(i) for cubic ,orthorhombic, tetragonal systems
* * * *
a  b  c  a
* *
* *
* *
a  b  0; a  c  0; b  c  0
 
 
b c
| b || c | sin( / 2)aˆ
bcaˆ
*
a        
 
a  (b  c ) a  | b || c | sin( / 2)aˆ a  bcaˆ
bcaˆ
1

 aˆ
o
abc cos 0
a
 *  * aˆ aˆ 1
a a    2
a a a
Similarly,
 *  * bˆ bˆ 1
 *  * cˆ cˆ 1
c c    2
b b    2
c c c
b b b

1
*
*
*
*
*
*
 2  ( ha  kb  lc )  ( ha  kb  lc )
d hkl
h2 k 2 l 2
 2 2 2
a
b
c
1
h2 k 2 l 2

 2 2
2
d hkl
a
b
c
(ii) for the hexagonal system
1

d hkl
4( h 2  k 2  hk ) l 2
 2
2
3a
c
* *
* *
1
*
*
 ( ha  kb  lc )  ( ha  kb  lc )
2
d hkl
* * * * * *
 a  c ; b  c ; a  b  60o
* *
* *
a  c  0; b  c  0
4
4
4
 *  * 2aˆ 2aˆ
o
a a 

 2 aˆ  aˆ  2 cos0  2
3a
3a
3a 3a 3a
 *  * 2bˆ 2bˆ
4 ˆ ˆ
4
4
o
b b 

 2 b  b  2 cos 0  2
3a
3a
3b 3b 3a
4
4
 *  * 2aˆ 2bˆ
ˆ
a b 

 2 aˆ  b  2 cos 60 o
3a
3a 3b 3a
* *
4 1
2
 2   2  b a
3a 2 3a
* *
* *
1
*
*
 ( ha  kb  lc )  ( ha  kb  lc )
2
d hkl
 * 2 * *
1
* *
2 * *
2 *
 h a  a  2hka  b  k b  b  l c  c
2
d hkl
4
2
4
2
2
2 1
h
 2hk 2  k
l 2
2
2
3a
3a
3a
c
4(h 2  hk  k 2 ) l 2

 2
2
3a
c
6-4. Angle between planes (h1k1l1) and (h2k2l2)
*
*
* *
Gh k l  Gh k l  Gh k l Gh k l cos 
*
*
Gh k l  Gh k l
cos    *  *
Gh k l Gh k l
1 11
2 2 2
1 11
1 11
1 11
2 2 2
2 2 2
2 2 2
for the cubic system
*
*
*
*
*
*
(h1a  k1b  l1c )  (h2 a  k2b  l2 c )
cos  
h12  k12  l12 / a 2 h22  k22  l22 / a 2
2
2
2
h1h2 / a  k1k 2 / a  l1l2 / a

h12  k12  l12  h22  k22  l22  / a 2
h1h2  k1k2  l1l2

h12  k12  l12  h22  k22  l22 
6-5. The relationship between real lattice and
reciprocal lattice in cubic system：
Real lattice
Simple cubic
BCC
FCC
Reciprocal lattice
Simple cubic
FCC
BCC
Example：f.c.c b.c.c
(1) Find the primitive unit cell of the selected
structure
(2) Identify the unit vectors
1
a( xˆ  yˆ )
2
1
a( yˆ  zˆ)
2
1
a( xˆ  zˆ)
2
1
1
1
a(  xˆ  yˆ  zˆ) a( xˆ  yˆ  zˆ)
a( xˆ  yˆ  zˆ)
2
2
2
 a
 a
 a
a  ( xˆ  zˆ)
b  ( xˆ  yˆ )
c  ( yˆ  zˆ)
2
2
2
a
a
a
  


V  a  (b  c )  ( xˆ  zˆ )   ( xˆ  yˆ )  ( yˆ  zˆ )
2
2
2

2
a
a
a
 ˆ ˆ

 2 ( x  y )  2 ( yˆ  zˆ )  4 ( zˆ  (  yˆ )  xˆ )
a
a2
a3
a3
  
V  a  (b  c )  ( xˆ  zˆ)  ( xˆ  yˆ  zˆ)   2 
2
4
8
4
Volume of F.C.C. is a3. There are four atoms
per unit cell!  the volume for the primitive
of a F.C.C. structure is ?
a
a
 
( xˆ  yˆ )  ( yˆ  zˆ)
b c
*
2
a     2
a3 / 4
a  (b  c )
a2
( xˆ  yˆ )  ( yˆ  zˆ )
( xˆ  yˆ )  ( yˆ  zˆ )
4


3
a /4
a
zˆ  (  yˆ )  xˆ xˆ  yˆ  zˆ


a
a
Similarly,
 * xˆ  yˆ  zˆ
 *  xˆ  yˆ  zˆ
b 
c 
a
a
xˆ  yˆ  zˆ xˆ  yˆ  zˆ  xˆ  yˆ  zˆ
,
,
 B.C.C.
a
a
a
See page 23
Here we use primitive cell translation vector to
calculate the reciprocal lattice.
When we are calculating the interplanar spacing,
the reciprocal lattices that we chosen is different.
Contradictory?
Which one is correct?
Using primitive translation vector to do the
reciprocal lattice calculation:
Case: FCC  BCC
xˆ  yˆ  zˆ xˆ  yˆ  zˆ  xˆ  yˆ  zˆ
,
,
a
a
a
*
*
c
b
* *
* *
*
*
1
 (ha  kb  lc )  (ha  kb  lc )
2
d hkl
* *
* *
2 * *
 h a  a  hka  b  hla  c
* *
 *
* *
2 *
 hkb  a  k b  b  klb  c
* *
* * 2 * *
 hlc  a  klc  b  l c  c
*
a
 *  * xˆ  yˆ  zˆ xˆ  yˆ  zˆ 3
a a 

 2
a
a
a
 *  *  *  * xˆ  yˆ  zˆ xˆ  yˆ  zˆ  1
a b  b a 

 2
a
a
a
 *  *  *  * xˆ  yˆ  zˆ  xˆ  yˆ  zˆ  1
a c  c a 

 2
a
a
a
 *  * xˆ  yˆ  zˆ xˆ  yˆ  zˆ 3
b b 

 2
a
a
a
 *  *  *  * xˆ  yˆ  zˆ  xˆ  yˆ  zˆ  1
b c  c b 

 2
a
a
a
 *  *  xˆ  yˆ  zˆ  xˆ  yˆ  zˆ 3
c c 

 2
a
a
a
* *
* *
*
*
1
 (ha  kb  lc )  (ha  kb  lc )
2
d hkl
3h 2 hk hl hk 3k 2 kl hl kl 3l 2
 2  2 2 2 2  2 2 2 2
a
a
a
a
a
a
a
a
a
1
 2 [3( h 2  k 2  l 2 )  2( hk  kl  hl )]
a
h2 k 2 l 2 h2  k 2  l 2
not 2  2  2 
Why?
2
a
b
c
a
(hkl) defined using unit cell!
(hkl) is defined using primitive cell!
(HKL)
Example
 Find out the relation between the (hkl) and
  
[uvw] in the unit cell defined by a, b , c and the
(HKL) and [UVW] in the unit cell defined by
  
A, B, C .

 

A  1a  2b  0c

 

B  1a  1b  0c
 


C  0a  0b  1c

B

b

a


 A   1 2 0  a 
  
  
In terms of matrix  B     1 1 0  b 

  


 C   0 0 1  c 

A
Find out the relation between (hkl) and (HKL).
Assume there is the first plane
intersecting the a axis at
a/h and the b axis at b/k.
A
In the length of |a|, there
are h planes. In the length
of |b|, there are k planes.
B
b
2k
a
How many planes can
be inserted in the length
b/k
a/h
h
|A|? Ans. h + 2k
A/(h+2k)
 H = 1h + 2k + 0l
Similarly, K = -1h + 1k +0l and L = 0h + 0k + 1l
H
K
L   h k
 1  1 0
 H   1 2 0  h 


  
 
or
l  2 1 0 
 K     1 1 0  k 
 0 0 1
 L   0 0 1  l 


  
 



a  axˆ; b  ayˆ ; c  azˆ
 a
 a
 a
A  ( yˆ  zˆ ); B  ( xˆ  zˆ ); C  ( xˆ  yˆ )
2
2
 2

 A
0 1 1  a 

  1 
  
 B    1 0 1  b 
 C  2  1 1 0  c 

 
 
 h    1 1 1  H 
H 
 0 1 1  h 
  
 
  1
 
 k    1  1 1  K 
 K    1 0 1  k 
 l   1 1  1 L 
 L  2  1 1 0  l 
  
 
 

 
1
1
1
H  ( k  l ); K  ( h  l ); L  ( h  k )
2
2
2
h   H  K  L; k  H  K  L; l  H  K  L
1
1
2
2
2
[3( H  K  L )  2( HK  KL  HL )]  2
2
a
d hkl
1
2
2
2
 2 [3( k  l )  3( h  l )  3( h  k ) )
4a
 2( k  l )(h  l )  2( h  l )(h  k )  2( k  l )(h  k )]
 3(2h2  2k 2  2l 2  2hk  2kl  2hl)
 2(h  k  l  3hk  3kl  3hl)
2
2
2
1
1 2
2
2
2
 2 [4h  4k  4l ]  2 [h  k 2  l 2 ]
4a
a
There are the same!
Or
1 2
1
2
2
(
h

k

l
)

2
2
d hkl a
 (H  K  L)  ( H  K  L)  ( H  K  L)
 H 2  K 2  L2  2 HK  2 HL  2 KL
2
2
H 2  K 2  L2  2 HK  2 HL  2 KL
H  K  L  2 HK  2 HL  2 KL
2
2
2
 3( H  K  L )  2( HK  HL  KL)
1
1
2
2
2

[
3
(
H

K

L
)  2( HK  KL  HL)]
2
2
d hkl a
2
2
2
We proof the other way around!
2