Transcript Crystallography 6

Reciprocal lattice
How to construct reciprocal lattice
Meaning of reciprocal lattice
Relation between reciprocal lattice and diffraction
Geometrical relation between reciprocal lattice and
original lattice
 How to construct a reciprocal lattice from a crystal
(1) Pick a set of planes in a crystal
parallel
Plane
set 2
d2
d1
d
d 2*
Plane set 1
Plane
set 3
d3
d1* : d 2*  (1 / d1 ) : (1 / d 2 )
k
k
*
*
d1  ; d 2 
d1
d2
*
1
d
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*  d 010
; c *  d 001
*
*
*
*
d hkl
 hd100
 kd 010
 ld 001
 ha *  kb*  lc *
Integer
Lattice points in real space
ruvw  ua  vb  wc
Reciprocal lattice cells for cubic crystals:
Simple cubic:
z



a  axˆ; b  byˆ ; c  czˆ
y abc
x
d*002
z
z
(100)
*
d100
 1/ a
a
y
d
*
001
 1/ a
x
d
x
x
(001)
a
(010)
y
a
*
010
 1/ a
(110)
y
y
x
2a / 2
*
d110
 2/a
(002)?
*
d 001
*
d100
*
d 010
*
d110
Simple cubic
Base centered cubic (BCC):
z



a  axˆ; b  byˆ ; c  czˆ
y abc
x
*
d 200
 2/a
*
d 020
 2/a
*
d 002
*
d 200
*
d 020
d
*
d 002
 2/a


*
d110
 1 / 2a / 2  2 / a
*
110
*
d 220
*
d 011
*
d 002
*
d 011
*
d101
*
d101
*
d 200
*
d 020
d
(111)
*
d 220
(111)
*
d 002
O
*
d 222
O
*
d 020
*
d 200
(111)
*
d 222
O
*
110
(222)
*
d 220
FCC
corner
Up and down
F and B
L and R
You should do the same for a FCC and show it forms
a BCC lattice! (Homework!)
Vector: dot and cross product
v.u = |v||u|cos
Projection of v onto u and times
each other (scaler)!
vu = |v||u|sin w
|v||u|sin is the area of the
parallelogram. w  v and u
v

|v|cos

u
|v|sin
u
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c*
|c*|ccos,
=

 cc* = 1
*
|c*|
c
*
b  a  0 and b  c  0; b // c  a
*
: c  c*.
*
= 1/d001  ccos = d001
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
*
 The Weiss zone law or zone equation:
A plane (hkl) lies in a zone [uvw] (the plane contains
the direction [uvw]). d*hkl  (hkl)  d*hkl ruvw = 0 
(ha*  kb*  lc* )  (ua  vb  wc)  0
 hu  kv  lw  0
d*hkl  ruvw  d*hkl  (r1  r2 )  d*hkl  r2
nth plane
ruvw
Define the unit vector in the d*hkl
d*hkl
direction i,
d*hkl
i  *  d*hkl d hkl ; | d*hkl | 1
; ii 1
d hkl
| dhkl |
r2  i  ndhkl  r2  d*hkl dhkl  ndhkl  r2  d*hkl  n
d*hkl  ruvw  n  hu  kv  lw  n
r1
r2
uvw
 Reciprocal Lattice: Fourier transform of the spatial
wavefunction of the original lattice
 wave process (e. g. electromagnetic) in the crystal
Crystal: periodic
Physical properties function of a crystal  (r )   (r  T)
Crystal translation vector
Periodic function  Exponential Fourier Series
 (r)  k e2ikr   (r  T)  k e2ikr e2ikT
k
k
 e2ikT  1
T  ua  vb  wc
u, v, w: integer
Translation vectors of the original crystal lattice
 e
2ikT
 1 for all T
*
*
*
k

h
a

k
b

l
c
If
h, k, l: integer
Vectors of the reciprocal lattice
 k  T  (ha*  kb*  lc* )  (ua  vb  wc)
 (hu  kv  lw) always integer
 e2ikT  1
If k  (reciprocal) lattice ; T  original lattice!
Vice versa!
In crystallography
bc
ca
ab
*
*
a 
; b 
; c 
V
V
V
*
In SSP
bc
ca
ab
*
*
a  2
; b  2
; c  2
V
V
V
*
e
2ikT
ikT
e
k (in general): momentum space vector;
G: reciprocal lattice points
Proof: the reciprocal lattice of BCC is FCC
Use primitive translation vectors only
BCC
FCC
z
x
y
corner
F and B
1
a( xˆ  yˆ  zˆ )
2
1
a( xˆ  yˆ  zˆ )
2
1
a( xˆ  yˆ  zˆ )
2
Up and down
L and R
1
a( xˆ  yˆ )
2
1
a( yˆ  zˆ )
2
1
a( xˆ  zˆ )
2
BCC
1
1
1
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
a  a( x  y  z ); b  a( x  y  z ); c  a( xˆ  yˆ  zˆ )
2
2
2
bc
a 
V
xˆ  yˆ  zˆ
zˆ  xˆ  yˆ
*
yˆ  zˆ  xˆ
1
1
a2
3
b  c  a( xˆ  yˆ  zˆ)  a ( xˆ  yˆ  zˆ)  (2 yˆ  2 zˆ) ; V  a / 2
2
2
4
bc 1
a 
 ( yˆ  zˆ )
V
a
*
b* 
ca 1
 ( xˆ  zˆ )
V
a
c* 
ab 1
 ( xˆ  yˆ )
V
a
The vector set is the same as the FCC primitive
translation vector.
Unit of the reciprocal lattice is 1/length.
Mathematics of Interference
Sum of two waves:
R  A1 cos(t  1 )  A2 cos(t  2 )
assume A1 = A2 = A
R  A[cos(t  1 )  cos(t  2 )]
1
1
cos a  cos b  2 cos (a  b) cos (a  b)
2
2
1
1 2
R  2 A cos (1  2 ) cos( t   )
2
2 2
new amplitude AR
R
Geometrical analysis of Interference

A2

AR
2
 2 1

2 2


A1  A2
1
1  2
2 A cos (1   2 ) cos( t   )
2
2 2
2

1 A1
2
2

1
2
A
 1 
t term  rotation vector
1  2
2
1
A cos (1  2 )
2
Complex Wave Representation of Interference
A1 cos(t  1 )  A1ei (t 1 )

A2 cos(t  2 )  A2ei (t 2 )

R  A1ei (t 1 )  A2ei (t 2 )  A1ei1  A2ei2 eit
AR2  A1ei1  A2ei2
Rˆ  AR eiR
2
 A1ei1  A2ei2 A1ei1  A2ei2



 A12  A22  A1 A2ei (1 2 )  A1 A2ei (2 1 )
 A12  A22  2 A1 A2 cos(1  2 )
assume A1 = A2 = A
AR2  2 A2  2 A2 cos(1  2 )  2 A2 (1  cos(1  2 ))
cos2a  2 cos2 a  1
1
2
2
2 1
AR  2 A  2 cos (1  2 )  AR  2 A cos (1   2 )
2
2
Diffraction conditions and reciprocal lattices:
Theorem: The set of reciprocal lattice vectors G
determines the possible X-ray reflections.
A( x, t )  A0 cos(kx  t   )
k  2 / 
A(r, t )  A0 cos(k  r  t   )
| k | 2 / 
r
k
k
r
k
http://en.wikipedia.org/wiki/Plane_wave
Complex exponential form
U (r, t )  A0ei (krt  )
 A0 cos(k  r  t   )  iA0 sin(k  r  t   )
 A(r, t )  iA0 sin(k  r  t   )
U (r, t )  A0ei (krt  )  U0ei (krt )
U0  A0ei
Complex number
What happen to the time dependent term?
U (r, t )  U0ei t eikr
X-ray wavelength ~ 0.1 nm   ~ 3x1018 1/s
Detectors get the average intensity!
Detectors measured the intensity only!
| ei t |2  1
| eiA |2 | cos A  i sin A |2  cos2 A  sin 2 A  1
A lot of time, examining e ikr is enough!
Path difference
Phase angle
= (2/)rsin
= kr
Similarly,

 r cos(  ) dV
2
r
Phase angle
 k '  r
k'
k
k'
O
exp(ik  r)

exp(ik '  r)
F   exp(ik  r )n(r ) exp(ik '  r )dV
  n(r ) exp[i (k  k ' )  r ]dV   n(r ) exp(ik  r )dV
   nG exp[i (G  k )  r ]dV
G
 0 for G = k
otherwise, = 0
k  k '  k
Fourier expansion n(r)
The diffraction condition is G = k.  k + G = k’
k  G  k '  (k  G)2  (k ' )2  k 2
(k  G ) 2  k 2  2k  G  G 2  k 2  2k  G  G 2  0
G: reciprocal lattice, -G: reciprocal lattice? ____
2k  G  G 2  0  2k  G  G 2
| k | k  2 / 
| G | 2 / dhkl
Bragg condition?
| k | k  1 /   k
G
k
or | G | 1/ d
k
hkl
 2 
2k  G  2  | G | cos(90   )  G 2
  
2
 2 
 2d hkl sin   
 2  sin  
d hkl
  
dhkl
(hkl)
plane
Bragg law
G = k.
More geometric relation between direct lattice and
reciprocal lattice:
e1, e2, e3: contravariant basis vector of R3
covariant basis vectors e1, e2, e3 (reciprocal lattice)
e2  e3
e 3  e1
e1  e 2
e1  1 2 3 ; e2  1 2 3 ; e3  1 2 3
e e e
e e e
e e e
ei and ei are not normal, but mutually orthonormal:
e i  e j   ij
For any vector v:
q1  v  e1 ; q 2  v  e 2 ; q 3  v  e3
q1  v  e1 ; q2  v  e 2 ; q3  v  e3
v can be expressed in two (reciprocal) ways:
 v   qi ei  qi ei  q1e1  q2e 2  q3e3

i
i
1
2
3
 v   q ei  q ei  q e1  q e 2  q e3
Einstein’s summation
convention, omitting 
No proof here, but you can check whether these relation
is correct or not?
Use BCC or FCC lattice as examples, next page!
v   qi e i  qi e i  q1e1  q2 e 2  q3e 3

i
i
1
2
3
v   q ei  q ei  q e1  q e2  q e3
Use the BCC lattice as examples
1
1
1
a1  a( xˆ  yˆ  zˆ ); a2  a( xˆ  yˆ  zˆ ); a3  a( xˆ  yˆ  zˆ )
2
2
2
1
1
1
ˆ
ˆ
b1  ( y  zˆ )
b2  ( x  zˆ ); b3  ( xˆ  yˆ )
a
a
a
Assume v  2 xˆ  yˆ  zˆ
1
a a
q1  a1  v  a ( xˆ  yˆ  zˆ )  (2 xˆ  yˆ  zˆ )  a    a
2
2 2
1
q2  a2  v  a ( xˆ  yˆ  zˆ )  (2 xˆ  yˆ  zˆ )  2a
2
You can check
1
q3  a3  v  a ( xˆ  yˆ  zˆ )  (2 xˆ  yˆ  zˆ )  0
the other way
2
around.
1
1
2 xˆ  yˆ  zˆ  v  a ( yˆ  zˆ )  2a ( xˆ  zˆ )
a
a
q j  v  e j  q i (ei  e j )  (ei  e j )q i  g ij q i
where g ij  ei  e j
Similarly, q j  v  e j  qi ei  e j  (ei  e j )qi  g ij qi
Prove
g ij g jk   ki
ei  e j  ij  ei  g jk e k  g jk ei  e k  g jk g ik
gij: metric tensor in direct lattice
a, b, c and , ,  : direct lattice parameters
(standard definition of the Bravais lattice)
g11  a  a  a 2 ; g12  a  b  ab cos  ;  etc.
 a2

g ij   ab cos 
 ac cos 

ab cos 
b2
bc cos 
ac cos  

bc cos  
c 2 
det|gij| = V2.
V 2  a 2b 2 c 2 (1  2 cos  cos  cos   cos 2   cos 2   cos 2  )
 a* 
a
 *
 
 b   g ij  b 
 c* 
c
 
 
 a* 
 a* 
a
 *
 *
 
1
ij
 b   g ij  b   g  b 
 c* 
 c* 
c
 
 
 
Matrix inversion:
a

M  d
g

a b

d e
g h

c

f
i 
b
e
h
1
c
A B

1 
f 
D E
det(M ) 
k 
G H
e
A  
h
b
D  
h
f

k
b
G  
e
c

f
c

k 
T
C
A D G


1 
F 
B
E
H


det(
M
)
C F K 
K 


d f 

B  
g k 
a c

E  
g
k


d
C  
g
a
F  
g
a
H  
d
 a b

K  
d e
c

f
e

h
b

h 
Inverting the matrix gij.

bc sin 2  / a
c(cos  cos   cos  ) b(cos  cos   cos  ) 


abc
ij
2
g  2  c(cos  cos   cos  )
ac sin  / b
a (cos  cos   cos  ) 
V 

2
b
(cos

cos


cos

)
a
(cos

cos


cos

)
ab
sin

/
c


gij : metric tensor in reciprocal lattice
a*, b*, c* and *, *, *: reciprocal lattice parameters
*2
* *
*
* *
*

a
a
b
cos

a
c
cos



2
g ij   a *b* cos  *
b*
b*c* cos * 
 * *

*
* *
*
*2
 a c cos  b c cos 

c


These two are the same!
One gets relation like
bc sin  * ac sin  * ab sin 
a 
;b 
; c 
V
V
V
2
bc
sin

ac
sin

abc
(cos  cos   cos  )
* *
*
*
 a b cos  
cos  
V
V
V2
(cos  cos   cos  )
 cos  * 
sin  sin 
*
Similarly,
(cos  cos   cos  )
(cos  cos   cos  )
*
cos  
; cos  
sin  sin 
sin  sin 
*
……………..
d-spacing of (hkl) plane for any crystal system
d
*
hkl
d
*
hkl
2 *2
2 *2
2 *2
 (ha  kb  lc )  (ha  kb  lc )  h a  k b  l c
*
*
*
*
*
*
 2hka*b* cos  * 2klb*c* cos  *  2lhc*a * cos  *
2 2
2
2 2
2
2 2
2
1
2 b c sin 
2 a c sin 
2 a b sin 
h
k
l
2
2
2
d hkl
V
V
V2
abc 2 (cos  cos   cos  )
a 2bc(cos  cos   cos  )
 2hk
 2kl
2
V
V2
ab 2 c(cos  cos   cos  )
 2lh
V2
Example：FCC  BCC
(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
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
Interplanar spacing (defined based on unit cell)

1
h k l 
Cubic:

2
d hkl
a2

1
h2  k 2  l 2
Tetragonal:

 2
2
2
d hkl
a
c
1
h2 k 2 l 2
 2 2  2
Orthorombic:
2
d hkl a
b
c
2
2
2
1
4 h  hk  k
l

 2
Hexagonal:
2
2
d hkl 3
a
c
2
2
2
Get the metric tensor!
Perform the inversion of the matrix!
Comparing the inversion of the metric tensor in
direct lattice with the metric tensor in reciprocal lattice
 Geometrical relation between reciprocal lattice and
direct lattice can be obtained!