Wave diffraction and the reciprocal lattice
Dept of Phys
M.C. Chang
• Braggs’ theory of diffraction
• Reciprocal lattice
• von Laue’s theory of diffraction
Braggs’ view of the diffraction (1912, father and son)
Treat the lattice as a stack of lattice planes
25
1915
• mirror-like reflection from crystal planes when
2dsinθ = nλ
• Difference from the usual mirror reflection:　λ > 2d, no reflection
λ < 2d, reflection only at certain angles
• Measure λ, θ → get distance between crystal planes d
2dsinθ = nλ
Powder method
For more, see
www.xtal.iqfr.csic.es/Cristalografia/parte_06-en.html
• Braggs’ theory of diffraction
• Reciprocal lattice
• von Laue’s theory of diffraction
Fourier transform of the electron density of a 1-dim lattice
ρ(x)
Lattice in
x real space
a
ρ ( x) = ∑ eikx ρ (k ), k = n
k
2π
, n∈Z
L
ρ ( x + a) = ∑ eikx eika ρ (k ) = ρ ( x)
∑a e
Therefore, eika = 1
→ ak = 0 for ∀k
k
→ k =n
ρ ( x) =
n =−∞
-g
=0
k
eingx ρ n
ρn
-2g
ikx
2π
≡ ng
a
∞
∑
k
0
g
2g
Lattice in
momentum space
k (reciprocal lattice)
simplest example
important
Reciprocal lattice (倒晶格 )
(direct) lattice
primitive vectors a1,a2,a3
Def. 1
reciprocal lattice
primitive vectors b1,b2,b3
b1 ⋅ a1 = 2π , b1 ⋅ a2 = b1 ⋅ a3 = 0,
b2 ⋅ a2 = 2π , b2 ⋅ a3 = b2 ⋅ a1 = 0,
b3 ⋅ a3 = 2π , b3 ⋅ a1 = b3 ⋅ a2 = 0.
Def. 2
b1 ∝ a2 × a3 because of orthogonality,
then use b1 ⋅ a1 = 2π
to determine the constant.
a2 × a3
,
a1 ⋅ (a2 × a3 )
a3 × a1
,
b2 = 2π
a1 ⋅ (a2 × a3 )
a1 × a2
.
b3 = 2π
a1 ⋅ (a2 × a3 )
b1 = 2π
• The reciprocal of a reciprocal lattice is the direct lattice
(obvious from Def.1)
Ex: Simple cubic lattice
z
z
2π/a
a
y
y
x
x
a1 = axˆ ,
a2 = ayˆ ,
b1 = 2π
a2 × a3
2π
xˆ ,
=
a1 ⋅ (a2 × a3 ) a
b2 = 2π
a3 × a1
2π
yˆ ,
=
a1 ⋅ (a2 × a3 ) a
b3 = 2π
a1 × a2
2π
zˆ.
=
a1 ⋅ (a2 × a3 ) a
a3 = azˆ.
a1 ⋅ (a2 × a3 ) = a
3
⎛ 2π ⎞
b1 ⋅ (b2 × b3 ) = ⎜
⎟
⎝ a ⎠
3
BCC lattice
FCC lattice
z
z
4π/a
y
a
y
x
x
a
a1 = ( xˆ + yˆ ) ,
2
a
a2 = ( yˆ + zˆ ) ,
2
a
a3 = ( zˆ + xˆ ) .
2
a3
a1 ⋅ (a2 × a3 ) =
4
b1 = 2π
a2 × a3
4π 1
=
( xˆ + yˆ − zˆ ) ,
a1 ⋅ ( a2 × a3 )
a 2
b2 = 2π
a3 × a1
4π 1
=
( − xˆ + yˆ + zˆ ) ,
a1 ⋅ ( a2 × a3 )
a 2
b3 = 2π
4π 1
a1 × a2
=
( xˆ − yˆ + zˆ ) .
a1 ⋅ ( a2 × a3 )
a 2
1 ⎛ 4π ⎞
b1 ⋅ (b2 × b3 ) = ⎜
⎟
2⎝ a ⎠
3
Two simple properties:
1.
R = n1a1 + n2a2 + n3a3 (n1 , n2 , n3 ∈ Z ) ∈direct lattice
G = k1b1 + k2b2 + k3b3 ( k1 , k2 , k3 ∈ Z ) ∈ reciprocal lattice
⇒ G ⋅ R = 2π (n1k1 + n2 k2 + n3k3 ) = 2π × int eger.
(∴ exp(iG ⋅ R)
2.
)
is always equal to 1
Conversely, assume G．R=2π×integer for all R,
then G must be a reciprocal lattice vector.
if
G ⋅ a1 = 2π h,
G ⋅ a2 = 2π k ,
G ⋅ a3 = 2π l ,
then
(h, k , l ∈ Z )
G = hb1 + kb2 + lb3 (≡ Ghkl ).
important
If f(r) has lattice translation symmetry, that is, f(r)=f(r+R)
for any lattice vector R, then it can be expanded as,
f (r ) = ∑ eiG⋅r f G , where G is the reciprocal lattice vector.
G
Pf:
Fourier expansion,
f (r ) = ∑ eik ⋅r f (k )
k
f (r + R) = ∑ eik ⋅r eik ⋅R f (k ) = f (r )
k
Therefore, e
ik ⋅ R
= 1 for ∀ R
k = hb1 + kb2 + lb3
= Ghkl
∀ h, k , l
The expansion above is very general, it applies to
• all types of periodic lattice (e.g. bcc, fcc, tetragonal, orthorombic...)
• in all dimensions (1, 2, and 3)
All you need to do is to find out the reciprocal lattice vectors G.
Summary
The reciprocal lattice is useful in
• Fourier decomposition of a lattice-periodic function
• von Laue’s diffraction condition k’ = k + G (later)
Direct lattice
Reciprocal lattice
cubic (a)
cubic (2π/a)
fcc (a)
bcc (4π/a)
bcc (a)
fcc (4π/a)
hexagonal (a,c)
hexagonal (4π/√3a,2π/c)
and rotated by 30 degrees
(See Prob.2)
important
Geometrical relation between Ghkl vector and (hkl) planes
( h, k , l ) planes ⊥ Ghkl ≡ hb1 + kb2 + lb3
⎧
⎪⎪ v1 =
⎨
⎪v =
⎪⎩ 2
Pf:
m
m
a1 − a3
h
l
m
m
a2 − a3
k
l
m/l
a3
v1
a1
⇒
Ghkl ⋅ v1 = 0
v2
a2
m/h
Ghkl ⋅ v2 = 0
∴ Ghkl ⊥ (h, k , l )-plane
• When the direct lattice rotates, its reciprocal lattice rotates
the same amount as well.
m/k
important
Inter-plane distance
Ghkl
(hkl) lattice
planes
R
dhkl
Ghkl ⋅ R = 2π ( hn1 + kn2 + ln3 )
= 2π n
⇒
Given h,k,l, and n, one can
always find a lattice vector R
arXiv:0805.1702 [math.GM]
(n could be any integer)
Gˆ hkl ⋅ R = 2π n / | Ghkl |
∴ inter-plane distance d hkl = 2π / | Ghkl |
For a cubic lattice
Ghkl = hb1 + kb2 + lb3
2π
( hxˆ + kyˆ + lzˆ )
a
a
∴ d hkl =
h2 + k 2 + l 2
=
• In general, planes with higher index have smaller inter-plane distance
• Braggs’ theory of diffraction
• Reciprocal lattice
• von Laue’s theory of diffraction
Scattering from an array of atoms (Von Laue, 1912)
• The same analysis applies to EM wave, electron wave,
neutron wave… etc.
First, scattering off an atom at the origin:
1914
eikr
scattered wave ψ (r ) ∼ f a (θ )
for large r
r
• Atomic form factor: Fourier transform of atom charge distribution n(ρ)
原子結構因數
f a (Δk ) = ∫ dVe− iΔk ⋅ρ n( ρ ), Δk ≡ k '− k
The atomic form factor
f a (q ) = ∫ dVe − iq⋅ρ n( ρ )
(See Prob.6)
10 electrons
tighter
~ Kr 36
~ Ar 18
http://capsicum.me.utexas.edu/ChE386K/docs/29_electron_atomic_scattering.ppt
• Scattering off an atom not at the origin
eik |r − R| ik ⋅R
e
scattered wave ψ (r ) ∝ f a (Δk )
|r −R|
| r − R |≈ r − rˆ ⋅ R
1
1
≈ + O ( r −2 )
|r −R| r
R
eikr − iΔk ⋅R
e
∴ψ (r ) ∝ f a (Δk )
r
• Two-atom scattering
(
eikr − iΔk ⋅R1
e
ψ (r ) ∝ f a
+ e − iΔk ⋅R2
r
A relative
phase w.r.t.
an atom at
the origin
)
N-atom scattering: one dimensional lattice
ψ ≈ e − iΔk ⋅0 + e − iΔk ⋅a + e −iΔk ⋅2 a +e −iΔk ⋅3a + e −iΔk ⋅4 a
2π
N −1
ψ ≈ ∑ e− iΔk ⋅na
|ψ|2
n =0
1 − exp(−iN Δk ⋅ a ) N >>1
⎯⎯⎯
→ ∑ δ Δk ⋅a ,2π h
1 − exp(−iΔk ⋅ a )
h
For large N , ψ is nonzero only when
=
Δk ⋅ a = 2π h (h is an integer )
(see Prob. 4)
http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html
aΔk
important
N-atom scattering (3D lattice, neglect multiple scatterings)
For a simple lattice with no basis,
ψ (r ) ∝ f a ∑ e − iΔk ⋅R , R = n1a1 + n2 a2 + n3a3 .
R
The lattice-sum can be separated,
∑e
− iΔk ⋅ R
R
⎞
⎛
⎞⎛
⎞⎛
− iΔk ⋅n3 a3
− iΔk ⋅n1a1
− iΔk ⋅n2 a2
= ⎜∑e
⎟⎟
⎟⎜ ∑ e
⎟ ⎜⎜ ∑ e
⎝ n1
⎠⎝ n2
⎠ ⎝ n3
⎠
≠ 0 only when
Δk ⋅ a1 = 2π h,
Δk = Ghkl
Δk ⋅ a2 = 2π k ,
Laue‘s diffraction
condition
Δk ⋅ a3 = 2π l.
⇒
∑e
R
− iΔk ⋅ R
= N ∑ δ Δk ⋅a ,2π hδ Δk ⋅a ,2π k δ Δk ⋅a ,2π l = N ∑ δ Δk ,G
hkl
1
Number of atoms
in the crystal
2
3
Ghkl
hkl
important
Previous calculation is for a simple lattice, now we
calculate the scattering from a crystal with basis
dj : location of the j-th atom in a unit cell
Eg.,
a
d1 = 0, d 2 = ax
atomic form factor
for the j-th atom
⎛ p
− iΔk ⋅( R + d j ) ⎞
ψ (r ) ∝ ∑ ⎜ ∑ f aj e
⎟
R ⎝ j =1
⎠
p
⎛
⎛
⎞
− iΔk ⋅d j ⎞
− iΔk ⋅ R
= ⎜∑e
⎟
⎟ ⎜ ∑ f aj e
⎝ R
⎠ ⎝ j =1
⎠
= N ∑ δ Δk ,G ⋅ S (Δk )
hkl
Ghkl
Structure factor
(of the basis)
p
S (h, k , l ) ≡ ∑ f aj e
j =1
− iGhkl ⋅d j
Example:
The structure factor for fcc lattice
(= cubic lattice with a 4-point basis)
p
S (h, k , l ) = ∑ f aj e
d1 = 0,
a a
a a
a a
d2 = 1 + 2 , d3 = 2 + 3 , d4 = 3 + 1
2 2
2 2
2 2
− iGhkl ⋅d j
j =1
Cubic lattice
Ghkl = hb1 + kb2 + lb3
S (h, k , l ) = f a ⎡⎣1 + e − iπ ( h + k ) + e − iπ ( k +l ) + e − iπ ( l + h ) ⎤⎦
= 4fa when h,k,l are all odd or all even
= 0 otherwise
Eliminates all the points in the
reciprocal cubic lattice with S=0.
The result is a bcc lattice, as it
should be!
Reciprocal of cubic lattice
Atomic form factor and intensity of diffraction
fK ～ fCl
cubic lattice with
lattice const. a/2
fK ≠ fBr
fcc lattice
h,k,l all even or
all odd
Summary
• Find out the structure factor of the honeycomb structure, then draw its
reciprocal structure. Different points in the reciprocal structure may have
different structure factors. Draw a larger dots if the associated |S|2 is larger.
Laue’s diffraction condition
k’ = k + Ghkl
• Given an incident k, want to find a k’ that satisfies this condition
(under the constraint |k’|=|k|)
• One problem: there are infinitely many Ghkl’s.
• It’s convenient to solve it graphically using the Ewald construction
(Ewald 構圖法)
More than one (or none)
solutions may be found.
k
k’
G
Reciprocal lattice
http://capsicum.me.utexas.edu/ChE386K/docs/28_The_Laue_Experiment.ppt
Laue’s condition = Braggs’ condition
• From the Laue condition, we have
k ⋅ Ghkl = −
Ghkl
Ghkl
2
k
• Given k and Ghkl, we can find the diffracted wave vector k’
Ghkl
k
θ
θ’
k’
a (hkl)-lattice plane
• It’s easy to see that θ = θ’ because |k|=|k’|.
By using 2k sin θ = Ghkl =
and
k=
2π
λ
2π
n
d hkl
Integer multiple
of the smallest
G is allowed
,
⇒ 2d hkl sin θ = nλ.
Bragg’s diffraction condition
Another view of the Laue condition
k ⋅ Gˆ hkl
G
= hkl
2
Ghkl
k
∴ The k vector that points to the plane bi-secting a Ghkl vector
will be diffracted.
Reciprocal lattice
Brillouin zone
Def. of the first BZ
A BZ is a primitive unit cell
of the reciprocal lattice
Triangle lattice
direct lattice
reciprocal lattice
BZ
• The first BZ of fcc lattice (its reciprocal lattice is bcc lattice)
4π/a
• The first BZ of bcc lattice (its reciprocal lattice is fcc lattice)
z
4π/a
y
x