Microsoft PowerPoint - chap02.pptM.C. Chang
Braggs’ view of the diffraction (1912, father and son)
Treat the lattice as a stack of lattice planes
• mirror-like reflection from crystal planes when 2dsinθ = nλ
• Measure λ, θ → get distance between crystal planes d
• Difference from the usual mirror reflection: λ > 2d, no reflection
λ < 2d, reflection only at certain angles
1915
25
simplest example
Fourier transform of the electron density of a 1-dim lattice
a
ρ(x)
x
2( ) ( ), ,
x a e e k x
e
Lattice in real space
Def. 2
1 1 2
π
1 2 3
1 2 3
2 3 1
1 2 3
3 1 2
1 2 3
2 0
2 0
2 0
Def. 1
• The reciprocal of a reciprocal lattice is the direct lattice (obvious from Def.1)
important
a a a a
b b b a
a ab x y z a a a a
a ab x y z a a a a
a ab x y z a a a a
ππ
ππ
ππ
b b b a π ⋅ × =

x
y
z
R n a n a n a n n n Z
G k b k b k b k k k Z
G R n k n k n k
= + + ∈ ∈
= + + ∈ ∈
⇒ ⋅ = + + = ×
1 1 2 2 3 32 2
( , , )
( , , )
G hb kb lb G
π
π
π
then G must be a reciprocal lattice vector.
Two simple properties:
1.
2.
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,
, where G is the reciprocal lattice vector.( ) iG r G
G
Pf:
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.
1 2 3
f r R e e f k f r
e R
The reciprocal lattice is useful in
• Fourier decomposition of a lattice-periodic function
• von Laue’s diffraction condition k’ = k + G (later)
hexagonal (a,c)
Summary
1 2 3( , , ) planes hklh k l G hb kb lb⊥ ≡ + +
Geometrical relation between Ghkl vector and (hkl) planes
Pf:
a1
a2
a3
m/h
m/k
m/l
v1
v2
1
2
0
0
( , , )-plane
hkl
hkl
hkl
m mv a a h l m mv a a k l
G v
G v
= − = −
⋅ = ⇒
⋅ =
∴ ⊥
• When the direct lattice rotates, its reciprocal lattice rotates the same amount as well.
important
ˆ 2 / | |
hkl
G R
hkl
hkl
• In general, planes with higher index have smaller inter-plane distance
Given h,k,l, and n, one can always find a lattice vector R arXiv:0805.1702 [math.GM]
important
Scattering from an array of atoms (Von Laue, 1912)
• The same analysis applies to EM wave, electron wave,
neutron wave… etc.
ψ θ∼
First, scattering off an atom at the origin:
( ) ( ), 'i k af k dVe n k k kρ ρ− Δ ⋅Δ = Δ ≡ −∫
• Atomic form factor: Fourier transform of atom charge distribution n(ρ)
1914

http://capsicum.me.utexas.edu/ChE386K/docs/29_electron_atomic_scattering.ppt
• Two-atom scattering
( )
| |
a
r R r r R
O r rr R
ψ
ψ
er f e e r
ψ − Δ ⋅ − Δ ⋅∝ +
A relative phase w.r.t. an atom at the origin
ψ ≈ +− ⋅ − ⋅e ei k i k aΔ Δ0 + − ⋅e i k aΔ 2 + − ⋅e i k aΔ 3 + − ⋅e i k aΔ 4 http://hyperphysics.phy-astr.gsu.edu/hbase/hfram
e.htm l
aΔk
(see Prober . 4))
N
2 ,
2 ,
2 .
hkl
hkl
i k R k a h k a k k a l k G
hklR G
≠
⇒ = =∑ ∑ ∑
ψ ( ) , .r f e R n a n a n aa i k R
R
The lattice-sum can be separated,
Number of atoms in the crystal
3 31 1 2 2
1 2 3
i k n ai k n a i k n ai k R
n n nR
∑ ∑ ∑ ∑
Laue‘s diffraction conditionhklk GΔ =
important
1
=
∝
∑ ∑
dj : location of the j-th atom in a unit cell
Eg.,
Previous calculation is for a simple lattice, now we calculate the scattering from a crystal with basis
1
, ( ) hkl
k
j
atomic form factor for the j-th atom
d d ax1 20= =,
=
≡ ∑
important
Reciprocal of cubic lattice = 4fa when h,k,l are all odd or all even
= 0 otherwise
G hb kb lbhkl = + +1 2 3
( ) ( ) ( )( , , ) 1 i h k i k l i l h aS h k l f e e eπ π π− + − + − + = + + +
1 ( , , ) hkl j
p iG d
=
= ∑
d a a d a a d a a 1
2 1 2
3 2 3
4 3 1
=
= + = + = +
,
, ,
Example: The structure factor for fcc lattice (= cubic lattice with a 4-point basis)
Eliminates all the points in the reciprocal cubic lattice with S=0. The result is a bcc lattice, as it should be!
Cubic lattice
fcc lattice h,k,l all even or all odd
fK fCl
fK ≠ fBr
cubic lattice with lattice const. a/2
• 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.
Summary
• One problem: there are infinitely many Ghkl’s.
• It’s convenient to solve it graphically using the Ewald construction
• Given an incident k, want to find a k’ that satisfies this condition
(under the constraint |k’|=|k|)
k
k G G hkl
Ghkl
k
• Given k and Ghkl, we can find the diffracted wave vector k’ Ghkl
k k’ a (hkl)-lattice planeθ θ’
• It’s easy to see that θ = θ’ because |k|=|k’|. 2By using 2 sin
2and ,
Integer multiple of the smallest G is allowed
∴ The k vector that points to the plane bi-secting a Ghkl vector will be diffracted.
ˆ 2 hkl
hkl Gk G⋅ =
Reciprocal lattice
reciprocal lattice
Triangle lattice
direct lattice
A BZ is a primitive unit cell of the reciprocal lattice
• 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)
x y