Wave diffraction and the reciprocal lattice

Transcription

1 Wave diffraction and the reciprocal lattice Dept of Phys M.C. Chang

2 Braggs theory of diffraction Reciprocal lattice von Laue s theory of diffraction

3 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 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

4 2dsinθ = nλ

5 Powder method For more, see

6 Braggs theory of diffraction Reciprocal lattice von Laue s theory of diffraction

7 Fourier transform of the electron density of a 1-dim lattice ρ(x) a ikx 2π ρ( x) = e ρ( k), k = n, n Z L ikx ika ρ( x+ a) = e e ρ( k) = ρ( x) ika Therefore, e = 1 2π k = n ng a ρ( x) = k n= k e ρ n ingx ρ -2g -g 0 g 2g n x k Lattice in real space k ae k ikx = 0 a = 0 for k k Lattice in momentum space (reciprocal lattice) simplest example

8 Reciprocal lattice ( 倒晶格 ) important (direct) lattice reciprocal lattice primitive vectors a 1,a 2,a 3 primitive vectors b 1,b 2,b 3 Def. 1 b1 a1 = 2π, b1 a2 = b1 a3 = 0, b2 a2 = 2π, b2 a3 = b2 a1 = 0, b a = 2π, b a = b a = b a a because of orthogonality, then use b a = π to determine the constant. Def. 2 a2 a3 b1 = 2π a1 ( a2 a3), a3 a1 b2 = 2π a1 ( a2 a3), a1 a2 b3 = 2π a ( a a ) The reciprocal of a reciprocal lattice is the direct lattice (obvious from Def.1)

9 Ex: Simple cubic lattice z z a y 2π/a y x a1 = axˆ, a2 = ayˆ, a3 = azˆ. a ( a a ) = a x b 1 b 2 b 3 a a 2π = π = ˆ a1 ( a2 a3) a a a 2π = π = ˆ a1 ( a2 a3) a a a 2π = π = ˆ a ( a a ) a x, y, z π b1 ( b2 b3) = a 3

10 FCC lattice z BCC lattice z a x y 4π/a x y a a1 = ( xˆ+ yˆ ), 2 a a2 = ( yˆ + zˆ ), 2 a a3 = ( zˆ + xˆ ). 2 a a1 ( a2 a3) = 4 3 a a 4π 1 b1 = π = ( x ˆ + y ˆ z ˆ ) a1 ( a2 a3) a 2 a a 4π 1 b2 = π = x+ y+ z a1 ( a2 a3) a 2 a a 4π 1 b3 = π = ( x ˆ y ˆ + z ˆ ) a ( a a ) a , ( ˆ ˆ ˆ) 3 1 2, π b1 ( b2 b3) = 2 a 3

11 Two simple properties: R= na 1 1+ n2a2+ na 3 3( n1, n2, n3 Z) direct lattice G= kb 1 1+ k2b2+ k3b3( k1, k2, k3 Z) reciprocal lattice G R= 2π( nk 1 1+ n2k2+ n3k3) = 2π integer. exp( ig R ) is always equal to 1 ( ) Conversely, assume G.R=2π integer for all R, then G must be a reciprocal lattice vector. if Ga 1 = 2 πh, Ga 2 = 2 πk, Ga 3 = 2 πl, (,, Z) then G= hb + kb + lb ( G )

12 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, ig r f( r) = e f G, where G is the reciprocal lattice vector. Pf: G Fourier expansion, ik r f( r) = e f( k) k ik r ik R f ( r + R) = e e f( k) = f( r) Therefore, e k ik R = 1 for R k = hb1+ kb2 + lb3 = G h, k, l important 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.

13 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)

14 important Geometrical relation between G vector and () planes (,, ) planes G hb1+ kb2 + lb3 Pf: m m v = a a h l m m v = a a k l G v = 1 G v2 = 0 0 m/h m/l v 2 a v 3 1 a 1 a 2 m/k G ( h, k, l)-plane When the direct lattice rotates, its reciprocal lattice rotates the same amount as well.

15 Inter-plane distance important () lattice planes G R d For a cubic lattice G R = 2π hn + kn + ln ( ) = 2 π n ( n could be any integer) Gˆ R= 2 π n/ G inter-plane dista nce d = 2 π / G 2π = a G = hb + kb + lb d = ( hxˆ+ kyˆ+ lzˆ ) a h + k + l 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: [math.gm]

16 Braggs theory of diffraction Reciprocal lattice von Laue s theory of diffraction

17 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 ikr e scattered wave ψ( r ) fa ( θ) for large r r Atomic form factor: Fourier transform of atom charge distribution n(ρ) 原子結構因數 ( ) i k Δ ρ f Δ k = dve n( ρ), Δk k ' k a

18 The atomic form factor iq fa ( q) dve ρ = n( ρ) (See Prob.6) 10 electrons tighter ~ Kr 36 ~ Ar 18

19 Scattering off an atom not at the origin ik r R e scattered wave ψ ( r) fa ( Δk) e r R r R r rˆ R O( r ) r R r ikr e ψ ( r) fa ( Δk) e r Δ i k R ik R A relative phase w.r.t. an atom at the origin R Two-atom scattering ikr e ψ ( r) fa e + e r ( Δ i k R Δ i k R) 1 2

20 N-atom scattering: one dimensional lattice iδk a + e iδk 2 a + e iδk 3 a + e iδk 4a iδk 0 ψ e + e ψ N 1 n= 0 e Δ i k na 1 exp( inδk a) 1 exp( iδk a) N >> 1 = Δka,2π h For large N, ψ is nonzero only when Δk a = 2 π h ( h is an integer ) (see Prob. 4) h δ 2π ψ 2 aδk

21 N-atom scattering (3D lattice, neglect multiple scatterings) important For a simple lattice with no basis, i k R ψ ( r) f e, R = na + n a + n a. R a Δ R The lattice-sum can be separated, R Δ i k R e e e e Δ i k n1a1 i k n2a2 i k n3a3 = Δ Δ n1 n 2 n 3 0 only when Δk a1 = 2 π h, Δk a2 = 2 π k, Δk a = 2 πl. 3 Δ i k R e N δ δ δ N δ Δka 1,2 πh Δka 2,2 πk Δka 3,2 πl ΔkG, G = = Number of atoms in the crystal Δ k = G Laue s diffraction condition

22 Previous calculation is for a simple lattice, now we calculate the scattering from a crystal with basis important d j : location of the j-th atom in a unit cell Eg., a d = 0, d = ax 1 2 p Δ i k ( R+ d j ) ψ ( r ) faje R j= 1 = p Δ i k R Δ i k dj e faje R j= 1 = N δ S( Δk) kg, G hl k Δ atomic form factor for the j-th atom Structure factor (of the basis) ig d j S (,, ) fe aj p j= 1

23 Example: The structure factor for fcc lattice (= cubic lattice with a 4-point basis) d1 = 0, a1 a2 a2 a3 a3 a1 d2 = +, d3 = +, d4 = G = hb1+ kb2 + lb3 S (,, ) = f a 1+ e + e + e iπ( h+ k) iπ( k+ l) iπ( l+ h) = 4f a when h,k,l are all odd or all even = 0 otherwise ig d j S (,, ) = fe aj p j= 1 Cubic lattice Reciprocal of cubic lattice Eliminates all the points in the reciprocal cubic lattice with S=0. The result is a bcc lattice, as it should be!

24 Atomic form factor and intensity of diffraction f K ~ f Cl cubic lattice with lattice const. a/2 f K f Br fcc lattice h,k,l all even or all odd

25 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.

26 Laue s diffraction condition k = k + G Given an incident k, want to find a k that satisfies this condition (under the constraint k = k ) One problem: there are infinitely many G s. It s convenient to solve it graphically using the Ewald construction (Ewald 構圖法 ) k k G More than one (or none) solutions may be found. Reciprocal lattice

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28 Laue s condition = Braggs condition From the Laue condition, we have k G = G 2 k G Given k and G, we can find the diffracted wave vector k G k k θ θ a()-lattice plane It s easy to see that θ = θ because k = k. 2π By using 2ksinθ = G = n d 2π and k =, λ 2d sin θ = nλ. Bragg s diffraction condition Integer multiple of the smallest G is allowed

29 Another view of the Laue condition k Gˆ = G 2 k G The k vector that points to the plane bi-secting a G vector will be diffracted. Reciprocal lattice

30 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

31 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 x y