Reciprocal space. Ernesto Estévez Rams. IUCr International School on Crystallography, La Plata, 2014.
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1 Reciprocal space Ernesto Estévez Rams Instituto de Ciencia y Tecnología de Materiales (IMRE)-Facultad de Física Universidad de la Habana IUCr International School on Crystallography, La Plata, 2014.
2 2 / 40 Definition While constructing the stereographic projection we saw that the angle between planes could be also measured as the angles between the corresponding normal.
3 3 / 40 Definition We can add metric information to the normals by asociating the norm of the vector somehow to the interplanar distance.
4 3 / 40 Definition We can add metric information to the normals by asociating the norm of the vector somehow to the interplanar distance. The reciprocal vector r hkl is defined as a vector normal to plane (hkl) with length 1/d hkl
5 4 / 40 Definition r hkl ˆ n hkl r hkl = 1 d hkl
6 5 / 40 Equation of a plane (u, 0, 0) (0, v, 0) (0, 0, w) u = u a v = v b w = w c
7 6 / 40 Equation of a plane n = ( u v) ( v w) n = ( v w) + ( w u) + ( u v) n = vw( b c) + uw( c a) + uv( a b)
8 7 / 40 Equation of a plane n ( r u) = 0 n r = n u n = vw( b c) + uw( c a) + uv( a b) r = x a + y b + z c vwx( b c) a + uwy( c a) b + uvz( a b) c = uvw( b c) a
9 8 / 40 Equation of a plane vwx( b c) a + uwy( c a) b + uvz( a b) c = uvw( b c) a
10 8 / 40 Equation of a plane vwx( b c) a + uwy( c a) b + uvz( a b) c = uvw( b c) a [ ] [ ] [ x b c u a ( a + y c a b c) v a ( b + z a ] b b c) w a ( c = 1 b c)
11 Equation of a plane vwx( b c) a + uwy( c a) b + uvz( a b) c = uvw( b c) a [ ] [ ] [ x b c u a ( a + y c a b c) v a ( b + z a ] b b c) w a ( c = 1 b c) a = b c a ( b c) b c a = a ( b c) c = a b a ( b c) 8 / 40
12 9 / 40 Equation of a plane a = b c a ( b c) b c a = a ( b c) c = a b a ( b c) a a = 1 a b = 0 a c = 0 b a = 0 b b = 1 b b = 0 c a = 0 c b = 0 c b = 1
13 10 / 40 Equation of a plane [ ] [ ] [ x b c u a ( a + y c a b c) v a ( b + z a ] b b c) w a ( c = 1 b c)
14 10 / 40 Equation of a plane [ ] [ ] [ x b c u a ( a + y c a b c) v a ( b + z a ] b b c) w a ( c = 1 b c) (x a) (h a ) + (y b) (k b ) + (z c) (l c ) = 1 Equation of the plane hx + ky + lz = 1
15 11 / 40 Some history Gibbs ( ) Gibbs introduced the reciprocal base { a j } from the condition a i a j =K δ ij
16 12 / 40 Reciprocal vector (x a) (h a ) + (y b) (k b ) + (z c) (l c ) = 1 can be written as where hx + ky + lz = 1 r xyz r hkl = 1 r hkl = h a + k b + l c if r hkl ˆ n hkl r hkl = 1 d hkl
17 13 / 40 Reciprocal vector r hkl ( r p r hkl ) = 0 p r hkl 2 = r r hkl = 1 Now by construction p = 1 r hkl 2 p r hkl = d hkl r hkl = 1 d hkl
18 14 / 40 Reciprocal vector Hexagonal crystal basis Find the reciprocal basis. Write the reciprocal metric tensor G. Calculate G G.
19 15 / 40 Reciprocal vector a = b = a c = 1 c α = β = π 2 γ = π 3
20 16 / 40 Reciprocal vector G = a a a a /c 2
21 17 / 40 Reciprocal vector Tetragonal crystal basis Find the reciprocal basis. Write the reciprocal metric tensor G. Calculate G G.
22 18 / 40 Reciprocal vector FCC crystal basis Find the reciprocal basis.
23 19 / 40 Reciprocal vector FCC crystal basis
24 20 / 40 Reciprocal vector FCC crystal basis
25 21 / 40 Reciprocal vector FCC crystal basis
26 22 / 40 Reciprocal vector FCC crystal basis
27 23 / 40 Reciprocal vector FCC crystal basis BCC reciprocal basis
28 24 / 40 Reciprocal vector G G = I
29 25 / 40 Reciprocal vector Hexagonal crystal basis From the reciprocal metric tensor, find the interplara distance of the plane (hkl).
30 26 / 40 Reciprocal vector 1 d 2 hkl = r r = (h k l) a a a a /c 2 h k l
31 27 / 40 Reciprocal vector 1 d 2 hkl = r r = 4 h 2 + hk + k 2 3 a 2 + l2 c 2
32 28 / 40 Coordinates transformations r = ( a b c) from where x y z Basis transformation x y z = F = ( a b c ) x y z x y z = ( a b c ) F x y z ( a b c ) = ( a b c) F 1
33 29 / 40 Metric tensor transformations r r = (x y z) G x y z from where calling M = (F T ) 1 Metric tensor transformation = (x y z ) G x y z (x y z) F T G F = x y z G = M G M T
34 30 / 40 Metric tensor transformations a b c = G a b c a b c = G a b c
35 31 / 40 Transformation in reciprocal space r = (h k l) (h k l) G 1 a b c a b c = (h k l ) = (h k l) G 1 F T a b c a b c (h k l) G 1 F T G = = a b c
36 32 / 40 Transformation in reciprocal space r = (h k l) a b c = (h k l) G 1 F T G a b c Now and finally G = M G M T = (F T ) 1 G F 1 Reciprocal coordinates (h k l ) = (h k l) M T = (h k l) F 1
37 Point symmetry transformation in reciprocal space 33 / 40 Let R be a point symmetry operation acting over the coordinates of the atoms, then according to the transformation rules already derived (h e k e l e ) = (h k l) R 1 and (h e k e l e ) is a plane symmetry related to (h k l). Therefore: Point symmetry in reciprocal space R = R 1
38 Point symmetry transformation in reciprocal space 34 / 40 Example: Three fold axis 3 [0 0 1] : (x y z) ( y x y z) (y x x z)
39 Point symmetry transformation in reciprocal space 34 / 40 Example: Three fold axis 3 [0 0 1] : (x y z) ( y x y z) (y x x z) R =
40 Point symmetry transformation in reciprocal space 34 / 40 Example: Three fold axis 3 [0 0 1] : (x y z) ( y x y z) (y x x z) R = R =
41 Point symmetry transformation in reciprocal space 34 / 40 Example: Three fold axis 3 [0 0 1] : (x y z) ( y x y z) (y x x z) R = R = (h k l) ( h k h l) (k h k l)
42 35 / 40 Symmetry group in reciprocal space Symmetry group in reciprocal space {R } = {R}
43 36 / 40 Symmetry group in reciprocal space Grupo espacial C2/m Symmetry operation: x y z x y z x ȳ z x ȳ z
44 37 / 40 Symmetry group in reciprocal space Grupo espacial C2/m Symmetry operation: x y z x y z x ȳ z x ȳ z Equivalent planes: h k l h k l h k l h k l
45 38 / 40 Symmetry group in reciprocal space Grupo espacial P 222 Symmetry operation: x y z x ȳ z x y z x ȳ z
46 Fourier transform and reciprocal vectors 39 / 40 Fourier transform Γ [f( r)] = f( r ) = F ( r ) f( r) exp ( 2πi r r)d r where r = x a + y b + z c r = x a + y b + z c and therefore r r = x x + y y + z z
47 40 / 40 Fourier transform of the lattice Definition (Dirac comb) L = u,v,w= δ( r u a v b w c) Γ [L( r)] = = u,v,w= u= = e ( 2πiu r a) δ( r u a v b w c) exp ( 2πi r r)d r u,v,w= v= exp ( 2πi r (u a + v b + w c)) e ( 2πiv r b) w= e ( 2πiw r c)
48 41 / 40 Fourier transform of the lattice Dirac comb 1 a n= exp (2πinx/a) = m= δ(x ma)
49 41 / 40 Fourier transform of the lattice Dirac comb 1 a n= exp (2πinx/a) = m= δ(x ma) Γ [L( r)] = e ( 2πiu r a) e ( 2πiv r b) u= v= w= e ( 2πiw r c)
50 41 / 40 Fourier transform of the lattice Dirac comb 1 a n= exp (2πinx/a) = m= δ(x ma) Γ [L( r)] = e ( 2πiu r a) e ( 2πiv r b) u= v= w= e ( 2πiw r c) L ( r ) = δ( r a h)δ( r b k)δ( r c l) h,k,l= = h,k,l= δ( r h a )δ( r k b )δ( r l c )
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