Experimental Determination of Crystal Structure Branislav K. Nikolić Department of Physics and Astronomy, University of Delaware, U.S.A. PHYS 624: Introduction to Solid State Physics http://www.physics.udel.edu/~bnikolic/teaching/phys624/phys624.html 1
Principles of Diffraction How do we learn about crystalline structures? Answer: Diffraction: Send a beam of particles (of de Broglie wavelength or radiation with a wavelength λ a comparable to characteristic length scale of the lattice ( a twice the atomic or molecular radii of the constituents). λ=h p EXPERIMENT: Identify Bragg peaks which originate from a coherent addition of scattering events in multiple planes within the bulk of the solid. 2
Principles of Diffraction in Pictures Figure 1: Scattering of waves or particles with wavelength of roughly the same size as the lattice repeat distance allows us to learn about the lattice structure. Coherent addition of two particles or waves requires that condition), and yields a scattering maximum on a distant screen. 2d sinθ = λ (the Bragg 3
Available Particles for Diffraction Experiments 4
Bad Particles for Diffraction Not all particles with de Broglie wavelength will work for this application For example, most charged particles cannot probe the bulk properties of the crystal, since they lose energy to the scatterer very quickly: For non-relativistic electron scattering into a solid with The distance at which initial energy is lost is: NOTE: Low energy electron diffraction can be used to study the surface of extremely clean samples. λ a de 4πnq e mγv q ln dx mv qeω v 2 2 3 2 2 2 0 h a= λ= = E E= p 8 12.3 10 cm/ 50eV δe = E n = x = 23-3, 10 cm δ 100Å a 2Å 5
Electron Probe Sees only Surface CONCLUSION: Figure 2: An electron about to scatter from a typical material. However, at the surface of the material, oxidation and surface reconstruction distort the lattice. If the electron scatters from this region, we cannot learn about the structure of the bulk. CONCLUSION: Use neutral particles or electromagnetic radiation which scatter only from nuclei NEUTRONS or X-rays. 6
Neutron Scattering Experiments Neutrons scatter almost completely isotropic: Elastic scattering gives precise information about the static lattice structure while Inellastic scattering allows one to study lattice vibrations. Antiferromagnet Spin-Spin Interactions Like NaCl MnO 7
Sources of Photons (EM Radiation) Interaction of X-rays with condensed matter: charged particle (electron density) vibrates at the frequency of the incoming radiation. 8
Classical Theory of Diffraction Three basic assumptions: 1. The operator which describes the coupling of the target to the scattered "object" (in this case the operator is the density) commutes with the Hamiltonian realm of classical physics. 2. Huygens principle: Every radiated point of the target will serve as a secondary source spherical waves of the same frequency as the source and the amplitude of the diffracted wave is the sum of the wavelengths considering their amplitudes and relative phases. 3. Resulting spherical waves are not scattered again. For example, in the fully quantum theory for neutron scattering this will correspond to approximating the scattering rate by Fermi golden rule, i.e., the so-called first-order Born approximation. 9
Setup of Scattering Experiment R r A = A e P i( k0( R+r) ω0 t) 0 (incident wave) e Hygens: AB ( R ) drap ρ( r) R r A R A e d B i( k 0R+ kr ω0t ) ( ) rρ( r) 0 ik ( R r) 0 e R r i ( k k ) r 10
The Role of Fourier Transforms in Diffraction Pattern Analysis R At very large, i.e., in the so-called radiation or far zone: i( k0r+ kr ω0t ) A0 e AB ( R ) drρ( r) e R In terms of the scattered intensity I B A B i( k k ) r A 2 0 i ( 0 ) I B d ρ ( ) e 2 R k k r r r A 2 0 i A0 I B ( ) K r K d rρ ( r ) e = ρ ( K ) 2 2 R R 2 0 2 Fourier transform of the density of scatterers 11
Phase Information is Lost! I( K) ρ( K) 2 i i K ρ( ) = Kr K drρ( r) e = ρke θ From a complete experiment, measuring intensity for all scattering angles, one does not have enough information to get density of scatterers by inverting Fourier transform Instead guess for one of the 14 Bravais lattices and the basis, Fourier transform this, fit parameters to compare to experimental data. x k π From the Fourier uncertainty principle : Resolution of smaller structures requires larger values of K (some combination of large scattering angles and short wavelenght of the incident light). 12
Patterson Function 2 ( ) ( ) ( ) i K r ikr I K ρ K ρ r e dr ρ ( r ) e dr r r + r ikr I ( K ) e dr ρ ( r) ρ ( r r) dr + The Patterson function is the autocorrelation function of the scattering density (it has maximum whenever r corresponds to a vector between two atoms in the structure): P ( r ) = ρ ( r) ρ ( r + r) dr 13
Vocabulary: Patterson Function, Structure Factor, Pair Correlation Function 2 Patterson: P( r ) = Nf δ 0, r + N ρ( r) ρ ( r + r) dr, ρ ( r) = ρatom ( r rj ) N j i 2 Pair Correlation: ( ) ( ) ( ) V f g r = ρ r ρ r + r d r N ik Structure Factor: I( ) S( )=1+ g( r) e r K K d V r 14
Scattering From 1D Periodic Structures Density of periodic crystal: ρ( x + ma) = ρ( x) ρ( x) = ρ e e ign ( x+ ma) ignx ignma ( x ma) ne ne e ( x) n n ig ma n = 1 G = n 2nπ a n n ig x ρ + = ρ = ρ = ρ n 15
Scattering from 3D Periodic Structures ρ Generalization to three-dimensional structures: ( r+ r ) = ρ( r), r = na + n a + n a ; n, n, n Z n 1 1 2 2 3 3 1 2 3 G rn = 2 mπ, m Z G= hg + kg + lg hg + kg + lg na = 2 mπ, m Z ( ) 1 2 3 1 2 3 1 1 g a = 2 π, g a = g a = 0 1 1 2 1 3 1 16
Reciprocal Lattice (in Reciprocal Space) ( g, g, g ) The orthonormal set 1 2 3 forms the basis of the reciprocal lattice: a a a a a a g = 2 π, g = 2 π, g = 2π ( ) ( ) ( ) 2 3 3 1 1 2 1 2 3 a1 a2 a3 a1 a2 a3 a1 a2 a3 http://www.matter.org.uk/diffraction/geometry/sperposition_of_waves_exercises.htm (2 π) (2 π) Ω BZ = g1 ( g2 g3) = = a ( a a ) Ω 1 2 3 3 3 Real-space and reciprocal lattice have the same point group symmetry (but do not necessarily have the same Bravais lattice: example FCC and BCC are reciprocal to each other with point group symmetry ). O h PUC 17
Scattering intensity for a crystal: Laue i A ρ( ) = Gr r ρ e IB( K) d ρ e G r R G 0 i( K G) r 2 G G 2 V, = i( ) G K K G r e dr = V lattice volume 0, G K ( ) A0 2 2 I B K V 2 R ρ δ G This is called Laue condition for scattering. The fact that this 2 is proportional to V rather than V indicates that the diffraction spots, in this approximation, are infinitely bright (for a sample in thermodynamic limit) when real broadening is taken into account, I ( ) B K V G, K 18
Freidel Rule I hkl ρ hkl 2 ρ ( r) R ρ = ρ G I I = I h k l hkl hkl k k = G For every spot at 0, there will be one at 0. Thus, for example, if we scatter from a crystal with a 3-fold symmetry axis, we will get a 6-fold scattering pattern. The scattering pattern always has an inversion center even if none is present in the target! * G G k k = G G 19
Graphical Laue If, and only if the three vectors involved form a closed triangle, is the Laue condition met. If the Laue condition is not met, the incoming wave just moves through the lattice and emerges on the other side of the crystal (neglecting absorption). 20
Graphical Laue: Ewald sphere Use powder X-ray Diffraction (powdered sample corresponds to averaging over all orientations of the reciprocal lattice will observe all peaks that lie within the radius 2 k 0 of the origin of reciprocal lattice. Figure 1: The Ewald Construction to determine if the conditions are correct for obtaining a Bragg peak: Select a point in k-space as the origin. Draw the incident wavevector to the origin. From the base of, k spin (remember, that for elastic scattering k = k 0 ) in all possible directions to form a sphere. At each point where this sphere intersects a lattice point in k-space, there will be a Bragg peak with 0. In the example above we find 8 Bragg peaks. If however, we change by a small amount, then we have none!. k 0 k 0 G= k k k 0 21
Miller Indices G r = 2 πm q= ( n pg ) a + ( n pg ) a + [ n + p( g + g )] a, G q= 2πm d m n 2π m = G 1 3 1 2 3 2 3 1 2 3 1 1 1 : : = h: k : l ( hkl ), h h u v w g : g : g = h: k : l Miller indices G 1 2 3 Conventions: hkl, [ hkl ],( hkl ),{ hkl}, hkl 22
Bragg vs. Laue = Reciprocal vs. Real Space Analysis K = K = k k = 0 G h kl 4π 2π K = 2 k 0 s in θ = s in θ = λ = 2 d h kl s in θ λ d h kl 23
Brillouin Zone Interpretation of Bragg and Laue Diffraction Conditions 2 2 2 2 2 0 hkl 2 k = k = G + k = G + k + G k ( ) ( ) ( ) We want to know which particular wave vectors out of many (an infinite set, in fact) meet the diffraction (Bragg & Laue) condition for a given crystal lattice plane. If we construct Wigner-Seitz cells in the reciprocal lattice, all wave vectors ending on the Wigner-Seitz cell walls will meet the Bragg condition for the set of lattice planes represented by the cell wall. G h kl hkl G h kl 2 + k = hkl 0 24
3D Brillouin Zones Constructing Brillouin zones is a good example for the evolution of complex systems from the repeated application of simple rules to simple starting conditions - any 12-year old can do it in two dimensions, but in 3D, Ph.D. thesis in 1965 25
Reciprocal vs. k-vectors Arbitrary wave vector k can be written as a sum of some reciprocal lattice vector G plus a suitable wave vector k ; i.e. we can always write k = G + k and k can always be confined to the first Brillouin zone, i.e. the elementary cell of the reciprocal lattice. 3 (2 π ) d k = N Ω BvK: Ψ ( r) = Ψ ( r+ N a ) nk nk j j ± 2π n ± 2π n x y ± 2π n z k =,,, Lx = N1a Lx Ly L z U( r) U( r R) U( r) U( r) e i = + = Gr 3 PUC G 26
Fourier Analysis in Solid State Physics of Crystals Periodic Function, with the same periodicty as the lattice, expanded in Fourier series: U( r) U( r R) U( r) U( r) e i = + = Gr Arbitrary wave function expanded in plane waves: Ψ ( r) Bloch wavefunctions (eigenstates of crystal Hamiltonian) expansion: G = i C e k r k k Ψ = = =Ψ ( r) = u ( r+ R) i( k ) i i ( ) C e G r C e Gr k e r k r k G k G k+ G( r) G G u k k 27
Nearly-free-electron-like? 28
Crystal Electrons in the BZ-realm k G k 0 ± U G All wave vectors that end on a BZ, will fulfill the Bragg condition and thus are diffracted π π states with is Bragg reflected into state with (and vice versa) + a Wave vectors completely in the interior of the 1. BZ, or well in between any two BZs, will never get diffracted; they move pretty much as if the potential would be constant, i.e. they behave very close to the solutions of the free electron gas. a 29
Crystal Electrons in the BZ-realm 30
Scattering From a Lattice with a Basis Need structure factor S and form factor f, respectively. 31
Structure and Form factors 2 1 i 1 hkl Ihkl ρhkl, ρhkl = drρ( r) e = drρ( r) e V V cell ρ ρ hkl hkl 1 i ( n + α + ) = G r r r drρ( r) e since r = r cell n + rα + r V 1 = V N, N, N cell 1 2 3 α Atomic Scattering Form Factor S i hkl α e dr ρ ( r ) e G r ighklr α = f One atom per unit cell G r ig r cells ighkl r f = dr ρ ( r ) e Z I Z α ρ hkl 1 i S hkl α = G r e fα = V V c α α Structure Factor hkl c hkl 2 32
Extinctions r = u a + v a + w a α α 1 α 2 α 3 Shkl = f exp 2π i hu + hv + wu r α 1 2 ( ) α α α α 1 1 1 = ( 0,0,0 ) r = (,, ) 2 2 2 f f X ray Fe neutron Fe f X ray Co f neutron Co ( iπ ( h+ k+ l) 0, h+ k + l odd S 1 ) hkl = f + e = 2 f, h+ k + l even Position of Bragg reflection: Shape and dimension of the unit cell Intensities of reflections: Content of the unit cell 33
Structure Factor Revisted: Quantum Mechanical Case Example: Diffraction of electron on crystalline potentialλ a= ikr e U ( r) = U α ( r rα ) e transition from Ψ k = to Ψ k = V α Quantum-Mechanical Probability Amplitude for this transition: ˆ * k k = Ψk Ψ k = r Ψ k ( r) ( r) Ψk ( r) A U d U 1 1 1 1 V i( k k1) rα i( k k1)( r rα ) Ak = e e U d U S 1k α α = α α 8 10 cm E 0.1eV e ik r 1 V ( r r ) r ( G) ( G) 1 i ( α ) Uα ( ) = G r r G dr e Uα ( r rα ) 1 i V0 S( G) = e Gr N Structure factor is completely determined by geometrical properties of the crystal. α α 34
Structure factor: Conclusion Any matrix element that describes a transition between two electronic states under the action of crystalline potential will contain a structure factor. Crystal potential does not have to be necessarily expressed in terms of sum of the atomic potentials; furthermore, the transition do not necessarily involve external electrons everything is valid also for transition between electronic states of a crystal itself. Extracting of structure factor reflects how spatial distribution of ions affects dynamics of processes in crystals. Example: ρ( r) = ρ( r rα ) ρ( k) = S( k) ρ0( k) for electrons α ρ( r) ρ( r) S( k) = ρ( k) ρ( k) 1 for molecules 35
Experimental Techniques: Rotating Crystal 36
Experimental Techniques: Powders 37