Quantum Condensed Matter Physics Lecture 5
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1 Quantum Condensed Matter Physics Lecture 5 detector sample X-ray source monochromator David Ritchie QCMP Lent/Easter
2 Quantum Condensed Matter Physics 1. Classical and Semi-classical models for electrons in solids (3L) 2. Electrons and phonons in periodic solids (6L) Types of bonding; Van der Waals, ionic, covalent. Crystal structures. Reciprocal space, x-ray diffraction and Brillouin zones. Lattice dynamics and phonons; 1D monoatomic and diatomic chains, 3D crystals. Heat capacity due to lattice vibrations; Einstein and Debye models. Thermal conductivity of insulators. Electrons in a periodic potential; Bloch s theorem. Nearly free electron approximation; plane waves and bandgaps. Tight binding approximation; linear combination of atomic orbitals, linear chain and three dimensions, two bands. Pseudopotentials Experimental probes of band structure (4L) 4. Semiconductors and semiconductor devices (5L) 5. Electronic instabilities (2L) 6. Fermi Liquids (2L) QCMP Lent/Easter
3 The reciprocal lattice and diffraction Reciprocal lattice concept arises from scattering of waves by crystals. Builds on Fraunhofer diffraction from a grating generalised to scattering from a 3D periodic lattice. Consider scattering of a plane wave off a single atom or more generally the basis forming the unit cell Incoming wave of wavevector is incident on potential centred at R i At large distances scattered wave is circular Total field taken as a scalar ψ k 0 ik0 r R 0 ( ) i ik r Ri e + f r Ri Details of scattering buried in Form Factor This is a function of scattering angle, type and arrangement of atoms etc. Total scattered intensity assumed small e f Bragg scattering k 0 k k 0 G k QCMP Lent/Easter
4 The reciprocal lattice and diffraction At large distance from the scattering centre Defining scattered wavevector momentum transfer The waveform is given by: Effective scattering amplitude k r R kr k Summing over identical lattice sites the scattered intensity is proportional to the differential scattering cross-section: We add terms with different phases i leading to cancellation unless Bragg condition is satisfied for all with an integer If two vectors satisfy the Bragg condition then so will their sum hence the special values of satisfying this lie on a reciprocal lattice q G Primitive vectors of the reciprocal lattice in terms of real space lattice primitive vectors: b ( rr ) i r 0 i 0 0 r k = k 0 ( r ) q= k k0 k0 R i k0 r 0 ψ e e + ce f f( θ ) = f exp[ iq R ] iqr i i ik r e dσ 2 = f( θ ) = f exp[ i i ] i dω R q R q R q R = 2π m R a a = 2π and cyclic permutations QCMP Lent/Easter 2019 a1 a2 a3 5.4 i 2 m r i
5 Diffraction conditions For elastic scattering conservation of energy requires 0 and Bragg condition requires where is a reciprocal lattice vector Combining these two conditions k ko = G G k G/2 = G/2 G ( ) 2 This defines a plane perpendicular to which intersects at its midpoint The set of all such planes defines the incident wavevectors that satisfy the conditions for diffraction k = G k Ewald construction in reciprocal space Reciprocal lattice points shown k 0 incident wavevector with origin chosen (point a) so it ends on a reciprocal lattice point Sphere radius k 0 drawn about origin of k 0 Diffracted beam forms if sphere intersects any other reciprocal lattice point Origin of 0 is on perpendicular bisector of Angle is the Bragg angle and we obtain θ k 2 k sinθ = G G θ k 0 k a G 2θ k 0 QCMP Lent/Easter
6 From last slide The spacing between parallel lattice planes perpendicular to G = hb1+ kb2 + lb3 k = 2 π / λ is given by Given that we can write θ Diffraction conditions 2 k sinθ = G Where is the angle between the incident beam and the crystal planes and equals half the angle of deflection The indices defining an actual crystal plane may contain a common factor and we can generalize this equation to give the conventional form of Bragg s law: 2dsinθ = nλ ( ) 2 π / d hkl = G ( ) d ( hkl ) d ( hkl ) 2 2 π / λ sinθ = 2 π / 2 sinθ = λ n QCMP Lent/Easter
7 High resolution x-ray diffraction Find thickness and composition of SiGe thin films on Si substrates From Bragg s law thickness ( 2cos ) 1 t = λ ω ω Vertical lattice constants found from c sin c L S, c Hence concentration of Ge found = ω sinω S L c L S detector sample X-ray source monochromator sample ω ω X-ray source detector View from above QCMP Lent/Easter
8 X-ray analysis of superlattice structure QCMP Lent/Easter
9 Diffraction conditions and Brillouin zones The set of reciprocal space planes satisfying the Bragg condition is constructed by finding those planes which are perpendicular bisectors of every reciprocal lattice vector. The planes so constructed divide reciprocal space up into cells. The cell closest to the origin is called the first Brillouin zone. The n Brillouin zone consists of all the fragments exterior to the ( n 1) th plane but interior to th the plane. n G th The first Brillouin zone is the Wigner-Seitz cell of the reciprocal lattice which has an important role in discussion of electronic states in a periodic potential. Reciprocal lattice point Reciprocal lattice vector vector perpendicular bisector 1 st Brillouin zone 2 nd Brillouin zone The volume of each Brillouin zone (adding up the fragments) is equal to the 3 volume of the primitive unit cell of the reciprocal lattice, which is π /Ω where is the volume of the primitive unit cell of the crystal. (2 ) cell Ω cell QCMP Lent/Easter
10 b1 = 2π a a Brillouin zone in 3D 2 3 Using etc we can calculate primitive vectors in reciprocal a1 a2 a3 space for a number of lattices Reciprocal lattice of simple cubic lattice is also simple cubic Reciprocal lattice of face centred cubic is body centred cubic Reciprocal lattice of body centred cubic is face centred cubic Example: face centered cubic lattice Primitive vectors of FCC lattice Reciprocal lattice of FCC is BCC Brillouin zones shown QCMP Lent/Easter 2019 (Taken from Kittel) 5.10
11 Lattice dynamics and phonons 1D monatomic chain Consider a row of identical atoms distance a of constant horizontally apart connected by springs K and free to move th n r na u un m = Ku ( 2 n+ 1 un) + Ku ( n 1 un) t u () t = u cos( qr ω( q)) t Displacement of atom at point n is given by, equation of 2 motion: We guess wave solution: n o n wavelength λ = 2π / q and period T = 2 π / ω( q) Substituting into equation of motion we obtain = n 2 2 mω q = K qa = K qa ( ) 2 (1 cos( )) 4 sin ( / 2) Hence dispersion relation between frequency and wavevector ω 12 / ( q) = 2( K / m) sin( qa / 2) QCMP Lent/Easter
12 Lattice dynamics and phonons 1D monatomic chain From last slide Periodic in with period Long wavelength modes linear dispersion Same as for a wire with tension and density compressive waves with velocity These waves behave like sound waves - acoustic mode 12 / ( ) 2( ) sin( / 2) ω q = K / m qa q 2 π /a ( qa 0) ω ( q) = qa K / m Ka m/ a v= ak ( / m ) / q 12 For larger dispersion is periodic, phase shift between neighbouring iqa atoms given by e So for qa = 2π neighbouring atoms move in phase, for qa = π they move in anti-phase π π We simplify things by only considering a range of a q a This corresponds to the first Brillouin zone QCMP Lent/Easter
13 Lattice dynamics and phonons 1D diatomic chain The monoatomic chain contains only acoustic modes. The spectrum becomes more complicated with more atoms per unit cell. Assume two different atoms with different masses and spring constants There are two equations of motion 2 una ma = Ku ( ) ( 2 nb una + K un, 1 B una) t 2 unb mb = K( u 2 n+ 1A unb ) + Ku ( n, A unb ) t The solution is quite complicated so look at a limit Suppose the atoms are the same mass so A B and are quite strongly bonded in molecular pairs so K K Every molecule will have a mode where the atoms oscillate out of phase at 2 a frequency ω o = 2K / m QCMP Lent/Easter m = m
14 1D diatomic chain different spring constants The coordinate undergoing oscillation is q = 0 u ( q= 0) = u u opt A B We have assumed this is at where each molecule undergoes the oscillation in phase with the next K K Since the restoring force and frequency is largely independent of There are two branches to the dispersion curve The acoustic branch the (low) frequency vanishes linearly with wavevector. The optical branch has finite (high) frequency as q 0 and is referred to as optical because of interaction with light Atomic displacements for two modes quite different Acoustic Optical ππ/a 2 a 2 q Note that a 2 is the distance between two molecules QCMP Lent/Easter
15 1D diatomic chain different masses K = K ma mb 2 2 una unb ma = Ku ( 2 nb + un, 1B 2 una), mb = Ku ( 2 n+ 1A + un, A 2 unb ) t t a αβ, u = αexp( inka)exp( iωt), u = βexp( inka)exp( iωt) Using the equations of motion assuming but With lattice constant, substituting in travelling wave solutions with amplitudes on alternate planes: na and solving the simultaneous equations leads to m m ω 2 K( m + m ) ω + 2 K (1 cos ka) = A B A B 2 ω k = ±π / a ka ka k a May be solved for - see diagram Assuming or cos = 1 / 2 we obtain the limiting results shown Notice there are no solutions for ω between 2 K / m and 2 K / m A B If we look there for real ω solutions we find k is complex so wave is damped QCMP Lent/Easter nb 1 1 2K + m A m ka B K 2m + 2m A B ω Optical phonons m A > m B acoustic phonons π / a 2 K / mb 2 K / ma k
16 Lattice dynamics and phonons 3D crystal ω( k) Extending to 3D requires a dispersion relation describing waves propagating in different directions As well as compressional waves there are transverse waves resulting in three branches of phonons, two transverse and one longitudinal. There are always acoustic modes and for a solid with atoms per unit cell there will be 3( m 1) optical modes again split into two transverse and one longitudinal. Phonon dispersion in Ge - each panel a different direction Diamond structure 2 atoms per unit cell Measured by inelastic neutron scattering 3 m QCMP Lent/Easter
17 Inelastic neutron scattering Incident neutron transfers some of its (well controlled) energy ε and momentum k to phonon Measurement made of outgoing neutron energy and momentum ε ( k ), k Most common method for measuring phonon dispersion curves E( Q) Can also measure phonon lifetimes Quantitative technique, can measure throughout Brillouin zone Neutrons produced by nuclear reactor so large facilities required. Q= k k + G E( Q) = ε( k) ε( k ) k ε ( k) = 2m k, ε ( k) Incident n Multi axis crystal spectrometer at the National Institute for Standards and Technology Centre for Neutron Research in Phys Rev Lett 110, (2013) Gaithersburg, Maryland, USA said to be the most intense QCMP Lent/Easter 2019 neutron beam in the world k, ε ( k ) Q, E( Q)
18 Summary of Lecture 5 Reciprocal lattice and diffraction Diffraction conditions High resolution x-ray diffraction Diffraction conditions and Brillouin zones Brillouin zones in 3D Lattice dynamics and phonons 1D monatomic chain 1D diatomic chains different spring constants 1D diatomic chains different masses Inelastic neutron scattering for determination of phonon dispersion QCMP Lent/Easter
19 Quantum Condensed Matter Physics Lecture 5 detector sample X-ray source monochromator The end QCMP Lent/Easter
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