Lecture 7

Crystal Vibrations

… and Neutron Scattering 7.1 Vibrations of crystals with monatomic basis 7.2 Two atoms per primitive basis 7.3 Quantization of elastic waves 7.4 Phonon momentum 7.5 Inelastic neutron scattering for phonons

References: 1) Kittel, Chapter 4 2) Marder, Chapter 13 3) Ashcroft, Chapters 22, 24 4) Burns, Chapters 12 5) Ziman, Chapter 2 6) Ibach, Chapter 6 Lecture 7 1

Longitudinal and transverse waves

Planes of atoms as displaces for longitudinal and transverse waves

Lecture 7 2

1 Continuous Elastic Solid

We can describe a propagating vibration of amplitude u along a rod of material with Young’s modulus Y and density ρ with the wave equation:

∂ 2u Y ∂ 2u = for wave propagation along the x-direction ∂t 2 ρ ∂x2

By comparison to the general form of the 1-D wave equation: ∂ 2 ∂ 2 u 2 u Y So the wave speed is independent of = v we find that v = ∂t 2 ∂x2 ρ wavelength for an elastic medium!

v ω ω is called the ω = 2πf = 2π = kv (k) λ dispersion relation of the solid, and here it is linear (no dispersion!) ω group velocity = d vg dk k Lecture 7 3

7.1 Vibrations of Crystals with Monatomic Basis

By contrast to a continuous solid, a real solid is not uniform on an atomic scale, and thus it will exhibit dispersion. Consider a 1-D chain of atoms: M a In equilibrium: + + s −1 s s 1 s p

Longitudinal wave:

us−1 us us+1 us+ p

p = atom label ± = ( − ) p = 1 nearest neighbors For atom s Fs ∑c p us+ p us p = ± 2 next nearest neighbors (for plane s) p cp = force constant for atom p

Lecture 7 4

2 The equation of motion of the plane s

Elastic response of the crystal is a linear function of the forces (elastic energy is a quadratic function of the relative displacement)

The total force on plane s comes from planes s ± 1

FS = C (u S+1 – uS) + C (uS - 1 – uS) C – the force constant between nearest-neighbor planes (different for transverse and longitudinal waves)

2 Equation of motion d uS M = C(u + + u − − 2u ) of the plane s: dt 2 S 1 S 1 S Solutions with all displacements having time dependence exp (-iωt). Then d 2u S = −ω 2u dt 2 S 2 This is a difference equation in − Mω u = C(u + + u − − 2u ) S S 1 S 1 S the displacements u = isKa ±iKa uS ±1 Lectureue 7 e 5

Equation of Motion for 1D Monatomic Lattice

∂ 2 us () Applying Newton’s second law: F = M = c u + − u s ∂ 2 ∑ p s p s t p

For the expected harmonic i(kx −ωt) u = ue s x = sa position of atom s traveling waves, we can write s s

− ω 2 i(ksa −ωt ) = ( i(k (s+ p)a−ωt ) − i(ksa −ωt) ) Thus: Mu ( i ) e ∑ c p ue ue p

− ω 2 i(ksa −ωt ) = i(ksa −ωt ) ( ikpa − ) Or: M e e ∑ c p e 1 p

2 ikpa − ω = ( − ) Now since c-p = cp by symmetry, So: M ∑ c p e 1 p − ω 2 = ( ikpa + −ikpa − )= ( − ) M ∑ c p e e 2 ∑ 2c p cos( kpa ) 1 p>0 p>0

Lecture 7 6

3 Dispersion relation of the monatomic 1D lattice

2 4 The result is: ω 2 = − = 2 1 ∑c p 1( cos( kpa )) ∑c p sin ( 2 kpa ) M p>0 M p>0

Often it is reasonable to make the 4c nearest-neighbor approximation (p = 1): ω 2 ≅ 1 sin 2 ( 1 ka ) M 2

ω The result is periodic in k

and the only unique 4c1 solutions that are physically M meaningful correspond to π π values in the range: − ≤ k ≤ a a k

π π π π − 2 − 0 2 a a a a Lecture 7 7

Theory vs. Experiment

In a 3-D atomic lattice we expect to observe 3 different branches of the dispersion relation, since there are two mutually perpendicular transverse wave patterns in addition to the longitudinal pattern we have considered

Along different directions in the reciprocal lattice the shape of the dispersion relation is different

Note the resemblance to the simple 1-D result we found

Lecture 7 8

4 Counting Modes and Finding N( ωωω) v A vibrational mode is a vibration of a given wave vector k (and thus λ), ω = hω frequency , and energyv E . How manyv modesv are found in the interval between ( ω , E , k ) and ( ω + d ω , E + dE ,k + kd ) ? v # modes dN = N(ω)dω = N(E)dE = N(k)d 3k

We will first find N(k) by examining allowed values of k. Then we will be able to calculate N( ω) First step: simplify problem by using periodic boundary conditions for the linear chain of atoms: We assume atoms s and s+N-1 s+N have the same L = Na displacement—the lattice s has periodic behavior , where N is very large s+1

x = sa x = (s+N)a Lecture 7 s+2 9

Step one: finding N(k)

Since atoms s and s+N have the same displacement, we can write:

= i(ksa −ωt) i(k (s+N )a−ωt) ikNa us us+N ue = ue 1 = e

This sets a condition on 2πn allowed k values: kNa = 2πn → k = n = ...,3,2,1 Na

π π independent of k, so So the separation between ∆ = 2 ∆ = 2 allowed solutions (k values) is: k n the density of Na Na modes in k-space is uniform # of modes 1 Na L Thus, in 1-D: = = = interval of k − space ∆k 2π 2π

Lecture 7 10

5 Next step: finding N( ωωω)

Now for a 3-D lattice we can apply periodic boundary N3c conditions to a sample of N 1 x N 2 x N 3 atoms:

N2b # of modes N a N b N c V N a = 1 2 3 = = N(k) 1 volume of k − space 2π 2π 2π 8π 3

Now we know from before that v V v we can write the differential # dN = N(ω)dω = N(k)d 3k = d 3k of modes as: 8π 3

We carry out the integration in k-space by using a v 3 = = [ ] “volume” element made up d k (surface area ) dk ∫ dS ω dk of a constant ω surface with thickness dk: Lecture 7 11

Finding N( ωωω)ω)))

Rewriting the differential number V dN = N(ω)dω = dS ω dk of modes in an interval: 8π 3 ∫ V dk V 1 We get the result: N(ω) = dS = dS π 3 ∫ ω ω π 3 ∫ ω ∂ω 8 d 8 ∂k

A very similar result holds for N(E) using constant energy surfaces for the density of electron states in a periodic lattice!

This equation gives the prescription for calculating the density of modes N( ω) if we know the dispersion relation ω(k).

Lecture 7 12

6 7.2 Two Atoms per Primitive Basis

Consider a linear diatomic chain of atoms (1-D model for a crystal like NaCl): a

In equilibrium:

M1 M2 M1 M2 M1 < M 2 Applying Newton’s second law and the nearest-neighbor approximation to this system gives a dispersion relation with two “branches”:

2 2/1  M + M    M + M  4c2  ω 2 = c  1 2  ± c2  1 2  − 1 sin 2 ( 1 ka ) 1 M M  1  M M  M M 2  1 2    1 2  1 2 

ω ω -(k)
0 as k
0 acoustic modes (M 1 and M 2 move in phase) ω ω ω +(k)
max as k
0 optical modes (M 1 and M 2 move out of phase)

Lecture 7 13

Optical and acoustical branches

Two branches may be presented as follows:

M1 < M 2

1/2 (2a/M 1) gap in allowed frequencies 1/2 (2a/M 2)

If there are p atoms in the primitive cell, there are 3p branches to the dispersion relation: 3 acoustical branches and 3p-3 optical branches

Lecture 7 14

7 Optical and acoustical branches

Lecture 7 15

7.3 Quantization of Elastic Waves

The energy of a lattice vibrations is quantized The quantum of energy is called a phonon (analogy with the photon of the electromagnetic wave)

Energy content of a vibrational mode of frequency ω is an integral number of energy quanta h ω . We call these quanta “phonons”. While a photon is a quantized unit of electromagnetic energy, a phonon is a quantized unit of vibrational (elastic) energy.

Lecture 7 16

8 7.4 Phonon Momentum

v ω Associated with each mode of frequency v is a wavevectork , which leads to the definition of a “crystal momentum ”: h k

Crystal momentum is analogous to but not equivalent to linear momentum. No net mass transport occurs in a propagating lattice vibration, so a phonon does not carry physical momentum

But phonons interacting with each other or with electrons or photons obey a conservation law similar to the conservation of linear momentum for interacting particles

Lecture 7 17

Conservation Laws

Lattice vibrations (phonons) of many different frequencies can interact in a solid. In all hω + hω = hω 1 2 3 interactions involving phonons, energy must be v v v r h + h = h + h conserved and crystal momentum must be k1 k2 k3 G conserved to within a reciprocal lattice vector: v ω k 1 1 v ω 3 k3 Schematically:

v ω 2 k2

Compare this to the special case of elastic v v r Just a special case scattering of x-rays with a crystal lattice: k ′ = k + G of the general conservation law!

Photon wave vectors Lecture 7 18

9 Back to Brillouin Zones

The 1st BZ is the region in reciprocal space containing all information about the lattice vibrations of the solid v Only thev k values in the 1st BZ correspond to unique vibrationalv modes. Any k outside this zone is mathematically equivalent to a value k 1 inside the 1st BZ

This is expressed in terms of a general translation vector of the reciprocal lattice: ω r 4c1 v v r M G = + k k1 G k

π π π π π π π π − 4 − 3 − 2 − 0 v 2 v 3 4 a a a a k1 a a k a a

Lecture 7 19

7.5 Neutron scattering measurements

What is a neutron scattering measurement? - neutron source sends neutron to sample - some neutrons scatter from sample - scattered neutrons are detected

2 h2k 2 h2k Conservation of energy: i = f ± ∆E 2M 2M When a phonon of wavelength | K| is created by the inelastic scattering of a photon or neutron, the wavevector selection rule: r r r r k + K = k + G creation of a phonon rf r ri r = + + Lecture 7 20 k f ki K G annihilati on of a phonon

10 Why neutrons?

.9 044 Wavelength: λ = E - At 10 meV, λ=2.86Å ⇒ similar length scales as structures of interest

Energy: - thermal sources: 5-100meV - cold sources: 1-10meV - spallation sources: thermal and epithermal neutrons (>100meV)

can cover range of typical excitation energies in solids and liquids!

http://www.ncnr.nist.gov/summerschool/ss05/VajklectLectureure.pdf 7 21

Energy and Length Scale

Lecture 7 22 http://www.ncnr.nist.gov/index.html

11 Effective Cross Section

Cross Section, σσσ: an effective area which represents probability that a neutron will interact with a nucleus σ varies from element to element and even isotope to isotope

Typical σ ~ 10 -24 cm 2 for a single nucleus

One unit of cross section is a 1 barn= 10 -24 cm 2 … as in “it can’t hit the size of the barn”

Total nuclear cross section for several isotopes http://www.ncnr.nist.gov/index.html Lecture 7 23

Neutron Scattering

Number of scattered neutrons is proportional to scattering function, S ( G, ω)

Scattering: elastic ; quasielastic ; Inelastic

Neutrons are sensitive to components of motion parallel to the momentum transfer Q

Angular width of the scattered neutron beam gives information on the lifetime of phonons Lecture 7 24

12 Phonon dispersion of bcc-Hf

LA-Phonon

TA-Phonon

Trampenau et al. (1991)

Lecture 7 25

Inelastic neutron scattering data for KCuF 3 measured

Bella Lake, D. Alan Tennant, Chris D. Frost and Stephen E. Nagler Nature Materials 4, 329 - 334 (2005) Lecture 7 26

13 2D: Inelastic Scattering on the Surfaces

Thermal Energy Helium Atom Scattering

Lecture 7 27

Time-of-Flight Spectra and Dispersion Curves

Time-of-flight spectrum for He atoms scattering from an LiF(001) surface along the [100] azimuth. The sharp Lecture 7 28 peaks are due to single surface phonon interactions (From Brusdeylins et al, 1980)

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