Chapter 4: Crystal Lattice Dynamics

Debye

January 30, 2017

1 An Adiabatic Theory of Lattice Vibrations 3

1.1 The Equation of Motion . . . . . . . . . . . . . . . . . . 6

1.2 Example, a Linear Chain . . . . . . . . . . . . . . . . . . 8

1.3 The Constraints of Symmetry . . . . . . . . . . . . . . . 12

1.3.1 Symmetry of the Dispersion . . . . . . . . . . . . 12

1.3.2 Symmetry and the Need for Acoustic modes . . . 15

2 The Counting of Modes 18

2.1 Periodicity and the Quantization of States . . . . . . . . 19

2.2 Translational Invariance: First Brillouin Zone . . . . . . 19

2.3 Point Group Symmetry and Density of States . . . . . . 21

3 Normal Modes and Quantization 21

3.1 Quantization and Second Quantization . . . . . . . . . . 24

4 Theory of Neutron Scattering 26

4.1 Classical Theory of Neutron Scattering . . . . . . . . . . 28

4.2 Quantum Theory of Neutron Scattering . . . . . . . . . . 30

4.2.1 The Debye-Waller Factor . . . . . . . . . . . . . . 35

4.2.2 Zero-phonon Elastic Scattering . . . . . . . . . . 36

4.2.3 One-Phonon Inelastic Scattering . . . . . . . . . . 37

2 A crystal lattice is special due to its long range order. As you ex- plored in the homework, this yields a sharp diraction pattern, espe- cially in 3-d. However, lattice vibrations are important. Among other things, they contribute to the thermal conductivity of insulators is due to dispersive lattice vibrations, and it can be quite large (in fact, diamond has a ther- mal conductivity which is about 6 times that of metallic copper). in scattering they reduce of the spot intensities, and also allow for inelastic scattering where the energy of the scatterer (i.e. a neutron) changes due to the absorption or creation of a phonon in the target. electron-phonon interactions renormalize the properties of elec- trons (make them heavier). superconductivity (conventional) comes from multiple electron- phonon scattering between time-reversed electrons.

1 An Adiabatic Theory of Lattice Vibrations

At rst glance, a theory of lattice vibrations would appear impossibly daunting. We haveN1023ions interacting strongly (with energies of about (e2=A)) withNelectrons. However, there is a natural expansion parameter for this problem, which is the ratio of the electronic to the 3 ionic mass: mM 1 (1) which allows us to derive an accurate theory. Due to Newton's third law, the forces on the ions and electrons are comparableFe2=a2, whereais the lattice constant. If we imagine that, at least for small excursions, the forces binding the electrons and the ions to the lattice may be modeled as harmonic oscillators, then

Fe2=a2m!2electronaM!2iona(2)

This means that

ion! electronmM

1=2103to 102(3)

Which means that the ion is essentially stationary during the period of the electronic motion. For this reason we may make an adiabatic Figure 1:Nomenclature for the lattice vibration problem.sn;is the displacement of the atomwithin the n-th unit cell from its equilibrium position, given byrn;= r n+r, where as usual,rn=n1a1+n2a2+n3a3. 4 approximation: we treat the ions as stationary at locationsR1;RNand deter- mine the electronic ground state energy,E(R1;RN). This may be done using standard ab-initio band structure techniques (DFT,

GGA, etc.).

we then use this as a potential for the ions; i.e.. we recalculate Eas a function of the ionic locations, always assuming that the electrons remain in their ground state.

Thus the potential energy for the ions

(R1;RN) =E(R1;RN) + the ion-ion interaction (4) We will dene the zero potential such that when allRnare at their equilibrium positions,= 0. Then H=X nP

2n2M+(R1;RN) (5)

Typical lattice vibrations involve small atomic excursions of the or- der 0:1Aor smaller, thus we may expand about the equilibrium position of the ions. (frni+snig) =(frnig) +@@r nisni+12 2@r ni@rmjsnismj(6) The rst two terms in the sum are zero; the rst by denition, and the second is zero since it is the rst derivative of a potential being evaluated at the equilibrium position. We will dene the matrix mj ni=@2@r ni@rmj(7) 5 From the dierent conservation laws (related to symmetries) of the system one may derive some simple relationships for . We will discuss these in detail later. However, one must be introduced now, that is, Figure 2:Since the coecients of potential between the atoms linked by the blue lines (or the red lines) must be identical,mj ni= (mn)j 0i. due to translational invariance. mj ni= (mn)j

0i=@2@r

0i@r(nm)j(8)

ie, it can only depend upon the distance. This is important for the next subsection.

1.1 The Equation of Motion

From the derivative of the potential, we can calculate the force on each site F ni=@(frmj+smjg)@s ni(9) so that the equation of motion is mj nismj=Msni(10) 6 If there areNunit cells, each withratoms, then this gives 3Nrequa- tions of motion. We will take advantage of the periodicity of the lattice by using Fourier transforms to achieve a signicant decoupling of these equations. Imagine that the coordinatesof each site is decomposed into its Fourier components. Since the equations are linear, we may just consider one of these components to derive our equations of motion in

Fourier space

s ni=1pM u i(q)ei(qrn!t)(11) where the rst two terms on the rhs serve as the polarization vector for the oscillation,ui(q) is independent ofndue to the translational invariance of the system. In a real system the realswould be composed Figure 3:ui(q)is independent ofnso that a lattice vibration can propagate and respect the translational invariance of the lattice. of a sum over allqand polarizations. With this substitution, the equations of motion become

2ui(q) =1pM

Mmj nieiq(rmrn)uj(q) sum repeated indices: (12) 7

Recall that

mj ni= (mn)j

0iso that if we identify

D j i=1pM Mmj nieiq(rmrn)=1pM Mpj

0ieiq(rp)(13)

whererp=rmrn, then the equation of motion becomes

2ui(q) =Dj

iuj(q) (14) or Dj i!2j i u j(q) = 0 (15) which only has nontrivial (u6= 0) solutions if detD(q)!2I= 0. For eachqthere are 3rdierent solutions (branches) with eigenvalues (n)(q) (or rather!(n)(q) are the root of the eigenvalues). The de- pendence of these eigenvalues!(n)(q) onqis known as the dispersion relation.

1.2 Example, a Linear Chain

Figure 4:A linear chain of oscillators composed of a two-element basis with dierent masses,M1andM2and equal strength springs with spring constantf. Consider a linear chain of oscillators composed of a two-element basis with dierent masses,M1andM2and equal strength springs with spring constantf. It has the potential energy =12 fX n(sn;1sn;2)2+ (sn;2sn+1;1)2:(16) 8 We may suppress the indicesiandj, and search for a solution s n=1pM u (q)ei(qrn!t)(17) to the equation of motion

2u(q) =Du(q) whereD=1pM

Mp;

0eiq(rp)(18)

and, m;n;=@2@r

0;@r(nm);(19)

where nontrivial solutions are found by solving det

D(q)!2I= 0.

The potential matrix has the form

n;1 n;1= n;2 n;2= 2f(20) n;2 n;1= n;1 n;2= n1;2 n;1= n+1;1 n;2=f :(21) This may be Fourier transformed on the space indexnby inspection, so that D =1pM Mp

0eiq(rp)

0 2fM 1fpM

1M21 +eiqa

fpM

1M21 +e+iqa2fM

21
A (22) Note that the matrixDis hermitian, as it must be to yield real, physi- cal, eigenvalues!2(however,!can still be imaginary if!2is negative, indicating an unstable mode). The secular equation detD(q)!2I=

0 becomes

4!22f1M

1+1M 2 +4f2M

1M2sin2(qa=2) = 0;(23)

9 with solutions 2=f1M 1+1M 2 fs 1M 1+1M 2 2 4M

1M2sin2(qa=2) (24)

This equation simplies signicantly in theq!0 andq=a!limits. Figure 5:Dispersion of the linear chain of oscillators shown in Fig. 4 whenM1= 1, M

2= 2andf= 1. The upper branch!+is called the optical and the lower branch is

the acoustic mode. In units wherea= 1, and where the reduced mass 1== 1M 1+1M 2 lim q!0!(q) =qarf

2M1M2limq!0!+(q) =s2f

(25) and (q==a) =p2f=M2: !+(q==a) =p2f=M1(26)

As a result, the + mode is quite

at; whereas themode varies from zero at the Brillouin zone centerq= 0 to a at value at the edge of the zone. This behavior is plotted in Fig. 5. 10 It is also instructive to look at the eigenvectors, since they will tell us how the atoms vibrate. Let's look at the optical mode atq= 0, +(0) =p2f=. Here, D=0

2f=M12f=pM

1M2 2f=pM

1M22f=M21

A :(27) Eigenvectors are non-trivial solutions to (!2ID)u= 0, or 0 = 0

2f=2f=M12f=pM

1M2 2f=pM

1M22f=2f=M21

A0 u1 u 21
A :(28) with the solutionu1=pM

2=M1u2. In terms of the actual displace-

ments Eqs.11 s n1s n2=rM 2M 1u 1u 2(29) orsn1=sn2=M2=M1so that the two atoms in the basis are moving out of phase with amplitudes of motion inversely proportional to their masses. These modes are described as optical modes since these atoms, Figure 6:Optical Mode (bottom) of the linear chain (top). if oppositely charged, would form an oscillating dipole which would couple to optical elds witha. Not all optical modes are optically active. 11

1.3 The Constraints of Symmetry

We know a great deal about the dispersion of the lattice vibrations without solving explicitly for them. For example, we know that for each q, there will bedrmodes (wheredis the lattice dimension, andris the number of atoms in the basis). We also expect (and implicitly assumed above) that the allowed frequencies are real and positive. However, from simple mathematical identities, the point-group and translational symmetries of the lattice, and its time-reversal invariance, we can learn more about the dispersion without solving any particular problem.

The basic symmetries that we will employ are

The translational invariance of the lattice and reciprocal lattice. The point group symmetries of the lattice and reciprocal lattice.

Time-reversal invariance.

1.3.1 Symmetry of the Dispersion

Complex Properties of the dispersion and EigenmodesFirst, from the symmetry of the second derivative, one may show that!2is real. Recall that the dispersion is determined by the secular equation detD(q)!2I=

0, so ifDis hermitian, then its eigenvalues,!2, must be real.

D j i=1pM Mpj

0ieiq(rp)(30)

1pM Mp;;j

0;;ieiq(rp)(31)

12 Then, due to the symmetric properties of the second derivative D j i=1pM M0;;i p;;jeiq(rp)=1pM Mp;;i

0;;jeiq(rp)=Dij(32)

Thus,DT=Dy=DsoDis hermitian and its eigenvalues!2are real. This means that either!are real or they are pure imaginary. We will assume the former. The latter yields pure exponential growth of our Fourier solution, indicating an instability of the lattice to a second- order structural phase transition. Time-reversal invariance allows us to show related results. We as- sume a solution of the form s ni=1pM u i(q)ei(qrn!t)(33) which is a plane wave. Suppose that the plane wave is moving to the right so thatq=^xqx, then the plane of stationary phase travels to the right with x=!q xt:(34) Clearly then changing the sign ofqxis equivalent to takingt! t. If the system is to display proper time-reversal invariance, so that the plane wave retraces its path under time-reversal, it must have the same frequency when time, and henceq, is reversed, so !(q) =!(q):(35) Note that this is fully equivalent to the statement thatDij(q) = D ij(q) which is clear from the denition ofD. 13

Now, return to the secular equation, Eq. 15.

Dj i(q)!2(q)j i j(q) = 0 (36) Lets call the (normalized) eigenvectors of this equation. They are the elements of a unitary matrix which diagonalizesD. As a result, they have orthogonality and completeness relations X ;i(n) ;i(q)(m) ;i(q) =m;northogonality (37) X n(n) ;i(q)(n) ;j(q) =;i;j(38) If we now take the complex conjugate of the secular equation Dj i(q)!2(q)j i j(q) = 0 (39)

Then it must be that

j(q)/j(q):(40) Since thefgare normalized the constant of proportionality may be chosen as one j(q) =j(q):(41) Point-Group Symmetry and the DispersionA point group operation takes a crystal back to an identical conguration. Both the original and nal lattice must have the same dispersion. Thus, since the recip- rocal lattice has the same point group as the real lattice, the dispersion relations have the same point group symmetry as the lattice. 14 For example, the dispersion must share the periodicity of the Bril- louin zone. From the denition ofD D j i(q) =1pM Mpj

0ieiq(rp)(42)

it is easy to see thatDj i(q+G) =Dj i(q) (sinceGrp= 2n, where nis an integer). I.e.,Dis periodic in k-space, and so its eigenvalues (and eigenvectors) must also be periodic. (n)(k+G) =!(n)(k) (43) j(k+G) =j(k):(44)

1.3.2 Symmetry and the Need for Acoustic modes

Applying basic symmetries, we can show that an elemental lattice (that withr= 1) must have an acoustic model. First, look at the transla- Figure 7:If each ion is shifted bys1;1;i, then the lattice energy is unchanged. tional invariance of . Suppose we make an overall shift of the lattice by an arbitrary displacementsn;;ifor all sitesnand elements of the 15 basis(i.e.sn;;i=s1;1;i). Then, since the interaction is onlybetween ions, the energy of the system should remain unchanged. E=12 X m;n;;;i;j mn;;j

0;;isn;;ism;;j= 0 (45)

12 X mn;;i;j mn;;j

0;;is1;1;is1;1;j(46)

12 X i;js

1;1;is1;1;jX

mn; mn;;j

0;;i(47)

Since we know thats1;1;iis nite, it must be that

X m;n;; mn;;j

0;;i=X

p;; p;;j

0;;i= 0 (48)

Now consider a strain on the systemVm;;j, described by the strain matrixm;i ;j V m;;j=X ;im ;i ;jsm;;i(49)quotesdbs_dbs6.pdfusesText_12

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