Theory.classical Harmonic Crystals.dell.10 12(2)

8/10/2019 Theory.classical Harmonic Crystals.dell.10 12(2)

1/24

Classical Theory of Harmonic Crystals

Two Assumptions in what follows:1) Mean equilibrium position of each ion is a BL site, and denotes

this position.2) Excursions of each ion away from its equil. pos. are SMALL

compared with interionic spacing. (define more precisely in what follows)

Introduction

Relax rigid lattice assumption of all preceding discussion. Investigate motion oflattice ions/atoms, etc. about equilibrium lattice sites.

Now: Denote position of ion whose mean position is by .

At any given time , where is deviation from equil. pos.

Leads to Harmonic approximation

to make life easier

!

R

!

R!

r (!

R)!

r (!

R)=!

R+!

u(!

R) !

u(!

R)

Internal Energy:

1) Initially, assume only pairwise interactions, i.e., pair of ions separated by ,

contributes to P.E. of crystal.

2) Static approx. (like last chapter)

!

r

!(!

r)


2/24

Classical Harmonic Crystal

Total P.E. is just sum of P.E.s of all distinct pairs.

, where

But all BL sites are equivalent, so and thus

All BL vectors

P.E. at site Rdue to all

other ions

U = 1

2!

!

R!!

"R( )!

R,!

"R

# = 12

!

!

"R( )!

"R

# !(!

!R ) = !!

R"!

!R( )!

R#!

!R

$

!

!

!R( ) = N!!

!R( )!

!R

"

U =N

2

!

!

R!!

"R

( )!

R#!

"R$ =

N

2

!

!

R

( )!

R#0$

R - R is also BL vector and we

are summing over all BL vectors.

!

R

Now relax static approx.Atom whose average pos. is will generally be found at

Thus

!

r (!

R)

!

!

R !

r(!

R)

!

u(!

R)

U=1

2!

!

r (!

R)!!

r (!

"R)( ) =!

R

!

"R

# 1

2!

!

R!!

"R +!

u(!

R)! !

u(!

"R)( )!

R

!

"R

#

P.E. now depends on dynamical variables .

Hamiltonianfor dynamical system, H = T + U, given by

!

u(!

R)

H =

!

P(!

R)!" #

$2

2M+U

!

R

%

1


3/24


Harmonic Approximation!

u(

!

R)1)

Assume are small in sense that is small for all atom pairs

that have appreciable interaction, .

2) Expand P.E. in 3D Taylors series of the form

!

u(

!

R)!

!

u(

!

"R)

!

!

R!!

"R( )

f( !

!+!

a) = f( !

!)+!

a!

!f( !

!)+1

2

(!

a!

!)2f(

!

!)+1

3!

(!

a!

!)3f(

!

!)+....

Apply this to each term of with , and :1

!

u(!

R)! !

u(!

"R) =!

a!

R!!

"R =!

!

U=1

2

!(!

R!!

"R )+!

u(!

R)! !

u(!

"R )( )!

#!(!

R!!

"R )+

1

2

!

u(!

R)! !

u(!

"R )

( )

!

#$

%

&

'

2

!(!

R!!

"R )+O(!

u3)+....

(

)*

+*

,

-*

.*!R !"R

/

Zeroth

order

First order

2nd order

!"

=

0

)(2

R

RN

!

!

# from previous static result

Rewriting on next pg


4/24


( ) ( )[ ]! !!" ##

++#$%#$+#$%#$+=0

32 ...)()()()(4

1)()()(

2

1)(

2 R RRRRuORRRuRuRRRuRuR

NU

! !!!!

!

!!!!

!

!

!

!!!!

!

!

!

!

&&&

Linear in :

Coeff. of is ; but this is just minus (F = grad P.E.), the force on atom at R

exerted by ALLother atoms when at their equil. positions;

This 0 --- no net force in equil. -- Ditto for atom at R#

)()( RuRu !"!

!

!

!

)(Ru!

!

!"

"#$R

RR!

!!!

)(%

Eval. at

thisR

orderLinear Term = zero

Harmonic P.E. usually written in more general form:

( ) ( )!

!!

!

!

"""

rr

rrRuRuRRRuRuU v

zyxRR

harm

##

#=$%$%$%= &

=

$

)()(where,)()()()()(

4

1 2

,,,

.

!

!!!!!!

!!

Frequently ignore Uequil.(constant, indep. of us and Ps), and then U = Uharm.Starting point for essentially all theories of lattice dynamics. Further corrections $u3and u4

are known as anharmonic terms.

Lowest-order non-vanishing correct to equil. is quadratic! The harmonic approx.

retains onlythis term, U = Uequil.+ Uharm.,where

)()()(2

1

,,,

. RuRRDRuU

zyx

RR

harm !!"= #=

!

!!!!

!!

$

$

$

Relaxes restrictionto only pairwise

interactions


5/24

)()()(2

1

,,,

. RuRRDRuU

zyxRR

harm !!"= #=

!

!!!!

!!

$

$

$


This is equivalent to prev. expression for Uharm.(assuming pairwise interactions) if we take

)()()( RRRRRRDR

RR!""!!"=!" #

!!!

!!!!!!

!

!!

$$$ %%&

2

Not in sum over R%

Eq. represents general interaction (through D!) between displaced ion at R

with displaced ion at R . It need not be simple, direct pairwise interaction. Could

take place via another mechanism, e.g., distortion of electronic arrangement around

ions (contributes to P.E. of xtal), with arrangement depending on ionic configuration.Also Coulombic interaction among ions is not simply pairwise, e.g., displacement of

ion idirectly affects ionj, but also affects ionj through interacting with ions k, l, m, !.,

and theirdirect interaction withj.

Generally very complex.

2

Demonstrating the equivalence is astraightforward but tedious exercise


6/24


To simplify, make Adiabatic Approximation:

Electronic arrangement, and hence contrib. of valence electrons to total energy

depends in detail on arrangement of ion cores. When ions displaced from equilib.,

electronic wavefunction can be deformed --- hard to calculate so we make

adiabatic approx.

Note that electron velocities in atom/ion are typically much greater than the

Velocities of ions in solids: velec.&106m/sec; vion&10

3m/sec. So, we assume that

at any instant the electrons have adjusted to the slow motion of the ions and are in

Their ground state configuration they can follow the ionic motion.

Still a tough problem


7/24


Specific Heat of Classical Crystal (want to calc. u(T) = U(T)/V)

Must average over all possible configurations, giving each a weight proportional to exp[-E/kBT),Where E is the energy of a particular configuration of all ions.

From Classical Stat. Mech., energy density is

Tkwithed

Hed

VV

Uu

BH

H

/1,1

=

!

!==

""

#

#

$$

$Here H is the classical

Hamiltonian of the system

And d'is the volume element in crystal Phase Space, .)()()()(,

!! "=#

RR

RdPRduRPdRudd!!

!!!!!!

3

Write in more compact form

][ln1

! "#

$

$"= HedV

u %

% x

xu

xu

xu

x !

!=

!

! )(

)(

1)(ln

Now

[ ]! +=R

harmU

M

RPH

!

!!

.2

2

)( Ignoring Uequil. Be careful withnotation in the

following: u and!

u(!

R)Make change of variables

)()();()(

)represents()()(;)()(

2

3

2

1

2

3

2

1

21

21

21

RPdRPdRPRP

udududdududuudRudRudRuRu zyxzyx

!!!!!!

!

!

!

!

!

!

!

!

!

!!

!!!

!!

==

===

""

"""""


8/24


Dulong-Petit Law

!"

!#$

!%

!&'

(()

*

++,

-../=0 12 3 31

..

R

NHRuDRu

M

RPRPdRuded

!

!!

!

!!!

!

)()(2

1

2

)(exp)()(

23

44

55

Integral is indep. of T (()

( ) [ ].)ln(ln31]ofindep.term[ln1 3 constNVV

u N

+!"

"!=#

"

"!=

! $$

$$$

Here N is # ofBravais Lattice

points

TnkV

Nu

B3

3==

!

n = # ions per

unit volume

With some manipulations

BV nkT

uc 3=

!

!=

Does not agree at all with experiment (particularly at low T)

1) at low T, cVdrops well below this value, " 0 as T "0.

2) Even at higher T substantial discrepancies (largely because of Harm. Approx.)

Need for Quantum Theory


9/24

Need quantum theory, but must understand classical problem ---- Harm. crystalrepresents special case of general class of problems concerning small oscillations.

(quadratic in Pand u). General solution for N ions is represented as superposition(or linear comb.) of 3N normal modes of vibration, each with its characteristic

frequency, #.Carry over to quantum mechanics solution of harmonic osc. So analysis of normal

mode problem for lattice ions is helpful.

Initially consider only 1D problems to get ideas across absent the

overwhelming notation.

Normal modes of 1D monatomic BL

Consider linear chain sketched below; BL vectors are just R = na, with n an integer.Let u(na) be displacement along the line of the ion whose equil. position is na.

Classical Harmonic Crystal (cont.)

Dropping vector notation and considering only longitudinal oscillations.

u(na)u(n-2)a

static

dynamic

na (n+1)a(n-1)a (n+3)a(n+2)a(n-2)a(n-3)a


10/24


For pairwise interactions

Uharm. = 14

u(!

R)!u(!

"R)#$ %&!"(!

R!!

"R )!

R!

"R,!=x,y,z

' u!(!

R)!u!(!

"R)#$ %& e.g., R = (n+1)a,R= na

Further assume only nn interactions, (R-R = a )

Therefore only non-zero terms are of the form u(na) u([n+1]a), and we

drop the subscripts since its 1D.

Uharm.

=

1

2K u(na)!u([n +1]a)[ ]

2

n

" , where K =#2!

#x2x=a

= $$! (a)

Factor of 2 because and are

identical for nn interactions

!

R

!!

!R

" $(x) is P.E. oftwo ions a

distance x

apart

Equations of motion

M!!u(na)= !"Uharm.

"u(na)= !

"

"u(na)

K

2u( #n a)! u [ #n +1]a( )$% &'

2

#n

()*+

,-.

= !K u( #n a)! u [ #n +1]a( )$% &'"u( #n a)

"u(na)!"u [ #n +1]a( )

"u(na)

$

%/

&

'0

#n

(

A

A = 0 unlessn=n (and then A=1)

%nn B

B = 0 unlessn = n+1 (and then B = 1)

%n,n+1


11/24


So!"Uharm.

"u(na)= !K u(na)! u [n+1]a( )#$ %&+K u([n !1]a)! u na( )#$ %&, and

M!!u(na) = !K 2u(na)!u [n!1]a( )!u [n+1]a( )#$ %& Same Eq. of motion as for linearchain of masses connected by

perfectly mass spring of spring

const. KBoundary Conditions

For finite # of ions N must specify how ions at ends are to be treated.

Assume very large N (end effects relatively unimportant) and useBorn-von Karmen (cyclic) boundary conditions.

a

Assume osc. solns. of form

u(na, t)!ei(kna"!t)

Join ions at either end of chain by same spring that connects internal ions.

a

Na

0

a 2a

Boundary

u(Na, t)= u(0,t); eikNa

= e0=1

so kNa = 2!m, and k =2!

a

m

N

Boundary

cond.Allowed values

of k

Linear chain


12/24


Very Important Physics

If k is changed by 2#/a(i.e., k!k 2#/a) u(na,t) is unchanged .

1)(2

)()(=!

"#

$%&

'

knaina

ai

knaiknaieeee

(

aka

2

2;4

!

"

!" ===

5

42,

2

52

2;

2 a

kand

aaakso

ak ===+==!

"#

""""

Amplitude at lattice sites isthe same in both cases

Example: ()= 4a)

a

a

What does this mean??

We can change k by ANY RECIPROCAL LATTICE VECTOR (an integral multiple of2#/ain this simple case) WITHOUT CHANGING THE PHYSICAL SOLUTION!!!!

(a displacement of the ion in this case)


13/24


Summary and Additional Comments

From the preceding, there are just N values of k that provide solutions to theeq. of motion and satisfy the boundary conditions. Values of k span a range of 2*/a

In reciprocal space. We pick them to lie between -*/a and +*/a to be consistent withThe usual definition of the first Brillouin zone in 1D (bisect nn. vectors of RL, which

have length 2*/a.

Now

Substitute assumed solutions for u(na,t) into eq. of motion.( ) ( )

[ ]

( )kaM

Kor

eeeeeKeM

tanutanutnauKtnauM

tknaiikatknaiikatknaitknai

cos12

2

,]1[,]1[),(2),(

2

)()()()(2

!=

!!!=!"

+!!!!=

!+!!!!

#

# ####

!!

There are harmonic

solutions provided this is

satisfied

or, and taking the positive rootkaka

2

1sin

2

cos1using

2/1

=!"

#$%

& '

( ) kaM

Kka

M

Kk

2

1sin2cos1

2)( =!="

Solutions are periodic with period 2!/a

ka21sin

k2! / a


14/24

Classical Harmonic Crystal (cont.)Actual displacements are given by real or imag. parts of u(na, t)

Took positive root because +is usually considered to be positive and because +is an even

funct. of k; above solutions for k and -+(k) are the same as those for -k and +(k) .

!"#

$$%

)sin()cos(),(

tkna

tknatnau

&

&

cos(-x)cos(x)since

sametheare)2

1sin2cos()

2

1sin2cos(

=

!!++ tkaM

Kknaandtka

M

Kkna

.

Negative root

doesnt add

anything

Now there are N distinct values of k (between -*/a and +*/a, each with a unique frequency; so

There are 2N independent solutions (but only N normal modes) But sine solution is cosine

solution shifted in time by */2+-- can take linear combination of the two normal modes. Arbitrary

Motion of the chain is obtained by specifying N initial positions and N initial velocities so this

is a complete solution. These solutions are waves propagating along the chain with phase

velocity c =+

/k and group velocity v =,+

/,

k.+(k)

*/a-*/a

M

K4ka

M

Kk

2

1sin2)( =!

k > a, or k


15/24

Classical Harmonic Crystal Lattice with basis

Normal modes of 1D BL with a basis

One-D lattice with 2 atoms/prim cell. Equilibrium pos. na and na +d.Assume identical ions (same mass M), but d &a/2, so force between neighboring

ions depends on whether their separation is d or a d. Assume also onlynninteractions.

(n-1)a (n+3)a(n+2)a(n+1)ana

d a - d Denotes unit

cell; cells

separated by a

Now denote by u1(na) the displacement of ions that oscillate about equilibrium siteat na , and by u2(na) the displacement of ions that oscillate about equilibrium site

at (na + d) .

Recall general form for P.E. (generalized a bit and specific to 1D no need for -sum).

Added sum over s (basis will yield 2 sums , one for u1s and one for u2s.)

Again a factor of two will arise from assumption of nn interactions and double sum over

Rand R (sums are equivalent).As we assumed only nn interactions,

[ ] [ ])()()()()(

4

1

2,1

.RuRuRRRuRuU

s

RRs

harm !"!"!!!"=

#=

!

!!!!!!

!!

$

dad x

uG

x

uKdaordRR

!"

"=

"

"=!=#!

2

2

2

2

;);(,!!

K > G

dandan =!!+! )1()1(

withina cell

dadanna !=+!! )1(

betweencells


16/24

d a - d

(n-1)a (n+3)a(n+2)a(n+1)ana

[ ]221 )()(

42 naunauK

!

1) Single out ion in cell na, and examine interaction with ion in same cell.

2) Single out ion in cell na, and look at interaction with ion in cell (n + 1)a.

[ ]212 ))]1([)(4

2 anunauG

+!

[ ] [ ]!! +"+"=nn

harmanunau

Gnaunau

KU

2

12

2

21

. ))1([)(2

)()(2

So

within a cell between cells

Equations of Motion (2 sets)

[ ] [ ]

[ ] [ ] 1,1221

2

12

2

21

11

.

1

))1([)()()(

))1([)(2

)()(2)()(

)(

+!!!

!

!!

""

""

+!#!#!#!#=

$%&

'()

+!#!+!#!*

*#=

*

*#=

nn

nn

nn

nn

harm

anuanuGanuanuK

anuanuG

anuanuK

naunau

UnauM

++

!!

( )1,

)(

]1[

)(

)(

+!

!

=

"

+!"

=

"

!"

nn

nn

nau

anu

nau

anu

#

#

Singled out site at na; nninterac. ( spring const. G)

with cell at n-1


[ ] [ ]))1([)()()()( 21211 anunauGnaunauKnauM !!!!!=!!1


17/24

[ ] [ ]))1([)()()()(

)( 12122

.

2 anunauGnaunauKnau

UnauM

harm

+!!!!="

"!=!! 2


at site na at site (n+1)a

Solutions of form:

)(

22

)(

11

),(

),(

tknai

tknai

etnau

etnau

!

!

"

"

#

#

=

=

$1and $2 are amplitude and

phase of u1and u2 (to be

determined)

Apply B-vK b.c.

Nm

ak !2= m is an integer, N is number of unit

cells in chain (there are 2N ions)!

Now substitute us into Eqs. of Motion

1

2

( ) 021

2=+++!

!

""# ika

GeKGKM

( ) 012

2=+++! ""#

ikaGeKGKM

Rewriting

( ) 021

2=+++!

!

""# ika

GeKGKM

( ) 02

2

1 =+!++ "#" GKMGeK

ika

Pair of homogeneous linear eqs.- 2unknowns (.1 and .2)

(Solutions if det. of coeff. vanishes)

This condition yields dispersion relation

kaKGGKMM

GKcos2

1 222++

+

=!3


18/24


Look at amplitudes of displacements:

From and 32

ika

ika

ika

ikaika

GeK

GeKthus

GeKkaKGGKbut

kaKGGK

GeK

GKM

GeK

+

+

=

+=++

++

+

=

+!

+

!=

!

!

1

2

22

222

1

2 cos2,cos2)(

"

"

#"

"

Consider two limiting cases:

1) k &0, so ( )2

1cos

2

kaka !"

( )( )

( )22

2

2221

222

1ka

GK

KG

M

GK

M

GKkaKGKGGK

MM

GK

+

!

+

+

=!++

+

"#

(K + G) 2

Positive Root

Use binomialexpansion on

radical

( ) !"

#$%

&+

+

'+

+

( ....)(2

1 22

2ka

GK

KG

M

GK

M

GK)

4

( )222 )(

)(2;....)(2 kaO

M

GKsoka

GKM

KG

M

GK!

+

="#

$%&

'+

+

!+

( ++

))

Binomial expansionagain; also note:

const. plus neg.

quadr. term. (ZB

value < Z. centerNegative Root

( ) ka

GKM

KGsoka

GKM

KG

M

GK

M

GK

)(2....;)(

2

22

+

!+

+

+

+

"

+

!""

##Linear in k, and.2= + .1

Long )modes ".2= +.1 for +-; and .2=.1 for ++

5

.2= - .1


19/24


2) k = */a, (cos *= - 1)

[ ]M

GK

M

GKKGGK

MM

GK !

+

=!+

+

" 2

1

2222

1#

Positive Root

M

Kso

M

K

M

GK

M

GK 2;2

2==

!

+

+

"++

## .2= - .1

Negative Root

M

Gso

M

G

M

GK

M

GK 2;2

2==

!

!

+

"!!

## .2= + .1

(k)

k = 2*/)0-2*/a -*/a 2*/a*/a

1stBZ

M

K2

M

G2

M

GK )(2 +

kaGKM

KGSlope

)(2 +=

6

For k = */a, eika= cos(*)+

i sin(*) = cos(*) = -1

So ( K > G )

ika

ika

GeK

GeK

+

+

=!

1

2

!

!

1

1

2!! =

!

!

=

GK

GK

"

"

Motion at k = */a: (eikan= ei*n; )=2a)

d a - d

d a - d

.1 corresponds to ; .2 corresponds to

++ optical modes

+

- acoustic modes

Within cell ions are out-of-phase

Within cell ions are in-phaseIn long )limit ()>>a) motion is the samecell-

to-cell

Note: for K = G, no GAP at BZ boundary; just

monatomic lattice with spacing 2a. (left as

exercise)


20/24


21/24


22/24

Classical Harmonic Crystal 3D

Extension to 3D

Monatomic BL first

Uharm.

=

1

2

!

u(!

R)"

D(!

R!!

"R)!

R!

"R

# !

u(!

"R)

Matrix/Tensor D!

Make use of symmetries to derive prop. of D

(RR)

D!(RR) = D!(RR)

D!(RR) = D!(RR)

%R D!(R) = 0

See A&M

3N Eq. of Mot., one for each 3 components of displ. of N ions

)()()(

)(,

.

RuRRDRu

URuM

R

harm

!!"=#

#"= $

!

!!

!""

! %

%

%

Assume harm. Solutions: )(),( tRkietRu !" #=!!

!

!

!

$ is complex polarization vector

Use B-vK b.c. s: 321),()( NNNNRuanRu ii !!==+!

!!

!

! Very large integers

Allowed values of k3

3

5

2

2

2

1

1

1b

N

nb

N

nb

N

nk

!!!!

++=

Remember, only values ofk lying inprimitive cell of RL give distinct

solutions

Next Page


23/24


So e-val. Eq. becomes ssss kkDM !"!!# !

!

!

"

!

)()(2 ==

3 normal modes with wave vector kwill have pol. vectors$s(k)

and frequencies ; when k 00,M

kk

s

s

)()(

!

! !" = akkck

ss >>! "# ,)()(!!

As in 1D case

In general, in 3D must consider relationships among directions of pol. vectors, $,

and directions of propagation vectors,k. Isotropic media

Choose one $sparallel to k (longitudinal); other two are perpendicular to k (transverse)

Anisotropic media (not so simple)

For k along symmetry axes (3-fold, 4-fold, 6 fold) one pol. can be chosen along k, and

the other two perpendicular (common in cubic matls). Refer to long. And transverse

modes even for general direction where they are mixed not purely long. or transverse.

Subst. assumed solutions into eqs. of motion:

Rki

R

eRDkDwherekDM

!!

!

!"!"

!

"

!

!"== )()(,)(2 ##$D(k) is even function of kand a real,

symmetric matrix.(from above

symmetries)

From matrix algebra - real symm. matrix has 3 real e-vectors that satisfy

)()()()( kkkkDsss

!!!!"

!"! = e-val. Eq. --- diagonalizes matrix D(k)); $1, $2, $3

e-vectors can be made orthonormal 3,2,1,,)()( =!= !! sskk ssss "##!

!

!

!


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Ions vibrate

relative to

one another

in cells

Cells vibrate

as a unit

!s(!

k)!

3D Lattice with a Basis(Introduce index to refer to ions in the basis) OPTICAL MODES

For each k, 3p normal modes, where p is number of ions in basis

Frequencies (s = 1, 2, !.p) are all functions of k with periodicity of RL.

Threeof 3p branches are acoustic -- goes to 0 linearly ask 0.

Other 3(p-1)branches are optical -- const. as k 0.

!s(!

k)

!s(!

k)

Schematic for p = 2ZB

!s(!

k)

!

kZB

TTL

T

L

T

Along symmetry directions incubic crystals, 2 transverse

modes are degenerate and lower

in freq,. than long. mode

Why are freq. of L modes higher than T?Additional contrib. to restoring force for L.

Optical due to long range Coulomb

interactions;acoustic due to elastic

restoring forces being smaller for T than for L.

Important points:

1)

Original Eqs. of motion: p atoms/cell, N cells, 3pN equations

2) Reduce to set of eqs. in k-space; only 3p equations, (for each allowed k, N values of k.

3) Results in terms of set of independentSHOs for each k (can always diagonalize matrix D(k))

4) Generate dispersion relations .!s(!

k)

Theory.classical Harmonic Crystals.dell.10 12(2)

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Transcript of Theory.classical Harmonic Crystals.dell.10 12(2)