Applications of Group Theory & Phase transitions Mike ...cloud.crm2.univ-lorraine.fr/pdf/Bogota2018/Glazer_Bogota...Mike Glazer MathCryst School Bogota 2018 Landau theory of phase

Mike Glazer MathCryst School Bogota 2018

Applications of Group Theory & Phase transitions

Mike GlazerUniversities of Oxford and Warwick


PHASE

TRANSITIONS

DISRUPTIVE

NON-

DISRUPTIVE

FERROIC

NONFERROIC

NONDISRUPTIVE

FERROELASTIC

NONFERROELASTIC-

FERROIC

TRANSLATIONAL

NONDISRUPTIVE

ISOSTRUCTURAL

3BaTiO 3 4Pm m P mm

3Cu Au 3 3Fm m Pm m

3 3Ce 6 / 6 /P mmc P mmc

1 4QUARTZ 3 21 6 22P P

3Ti, Zr, Hf P6 / 3mmc Im m

A crystallographic classification of phase transitions


Zn

Cu 0.5(Cu+Zn)

0.5(Cu+Zn)

Displacive

Order-disorder


EhrenfestHysteresis

L=00 0

p

GL T T S

T

= − =

1st order

2nd order, continuous

T0

T0

T0

T0

T0

T0

Paul Ehrenfest1880-1933


Landau theory of phase transitions

Lev Landau 1908-1968

2 4 6

0

1 1 1( ) ....

2 4 6= + + + G G a T b c

0 order parameter = 0 for T>T =

2 4

0

1 1For 0 assume ( )

2 4b G G a T b = + +

Find stable solution for 30 ( )

dGa T b

d= = +

0Assume ( ) '( )a T a T T −

( )

1 112 22

0

( ) '0 and

a T aT T

b b

= = − = −

Critical exponent b

Assume group-subgroup relationship,define quantity called the order parameter


0G G−

2 0T T=

0T T

0T T

0T T

0( ) '( )a T a T T −


0G G−

2

0T T=

0T T0T T

0T T

i.e. with b > 0 we get a 2nd order transition


b < 02 4 6

0

1 1 1( )

2 4 6G G a T b c= + + +

Find stable solution for

3 50 ( )dG

a T b cd

= = + +

0Assume ( ) '( )a T a T T −


0G G−

2 0T T=

0T T

0T T

0T T

2 4 6

0

1 1 1( )

2 4 6G G a T b c= + + +


0G G−

2

No minimum0T T=

0T T

0T T

0T T

2 4 6

0

1 1 1( )

2 4 6G G a T b c= + + +


0G G−

2

0T T=

cT T=

cT T

0cT T T

i.e. with b < 0 we get a 1st order transition

2 4 6

0

1 1 1( )

2 4 6G G a T b c= + + +


T

P

( )1

40Prove that if 0 then b T T= −

Critical Point

C

gas

liquid

solid

ice water steam− −


cT

( )1

40Prove that if 0 then b T T= −

Critical Point


3( )dG

E a T P bPdP

= = +

21( ) 2

dEa T bP

dP= = +

0 0

1 0 ( ) '( )T T P a T a T T = = = −

( )

1122

0 0 0

' 1 2 '( )

aT T P T T a T T

b

= − = −

Dielectric susceptibility

Li2Ge7O15:0.7%Bi

Curie-Weiss Law (Curie Law in magnetism)

A.K. Bain, Prem Chand & K. VeerabhadraRao, Ferroelectrics- Physical Effects(2011)


2 4 6 2 2

0

1 1 1 1 1( )

2 4 6 2 2G G a T b c C= + + + + +

Strain coupling

Exercise


2 4 6 2 2

0

1 1 1 1 1( )

2 4 6 2 2G G a T b c C= + + + + +

210

2

dGC

d= = +

2

2C

= −

2 4 6 4 4

0

1 1 1 1 1( )

2 4 6 4 8G G a T b c

C C

= + + + − +

22 4 6

0

1 1 1( )

2 4 2 6

= + + − +

G G a T b c

C

Strain coupling

Can be positive or negative


Effect of strain on domain growth and phase transition order

How many domains? How many ferroelastic and ferroelectric domains?

BaTiO3 : P4mm3Pm m


Effect of strain on domain growth and phase transition order

BaTiO3 : P4mm3Pm m

486

8i = =

Number of ferroelectric states in each ferroelastic state = 2

Number of ferroelastic states = 3


What’s it all about?

Eigenvalues

Eigenvectors


S3

65 = 6-1

SCHOENFLIES VS INTERNATIONAL


Symmetry OperationsSchoenflies International

Rotation -

inversion

S65

S64 = C3

2

S63 = i

S62 = C3

S6

S35

S34 = C3

S33 = s

S32 = C3

2

S3

3

32 = 32

33 = 1

34 = 3

35

6

62 = 3

63 = m

64 = 32

65


Class

bxxa 1−=

Each element is said to be in a class if there is a similarity transform

a is said to be conjugate to b

a is always conjugate to itself

e.g. 2/m (C2h)

2112__

11 === −− mmmmSame

class


110 ( )dm s

_

110

( )dm s

100 ( )vm s

010 ( )vm s

)(2)(4 2

22

4CC =

)(1 E1 1 3

110 001 110 110 010 0014 4m m m m− −= =

But in 4mm (C4v) the elements are

)(4 4C

)(4 33

4C

5 classes

1

4 43

2m100 m010

m110 m110

_


Representations of a group

•A set of matrices G, each corresponding to an operation in the group

e.g. consider 2/m (C2h)

_

001 2 0011( ) 2 ( ) 1( ) ( )hE C i m s

+

+

+

100

010

001

−

−

100

010

001

−

−

−

100

010

001

−

+

+

100

010

001


Let Y = Basis Function

Suppose

11 =R 22 =R

Consider

2121 = RRR

But R1 operates on each element of Y individually

2 is a matrix that permutes elements of Y

1221 RRR = 1221 =RR

Better to use transposed matrices =R


Note that in group theory it is usual to specify operations according to

Cartesian axes rather than crystal axes as used so far

−

100

0cossin

0sincos

These representations are 3-dimensional representations and can in

general be decomposed into so-called irreducible representations


Irreducible representations

AXXA 1' −=

Consider matrices A, B, C and D forming a representation of a group

Suppose we perform a similarity transformation:

BXXB 1' −=

CXXC 1' −=

DXXD 1' −=

DAB = if Now

BXAXXXBA 11'' −−= ABXX 1−= '1 DDXX == −

So the set of matrices A’, B’, C’ and D’ is also a representation of the group


Suppose A can be written in matrix form e.g. 6 X 6 matrix

This is a 6-dimensional reducible representation

2X2

1X1

3X3

6 X 6

AXXA 1' −=


2X2

1X1

3X3

Suppose A’ can be written in matrix form as a BLOCKED matrix e.g.

These form 3 irreducible representations 1d, 2d and 3d


2X2

1X1

3X3

6 X 6

2X2

1X1

3X3

AXXA 1' −=

A A’

Reduction of Reducible

Representations


Characters

In general, it is difficult to do this process to find the correct matrix X, unless the

representations are 1-dimensional. Fortunately, the traces of the matrices,

called CHARACTERS , can be used, since they behave in the same way and

are by definition 1-dimensional.

_

001 2 0011( ) 2 ( ) 1( ) ( )hE C i m s

+

+

+

100

010

001

−

−

100

010

001

−

−

−

100

010

001

−

+

+

100

010

001

1 2 1 m

3 1− 3− 1


Character Tables

group oforder ......2

3

2

2

2

1

2 hlllli =+++=

rep eirreduciblth i' ofdimension Let (1) =il

ji

jihRR ij

R

ji

=

===

for 0

for )()(ity Orthogonal (2)

identical are class

same in the matrices all of characters therep,any In (3)

classes ofnumber reps eirreducibl ofnumber (4) =

reps d-1four are there

repsfour bemust there(4) From

reps d-1four or rep d-2 oneeither (1) From

classes 4 are thereand 4h )(C 2/min e.g. 2h

=


)(1 E )(2 2C )(1 i )( hm s

gB 1 1− 1 1−

uA 1 1 1− 1−

uB 1 1− 1− 1

g = gerade (even) u = ungerade (odd)

gA 1 1 1 1

)(/2 2hCm

Totally symmetric

representationhttp://www3.uji.es/~planelle/APUNTS/TGS/taules_TG_oxford.pdf


)(3 3vCm

)(1 E

)(3

)(3

2

3

2

3

C

C 110

120

210

( )

( )

( )

v

v

v

m

m

m

s

s

s

Order h = 6

Number of classes

(conjugate elements) = 3

6211 (1) From 222 =++

reps 3 classes 3 (4) From =


0))(1(3))(1(2)2)(1(1 =++= zyEA

0))(1(3))(1(2)2)(1(2 =−++= zyEA

232 −=+ zy 232 −=− zy0 1 =−= zy

Check group oforder 6)0(3)1(22 222 =+−+

ji

jihRR ij

R

ji

=

===

for 0

for )()(ity Orthogonal (2)

)(1 E )2(32 3C )3(3 vm s)(3 3vCm

1A 1 1 1

2A 1 1 1−

E 2 y z

Totally symmetric rep


−= 2 1Group { , , ,............... }n

n n n nC E C C C

There are n irreducible reps each of which is 1-dimensional

EC n

n =

1)( =n

nC

nm

eCmC nmi

nn

,......3,2,1

)( /2

=

=

IR 'th the in of Character

Characters in Cyclic Groups


For group 3 (C3) we can write the effect of rotation thus

= 3/2

3)(3 ieC

*)(3 3/23/42

3

2 == − ii eeC

In this group there h = 3 and so there are three 1-d irreducible reps

)(1 E )(3 3C )(3 2

3

2 C)(3 3C

A 1 1 1

1 1 *

2 1 *

E =


)(1 E )(3 3C )(3 2

3

2 C)(3 3C

A 1 1 1

E 23

2cos2

3

2cos2

Or for analysis of physical problems we can use

http://www3.uji.es/~planelle/APUNTS/TGS/taules_TG_oxford.pdf

https://tinyurl.com/y83xyynn


Reduction of reducible reps: Magic formula

typeof reps eirreducibl ofnumber =n

igi classin elements ofnumber =

=i

i RRgh

n )()(1


=i

i RRgh

n )()(1

_

001 2 0011( ) 2 ( ) 1( ) ( )hE C i m s

0)]1)(1(1)1)(3(1)1)(1(1)1)(3(1[4

1=+−+−+=gA

n

0)]1)(1(1)1)(3(1)1)(1(1)1)(3(1[4

1=−+−+−−+=gB

n

1)]1)(1(1)1)(3(1)1)(1(1)1)(3(1[4

1=−+−−+−+=uA

n

2)]1)(1(1)1)(3(1)1)(1(1)1)(3(1[4

1=+−−+−−+=uB

n

uured BA 2+=

)(1 E )(2 2C )(1 i )( hm s

gB 1 1− 1 1−

uA 1 1 1− 1−

uB 1 1− 1− 1

gA 1 1 1 1

)(/2 2hCm

+

+

+

100

010

001

−

−

100

010

001

−

−

−

100

010

001

−

+

+

100

010

001

)2/m(Cin tion representa reducible a had Earlier we 2h


• Find group symmetry for the crystal/molecule being studied

• Mathematical functions to describe property in question

•Apply each symmetry operator in group to each function to obtain a

Reducible Representation

•Reduce Representation to a sum of Irreducible Representations

•Understand the meaning of each Irreducible Representation, e.g. using

Projection Operator Techniques

•Do experiment to find eigenvalues

•Interpret using eigenvectors

Procedure


Vibrations of H2O

)(2 2C

x

y

z


Vibrations of H2O

)(2 2C

x

y

z

( )y

y vm s


)(2 2C

x

y

z

( )y

y vm s

( )x

x vm s

)(2 2vCmm

VIBRATIONS OF H2O


)(2 2C ( )v

y

ym s ( )x

x vm s

2A 1 1 1− 1−

1B 1 1− 1 1−

2B 1 1− 1− 1

1A 1 1 1 1

)(2 2vCmm )(1 E


x3

y3

z3

x2

y2

z2

x1

y1

z1


1

1

1

1

1

1

1

1

1

)(1

3

3

3

2

2

2

1

1

1

333222111

z

y

x

z

y

x

z

y

x

zyxzyxzyxEx3

y3

z3

x2

y2

z2

x1

y1

z1

9=


1

1

1

1

1

1

1

1

1

)(2

3

3

3

2

2

2

1

1

1

3332221112

z

y

x

z

y

x

z

y

x

zyxzyxzyxC

−

−

−

−

−

−

x3y3

z3

x2

y2

z2

x1

y1

z1

1−=


1 1 1 2 2 2 3 3 3

1

1

1

2

2

2

3

3

3

( )

1

1

1

1

1

1

1

1

1

y

y vm x y z x y z x y z

x

y

z

x

y

z

x

y

z

s

−

−

−

x3y3

z3

x2

y2

z2

x1

y1

z1

1=


1 1 1 2 2 2 3 3 3

1

1

1

2

2

2

3

3

3

( )

1

1

1

1

1

1

1

1

1

x

x vm x y z x y z x y z

x

y

z

x

y

z

x

y

z

s

−

−

−

x3y3

z3

x2

y2

z2

x1

y1

z1

3=


=i

i RRgh

n )()(1

3 1 1- 9

)( )( C E vv2

red

yzxz

ss

331194

1n 1A

=++−=

131194

1n 2A

=−−−=

231194

1n 1B

=−++=

2Bn ????=

)(2 2C ( )v

y

ym s ( )x

x vm s

2A 1 1 1− 1−

1B 1 1− 1 1−

2B 1 1− 1− 1

1A 1 1 1 1

)(2 2vCmm )(1 E


=i

i RRgh

n )()(1

3 1 1- 9

)( )( C E vv2

red

yzxz

ss

331194

1n 1A

=++−=

131194

1n 2A

=−−−=

231194

1n 1B

=−++=

331194

1n 2B

=+−+=

2121 323 BBAA +++=red

)(2 2C ( )v

y

ym s ( )x

x vm s

2A 1 1 1− 1−

1B 1 1− 1 1−

2B 1 1− 1− 1

1A 1 1 1 1

)(2 2vCmm )(1 E


2121 323 BBAA +++=red

)(2 2C ( )v

y

ym s ( )x

x vm s

2A 1 1 1− 1−

1B 1 1− 1 1−

2B 1 1− 1− 1

1A 1 1 1 1

)(2 2vCmm )(1 E

Tz

RzTx Ry

Ty Rx

212 BAvib +=

x3y3

z3

x2

y2

z2

x1

y1

z1

3N-6 vibrations

To find out the types of vibration we use projection operators


Projection Operators

• If a number of normal modes transform as a particular irreducible representation projection operator techniques only give a linear combination of the eigenvectors involved. These are called symmetry-adapted vectors.

R

RRV )(

• Define the character projection operator:


R

RRV )(

1 2A

z y xV E m m= + + +2 2

A

z y xV E m m= + − −1 2

B

z y xV E m m= − + −2 2

B

z y xV E m m= − − +

A1 modes

0)or or ( 3211 = xxxV

A

0)( 122111 =−+−= xxxxxV

A

x3y3

z3

x2

y2

z2

x1

y1

z1

)(2)( 21122111 yyyyyyyV

A−=+−−=

0)( 31 =yV

A

)(2)( 2111 zzzV

A+=

33 4)(1 zzVA

=For A1 modes

oxygen does not

move in x and y



A1 modes


R

RRV )(

B2 modes

0)(2 =xVB

x3y3

z3

x2

y2

z2

x1

y1

z1

)(2)( 21122112 yyyyyyyV

B+=+++=

33 4)(2 yyVB

=

)(2)( 2112 zzzV

B−= 0)( 3

2 =zVB

1 2A

z y xV E m m= + + +2 2

A

z y xV E m m= + − −1 2

B

z y xV E m m= − + −2 2

B

z y xV E m m= − − +


B2 mode


Triangular molecule

E 32C23C hs

32S vs3'

1A'

2A

'E"

1A"

2A"E

hD3

1

1

2

1

1

2

1 1 1 1 1

1 1− 1 1 1−1− 0 2 1− 0

1 1 1− 1− 1−

1 1− 1− 1− 11− 0 2− 1 0

zRxyT

zT

yxRR

z

x

y

62m


C1: 1

C2: 2010, 2110, 2100

C3: m210, m1-10, m120

C4: -6+001, -6-

001

C5: 3-001, 3+

001

C6: m001

62m


1 0 3 1- 0 9

3 2S 3C 2C E v3h23

red

ss

1 2

3

130330912

1n 1A'

=+++−+=

130330912

1n 2A'

=−++++=

261812

1n

E'=+=

030330912

1n 1A"

=−+−−+=

130330912

1n 2A"

=++−++=

161812

1n

E"=−=

""'2'' 221irr EAEAA ++++=


0 1- 2 0 1- 2

1- 2

1- 1 1-

2

1- 1

1 2

1- 1 1

2

1- 1

3 2S 3C 2C E v3h23

xy

y

x

T

T

T

ss

−−

−→

−

2

1

2

32

3

2

1

cossin

sincos

For example, 3-fold rotation C3 on Tx and Ty

=E’

''1 EAvib +=


A’1 mode


These are all the same mode as molecular orientation does not matter

Bending mode

Asymmetric modeThese two can be taken to be E’ as

the two partners are orthogonal

motion


Kronecker Products

Let G be a group with subgroups H and K such that

khhkK then k H,h (1) =

hkg as expressed becan G g all (2) =

{E} K H (3) =

G is the outer direct product of H and K

KHHKG ==

H and K are invariant subgroups of G

Number of classes in G = number of classes in H x number of

classes in K

If H and K are not commutative – semi-direct product of H and K


KHG =

G oftion representa andK and H of tionsrepresenta are and If ijji =

jiij = Kronecker product

bacge = ..

=

baba

baba

bb

bb

aa

aa

2221

1211

2221

1211

2221

1211

=

2222212222212121

1222112212211121

2212211222112111

1212111212111111

babababa

babababa

babababa

babababa

)()()( bac =


Kronecker Square

ji

i

oftion representa therequire wei.e.

tionrepresenta particular a of square i.e.

VV tohappens what know wish toWe

V. space ain function a have weSuppose

We can produce two functions

symmetric 2

1 +

ij

ijji

ricantisymmet 2

1 −

ij

ijji


The result for characters is

)()(2

1)( 22

2 RRR =

For a 1-dimensional representation

)()( 22 RR

=

2v 2 2

E 2 m m

. . in C (1 1) (1 1) (-1 -1) (-1 -1)

z y x

e g A A =

1 1 1 1 =

1A=


2-dimensional example --- E” representation

3h 3 2 h 3 v

2

D E 2C 3C 2S 3

2 -1 0 -2 1 0

4 1 0 4 1 0

s s

2 (R ) 2 -1 2 2 -1 2

Symmetric square

Antiymmetric square

3 0 1 3 0 1

1 1 -1 1 1 -1

Symmetrised square = A’1+ E’

Antiymmetrised square = A’2

62m


Selection Rules

0|| ifonly possible Transition

iHf

state ground for symmetric totally be i| of tionrepresenta Let i

elementmatrix zero-nonFor

0 ity orthogonal otherwise |

i| →=

fH

tionrepresenta symmetric totally the i||f =

H

= H

|f if possibleonly is This

General Principle:

The Kronecker product of the representations of the three terms

must contain the totally symmetric representation


Infra-red Absorption

onperturbatidependent -time moment dipole

with couples field py thespectrosco absorptionIn

→p

Ε

p.E=

'H

0|'| if possible Transition

iHf

acquiredenergy additional '=

H

==

ifeifiHf |||||'| r.Ep.E

}||||||{ ++= izfEiyfEixfEe zyx


Vibrational spectraSolution of harmonic oscillator problem in quantum mechanics

hnEn )2

1( +=

2/2

)()( −= eHA nn

constant gnormalisin=nA

polynomial Hermite)( =nH

xm

2/1

=

24)( 2)( 1)( 2

210 −=== HHH

nodes no symmetric totally is )(0 H

symmetric totally is state ground


Dipole transition has symmetry properties of translation

Therefore we must consider functions such as

etc. || iTf x

dxxf )(function Consider

= 0)( dxxf

+

-

ricantisymmet odd,

)(xff(x) R −⎯→⎯

0)( dxxf

++

symmetricy totall

)(xff(x) R⎯→⎯


symmetric totally is 0 dT ixf

species symmetric totally contains

then assymmetry same has If ixfxfi TT

An infra-red transition is allowed if the product of the

states before and after transition transforms with

respect to the relevant symmetry operations like a

coordinate axis.


e.g. H2O molecule

)(2 2C ( )v

y

ym s ( )x

x vm s

2A 1 1 1− 1−

1B 1 1− 1 1−

2B 1 1− 1− 1

1A 1 1 1 1

)(2 2vCmm )(1 E

z

Rz

x Ry

y Rx

Modes A1 B1 B2 are infra-red active

Thus all three modes seen in IR spectrum


E 32C23C hs

32S vs3'

1A'

2A

'E"

1A"

2A"E

hD3

1

1

2

1

1

2

1 1 1 1 1

1 1− 1 1 1−1− 0 2 1− 0

1 1 1− 1− 1−

1 1− 1− 1− 11− 0 2− 1 0

zRxyT

zT

yxRR

'

molecule Triangular

1 EAvib += active red-infra is 'only E


Raman Scattering

2|k| 0 =

In a crystal the electric field of a monochromatic

wave propagating with wave-vector

].[0 0tieEΕ

−=

rk0

This induces a dipole moment by exciting radiation

rlity tensopolarisabi == Ep


].[0 0tieEΕ

−=

rk0

E


+

-

rlity tensopolarisabi == Ep

The polarisability is modulated by vibrations

Let Qj = normal coordinate of the jth mode

E


....2

1'

'0'

2

0

0 +

+

+= jj

j j j jj

j

j

QQQQ

QQ

][ ti

jjjeAQ

−=

.rk j

...'])()[(0

0

][0

00000 +

+=

−−

ti

j

j

j

ti jeEAQ

eEp

.rkk.rk j

j 0frequency oflight scattered

jkkk s += 0

js += 0

js += 0

onannihilatiantiStokes

jkkk s −= 0

js −= 0

creationStokes

js −= 0


o o+1o−1

jkkk s += 0

js += 0

js += 0

onannihilatiantiStokes

jkkk s −= 0

js −= 0

creationStokes

js −= 0


0|| if b

Consider symmetry of polarisability tensor

zzzyzx

yzyyyx

xzxyxx

field electricincident of

direction is when direction in inducedon polarisati yxxy =

yxyx EP = yxxy TT like is


e.g. H2O

1 1 1 1

2 -1 -1 1

1 -1 -1

-1 1 -1

x y

z

xz

xy

T T

m

m

=

=

=

=

2 like s transform Axy

212 BAvib +=

All modes are Raman active

Note: to observe B2 mode light must be polarised parallel to

y direction and will be radiated polarised parallel to z. This is

particularly important for crystals.


Triangular molecule

'1 EAvib +=

E’ is infra-red active; A1 and E’ are Raman active

A mode is Raman active if the product of the states before

and after transition transforms with respect to the relevant

symmetry operations like the product of a pair of

coordinate axes.

Note: a totally symmetric mode is always Raman active


)(2 2C ( )v

y

ym s ( )x

x vm s

2A 1 1 1− 1−

1B 1 1− 1 1−

2B 1 1− 1− 1

1A 1 1 1 1

)(2 2vCmm )(1 E

z

Rz

x Ry

y Rx

E 32C23C hs

32S vs3'

1A'

2A

'E"

1A"

2A"E

hD3

1

1

2

1

1

2

1 1 1 1 1

1 1− 1 1 1−1− 0 2 1− 0

1 1 1− 1− 1−

1 1− 1− 1− 11− 0 2− 1 0

zRxyT

zT

yxRR

Subgroup to Group

A1+B2=E’


Infra-Red Raman

Equilateral E’ A1+E’

Isosceles 2A1+B2 2A1+B2

D3h

C2v

A1+B2=E’

Correlation Table


Lattice Vibrations (Sorry Massimo!)( )=

ls

l

ssuMT 2

2

1 crystal ofenergy Kinetic

componentsCartesian cell, in the atoms cells, sl

Potential energy V is expanded in a power series of displacements

( ) ( ) ( ) ( ) ( ) ...2

1 '

'

''

'

'0 +++= l

s

l

s

slls

ll

ss

l

s

ls

l

sVV b

b

b

uuu

.....210 +++= VVVV

( )( )

ionconfigurat mequilibriu 0

0

=

=

l

ss

l

u

V

( )( ) ( )

0

'

'

2'

'

=

l

s

l

s

ll

ssuu

V

b

b

If higher order terms neglected, harmonic approximation

b

b

along displaced is

)''( atom when the)( atomon direction in the force theis ofcomponent slls


irrelevant is and termstatic is 0V

( ) m.equilibriufor zero is and

),( atom on the acting force theof negative theis because 01 slV l

s=

( ) ( ) ( ) ( ) ++=

b

bb ls ls sl

l

s

l

s

ll

ss

l

ss uuuMVH''

'

'

'

'

2

02

1

2

1

is crystalfor n Hamiltonia Total

( )( )

( ) ( )−=

−=

''

'

'

is )( atomfor motion ofequation The

sl

l

s

ll

ssl

s

l

ss uu

VuM

ls

This gives an infinite number of coupled differential equations to solve. This can be simplified by making use of lattice periodicity i.e. the motions in different unit cells must be identical in amplitude and direction but not necessarily in phase.


( ) cellunit th ' ofector position v is )( )(.2/1 lleeUM liti

ss

l

s rurk−−=

( ) )(.2/12 liti

ss

l

s eeUMu rk −−−=

( ) )'(.2/1

'

''

'

'

)(.2/12 liti

ss

sl

ll

ss

liti

ss eeUMeeUM rkrk −−−− −=−

='

'

2 ),'(s

ss UssDUb

bb k

( ) )]'()'(.['

'

2/1

')(),'( lli

l

ll

ssss eMMss rrkkD

−=

matrix Dynamical ),'( =kD ss

So problem is reduced to solution of 3n equations, since as s runs over 3n values. D is a 3n X 3n matrix.


0),'( '

2 =− ssssD bb k

Non-trivial solution given by

Normal coordinates

3n values of 2 at given k k2

Problem can be made even simpler. T is diagonal, whereas V contains cross-products. So replace

s

lu by new parameters

These correspond to independent oscillators

Normal modes


So, why is this too difficult in practice?

Consider just 2 atoms and the possible force constants.




Internuclear force constant




Electron-electron force constant e.g. van der Waals




Nuclear-electron force constant




Nuclear-electron force constant




Polarizability




Polarizability

Applications of Group Theory & Phase transitions Mike ...cloud.crm2.univ-lorraine.fr/pdf/Bogota2018/Glazer_Bogota...Mike Glazer MathCryst School Bogota 2018 Landau theory of phase

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