Ordered lattie site

[14℄Nowwewanttosolvetheequationofmotiongivenin(5.23). Thesituation

is very similar to the usual three dimensional lattie disussed in subsetion

(3.2),sowewilltrythesameansatz

u l,m,n = U e ^iωt e ^{i k · L} , v l,m,n = V e ^iωt e ^{i k · L}

^and

w l,m,n = W e ^iωt e ^{i k · L} ,

^(5.34)

where again

U

^,

V

^and

W

^{are onstants telling the amplitude of the wave in}

dierentdiretions,

k = (k 1 , k 2 , k 3 )

^{isthewavevetor,}

L _l,m,n = l a 1 +m a 2 +n a 3

is the translation vetor telling the loation of the lattie site and

ω

^{is the}

frequenyofthewave. Notethattheabsolutevaluesoftheprimitivetranslation

vetors

a 1

^,

a 2

^and

a 3

^{areequal,asexplainedin thebeginningofthishapter,}

so

k · L = a (lk 1 + mk 2 + nk 3 )

^,where

a

^{istheabsolutevaluedisussed.}

Theansatzresultsintheequation

− ω ² U e ^iωt e ^{i k · L}

= − ω ² ₀ 2 − e ^{iak 1} − e ^{−iak 1}

U e ^iωt e ^{i k · L} − i Kω

2 e ^{iak 1} + e ^{−iak 1}

V e ^iωt e ^{i k · L} +i Kω

2 e ^{iak 1} + e ^{−iak 1}

W e ^iωt e ^{i k · L}

+ K ²

4 4 − 4e ^{iak 1} − 4e ^{−iak 1} + 2e ^{2iak 1} + 2e ^{−2iak 1}

U e ^iωt e ^{i k · L}

+ K ² 4

2e ^{iak 1} − e ^{i(ak 1 +ak 2 )} − e ^{i(ak 1 −ak 2 )}

V e ^iωt e ^{i k · L}

+ K ² 4

2e ^{−ia 1 k 1} − e ^{i(−ak 1 +ak 2 )} − e ^{−i(ak 1 +ak 2 )}

V e ^iωt e ^{i k · L}

+ K ² 4

2e ^{ia 1 k 1} − e ^{i(ak 1 +ak 3 )} − e ^{i(ak 1 −ak 3 )}

W e ^iωt e ^{i k · L} ,

+ K ² 4

2e ^{−iak 1} − e ^{i(−ak 1 +ak 3 )} − e ^{−i(ak 1 +ak 3 )}

W e ^iωt e ^{i k · L} + O (θ ³ ),

(5.35)

where we have dened the onstants

ω 0 = r ζ

m

^and

K = ζθ

^{to shorten the}

notation.

Dividingwith

e ^iωt e ^{i k · L}

^{andreognizingtheosinesyields}

whihafter olletingtermswiththesameonstantbeomes

0 =

Togetridofthethreeonstants

U

^,

V

^and

W

^{weneedmoreequations. Thuswe}

shallonsidertheequationsfor

v ¨ l,m,n

^and

w ¨ l,m,n

^{. Lookingatour}Hamiltonians wesee that they aresymmetri under thehange

u → v

^,

v → w

^and

w → u

^.

Thereforetheequations ofmotionfor

¨ v l,m,n

^and

w ¨ l,m,n

^{arederivedandsolved}

ompletelysimilarlyto

u ¨ l,m,n

^{,so theymustyieldthesamesolution,exeptfor}

hanging

U

^,

V

^and

W

^aswellas

k 1

^,

k 2

^and

k 3

^{in thesamemanneras}

u

^,

v

^and

w

^{anbehanged. Thusstartingwith}

¨ v l,m,n

^{weshould gettheequation}

0 =

andfromtheequationfor

w ¨ l,m,n

^weget

Sineallthreeoftheseequationshavethesameonstants

U

^,

V

^and

W

^,wean

ombine the equations to matrix form. The matrix will bequite large, so we

willwritealloftheequationsas

AU + BV + CW = 0,

^(5.41)

Nowweanwritetheequationsas



solvetheeigenvaluesandeigenvetorsof thematrix. Justasin thealulation

foroptial and aoustiwavesin (3.1.2), there are nontrivialsolutionsto this

matrixequationonlyifthedeterminant

det





A B C D E F G H J



 = 0.

^(5.54)

This gives us another equation for the system. Solving

ω

^{from this equation}

givesusthespetrumofvibrationsandsubsequentlythedensityofstates.

Sinetheentriesofthematrixarequitelarge,thedeterminantisenormous.

Thereforewewillexpandthesinestoseondorderin

k 1

^,

k 2

^and

k 3

^,sothatfor

example

sin ² (ak 1 )

^beomes

a ² k ₁ ²

^andthedeterminantequationbeomes

− ω ⁴ 3K ² − ω ²

+ a ² ω ₀ ² ω ² K ² − ω ²

k ₁ ² + k ² ₂ + k ² ₃

= 0.

^(5.55)

Thisisareasonableapproximation,sinein lowtemperaturesthewavevetors

k

shouldbesmall.

Theisofrequenysurfaesin thisequationaresphereswithradius

r = q

k ₁ ² + k ² ₂ + k ² ₃ ,

^(5.56)

sothisis oneoftheisotropi situations disussedin theend of thesubsetion

(3.2.2). This is natural sine we assumed the system to be isotropi in the

beginning of this hapter. Therefore the density of states an be alulated

withthevolumeinsidetheisofrequenysurfaes.

Sinethedeterminantequalszeroweget

r = s

ω ² 3K ² − ω ²

a ² ω ₀ ² (K ² − ω ² ) .

^(5.57)

Thus thevolumesofthespheresare

4πr ³

3 = 4π 3ω ³ ₀ a ³

ω ² 3K ² − ω ² (K ² − ω ² )

! 3/2

.

^(5.58)

Here we anreognize the veloity of sound

v c = r ζ

m a = ω 0 a

^{for arystal}

orrespondingtooursystemshownin (3.89).

Ifweproeededsimilarlytothealulationsinsubsetion(3.2.2),thedensity

ofstates

g

^{would be}

V / (2π) ³

^{times thederivativeof thisvolumewith respet}

to

ω

^,where

V

^{isthevolumeofthewhole glass. Howevernowwehavesplit the}

lattieinto twokindsofpartiles,orderedanddisordered,andbothofthetwo

stateshastobedividedbytwo,sothatfortheorderedpartiles

The purpose of the absolute value is to keep

g o

^{positive. Expanding the}

de-nominatorand omittingthe

K ⁴

^terms,sine

K

^is proportionalto thesmall

θ

^,

simpliestheexpressioninto

g o (ω) = V ω ²

Thedensityofstateshasadivergeneat

ω o,div = K.

^(5.61)

ThustheglassreallyhasbehavioursimilartovanHovesingularities. Realling

that

K = ζθ

^{wesee thatthe} singularityisrelated tothenonommutativityof ourmodel. Themoredisorderedthesystemis,thelargerthenonommutativity

parameter

θ

^{shouldbe,andthereforethehigherthefrequenyofthe}singularity shouldbe.

5.2.2 Disordered lattie sites

[14℄ Now wewant to solvethe equation of motion(5.32), whih desribes the

partiles in the disorderedlattie sites. The proedure is the sameas for the

ordered sites. Pluggingin thesame ansatz(5.34)as for theordered partiles

anddroppingthe

O θ ³

Dividingwith

e ^{i k · L} e ^{iak 1}

^{givestheequation}

ω ² − ω ₀ ² 2 − e ^{iak 1} − e ^{−iak 1} U +iω K

2

4 − e ^{iak 2} − e ^{−iak 2} V

− iω K 2

4 − e ^{iak 3} − e ^{−iak 3} W = 0.

^(5.63)

Expressedwithosinesitis

ω ² − 2ω ² ₀ (1 − cos (ak 1 )) U +iωK { 2 − cos (ak 2 ) } V

− iωK { 2 − cos (ak 3 ) } W = 0.

^(5.64)

Again the equation of motion for

v ¨ l+1,m,n

^{should leadto the sameresult,}

exept that

U

^{should be hanged to}

V

^,

V

^{should be hanged to}

W

^and

W

shouldbehangedto

U

^,and

k 1

^,

k 2

^and

k 3

^{shouldbepermutedsimilarly. Thus}

weget

ω ² − 2ω ₀ ² (1 − cos (ak 2 )) V +iωK { 2 − cos (ak 3 ) } W

− iωK { 2 − cos (ak 1 ) } U = 0.

^(5.65)

For

w ¨ l+1,m,n

^thepermutationshould bedonetwie,leadingto

ω ² − 2ω ₀ ² (1 − cos (ak 3 )) W +iωK { 2 − cos (ak 1 ) } U

− iωK { 2 − cos (ak 2 ) } V = 0.

^(5.66)

Colletingthethree equationsleadstothematrix





ω ² − 2ω ₀ ² (1 − cos (ak 1 )) iωK (2 − cos (ak 2 )) − iωK (2 − cos (ak 3 ))

− iωK (2 − cos (ak 1 )) ω ² − 2ω ₀ ² (1 − cos (ak 2 )) iωK (2 − cos (ak 3 )) iωK (2 − cos (ak 1 )) − iωK (2 − cos (ak 2 )) ω ² − 2ω ² ₀ (1 − cos (ak 3 ))



 ,

(5.67)

whih hasnon-trivialsolutionsonly ifitsdeterminantequals0. Expanding in

k

yields

− ω ⁴ 3K ² − ω ²

+ a ² − K ² ω ⁴ + K ² ω ² ω ₀ ² − ω ⁴ ω ₀ ²

k ₁ ² + k ² ₂ + k ² ₃

+ O k ³ .

(5.68)

Omittingthe

O k ³

termsandsettingthedeterminanttozerogives

k ² ₁ + k ² ₂ + k ₃ ² = ω ² 3K ² − ω ²

a ² (K ² ω ² ₀ − (K ² + ω ₀ ² ) ω ² ) .

^(5.69)

Thedensityofstatesanagainbealulatedwiththevolumeoftheisofrequeny

spheres. Using

p k ₁ ² + k ₂ ² + k ² ₃

^astheradius

r

^{ofthesphereleadstothevolume}

4π

3 r ³ = 4πω ³ 3K ² − ω ² 3/2

3a ³ (K ² ω ₀ ² − (K ² + ω ² ₀ ) ω ² ) ^3/2 .

^(5.70)

uphalf ofthelattie,weseethat thedensityofstates

after disardingthe termwith

K ⁴

^{in thenominator and using theveloityof}

sound

v c = ω 0 a = p

ζ/ma

^{in a}orrespondingrystal.

Nowthedensityofstateshasadivergeneat

ω d,div = K

p K ² /ω ₀ ² + 1 .

^(5.73)

Expandingin

K

^{letsuswritethedivergeneas}

ω d,div = K + O K ³

,

^(5.74)

whihisthesameresultasfortheorderedpartiles. Thereforeweanonlude

that making everyseond partile of a lattie disorderedgivesthe density of

statesofallpartilesthesamepeakfrequeny. Thisisdesirableforourmodel,

sinethebosonpeakappearsonlyatonefrequenyforeveryglass.

In document A Noncommutative Glass Model and the Boson Peak (sivua 40-46)