Quantum Harmonic Crystals.10_27

8/10/2019 Quantum Harmonic Crystals.10_27

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Quantum Theory of Harmonic CrystalsIntroduction

Complex interacting many-body problem reduced to a set of simple independentcollective excitations; results from periodic lattice and identical cells

Classical situation we have shown that classical coordinates, us(R),and momenta,Ps(R), can be transformed to new set in which eqs. of motion are those of an

assembly of independent Simple Harmonic Oscillators (SHOs) of frequencyfor each value of k.

)(ks

!

!

Quantum mechanically, allowed energies of a (1D) simple harmonic oscillator aregiven by , where !is the frequency of the oscillator (more later).!!)(

21

+= nEn

Nowwe use general QM result for energy density

Here Eiis energy of ith stationary state of the crystal, is the statistical weight of state Ei

at temperature T, and "is over all stationary states.

BT

i

Ei

E

i

ke

eE

V

ui

i

1,

1=

!!

!

"

#

$$

$

%

&=

'

'(

(

))

)

iEe !"


2/30

Quantum Theory of Harmonic CrystalsNormal Modes vs. Phonons

We can describe energy of the system in terms of the excitation number nksof normal modesof wave vector kin branch s. (Classical Picture)

An equivalent corpuscular picture is to designate the quanta of excitation (normal modes

above) by the term PHONONS and to treat the phonons as indistinguishable quantum particles

(bosons in this case).nksis the number of phonons of wave vector kand type s.

(Quantum Mechanical Picture)

To calculate u, introduce the Partition Function

.)ln(1

! "=i

EieV

f #

!"#$ &' #(') &("&

!"

"#=

fu

En!ks

!" #$= (n!

ks+

12)"!

s(!

k)!

ks

% Is the energy of this configuration.

Here represents a configuration, i.e., the excitation number of the normal modes at k.

For each k

For a particular k energy can only take on values

n!ks

!" #$

!!s(!

k)

2,3!!

s(!

k)

2,5!!

s(!

k)

2,..........

Evaluate f:

Rewrite as

f =1

V ln exp !!En!

ks"# $%

"#

$%

n!ks

"# $%&

"

#''

$

%((1


3/30

Quantum Theory of Harmonic Crystals

n!ks

!" #$ n!ks

1In Eq. specifies particular SETof allowed values of , e.g.,

n!k1s

=100,n!k2s

= 75,n!k1s

= 50,........... or n!k1s

=10,n!k2s

=1,n!k1s

= 0,.......

So we can write !in as,1

exp !! n!ks+

12( )!!s

!

k( )"# $%!

ks

&n!ks

"# $%

' 2

Can exchange order of !and ", provided !is over just allowed values of nks (0,1,2,3#..)

exp !! n!ks+

12( )!"s

!

k( )"# $%n!

ks

&!

ks

'2 3

Students verify that every term occurring in also occurs below (exactly once)2

exp !! 12( )!"s

!

k( )"# $%+ exp !! 32( )!"s!

k( )"# $%+ exp !! 52( )!"s!

k( )"# $%+........( )!

ks

&

But the series in ( ) above is a convergent geometric series of ratio exp !!

2!"

s

!

k( )"

#$%

&'

So

3 =

exp !!

2!"

s

!

k( )"

#$%

&'

1! exp !!!"s

!

k( )"# %&!

ks

( and thus f = 1V

ln

exp !!

2!"

s

!

k( )"

#$%

&'

1! exp !!!"s

!

k( )"# %&!

ks

(


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And "# *+, - .*+

u = !"f"!

= !1V

"f"!

ln

exp !!

2!!

s

!

k

( )

#

$%

&

'(1! exp !!!!s

!

k( )#$ &'!

ks

)

*

+,,

-,,

.

/,,

0,,

/01&2 '3& *+453'62+&7 "+8

&"92 8201:"6:2u =

1

V!!

s(!

k) ns(!

k)+ 12

!"

#$

!

ks

%&'(

)*+

4

Where is the Bose-Einstein distrib. function,the mean excitation number of mode ks,

or the ave. # of phonons of type ks,at temp. T

ns(!

k)=1

e!!!s

!

k( )!1

5

Therefore

u = uequil. + 1V

12!!s(

!k)

!ks! + 1V

"!s(

!

k)e

!!!s(!

k) "1!ks!

;


5/30


General expression

cV =

1V

!!T

!!s(

!

k)

e!!!s(

!

k) "1#$% &

'(

!

ks

) 6 Depends ondetails of $s(k)

Limiting cases

High T: (arg. of exponent in is small --- expand in series, andthen use binomial expansion.) Take der. after doing this.

1

!>> !"

s(!

k)

6

cV =

3N

VkB = 3nk

B

Dulong and Petit

(classical result); additional terms in series

expansion give Quantum Corrections

Low T: First convert !to integral (dense set of allowed k-values).kBT


6/30


Comments (low T):1) at low T phonons with kBT


7/30

Quantum Theory of Harmonic Crystalslow T (cont.):

! "# #$

$=

s spacekallk

sV

se

kkckd

Tc

1

)(

)2( )(3!

"

!"

!

%&'

With previous simplifications at very low T:

F'*18 "+E*2

2*2G2+&

Use spherical polar coord. dk k2dkd"; and let x = # cs(k)k!

( )( ) 330

3

23

4

)(

1

43

11

,12

3

kc

d

cwheree

dxx

c

Tk

Tcss

x

B

V !" !" #

=

$%

&

'(

)

*+,

-./

0

1

1=

2

33

15

4!

34 15

4$+6 7%,*+8&98 +$:;


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Quantum Theory of Harmonic Crystals - Debye

Intermediate T - Debye Interpolation Scheme:1) Replace ALL branches of vibrational spectrum with three branches,

each with same disp. rel.

2) Replace Integral over 1stBZ with integral over sphereof radius kDchosen

to contain exactly N allowed values of k(N= # of ions in xtal). Leads to

ckk =)(!

!

H'*A 'B

I#& JD

H'*A 'B K2L$2

#


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!!"

=

##$

%

&&'

(

")

)=

D

B

B

D k

Tk

ck

B

Tk

ck

B

k

ckV

e

dkkTk

ce

k

e

dkk

T

c

c0 2

42

2

0

3

2

)1(

)(

2

3

12

3!

!

!

!

!

** +

2$A9B

1#$ &C*D$ "&9 C$ >+7E$9

c

kkwithxdkkx

Tk

ckDB

D

B !

! !== ;oftermsinandwritingand,

=H1+ &20G# 'B #1+E*2

2G 1F

!*

15

4,,

4!

="#$

$


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High T: :expansionbinomialandexp.forexpansionser.--1)(xintegrand, DT

B

T

D

B

T

x

x

D

BV nk

xTnk

e

dxexk

Tnkc

DD

33

9)1(

9

/

0

33

/

0

2

43

3

=!"

#$%

&''(

)**+

,-

./'

'(

)**+

,-

=

--

0 S*"##1="* U2#3*&

Physical Interpretation and comments:1) kD measure of inverse interparticle spacing

2) !D measure of maximum phonon frequency

3) $D temperature abovewhich ALL modes begin to be excitedthermally, and below which all modes begin to be frozen out .A Temperature that

VERY APPROXIMATELYseparates classical from quantum regimes.

4) NOTE: $D%&D= ckD, where c is sound velocity, the slope of&(k) vs. k.

As we have seen, in long wavelength limit &(k) = a[K/M]1/2. So $D %[K/M]1/2. This

means that hard materials (large K) like diamond have high Debye temperatures

(particularly if they have atoms with light masses like Carbon; while soft materials

like Pb and In have low Debye temperatures (particularly if they are comprised of

heavy mass atoms. This is directly reflected in the frequencies of optical modes.

)(k!

!

!

k/0

Dk

!

!%&


11/30

Quantum Theory of Harmonic Crystals - EinsteinEinstein Model:

Previous discussion for monatomic crystals. In Debye model of polyatomic xtals, optical

branches of the!

(k) spectra represented by high k-values of the same linear expression!(k) = ck , whose low k-values give acoustic branch.More realistic to apply Debye model only to

acoustic branches, and represent optical branches by single frequency (indep. of k). (Einstein )

Pictorial comparison: (Debye vs. Einstein 2D sq. diatomic latt. primitive cellseparation = a)

N%O& NP .%O&

Q,'"&< 0+&9"#

P$C6$ "7+"


12/30


13/30

Quantum Theory of Harmonic Crystals Density of Normal Modes

!! dg )(

Often have to deal with lattice properties that depend on summing over all kand s. (Also

in electronic properties), e.g.,

Where is some function of frequency, which itself is a function of k. Convenient

to transform to equivalent frequency interval and integrate over frequency.

( ) ( ) ( ),)(

21)(1

3

,!"! =s

s

sk

s kQkdV

kQV

!

!

!

!

#

$

#

)(kQs

!

!

13

Introduce density of normal modes - phonon density of states (DOS) g(!) defined such

that is number of normal modes (phonons) in the infinitesimal range d&between

&and &and &+ d&, divided by volume of crystal.

Then integrals like can be written13

.)()(!= """ dQgq 14?(1+9 'B X "# B3+=&A

'B!'+*$ (202

Comparing and we can show that13 14

g(!)=

d!

k

2!( )3 ! "!"s(

!

k)( )"s# . 15

Y*3E 1+&' Z

1+&20=("+E2 '0820 'B ""+8

1+&2E0"*O "+8 3#2


14/30

Quantum Theory of Harmonic Crystals Density of Normal Modes

A&M provide another useful representation (for anisotropic mode dispersion);

Use g(!)d!is just # of allowed modes (ks) in freq. range between !and !+ d!divided by vol.

of xtal. But this is just vol. of k-space cell (axes along prim. BL axes) with &'&s(k) '&+d&divided by vol. of k-space per allowed k. Defines shell in k-space. Convert to surface

integral and get

( ) )(

1

2)(

3k

dSg

ss surface

s!!

!"

!

#=$ %

@+&2E0"6'+ 1# ':20 &("&

#30B"=2 1+ P0#& JD '+ )(1=(

&s(k) = &

Because &s(k)periodic there are values of k (typically at BZ boundaries) for which the

denominator (group vel.) vanishes Van Hove Singularities in DOS. -- also

happens in electronic case. Important in phonon and electronic properties (optical in

particular).

We can do everything weve done previously (total energy, specific heats, etc. in terms of

level density (usually called the density of states (DOS), e.g., Debye approx.

Take all three branches to have same dispersion,&s(k) = ck, and all wave vectors

assumed to like within sphere of radius kD

gD(!) = 3

d!

k

2!( )3

! "! ck( )k

"**')28 #&"&2# 1+

I#&JD

16

17


15/30

Quantum Theory of Harmonic Crystals Quantizing normal modes

Now examine in more detail the concept of PHONONS (more than justEn= (n + ")#$).

First examine the Hamiltonian for a 1D simple harmonic oscillator and consider

the OPERATOR APPROACH and commutation relations for obtaining theallowed energies and eigenfunctions.

Then generalize this to 3D and a periodic lattice to consider how to apply thesimple 1D results to phonons.

This involves the CREATION and ANNIHILATION operators .a+

and a


16/30

Review: Simple Harmonic Oscillator

H =p2

2m +V(x) =p2

2m +1

2 m!2

x2

,where!=K

m ,

with K the spring constant and ! the characteristic frequency.

Schroedinger Eq.

Hamiltonian

p2

2mu+

1

2m!

2x

2u =Eu

Commutator Bracket [p,x]= (px!xp)=

!i!

Rewrite Hamiltonian

!"

#$%

&+

+

'(

=++(=2

1

2

)(

2

)(

2))((

2

1)

)

)

)

)

)

))

!!!

!

m

pixm

m

pixmpixmpixm

mH

Define Annihilation and Creation Operators

!

!

!

!

!! m

pixma

m

pixma

2

)(;

2

)(

"

=

+

= +

So

!"

#$%

&+=

+

2

1 aaH '! Very simple form


17/30

Simple Harmonic Oscillator (Cont.)

Properties of Creation and Annihilation Operators

],[0],[;1],[ aaaaaa === +++

And, using commutator identity

1

23

2

],[

:3],[

2],[

,],[],[],[

!

++

++

++

=

=

=

+=

nn

naaa

aaa

aaa

thatshowcanCBACABCBA

From properties of operators, and can show that,nnn

uEuH =

H(a+

un)= [ H,a

+

]un+ a

+

( Hun), and that [ H,a

+

]= !!a+

thus H(a+

un)= (E

n+!!)(a

+

un)

))(()( nnn uaEuaH !!"

=

similarlyShows that if unis e-f of with e-val En, (a

+un)is also e-f of , but

with e-val En+H H

!!

In General, (a+un)is e-f of with e-val En+ , and (aun)is e-f

of with e-val En! .

H !!n

H !!n


18/30

Now assume that system has ground state, u0, with energy E0, and consider

))(()( nnn uaEuaH !!"

=

Clearly .generatetothisusecanweand,0 00 uua =

.,)2

1exp()( 0

2

00 ionnormalizatisNwherexm

Nxu!

!

"=

Use to to generate u1, u2, etc.+

a

Can get e-values without knowledge of explicit e-fs.Hu0 = !!(a

+a+1

2)u0 =

1

2!!u0,!E0 =

1

2!! .

Generate excitedstates with :

nth excited state is

so e-vals are:

+

a

).2

1(, 0 nEanduau nn +=!

+

"!

).21( += nEn !!

Can Generate thenormalized e-f

s from

)

2

1exp(where,

!

1get

:1*and,

2

2/1

00

0

xmm

uuan

u

dxuuuacu

n

n

nnnn

!!

!

"

!

#$

%

&'

(

)==

==

+

+*

*#

+

+

Simple Harmonic Oscillator (Cont.)


19/30

Simple Harmonic Oscillator and Phonons

Summary

[ ]

( )

1

!

1

)(1,

)(

2

*

0

21

21

=

=

=

+=

=

+=

!

+

+

+

n

nmmn

n

n

nn

u

dxuu

uan

u

unuH

aa

aaH

#

$

$

!

!

SHO energy levels

!!

+ - >

+ - N

+ - V

+ - I!!

How do we apply this to phonons?

Normal modes

!!"

""#+=RRR

RuRRDRuRPM

H!!!

!!!!

)()()()(2

1212

$$

?"92 &""# "+8 (""#&'

L2 "+E3*"0 B025A "+8


20/30

Commutation Relations


[ ] [ ])(),(0)(),(

)(),(

RPRPRuRu

iRPRu RR

!==!

=!!

!!!!!!

"

!!!!!

""

"" ##

Can show that

RLVkN

RLVke

R

Rki

is,

not,0=!

!

!!

Leads to

[ ] [ ]+!

+

!

!

+

!

==

=

kkk

kkkk

aaaa

aa

k

!!!

!!!!

!

,0,

, "

ExpressPanduin terms of aand a+

( )! +"+=k

Rki

kk ekaa

kMNRu

!

!!

!

!!!

"

"!!)(

)(2

1)( #

$

( )! +"""

=

k

Rki

kk ekaa

kM

N

iRP

!

!!

!

!!!

"!!)(

2

)()( #

$

19

M#2

0=! R

Rkie

!

!!

!! "## =)()( kk!!

)()( kk!!

!= ""

)()( kk!!

!= ""


21/30


Substitute Eqs. for Panduin terms of aand a+into Hamiltonian19 18

Kinetic Energy ( )( )kk kkR aaaakRPM k!

!!!

! !

!

"

!

!

++

! !!= "" )(41

)(2

1 2#

Potential Energy ( )( )++! !+ !" kk

kk aaaak

k

!

!

!!!

!" )(

4

1#

Add these to get H (use )kk

aaaa

k

k

k

k !!

!

!

!

! !! ++" ="

( )! ++

+=

k

kkk kaaaakH

!

!!!!

!

" )(21 "

Nowuse , and include branches (s)1, =+

kk aa !!

! "#

$%&

'+=

+

sk

sks skaakH

!

!!

!"

2

1)( ( 20

Sum of 3N indep. Harmonic osc. (N values of k and 3 branches)Eigenvalues are sum of individual eigenvals. and e-functions are products. Therefore

specify eigenstates of system by set of 3N quantum numbers, nks, one for each of 3N

independent harmonic osc. Hamiltonians

! "#

$%&

'+=

sk

sks nkE

!

"

!#

2

1)((


22/30

QuantumHarmonic Crystals Measuring Phonon Frequencies andDispersion

Two primary Probes used: 1) neutrons; 2) photons (inelastic scattering processes)]'&2^ 'Cc=G - >AV +G7


23/30


$%&'() ") *+ ,-.'#

$%&'(/ "/ *+ 01#

> V d e c I>CIc

CIe

CId

CIV

CI>C c

C e

C d

C V

>

V

d

e

?$


24/30

QuantumHarmonic Crystals Crystal momentum

S"+ #(') $S,


25/30


Fundamental physics -- symmetries of Hamiltonian imply conservation laws

(this symmetry (invariance of Ham under translations by BL vector --- k + K is

conserved. Write as

Knkppsk

sk

!"

!"

!!

!

! +!"="# $


26/30


-./012 34145

Crystal in some state with set of phonon occ. #s.[nks]; neutron with initial mom.p, energy E = p

2/2Mn.

6/.12 34145

Crystal in state characterized by [n*ks]; neutron withp(and energy E*= p*2/2Mn.

Conservation of Energy Assume harmonic approx. for xtal

skxksk

sk

sks

sk

sks

sk

sks

nnnnkEEor

nkEEnk

!!!

!

!

!

!

!

!

!"

!"

!"

!"=##!=!"

"+"=+

$

$$

;)(

)()(

%

%%

Conservation of Crystal Momentum

Knkppor

Knkpnkp

sksk

sksk

sksk

!"

!"

!!

!"

!"

!!"

!

!

!

!

!

!

!

+!"="#

#+#=+

$

$$


27/30


28/30


=*8$:F

!

When we subst. for kin energy expression, we can ignore K, because &s(k)is periodic

function in reciprocal space, with period = shortest RLV (K) in direction of interest.

So, procedure is to solve momentum eqs. For k K (equiv. to k), and subst. in energy eq.

for &s(k) gives &s((pp)/ )) single eq. for abs. and single eq. for emission.

!"

#$%

& '('=

(

!"

#$%

& '(+=

(

!

""

!

!

""

!

pp

M

p

M

p

pp

M

p

M

p

s

nn

s

nn

)

)

22

22

22

22

Absorption

Emission

We knowpandp2/2Mn; measure 3 components of final neutron momentum and final

neutron energy. In general, three components and energy define 3D surface, and by

specifying a direction (detector angle) determine point on this surface. See neutrons

scattered by one-phonon processes at a few discrete energies, E(. Use spectrometer to

measure as function of neutron wavelength (+*= 2,/q*); Determines |p*|; knowing direction,

can construct E*- E, andp

- p = k . ((= (p

2 - p2)/2Mn -- plot this vs. k = (p - p)/ .

(homework problems on graphical solutions for simple cases). Determine kand &s(k) from

whole series of angles, orientation of crystal and neutron energies.

! !

!


29/30

QuantumHarmonic Crystals Phonon Frequencies from InelasticLight Scattering

Photon Scattering: (Inelastic Light Scattering)

Usually Visible Light. 1) Raman Scattering (Optical Phonons); 2) BrillouinScattering

(acoustic phonons). Can only measure phonons at very small wave vectors (near k = 0)

What is k for a visible photon (wavelength 500 nm)? Wave vectors for photons inside media

(solids) differ from those in vacuum (index of refraction, n; qin= nqout ).

!!+ !!s(!k)= " !!

!n"q +!

"k+ !

"K= !n

"!q

! !! = !!" !!s(!k)

"n!!q = "n

!q ""

!k+"

!K

7 823%4%+ 5670 70,4%8 *+ 76,99-

" 823%+%+ 5670 70,4%8 *+ 76,99-

XC:*+,'*9 *) *9$ ,#*9*9

+7

+79

"[

H%7::7*9 *) *9$ ,#*9*9

+7

+79

"

[

k"RA 2+20E$ 'B `Y? > G2HZ Y('&'+ 2+20E$ "L'3&

V 2HO 4o\ n o\pZ q5qn q5pq7 Z#'

&01"+E*2 1# "


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Quantum Harmonic Crystals.10_27

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Transcript of Quantum Harmonic Crystals.10_27