Lecture 4 microscopic theory

Lecture 4 Microscopic Theory

•The 2-Electron Problem •Second Quantization: •Annihilation and Creation Operators •Solution of the 2-electron Schroedinger Equation •Cooper Pairs •The many-electron problem-BCS Theory •Solution of the Many-particle Schroedinger Equation by the Bogoliubov-Valatin Transformation •The BCS Energy gap

Even number of electrons/unit cell

Band picture - electrons in momentum space

electrons in a periodic potential form Bloch waves and energy bands

Bloch waves

n,k (r ) e

ik r u

n.k (r ) Energy eigenvalues

n (k )

Odd number of electrons/unit cell E

metal insulator semiconductor

E

energy gap

Repulsive interaction between electrons is a perturbation

Fermi sea

Fermi liquid of “independent” Quasiparticles (Landau, 1956)

Insulator, Semiconductor

Metal

Phonon Coupling The Cooper Pair Problem

+ + + +

+ + + + Analogy

+ + + + 2 Bowling Balls on a

- + + + + MATTRESS

Cooper Pairing

Many electron system

+ + _ + +

† †

1122

21 ,kqkkqk

kkqCCCCVH

Consider a subset of the many – electron system , i.e. a Cooper pair, with 2 free electrons with antiparallel spins (for parallel spins, exchange terms reduce the phonon-mediated attractive electron-electron interaction). With no interaction,

2211 ..

2121 ,,,xkxkxxkk

i

e

Assume ϵF – ωD < ϵk , ϵk ± q < ϵF + ωD so that H ̎ is predominately attractive

† †

(here we have let k’ replace k2 and k replace k1).

Consider two free electrons, and introduce center of mass coordinates:

x = x1 – x2

q kk

kqkkqk CCCCVH'

''''

)(2

1

);(

21

)(

2122211

xxX

xxkk xkxk

i

e1

xXx

xXx

2

1

2

1

2

1

kKk

kKk

kkkkkK

2

1

2

1

)(2

1

2

1

2121

''1

'')(2

1

'H'

- , 0

4

11

22

),,,(

22

2

2

1

21

222

2

2

1

)(

ninteractioelectron -electron theIntroduce

thatsoConsider

: is state thisofenergy The

Hpm

Hppm

H

mm

k

m

k

ei

kk kkK

kK

xXkK xkXK

'

'

'

)('

'

)(

21

0',''',)(

0)(

0)(

2121

21)(

k

kkk

k

ii

Hgg

egHedd

H

k k

ie

ieg

ieg

kkkk

xx xxk

k

xxk

xkxk

kxk

kx

form the of oneigenfunti an forLook

m

F

cgdVg

V

Hgdg

DFmFk

mk

m

andF

and

2

2

)'()'(')()(

''H'

)(

0''')'()'(')()(

22

2

1

: where,m

K F

Kbetween i.e. D

F

between sea fermi theof top the tostateselectron -one theConfine

stateselectron -2 ofdensity

Fm

F

m

F

F

m

F

m

F

m

F

d

V

dV

dV

Cg

22ln

2

2ln

'

'1

'

1'1

0'

)'('1

)(

2

2

2

2

2

2

result. sobtain thinot could

n calculatioon perturbati Vin seriespower a as written benot may

pair bound

0 e)(attractiv 0

1

1

2

1

22

F2

V

VFe

D

VFe

Fm

- - 2 1

The region of increased positive charge density propagates through the crystal as a

quantized sound wave called a phonon

The passing electron has emitted a phonon

A second electron experiences a Coulomb attraction from the increased region

of positive charge density created by the first electron

BCS Theory – a Brief Treatment

For many electrons, we need to make sure the many-particle wave function is anti symmetric.

We can write in general that the Hamiltonian is:

† †

† †

sksksqk

qk

sqkq

sksksqk

qsksk

sqkq

CCCCVHH

CCCCVHH

,',',

,

,0

,','','

,',',,

,0

2

1

2

1

:case) Cooper the in (as -k'k which for sonly state consider us Let

the are s' s,Here indices. spin

Summing over s, it can be shown (using anticommutator relationships for the annihilation and creation operators) that:

† †

† †

Here we have chosen S ↑ , S´↓ (to minimize the energy as before), and summed over S,

We have also assumed that Vk,k’ = V-k,-k’

Note that the eigenstates for H0 are just the Block waves uk eik.x in

the crystal.

k'k,

-

(1.)

kkkkkk

k

kkkkk

CCCCV

CCCCH

'''

k

kC

kC

kH

0 taken and

†

Eq. (1.) is the BCS Hamiltonian

There are in general 2 approaches to solve the many-particle

Schroedinger equation (see, e.g. TINKHAM):

1. variational approach to minimize the energy

2. solution by a canonical transformation (the Bogoliubov/ Valatin transformation).

We will illustrate the second approach here.

Bogoliubov diagonalized the Hamiltonian for the liquid helium superfluid condensate by introducing 2 new operators:

kc

kc

kc

kc

kkkk

cvcu

cvcu

kkkkk

kkkkk

,,,

0''

for solve and (2.)invert then We

i.e. ate,anticommut also s' theshown that becan It

and

The Bogoliubov/ Valatin

transformation. (2.)

†

†

†

†

†

Substituting these C’s into (1.) gives as the kinetic energy term

HT (1st set of terms):

† †

† †

Take mk = m-k = 0 for the ground state.

Next we consider the potential energy term Hv (second set of terms

with V)

kkkk

k

kkkkkkkkkkkkT

km

km

vummuvvH

and Here

)(22 222

2

1

2

1

2

1 ,

2

1

' '''

such that, k

x variablenew a introduce weand

sorder term th 4 eneglect th now We

0 termsdiagonal offorder th 4

',

22'''

2

Then,

.V

Hin termsdiagonal-off ingcorrespond

the T

Hin termsdiagonal-off t theinsist tha weH, ediagonaliz To

kkkk xvxu

kk kC

kC

kC

kC

kkV

VH

kkkkkkk

vk

uk

vk

ukk

V

kkkkkk

vk

uk

cancel

†

†

†

†

† †

This gives:

022

1

2

4

1

k2

2

1

2'4

1

'

(3.) 04

12

4

12

toleads (4.) and (3.) from

(4.) which,

by given k

quantity new a define weNow

'

2

1

2

''

2

1

2

kkx

kx

kk

xkk

Vk

xVxxk

kkkkkk

constant). electrons ofnumber the(keeping potential

chemical thebe energy to of zero thechoose now We

moment. ain

case special afor thisdo will Wesolved. becan thisknown, is kk'

V If

2

1

2

''

'

''2

1

give, now (5.) & (4.) and

22

2

gives for Solving

(5.)

kk

k

k kk

Vk

kk

k

kx

kx

2

1

,

21

(...)222

,0

221

kx

kk

x

kk

vk

uk

v

and

kmSo

kC

kC

kC

kx

or

)degeneracy (Spin

choose F

Eenergy Fermi k

For

N

:that sons,interactio of absence in terms latter the neglect We

N

isN of value nexpectatio the

km state,ground the in

kk

CN Consider

† †

involved.

phonon theofenergy the,q

than less )F

E to(relative k

choose n,interactiophoton -electron in theorigin its having kk'

VFor

root. square the thereforechoose we

0kk'

V when caseelectron free the toreduce To

2

1

22k

2

k want We

021 2

1 choose ,

kFor

negative

k

kx

kx

kx

FE

We take

Vkk’ = V, constant if |εk|< ħωD

= 0, otherwise

Here ωD is the Debye Frequency

Here ∆k can be evaluated as

2

1

''2

'''

2

1

kk

kkkk dD

Here we will take the “density of states” D(εk’) as a constant, D(EF).

Consider Vkk’ Vkk’

ħωD

εk

So we need only evaluate

V

1

2

122

)(

1sinh

2

1

F

D

F

EVD

dEVDD

D

gives This

This is the BCS gap energy in the density of states

For weak coupling (V small), this can be written as:

For the ground state wave function and finite temperature effects,

See TINKHAM.

)03.0~

2)(

1

eV

e

DD

EVD

DF

( of 1% ~ Typically,