1. 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

2. 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) eik r
un.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

3. Phonon Coupling
The Cooper Pair Problem
+ + + +
+ + + + Analogy
+ + + + 2 Bowling Balls on a
- + + + + MATTRESS
Cooper Pairing
Many electron
system
+ +
_
+ +

4. † †
1122
21 ,
kqkkqk
kkq
CCCCVH 
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 ,,, xkxk
xxkk 
 i
e

5. 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
)(
212
2211
xxX
xxkk xkxk

 i
e1
xXx
xXx
2
1
2
1
2
1



6. 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
22
2
2
2
1
)(
ninteractioelectron-electrontheIntroduce
thatsoConsider
:isstatethisofenergyThe
Hp
m
Hpp
m
H
mm
k
m
k
ei









 
kkkkK
kK
xXkK xkXK


7. 
 




 




'
'
'
)('
'
)(
21
0',''',)(
0)(
0)(
2121
21)(
k
kkk
k
ii
Hgg
egHedd
H
k k
i
e
i
eg
i
eg
kkkk
xx xxk
k
xxk
xkxk
k
xk
kx




formtheofoneigenfuntianforLook

8. 















m
F
cgdVg
V
Hgdg
DFmF
k
m
k
m
and
F
and
2
2
)'()'(')()(
''H'
)(
0''')'()'(')()(
22
2
1
:where,
m
K
F
Kbetweeni.e.
DF
betweenseafermitheoftopthetostateselectron-onetheConfine
stateselectron-2ofdensity






9. 

























Fm
F
m
F
F
m
F
m
F
m
F
d
V
dV
dV
C
g
22
ln
2
2
ln
'
'1
'
1
'1
0
'
)'(
'1
)(
2
2
2
2
2
2









10. result.sobtain thinotcould
ncalculatioonperturbatiVinseriespoweraaswrittenbenotmay
pairbound
0e)(attractiv0
1
1
2
1
22
F2










V
V
Fe
D
VFe
Fm





12. - -
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

21. 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)Cooperthein(as-k'kwhichforsonly stateconsiderusLet
theares's,Here indices.spin

22. 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 k
C
k
C
k
H 
0
takenand
†

23. 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:

24. k
c
k
c
k
c
k
c
kkkk
cvcu
cvcu
kkkkk
kkkkk






,,,
0
''
forsolveand(2.)invertthenWe
i.e.ate,anticommutalsos'theshown thatbecanIt
and




The Bogoliubov/ Valatin
transformation. (2.)
†
†
†
†
†

25. 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
k
m
k
m
vummuvvH





 


andHere
)(22 22
2

26.     
2
1
2
1
2
1
,
2
1
' '''
such that,
k
xvariablenewaintroduceweand
sorder termth4eneglect thnowWe
0termsdiagonalofforderth4
',
22
'''
2
Then,
.
V
Hintermsdiagonal-offingcorrespond
the
T
Hintermsdiagonal-offt theinsist thaweH,ediagonalizTo














 









kkkk xvxu
kk k
C
k
C
k
C
k
C
kk
VVH
kk
kkkkk
v
k
u
k
v
k
u
kk
V
k
kkkkk
v
k
u
k

cancel
† †
† † † †

27. This gives:
022
1
2
4
1
k
2
2
1
2
'4
1
'
(3.)0
4
1
2
4
1
2
toleads(4.)and(3.)from
(4.)which,
bygiven
k
quantitynewadefineweNow
'
2
1
2
''
2
1
2







 





















kk
x
k
x
k
k
x
kk
V
k
xVxx
k
kkkkkk



28. constant).electronsofnumberthe(keepingpotential
chemicalthebeenergy toofzerothechoosenowWe
moment.ain
casespecialaforthisdowillWesolved.becanthisknown,is
kk'
VIf
2
1
2
''
'
'
'2
1
give,now(5.)&(4.)and
22
2
givesforSolving (5.)




 




kk
k
k kk
V
k
kk
k
k
x
k
x




29.  
 
 
2
1
,
21
(...)222
,0
221













k
x
k
k
x
k
k
v
k
u
k
v
and
k
mSo
k
C
k
C
k
C
kx
or
)degeneracy(Spin
choose
F
EenergyFermi
k
For
N
:thatsons,interactioofabsenceintermslattertheneglectWe
N
isNofvaluenexpectatiothe
k
mstate,groundthein
k
k
CNConsider

† †

30.  
involved.
phonontheofenergythe,
q
thanless)
F
Eto(relative
k
choosen,interactiophoton-electronin theoriginitshaving
kk'
VFor
root.squarethethereforechoosewe
0
kk'
VwhencaseelectronfreethetoreduceTo
2
1
22
k
2
kwantWe
021
2
1
choose,
k
For





negative




k
k
x
k
x
k
x
F
E

31. 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
k
kkk dD


 

 


32. 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
1
22
)(
1
sinh
2
1




















 
F
D
F
EVD
dEVD
D
D








givesThis

33. 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
D
F




(of1%~Typically, 

