.
So, having ah laid the foundation of BCS theory
that is how ah cooper established that there
could be a electron electron attractive interaction
ah mediated by phonons.
Ah, let us go over to start BCS theory and
before we start, let me tell that ah it involves
some ah mathematics that is algebra ah which
you should do ah because every step cannot
be shown in the in the class. So, ah go through
the steps ah yourself before you convinced
that the results that we are coating are correct.
So, ah postulate ah many body ground state
as ah this is a sum over. So, this is like
psi 0 and this is all ah K greater than KF
and there is a gK, CK up dagger C minus K
down dagger acting on the 0.
Here 0 is not vacuum, but it is the ah filled
filled pharmacy the gK is some amplitude of
the wave function and CK up dagger C minus
K down daggers at 2 ah ah a pair is the pair
created ah with ah momentum K ah and minus
K with an up and down spins and this is a
many ah minus body wave function ah superconducting
many minus body wave function and then we
have ah say m electrons and we want to choose
N of them to make N by 2 pairs and the number
of ways, it can be done 
is ah m factorial divided by m minus N by
2 ah factorial and N by 2 factorial .
For m equal to ten to the power twenty three
ah this combination is equal to ten to the
power twenty to the power twenty . So, in
principle we have to solve for these many
gK's in order to be able to write down a proper
many body state and which is an impossible
task what can be done is that one can treat
the problem statistically and in order to
do that ah one can also take a grand canonical
ensemble such that we do not keep the particle
number of fixed and instead talk about an
average number of particles . So, we shall
talk ah in grand canonical ensemble 
and ah talk about ah average noise, we call
it as N bar instead of the number of particles.
So, BCS ground state BCS rolled down the ground
state as a psi g it is equal to a product
of all these K one K 2 and K N or say ah m
and this is uK plus vK CK up dagger C minus
K down dagger and acting on a phi 0 . So,
this is the ah filled pharmacy as we have
been talking about. So, the probability of
a pair to exist . So, a pair formed off K
up and minus K down to exist is given by v
K square and ah that this is unoccupied is
given by u K square that is right it has subscript
and of course, the normalization says that
the u K square plus v K square should be equal
to one which means that the probability that
a K up and minus K down would be occup either
occupied or unoccupied and the total probability
is equal to 1.
take 2 states ah K one and K 2 . So, ah the
amplitude with ah uK 1 uK 2 represent no pairs
in these 2 states ah uK 1 vK 2 ah one pair
K 2 and minus K 2 and vK 1 vK 2 ah clearly
distinguish your use and v's there are 2 pairs
for K 1 minus K 1 and K 2 minus K 2 both are
occupied . So, these are the notations ah.
So, a pair would be unoccupied with a probability
uK mod square and a pair would be ah occupied
would be given by v K square mod vK square
.
So, average number of particles 
N bar this is equal to ah. So, the sum over
K sigma and nK sigma which is an operator
and this is equal to we can write this as
ah psi g sum over K CK up dagger CK up that
is the number operator for up spin and now
we will add also some over spins.
Ah so, CK down dagger CK down and this expectation
has to be taken between the ground state that
we have written here. So, that is the ground
state postulate of the VCS ground state. So,
we have to take this thing here and this can
be written as ah because there is no ah preference
over one spin on another. So, we can simply
write this as twice of psi g sum over K CK
up dagger CK up and ah psi g this I am writing
it once, but then ah later on I will skip.
So, this will be a phi 0 uK star plus v K
star C minus K down CK up that is that is
the psi g here the brass psi g and now I have
a CK up dagger CK up and now I have a uK plus
a vK CK up dagger C minus K down dagger and
now I will have terms which are ah.
So, this is the same case as it is there in
the operator nK sigma and now, I will also
have all the other terms in which K is not
equal to ah or rather ah K is not equal to
l all the other indices which are not same.
So, you l star plus a v l star C minus l down
C l up ah and now I will have a ul plus ah
v l C l up dagger C minus l down dagger and
there is a phi 0 here ok. So, this corresponds
to K naught equal to l and the top 1 is for
K equal to l . So, in principle uK and vK
are complex quantities. So, lets write that
uK and vK are generally complex 
that is why the stars are written separately
and. So, what we have done here is that we
have used ah a phi psi as a phi a dagger psi
ok.
Now, look at the term l not equal to K that
is a the term that is written later now you
can understand that ah this will have a ul
mod square then you l star v l with a cl up
dagger C minus l down dagger plus ah ul vl
star and C minus l down C l up and plus a
vl mod square C minus l down C l up C l up
dagger C minus l down dagger now you can understand
that ah.
So, this is coming from the product of this
terms and so, they are four terms which are
here now you can see that this term gives
you 0 because ah it will create a pair and
so, we will change the occupancy of pairs
in the ground state. So, it will have 0 expectation
value. So, this term is not equal to 0, but
then when you take it between this then that
is equal to 0 this is what I mean and similarly,
this term will also yield equal to 0 when
taken between the ah the field for me see
now this simply adds a normalization that
is it creates a pair and then it annihilates
a pair.
So, ultimately what happens is that. So, l
not equal to K gives you u l square plus ah
v l square equal to one and now let us look
at the l equal to K this if you look at it
carefully you will have a term which is a
uK mod square CK up dagger C ah K up acting
on phi 0 would give me 0 and ah u K vK because
this this tells you that there are no states
to annihilate ah for K greater than K f and
u kvk the cross minus terms both uK star vK
and u kv K star will give 0 for the same reason
as its told above.
So, the only term that contributes is a term
with v K square . So, this N average which
is equal to twice of K v K square . So, that
is a result. So, 2 comes because of the pair
and vK square is the probability of occupied
states .
Now, second thing is a fluctuations in N bar
. So, that is given by N minus N bar square
which is equal to N bar square N square minus
ah or minus twice of N N bar plus the N bar
square which is equal to a N square bar minus
a N bar square . So, this if you again ah
repeat the same calculation and use the same
logic to cancel out terms this is given by
four K u K square v K square and this is positive
definite . In fact, vK as a function of K
goes from one to 0 whereas, uK as the function
of K goes from 0 to one ah. So, in a all happening
in an energy range which is given by ah K
t c. So, if you ah write down the variation
of this . So, this is my . So, this is my
epsilon K or psi K which is equal to epsilon
K minus mu and the vK drops from ah 0 to 1
and this is my superconducting gap delta this
is minus delta 2 plus delta and this is ah
my .
So, this thing is my minus h cross omega d
to this as h cross omega d. So, in this ah
range ah v K becomes from one to 0 and so,
the sum above goes as ah T over or TC over
TF whole to the power N bar . So, this is
the ah practically if you want to estimate
ah the fluctuation in N. So, this goes as
that . So, let us write down a many a BCS
many minus body ground state .
Oh, rather, let us write down a ah Hamiltonian
and the Hamiltonian is a single particle term
epsilon K is psi K sigma C K sigma dagger
CK sigma plus ah K l vK l CK up dagger C minus
K down dagger C minus l down C l up that is
the Hamiltonian and the of course, we know
that ah psi K equal to epsilon K minus mu
we have already argued that BCS theory cannot
be obtained by doing a perturbation theory
of any order.
So, we will do a variational calculation instead
ah. So, and the variational calculation with
these small uK vk's which are the occupancies
will be used as a variational parameter . So,
what we have to do is that we have to take
this variation ah and it is a K sigma psi
K sigma ah N K sigma plus ah K l a vK l CK
up dagger C minus K down dagger C minus l
down C l up that is the ah. So, we have to
take a variation of this and put this equal
to 0.
So, let us look at the kinetic energy term
or the single particle the first term here.
So, this is the kinetic energy and this is
the potential energy and this is given by
let us call this as kinetic energy operator
which is a psi g sum over K sigma sigma N
K sigma this you should work out and get this
thing as almost we have ah gotten this when
we did the average number this comes out to
be 2 sum over K v K square and a psi K and
similarly for the potential energy .
We have v this is e equal to a psi g and then
ah there is a sum over K l vK l CK up dagger
C minus K down dagger C minus l down C l up
and this and this comes out to be K l vK l
uK star v l star u l vK . So, these are the
kinetic energy and the potential energy these
are the expectation values of those ah with
of course, the constraint as u K square plus
v K square equal to one now because of this
constraint one can actually take a pair .
And reduce ah so, a pair of variables like
this 
and reduce the number of variables from 2
to one see it is always very difficult to
do a variational calculation with 2 variational
parameters then you have to look for the minimum
in a space in a 2 minus dimensional space
it is much easier to look for a minimum in
a 1 dimen on a 1 minus dimensional line and
that is now given by this single variable
theta K .
So, we can take uK equal to sin theta K and
vK equal to cos theta K you can take the other
combination that is uK equal to cos theta
K and v K equal to sin theta k, but it seems
that this combination works better . So, now,
what we do is that we take a variation with
respect to theta K of this . Now I will use
a dummy variable K prime one plus cosine 2
theta K once again this algebra you should
do because we have come from a vK square which
we have written as 1 plus cosine ah 2 theta
K and there is a K prime l . So, there is
a vK prime l sin 2 2 theta K prime sin 2 theta
l and put this equal to 0, in order to do
a variational calculation and see that what
is the extremum value of theta K which minimizes
the energy .
So, minus 2 psi K sin 2 theta K after you
do this derivative. So, it is l and there
is a vK l cos 2 theta K sin 2 theta l equal
to 0 if we define delta K equal to minus half
of sum over l vK l sin 2 theta l, then ah
using this and putting it into this equation
one gets a ah nice equation such as 2 psi
K sin 2 theta K equal to sum over l vK l cos
2 theta K sin 2 theta l equal to minus 2 delta
K cos 2 theta K plus 2 psi 2 psi K sin 2 theta
K is equal to minus 2 delta K cos 2 ah theta
K now this equation gives us tan 2 theta K
if i ah divide ah or rather ah bring this
below it is equal to a minus delta K by psi
K .
So, this can also be written as ah sin 2 theta
K cos 2 theta K which is equal to minus delta
K by psi K . Now, if we use the definitions
that 2 u K vK which is equal to sin 2 theta
K this is equal to a delta K divided by ah
psi K square plus a delta K square . So, this
is my ah sin 2 theta K definition of sin 2
theta K which is also equal to twice of uK
vK and also v K square minus u K square which
is equal to cosine 2 theta K which is equal
to psi K ah divided by root over psi K square
plus delta K square, there is a little bit
of understanding that needs to be done here
is that this particular choice, we could have
taken the other choice also that is a vice
versa, but this choice fits all the definition
and we have taken the cos 2 theta K to be
negative because if psi K is large which means
that epsilon K is much much greater than the
chemical potential mu then vK should go off
to 0 which is a apparent from this diagram.
So, that is why the cosine 2 theta K is taken
with an ah negative sign alternately we could
have taken the sin 2 theta k, but that would
not have satisfied the conditions ah or the
boundary conditions that we have . So, hence
the quantity delta K assumes a form that delta
K equal to minus half of sum over l v kl sin
2 theta K and hence this is equal to minus
half sum over l v kl delta K divided by putting
the value of sin 2 theta K here is ah delta
l square plus psi l square. So, that is the
definition of ah delta K . So, a trivial solution.
So, we have to solve for these in order to
solve for ah delta K ah a priori let us say
that the delta K is really the energy gap
or the superconducting energy gap and ah we
have to solve for it in order to find that
what is the or how does the gap vary with
different parameters especially say v and
psi l or how how does that enter into the
expression of the gap.
Ah in order to see that we can also ah look
at this expression and see that ah the ah
trivial solution is ah delta equal to ah rather
this is delta equal to 0 is the trivial solution
. Now you see delta is there on both the sides
of the equation here it is just a standalone
delta K and here it is sum over l delta l
and K is not equal to l. So, for a given momentum
value l delta l is the value and delta K has
to be computed by summing over all those delta
l, but since we are solving for delta l we
do not know what it is. So, we cannot. So,
the unknown quantity appears on both the sides.
So, a trivial solution as we said delta equal
to zero. So, what does delta equal to 0 gives
it gives that 2 uK vK equal to 0 because you
see that 2 uK vK equal to delta k. So, if
ah delta K equal to 0 2 K uK vK equal to 0
from this equation.
So, 2 uK vK equal to 0 and that tells that
ah we also have ah vK square minus u K square
equal to minus one , but then we have the
normalization condition is that uK mod square
plus vK mod square it is equal to one that
it tells that vK square equal to 0. So, at
delta equal to 0 implies that vK square equal
to 0. So, there are no pairs and hence this
should correspond to the normal state . So,
the trivial solution is important because
it talks about the normal state, but at the
same time it gives a meaning to delta now
delta can be used as an order parameter for
the superconducting transition because at
normal state delta is equal to 0 and delta
is not equal to 0 for the superconducting
state. So, now, go to the original ah coopers
proposition .
That vkl equal to minus v if ah psi K minus
psi l is less than h cross omega d thus delta
K equal to half v sum over l sin 2 theta l
for psi K minus psi l is less than h cross
omega d equal to 0 otherwise . So, this tells
that as if delta does not depend upon K it
simply depends upon it just a number . So,
that is even more convenient because delta
is now can we a truly thought as a number
which when is nonzero will give a superconducting
state; however, when it is 0 it will give
rise to a normal state . So, then ah we have
delta l which has to be put here yeah . So,
this ah says that if it is a sum over all
le delta ah loses the K dependence .
So, it is just a just a number and has no
K dependence. So, we can write this as ah
delta equal to ah v by 2 delta sum over l
one divided by root over delta square plus
psi l square and we can cancel delta from
both sides and this gives rise to an equation
which is ah v by 2 l and a root over delta
square plus.
So, there is no delta l. So, it is equal to
psi l square now this is the equation for
delta you may not see delta on the ah left
hand side to solve for, but there is delta
in the right hand side and in the denominator
and in the square root of a denominator that
tells that it is a highly no linear equation
and you have to solve it either by a root
finding method or ah one of the root finding
methods such as Newton, Rapson or ah bisection
method if you want to do it using a computer
that is numerically .
We can also solve this problem ah analytically
by converting this sum sum over l to an integral
which is of this form remember the v is the
strength of the attractive potential . So,
it is the once when we ah convert a summation
into an integral we need to bring the density
of states or we can ah write down the density
of states like this and ah do this integral
such as ah.
So, there is a 2 plus delta square plus psi
square by psi is the variable . Now, the integral
will be from 0 to h cross omega d this is
what we have said earlier and cooper had explained
that how the pairs have to be formed within
an energy shell of h cross omega d from the
fermi surface. So, it is measured from the
fermi surface and the now N of psi the detailed
feature of N of psi is not required because
we know that this whole ah phenomena is occurring
at the fermi level.
So, we can write this as N of epsilon f by
2 0 to h cross omega d d psi and a root over
delta square plus psi square and that tells
us that ah this is equal to 2 by N v ah and
this is 0 to ah h cross omega d d psi ah root
over delta square plus psi square and this
is equal to 2 sin hyperbolic inverse psi by
delta and from 0 to h cross omega d if we
ah put these values and rearrange then we
will get ah delta equal to h cross omega divided
by sin hyperbolic one divided by N epsilon
f v now since we are talking.
About a for week coupling super conductors
N epsilon fv is much smaller than one thus
delta assumes a form which is 2 h cross omega
d exponential minus one by N ef of V .
So, we are ah getting a similar expression
for the energy gap as we have done by solving
the 2 particle ah Schrodinger equation this
tells that this much of energy has to be supplied
in order to break a pair ah and go from a
superconductor to a normal state and in BCS
theory this energy gap ah is is a scale which
is given ah by the temperature scale. So,
delta is of the order of K T C and. So, a
K T C becomes ah equal to 1.14.
It is actually 2 here as we have written there
and very accurate calculation shows that this
is equal to 1.14 and so, the TC expression
is obtained from here we will just do it in
a minute . So, this gives the how the energy
gap depends upon the phonon energy spectrum
and ah how the density of states at the fermi
level come into the picture and ah the strength
of the attractive interaction is also there
ah which is v and we also have an epsilon
f multiplied by v is ah much smaller than
one which is ah relevant for a weak coupling
superconductor .
So, let us give you all these occupation ah
probabilities or ah what are also called as
coherence factors that we have a vK square
ah equal to half of one minus psi K by EK
ah which is equal to half of one minus psi
K divided by root over delta square plus psi
K square. So, that is ah vK and uK square
is simply equal to one minus vK square which
is equal to half one plus psi ky K . So, these
are important ah because these decide the
behavior ah of the gap the K dependence of
the gap now let us go to the finite temperature
.
So, we have seen that ah lets. So, let us
write down the BCS Hamiltonian once more ah
which is a mean field BCS Hamiltonian . So,
this is equal to ah psi K C K sigma dagger
C K sigma and in the mean field picture we
have a minus delta K CK up dagger C minus
K down dagger plus a delta K star a C minus
K down CK up this would be obtained if you
take the mean field decoupling of the p e
tem of the potential energy term .
These kind of d couplings we have done earlier
and this equation ah or rather, this Hamiltonian
can be diagonalized 
using a Bogoliubov Valatin transformation
where the C operators are transformed into
quasi particle operators of this form CK up
equal to u K star gamma K 0 plus ah vK gamma
K one and the C minus K down dagger equal
to minus vK gamma K 0 plus uK gamma K one
and ah. So, gammas are quasi particle operators
.
Ah. So, they have. So, gamma gamma dagger
the have usual anti commutation relations
as the CK ck daggers now a generic form of
the gap .
Now, we call delta as the gap because we have
established that delta is nonzero for the
superconducting state and 0 for the normal
state . So, this is equal to a minus ah you
can call it l and a vK l and a C minus l known
C l up . So, a little bit of algebra in terms
of this gamma operators will yield vK l ah
u l or uK star vK and one minus gamma l 0
dagger gamma l 0 minus gamma l 1 dagger gamma
l 1 and this at finite temperature is given
by ah each one of them will be given by a
fermi distribution function which is ah or
this is ah E l and the E l .
Thus ah one minus gamma l 0 dagger gamma l
0 minus gamma l one dagger gamma l one it
is equal to one minus twice of ah l . So,
to say . So, delta K putting it back into
this equation the gap equation it is equal
to ah minus of v K l ah u l star ah u l. So,
this should be l actually ul star ah v l u
l star vl and the one minus 2 ah f E l and
this is nothing, but tan hyperbolic beta E
l by 2.
So, ah with v K k prime equal to minus v which
is coopers assumption I get this Equation
as one over v equal to half sum over K tan
hyperbolic beta E K by 2 by E K again converting
that sum into the integral and using the density
of states to have a value that is at the fermi
level .
So, this is equal to a one N this and a 0
to a beta C h cross omega C by 2 a tan hyperbolic
x by x dx where x equal to beta ah e K by
2 this integral as a standard value which
is given by log of 2 e to the power gamma
by pi beta C h cross omega C where beta C
is equal to one over K T C and ah gamma is
the Euler constant which has a value 0.577
and hence this 2 E to the power gamma by pi
has a value 1.13 or 14 ah around 14 and that
if you simplify this it comes out as K T C
equal to 1.14 h cross omega C exponential
minus 1 by N epsilon f v, this is exactly
the formula for ah TC that we have got. So,
this is the how TC varies with the phonon
frequency and the density of states at the
fermi level and v . So, thus delta 0.
So, that is a value of the gap at t equal
to 0 it is equal to 2 divided by 1.14 which
is equal to 1.67 nearly and that tells you
that this is a feature of ah BCS super conductivity
that twice of delta divided by K TC it is
equal to 3.52 this is followed by all weak
coupling phonon mediated 
superconductors ok. So, this is the a feature
or rather a property of this that the twice
of the energy gap verse ah divided by TC or
K TC should be a number which is ah three
point five 2 ah it deviates strongly for the
high temperature 
superconductors .
So, let us summarize what we have seen so
far we have generically talked about ah the
properties of superconductor 
and those properties include ah the Meissner
effect that is the electromagnetic response
then the thermodynamic response and ah the
conducting properties or rather the resistivity
how it suddenly drops to 0 below certain ah
critical temperature or threshold temperature
and so on.
And then we have gone on to talk about what
is the origin of attractive interaction 
and this tells explicitly that there is a
phonon part involved the role of phonon is
very apparent because of the isotope effect
that we have seen because the Debye frequency
actually ah scales as the ionic mass . So,
the lattice ah is ah or rather the TC scales
with ah the ionic mass. So, the involvement
of the lattice is very clear and there we
can actually for us narrow energy range we
can get an attractive interaction between
the electrons.
So, the wave function of the electrons will
have to actually select this energy range
to form a bound pair and then ah we have written
down a many minus body ground state and have
taken a Hamiltonian which is a generic Hamiltonian
where ah the an interaction is taking place
between 2 particles for a particular form
of this interaction term that is v kk prime
equal to a minus v for the epsilon psi K minus
i K prime to be falling in this energy interval
h cross omega d we find that the gap ah or
rather we have been able to write down ah
the equation of a gap and also at finite temperatures
have able to write down the temperature dependence
or rather the ah the TC how the TC that threshold
temperature the critical temperature depends
on these quantities such as ah v that is a
attractive interaction the density of states
and the phonon frequency call it omega d or
omega C.
So, this is in a nutshell what is super conductivity
all about and. So, as you either ah destroy
superconductivity by using temperature or
thermal effects or you destroy superconductivity
by applying ah magnetic field the state goes
on to a ah ah normal state a metallic state
which is apparently not the case for the high
minus temperature superconductors and they
have many other complications ah there are
very little consult consensus whether phonons
are involved into this pairing mechanism or
there is something else that is involved in
a any case.
And also these behavior or rather this 2 delta
by K T see that we have found to be ah 3.52
is a value there is much higher maybe 5 between
5 and 6 and which are not at all ah called
as the weak coupling superconductors and these
high minus temperature superconductors they
belong to a class of a non weak coupling rather
strong coupling superconductors .
