So, let us have a quick review on the topics
that have ah been discussed, so, far in the
context of greens functions. So, the greens
function consists of time order product of
a creation and an annihilation operator followed
by n interaction terms, which are all written
at distinct times that is the first point
ah. So, this was just to say that it is of
the form of ah a phi 0 ah and T and then a
C k ah t and a C k prime t prime dagger, and
then there are H ah primes or H interaction
ah terms that are at T 1 and all the way up
to h prime t n and ah this is the expectation
value of the time order product is taken.
So, this is 1 ah annihilation term and the
ah there is a creation term and this thing
ah equal to. So, this along with a minus i
is ah is same as ah is same as G ah k let
us say that let us make this ah same. So,
it is ah G k a and p minus t prime.
So, this is what it is proportional to with
a factor of minus i ah will see all that ah
in a short while. Now each order of H prime
ah included here in this expression ah will
give a sort of that order of the greens function.
So, if I include 1 H prime I will get a first
order greens function, if I include 2 ah h
primes I will get a second order greens function
and so, on.
So, these are ah in principle we have infinite
number of these ah H primes. So, a full greens
function contains ah infinite number of H
primes, ah what ah are the implications of
a each of these H primes is what we are going
to come next?
So, a fully interacting greens function contains
ah a 1 creation and 1 annihilation and n of
those H primes. So, each of the the third
point is each of the interaction terms H prime
itself is a collection of to creation and
to annihilation operators. This is what we
have learned when we are dead second quantization,
that each interaction term at least as ah
2 ah creation and 2 annihilation operators
hm.
So, there could be higher order of interaction
terms ah, but we are mostly going to be talking
about interaction terms, where ah 2 particles
come and interact ah and they exchange their
ah you know spin momentum etcetera and they
go on as 2 different particles there could
be you know production of particles as a ah
result of scattering, but let us not talk
about that immediately will ah sort of only
be constrained with ah each of these H primes
having 2 creation and to annihilation operators.
So, H prime ah has C dagger C and C dagger
C with a momentum and spin indices as case
ah requires. So, the expectation value of
the time ordered product is ah taken with
respect to the non-interacting ground state,
this is what we have learned ah it is actually
the ground state of the many body system,
but then we learned how to construct a non-interacting
ground state from an interacting ah many body
ground state, and ah we have shown that there
is a ah phase factor ah that comes, ah which
is which connects the many body ground state
as well to the to that of the ah non interacting
ground state. And properly done we can simply
take the time ordered product to be taken
between or the expectation to be taken between
the non-interacting ground states.
All the operators point number 5 all the operators
are written in the interaction representation.
This is important, because ah these greens
function is fully in the interacting interaction
representation, in which both the wave functions
ah and operators they carry the time dependence.
Somehow ah it is only important for us that
the operators carry the time dependence, but
nevertheless we should write those ah operators
the creation and the annihilation operators
to be in the interaction representation.
In fact, the difference between the interaction
representation and the ah the Heisenberg representation
is that that the Heisenberg representation
ah has similar time evolution as that of the
non-interacting system for the interaction
representation.
So, ah now the non-interacting so, the interacting
greens functions are written in terms of the
products of non-interacting greens function
at the number operators. This is done by doing
a normal order product and each of the or
rather the interacting of the full greens
function is split up into ah several of the
non-interacting greens functions. And all
the number operators and then of course, the
the interaction vertex ah what we mean by
that is, what we are going to come next the
interaction vertex, ah will appear and or
the strength of the interaction will appear
and the fully greens function will contain
information about ah the nature of the interaction
vertex or their strength of the interaction?
Such products are most conveniently computed
using Feynman diagrams. So, we are next going
to discuss Feynman diagrams, but before that
let us write down 1 example of ah a greens
function and see that what it means or how
it can be split up decomposed into the non-interacting
greens function?
So, we start with an example 
and ah let us write down ah apart without
those ah factors of minus i etcetera will
simply write down a time order product and
ah this means this ah left angular bracket
means that there is it is taking with respect
to phi naught 1 can write it also explicitly
ah, but it is understood.
So, it is a C k t C k 1 d agger t 1 ah C k
2 dagger t 1 and ah C k ah 2 ah plus q plus
ah C k ah 2 plus q ah t 1 and C k 1 minus
q ah t 1 and ah C k ah dagger t prime and
so, this is the 1 that we are ah going to
have.
So, let us just write it as ah so, there will
be nothing here and there will be nothing
here, but as I said the angular brackets denote
that we are taking the expectation with respect
to the non-interacting greens function. Now
what we have done is that we have simply written
this as the simply this is the H prime t 1.
So, we have written down a first order greens
function ok ah containing 1 H prime 1 ah 1
annihilation operator here and a creation
operator here.
So, that is ah our that is the term that we
need to calculate. So, let us ah write down
or rather number these ah operators. So, that
we can take the proper combination in order
to write the non-interacting greens function.
And so, this is 1 this is 2 let us let me
write a different ah color ah for that. So,
that. So, this is 1, ah this is 2, this is
3, this is 4, this is 5, hm and this is 6.
So, what are the combinations allowed? So,
possible combinations so, a will be a is 1
2 ah 3 4 and 5 6 that is combination number
1. So, we can take a ah combination of of
1 and first and second term third and fourth
term and 5th and 6th are also we can take
ah first and third term.
So, the combinations are being formed by 1
creation and 1 annihilation operator. So,
1 3 2 4 and 5 6. So, we have a third combination,
which is ah 1 3 ah 2 5 and 4 6. So, that is
the third combination and ah d ah to be ah
1 2 3 4 or rather 3 5.
1 2 3 5 and ah 4 and 6 ah fifth combination
is 1 and 4 ah 3 and or ah 2 and 4 1 and 6
sorry 1 and 1 and 6 ah 2 and 2 and 5 and ah
3 and 3 and ah 4, so, 1 and 6 2 and 4 and
3 and 5.
So, this is the fifth combination and there
will be another 1 there will be another one,
which is 1 6 and 2 5 and 3 4 ok. So, these
are the 6 combinations and the 6 combinations
are coming, because of we have ah 3 ah operators
3 creation operators and 3 annihilation operators.
So, there are 3 factorial combinations possible.
So, all these 6 combinations for just for
the first order greens function have to be
evaluated.
So, let us first write down ah each one of
them and specifically ah we shall take some
time and treat this as a tutorial such that
you learn, hm these ah these decoupling and
writing it writing the full brains function
the first order greens function as in terms
of the non-interacting greens function.
So, the first term is ah according to the
combinations ah prescribed 1 2 3 4 and 5 6.
So, will take a combination of C k t C k 1
t 1 then C k t 2 dagger and C k a 2 and then
C k 1 minus k, and then C k dagger remember
that that is not going be the greens function
unless you multiply it is going to be a non-interacting
greens function if you multiply it with a
factor of i, because the greens function itself
is defined with a factor of minus i.
So, a becomes equal to T ah C k t and C k
1 dagger t 1 right and so, this one ah and
then it is t C k 2 plus q ah t 1 ah C k 2
dagger t 1 and and a C time ordered of C k
1 minus q t 1 and C k dagger t prime and these
are the 3 combinations that I get.
And so, I will have to multiply each one of
them by a negative sign and also do not forget
to ah bring in the negative sign that comes,
because of the swapping of the operators that
is very important. So, setting that aside
for the moment we would write this 1 as delta
k k 1 and G 0 ah k ah t minus t 1 ok. So,
just remind you that a factor of I is what
we have dropped so, far, but will ah get it
back ah once when we do ah more thorough calculations.
So, this is the first term. So, your k has
to be equal to k prime that is the ah the
conservation says and ah this is a G 0 with
order greens function ah which is a function
of k and t minus t 1.
Now, this one necessarily demands that ah
because k 2 plus q has to become equal to
k 2, which means that this is necessarily
a delta q equal to 0 and hm and a delta q
equal to 0. And now this is going to be a
number operator, because it is at same time
at time t equal to t 1. So, it is not a propagator
or a greens function that propagates the system
from sometime t 2 sometime t prime. So, it
just loops back to the same ah time. So, will
write it with a n and this ah ispsi k, where
xi k could be epsilon k minus mu where epsilon
k is this ah other single particle energies.
Similarly so, this is the second term and
similarly the third term is here. So, that
is again equal to delta q equal to 0 and a
G 0 ah k 1 and t 1 minus t prime. This is
how we have now written ah it ah the full
greens function at the first order in terms
of the non-interacting greens function G 0
and a number operator which is n of xi k and
ah another G 0, which is ah there? And how
to compute this is what we will see and that
is where the Feynman diagrams play a role,
but before that let us writes down all the
terms. The second term so, ah again I want
to remind you that I am treating it as a tutorial
you should do each and every term by hand
and satisfy yourself that this ah is what
is coming?
Because this forms the basis of what we are
going to do next ah with the Feynman diagrams.
Similarly for b we have ah T C k t C k to
dagger t 1 T C k 2 plus q t 1 C k 1 dagger
t 1 and T C k 1 minus q t 1 C k dagger t prime.
So, this tells ah that it is equal to delta
k k 2 ah sorry this is delta k and k 2 naught
k prime and ah G 0 k t minus t 1. And similarly
this one will have ah delta of k 1 equal to
k 2 plus q and this will be ah simply a G
0 ah and ah k 1 and ah again ah this will
give me a so, this is from ah t 1 ah 2 t 1,
which means it is time that ah if the time
is repeated ah provided I have written it
correctly yeah that I have written it correctly.
So, this will be ah, but this is not ah at
and given ah q equal to 0. So, this will be
like ah G 0 and k 1 and this will be like
ah a t 1 ah k 2 plus q ah will come back to
this. And will have a delta k 1 minus q is
equal to k and there is a G 0 ah and there
is a ah k and there is a t 1 minus t prime.
And so, these ones so, let us write down the
number C, T C k t C k 2 dagger t C k 1 minus
q t 1 C k 1 dagger t 1 and T C k 2 plus q
t C k dagger t prime and so, on.
So, this tells me that this is equal to delta
k k 2 again ah this is at the same time. So,
this is equal to a number operator and apsi
k say for example, ah and again this is ah
this is ah for ah. So, this is k equal to
k 2 and this is for. So, this is actually
k 1.
So, this is delta k q equal to 0 and there
is a number operatorpsi k 1 and this will
be ah delta ah k k 2 plus q and there is a
greens function, ah which propagates k ah
t minus t prime. So, these are ah the 3 ah
terms there and similarly for d, ah we have
t C k t ah C k 1 dagger t 1 ah and t C k 1
minus q k 1 minus q ah t 1 and C k 2 dagger
t 1 and similarly T C k 2 plus q t 1 and C
k dagger t prime and so, on.
So, this is equal to delta k k 1 and G 0 once
again to remind you that this G 0 will have
to add an i. So, that without that it does
not become G 0 we are ah writing it loosely
without that factor of i. So, this is k t
minus t 1 ah and now this is of course, ah
will ah have a delta k 1 minus q, which is
equal to ah k 2 ah. So, this is k 2 and this
will give me ah a so, just like the first
1 it will give me a ah so, it will be like
a n ah xi k 2.
Similarly this will also be k 2. So, this
will also be let me correct it here this was
written mistakenly. So, this will be like
n and xi k 1. So, similarly it will be xi
k 2 and then this will be like ah delta of
k k 2 plus q ah G 0 and k ah t 1 minus t prime.
So, that is ah the fourth 1 and similarly
the fifth 1 can be written as something interesting
about the fifth 1 which will come.
So, we have a C k t and C k dagger t prime
ah C k 2 plus q ah t 1 ah C k 1 dagger t 1
t C k 1 minus q t 1 and C k 2 dagger t 1.
So, these are the 3 terms ah this term will
give me a delta k ah. So, this ah this is
anyway k ah. So, delta k k um and there is
nothing, but there is a G 0 ah which is at
k 2 t minus t prime.
Similarly this will be k 1 to k 2 plus q and
again it will be like this k 1 and this will
be like ah delta of k 1 or k 2 k 2 k 1 minus
q and ah a a G 0 and ah k 2 and this is at
the same time. So, this will actually be a
so, this will actually be a n which is xi
k 2 anyway. So, these are 5 terms and to write
down the sixth term we have T C k t C k dagger
t prime ah T C k 2 plus 2 hm a t 1 C k ah
dagger t 1 and ah C k 1 minus q ah t 1 and
ah C k dagger C k dagger t 1 and again ah
this is equal to a G 0 ah k t minus t prime
there is nothing to write here. So, let me
it is this part.
And right this one has to be written as ah
a delta this is k 2. So, this is equal to
q equal to 0 this and a k 2 and similarly
this one is also delta q equal to 0 ah psi
k 1 and so, on.
So, it is 3 5 ah 3 and 5 ah k 2 and ah k 1
minus q. So, this is actually k 2 ah. So,
this is the ah last 1 is 3 5 3 is t 1 this
t 1 this k 2 and k 1 minus ah k 2 3 is k 2
and this k 1 minus q this 2 ok. So, this is
ah k 1 minus k 2 equal to. So, so this is
anyway.
So, these are the terms that are that need
to be written down ah including the 0th order
greens functions. And now if you write down
second order greens function you will have
many more terms and ah we can write down and
then there will be a second order greens function
will have how many terms a second order. So,
how many ah creation and annihilation operator
to H prime will give me ah 4 4 8 ah 4 creation
operators and for annihilation operators ah
and 1 will come from the greens function definition.
So, there are 5 ah creation and 5 annihilation
operators. So, this will contain 5 factorial
terms.
Which means that ah equal to 5 into 4 into
3 into 2, which is equal to 20 into so, it
is 120 terms that will be there a second ordered
itself and the third order you can understand
that it will go as. So, third order will have.
So, 67 so, 7 factorial number of terms so,
the question is that how actually to compute
all these terms, even if we say want to write
down the first order ah greens function that
is containing 1 ah H prime.
We got 6 terms and maybe some terms are 0
we do not know as yet, but even then the nonzero
terms how to compute them ah that is the question
and the computation of these terms are done
by a Feynman diagrams.
So, this is ah will introduce Feynman diagrams
here 
and tell you the basic features of the Feynman
diagram. So, so Feynman introduced this idea
of representing ah the terms in these above
discussions or above expressions by diagrams
and these diagrams are actually ah very very
important in a number of physical processes,
ah which are themselves I mean interesting
ah scattering problems so, the rules of the
diagram.
So, rule number 1 is that a solid line with
an arrow 
pointing 
from smaller time t prime, which is a smaller
time 
to t, which is larger time ah for free electron
greens function ah with with a momentum information
ok.
So, basically that tells that this is like
this with a t prime going towards a t and
there is a momentum information. So, this
line carries a momentum ah p. So, the system
ah or rather the free electron greens function
ah propagates the free system from a smaller
time t prime to a larger prime t ah and a
corresponding to a given momentum p.
So, this is written as ah this is equal to
G 0 p t minus t prime, this is the notation
for that this is for the electron greens function,
what about the phonon or the photon greens
function? Let us just write it for the ah
photon ah or phonon. So, a dashed line 
with no arrow 
for the phonon greens function and no arrow
because 
because ah phonons can move in either direction
on the line that is in time.
Sometimes this so, it is this line 
and there is a t prime to t and similarly
there is a say a q momentum will not put a
arrow, ah sometimes this is also written it
with a springy line ah like a spring.
So, this is t prime and t it is also written
like that and it is written as d 0 ah q ah
t minus t prime. So, these are the electron
and phonon ah non interacting greens function
ah third very importantly.
The electron density n of xi p is shown via
a loop to imply that it starts and ends at
the same time time.
So, it just shows like this we show it by
an arrow this this is p and this is time ah
it starts from the same time, I mean it starts
from time t and comes back to the same time
and this is written aspsi p this is what we
were writing earlier.
Forth any interaction term v q is represented
by a wiggly line so, such as this. So, this
is it may carry a momentum q. So, this is
the ah v of q v as a function of q it could
have a energy information, ah which means
that it will be a v q omega. So, for example,
ah we can write down ah 1 of the terms ah
in this ah which had ah, so a. G 0 p v of
q which comes from an int and there is a G
0 p minus q and G 0 p will be shown as a line
and then there there is a wiggly line.
So, it comes from ah we have not written the
time information, but the time is actually
t 2 ah t minus t 1 and t 1 minus t 2 etcetera.
So, these are this this and so on. So, this
is say p this is p minus q and this is again
p. So, an electron was coming ah with momentum
p ah and it is scatters of ah by a potential
v of q, which takes away a momentum q here.
So, that is why the wiggly line is represented
by wiggly line is ah the v q which carries
a momentum.
Then so, this is the this the leftmost side
is the propagated here ah the v q is by the
wiggly line and p minus q is the propagated
here and then there is a propagator there.
So, the total greens function or the total
expression for the ah interacting greens function
at the first order is written as a G 0 and
then there is a wiggly line and then there
are you know ah sort of ah is a propagator
from p ah sometime, t 2 t 1 or ah. So, this
is like ah t 2 hm or t prime, hm which we
have not put it here. So, maybe t 1 and then
maybe t 2 and then maybe t and so on. So,
these are ah the way to write down this each
one of those ah terms, that we have got.
And. So, rule number 5 is that at the vertex
at the vortex, there is a momentum and energy
conservation.
So, what I mean by vertex is these points
ah that is a vertex and this is a vertex there
are energy and momentum conservation. So,
what we have to do is that? So, integrate
over all the intermediate momenta and some
over all ah integrate over all.
So, integrate over all intermediate ah let
us not write it as momenta as energies 
and some overall intermediate . So, basically
ah it is a you have to integrate ah over ah
all let us not introduce this word some ah
and 
and over all intermediate momentum.
So, that is the rule number 5, that is rule
number 5 and the rule number 6 immediately
it will not be clear, but nevertheless will
put it any way ah multiply by a factor of
i to the power n, because ah each greens function
ah to convert a line into a greens function
or each of this time ordered product into
a greens function we need to multiply it with
i.
So, each of the terms will have to be multiplied
with i and hence n of those terms will have
to be multiplied with i to the power n. And
then we will have to also multiply it with
a minus 1 whole to the power F, where F are
the number of closed for me on loops. So,
these are the number operators. So, if we
have 1 number operator then will have to multiply
it by minus 1 to the power 1, if there are
2 of them then it is ah there will be no sign
because minus 1 whole to the power 2 it will
have to be multiplied by and so, on.
So, these ah all these rules put together
from 1 to 6 as written there will help us
in writing down these greens functions the
fully interacting greens function in terms
of the non-interacting greens function and
then by ah summing over ah the all the intermediate
momentum and energy will get the ah finally,
the interacting or fully interacting greens
function.
So, ah let us ah write down at least 1 ah
term ah we have ah written down 1 term already
let us write down another term, which is ah
which was written down by us.
So, example is ah time ordered C k t C this
is the third term ah in the example that we
have done T and ah we have ah T C k 1 minus
q t 1 and C k 1 dagger t 1 and C k 2 plus
q t and C k dagger t prime.
So, this term can be written as ah a line
which will go like this from a t prime to
t and then ah there will be a term which is
ah and there is an interaction term. Of course,
that we are not writing it explicitly, but
the agent will come with a with a v q here
ah and this ah. So, this will be here. So,
this is a because this will correspond to
delta q equal to 0. So, this interaction term
corresponds to ah v q equal to 0 and this
corresponds to a. So, this is 1 greens function,
taking it from ah some t prime to some intermediate
t 1 and this another.
So, this is that greens function. So, this
is that greens function here, this is that
again that greens function which is here and
this is that term which is here ok? So, this
is the so, this is a fermionic bubble. So,
there is 1 fermionic bubble. So, will have
to multiply it by minus 1 to the power 1 and
since there are ah 3 such terms will have
to multiply it by ah i to the power ah n ah
which is what we have prescribed ah I to the
power n. So, it is i to the power 3 and ah
rather there are 2 greens functions. So, will
have to write it with i square ah and there
will be a minus sign coming because of this.
So, this is ah momentum is k here momentum
is. So, this does not carry any momentum.
So, this is ah and this has momentum ahpsi
k 1 and this is this continues with the momentum
k. So, a free particle greens function propagates
from minus t, I mean from t prime to t 1 with
a momentum k, then there is a ah interaction
vertex which carries no momentum it is connected
to a fermionic bubble and then it carries
on with ah. So, the the greens function or
the propagator carries on with a momentum
k from time t 1 2 t, will do more examples
of ah feynman diagrams and will ah in the
next ah discussion will ah try to compute
them by computing the internal ah summing
over the internal moment and the energies
will calculate and for a specific case.
