 So, and I can tell you right away what this one particle irreducible is. So, we already saw that disconnected simply means two diagrams that do not talk to each other right. I can have, but I can simultaneously have something else going on here ok. So, this is and so there is a multiplication between the two, but they are disconnected. Connected diagrams can be of three types and for argument sake I am taking some from some book, but this is of course, connected because everything is connected to everything. Then we have no specific theory I am writing, but I am just drawing some graphs like graph theory. These are all connected, but the point is we say oh this only has a two point function on the external leg that just gets reinterpreted as the if you that just gets reabsorbed into the mass of the particle. We just say the correct propagator is such that these things are all absorbed into the. So, whatever higher order effects generate things like this where there are only two external legs, we just reabsorb it into the propagator and we do not consider it as new physics. This thing is said that it can be cut into two ok. So, becomes disconnected by cutting one line. So, we do not like this we said that is too simple, but this last one is what we call one particle irreducible, one particle meaning one line something that cannot be made disconnected by cutting one line is called one particle irreducible, any one internal line. So, for this reason it is useful to keep track of what is connected disconnected and what is one particle irreducible. And it also goes to the core of this locality and cluster decomposition. So, if physics is local ultimately then you want that if you start with a larger greens function endpoint function and if I separate first five and the next seven particles and send them far away it should just become product of a five point and a seven point greens function that property is assumed of. So, it can be proved that so long as you write your Lagrangian as local products of fields and derivatives and just to not stress you with unbalanced Lorentz indices something like this products of is just a generic example of products. So, this has one field and two derivatives, but you could have powers of you know just make it square if you like. So, that it is no charges are hanging, but the point is so long as this is a product of at the same space time point ok. So, I will not try to write the definition of cluster decomposition because to be precise one has to write a lot, but you got the idea that if I separate out particles then they become products of individual scatterings. So, this kind of locality of the theory again goes back to the possibility that higher order greens functions contain products of lower order ones, but they are just nuisance they are because they follow because after all it is a local field theory. And the only nontrivial thing that we have to worry about while renormalizing is to worry about the one particle irreducible diagrams that to there will not be too many of them. If there are too many of them then you are sunk in the sense that you have a large number of unrenormalizable terms and the theory reduces in its ability to predict. At present we have theory of pions and nucleons in terms of what is called a chiral Lagrangian. And this chiral field theories you have to include all possible higher order terms consistent with symmetries of the theory. And each coefficient then gets renormalized if you do higher order scattering. So, you have to just determine each coefficient by doing scattering at the required energy. And so, its predictivity goes because you cannot predict what the coefficients are. So, if the theory has only finite number of one particle irreducible diagrams needing renormalization then you have a renormalizable theory. And then you have lot of predictive power like the standard model has bit which gives us confidence to build a machine like LHC. So, those are the advantages of this and that is why this kind of formalism is developed. So, now you may ask. So, last time we introduced what is called effective potential and that is what we are going towards now or effective action. So, so far we got a functional which is a functional of some auxiliary thing j. So, what do we deal with what are we dealing with? So, instead of w of j we would like gamma phi a functional of space time field phi which can have such that we get the dual interpretation by which I mean class expectation value right, vacuum expectation value of the quantum field phi. So, we want to construct an energy and that gamma has a interpretation of governing the dynamics of phi. So, in particular we expect to recover S of ordinary phi the ordinary action of the ordinary phi as the lowest order the zeroth order approximation in gamma I should not write gamma 2, but whatever ok. So, so this is what we would expect. If there is no quantum if there are no quantum corrections then the field theory should just remain classical field theory we will see. So, if you do the quantum action principle for the that is this gamma for our free field theory we will simply recover the same equations of motion for the correctly defined classical field which can now be interpreted as vacuum expectation value of the quantum field ok. So, we are in search of such a formalism and of course, gamma will have gamma will be of the form T minus V where T is kinetic term, but I should say kinetic and derivative terms and V a polynomial well power series in phi alone not the derivatives. So, that is what we call the potential and this is what we call the effective potential. So, this is called effective action and this is called effective potential because if you deal with. So, after you construct this. So, these are exactly like I have been trying to say in thermodynamics you have some gas and whatever complicated dynamics it has the variables that you see for the gas as a whole are pressure, volume, temperature etcetera. Volume is an extensive parameter if you change if you add something then the volume also increases pressure is only local it does not if you have more gas the pressure at that point does not change. So, those are intensive versus extensive variables and you can construct free energy of the system and ask what is its dependence on volume. So, this phi acts like some kind of an extensive variable you ask for it ask for the dependence of field on well this is an intensive variable you ask for its dependence on this particular collective variable phi and it tells you the answer if you switch off the derivatives then it tells you the energy it has and it can be then thought of as a question of suppose I constrained my phi to acquire a part it need not even be a vacuum I could just take. So, typical plot of V effective is something like this ok. What this means is this is the minimum this will be actually the vacuum expectation value because that is where V is 0. So, we take this is how it is done right whoever has seen simple examples of symmetry breaking and so on you vary with respect to phi and wherever you find minimum that is the vacuum expectation value. You may ask what are the what is the meaning of this part of the curve well the answer is suppose I constrained my phi to have this value somehow I arrange it because after all I have that external current J in hand. So, I could crank up the J in such a way that phi gets this value what will be the energy of the system for that value of phi. So, we trade the auxiliary variable J for the actual physical variable phi then we can ask such questions and then we can constrain the value of phi to be where we like or we can make it both space and time dependent then we will have an action for it and we can ask if my phi has this particular space time behavior what is the value of the action and so on. So, that will be the meaning of that. So, the question is how do we go from this description w of J to gamma phi and it is a very interesting and elegant mathematical connection because it uses the Legendre transform. What we observe is that sorry the two point function that we had is something that has two phi's in it G 2 is expectation value of two phi's. So, somehow if we vary this G with respect to phi once we should have information corresponding to one phi left. So, the way this is done is we start. So, I will now try to follow the calculation I have. So, we go back to this free theory example. So, we have to motivate firstly what this phi is. So, we take this w of J which is w 0 times e raise to minus i times this propagator i over 2 times integral d 4 x 1 d 4 x 2 J x 1 delta of x 1 minus x 2 times J of x 2. Now, we see that if I do d w by d J then I get down minus i over 2 integral d 4 x 2 and this is x 1. So, n times the w J back. Therefore, we divide out by w 0 we had that little annoyance there. So, now we divide out by w 0. So, now we say therefore, 1 over w J times d by d J of w that is carrying this to this side which becomes d by d J of log of w and remember log of w we define to be Z aside from an i one should put an i there. So, I can put i here from the beginning. So, I will put supply there is a minus i here. So, if I keep i in the denominator that is bad. So, let us do this and that will produce correctly it will remove the minus i and that would leave a minus sign does not matter what will leave that sign it is ok. So, it is equal to minus d by d J of Z, but now we note that that box plus m square acting on delta was equal to minus delta 4 and therefore, this x 1 we can write, but since it is a symmetric function I can even write x 2 in anticipation of what we are going to do here. This expression is a yeah. So, x 1 is fine correct. So, if I this thing is a function of x 1 it is a functional of a function of x 1. So, if I hit this now which is I mean this refined one the one without the w. So, if I therefore, box x plus m square x 1 plus m square on this I may withdraw the minus sign later by changing the convention, but right now to avoid ok. So, if I hit it with this then this you know is nothing, but what is in this curly bracket this box plus m square can be carried inside the x 2 integral it will produce a negative delta function here remove this minus sign and yeah and leave behind J right and set it equal to x 1 with perhaps an I left over certainly it is proportional to J of x 1 this is and I have to decide whether there is an I or I have lost track of what sorry what is happening here. My book says minus I right. So, this is the correct expression. So, we will define there is no I anywhere because I put an I over here and that tell is correctly with this definition right because we set w to be equal to e raise to I times z. So, very good. So, I am going to now announce that this expression is what I call phi classical because indeed that is the one that was going to obey this Klein Gordon equation the from the original action which was kg action what is the equation of motion we get well we get box minus m squared box plus m squared phi equal to minus j I guess how do all the Lagrange equations work. So, we get d. So, we get a minus d l by d phi aside from the first term d by d of d l by d d right equal to 0. So, if I put this here I would have got minus j here. So, I get plus j here. So, this is fine right. So, when I have this this is exactly what should be called phi classical. So, phi classical is nothing, but variation of that log of the w with respect to the external current carry out. So, now what we do is we declare this phi to be the real the actual variable in terms of which I want to deal with the theory and I want to throw away the j. So, what we do is we do a Legendre transform. So we define gamma phi to be where phi classical x is d z by d j. Now here we have to remember. So, phi classical equal to aside from a minus sign later the minus sign disappears. So, I mean in Ramon's book the minus sign is not there. So, how do I get rid I will call this minus phi classical here. So, I want this definition. So, phi classical is just d z by d j right. And the meaning of this expression is that this gamma is a functional of phi, but you see here j's the point is wherever you see j you have to invert this equation. So, this implies that j of phi ok. So, we have an expression phi classical equal to which will come out to be an expression in j z is a some functional of j. So, if I vary with respect to j I will have some functional of j I have to invert this and express j in terms of phi classical and then plug that j everywhere. So, this j and this j will be this j substituted there that is when gamma will be defined in its correct domain phi classical right. This is the usual Legendre transform trick, but that is what we are now going to propose. So, now we can see we can actually prove that we recover the effective action we will recover by systematically plugging this in we will exactly get what we expect we will get just s because there are no quantum corrections and it is in two steps. So, in the present case we check the utility of this in the free case we find gamma phi. So, first I have to write z of j, but z of j was nothing, but minus j delta j right that is the z j and I have to write minus integral j phi symbolically this is the starting point, but j is nothing, but I substitute from here j is box plus m square acting on phi. So, that is the process of re-expressing j in terms of phi classical. So, here instead of having to quote invert this and all that we can simply read this equation backwards to express j in terms of phi classical and this will be 1 phi 1 and then delta 1 2 and then box 2 plus m squared times phi 2 right in shortened notation that is what the first term becomes and minus again j phi I replace by box plus m squared phi times phi this will be 1 and this will be 2, but well they are local they are the same. So, I do not have to do 1 and 2. So, if we make this act on the delta we will get a delta function because there are second this is second derivative if we transfer both derivatives in succession the sign will not change and it will act on this, but that will produce a delta 4 function and that will set the 2 that will set the argument the same for both. So, what I am saying is that this can be reduced to being simply phi box plus m square phi with all 1 I will do it in a minute with a half factor, but the second one and minus delta function. So, this will become plus half and there is a minus this same thing yeah, but we do it twice because it is a second order derivative. So, that sign does not change and so, we will recover exactly half, but with a sign aside from a sign we will just recover the classical action. So, now let us just see what we are saying over there I think all of you got the idea, but let us write it out. So, what we are claiming is that if I have integral d 4 x 1 d 4 x 2 box 1 plus m 1 square box 1 plus m square phi 1 times this delta 1 2 and then box 2 plus m square phi x phi 2 what I claim is that integral delta x 1 minus x 2 and let me put in the d 4 x 2 here. So, d 4 x 2 delta x 1 minus x 2 times this box operator which is d t square minus grad square plus m square on acting on this phi of x 2 these are both 2 I transfer them both to this side. So, I will get some boundary term which I am going to throw anyway. So, right so, if I turn it once if I turn it twice boundary terms will be of the form integrated values of delta. So, there could be one derivative of delta minus x 1 minus x 2 times of phi evaluated on the boundary of the domain whatever the domain is how do we write a domain some call it d. We will claim that this is 0 the phi is such that the asymptotically phi and its derivatives all go to 0. So, having transfer this derivative twice on this side I will get a plus sign d 4 x 2 times box minus box plus m square and this time 2 acting on delta of x 1 minus x 2 and phi of x 2 sorry this is. So, I this is what was untouched I am sorry. So, I want to write d 4 x 1 times box 1 plus m square phi 1 and then from the d 4 x 2 I got this, but now this became minus delta 4. So, then using up the x 2 integral it just became box plus it just became phi of x 1 and you can then do a partial transfer of a derivative. So, the minus sign should eventually go. So, this is correct. So, right. So, this manipulation all leads to this basically and therefore, we are left with minus a half of this integral, but now I transfer one derivative not both to get my usual form with a plus sign, but that would change the relative sign, but which is exactly my classical action. So, modulo some sleight of hand I may have done with minus signs this is the this is the basic concept involved. So, this is the idea of effective action and we can actually carry out the manipulations and see that for free field theory such a abstractly defined process actually gives you back your original action. So, this is the grand definition with all this legal text attached to it because without that it does not make sense. Lot of people end up writing Hamiltonian as function of velocities. So, you have to remember that whenever you would say h equal to p q dot minus l well how does it work well it works because you have to say q dot is evaluated at p's and this q dot by inverting the definition p equal to d l by d q dot. So, this has to be transferred to read q dot as function of p in it has to be inverted to give q dot as function of p and then substituted here that is when it becomes Hamiltonian until then it is not Hamiltonian function. So, by inverting when you put it back you get the domain of dependence of gamma correctly domain of dependence of gamma is not J. So, you have to replace J by its dependence on phi classical then you recover everything right.