 So, let us begin with a little bit of recapitulation of what we have been doing. What we have done is define this external force method. So, for the case of harmonic oscillator we defined in the Dirac picture to be equal to integral well actually we did only. So, we did configuration space this would be the usual Lagrangian and to this we add. So, this introduction of f is essentially a trick in the presence of f it looks like this because queues are going to get integrated out. So, there will be no functional dependence left. So, J is the book keeping function that will keep track of what is happening and when we do then we do one more modification which is that change to vacuum to vacuum. So, I will just skip two things together and then for vacuum to vacuum amplitudes which we call W of J of f here which is this X f in infinity X i minus infinity in the presence of this f as equal to you know you have to insert X f infinity X f T f and so on. So, you can carry this out advance from any finite time T f to infinity on both sides. So, after all of that is done it becomes W 0 times that f d f. So, it becomes the same as W you would have obtained with you know shifted values of Q, but you just get usual W, but then e raise to i times and then this integral d T d T prime f of T d of T minus T prime f of T prime where d becomes the Feynman propagator is equal to this language what is it theta of T minus T prime e raise to minus i omega T is that right omega T minus T prime and plus theta of T prime minus T e raise to plus i omega T minus T prime 1 by 2 is here correct. Now, this whole thing just carries over to field theory where instead of T we have X Y Z and T. So, in in Q f T we will have W of J where now J is function of X which is the so called vacuum to vacuum amplitude and that has the form W of 0 times e raise to i over 2 and integral d 4 X and now this we wrote in terms of in momentum space probably directly yeah we wrote in terms of d 4 p. So, minus i over 2 is there a minus sign here also probably. So, minus i over 2, but now we write this in momentum space for ease I will just explain quickly what happens. So, then it is d 4 p and then J tilde of p J tilde of minus p divided by p square minus m square plus i epsilon where the Fourier transform for any function. So, J tilde p is with defined with e raise to minus i p X. So, this is very similar to this we could have written d 4 X d 4 X prime that check that integral d 4 X d 4 X prime J of X and then delta Feynman X minus X prime and then J of X prime becomes equal to this sorry d 4 that J tilde p. So, that is same form as this just that it is converted to this the important property we use is that the Feynman propagator is function only of the difference of the arguments. So, there is translation in variance. So, in 2 X, X are not really required 1 p is enough in the momentum space ok. So, there is a delta function. So, use is translation in variance. So, this is the sort of starting point of the next part and then we had said that any Green's function can be defined. So, which is actually so Ramon does not somehow write this t product definition, but I think that is what it has to be. So, G n of X 1 X 2 G we call this. Now, that becomes how do you evaluate the t product of these things? Well we know that that is equal to integral of d phi e raise to i s of phi and plus i integral J phi and then these insertions so this is actually same as this. This you know this is the definition if you like we know that if in the path integral if you insert certain operators you will automatically get the expectation value of the t product between these end points, but that is what we are calling Green's function now, but clearly this can then be got by using this trick by using this expression because all we have to do is vary with respect to i times J. So, this is equal to 1 over i times variation with respect to X 1 of the simple amplitude in presence of J and I have to put this because this has no J in it. So, after doing this one evaluates it at J equal to 0. So, this is J, but evaluated at J equal to 0. This amplitude will not contain any phi's this amplitude is this right. It does not contain any phi because phi got integrated out, but there is a J dependence. So, if I vary this with respect to J it will bring down a phi at X 1 it will bring down a phi. So, I can populate this numerator by the required operators by acting by d by d J on this right very standard trick for generating functions. And so, this is what we finally, say is our Green's function g n is this. So, this is how far we got and then we want to. So, we check then that this delta f emerges more or less automatically. If you start with the free theory the free just as we started here with the harmonic oscillator which serves as a emblem for what is the word which it is emblematic of the somewhere where we wrote S right. So, it is the free propagator square phi square. So, that looks very much like half q dot square plus half omega square q square because this is half phi dot square minus half d by d phi by d x square right the Laplacian square. So, this is what the free action is and from that we generate this delta Feynman which we can check satisfies which works as Green's function of the classical differential equation. Here Green's function in the traditional sense no quantum mechanics just in the partial differential equation Green's function of the PDE and by the conventions that Raman uses it is equal to minus delta 4. So, this can also be checked. So, you can just see that that is what that is exactly what you get. So, far so good. Now, we can do a very simple calculation in this particular case. So, for the free theory there are no interactions just the Klein Gordon is that if I calculate G 2. So, G 1 will be 0 right because if I vary with respect to G what have a J. So, G 1 will be equal to 1 over i d by d J of this W J, but that will be to be evaluated at J equal to 0, but that will bring down when I vary with respect to J I am left with a delta and a J right. So, we are left with because this is in the exponent this whole thing is in the exponent if you vary this W of J with respect to J well we should write in real. So, these two are same right. So, this is same as this. So, if I vary with respect to this I will bring down this remainder piece down with a d 4 x prime ok, but I have to evaluate it at J equal to 0. So, while W of J equal to 0 is some sensible answer of the free of the free un unforced theory this J is going to make it equal to 0. So, this equal to 0 right at J equal to 0. So, our next hope is that we calculate G 2 not 1 sorry, but that variation. So, that then exactly produces the propagator because I have varied with respect to first J this will come down then I vary with respect to second J that will get cancelled and this delta times this W will remain and then of course, you will also have J acting on this W itself, but that term will go to 0 for the same reason that this first term went to 0 here and the only term that remains non 0 when you send J 2 0 is this one aside from the W of 0 itself which should be really factored out and there is a minus sign in the exponent right. So, that is why there is no minus sign. Yeah this has yeah there is this minus sign. So, when you vary you would have had that minus sign over here. So, when you vary second time and you get a I to come down. So, you are varying with respect to I in the denominator to cancel that I, but when you do it the second time there will be a I in the denominator which will go to numerator with that minus sign it will become plus I. So, now, we calculate delta of the G 4 by the same method. So, we can see that G 3 has to be 0 any odd G. Now, the ok, so the point of doing this exercise is that we begin to see the particle interpretation of this big formalism. We have plugged in a field phi into the action, but what we come up with where did we write the Lagrangian here. So, we are dealing with a field theory, but after we do this end point function we eventually come out with a particle like interpretation because we already have this G 2 which is the Feynman propagator which as you know from your quantum 3 has basically this kind of behavior it propagates positive frequency particles forward negative frequency particles backwards. So, one often draws this G 2 as a line joining the points x 1 and x 2 create a particle here destroyed there or destroy a antiparticle there and create it here. We will do it in a minute, but just to tell you in advance why we are doing this. So, we are doing this to see that the particle like interpretation emerges automatically from the Green's functions of this theory. So, G 4 then is going to be you can see actually what happens if I vary with a vary 4 times the point is that I when I vary with first 2 of the arguments I will get a delta f down. When I vary with the next J these things cannot give any contribution. So, again more delta have to be brought down from the exponent of W. So, what will happen is that I will get I times delta f x 1 minus x 2 and I times delta f times x 3 minus x 4 right this is the only I mean this is the there are 2 other terms of this kind, but essentially you get pair wise products you cannot get anything new from the what comes down already what comes down in the numerator or the main line because you can vary it only twice and that just produces a delta. The next non trivial answers come when you vary again twice one more delta comes down except that all the orderings are possible. So, you can get this plus I times delta x 1 minus. So, since you are going to differentiate with respect to all the 4 and set J equal to 0 in the end temporarily you can get various combinations down. So, you will also have this. So, all the permutations happen this is happening because we have a free theory there are no interactions ok. So, no complicated connections between these lines get generated, but what we do is we represent this diagrammatically. So, when you do d by d j 1 let us say you will bring down an integral delta. Now, you all you know is you have a d 4 x prime x 1 minus x prime times J of x prime and times the w of j. When you vary twice and have not set J equal to 0 you could end up with d 4 x double prime delta of x 2 minus x double prime integral d 4 x prime delta of x 1 minus x prime times w right and the J is of course, the J x and so on. So, you next you do d by d j 3 the point is when I now hit with a d j 3. So, at this point itself I have two possibilities one is this where this has come down from the exponent or the j 2 could just hit this. So, I will have this plus simply delta x 2 minus x 1 times w. So, both the terms occur when I do this, but of course, this term would get set to 0 if I set J equal to 0, but now I am going to act with more j's. Now, those j's can remove this or this or bring down a more j and so on. So, in the end you end up with all the combinations like this ok. So, this is also an exercise just check this in detail exercise 2. So, the g 2 we interpret simply as creation of a particle at x 1 and going to x 2. So, I will write it like this particle with t 2 greater than t 1 or like this antiparticle if t 1 is greater than t 2. So, that is the detailed interpretation of this I mean this is what the Feynman propagator actually gives and we think of it as a particle being created here and transported there or antiparticle created at t 1 and sent to t 2 ok. Then the g 4 has the interpretation that it is equal to x 1, x 2 times. So, there is a cross between the 2 x 3, x 4. So, these amplitudes are just multiplied to each other the 2 product of the 2 deltas plus sign. So, this plus sign is in the sense of linear superposition. So, in quantum mechanics this 4 point amplitude is a sum of 3 possible ways it can happen with equal weightage. So, either x 1 goes to x 2 and x 3 goes to x 4 or x 1 goes to x 3 and x 2 goes to x 4, x 1 goes to x 4 and x 2 goes to x 3. Now there is one advance lesson here which the book does not spend time explaining at this point, but it just drops the answer and I have also chosen to postpone it because we are not done interacting here any interacting theory yet. But the point is that the fact that we come out with this kind of a combinatoric answer sum of various products of some more basic Green's functions. So, what has happened is this of course, we already saw, but the more interesting remark is that G 4 is simply linear combination of products of G 2. In fact, in a free theory it is somewhat boring even if you do G 56 you will only get products of 2 point functions because there is nothing else that can happen. So, due to this being a free theory all higher G 2 m will be sums of products of M G 2's. It can be shown that the combinatorics of it works out correctly. So, in the more general case new irreducible vertices however, we still get this products of the lower Green's functions anyway in the higher one. In this redundancy of seeing the lower ordered ones appear in higher order ones can be removed by a very clever trick and that trick is that you look at that you take a log of this. The way these are the combinatoric factor comes out exactly correct so that. So, let me write 3 a and 3 b. So, due to b if you do this then because there is a exponent of whatever is z there will be various powers of the terms of z which will get multiplied among each other to produce W in exactly the same way that the lower ordered Green's functions multiplied to produce higher order ones. So, the it can be shown that. So, this will be done in the future. So, it can be shown that z j is the generating functional of the connected diagrams. So, this is one diagram. So, this g 4 is sum of 3 diagrams, but this diagram is disconnected because it has some 2 parties not talking to the other 2 it is like it has fallen into. So, this is a disconnected diagram this is disconnected, but one can show that if you take if you take the log of W which is this z then you get only the connected ones and when you exponentiate this connected diagram generating function then you will generate all the product pieces in exactly such a way as to get the W with products of lower order diagrams multiplying it. So, in our case it is somewhat obvious. So, in the simple case of free theory we already see this because W of j equal to W 0 times e raise to minus i times z with z equal to I have to return that one half factor one half integral j j tilde j tilde. So, or well one half d 4 x 1 d 4 x 2 right I should use Ramon's notation, but may as well write down everything. So, j x 1 delta f x 1 minus x 2, but this we already know is g 2 and we know that exponentiating just g 2 is going to produce the W. So, z this log of W in the case of the free field theory is simply in simple language it is equal to one half j g 2 j that is all there is and because it is a free theory that is the only non-trivial diagram you have and its exponentiation will generate all the possible diagrams of W ok. But when you have non-trivial physics going on then you will find newer g's at a higher level and that is the aim of my part of the this set of lectures to show that that is what we get, but we will be going through a few more things in between. So, now we introduce the Green's function last time. One last comment about this is that there is a momentum space version of this business of connected and I should tell you that I am giving you a only a preliminary discussion. This discussion goes better and again get better and better as you spend more time.