 So, now we come back to functional method and how to compute Green's functions in quantum field theory. So, we begin by introducing the vacuum to vacuum functional in QFT and to that end we need the Lagrangian density in relativistic quantum mechanics which is in the m squared will be. So, we can put minus epsilon by 2 phi squared plus I guess right because we want I I. So, at this level it is minus. So, if I multiply by I it should become minus. So, it is plus I. So, one analogy is that this is exactly analogous to the harmonic oscillator with enhancement of space and time points. So, compare space and time coordinates. So, compare the fact that I have d Q of t only and then e raise to I integral d t and then this Q dot square and minus the half factor is there, but half Q dot square minus omega square Q square. So, from this we are going to integral d the configuration space variable I call phi, but then put t x 1 x 2 x 3 and do an e raise to I and now I do a d 4 x and now I do phi dot square, but minus gradient phi is square minus m square phi square. So, instead of considering a one coordinate problem in one time direction I consider a one coordinate problem, but as a function of four dimensional space time, but with the also with the Minkowski and Metric that is the main difference. And then the analogy is complete because the omega square term looks like the mass square term and all the methods will all the technical calculations will look identical there is not much difference. At this point I could also comment a little bit about this why we put the Minkowski invariance because we want Minkowski invariant S matrix here in the end of course, but if you go to the constructive approach of Weinberg what he basically says is that. So, remark on. So, here what we do is we start with you remember that we argued that if you have a weakly coupled system it always boils down to introducing a and a daggers corresponding to single particle set of Eigen values alpha, but now we implement the so called the word implement Munker invariance by identifying this Eigen values at alpha with momentum k and sigma spin ok. These are single particle attributes. So, single particle momentum and single particle spin and how does that connect well recall in Poirier group we have has all the j mu nu j rho sigma commutators and it also has the p mu p nu and this is equal to 0 and we also have j mu nu p rho commutators. This will this j mu nu p rho because j mu nu is rotation generator it will only rotate the p's into each other. These are the 3D part is the usual rotation group and thus is the Lorentz bus. So, this is the full algebra this is the algebra of the Poirier group it has it has precisely two Casimir invariance. So, into so first of all it has exactly 4 mutually commuting observables which are the p's which become the k mu yeah which become k mu, but remember that this will be using only the space like k because there are two Casimir invariance. So, the if you did not have the translations then you know that there will be two observables one is the rotation j 3 I mean it is equivalent to SU 2 cross SU 2. So, there will be two independent observables one you can say is your third axis rotation, but once you put in the p mu's the p mu's never leave any of the j mu nu invariant. So, the only mutually commuting observables available in Poirier group are the k mu. However, there are two Casimir invariance which are the two who knows one is j squared no no not quite sorry not j squared one is the one is some of the p squares this is one Casimir invariant it will commute with everything because it is a Lorentz invariant the j's will not touch it and the p's will also of course, commute with it. So, this is one invariant which we call m square what is the other Casimir invariant for this anyone else knows it is not just spin. So, there is a 4 vector called Pauli-Lubanski vector. So, it took a long time it was like in mid forties or sometime that and involved Pauli. So, you construct a W mu which is equal to epsilon mu nu rho sigma times j nu rho p sigma now you remember has this been done before no. So, now you learn it. So, we may as well break the field theory for a bit to just recover this well known facts of relativistic quantum mechanics that these the one Casimir invariance comes simply from this and the other one is constructed out of this W. So, W's mu W mu turns out to have magnitude m square times j into j plus 1 where j is the would be Casimir invariant of only the rotations the j into j plus 1 would have been there is exactly the same as would have been there if you had not put in translations. Just to explain quickly what this W mu is a really mysterious thing and you may wonder how the thought up of it. The point is if you look at it in the particles own frame of reference. So, for a massive particle at least in the rest frame p mu equal to m 0 0 right m n 0 vector. So, you choose the sigma to be 0. So, then W right. So, once we said this the sigma becomes has to be necessarily 0 once that is 0 and you have this completely anti-symmetric vector this can never be 0 right. So, it will become W i equal to epsilon i j k j j k times n, but this is our friend the angular momentum I do not know what reasoning they used, but yeah it is very clever construction yes. So, the important and really really the most important statement at this point is that there is nothing that instructs you to put spin in quantum mechanics. Lot of people think that since Dirac equation came from special relativity and it contains spin therefore, we have spin. So, all completely wrong ideas because you can write a perfectly valid relativistic wave equation which is Klein Gordon equation there is no spin to be seen in it. So, you can perfectly do relativity without any spin, but lot of people seem to think that Dirac proved that spin has to be there in relativity without spin we could not have had a relativistic wave equation completely wrong. You have perfectly valid relativistic wave equations without any spin in them. In fact, there is nowhere that you learn from within the formalism that you need to put in the so called intrinsic spin. You are forced to assume due to observations in nature that J has two parts where L can be constructed from our postulate number 4 which said that all observables can be constructed out of the canonical ones. So, L is of course, constructed as R cross P, but this S cannot be constructed from any canonical variable you have to take it on face value you have to put it out of your pocket ok. So, not constructable out of this has to be put by hand, but once you put it and there is the beautiful plus sign whatever that plus means once you put it the algebra is exactly same as that of J and we know in fact, that the algebra of J is complex and allows for both integer and half integer point values. And so, that fortunately matches what we observe in nature it allows you to introduce spin half, but that is all there is and the square of this w mu w mu will come out to be m square times J squared and so, that is the other Casimir invariant. So, you have to take the other Casimir invariant and factor out the m then you get. So, whatever this J is in the particle's own rest frame it will of course, become just S only the intrinsic spin will remain you sit at the location of the particle in its own rest frame there will be no orbital angular momentum. What remains is by definition what is spin and so, the other Casimir invariant has value m squared S into S plus 1 and that is what we put the auxiliary not canonically obtainable, but observed in nature observable is what is spin. So, in the constructive approach then it goes like this and you choose the a dagger to be like this and at this point you argue that you need to construct fields why we got on to this point was this this relativistically invariant difference derivatives we put because if we just went by the analogy then we should have just put the other coordinates that we introduced should also be just that square, but instead we put this because we are trying to implement the Poincare group and there you impose you need to reorganize we can show unitarity causality and Lorentz invariance. We need to assemble so, recipe as sub parts and there would be some species label it could be a spinner electromagnetism whatever it is ok. So, but schematically we write it as sum over these alphas well we can now write k and sigma that does not matter. So, a of k sigma and then we need to put in some u of k sigma with some Lorentz indices a. So, this a are except that they are not necessarily just a mu not just space time once, but they could be spinner ones, but they represent the Lorentz group in the spinner rep. So, somehow it is a Lorentz. So, and we are forced to put these are a functions of x. So, the an a I meant to put that a outside right. So, certainly a will be related to the value of this sigma ok. So, sigmas will be related to it will be some other representation, but it will be a representation, but same matching it. And the recipe c is to construct interaction Hamiltonian. So, this part c is a recipe about constructing which we will see later. So, to be so, if you will remind me I will tell you when we get to that point, but the point is that in this way of doing things you never write the free wave equation lot of. So, this reason why I am telling you is that historically because Dirac pulled his equation like out of the hat it left people stunned for almost 20 years you know people could not work they could not sleep because they thought how do we now write equations for higher spin. So, the causal statement is that yeah causality statement is that only if you assemble them like this will you get that support on the light cone right if you take their if you take their commutators. If you calculate their Feynman propagator only if they have been assembled in this form I should not say Feynman propagator. So, actually it is the two point function you calculate their commutator that commutator will be relatively covariant only if you have assembled them as space time fields like this with both this and its conjugate part included yes. So, it will correctly capture this. So, these are the a by a by themselves are just they are nuts and bolts right. In fact, Feynman is supposed to have commented somewhere how can you have a creation operator for an electron because you are violating electric charge creating charge out of the vacuum. Well I am not anybody to correct Feynman, but it is just a calculation device it does not actually create anything. So, by themselves they are rather bare they begin to acquire a physical or more compatible physical appearance only if you assemble them like this that is the statement. And there is very beautiful reason why you of how you assemble the etch interaction and how the unitarity follows, but that is a much longer story and it is a different path. Right now we want, but I just wanted to say the most important remark is that this bit the so called free field theory is not required in that constructive approach because it is already solved for you already put these quantum numbers. So, and they satisfy the and you only put the k because it is on shell the n I have erased, but we showed that there are 2 Casimir invariants and 4 mutually commuting observables out of the 2 Casimir invariants 1 is here I mean square root of the other Casimir invariant got used up in reducing from this 4 to 3 it is on shell. So, that is how these are listed. So, we exhausted all there is to this beautiful differential equation the Klein-Gordon equation over which often half of the quantum 3 course is spent trying to tell you that it has bad interpretation it gives you negative probabilities. All this is not really required at all it is not part of quantum theory at all you do not have to worry. If you start quantum theory right this is the process implement classically observed symmetries on the Hilbert space by constructing the operators that have the same commutation relations as the classical ones have the Lie algebra that is the realize. And with an advance assumption that we will recover these operators out of the canonical variables of the system which we eventually do we can recover them out of the phi and pi phi and the canonical conjugates, but the point is that the free equation is already solved and there is no further content to Klein-Gordon equation then imposing the energy momentum relation that is all there really is. All the rest there is is the spin content which you have to carefully construct into the construction of the functions u they are either the spinner functions or the polarization vectors of the electromagnetic you know epsilon polarized. So, that that is so and of course, e raised to i k which is solution of that it is already built in. So, the free theory is already solved there is nothing to be done about it and that is where we have a bit of a contrast because here the free propagator appears and it has the free wave equation and you etcetera. So, for the quantized field all everything goes through exactly as before. So, I can hurry up a little bit and say that therefore, we write the w of j now now we switch to the notation j current it will be equal to w 0 times e raised to minus i over 2 integral of well and the same notation j 1 delta f now delta f 1 2 j 2 where so and then we get. So, delta Feynman 1 2 ok. So, let us just specify how this is with respect to the Klein Gordon equation it is equal to d mu d mu minus m squared delta f equal to minus delta 4 sin is not right, is not it? Box plus m square I have to write because box in my definition is d t square minus d x squared. So, d x square will minus d x square will give plus p squared. So, it should be because of the i it will act on e raised to i omega. So, because of the i is I have to put a plus m square here which is probably required here also. So, so far so good and there is a trailer of next time let me tell you that if you now do the same trick right yes. So, anyway so long as these signs remain the same it is fine correct. If you now use your machinery of varying with respect to external current often people put this 2 because you go to endpoint functions this w evaluated at j equal to 0 you will find Feynman propagator here and for if you do it for G 4. So, you will find that the odd ones are always vanishing because remember that after differentiation you have to set j equal to 0. So, if you differentiate and this expression is quadratic in j j and j. So, if you have varied j once this thing would have come down from the exponent with a j sitting in front. So, it will become 0 and you set j equal to 0. So, only the even numbered greens functions are non-zero and if you try to do this it will become factored into delta f 1 2 delta f 3 4. So, x 1, x 2, x 3, x 4, but plus alternatives. So, the I mean you can just check next time we will check it in more detail next time. So, the 4 point function will just boil down to a sum of terms each of which is again product of pair wise 2 point functions and this is how eventually the particle interpretation emerges because you always have this 2 point functions appearing everywhere. So, it is something propagating from one point to the other you can draw the Feynman propagator of the line and so eventually that is those are the building blocks. So, the building blocks of greens functions eventually are these ticks and blobs which are vertices that is what we will show in the next 1 or 2 lecture.