 is, which is firstly a real function because well that is clear in this form right because it contains this and its complex conjugate minus its complex conjugate. So, right so if you take take out right if you take out an I then it will become real. So, this is purely imaginary because you are taking expression minus its complex conjugate. So, it is purely imaginary and it is odd under exchange of x and y that it should be because the commutator has to be this. So, you can check that disintegration gives is odd and it is Lorentz invariant and it also solves the Klein Gordon equation. So, thus if you define this to be equal to I times delta of x minus y because it will just hit if you set up the kg operator in x coordinate or y coordinate it will go and hit the corresponding one and then it will just give 0 because it is the free field. Now, what we want to show is that the general one boils down to some kind of a form that looks very similar to this aside from an overall density that multiplies it which is the finer version of just doing putting a constant. We will see that it has a more detail and gives a k dependent density which multiplies it. Now, to do this we begin by. So, the k L theorem or k L representation cast the. So, the proof goes like this. So, we begin with after all we guaranteed everything only under a mid under a matrix element not as an operator equality. So, what we do now is that we rewrite phi x is equal to e raise to sorry e. So, this is by translation invariance. This translation invariance is actually cornerstone of all of the quantum field theory that is used in S matrix theory where. So, P as you recall is the P mu are the generators of translations according to Northeer's theorem right. So, the point is the origin of field theory should not matter. So, that is first statement. Therefore, if we take 0 phi x phi y taking the first term of the commutator is equal to 0 e raise to I p dot x phi 0 e raise to minus I p dot x times e raise to I p dot y phi y e raise to minus I p dot y 0 and what I do now is to replace this by it is momentum Eigen I insert complete set of states which are momentum Eigen states. So, this becomes equal to the translation operator acting on the vacuum will just all be 0s. So, that part becomes 1 phi 0. So, we get equal to summation over alpha phi 0 and phi 0 alpha e raise to minus I because only this P survives minus I p alpha x minus y alpha phi 0 0 which can be written as mod square of phi. So, as you know if we had only the free field then the only the one particle state would have contributed of course, this is P alpha. So, it is in terms of total momentum. So, it is not in terms of number Eigen states, but this summation would be much smaller if it was free field, but now it contains everything and of course, you have to put the terms with x going to minus x x changing x and y. So, that we get the other term with a minus sign. Now, we want to be cleverer with this and what we do is we want to write this in the covariant form because we want to tie it up with that thing. So, to go to the covariant form insert. So, then this will become equal to times sum over alpha e raise sum over. So, this we need outside. So, what we put inside is only the delta 4 of q minus T alphas times this mod square and e raise to nothing fine. So, to be able to tally later we since we have a 2 pi cube stuck here we put a 2 pi cube here and insert a 2 pi cube here. So, now, this has the form of d 4 k over 2 pi cube e to the minus and this P alpha very good, this P alpha I am now allowed to replace by just q the main point of this insertion. Therefore, it looks like we can rewrite our original expression times a rho of q times e raise to minus i q dot x minus y minus e raise to plus i q dot x minus y with rho q equal to. So, we of course, gathered everything else that was not looking nice into this thing which is a density which is obviously, positive definite not only that it has it is non trivial only for q with q in the positive light cone. So, because this delta function sets it equal to sum of the or any of the values of the P alpha which are all physical momenta. So, now comes the final rearrangement or relibling is to say that it is a function only of q square and is nonzero only for q 0 greater than 0 is the heavy side theta function. This is true because the further thing that it is function only of q square is because the left hand side is Lorentz invariant right it has Lorentz scalars which are Lorentz invariant product of 2 Lorentz scalars. So, commutator of 2 Lorentz scalars. So, this is a Lorentz scalar. So, whatever happens on the right hand side it has to come out a Lorentz scalar and if you look at this whole thing this is a Lorentz invariant measure this is a Lorentz invariant expression. So, this has to be Lorentz invariant. So, therefore, it should depend only on. So, is positive definite with support only for q 0 greater than 0. So, it is function only of the Minkowski square q square, but now we look at our expression here this had almost the same things in it except for the and. So, the theta q 0 becomes our this function epsilon k 0 we want to pretend that this is an integral over this d of q squared only and for q square greater than 0. So, let me just write it. So, I we had inserted in those that delta to make it real I am sticking to the notation I have seen. So, this is a step that says it amounts to changing the definition of your mass. So, you have to read it backwards that with this form which looks like that, but with k square the q square is fixed, but it is the mass values that keep changing and with a weightage factor that is this sigma which is extracted from this density and that is what this expression looks like. This probably needs a little more elaboration, but this is what we will claim yeah well this we have already done. So, this is for a this already derived free field function, but with different values of the invariant in it the invariant q square. So, this was delta of k square minus M square right. So, we trade the q 0 integration. So, if we go back we can think of this as the q 0 with a delta inserted instead of that delta we inserted in the free field we have this density, but we write a different dummy variable which we call since it is anywhere positive it is forced to be positive because it has support only when q square is positive and the theta function is there. So, we call it M squared right and whatever remains is an expression that has to satisfy that delta function it is corresponding delta function. So, it becomes as if it is the free field delta function with that value the instead of omega k's you will be using the q 0 replaced by M square. So, sorry the q 0 it will be the q k 0's inside these will be determined by the condition that they will refer to M prime square such that omega k which is the value of square root of k square plus M square is now to be read as and here it is yeah k square plus M square is to be read as M prime square. So, this will be the q 0 integration is now same as the k integration for the individual ones except that it is quote on shell on mass shell provided you put the mass value corresponding to this integration variable. So, it is just re-labeling of the integration variable and imposing the theta to take it from 0 upward. Now, the power of this formula is that it says that the most general one the interacting commutator is just like summing over the free particle ones. So, that particle picture kind of persists even in your interaction region except that you have to sum over a large number of them that corresponds to all the possible momentum Eigen states. This is what we tried to argue when we said this square root z thing that not only one particle states, but all kinds of other states that a pair so, long as charge is conserved. So, all the pair productions and everything else is generated by these interacting fields, but it is, but we insisted that the energy expression looks like it is sum over. So, this is compatible with justifies sum over k as the unaltered spectrum of the full Hamiltonian. Not only that we now claim that we can isolate the part corresponding to one particle states or the free particle state out of this and so, maybe I write over there. Note that this is nothing, but if you replace alpha by one particle state of all momenta, if you replace alpha by one particle state then this is nothing, but the wave function of the single particle. So, this is equal to then this is the physical mass and as we had agreed on the one particle state the interacting field has square root z times 1. So, the square is z and the delta comes from the commutator that we are taking and plus the remainder part which will look like integral now from m 1 squared some specific value where m 1 square is called threshold for higher excitations with more particles. Now, if we go back to so, this is the final statement. If we go back to our assumption that we will take only polynomial interaction terms remember we wrote the kinetic part and plus a u of phi or v of phi with only phi and not any derivatives. So, finally, this happen, but another finally, is that now if I take time derivative of this which should become and say this is T x this becomes this provided the Hamiltonian or the Lagrangian is only of the form or since we have been writing Hamiltonian is of the form phi dot square is well Hamiltonian we do not write. So, provided Lagrangian is of the form d mu phi d mu phi and minus m square phi squared and minus a u only of phi and no derivatives. Then it is clear that which ensures phi dot equal to phi right. However, badly interacting the theory is so, long as the interactions are only polynomials in phi the canonical momentum will simply be phi will come only from this term and will be equal to phi dot. So, if I now take dot of this I will get phi phi commutator, but which is 1 minus 1. So, that will set the. So, looking at the full expression here here this one hitting with this I will get a minus 1 here, but on the right hand side I have all these things which are right. So, I will get a minus 1 and on this side I will get same things because these are all delta of. So, they are also commutators like that right. So, I will get the ones there interpretation remains the same. So, the I am just arguing about the sign or you take t of y. So, that you get phi phi commutator. So, that we get 1 equal to z plus all those things become 1 because they are commutator canonical commutators. So, they become 1 or minus 1 depending on which order you differentiate it, but this is what it becomes. So, this is a very nice result which shows explicitly that 1 is equal to z plus some positive stuff. So, z has to be between 0 and 1. Now, this was a very important thing back in the days of renormalization because everybody was so scared that everything was diverging and there were infinities. So, shell and lemon were the ones to prove this representation first to reassure people that the wave function renormalization is going to be less than 1. But if you do it in quantum electrodynamics then it comes from the photon wave function renormalization comes from this graph because it is a a a mu a mu a mu and this piece when integrated has to be reabsorbed as just ordinary propagator with a redefinition z to the half at each end. This two particle content has to be absorbed in the free propagator well. So, you do not get any dot there you only get renormalization, but this thing is lot divergent it is not actually not finite at all. So, there is a psychological contradiction between the one loop perturbation theory answer for what z is and this rigorous proof which did not involve any assumptions about anything except that the interaction is purely polynomial that is the only main assumption involved. So, Schellen later proves a theorem that says even in quantum electrodynamics out of the three renormalization constants the this renormalization and renormalization of electric charge and of the electron mass though product of out of those three at least one has to be finite cannot be in finite and that goes back to being able to prove this for the single particle. So, this is only one field now you had several fields and some complicated things were happening, but he proved a non perturbative theorem that at least one out of the three renormalization constants remains finite even in quantum electrodynamics and then one fine day his plane crashed. So, otherwise Schellen was quite a I mean challenge to Schwinger and others because he was proving various theorems without having to take recourse to real perturbation theory, but well that is what the outcome of all these considerations is. So, field theory gets very subtle because you are really dealing with uncountable infinities at uncountably infinite number of space time points, but you can still carry out manipulations under integral signs and protected by some things or the other and you can prove some general things that are that remain valid and they give you quite a bit of insight into how field theory works and how it relates to the S matrix.