 So, what we were discussing was this asymptotic condition and what we want to say is that the point is that we use interaction picture and in this we say that the interaction Hamiltonian switches off in between ok. This is what we try to say interaction Hamiltonian turns on only in between. So, it is h 0 and h 0 here and h 0 plus h i over here. However, and this is a time. So, minus infinity to infinity, but you might wonder what decides what is h 0 and the answer is that for quantum field theory for relativistic particles ok. So, for relativistic particles what we expect is that the energy continues to if each particle has some kind of meaning on its own then we expect that the total energy plus right k square plus m square where n k is the number of particles with momenta k and and most importantly that m is the physical mass. In particular whatever you write has to be valid in the rest frame of at least some set of particles. So, you can boost to that frame and then those particles are at rest. So, it should just be equal to n times k equal to 0 times the just the physical masses, but if you change to another frame of reference some other k becomes nonzero may be set to 0. So, you would have to from relativistic invariance and provided there is a particle like picture expect that the total energy of the system would just look like this at any instant of time. And therefore, what we do is that we choose H 0 with the split. So, what we expect is that asymptotically as well the free Hamiltonian should be that which contains the physical masses of the particle. So, for equal to n now we are really writing for a scalar field with m square the physical mass. So, the full Hamiltonian could be and let us put some u of phi of x let us not put derivatives for the time keep it simpler take this kind of a Hamiltonian density. So, it could be like this and u would contain some couplings let me put g i g i and we also assume that there are no bound states well if they are then we will kind of leave them out ok. So, if they are the reasoning does not collapse. So, subtract them for the time being. So, so long as the existence of bound states is not going to change the physical mass of the particles this assumption is ok. So, the point is that we assume that the full Hamiltonian has a spectrum like this. So, what is it asymptotic assumption actually as with physical mass if you do not do this then the. So, this is required or this is one way of ensuring and the accepted way ok way of ensuring this is one way of and the accepted way of ensuring that the double limit exists for the unitary operator of time evolution. So, the interaction picture is defined by this making the split h 0 and h i and you know how it works it is i. So, quick recall that in the interaction picture it is i d by d t of psi t i is obtained by h i acting on psi t i and minus i d by d t of any operator is equal to h interaction. Now, there are two h interaction in the interaction picture, but actually the it turns out to be the same whether it you you are shorting or in interaction picture, but just to be complete let us put. So, let me put it like this does not matter keep it simple o i t where h well just to be sure let me put the i here that makes it then I do not have to make any caveats. So, h interaction in the interaction picture is same as h interaction in Schrodinger picture because yeah I mean this is what I remember. So, the point is this is how you define the interaction picture and then because of this and the time evolution and u is the greens function of this equation in the sense that it satisfies. So, you want to write is simply equal to u t prime psi t prime of i it is something that just propagates it from that time to this time and therefore, satisfies the differential equation i d by d t of u t t prime equal to and since it is an operator it will satisfy h interaction and this has opposite sign of the if this general rule for the operators. So, u is the unitary operator that implements time evolution in the interaction picture and the s matrix is defined as the limit when you take t and t prime to go to plus and minus infinity. So, the point is that the region where you can set h i equal to 0 becomes smaller and smaller as you go further and further away and this limit has to exist and the limit exists provided you assume that the spectrum remains the same and that the h 0 assumes the same spectrum as h as you go to plus and minus infinity. So, this is the sort of nuts and bolts thinking behind making these provisions. So, this is end of the recall. Now, with all this last time we saw that the field in the interaction region will be will have more content than the field in the asymptotic region. Oh yes, yes, yes thank you that is why I was getting confused yes. So, that is why the thank you and yes I remember that h 0 thank you h 0 in interaction picture is same as h 0 in Schrodinger picture and so, it does not matter which you put yes, but that is not true for the interaction Hamiltonian that is why I was getting right. So, the thing is the other important theorem of field theory is that. So, what we are next trying to prove is as I was beginning to say. So, we know that we argued that the phi free did we use the word phi free last time what did I use? Phi in yeah. So, the in and out states essentially creates a single particle and similar yes. So, what I meant to say and phi out at t equal to plus infinity create only one particle states. So, we say 1 phi in or out x on vacuum is equal to 1 whereas, so it creates the creation operator from this will create one particle out of this it is overlap with this will be 1 and then there is a delta function which will give you 1. On the other hand for the for phi at finite times in the presence of full Hamiltonian to have more content than this and therefore, we expect that the free one particle state is less than 1. So, we set and this is where we introduce the wave function renormalization. So, you have probably done renormalization or are going to do out of the. So, this already sets the story for what has to happen I just erase the Hamiltonian, but in the in the free Hamiltonian yeah here in the free Hamiltonian we have a mass parameter and we have these couplings. So, when we write this Hamiltonian formally we have so called unrenormalized masses and couplings, but eventually after you take account of interactions within the asymmetric formalism you have to renormalize you have to ignore some of the diagrams and renormalize the couplings. The mildest kind of renormalization that is required is to change the normalization of the field itself and that is already seen at this level you do not have to do any diagrammatic calculations to see that you actually need to renormalize the wave functions. So, thus we expect that there is a factor square root z as t goes to plus in minus infinity or plus infinity of phi in out of x times a square root z and that z we expect same for in and out which because after all the evolution is unitary. So, the normalization from this end to that end does not change. So, z is the same in both in and out regions, but this is what we expect where z is a number between 0 and 1 ok. It becomes one if it is free field theory and further the caveat is that the above is not an operator field although this is about quantized fields there is a word for it only as a weak limit. So, when we withdraw from making the full matrix full operator assumption to only for matrix elements what it means is that if you take higher powers of this statement then they may not hold because you then you have to do matrix multiplication of and you get a more complicated answer. So, do not expect that if you do phi square we go into z times phi in squared, but phi itself we expect this to happen because the phi squared operator then has intermediate and contributions from various other states. So, it may not hold right. So, now I want to show a slightly intricate derivation I hope that I get it all right. So, this is called this is not German umlau, but some Swedish umlau. So, what is the representation is about I will tell you soon well. So, the representation is about the commutator phi x phi y phi y in a form understandable from free field theory this is what the representation does. It rewrites the general interacting field commutator in terms of expressions that refer to free field quantities. First begin with phi in could be out also, but let us just do phi in. Consider this commutator. So, this is not time ordered product this is genuine commutator of phi x phi y which you know has to be 0 at equal time the canonical commutation relations say that this has to vanish at t equal to 0, but it is for general arguments x and y ok. So, we write this out to we have to first compute this. So, the easiest way to do it is to say I will split it up into phi in plus x plus phi in my. So, we will drop the in for the time being because ok. So, everything is in. So, phi plus x phi minus x phi y plus plus phi minus y well these are the positive and negative frequency parts. So, they contain a and the negative frequency part contains a dagger just since we have not done field theory fully together let me just say specifically in my normalization this is d 3 k over 2 pi cube 2 omega k everything square root times a k and times e raise to minus i omega k t plus i k dot x and phi minus is the dagger of this, this is plus sign and this is a dagger. So, plus just means positive frequency and the minus i occurs because of the Schrodinger choice of i h cross as the i i i i d by d t as being the Hamilton plus i d by d t as the energy. So, this is what we mean by the positive frequency part and our convention is that a dagger is equal to 1 a k a k dagger a prime dagger is equal to delta 3 k minus k prime. So, if you do. So, in these commutations this with this will give 0 because they both contain annihilation operator. So, we have to worry about phi x plus comma phi y minus which will be equal to integral d 3 k d 3 k prime and then all this stuff square root 2 omega k you can put 2 outside also if you like omega k omega k prime times e raise to minus i omega k t and this is t x t corresponding to x coordinate plus i k dot x times the thing from the positive negative frequency part which is plus i omega k prime t y minus i k prime dot y and then the commutator which just gives us a delta function. So, this just becomes equal to integral d 3 k over 2 pi cube 2 omega k times e raise to minus i k became same as k prime. So, it is now equal to t x minus t y and plus i k dot x minus y. So, putting it all together we will see. So, now we can just be a little casual write k dot x minus y and when we do the same thing with. So, here it is to be understood that k 0 is actually because there is only d 3 k. So, you have to understand the k 0 to be the positive square root of k square plus m square and the other part which is commutator of phi in plus sorry phi minus x with phi plus y will contain firstly these signs will be opposite, but it will contain sorry it will contain a dagger here and a there. So, that will put a minus sign right because the commutator is in the opposite order and we will and this sign will become plus. So, we write minus sign and e raise to plus i k dot x minus y with this i always find confusing. So, when you have only d 3 k integration then actually I always prefer to write out omega k, but it would become too long and that is written up there. So, now this looks a little awkward you cannot make much sense of it, but you can be clever and rewrite it as equal to as a 4 integral. So, you say that this is d 4 k over it is still 2 pi cube, but now we insert a delta of k square minus m square right to take care of the extra k 0 integral that you put. Now, I will just for the time being try that this is equal to this times e raise to minus i k dot x minus y and here of course, you do not have to understand the k 0 as here k 0 is free. If you do this then basically this sign will come out wrong. So, let us just check this. So, this is equal to integral. So, you can take out the d 3 k and 2 pi cube outside and then integral d k 0 of and now you have this delta function of right delta function of a polynomial that is the delta function at each root, but divided by the f prime evaluated at that root. So, it is equal to delta of k 0 minus omega k and some denominator and plus delta function of k 0 plus omega k and now what is the derivative? So, you have to put. So, it will be 1 over that is right 2 times omega k because you have to put modulus right. So, it is 2 times omega k in both. Now, if you really followed this through. So, that is why I said try. If you really followed this true you actually recover each term correctly because all you have to do now is to set do the d 3 k 0 integral set k 0 equal to plus omega sorry k 0 equal to omega k here which will give you this and in the second term k 0 equal to minus omega k. So, it will give correctly this with the k 0 understood as minus omega k and the d 3 k signs can always be flipped you know if it here it will be minus i k 0 plus i k dot x minus y when it becomes e raise to minus i omega k times t x minus t y we are fine, but when it becomes e raise to plus i omega k you are still stuck with plus i k dot x minus y, but that can be flipped to minus without costing anything because this d 3 k can be flipped with the and the Jacobian of going from k to minus k is 1. So, it does not change anything. So, that can be reversed and so you recover this term as well as this term, but this sign comes out wrong. So, to take care of that and the whole exercise now is if we go towards this then we have a Lorentz invariant expression. So, far we were just commuting and we were getting something a little unclear, but once you cast it in this and because there is a d 3 k integral it is it looks non-covariant, but now your d 4 k integral a delta function that involves only a Lorentz invariant product I mean magnitude of k and this is a Lorentz invariant inner product. So, the whole function becomes manifestly Lorentz covariant except that this sign came out wrong. So, we need minus sign or and to cure that we insert a function k 0 over mod k into this. So, and this we might call epsilon of whatever it is argument it is sign of sign of k 0 function which is also invariant under allowable Lorentz transformation. So, long as you do not do transluminal what is the word you know you do not try to cross the light cone. So, long as you do any normal Lorentz transformation a axis that is positive time axis will remain positive time axis. So, Orthochronus which is Lorentz invariant under which are the connected set you have disconnected sets if you take the full invariance group of the Minkowski inner product, but the connected one that you use physically is the so called Orthochronus one.