 Last class in this unit in which we plan to introduce Feynman diagram methods. We develop the equation of motion for the time evolution operator and today we will introduce the Dyson chronological operator. Now it will take us a few classes of you know development of the interaction picture formalism before we actually introduce these beautiful diagrams which are known after Richard Feynman. So, that will come toward the end of this unit or at least after we are half way through. So, I believe there are some six or seven classes in this unit. So, this is how an interaction picture state evolves with time or we discussed this in the previous class. The time evolution operator satisfies an equation of motion which is often called as the Schrodinger equation for the time evolution operator and we got its final solution in our previous class which consisted of these three operators and you have to remember that you have h 0 here, h 0 here as well, but the full Hamiltonian here. So, h 0 is the soluble part of the full Hamiltonian and the full Hamiltonian h is made up of the soluble part h 0 and an additional part which we wrote as h 1 which is the one which generates the complication and which is makes which is what makes the whole problem so complex. So, we had the equation of motion set up for the time evolution operator and we have a formal solution in which the Hamiltonian appears in these three locations here as h 0 which is the soluble part of the Hamiltonian, here we have the full Hamiltonian and here again we have the soluble part of the Hamiltonian. So, the part which is missing an h 0 is the one which is generating correlations and that is not amenable easily to perturbation theory which is why we need many body techniques to handle that. Now, what we are going to do today is to develop an integral equation for the time development operator because you can set up the equation of motion either as a differential equation as you see over it over here or as an integral equation and what we are going to do today is to develop an integral equation for the time development operator. So, mind you I have added a subscript i to remind us that this is the interaction picture time evolution operator and the primary references are these two books Reims and Fetra and Wallachia. The notation is slightly different in the two books here you have t and t prime whereas in Fetra and Wallachia you have t and t 0 but other that you know it is the both the books provide you essentially the same kind of treatment. So, this is the time evolution operator which this is the differential equation for the time evolution operator I just take i h cross on the other side and now you have got a derivative on the left. So, it is obvious that it is amenable to integration. So, integrate it from 2 t 0 to t and on the left hand side now you have an integral of a derivative. So, you can immediately integrate this put the limits and you have the difference of u at the upper limit and the lower limit. So, the lower limit it is u t 0 t 0 when t prime is equal to t 0 and at the upper limit you have got t prime equal to t. So, this difference is equal to this definite integral. So, it is a definite integral from a lower limit to an upper limit and this is the equation that you will satisfy. Now, we already know that the time evolution operator from any instant of time to the same instant of time is essentially the unit operator. So, you take this unit operator to the right and you get the solution for u t t 0 equal to 1 minus i over h cross and this definite integral. Now, this is nice, but in some sense it is a solution in another not really so, because you always a solution means that you are able to determine the left hand side in terms of the right hand side and how can you get the right hand side because the left hand side involves the u which is the left hand side under the integration. So, it formally correct, but it really is not very helpful. So, we are still stuck and we still have to figure out how we are going to resolve this. So, it is formally absolutely correct nothing wrong with it, but not very helpful at this point. So, we still have to figure out how we are going to deal with this. There is another issue over here that if you look at the upper limit of integration over here, it is also an independent variable. Time is the independent variable over which integration is carried out and it appears as an upper limit of the integration over here. So, these integrals are well known in mathematics and typically they are known as Volterra integrals in which the independent degree of freedom over which the integration is carried out appears as one of the limits of integration. And the mathematics of Volterra integrals is fairly well developed and those techniques can be used. By and large you can solve this specially when the what is being integrated are functions. We have to be a little extra careful in quantum theory. In physics you do rigorous mathematics as I always like to point out, but you do something more than that because the mathematics that you are doing has to confirm to the physical laws of nature. And in physics the relationship that we are working with involves not ordinary functions, but operators. So, the operator algebra will have its own consequences. And the immediate consequence that concerns us in quantum theory is that when you are working with operators we never take for granted that when you have two operators a and b they commute. They sometimes they do and sometimes they do not. So, we have to be very careful about it. So, for ordinary functions iterative solutions for such equations can be well developed. These are the Volterra integrals and the methods are available for that. In the present case very similar solutions some similar strategies can be developed and that is what we are going to do in today's class. But the inspiration comes from the mathematical formalism which is well developed and very well organized already. And we will keep an additional rider on in our mind that we must preserve the ordering. So, the order in which the operators come that has to be very carefully kept track of. And that is something that you do not always have to do when you are dealing with ordinary functions. So, here is the solution this is the formal solution. And this one will have a very similar solution, but now the variable t prime is the dummy variable which is used in the first equation. So, when you want to use a dummy variable for which will appear in the integration for u t prime t 0 you must use a different dummy variable and the one that I have used is t double prime. So, I have u t prime t 0 which is given by an exactly identical relation with the difference that now the integration is over t double prime. And the integration is over the limits t 0 to t prime which are the two parameters which come here. Over here the integration was over t prime which was the dummy label here. And this was from this limit t 0 which is the lower limit to this one which is the upper limit coming over here. So, now you have t prime t 0 equal to 1 minus i over h cross integral t 0 to t prime. And if you now put this the second equation in the first then the left hand side which is u t comma t 0 is now given by 1 minus i over h cross and now you have these terms. So, you have d t prime h i t prime and in place of this you can write this. Now, it is not necessary to really take notes and write these equations in your notebooks during the class because at the end of the class all of this is going to be uploaded a pdf will be uploaded on the course website. So, you will have access to that. So, you do not have to really write down anything just follow the discussion and see how the steps are developed and that is part of the reason I use these slides. So, that you can have all the material you do not have to write down anything in the class and you can concentrate on how the development of the formalism takes place. So, if you have any question you can always stop me, but otherwise just follow the discussion. So, now let us write these terms. So, you have got the first term which is here, you have got the second term coming from this and then you have got a unit operator here which does nothing really. So, the second term is minus i over h cross this integral t 0 to t d t prime and then you have this h i t prime over here right. So, that is the second term. What about the third term? You have got minus i over h cross coming twice. So, you get the square of minus i over h cross, you have got two integrations one over t prime the other over t double prime and the limits of t prime are t 0 to t. So, that is what you have over here the limits of t double prime are t 0 to t prime and then what is being integrated out are h i t prime and h i t double prime. So, h i t prime comes first and h i t double prime on its right. So, that order has to be maintained and then this u is also an operator. So, this comes at the end. So, the ordering of the operators must be respected you cannot swap them these are quantum operators. So, let us write them in a slightly neater form. So, you have got the integration over t prime integration over d t double prime then you have got h i t prime over here h i t double prime and u t double prime t 0. So, all the orders are written in the correct sequence the integration limits are taken care of and this is the form that we will analyze now. Observe however that you have got u. So, this is again not really a solution because you will have a similar expansion which is written in the very first equation at the top of the slide which will appear over here. So, you will actually end up getting an infinite series and then you can develop some approximation methods to see if you can truncate it somewhere. So, essentially you will you are looking at an infinite set of series. So, just remember what is at the left and what is at the right. So, h i t prime is at the left and h i t double prime is at the right. So, that is what I bring to the top of this slide here now and here you will have a similar expression. So, now you have another integration a third dummy variable which is t triple prime and we kept keep showing this for the rest of the course and never finish because there are infinite terms. So, what we will do is represent the remaining terms by these dots that is the tail of the infinite series. Now, in this infinite series you have got a first term, a second term, a third term and then the fourth term and the fifth and so on all the way to infinity. Now, we look at the third term and the third term does not have this u. So, the third term we can certainly discuss by itself. So, let us look at the third term and any physical quantity of physical quantity is a sum of two of its halves. So, I write this term and take half of it and write the same term take the half of it and add these two. Now, this will really this is a simple thing that we are doing, but it is going to give us the tools to write this whole term in a very beautiful manner and you will see it in the next few minutes. So, these are the two halves which we have added together to get the left hand side this is the third term other than that minus i over h cross and so on. So, this is the integration part of it. The two integration variables t prime and t double prime both are time variables, but they are quite independent of each other and they are to be treated like independent variables they are completely independent degrees of freedom and you can represent them by two variables which you represent graphically orthogonal to each other orthogonal because they are completely independent of each other. So, you have got t prime on the x axis t double prime on the y axis and you have the t 0 at the crossing which is the origin of this orthogonal Cartesian kind of coordinate frame of reference for time both are time one is t prime the other is t double prime. If you look at a diagonal in this along this diagonal the variable t prime is equal to t double prime along this diagonal above the diagonal t double prime is greater than t prime and below the diagonal t prime is greater than t double prime it is the other way round. You can also look at some special points like a particular value of t when t prime is equal to t is given by what happens on this vertical line. So, this vertical line corresponds to t prime equal to t and likewise you have a horizontal line over here at which t double prime is equal to t. So, this is a very simple neat graph you will find it in both the books by Reims as well as Fetcher and Wallacher and all it does is to describe how these two variables are dealt with as if they are completely independent of each other and what consequences they have on the integrations that we are considering in these two terms and to do that we make use of what is called as the Heaviside step function and you have probably met the Heaviside step function earlier. So, let me remind you it is a function whose value is 0 for an independent parameter up to a certain point and beyond that point its value is equal to 1. So, here it jumps there is a singularity here it jumps it jumps and that is the reason it is called as a step function it is just the way you climb a staircase. So, this is called as a step function and we will use the Heaviside step function to represent our operator functions in the integrand. So, this is what we are integrating out and in this term for example, t double prime must go from t 0 to t prime. So, it shows a certain order in the variable t double prime because t double prime must always be greater than or equal to t 0 the lower limit of integration and it must always be less than or equal to t prime which is the upper limit and if this relation is not satisfied then what is being integrated out should be considered to be 0. So, you can use the Heaviside step function to represent this because the Heaviside step function will have a value which is 0 below a cutoff and it will be unity above that cutoff. So, let us do that let us write. So, there are two terms of the right side and the first term I write using the Heaviside step function and what it will let me do is allow me to take this upper limit t prime to t because I do know that I have to carry out another integration over t prime which must go from t 0 to t. So, this limit will have to be stretched all the way up to t. So, I extend this limit t prime to t, but I have to receive respect the fact that t double prime must always be less than or equal to t prime because that is the integration constrained which is coming from these limits. So, I have a Heaviside step function over here. So, let me make it manifest here and this Heaviside step function will take care of this and allow me to extend the integration the top limit of the integration from t prime to t. So, now, we have got the first term which is strictly correct it is exactly in accordance with this exactly in accordance with the five that t double prime varies from t 0 to t prime and furthermore that t prime varies from t 0 to t because the upper limit of integration itself is a variable this is what made it a Volterra integral in the first place if you remember because they have the independent degree of freedom appear as limits of integration. So, this is how you make use of the Heaviside step function which is theta and this will be unity if and only if t prime is greater than t double prime. So, that takes care of it. So, now, you have the first term in which the limits are properly taken care of and now let us look at the second term which is the second half of the whole the whole is on the left side which we wrote as two halves the first half and the second half we now look at the second half and if the second half look at the upper limit of t double prime t double prime goes from t 0 to t prime. So, you have t double prime which must go from a lower limit t 0 to an upper limit t prime which is this and now you can let the upper limit go all the way to t because t prime must subsequently go from t 0 to t. So, you let that happen and then have the ordering taken care of by the Heaviside step function. So, now everything is fine and t prime and t double prime are dummy labels we can interchange them as long as you call you rename t prime as t double prime and rename t double prime as t prime you can write t double prime do not blame me for that, but that is going to let you write these terms in a very nice manner. So, that is why you do that. So, you now relabel t prime as t double prime and t double prime as t prime. So, when you do that you have this relation. So, instead of t prime you have t double prime instead of t double prime you have t prime here instead of this t prime you have got t double prime here instead of this t double prime you have got this t prime here and instead of the Heaviside step function t prime minus t double prime you have Heaviside step function t double prime minus t prime. So, everything is exactly the same as before all we have done is to interchange the labels. So, let us take it to the top now and what is interesting about this is the consequence of the Heaviside step function because that has got a discrete value which is either 0 or 1 depending on a cut off of the argument of the Heaviside step function. What it is essentially doing is preserving a certain ordering of the time operators what it is ensures that the operators containing the latest time stand farthest to the left. These operators stand farthest to the left and it always reminds me of the orders they give in the army or the NCC. Lamba dai chota bai ek line mein kadwar have you heard of it? Lamba dai the tallest man stands to the right and the shortest to the left because otherwise when soldiers line up in one row and they have different heights it looks like a very crooked skirt. Skyline but if you order them to stand so that the tallest one is to the right and the shortest one to the left. Lamba dai chota bai. Then it looks like a very neat ordering of how the soldiers line up that is how it is often done in the army so that your squad looks very nice. So, here you have the operators containing the latest time stand farthest to the left and that is ensured by the Heaviside step function but you can do it by writing the same in a different way using what is called as a chronological operator or the time operator and what the time operator will do is that it will order all the operators on which it is operating such that the operator with the latest time stands farthest to the left. It is exactly the same order and that is done by this operator which is called as the T operator. So, first we combine these two terms we have written half over here and you have got these two terms sitting over here and now we will exploit the fact that operators containing the latest time stands farthest to the left by introducing the T operator. This is called as the chronological operator or the T operator sometimes as the Dyson's chronological operator and so on. And what this operator is going to do is whatever operators it operates on and these operators have time appearing in the argument. So, you have got T prime here and T double prime and what this operator will do the T operator the chronological operator is that it will force these two operators to interchange their positions if need be if it is required that the operator with a later time stands farthest to the left and that is achieved by this time order operator the T operator. So, this integral you can write in terms of the T operator. So, now instead of the Heaviside step function theta I have the same exactly the same effect which is represented by the T operator. So, I introduce the Heaviside step function as an auxiliary step that we do not really need it what we need is the time ordering of these correlation operators that must be preserved and the Dyson chronological operator guarantees that it will be done and now you have if you remember you had an infinite series the first term the second term the third term and then the rest of them represented by these dots the third term was this which we wrote as two halves and what we find is that these terms can be combined in this manner by using the T operator. So, let us do that you can generalize this we had done this only for two operators but here you have to deal with this infinite series all these dots. So, you can generalize it and you can write all of these dots let n go all the way to infinity because those are the infinite terms which are sitting in these dots you can let n go to infinity and you have a very general form of the solution you have a very general form it is a very formal solution it is a complete solution it is absolutely accurate what it requires you to do is to keep track of the time ordering of these operators. You can let the integration variables go from t0 to t like over here both t prime and t double prime went from t0 to t and here we have generalized it to all of these n integration variables t1 t2 t3 all the way up to tn and we then let n go from 0 to infinity every time you have got this minus i over h cross but now you will have minus i over h cross raised to the power n right. So, you have that taken care of over here you do have to keep track of the time ordering. So, you have got the t operator over here which takes care of that and because there are factorial n time order rings over here which are available you have to over here you had divided it by 1 over factorial 2 1 over 2 as it was and here you have got 1 over factorial n because those are the number of different order rings that you can have. So, now you have got a very general solution in which the t operator has been made use of. So, let us work with this now you have got the solution this is what we had on the previous slide we are all together. So, we have this general solution now what we are going to do is let t0 go to minus infinity because this relation is valid no matter what the values of t0 and tr. So, we are going to let t0 go to minus infinity we can let later let t go to 0 because we have we are introducing adiabatic switching in such a manner that the Hamiltonian is the soluble part at 0 at minus infinity and it is a full Hamiltonian at t0. That is how we are developing the time evolution. So, we now let t0 go to minus infinity. So, all the lower limits which were written as t0 over here are now written as minus infinity look at it over here look at it over here look at it over here all the lower limits are now minus infinity. So, that is what we have done first and here we have these dots for the infinite series, but in an equivalent expression over here we do not have these dots to represent the infinite terms instead we have a general term which is the nth term and we let n go all the way from 0 to infinity. So, these two are completely equivalent forms one in which we use these dots to write the unwritten terms and in the this lower equation we have an exactly equivalent solution of final formal expression for the time evolution from minus infinity to t in which the t operator the dysynchronological operator is used and we now let n going from n to go from 0 to infinity. So, these dots are replaced by this upper limit of the summation which is infinite and that is the role it has. So, both the term both the expressions are completely equivalent and exactly equivalent to each other. So, here you have got the sum going from 0 through infinity I factor out the term corresponding to n equal to 0 which is the unit operator and the rest of the terms I bundle up together as sum over u n n going from 1 to infinity instead of 0 to infinity because the term corresponding to n equal to 1 n equal to 0 is already factor out over here. So, you have got 1 plus this sum over u n n going from 1 through infinity and u n this general expression is what you have here which is minus i over h cross to the power n 1 over factorial n you have got these n integrals each from a lower limit minus infinity to an upper limit t and whether t 1 is greater than t 2 or not and what would be the relative ordering of these two operators is determined by this t operator the chronological Dyson's chronological operator which guarantees that the operators containing the latest time will stand farthest to the left. So, what does it do for two operators if you have two operators a and b one at t 1 and the other at t 2 and if t 1 is greater than t 2 then a will stand to the left of b if t 2 is greater than t 1 then b will stand to the left of a that is what this operator is doing if t 1 is equal to t 2 is the same. So, it really does not matter. So, when t 1 is equal to t 2 then the interaction term the correlation term the Hamiltonian h i t 1 and h i t 2 will be equal to this if t 1 is equal to t 2 and it really does not matter how you write them. So, in general either for t 1 greater than t 2 or equal to t 2 this identity holds. So, both for t 1 greater than the t 2 as well as for t 1 equal to t 2 and we will use this in the general expression. So, I show it for just two operators, but you can do it for 3 4 or infinite it is the same. Now, what is interesting as a consequence of this is that you have this general result I think there is a little font problem over here which is why I had a carrot appearing on top of the t, but look at the argument here. Here the first argument is t 1 and here the second argument is t 2 and the right hand side I have got the first argument which is t 2 and the second argument which is t 1, but this operator t the chronological operator ensures that whichever is latest is going to stand to the left. So, it really does not matter if I write this as how it is written on the left hand side or how it is written on the right hand side because the t operator ensures the equivalence of the two is going to guarantee that whatever is the latest operator you know will stand to the left. So, we now look at the n th term which is this and there are how many factorial n time orderings take one of them any one of them let us take this one in which t 1 is greater than t 2 which is greater than t n. So, this is certainly one of the factorial n orderings that you may have and if you take one of these then you just go ahead and stick it in. So, you write t 1, t 2 all the way up to t n in this particular order, but then if you were to interchange it like if you have the j th argument here and the i th argument here it would not matter because the chronological operator t would guarantee that the latest operator will stand to the left. So, just what we showed for just two operators also holds good for n even in the limit n going to infinity. So, we will make use of this now and we turn on the perturbation adiabatically and that is where theoretical physicists have an edge over experimental physicists because experimentalists cannot switch on and off an interaction. Theoreticians represent that interaction by some term and they can develop an approximation in which they say that let me do it when this term is 0 and now let me do it when it is not 0 I will just switch it on. So, that is a mathematical switch and mathematics lets you do that. So, you have got this perturbation h 1 which is the difficult part you do not like it because this is the one which makes the problem you know you cannot solve the problem we have argued number of times that if you are looking for exact solutions even nobody at all is already too many. So, you cannot get exact solutions you have to develop a approximation methods there are these problems like in the many electron problem the atomic problem in which you have got n electrons you get a solution in the Hartree-Fock formalism which is good you have got a nice solution, but that is not exact because it leaves out something what does it leave out it leaves out the electron correlations it takes into account the exchange correlations, but it leaves out the coulomb correlations and we are now going to worry about how to account for the coulomb correlations. So, those are the ones that you cannot make use of by you cannot handle using perturbation theory. We showed in some of the our earlier classes including in the previous unit that the Hartree-Fock result is equivalent to what you would get if you were to treat the electron-electron interaction perturbatively if you go up to the first order perturbation theory and at that point I had commented that if you do second order perturbation theory if you try to get additional effects then the series does not converge. So, perturbation theory does not work. So, perturbative methods are not always successful and that is the reason you have to develop these new techniques which is why we are getting into Feynman diagram methods and you have this unfriendly part which you cannot handle using perturbative methods perturbation series does not converge and you have to find some mechanism to deal with this. This is the correlation part this is what is left out of the Hartree-Fock that is the one that we are trying to deal with. So, you have got a solvable part and you have got another part which is H 1 and what we do as theorists is to add a mathematical construct multiply this by e to the alpha t. Now nature will not let you do that but mathematics lets you do that. But eventually what you do as physicists must correspond to nature because that is what physics is about. So, we develop this mathematical technique we multiply H 1 by e to the power alpha t and in the end if we take the limit alpha going to 0 then you get the full Hamiltonian which is what the problem really is. We are looking for solutions for the full Hamiltonian. So, that is our target and we will get solutions to this problem by developing a formalism by inserting a mathematical device which is the switch control parameter. So, alpha is the switch alpha is the adiabatic switch it is a mathematical construct and what it does is if you take the limit t going to minus infinity then the full Hamiltonian collapses into the soluble part that we are happy with because we know the exact solution for that. So, in the limit we get in the limit t going to minus infinity alpha is a positive number and we will let the limit alpha go to 0 while staying positive. So, alpha is some tiny positive number which will eventually find its limit which is 0. So, in the limit t going to minus infinity you have got the soluble part of the Hamiltonian and as t goes to 0 you get the full Hamiltonian which is really the problem for which we are finding solutions. So, now we have got a very nice mathematical device and we switch on the perturbation adiabatically in a very quasi-static manner just the way you do things in thermodynamics and we are interested in end results such that they are independent of this mathematical device and we take the limit alpha going to 0. So, this is the summary of what we have we have got a soluble part this we are happy with we know the exact solution this is the full Hamiltonian that we want to address and this is the adiabatic switch which is the mathematical device to deal with this residual correlation which we are not able to handle in the Hartree-Fock approximation. We are doing this analysis in the interaction picture in which the Schrodinger picture operators transform to interaction operator interaction picture operators using this and the Schrodinger picture wave functions transform to the interaction picture wave functions using this relations. So, we discuss this in our previous class and in this interaction picture the correlation term the difficult part of the Hamiltonian in the interaction picture transforms to this, but this was the actual interaction that we wanted to work with, but now the interaction that we have to work with is not just h 1, but h 1 times e to the alpha t. In our formalism we have now introduced this e to the alpha t. So, let us plug this in and what do we get for the interaction picture Hamiltonian. So, you have got the same kind of relation over here, but this h 1 is now replaced by e to the alpha t times h 1. So, now time appears here, here and also here. So, it appears at three places over here it appeared only at two places. This is the relation that we developed the differential equation for the time evolution operator and the integral solution. Now, this is going to remain valid even if h 1 goes to e to the alpha t h 1. There is no reason why this would change. So, one can do the same thing and show it explicitly, but nothing in what we have done you know comes in our way of using this. So, we can we are going to use the both of these results with the difference that h i is now made up not just of this h 1, h 1 times e to the alpha t. So, that is the difference that we are now going to incorporate in the rest of the analysis. So, here we are we have got the interaction picture operators and state functions. You have got the interaction picture operator corresponding to the perturbation part or the correlation part or the part which is the unfriendly part as I sometimes call it. But the time development operator now in some sense must explicitly respect the fact that we have inserted a mathematical artificial term because I could use an alpha which is a tiny small number and you could use a different alpha may be a beta which is also a tiny small number right, but different from what I had chosen. So, whichever is being used must be kept track of and therefore, I put a subscript alpha on this u and I write this psi i t as u subscript alpha t comma t 0 psi i t 0. So, this alpha is going to keep track of the fact that some particular choice has been made whatever although in the limit at the end and what you started out with will not matter after you have taken the limit alpha going to 0, but not until then. So, at this point we have not taken the limit alpha going to 0 and we must therefore, keep track of what alpha was chosen. So, this time evolution operator now will be written with a subscript alpha to remind us that a particular switch, adiabatic mathematical switch alpha has been used. So, now this is the general solution that we had written earlier and we write it on in an equivalent form, but respecting the choice alpha where the full Hamiltonian is no longer h 0 plus h 1 rather it is h 0 plus h 1 times e to the alpha t. So, it is the same solution, but now you want the subscript alpha. So, let us take it further now and we are in particular interested in two limits t going to minus infinity when we know what the solution is that is the soluble part of the Hamiltonian. The other limit which is important to us is the limit in which we are really interested alpha going to 0. So, we have discussed what happens when t goes to minus infinity you get the soluble part right and the soluble part is this h 0 phi 0 is equal to e 0 phi 0 this is the solved Schrodinger equation you know the exact solution this is the time independent stationary eigenstate of the unperturbed Hamiltonian the h 0 the solvable part of the Hamiltonian. And how does this Schrodinger state evolve with time it evolves as e to the minus i h over h cross t this is the time evolution operator right and as t goes to minus infinity as t goes to minus infinity your Hamiltonian is actually h 0 right in the interaction picture you have got the full Hamiltonian over here, but this full Hamiltonian in the limit t going to minus infinity is h 0. So, you have got e to the minus i h 0. So, instead of this h you can write h 0 over here. So, that is what we have got over here. And now we ask this is how the Schrodinger picture wave function evolves with time we ask how would the corresponding interaction picture wave function evolve with time and we know the answer the interaction picture wave function would evolve as this as the Schrodinger picture wave function operated upon by e to the i h 0 over h cross t this is the solvable part of the Hamiltonian. And what it essentially means is that these two are exactly equal when t is equal to 0 it becomes independent of time. It becomes completely independent of time and just like the Heisenberg picture wave function right in the Schrodinger picture it is the wave functions which depend on time the operators do not in the Heisenberg picture it is the other way round right. So, that is what we have got here we will discuss now the limit alpha going to 0 because there are these two limits which are of interest to us t going to minus infinity we discuss that the other limit which is of interest to us is alpha going to 0. And when we do that we get a very important result in quantum field theory which is known as the Gelman and Rowe theorem. So, we will discuss this in our next class there is any question today I will be happy to take. If not goodbye for now and then we will meet in the next class.