 So, we did the Gellman and Lowe theorem in our previous class and we obtained a certain expression with reference to the adiabatic switching technique which we introduced in the previous few classes we have discussed this. We will also examine how it corresponds to the Rayleigh Schrodinger perturbation formalism. So, this is the Gellman and Lowe theorem just to remind you a few things that we did in our previous class. So, that we quickly recapitulate on that and then take it from there. Our question is that if we introduce a artificial mathematical parameter alpha with which we control the correlation or part of the Hamiltonian which you are not able to handle using normal methods of quantum theory for which you need some special techniques. So, alpha is this control parameter and our interest of course is in getting the eigenstates of the full Hamiltonian. The eigenstate of the full Hamiltonian actually will come down to the eigenstates of the unperturbed Hamiltonian as t goes to minus infinity because e to the minus alpha t will kill the h 1 term as t goes to minus infinity and then as t goes to 0 you have the full Hamiltonian which is at 0 plus h 1 that is the problem in which we are interested and we are trying to ask this question that if we know the eigenstates of the unperturbed Hamiltonian how do we get the eigenstates of the full Hamiltonian inclusive of the complexity which is there in the full Hamiltonian. So, we found that in the interaction picture the wave function at t equal to 0 can be obtained from the eigenstate of the unperturbed Hamiltonian phi 0 through this time evolution operator but then we carry a subscript alpha on this because the time evolution will depend on how you are turning on the perturbation. So, we are a little specific about this and our final results of course would be independent of alpha and the Gellman and Loth theorem tells us how such an eigenstate can be obtained it tells us that if this limit exists the limit of the ratio the limit as alpha tends to 0 and it does not worry about whether the limit of the numerator and the limit of the denominator exists independently. They may have certain divergences which may cancel each other but then if this ratio has got a well defined limit then the Gellman Loth tells us this theorem tells us that it will be an eigenstate of the full Hamiltonian. So, we discussed some aspects of this theorem and using that we showed that it tells us what the energy difference will be between the energy of the full Hamiltonian the eigenstate the eigenvalue of the full Hamiltonian and the eigenvalue of the unperturbed Hamiltonian. So, there is a certain energy difference and using the Gellman Loth theorem we showed that this energy difference this is the correction which we get and we also showed using algebraic methods that this particular ratio in the limit alpha going to 0 turns out to be equal to limit alpha tending to 0 and then it is a limit of what? It is a limit of the time derivative of the logarithm of this matrix element the time derivative taken at t equal to 0 then of course there is this you know I h cross scaling which we have already discussed. So, these are some of the things that we did in our previous classes and the question that we we had raised is what would be the correspondence of the delta E correction that we get from the adiabatic hypothesis with the corresponding correction from the Rayleigh Schrodinger perturbation theory which is our usual idea of getting corrections when we are not able to solve a quantum mechanical problem fully. So, we solve a part of it which is the unperturbed Hamiltonian and then we make a correction which is a perturbation. So, this will be examined in the context of what is the nature of this time evolution operator what exactly is the form of this time evolution operator and that is something that we have already discussed in our previous classes. So, let me quickly recapitulate some of those expressions here. So, this is the time evolution operator and we are inquiring what it is. So, we have earlier expressed it as an infinite series from n equal to 0 to infinity the term corresponding to n equal to 0 is just the unit operator and then you have the rest of the summation which is from n equal to 1 through infinity of un and each un is given by a number of terms in which there are the interaction picture Hamiltonian appears n times with n different time arguments, but then notice that there is this chronological operator over here which ensures that all the latest operators in time appear farthest to the left. So, that is the time ordering that is involved in the expression for un. Now, h i t 1 whichever you consider this please always remember that it will also include this e alpha t parameter because that is a mathematical you know factor that you have inserted. And the time argument t which we wrote generally as t will be different in each case because this is h i t 1. So, there will be an e alpha t 1 corresponding to this there is an h i t 2 over here. So, there will be an e alpha t 2 corresponding to this. So, all of these terms will have to be carried over very carefully. So, now let us first look at this matrix element which is a matrix element of the time evolution operator in this unperturbed states of the Hamiltonian. So, this is what it is and this time evolution operator is 1 plus this sum from 1 through infinity of u n and from the first term you just get the scalar 1 and then you get a n which is our abbreviation for the matrix element of u n in the unperturbed eigenstate of the eigenstate of the unperturbed Hamiltonian. Now, subsequently how to take the logarithm of this and this being a small quantity you can expand this logarithm and we discuss this as well in our previous class that the logarithm of this turns out to be given by this expression and now you understand why I used an abbreviation a n for this matrix element because the number of terms already fill up the space for us to write the expression and if you keep writing the entire matrix element it is just too much to write. So, this is some sort of a compact notation that we are developing and we have to keep track of what is what and we have a fairly complex expression over here because the energy correction now will be the time derivative taken at t equal to 0 of the logarithm expansion and you have got a number of terms over here each a n itself is quite a complicated factor it has got n different time integrals over different time domains and if you expand these terms summation over a n from n going from 1 through infinity you have a 1 a 2 and so on then you have got the squares of this then you have got the cubes of this and when you take the squares and the cubes and so on notice that you get a 1 square you also get a 1 a 2 you will also get a 2 a 1 when you take the cubes you will not only get a 1 cube a 2 cube a 3 cube you will also get a 1 a 2 a 3 then you will get a 2 a 1 a 3 in different orders and these are all operators so their positions are very important and in general the energy correction can be obtained to different orders depending on how many times the interaction appears in the operator whose matrix element is under consideration. So, there are nth order terms that you can get and this is really quite complicated you see that these terms come from a variety of combinations of a 1 a 2 and a 3 and in multiple different orders when you go to even higher order terms it becomes that much more complicated. So, it is quite a complex kind of mathematical system of equations that we have to work with and to get familiar with the techniques we will first work only with the first order terms just to see how that because that is the simplest that you can work with and the first order correction to energy would then be given by limit alpha going to 0 i h cross del by del t and only the first order term we take del by del t of this a 1 and this time derivative of course has to be taken at t equal to 0. So, that is the first order term that we shall consider. So, now let us see how we can work this out now this is the general expression for a n for an arbitrary value of n. So, for a particular value of n which is n equal to 1 we have this expression and now there is only one time integral over here h i appears only once with a dummy label which gets integrated out the dummy label is t 1 the limits of integration are from minus infinity to t. Now, in general you can carry out the transformation from a Schrodinger picture operator to an interaction or picture operator using this prescription which is the general expression. So, you can get this h i this is the interaction picture Hamiltonian for the interaction by carrying out a transformation on the Schrodinger picture operator h 1 by carrying out this transformation e to the i h 0 over h cross t h 1 and then you have e to the minus i h 0 h cross t. So, you are just applying this rule over here, but you can plug in the explicit form of h 1 which we already know in the second quantized form which we have done in great details in one of our earlier units the unit on in the unit on second quantization we wrote the n electron Hamiltonian in the second quantized creation and destruction operators. So, we will use this form. So, this is what h 1 is. However, it is h 1 is a little bit more than that because this is the raw h 1, but then we have introduced this adiabatic switch mathematical factor e to the alpha t. So, in addition to that you will have this e to the alpha t factor. So, that is something that you have inserted as a mathematical device. So, this is your h 1 along with this e to the alpha t and we have to carry out transformations of this operator to the interaction picture. So, we will have to carry out transformation of all these creation and destruction operators somewhere over here it comes as c i dagger c j here it comes as c i dagger c j dagger if you look at this this will be c k c l. So, the creation and destruction operators will come in a variety of different combinations because you are also summing over all of these i j k and l indices over here and i and j indices over here. So, they will come in a variety of combinations and we will and it does not matter in what order they come. So, we will take any one of them just to see how the creation and destruction operators transform to the interaction picture. So, we will consider just one such combination which is c i dagger c j there is nothing particular about particularly important about it it is just some set of two operators and we will see how they transform to the interaction picture. So, sure enough they will be sort of sandwich between this e to the i h 0 t and e to the minus i h 0 t and this is how they will transform. So, let us examine this term over here. So, this is the term we are going to examine now what do we find here in between this these two operators creation and destruction operators you can always plug in a unit operator. So, this is the unit operator which has been plugged in and now using the associative law you find that these three operators give you the transformation of c i dagger to the interaction picture and these three operators give you the transformation of c j to the interaction picture. So, what do we get we find that this combination of operators transforms to the interaction picture and what you get is the same kind of an operator. So, here you had c i dagger and c j here also you have got c i dagger and c j, but there is this subscript i which tells us that this is now in the interaction picture. So, the general form is retained, but these are different operators they are the operators in the interaction picture and just for the sake of brevity because of a notation is already complicated now that we know that we have carried out this transformation to the interaction picture. We have the time argument explicitly indicated in the parenthesis we know that these are interaction picture operators. So, I will suppress the subscript i and always I will remember that c i dagger t is an interaction picture operator because those are the ones which are time dependent. So, I will suppress the notation the subscript i will be notation in subsequent analysis. So, this is the operator that we have to transform to the interaction picture along with this e to the alpha t. So, you have the transformation to c i dagger c j t and now all of these operators c i dagger goes to c i dagger t c j dagger t c j dagger over here goes to c j dagger t in the interaction picture same thing with c k and c l. So, c k goes to c k t and this c l goes to c l t and we have this e to the alpha t coming in over here. So, now this is our interaction picture Hamiltonian corresponding to the 2 electron terms. So, these are the terms which we had difficulty with and this is the one which corresponds to the difficult part which you cannot solve using ordinary quantum mechanics with ordinary quantum mechanics you can only solve part of the problem which is at 0 h 1 is the difficult part. So, this is the interaction picture 2 electron term now let us ask what is the explicit form of this operator c k in the interaction picture this is already in the interaction picture this is the index subscript that we are suppressing and we are asking what is the explicit form of this annihilation operator because it will be a solution to the equation of motion for the annihilation operator. So, the equation of motion for the annihilation or the destruction operator is this which is i h cross del over del t of c i t, but c i t is nothing, but an operator which is obtained from c this is c k actually right. So, c k is c k t is the operator that you get in the interaction picture from the corresponding Schrodinger picture operator by these transformation operators according to the general prescription of getting an interaction picture operator from the Schrodinger picture operator. So, this is the equation of motion kind of thing this is the time derivative of c and this tells us its solution will give us the explicit form of c. So, let us work this out. So, this is the time derivative of a product of these three operators of which omega s is not time dependent, but this one is and so is this. So, if you take the partial derivative with respect to time and do it term by term, but when you do so make sure that the order of these operators is retained because you cannot assume that any pair of this automatically commutes. So, you have to keep track of the order. So, from the first term you get i as 0 over h cross when you take the time derivative then you have got these three operators and then omega s is not dependent on time, but then you have to take the time derivative of this which is minus i h 0 over h cross and then you have got this operator here. So, if you just rearrange these terms you can get rid of the h cross which cancels very nicely and you can rearrange these terms you have got an i here and i here. So, i square will give you minus over here. So, you have this term omega s 0 with a plus sign and you have got minus s 0 omega here you have these two operators in the reverse order, but with a minus sign this s 0 of course, commutes with e to the i as 0 over h cross. So, if you just rearrange these terms you find that you have got the commutator of the Schrodinger picture operator commutator with the unperturbed Hamiltonian sandwiched between these transformation operators. So, that is the result that you get. So, this is nothing but omega i t which is the interaction picture operator corresponding to the Schrodinger picture operator omega s. So, that is what you get because h 0 of course, commutes with e to the i as 0 and also with e to the minus i as 0 these are expansions in powers of h 0. So, to get omega i you have to solve this equation which is the time derivative of omega i is then given by the commutator of omega i with h 0 this is valid for any operator omega and therefore, it will be valid also for the destruction operator in which we were interested. So, the time derivative of the destruction operator will be given by the commutator of the destruction operator with h 0. Now, this is what we can easily determine because we know h 0 in terms of the creation and destruction operators. So, this is now h 0 which is written in terms of the creation and destruction operators and you can of course factor out this sigma j h cross omega j you can take it outside this bracket and then you have to find the commutator of C k with C j dagger C j and that can be determined very simply by using the fundamental commutation relations for the fermion creation and destruction operators. So, we know what they are these are the fermion operators for and they satisfy these commutation rules rather these are the anti commutation rules and using these are anti commutation rules you can work out all of these with this particular commutator of C k with C j dagger C j and you find that it really gives you a very simple term because these two terms cancel and you get only one operator C j and that too with a nice chronicle delta. So, that when you carry out the summations you will be able to contract the summation and you will be left with only one term. So, this is a very straightforward kind of analysis and you can work out all the intermediate steps one by one the pdf of all of these slides is available at the course web page. So, you can access that if you wish and now we know that you can plug in this delta j k C j over here and then sum over j. So, with this chronicle delta delta j k only the term in j equal to k survives and this is the solution that you get now. So, the time derivative of C is proportional to C. So, that has got a very simple solution. So, whenever you have any physical quantity whose time derivative is proportional to the actual amount of that quantity the solution involves an exponential function. So, that is a solution that you expect and that is precisely what you get. So, the destruction operator at time t is then related to the destruction operator in the Schrodinger picture and it is given by this exponential law as one would expect in any similar situation. You can take that joint of this equation and the result gives you how the creation operator behaves in the interaction picture. So, now you have got the creation operator in the interaction picture the explicit form of that you also have the explicit form for the destruction operator and you can use them in your interaction picture Hamiltonian corresponding to the two electron terms here you have the C i dagger t the C j dagger t and all of these will have these right hand sides will give their explicit forms. Likewise C k and C l t will be given by expressions of this kind. So, just carry over the indices carefully and now you get instead of C i dagger t you get C i dagger, but then you have this e to the plus i omega i t which is here. Likewise for the C j dagger t you have got the C j dagger here and then you have got the e to the plus i omega j t and from these two operators you have got these exponential functions, but then the exponent comes with a minus sign here because these are from the destruction operators. So, there are number of time dependent terms this is one this is the second this is the third and this is the fourth and these four come respectively from the four creation and destruction operators and then there is an additional time dependent term which is e to the alpha t which is coming from our adiabatic hypothesis that is the adiabatic switch that we have inserted as a mathematical device. So, there are these five terms which involve time and here you just have the two center integrals which you have sufficient experience with not only from the previous units, but also from earlier courses like in the Hartree fog you deal with these two center integrals all the time. So, these are the two center integrals and now everything over here is known. So, let us work with this and our interest is of course, in determining the first order correction which is given by the time derivative of a 1 at t equal to 0. So, here you have the first order correction and you have this integral remember that time the integration variable which is the dummy variable is t 1 the limits are minus infinity to the upper limit which is t which is not a dummy not here. So, h i t for this particular value of t is given by these operators the e to the alpha t has come from the adiabatic switch. So, this is the explicit form of a 1 which is this and here I have plugged in this h i t 1 from this equation here and that appears over here along with these four time functions exponential functions and along with this adiabatic term coming from e to the alpha t. So, everything has been carried forward and the integration variable is t 1 over here which is the integration variable, but the upper limit of the integration over here is t. So, here of course, you have got the space integrals and time integrals. So, the integration over time can be carried out separately. So, I factor out this integration over all the functions which involve time and these are over here this is this e to the power i omega i plus omega j minus omega l minus omega k t 1 which is here and then e to the alpha t. So, that function is here and this sum and difference of these four omegas is what I write as delta 1 again just for the sake of brevity. So, that I do not have to write all those four symbols every time I work with this term. So, these four terms which include a sum and difference of these omegas in this particular with appropriate indices. So, this is what I write as delta 1 and this time integral now is just integral from minus infinity to t of this e to the i delta 1 plus alpha t 1. Now, this is a very simple integration. So, what is the result? You can carry out this integration and you find that after integrating it is a definite integral from minus infinity to t. So, put the limits and this is the result that you get. What is also interesting is that you will subsequently be required to take the time derivative of this term with respect to time at t equal to 0. So, that will also come out in a very neat form. So, this is your result now all the integration over the time has now been carried out and this is the result that you get for a 1. So, let us bring it to the top of the next slide and here this is our general expression for the energy correction. The corresponding correction in the first order is this and you are not interested in just a 1, but in it is time derivative at a particular time which is at t equal to 0. So, this is a 1 as we got on the previous slide and if you take the time derivative of this with partial derivative with respect to time. So, this is the only term whose time derivative has to be taken and what will it give you? It will give you i delta 1 plus alpha times this term. So, this i delta 1 plus alpha and this i delta 1 plus alpha in the denominator these two terms will cancel each other you are left with only this term and what actually happens is that the partial derivative then turns out to be very simple over here you just get the factor h i t over here. So, this is what you have over here plus together with this is nothing but the interaction picture term for the two electron terms. So, this together with all those four exponential factors which are sitting over here in this delta 1 together with this alpha gives you essentially the interaction picture h i t. So, that is the term that you get and we will work with this term. So, h i t is this and now you have to take the time derivative at t equal to 0. So, what happens at t equal to 0? At time t equal to 0 you take the time derivative of this term at t equal to 0 and essentially you find that alpha really does not matter. When you take the time derivative with derivative with respect to time then alpha really does not matter and in the first order correction the mathematical artificial device that you had inserted alpha really does not matter but that is not the case when you work with higher order terms and in higher order terms you do have to keep track of these terms very carefully. So, let us begin to look at higher order terms but before we do that let me just quickly remind you that when you expressed A 1 as the matrix element of this time evolution operator with subscript 1. So, there is only one integration involved only one time parameter involved. The result is the matrix element of the interaction picture Hamiltonian this is what you get and this is a result that we will make use of when we deal with second and higher order terms. So, I just want you to remember this result at the back of your mind. So, your result then becomes independent of t 1 which gets integrated out and the result does carry the interaction term Hamiltonian with the argument t. So, that is the one that will show up in subsequent terms. So, let us now consider higher order terms and these are the higher order terms they have got a fairly complex structure we have considered them earlier. So, out of these very many higher order terms now like what we did earlier that to begin with we considered only the first order term let us now work only with the second order term. What will the second order term have? So, the second order term will have an a 2 from here and it will have an a 1 square from here. So, the second order correction will be given by limit alpha going to 0 which is here i h cross the partial derivative at t going to 0 at t equal to 0 partial derivative of what a 2 from here and half a 1 square from here. So, that is the second order correction that you will now have to determine and there are two terms of for which you have to take the partial derivative. So, I will separate these now. So, I have got the partial derivative of a 2 here and the partial derivative of half a 1 square over here. So, I have got this i h cross over 2 and then the partial derivative of a 1 square coming from the second term. So, I will now work with this the second order correction and to do that I now have to work with two terms one is this and the other is this. So, these are the two terms that we will now work with and we will take them up one at a time. So, when you have a complicated problem what do you do? You break it into pieces and then solve it bit by bit very patiently till you have solved all the pieces and then put everything together the way you do in a jigsaw puzzle. So, that is what we are doing over here. So, we now are going to work with only the first term which is in this box which is the limit alpha going to 0 of i h cross times the time derivative at t equal to 0 of the operator a 2 of the matrix element a 2 which is the matrix element of u 2. So, now we have these time evolution operators in equivalent forms one is an infinite series of this kind one is an infinite series which is a summation over infinite terms each term being given as a time ordered term of a finite number of terms here, but then n goes all the way from 0 to infinity. So, these are the two equivalent forms we have demonstrated the equivalence of this discussed it in one of our earlier classes. So, the second order term is what you get over here and instead of the variables t prime and t double prime I am going to use t 1 and t 2 the limits are from minus infinity to t prime over here. So, this is minus infinity to t 1 and this limit which was minus infinity to t prime is minus infinity to t. So, this is the second order term that we will now examine and of course, we have to keep track of the e to the alpha t, but e alpha t will come with e alpha t 1 and e alpha t 2. So, let us work with these two terms. So, the first term is what we are now working with. So, this is what we have got these are the integration limits. So, I have removed these circles here just so that you can see this expression clearly because I want to put two other circles over here one is this del over del t over here is what I want to highlight to draw your attention to and then this integral from minus infinity to t which is this of d t 1 h i t 1. So, this is the one I am now discussing. So, what is in this red box is what I am examining what I am going to examine now and that is the reason I have rewritten this equation in the middle again over here it is the same equation, but I am highlighting different aspects of that equation as you can see. And the reason to do it is because we have already determined this in our previous term for a 1 because in our earlier analysis which is there in just a few slides back we already have this result that when you take the time derivative of this matrix element of the first order term it gives you a result that you have to end up getting the matrix element of the interaction picture operator with the argument t which is the upper limit of integration here. The dummy variable t 1 gets integrated out. So, this result we have already seen and we can plug it in over here. So, we do not have to redo the whole thing. So, that gives us that a simple result that the partial derivative of a 2 which is this left hand side is now given by the matrix element in phi 0 and you have from this derivative and everything that is there in this red box you get the h i t which comes over here and then you have got this second integral which is from minus infinity to t d t 2 of h i t 2. So, now this is the term that we now have to examine. So, it has come no it is just the integration over the time. So, yeah I understand what you are saying you have got the space integral and you have got time integrals. So, the time integrals can be evaluated separately and that is the result we are using same thing in the previous one if you go back no let me go take you back to the slide 95 and remind you of how we got the result here this comes from the consideration of the time derivative of the time integral that is all there is to it the space integral is determined completely separately. So, all the terms corresponding to the space integration all the operators they hold on to their respective places and the time integration over t 1 is what gives you h i t. So, it is this result that we have used over here in this result over here. So, h i t and then the remaining time integration is still here, but I am keeping this h i t 2 to the right of this operator here. So, the time t comes as the upper limit of this integration and it comes as an argument over here. So, this is what we now have to determine and of course, we are interested in the derivative at a particular value of t which is t equal to 0 that is something that you should remind yourself of which means that this upper limit which is t is now 0 over here because this integration is from minus infinity to t. So, this value of t has to be 0. So, this argument of h i has to be 0 and this upper limit t has to be 0. So, this integral over t 2 is carried from minus infinity to 0 and now I make a simple substitution because now I do not have t 1 and t 2 2 time intervals to worry about. So, I drop the subscript 2 on t and I use t instead of t 2. So, I have got the same expression, but now there is only 1 t over here. So, this is what we have to determine. Now, what is this h i t at t equal to 0? This is the general expression for h i t at t equal to 0, this factor will give you unity, this factor will give you unity, e to the alpha t will give you unity and h i t equal to 0 is nothing but the 2 electron interaction term. So, you have that. So, this h i at t equal to 0 is nothing but h 1. Now, over here in the second time integral this is from minus infinity to 0 of h i t again. Now, here you have got a similar expression here, but this is now a dummy label which is integrated out this is not the t equal to 0. So, always remember which is a dummy label, a label is a dummy label in a particular context. So, now, t is the dummy label and it gets integrated out, it is the integration variable from minus infinity to plus infinity. So, this is what the interaction term Hamiltonian is and this is the integration from minus infinity to 0 of d t and you can carry out a very simple analysis over here because now you know that this phi 0 is actually an eigenstate of h 0 belonging to the eigenvalue e 0. So, you can replace this operator by the scalar e 0 and on this side you have to be careful because you have h 1 here and not h 0. So, here the e to the minus e 0 is now a scalar it is not going to operate on anything. So, it just factors out as a constant and the remaining operators whose matrix elements must be determined are these. Now, you have got a space integral this is actually a space integral and what is the space integral this is h 1 is operating on phi 0 star this is a Hermitian operator. So, it can operate to the left. So, you have got h 1 operating on phi 0 star and h 1 when it operates on phi 0 phi 0 is not an eigenstate of h 1. So, the result can nevertheless be expressed as a linear superposition of all the unperturbed states and all the unperturbed states are the phi m's with phi m going from 0 to infinity. So, this is the complete basis and these are normalized functions normalized base functions and the coefficient C m are nothing but the projection of this on a particular m th base element. So, these are the coefficients which are the matrix elements of the operator h 1 which is the interaction operator between two unperturbed states. So, this is the expansion of h 1 phi 0 and you can plug in this expansion over here. So, let us do that. So, this is an infinite expansion. So, this infinite expansion from m equal to 0 to infinity comes here you have got h 1 phi 0 over here and h 1 phi 0 on the left. So, one of them gives you a summation over n with the complex conjugate over here the other gives you a summation over m which is without the complex conjugate. So, there are two summation indices one is m and the other is n and now your space integral over here takes a rather simple form because both of these are eigenstates of h 0. So, both of these being eigenstates of h 0 you get the e to the i e m t because phi m is an eigenstate of h 0 belonging to the eigenvalue e m. So, you get e to the i m over h cross t this you can always factor out and then you have a space integral of phi n star with phi m which is nothing but the orthonormality integral. So, if n is equal to m you will get 1 and if n is not equal to m you will get 0. So, you have got the orthonormality over there that gives you the chronicle delta delta n m and then you can sum over one of these either m or n and then contract that. So, you are now left with this after you sum over this m with delta n m you are left with only one sum which is sum over n going from 0 through infinity and you have these two matrix elements and you have got a scalar here. Now, what are these two terms these are just complex conjugates of each other. So, you take the modulus squared. So, now, the whole expression is being simplified to a substantial extent not enough, but much better than what we started out with. Now, you have to take the limit of the derivative at t equal to 0. So, this matrix element which is here can be substituted by what you have on the right hand side let us do that. So, whatever was here on the right hand side comes here in this beautiful bracket you are following all the terms it is very simple substitution it takes a lot of time to write this because every time you write it you make a number of careless mistakes you will write t 2 in place of t 1 and then you are out for a toss, but do it carefully you will get it right. So, over here in the class I just want you to concentrate on the logic on how things are done and then you can work it out yourself. If you need any reference the PDF files of these slides are of course available at the course web page. So, here you now take this expression. So, this is the result from the right hand side which is borrowed here in this beautiful bracket. So, let us write it here at the top and what do you get over here. So, you separate this case integral from the time integral. So, you have got the time integral from minus infinity to 0, but what is time dependent over here this piece is not time dependent it is this piece which is time dependent and this piece which is time dependent and this piece which is time dependent. So, you carry out the integration over the time variable separately over here. And again you have got a very simple time integral just as you had in the earlier cases. It is just the time integral of an exponential function, everybody knows how to do it. So, now evaluate that time integral, put the limits minus infinity to 0, this is the result that you get, that is high school integration. So, let us use this result. So, this is the result of the time integration and we now get the partial derivative with respect to time at of the element a2 at t equal to 0, which is given by this term. At t equal to 0, this is what it turns out to be. So, you have got this, you have to take care of the powers of h cross. So, you have got 1 over h cross square over here, you have got h cross over here. So, you end up with 1 over h cross, you have a minus 1 sign, there is a minus sign over here. So, that takes care of it. So, you have to keep track of all these details, do it carefully, you will get it right. Reims write these results in natural units in which he puts h cross equal to 1 and then you do not see h cross and then it is very easy to make a mistake of the powers of h cross, which is why I have done this analysis inclusive of the factor of h cross. So, that you carried carefully, because when you do calculations of course, you have to plug it in. So, this is the result pertaining to this box over here. So, these are the two terms that we wanted to determine. We have succeeded in getting the first box, it has taken as a full class to do this. So, we postponed the determination of the second term for the next class. Too many terms, too many complexities, but everything finally comes down to somewhat simple terms. And the beauty of this whole analysis is that instead of writing a large number of terms, time derivatives, integrals and so on, we will then be able to express the importance of these terms using some very nice cartoons. So, those are the Feynman diagrams that you are looking for. That is what this unit is about, we have already had 4 classes in this unit and you have not seen those very nice pictures as yet, but they are coming. But it needs a lot of background to do that. So, we have already had 4 hours of background. Let us see when we get to them, eventually we will, that is what this unit is about.