 So, I just thought I will write all the formulas that we have derived so far and of course, you first derive this and then you derive this. So, we have to now we will now proceed with the Hartree-Fock perturbation theorem which means essentially H naught is Hartree-Fock Hamiltonian which is sum of the Fock operator. So, psi 0 0 becomes psi Hartree-Fock and all the psi k 0s are basically the determinants, the other determinants of H naught, all other determinants of H naught as among the MCN determinants that we discussed. So, we will have to start with that as the motive as that perturbation but before I do that I have to discuss the third sleight of rules. So, the last sleight of rule that we will discuss is the matrix element of the Hamiltonian between two determinants. So, let me call it psi 0 or let us say phi 0 does not matter which differ in two occupances, I was calling them psi I am just calling them phi it does not matter. So, this is the determinant and electron determinant this is also an electron determinant but now the difference between these two determinants are that two spin orbitals from these determinants which are A and B are replaced by virtual orbitals R and S is it clear just like A and R we had done. So, if you write phi 0 if I write phi 0 as chi 1, chi 2 in spin orbitals these are all in spin orbitals note that everything that we have first doing in spin orbitals we later spin integrate wherever possible and convenient. So, this phi A, B, R, S is then so let us say I have a general determinant I have identified chi A and chi B. So, I have also identified the ones around this and once around it does not matter I mean just for the sake of understanding. So, then we just write chi 1, chi 2 etcetera and wherever this chi A is there you replace this by R rest are all same where chi B is there you replace this by chi S rest are all same. So, I get the determinant A, B, R, S note that it is important to say what is being replaced by what in this case. So, I am replacing A by R B by S that is how the nomenclature is on the other hand if I replace A by S B by R the only thing that will happen is that this will be the negative of these determinants because this is essentially interchanging the two columns is it clear. So, I can write an equation that phi A, B, S, R is minus of phi A, B, R, S. So, if you replace A by S and B by R that is simply negative of A by R B by S because in the final determinant R and S are changing sign changing place okay. So, that is the reason that it is negative. So, similarly I can say this is minus phi of B A, R, S is it also clear. So, it is the same thing B to R A to S okay and of course this is same as plus phi A, B A, S, R. So, if I interchange both just like this two electron integrals anti-symmetry it has similar thing if I interchange both I come back to where I was okay. So, basically what is being replaced by what is now important in terms of sign the determinant is of course same you know it is physically it is the same just the negative sign does not mean it is I have a different wave function okay I have the same function but it is important to know the phase. So, what we are talking by A, B, R, S is that A replaced by R, B replaced by S. Oh sorry sorry I inverted A, B, R, S no no I had written A, B, S, R I think oh no one we don't know yeah yeah see what I wanted to write A, B, S, R is A, B, R, S, B, A I should write S, R here B, A, S, R and B, A, R, S. So, I started with A, B, S, R I did not start with A, B, R, S I think that was the it does not matter you understand what it means. So, if I interchange 2 pair here, pair here the chain side if I interchange both it becomes plus. So, of course right now I am writing A, B, R, S okay. So, the result is the following again I remind you the H has two parts now we are not talking of H naught plus V you know do not get confused we are talking of normal Hamiltonian which has two parts sum over H of I plus 1 by R I we are not talking of H naught plus V. So, this is not the Fock operator this is the normal H of I as we write. So, this of course you can write it in terms of Fock operator as well, but as long as you understand how to handle the sum of one particle operator then sum of two particle operators it is done then your sum of one particle and two particle may change. So, the first part does not contribute to this remember in the first type B the first part gave only one integral A, H, R right in this case it does not contribute very simply because H is a one particle operator. So, whenever you will put this here at least one of the spin orbitals between these and these will be different I had two differences. So, with the H of I you will be able to connect one difference for the second difference will directly integrate and hence it will become always 0. So, for example if I have an integral 1, 2 H of 1 plus H of 2 or let us say I call it A, B H of 1 plus H of 2 C, D or RS in the same nomenclature RS. So, A1 I think the nomenclature is clear these are coordinates A1 B2 H of 1 plus H of 2 R1 S2. So, if you take any one of them so let us say A1 B2 integration H of 1 one of them if you take R1 S2 then you can see it is 0 because H1 can connect A, H, R so it will have become A, H, R into B of BS and that is 0 because of the orthogonality here. So, just for showing two electron so it cannot connect but this is not 0 because H is able to connect. So, this integral may survive A of R AR is 0 but A HR is not 0 but H can only connect one difference but since there are two differences one of them will remain and that will make it 0. So, the first part has no contribution the second part would be of course able to connect and the result of the second part is the following. So, this type C result is basically A sorry there is no summation AB remember AB is what is contained in phi 0 that is replaced by RS AB anti-symmetrized RS that is all there is no summation nothing. So, remember how the progression is taking place in the case of type B there is the first one there is only a single term and then there is a summation over only one index if you remember AB anti-symmetrized RB there is a summation over one index in the type A there is a summation over one index for this summation over two indices for this in type B it became no summation only one term one index summation type C this becomes 0 no summation only one term. So, how the number is falling as you go and clearly by this logic if I have three occupancy difference nothing else will survive. So, anything else which is three occupancy difference it is 0 because by the same logic now 1 by R I j cannot connect three occupancies. So, one of the integrals one of the pair of orbitals will integrate to 0. So, this is where the slatter rule will actually end. So, if you understand up to this point this is very simple it is only one term the differences but remember it is an anti-symmetrized integral. So, do not forget it is AB 1 by R 1 to 1 minus P over 2 and RS. So, if you now understand this we can come back to this formula is it clear. So, I think you should be able to write all the slatter rules expectation value one difference two differences for both one and two electron operators. So, we will come back here. So, let us again go back and discuss this. So, our H naught is some of the Fock operator. So, I rewrite the expression some of the Fock operator which I am writing as H of I plus the rest of the Fock operator which I am calling V Hartree Fock I which is basically sum over the other occupied orbitals chi B star 2 1 by R 1 2 1 minus P 1 2 chi B 2 d tau the operator part. So, I call it V Hartree Fock I. So, the V is 1 by R I j minus V Hartree Fock I it is important to remember this because many times we have to feel that V is just 1 by R I j, but since the one electron operator already the H naught already includes this is very important to remember this. And I now write down the same formula E naught 0 now is some of the orbital energies. So, I use canonical equations. So, this is of course E naught 1 is psi Hartree Fock V psi Hartree Fock which is actually two terms psi Hartree Fock 1 by R I j psi Hartree Fock minus psi Hartree Fock V Hartree Fock psi Hartree Fock. So, we sum over V Hartree Fock. So, you will be able to now reduce these by slater rules this term as well as this term by type A. So, I have two types of slater rules there the type A there are two slater rules I can reduce it and write a long form which I did when one of the classes, but important thing is to realize is that if I add these two I get totally get Hartree Fock energy. Because simply this was psi naught psi Hartree Fock H naught psi Hartree Fock. So, you can see that if I write this up it becomes the Hartree Fock energy I have shown this also in the long form that is expand this by slater rule add you get the Hartree Fock energy. But when you expand remember this term was very important because this cancel the half that comes from this cancel actually brought the cancellation brought the plus half in the Hartree Fock energy when you did this. So, this part was very important to do. So, now what we will do we will first write psi 0 1 then we will write E naught 2. So, for psi 0 1 remember and for E naught 2 we need to sum over all the states of H naught except for the ground state. So, we have already identified those states of H naught as psi A R. All psi A R all psi A B R S and so on. Remember we did this exercise if you sum all of them plus Hartree Fock which is one determinant you get totally M C N we showed you that combination problem. So, this is M into N minus N into M minus N this is N C 2 into M minus N C 2 and so on. Number because we have to make sure that we take only pairs of A B. So, your number here will be adjusted by let us say A less than B or less than S. Note here you should not write equal to because if A is equal to B it is 0. I hope you can realize this because if A is equal to B that means what am I doing I am replacing the same. So, it will actually become identical either A equal to B or R equal to S. So, I have to write this as A less than B or less than S in terms of the number of determinants that we get. So, if I have just two orbitals or three orbitals for example this should be 1 2 1 3 2 3 right. So, only three of them would be replaced similarly by R S pairs all R S pairs. So, R also cannot be equal to S. So, then you have your psi 0 1. So, let us write the first order correction to the ground state using this. So, I have to now this K will be actually replaced by these determinants S A less than B because this is identical A greater than B and A less than B are negative of each other I just now showed. I just now showed if you interchange A B or R S they are they are negative of each other. So, they are basically same determinant they are not really any other functions. So, what are the unique W X R A determinants the unique sets that is all we require. So, psi 0 1 will now be replaced by we will not replace write K instead we will write these determinants. So, first we will be singly excited remember Hartree-Fock is anyway not there K not equal to 0. So, you do not have to bother. So, I first write all singly excited determinants. So, I just go back to the formulae of psi A R. So, it is a linear combination of all singly excited determinants first term and this is the coefficient. So, coefficient is psi A R V psi Hartree-Fock I am just writing the coefficient later for obvious reasons and this is E epsilon A minus epsilon R I will just explain to you. I already told you I prefer to write it in this manner because this I defined as a resolvent with this summation sorry with this summation. So, that this quantity can be written simply as R 0 V psi 0 0 just as you can write this as psi 0 0 V R 0 V psi 0 0. So, I define the resolvent. So, it is easier to write because you have this ket bra notation if I write this here this is lost because the ket bra that that is following psi k 0 psi k 0 is lost you cannot see this. So, you cannot have a resolvent. So, it is better to write it in this manner. So, exactly the same way this can be a part of the resolvent. The difference of these two energies is epsilon A minus epsilon R we have already discussed this. So, it differs from the E naught 0 by this the difference in the orbital energies. So, that becomes your first term and then the second term you continue because all excited states are there. So, you have to continue. So, the second term is A B R S again A less than B R less than R S psi A B R S sorry psi A B R S V psi hat to the power divided by epsilon A plus epsilon B minus epsilon R minus epsilon S right and we can derive this each of this. Now, we know how to get this by slater rule remember this will be conjugate of what I wrote. So, what will be the result of this integral it will be R S A B then you have a one electron part minus V R that is 0 just as H of I did not contribute. So, because it is a one electron part. So, there is a two electron thing which cannot contribute. So, this integral is actually R S A B remember I did this slater rules in terms of H, but in all my perturbation the rules are in terms of capital V. So, you have to take little precaution when you write capital V and H are different of course, but capital V also has a two electron and a one electron operator in a different way one electron operator has come not H of I that V Hartree Fock has come and that has to be subtracted not added. However, that one electron part will not contribute because you already said that it is a sum of one electron operator it cannot connect these two. So, these are let me let me analyze these two determinants these two integrals. So, one of them is psi A R V psi Hartree Fock of course, if you know this you know its conjugate how to do this. So, what is the result of this first let us start from here what is the result of that? Brutus theorem if it was Hamiltonian it would be 0 remember now what is the result of this? So, everybody thinks it is non-zero yes it is 0 I will show you why I can write this as psi A R there is a simple way to do it H minus H norm psi Hartree Fock correct which is psi A R H psi Hartree Fock minus psi A R H0 psi Hartree Fock note that this part is 0 by Brutus theorem note that this is also 0 because each of these determinants is an eigen function of H0. So, when H0 acts on psi Hartree Fock it has some value some of the orbital energies not E Hartree Fock some of the orbital energies times psi Hartree Fock remember what is my H0 H0 or some of Fock operators. So, H0 acting on psi Hartree Fock is E0 0 which was some of the orbital energies time psi Hartree Fock correct many people by mistake will write E Hartree Fock. So, I am again repeating it is E0 0. So, whatever it is it does not matter the psi A R's are orthogonal to psi Hartree Fock right. So, this also becomes 0. So, it is 0 for a for a for a for so I hope it is clear if you want to expand this V as 1 by Rij minus V Hartree Fock of course you have to get 0. So, please do that by long form slatter rule this is a quick proof but do this apply this slatter rule to the first part and the second part none of them will be vanishing in this case. So, each of this slatter rules will actually produce some numbers but when I subtract it should become 0 1 by Rij minus V Hartree Fock. So, please do that exercise but I just wanted to quickly show that it is again 0. So, the first thing to note that this set of terms do not contribute to the first order corrections coefficients are 0. So, the singly excited determinants do not contribute to the first order perturbed wave function for the ground state. So, that is a very important realization. So, the first set of determinants which will contribute are from W X R A determinants. So, we need to know what is the value we have already said that psi A B R S V psi Hartree Fock again this is very easy to prove you have a 2 electron part minus 1 electron part that is anyway 0. So, you have this as R S anti-symmetrized A B. So, that is the integral. So, this part will of course contribute with R S A B and so the whole result will then become A less than B in terms of orbitals. Now, I am writing in terms of orbitals please remember here I was writing in terms of this capital K was not orbital capital K by the states of the system. But now I am not writing in terms of states of the system states of the systems are defined by the orbitals. So, A less than B these are orbitals which are stuck here the rest of the orbitals I am not writing the original Hartree Fock orbitals are here the differences are only marked. So, A less than B psi A B R S and the coefficient of each of the W X R A determinants is R S anti-symmetrized A B divided by epsilon A plus epsilon B minus epsilon R minus epsilon X. Remember these coefficients can be positive or negative I had only remark that when you go back to E naught 2 E naught 2 will be always negative. But psi naught 1 is just a function. So, there is nothing called positive negative I mean these numbers can be either positive negative. So, it is just a linear combination of psi A B R S. But when I go back here I am now going to put this V and close it with psi 0 0 that psi naught 1 and that will become that will give me negative energy ground state energy will become negative as you will see we will do this here. So, I hope this is clear that we have this as the psi naught 1. So, if you want to calculate psi naught 1 the contribution of psi naught 1 comes from the W excited and because of this later rules there is nothing else. I hope that is also clear you cannot have triply excited because that everything is 0 2 electron 1 electron everything is 0. So, the only contribution and this is a very important realization for the there is a large amount of physics which is involved and you should read that are the article by Sinonoglu on pair correlation theory which I will refer to you is that the first order correction to the wave function comes only from W excited. So, the W excited determinants play a very important role in the correlation energy. Obviously, because psi naught 1 comes from W excited determinants E naught 2 will also have component from only W excited determinants. So, let me write down that component of E naught 2 now. So, let us write E naught 2 here so we can close this. So, I have already got psi naught 1 and we know that E naught 2 is nothing but psi 0 0 V psi naught 1 or we can use this long form it does not matter. So, psi 0 0 V psi A B RS is A B anti-symmetrized RS by the same way and again the one electron part V Hartree-Fock is 0 remember this is capital V, but the result is same because V Hartree-Fock is a one particle operator result is same. So, then you have E naught 2 as A less than V sorry A less than V or less than S, A B anti-symmetrized RS, RS anti-symmetrized A B divided by epsilon A plus epsilon B minus epsilon R minus epsilon. I hope all of you can see this because all I am doing is to write another psi 0 0 V with this and that gives me A B anti-symmetrized RS and the rest remains the same. You can see this from here also that this is A B anti-symmetrized RS, this is the RS anti-symmetrized A B divided by the orbital energies and that gives you the final E naught 2 which is basically what we call the MP2, MP2 energy. So, you have two anti-symmetrized integrals in terms of the spin orbital obviously this is conjugate of this. So, again we just note that this is always negative because the denominator is always negative, it is occupied minus virtual okay, pairs of occupied minus pairs of virtual. Remember this A less than V are less than S. We will have to of course discuss the importance of W-excited configuration in the correlation energy particularly for E naught 2. And in fact what is interesting to note which I am not doing that E naught 3 I just want to write it down. If I do E naught 3, E naught 3 also has W-excited configuration only. I call it configuration. This determinants, please remember that this determinants I am giving a name configuration, many people write configuration. In fact if you write the value write E naught 3 you will realize that that also has contribution only from W-excited configuration, only from E naught 4 singly excited and triply excited will start coming in a very complicated manner okay. So, that because series will become very different you know we are not writing the series. So, obviously even singly excited the reason singly excited starts to contribute because the singly excited directly with Hartree form through V is 0 by Brillouin's theorem or whatever because of mainly because of Brillouin's theorem. But singly excited will start to contribute to W-excited they will then contribute to the Hartree form because I told you the process of is that you go from Hartree form to all excited determinants. So, this is let us say singles, this is doubles you cannot go directly from here to here right because of Brillouin's theorem psi Hartree form V is 0. But you can go from here to here and then you can see third order what is the problem you can come back here because there has to be three processes and then come here you are again done because this is becomes 0. So, third order also you cannot do but when you do fourth order lot more things can happen you there are so many doubles. So, remember when I go to one doubles I can go to another doubles because W-excited is also not one configuration I can go from one W-excited to another W-excited by through V and then from that W-excited I can come back to another W-excited another one and then eventually come back here or that is one way or I can come from this W-excited back back to from here itself back to singles. So, that is one two then I go back to a W-excited and then come back here now this is allowed because I am not directly mixing you understand I go from one doubly ground to W-excited back to singly excited back to W-excited back to heart reform. So, that is E naught 4. So, I have to have four different V's so psi 0 0 V psi W-excited psi W-excited to psi singly excited psi singly excited to psi W-excited back to heart reform. So, the numerators will of course pile up but it is easy to see that the singly excited will start to contribute from the fourth order and so would be triply excited. So, there are several terms of course but I think I am not going to do E naught 3 E naught 4 but I just thought I will give a pictorial representation of what really will happen because this formula will become more and more complicated. So, all kinds of excited state will come not one. So, you will have a k k going to L L going to something back to 0 eventually psi 0 0 has to come back to psi 0 0 but through many scattering processes. So, there will be different excited states. So, all possibilities you have to consider but the point is that the W-excited determinants that is something that we now understand because second order energy is a very important energy at the lowest order this contributes a large amount. So, W-excited configurations are determinants when I said W-excited please remember it is with respect to psi heart reform now everything we are talking of a respect to psi heart reform it is a heart reform perturbation theory. So, W-excited configuration play a very important role in correlation energy that is a general thing and we will come back to this why. In fact, I had given you this Lovedin's article which was very nice but here this is a theme that is eventually there is a very important theme called pair correlation I will come to that the physics of this little later. So, this is actually a theme of pair correlation that the correlation mainly occurs because of two electrons getting excited together and this is what you see the psi A B R S is basically like two electrons getting excited. So, the entire theme is called pair correlation which is much more than just W-excited determinants but there are correlation coming due to two pairs going up three pairs going up and so on simultaneously. So, I will come back to this theory or rather this analysis and there is a very nice paper again just as I had given for Lovedin it is a Okta Sinanoglu he is a very brilliant turkeys physicist Okta Sinanoglu he is a brilliant I think he is one of the most brilliant physicists or to have worked in this area he almost worked out everything but he lost interest after that. So, he left the field in fact today what is couple cluster doubles or couple cluster is actually the genesis of couple cluster was by Okta Sinanoglu unfortunately he did not give the second quantization language work did not work out diagrams but he told everything that was there in couple cluster the physics and after that he lost he went to graph theory and started doing something else. So, there are people like that who do not push things till the end then of course the mercenaries of couple cluster came like Rod Bartlett and other people who actually kept pushing. So, but Sinanoglu is not of that type but I actually like his type of science because he is very very original. So, he wrote a nice review article just as I had given for parallel of Lovedin advances in chemical physics please remember this journal is very good volume 6 page 315 1964 just as Lovedin's had given volume 2 207 59 his tells me 58 I think it is still 59 the bibliography is 59. So, it might have been published in December 58 I do not know but anyway you see this Okta Sinanoglu is 64 I think the page numbers are pretty much correct I am pretty sure just see this article it is a very good article and those who have time and energy can read this article much of it is really written. So, not too many equations. So, that is very nice the physics will come flow why correlation energy is dominated by pairs pair correlation energy. So, that is important and you can see the first effect of this that the correlation energy comes in pairs the W excited determinant play a very important role. So, I hope it is clear just remember how to handle the V do not do not jump into a conclusion that V is H or V is 1 by R I j either way. So, although singly excited gets 0 it is it is a direct consequence of Brillouin's theorem, but it is basically because V can be written as H minus H naught and they are also the states of H naught. So, Brillouin's theorem combined with this it actually gives you 0 this part you can try to explain this by actually writing V as 1 by R I j minus V Hartree Fox. So, please do this exercise you know I am asking the exam actually to show by later rule I will not do it in the class I am just telling you. So, you must practice practice such things many of such algebra I will not necessarily do this algebra can also be written in many ways remember this is in terms of anti-symmetrize integral. So, if I ask you write this formula in terms of regular integrals what is the regular integral just A, B, R, S it is a regular integral. So, of course if you write this then each of the terms has to be expanded in the following manner. So, you have A less than B or less than S then each of these will be A, B, R, S so I am now writing only 1 bar. So, when I am writing 1 bar it is regular integral A, B, R, S minus A, B, S, R into the same A, B, R, S minus A, B, S, R, S, A, B, S that is okay they are identical it does not matter the integrals are identical, but it is better to write because we all use real orbitals. So, R, S, B, A or whatever divided by epsilon A try to expand this now is a numerator you will have 4 terms this into this, this into this, this into this and 2 sets of terms will be identical I hope you can see this for example A, B, R, S and R, S, A, B A, B, S, R and R, S, B, A will be identical why because R, S, B, A you can write as S, R, A, B by interchanging both the pairs correct. So, this A, B, S, R and this is actually S, R, S, R, A, B so they are identical similarly A, B, R, S and R, S, A, B will be identical. So, there is one sets of term which is called a regular integral another sets which is the exchange A, B, R, S with R, S, B, A, A, B, S, R with R, S, A so that is like an exchange so one pair has been exchanged when you multiply and see what you get okay because when you do summation you know because of dummy variable thing you can you can do lots of nice things but before that you must realize that I have A less than B or less than S so you cannot do a dummy variable summation because your A, B is not summed completely neither is R, S summed completely so you will have a problem. So, what you should first do before you can actually analyze this term in terms of regular integral is to write this with all A, B, all R, S. So, how will you write this? So, the first term itself E naught 2 in anti-symmetrize integral I want to write this as all A, B all R, S because then I can do lots of dummy variable interchange A, B less than R, S, A, B anti-symmetrize R, S, R, S anti-symmetrize A, B epsilon A plus epsilon B minus epsilon R minus epsilon S. It is not correct right so what do you do to make it correct? Minus why minus half, half just half, half into half 1 by 4. So, first write this expression as 1 by 4 then try to do this because lot of simplifications will come if I sum over all A, B, all R, S. Please do this again as in at home exercise. So, basically the question is write this write E naught 2 in terms of regular integrals over spinor. So, that is the I have to give you one assignment also. So, I think this let me give this as a formal assignment one. Second is spin integrate E naught 2 for closed shell systems. So, again I you know what I mean. So, these are all spin orbitals. So, you have already done lots of spin integration. So, spin integration will be easy after you do the regular integration, after you do the regular integral calculation then only spin integration will be easy. So, spin integrate E naught 2 for closed shell systems and derive expression in terms of special in terms of special orbitals. So, integrals over special orbitals. So, first write this it will become easy. Don't try to spin integrate directly from here I am just kind of telling you it is ok you can do it. So, I have not completed this because this is not completed I have written just in the long form. So, simplify this by recognizing that certain term there are two sets of terms one is like like regular a coulomb kind of thing one will have involve an exchange in the two way and then. So, first write this, but before that you use this expression. So, start from this expression it is it will be easier to write. So, starting from this expression do this one and then eventually spin integrate write everything in terms of special orbitals for closed shell systems. So, your summation will then become a and b will become n by 2 r s n will become m minus n by 2 summation index will also change. So, we will so we will complete today's class with this, but I will come back later as a theme I am going to discuss the importance of pair correlation. First the physics of pair correlation before I go to the C i the next topic because C i also you will see exactly the same thing that doubly doubly excited configurations are playing an important role. So, in general there is a theme that for correlation energy the electron pairs play a very important role and not one electron excitations one electron excitations could have played role, but because of the brilliance theorem you do not have the role and that is a direct consequence of heart reform the fact that we have varied the orbitals in the first order it essentially means I have eliminated some importance of the one electron excited determinants. So, in a way it is a consequence of the Hartree-Fock method theory. So, I am starting from Hartree if I start from any other orbital remember then perturbation then of course singly excited will play a role and we do not want to do that. So, from the Hartree-Fock when you start singly excited that is something that I want to reemphasize that this this is going because of out because of brilliance theorem otherwise of course it will it would have been there and there are other orbitals through which people actually try to sometimes start I do not think it is a good idea. So, it has never really taken shape in terms of programming, but there are lots of you know theoretical papers obviously Hartree-Fock perturbation is the best that you can do.