 I come back to this question of different ways of doing H naught. Let us say that, one of the overriding criteria is that, let us make E 0 0, E Hartree-Cock, one of the overriding criteria. One I said is to add with H naught that E naught 1 and then subtract in v, you will get it, but then the MP 1 is still 0, MP 1 will be still 0, but if you do look very carefully, I can look at not H naught plus a number, that number, but some operator, can I do like this? This is now an operator. Then of course, my v prime will now be v minus a and of course, you will clearly see that if you take an operator, then everything will change, but may I suggest that if I put this operator a such that a acting on psi Hartree-Cock is 0, now if I do that there are lots of ways to do that. Let us say I do a acting on psi Hartree-Cock is 0, if you do that then you find nothing changes, because when I do E 0 0, this is what remains, a is anyway 0, because a acting on psi Hartree-Cock is 0, this does not mean that a is a null operator, please remember, I am only saying that action of a on psi Hartree-Cock is 0, so it is not a null operator. So it acts on other determinants, it may give some value. If I do this, nothing will change, neither E 0 0 nor E 0 1. On the other hand, if I say a Hartree-Cock, a psi Hartree-Cock is E 0 1 psi Hartree-Cock, then something will change and I will get back the same thing as H0 plus v prime, except that I am saying H0 prime is not H0 plus E 0 1, but an operator whose eigenvalue with respect to psi Hartree-Cock is E 0 1. However a is not this number operator, because with respect to other determinants, it may give different value, but only when it acts on psi Hartree-Cock, it gives me E 0 1 psi Hartree-Cock. If I do this, I will simulate the previous prime thing and results will still not change, because everything with respect to psi Hartree-Cock, so basically psi Hartree-Cock H0 prime psi Hartree-Cock will become E Hartree-Cock, no doubt, but then again psi Hartree-Cock V prime psi Hartree-Cock will become 0, because this will become V and then minus a will give me E 0 1, so again it will become 0. So V prime will give me E 0 1 plus E Hartree-Cock, but I subtract E 0 1, again it will become same value. So that is bad that you already know. Then what did I do? I first ask the question, what is a good single determinant? If your sum over H of i is a good H0, then I would argue that if I just take the determinant with H i Eigen function, that would be also good determinant, but I found through the variation method that is not a good determinant. That is the result that we saw and because psi Hartree-Cock is a very good determinant, I am now searching in a retro manner what is a good H0. So I want to make psi Hartree-Cock Eigen function of that H0, nothing will be lost there, what has it got to be belongs there? It is valid for Hartree-Cock, but this particular thing of choosing not choosing H of i has nothing to do with Brilance theorem, because my psi Hartree-Cock is a good determinant. In case of non-interactive problem means if it is not there, if this is not there, then of course there is no problem, what is the Hartree-Cock? Then there is nothing to do. Yes, because psi Hartree-Cock is a good determinant. I think I have to talk to you, I am getting confused to your question. See, I initially said what is the best single determinant? We found it is psi Hartree-Cock. No, that we found that Brilance theorem is also a consequence of that. So you can say it is related to Brilance theorem. So your answer is not wholly wrong, because Brilance theorem is also I told is Hartree-Cock condition, but to say that it is because of Brilance theorem is not really a proper statement. It is a variation problem and we found that if I take the Eigen functions of H of i, those orbitals, they will not be the best, because otherwise I would not have done Hartree-Cock. Since that is not the best, I am also decided not to take this as H0 for the same reason, because my H0 should be a very good H0, dominant part of H. You understand, the Eigen function of sum over H of i is not a good wave function. That determinant is not a good determinant. So that is the reason I did Hartree-Cock and that is something I discovered rather after doing Hartree-Cock, that that is not the best. So the best is Fock operator Eigen values, not H. So then my H0 becomes some of the Fock operators. And then my V is just refined, that is all, 1 by Rij minus that Vr-Tree-Cock. That is all. There is no problem, but you have to remember that. So I want to tell you first that people have tried lots of tricks like this, but eventually it does not help. The sum and substance, I am not going to go through all the tricks. Eventually it does not help. It is within this Meuler-Plausser thing, it is impossible to get away from the fact that you do not get correlation energy at the first order. I think that is a very important fact that you really have to accept it, then only we will go to the second order, that the second order is where you will first start to get correlation energy. You can play around several things and in fact if you look at these articles by Diner, Claviery, Muldoo, they have done several such things, but you know it is not really very easy to do the recovery of correlation energy at the first order. We can, but our perturbation theory that we are devising is only for ground state. It works for ground state. Excited state perturbation is more complicated simply because you may have a multi-reference problem. We will discuss that later, meaning degeneracy. So there may be many eigenfunctions with the same eigenvalue. So you have to take a linear combination of all of them. In H0 itself, so that is the reason excited state is a different thing and then all the perturbation that we have done in 425 is also for ground state. We use excited states of H0, but to represent the ground state of H. So that is why it is called ground state perturbation theory. Excited state is not, I mean you cannot take one excited state of H0 and start to dig part out. That will be a wrong way to do it. They are excited determinants of H0. Of course, they are nothing to do with excited states of H0. That is correct. That is not a problem, but now what do you want to do? I am interested in ground. Yes. So that is first order. So first of all, I have to have is E0 1. I have to first, sorry, E1 0. H0 over what? Oh, that will be very bad. That will be very bad because who told you that this is a good approximation to excited state? We only discovered that for ground state psi hat refog is a good variation method is only for ground state. The upper bound theorem that we use only gives me ground state. Okay. On the other hand, many chemists use this as an approximation to excited state. I am aware of it. They will say singly excited. So psi r is a good approximation. It works to a large extent. If you do this, you will get some number. I am not saying the numbers will be terribly off, but suddenly as far as chemistry is concerned, I do not know if psi r is a good excited state approximation. There is no variation of optimist. I have only optimist for ground state. So I will get some number. I have no problem. Of course, if you do this, you will get some number, but I mean this number may be very poor, poor number if I add this and this. So what you are essentially telling adds psi r, h, expectation value with psi r. You will get some number. That is very easy to do. Some number you will get, but this is not really excited. Further, excited states of degeneracy, actual excited states of degeneracy problem that there are two states which can be very close in energy, and then it is much more difficult. You have to take linear combinations and so on. So that is other problem. This itself is a bad approximation, but other than that also even if you do this, there could be problem because of degeneracy in actual excited states. So it would have been very nice. Of course, what you are saying is that ground state, first excited states, each of them determinant would be an approximation to corresponding excited states, but unfortunately it does not work out. Only in the linear variation and I have done that in CI, you get an upper bound. So linear variation of CI is very important. In fact, we have discussed this for the McDonald theorem, but for the perturbation it is dangerous to do. So all that we are discussing in this course is only ground state perturbation theory. So having done this, we now kind of agree that whatever we are doing so far is very nice, but we have not done anything as far as correlation energy is concerned. We have still not recovered anything with correlation energy. So we have to do second order perturbation. And much as you would not like to have second order perturbation, it is little bit more clumsy, you have to do it if you want to do correlation. So that is all that I will do, the second order. I will not do third order because order by order as you go perturbation series becomes complex, but those who can will be able to see how to do. It is a matter of detail, but at least the second order perturbation we must do. We have done this as a general part of general perturbation theory again in the previous course. Those who have taken the course 4 to 5, I have done already second order, but I will redo it so that we can do for the heart reform perturbation. And I already did last time the first order perturbation theory. So we can start from that. So remember how I did it. We took H naught plus lambda v, then we wrote psi 0 0 plus lambda psi 0 1 and so on, lambda square. Note again that it is not important whether you write as half or you do not write as half. So it does not matter. So let us not worry about that is equal to E lambda which is E 0 0 plus lambda E 0 1. So lambda square is 0 2 and so on and then again psi 0 0. If I write half, they will only get scaled. The actual values will get scaled, but it does not matter. If you just add multiplied by number, nothing changes. So what we first did, we of course saw that this is 0th order is H naught psi 0 0 E 0 0 psi 0 0 which is of course something that we did. We also recovered the first order. First order again I write it H naught psi 0 1 plus v psi 0 0 equal to E 0 0 psi 0 1 plus E 0 1 psi 0 0 which I rewrote little bit nicely as H naught minus E 0 0 psi 0 1 plus v minus E 0 1 psi 0 0 equal to 0 and I somehow love this way of writing. I told you because it gives me easy way to partition H minus E psi equal to 0. So I am really partitioning H minus E psi equal to 0 first order. So this can be 0th order which is H naught minus. So 0th order of H minus E into first order of this, first order of H minus E into 0th order of this. So if I continue to do this, I can write the second order either by looking at lambda square term or by simply writing H minus E psi. So what will be the second order equation? H naught minus E 0 0 psi 0 2. Now everything has to be second order. So it will have 0 into 2 plus 1 into 1 v minus E 0 1 into psi 0 1. Now this will have second order here H minus E. What is second order of H minus E? H has no second order. H has only first order. Only E 0 0 has second order. So it will become minus E 0 2 psi 0 0. Right? Remember minus, do not forget minus because that is important because it is H minus E. So minus E 0 2, so that becomes your second order equation. And you can actually write third order, fourth order, everything. We will stop here. You remember by projecting this equation with psi 0 0 that is multiplying by psi 0 0 star and integrating, we got this term as 0. So this actually got cancelled. We had psi 0 0 v psi 0 0 equal to E naught 1. So that was the, I just want to remind you how did I get and we also notice that for E 0 1, I do not need psi 0 1. I only need psi 0 0 and v which is well defined. So let me see if I do the same thing, what do I get for E naught 2? Before I do this, let me also say that the perturbation equations are conveniently derived by a normalization which is called the intermediate normalization. I will spend a little time on this. Note that what is normalization? You normally say that if I integrate the wave function, the mod square of the wave function or complex conjugate multiplier by itself and integrate, it should be 1. That is the normal normalization. In the intermediate normalization, we are assuming something else that if I do the integration in the following manner, the 0th order with the full d tau, that should be equal to 1. So I take the exact wave function multiply by the conjugate of the dominant part of the wave function. Now again remember by my definition, this is the dominant part of the wave function which is psi heart repop because this is an eigenfunction of H naught and then if I integrate, that should give me 1. Psi is of course psi 0 0 plus psi 0 1 plus psi 0 2 etc. Remember lambda is gone because only at lambda equal to 1, my H of lambda becomes H. So just the lambda, lambda squares are gone. So I have written in this manner. Again as I said the factor half and 1 by 3 factorial I have already absorbed. So it is not a Taylor series, it is really a power series that I am writing. So my psi becomes psi 0 0 plus psi 0 1 plus psi 0. So how do I make this equal to 1? Remember my psi 0 0 star, psi 0 0 is already 1. So what I am actually in effect I am assuming that all the correction terms, all the correction d tau equal to 0, all corrections. So I will say 1, 2, 3 etc. So I will generally write correction k, kth correction for k equal to 1, 2, 3 etc. Now again you can argue that this is a sufficiency condition simply because you could have argued that this plus this sum could have been 0 but I think for a convenience I am suggesting that all corrections of the wave function, if I integrate with the 0th order wave function they must be 0. If that is so then automatically this normalization is followed. So this is the normalization that we will see. Later on we will see that it is very easy to do this because remember I had all the determinants of H0. All I need to do is to write these corrections in terms of all determinants except psi-hatry problem because all of them are orthogonal. So if my correction terms is a linear combination of first excited, second w excited, triply excited only and not hatry fog then automatically this will be followed because they are the determinants form an orthogonal space, orthonormal space sorry. So it is not very difficult to follow at all. So in fact that is what I am going to do but right now let me derive the equations using this and then we will stop today. So let me just derive the equation for E02. So note, I am writing in Dirac notation. In Dirac notation this means any psi 0 0 with psi 0 k equal to psi 0 k psi 0 0 equal to 0 for k equal to 1 2 3. Please remember only the corrections. So this is my restatement of intermediate normalization which automatically means that psi 0 0 psi 0 is equal to 1 which is the exact ground state. So this is the exact ground state of H because you have the psi 0 0 here so that gives you 1 and the rest becomes 0. This also means that if I take psi 0 0 because you may ask the question what happens to the full normalization. So of course the first term is psi 0 0 psi 0 1 which is 1 psi 0 0 psi 0 0. After that you have all cross terms psi 0 0 psi 0 1 psi 0 1 they will all become 0. But there are terms like this psi 0 1 psi 0 1 right plus psi 0 2 psi 0 2 the norm of the corrections right and so on. And also cross terms like psi 0 1 psi 0 2 etcetera etcetera lots of cross terms which are which are not orthogonal because they are only orthogonal to psi 0 0. All such terms will be there and in fact in particular I may mention that this will be actually nothing but I mean it cannot be 1 it is usually greater than 1 it cannot be 1 that is the important question. So when I use intermediate normalization I do not assume that psi 0 is fully normalized. So both I cannot do together. So intermediate normalization and normal normalization are incompatible can I use the word incompatible means both of them will not go together. So I am actually sacrificing that full normalization because the equations become simpler. So please remember however after I get psi 0 I can always re-normalize by writing the norm and remember also that if I am expanding these in terms of determinants there are lots of orthonormality that remains. So they will all become lots of interesting things will happen. It is very easy to write but the point is that I can re-normalize but when I re-normalize this one will vanish. One will become some coefficient so basically your psi will be some coefficient times psi 0 0 that is the trick catch that not only that I have make this equal to 0 this 0 but also when I have expanded this I have 1 here that is important otherwise this would not have happened. If this would have been some number then again this would not have happened. So when I re-normalize that is what is going to happen a number will come multiplying everything I can re-normalize so that is not a big problem but for convenience I am going to use intermediate normalization. So let me go back to the second order equation now which is written there and let me do the same projection with first psi 0 0 exactly like I did. So let me write down the projection psi 0 0 H naught minus E 0 0 psi 0 2 plus so I am writing now in Dirac notation all the integrals psi 0 0 V psi 0 1 minus E 0 1 psi 0 0 psi 0 1 I hope all of you have write it easily minus E 0 2 equal to 0 please note that this term this part I have split so I have written psi 0 0 V psi 0 1 minus E naught 1 is a number which comes out psi 0 0 psi 0 1 quite clearly you know the reason why I have written now because I can identify that this term is 0 because of intermediate normalization. So I am working explicitly with intermediate normalization so this is 0 and this is 0 anyway just as in the first case because this is 0 psi 0 0 Eigen function of H naught with E naught 0 as Eigen value so this is gone. So I have only two terms and then I can write E 0 2 very easily as psi 0 0 V psi 0 1 just as I wrote the E 0 1 as psi 0 0 V psi 0 0 it actually looks exactly similar except there is a second order so somewhere one more order has to increase so this is first order so it is no longer average value of V with respect to anything but a matrix element. What is interesting is that you can actually write an nth order equation and find out it by the same way by projection in psi 0 0 a general E 0 n which is psi 0 0 V psi 0 n minus 1 a more general equation I am not going to do this but I think it is trivial to see if you just do this go through this exercise write the nth order you will see everything else will fall off because of intermediate normalization everything else will fall off very easily and this will actually have just this term and last term will remain E naught n psi 0 0 psi 0 0 so that is the only thing that will remain. So of course you have to work with this but now you realize that to work with this I need psi 0 1 I have not bothered about it yet what is the correction to the wave function at the first order so that is something where I will start next time that what is the correction to the wave function at the first order if I know this then only I can get back to it or to and to do this I have to go back to my first order equation so I will let me write again the first order equation remind you so I will go back to the first order equation because I need psi 0 1 if I write Dirac notation consistently then everything should be k minus V E 0 1 psi 0 0 equal to 0 remember writing these equations are very very easy just scale down 1 you know 2 to 1 1 to 1 to 0 and this term is not there it is very easy to write so we will have to go back to that equation and then find how do I get psi 0 1 and we will do that tomorrow but let me first tell you that I already have a clue because I know my psi 0 1 is orthogonal to Hartree-Pock and I have all the eigen states of H naught so I will what I am going to do is to make a linear combination with all determinants psi a r psi a b r all singly doubly everything it should be full space except psi Hartree-Pock because I am going to force that condition so psi Hartree-Pock would not be there so I already know how to do it all that we have to find out is what are the combination coefficients when I expand what so that is what I am going to find out once I find out I am going to stick this here and I will get E naught 2 okay so that will be your MP2 correlation we will derive this everything right now remember I am writing in a more generic way and after that I will bring in Hartree-Pock I will expand in terms of spin orbital again right now there is no spin orbital here all this can be done with slatter rules so that is why slatter rule is very very important we will need as you go here to MP2 we will need one more slatter rule which I said I will defer the type C where you have one Hartree-Pock here another which is two occupancy difference I had done Hartree-Pock, Hartree-Pock, Hartree-Pock singly excited now doubly so two determinant differing in two spin orbitals and I will also say that is the end of the slatter rule because for all our atomic and molecular problem when you have three determinant everything is 0 the Hamiltonian matrix element is 0 because V has only two particle operators it cannot it cannot connect three three three differences so it becomes 0 so that type C that I will discuss is the only one so that is one more technical thing that I have to present and then I will go to the actual expression of MP2 correlation okay and then we will discuss a lot more things about perturbation theory how to actually apply and at some point of time I am also I also want to do what is called the second quantization it is a technique which is very routinely used in correlation theory so far we do not need it but the second quantization simplifies life many things you can write very nicely I will choose my time when I should do second quantization I am not decided but at least I will be finishing in this week the most of the perturbation theory.