 The particular way of doing interacting problem is what is known as configuration interaction. Again, many of you might have heard this acronym called CI, so this is actually called the configuration interaction. In fact, this is done much after Hartree-Fock normally, see first read Hartree-Fock which is only one electron, one determinant and then you go to this, but this is potentially much simpler to explain because it is very simple mathematics. So what is configuration? Configuration is determinant, interaction is a combination, so that is why it is called a linear combination of determinants or linear combination of configurations and this is called the configuration interaction and very popular quantum chemistry method called CI. Now depending on how I truncate, of course nobody can do all, this CI has different flavors CI SD, CI SD triples, all that we will do later. The point is that mathematically this is an exact form provided my basis is complete. So the question that you are asking the basis is not complete of course, you are right. So this was an example, but if I take that three to be a complete set potentially this is an exact or four to be the complete set potentially it was an exact. So if I have a complete basis that is one and within this complete basis all determinants, remember my basis may be complete, but I may choose that I am not going to take all determinants. I am going to only remove some determinants. So that choice is also there. So there are two levels of approximation, one is the basis, one is the number of determinants. I need not take MCN. Even if M is complete I need not take MCN for some reasons. So I can make approximation in the basis, I can make approximation in the number of determinants. But if you have a complete basis and within that if you take all determinants then this method is called full CI and I think it is obvious. Now full means you have taken all possible determinants in a complete basis. In a complete basis and then the full CI of course in a complete basis is exact. I repeat it that full CI in a complete basis is exact. This is an exact way of function because now there is no more error basis. So that is the problem. So complete basis can never be done because it is infinite. So this is really a statement which has no meaning not relevant because I really cannot do it an infinite dimensional problem. However what is normally practiced is a finite basis. This is a normal practice, finite basis and then I can take all determinants within the finite basic and then this is still called full CI in this basis in the finite basis. The name full CI just means that all determinants have been taken. So as long as you are taking all determinants in a given basis we will call it full CI because anyway in complete basis we will never get. So we will worry about full CI in the finite basis. So whenever we call full CI you have to specify the basis. So even full CI results are different. Remember if I do a problem of Strodinger equation solution full CI the results will be different because your basis is different. If you keep on increasing your basis your number of determinants are more in the full CI and then the results will be different. Of course if you do not take all determinants then again results will be different. So there are lots of approximations we can do. But let us understand that the exact is full CI in a complete basis. So I have complete basis I have all determinants do full CI that is an exact function yes well usually it is infinity. So that is what I am saying it is in practice never implemented that is the different matter. Limit is limit is when you do not find results changing you call it almost saturated. So that is all that is a different matter mathematically maybe the number is changing but it is changing so small that you do not care. So you say that it is a complete basis set limit but there is nothing called complete basis set limit is different from complete basis set yeah yeah yeah CBS limit is of course there that people have found out by calculations. So we will see that how to do the calculation we will see later and the energy does not change. So that will be on Hartree Fogg also full CI everything any CI calculation but right now we do not want to discuss CI I just thought I will tell you the form that a form of an exact wave function is possible that is what I wanted to tell you for an for an interacting Hamiltonian for a non interacting Hamiltonian of course any any one of this determinant is an exact function any one of the determinant is an exact function question is which is ground state which is excited state that is a different matter. So whichever has the lowest energy that can be also seen from the energy of those orbitals so one can find out the now what is important is that once you have understood that the interacting problem is a linear combination of determinant then we have to worry about how do I make good approximation so what is a good approximation is always to take a good basis even though it is finite it is a good basis it includes most of the physical effects of the complete basis and I cannot take all determinants also in many cases so I must take determinants which are physically important. So two levels of approximations that we will do in CI later for interacting problem is that one is the basis another is the number of determinant so the approximations that we will bring in CI is a the basis so what is a good basis so what do you mean by good basis good basis means with a less number of functions I should be able to generate the effects of the complete basis as good as possible so that is why you have to be very good so lot of people have worked on this basis this basis is not only a problem on CI all across quantum chemistry this is a very very important problem. Second is within this basis which determinants are important so if we know these we will only take those important determinants because eventually we cannot do exact calculation we will do approximate calculation so we want to take physically important determinants and not MCN because some of them may not really be important so these are the two very important parts which will define my CI even before that there is a very important question that we understand the value of a single determinant I told you chemistry comes in single determinant because if it is single determinant chemists are very happy you have N electron N spin orbital chemists don't understand all the CI they get confused right they will same confusion will happen with electron coordinate spin orbital what is happening you tell me which electron is in which spin orbital you have to give one answer such an answer is not possible but you have to satisfy an experimentalist right I mean even if you say electrons are indistinguishable okay tell me N spin orbitals N electrons for N electron problem give me N spin orbital so given everything when I am saying which determinant is important the approximation to one very important approximation that we want to do is that can I get one determinant which is very important one means single whenever I am saying determinant please remember it is later determinant which is most important among the single determinant I can have many many single determinant but can I get one of this determinant which is the most important because then at least I can go to the chemistry that determinant not MCN not even not even a partial set of determinant just one so that is the answer that we will get from by Hartree and Fock so this is the answer that we will get and that is will be called Hartree Fock approximation where we are actually going to 0 in only one determinant and not any set of determinant combination of determinants which is the most important so that is a question that I am asking simply because after doing all these mathematics we have to be as simple as possible so before we do better let us go back to the basics and ask the question just like in non-interacting Hamiltonian we had one determinant which was exact in this case of course we cannot get anything exact but can I get one determinant which is as close to exact as possible I hope you understand the question of course it is not exact so when I say most important it must be as close to the exact as possible so this answer is given by Hartree Fock method so I just kind of telling before we start the Hartree Fock there are lot of background work I have to do but I just want to tell physically what is the question that we are asking it is very important to ask the question for an interacting Hamiltonian this is never going to be exact but we are only trying to ask a question can I get a function can I get a determinant which is as close to one determinant which is as close to exact as possible because then I will have a very simple chemist interpretation n electron n spin orbitals so the problem is that for an interacting Hamiltonian which is sum over h of i note again when I write interacting Hamiltonian please remember very simple form I want to get one determinant which is as close to exact as possible of course you can immediately say forget about this make it a non-interacting problem and then I will get one determinant the question is is it as close to exact as possible that I don't know that is certainly one determinant I can easily get by completely neglecting it but that may not be the best determinant because we have no a priori reason to think that is the best determinant so how do I get that determinant will be the content of the Hartree-Fock method so before I start Hartree-Fock I just thought I will tell you I will again come back tomorrow and talk about it I think once the problem statement is clear what are you doing many people after doing Hartree-Fock not sure what are they doing why did they do it that is very important I think I first want all of you to understand what are we doing technical details if you don't means that's okay I don't care but you must know what are you doing what is the question that you're asking so the question that I'm asking is to give the chemist one single determinant even for an interacting problem and and would it be that determinant by simply neglecting this then I have an answer the answer is actually no so that is the reason Hartree and Fock did that's an answer that I can tell before I do it okay so I think we will start with that Hartree-Fock method now but before that there are certain background work that I have to do some little bit of mathematics which is essentially calculating matrix elements of the Hamiltonian I will have some rules later rules I will lay the lay them down because they will all be required and of course the most important thing will be answered by what is called the variation principle so I will again repeat the variation principle I've done it in 4 to 5 I will repeat variation principle some certain mathematical rules for getting the matrix elements before I actually can tackle this question some background work will be done from tomorrow okay so what I hope you have understood the question I think if you have any any problem with the question please ask you know that is most important in fact CI is much more easy it is just mathematics Hartree-Fock is more difficult because it has a physics how do I get the most important physics built in and if you don't ask that question then you will miss the essence of the Hartree-Fock so Hartree-Fock will be eventually very good yet not good enough because it is still a single determinant so so then the question that we will ask how good it is why is it not good enough that we have to do something better in fact the Hartree-Fock was developed in 1933 Hartree and Fock did long back why didn't quantum chemistry stop there you know that's the question so obviously it is not good enough so all those questions we will ask later not today we'll first see how can I do a best single determinant picture and then we'll see how can I improve this on the quantum mechanics the first paper came in this is a Heitler and London did valence bond theory in 27 1927 then Hartree and Fock did in the 30s this is molecular orbital theory okay I think things will become more interesting as you go to the real part so please try to follow and make sure that you ask questions okay I mean he asked a question good question everything was good question I mean everybody was confused the way you said because everybody will think just in change one and two index it is not that it is a coordinate change so it is a good question I would say so please feel free to ask questions that's very important no no it is a good question I mean I don't say anything which is a bad question okay okay thank you