 So, this must be the minimum energy, what does it contain, how to interpret this, let us interpret this. An electron 2 is sitting on chi j, I am integrating for a electron sitting in chi i, one electron sitting in chi i, the electron 2 is sitting in chi j and giving an average interaction, one particle interaction, this is the density remember chi j star chi j, it is just giving an average one particle interaction on an electron sitting with chi i t, this is the classical electrostatic interpretation for the Coulomb term, that I have an electron 1 which is sitting on chi i t, there are another electron sitting in chi j t, it is moving all across, I mean it is a classical interpretation, so electrons are not moving it is a density, so the entire density chi j star 2 chi j 2 is integrated between d tau 2, so integration essentially means it moves all across and what the electron 1 sees due to that, so this is a classical electrostatic force that the electron 1 sees due to the movement of 2, of course you can say what about 3, what about 4, I do not need to do, that has been achieved by the sum, again I repeat, so everything can be 2, so I just rename this electrons, they are in different spin orbital is it clear, so it is basically a classical electrostatic interaction that the electron 1 sees due to electron 2 and an one particle part has been extracted by average, I need a one particle part that has been extracted by d tau 2, essentially that means the 1 and 2 are dynamic but 1 C is a total average effect when 2 makes a complete movement, assume they are classical particle then 2 could have made a one full circle over electron 1, then on that circle after that what is the potential that the electron 1 sees due to electron 2 is what is captured here in this square bracket, so that is the, I am now going into the interpretative part of the heart reform which I will write maybe in the next class but let me first state, I will write that more clearly the interpretation but now I have another part exchange part which I am not able to interpret because now that is this is not a density, so electron 1 is was sitting in chi i 1, now this electron 1 is sitting in chi j 1, so this was chi i 1, this was chi j 2 now they have flipped, it has gone to chi j 1 and the d tau 2 interpretation is also very funny, it is on 2 on chi i and this is chi j star, so I do not know what interpretation I will give but that is the nature of the exchange, that is the nature of the anti symmetry that one that no electrons are not classical, so that is something that I have learnt from the beginning when I wrote the energy also, I could not give an exchange classical interpretation to exchange interaction, I am not able to give a classical interpretation to this operator whatever this operator is and then of course I have a further problem that this is not in the Eigen value equation structure yet, because this is not a number times chi i 1, if it is a diagonal it would have been number times chi i 1, I hope all is clear right, if this is a diagonal this j is equal to i, so it would have become number times chi i 1, but let us not worry about it right now, we will see how to do that, so that would be actually canonical, when this, so when lambda becomes diagonal we will actually get now I can tell you what is canonical, that will actually lead to canonical, how to achieve this I cannot make an approximation, if I approximate this then that is wrong, because then you are saying canonical is only an approximation that is not true, we will achieve this by exact means, so that there is a lot more to it, but right now we are stuck with the non-canonical, but our problem is that I cannot still even if it was canonical, I cannot make it into an Eigen value equation form because of this exchange part, so assuming that this is a diagonal I cannot still write it as an Eigen value equation form because of this guy, so this is already Eigen value, something times chi i t in the 1, something times chi i t in the 1, something times chi i t in the 1, I mean I would only assume the diagonal, but how about this part, and that was his question, and so the exchange part we always have a problem, however I will quickly show you that formally, mathematically I can derive an operator by making it, making this look like a chi i t in the 1, because that is what we want, the 1 should be on chi i t in the right on the right hand side, so let me analyze this term, so let me come back only to the exchange term of this, which is chi j star 2 1 by r 1 2, what is it chi i 2 d tau 2 chi j 1, no interpretation because chi j star 2 chi i 2 is no exchange, it is not a density, this is j raise to i sum over j of course, now what I will now do is strictly mathematical, and you may even consider that as a cheating, but it is mathematical, what I will do I will write this sum over j chi j star 2 1 by r 1 2, I will introduce between these an operator, which you are already aware of permutation operator P 1 2, permutation operator interchanges 1 and 2, so I will write this P 1 2, and then of course they have to be interchanged, so this guy will now become chi j 2, and then d tau 2, and this guy will go out as chi j 1, perfect I am no problem, you may only say what would you do permutation operator, that is the later part, I want to write a formally Eigen value equation, so what I did was simply nice mathematics, I just brought this inside P 1 2 between 1 by r 1 2 and this, and moment I do this P 1 2, this 2 will be equal only if I have interchanged 1 and 2 on the right, because P 1 2 acts only on the right, we have already done what is P 1 2 permutation operator, so I am going to bring the permutation operator here, and then write this term very easily, so now this term would be written as P 1 2 chi j 2 d tau 2, so still days, still days whatever, now we have something acting on chi i till day 1 equal to assume this is diagonal, we will talk about that number times chi i 1, now the question is what is this, this we know it is electrostatic interaction of 1 by r 1, this is also an electrostatic interaction, we can assume of 1 by r 1 to P 1 2, that is my operator, who cares, it is just that the operator is not a classical operator, so during that process of the Hamiltonian, I am saying that Hamiltonian has a 1 by r 1 2 for the classical Coulomb part, for the exchange part it is actually 1 by r 1 to P 1 2, that is all I am saying for only if you want to interpret, but that interpretation is up to you physical interpretation, but if I write it like this, then I can immediately write the Hartree-Fock equation in a nicer form, it is all about writing nicer form, all are right, the original one was right, so now what I do, I define an operator called f, this is something that you know, Fock operator, this operator is called Fock operator, f of 1, which is just the operator, h of 1, so I am not going to write the chi i till 1, they will be all out, plus sum over j, I am only writing the operator part, chi j star 2, of course this has to be integrated, 1 by r 1 2, 1 minus P 1 2, chi j till 2, I am going to do even more simplification, I do not want to write the Coulomb and the exchange, right, I will achieve this, 1 by r 1 to 1 minus P 1 2, 1 gives me the Coulomb, P 1 2 gives me the exchange, no problem, so eventually I am doing a classical interpretation of an operator called 1 by r 1 to 1 minus P 1, as if electron 2 sitting in chi j's thing, they are interacting via this potential, the potential I have only changed from 1 by r 1 to 2 particle potential has become 1 by, beautiful, if this is what it is, then all my Hartree-Fock equation is just f of 1 chi i till day 1 equal to sum over j lambda i j chi j till day 1, that is my general Hartree-Fock equation, is it, is it okay, now I have very, very simplified, I think I have to go through once more, necessary I will go through once more tomorrow, entire algebra, because this is the part which is, which is never done in the classes, they just come to that equation finally, nobody wants to derive this dot e t, so I have taken the risk of deriving, risk, the trouble no, the derivation I can do while sleeping also, so there is no trouble for me, the risk because it may put you off, so I have only taken the risk, no trouble for me, do not worry, I can keep on deriving every time you wonder, so eventually I get a very nice Hartree-Fock equation, that is called the non canonical Hartree-Fock equation, where f of 1 is defined as this, I have written everything now, these two lines write my Hartree-Fock equation, if provided I can make lambda diagonal, it will be a canonical Hartree, so I will see how to do that, that is a, that is the final part in the jigsaw puzzle, that I will leave, but both are right, we will see formally exactly you can make it canonical Hartree-Fock equation, so both are okay, but this is a general non canonical form which is yet not in the Eigen value equation structure, simply because this is not diagonal, otherwise it is equal to something and you know the diagonal values, if I can make it diagonal those are my orbital energies, all of you already know f chi equal to epsilon chi, everything is known to you all of you, so I am going to come to that little bit later, but that is the final piece which I will leave it as it is, and before I do that I have lot of other observations I have to make, so this then becomes a Hartree-Fock operator, so does it answer your question, how can I do it, mathematics allows me, not the physics, but this is the clever trick, if you want to write it in this form, but eventually when you integrate it will become complicated, the exchange part will spread, so that is the problem of the exchange part, that physically I am not saying that problem is simplified, but because the P12 is a non local operator, what is the problem, why it is spreading as you said for exchange, because of P12, so if you look at this operator, this is a non-local operator in a very way of saying, and this is where later on you will see the problem of density functional theory comes in, because this is density, no question, but unfortunately here it is a non-local, so when people want to write density functional theory with local exchange, it is an approximation, that is why LDA is an approximation, and this is the biggest problem in density functional theory, how can I do it without actually doing like Hartree-Fock, but Hartree-Fock there is no problem, we can do it, we have no problem, because it is only a question of an iterative solution, well I will come to the solutions and all that later, but I think first you should digest the equation, so remember I had a Lagrangian first when I started, it is a long process, I can repeat what we did, and you have to go, no okay, so you had a Lagrangian, and then we said that if I make these changes on chi I till, I did not bother about change in the, with respect to lambda, because anyway that will give you the equation, that was already known, if you change chi I first order change, first order change in Lagrangian should be 0, the Lagrangian explicitly included this, so then I made a first order change, so that was delta chi I 1, then they had delta L equal to 0, then I found that there are lots of terms, of course the complex conjugate part I had already knocked out, then within the Coulomb and exchange I found that both the terms are identical, so the half was knocked out, there is only one Coulomb and one exchange, with delta chi I 1 on the left, delta chi I 1 star 1 left, so since I have integrated the whole thing to be 0, the integrand was equal to 0, because for arbitrary delta chi I 1 if I do, this is called integrand, what is being differentiated, what is being integrated, that integrand was equal to 0, and that was my equation, and on this side the similar thing came, this was my integrand, then I wanted to write down the integrand that g of 1, which basically turns out to be of this form, where everything is now a function of 1, but where explicitly chi I till day 1 appears, so if I now say that integrand has an operator acting on chi I till day 1, what was my g of 1, so g of 1 is nothing but this operator, you remember the g of 1 I wrote that was integrand, it is operator f of i chi I till, for each i, so g of 1 will keep changing, so you remember my final equation as delta chi I star 1 g of 1, minus sum over j lambda i j delta chi I star 1 chi j 1, this is my original form, delta 1. Now I am saying this integrand is nothing but an operator times chi I till day 1, where the operator has the h plus the exchange and the Coulomb term, plus the exchange and the Coulomb term of the operator, which actually is contained in the g, which is basically given here, the interpretation of the exchange is the difficult part, we still do not know how to do it and how to actually mathematically implement that is very easy, that I will show next time, but so this was basically the equation that I said and this particular operator is called the Fock operator. One important thing however to note is that this Fock operator is not defined offline, the Fock operator is not defined offline, because to define Fock operator I need the chi j, this is a very important part to note, even if this was an Eigen value equation, normally what you will do, I have an operator just diagonalize, I will get chi I till day 1, but here you cannot get, because your operator is not defined, the definition of operator itself depends on this, which is supposed to be the output of this equation, so you can see that even the Hartree Fock, which looks like a now an operator Eigen value equation assuming this is diagonal canonical form, still not easy to do, because I do not have the operator form, so even the canonical Hartree Fock equation is really not an Eigen value equation, many time this is called the pseudo Eigen value equation and we will discuss in detail how do you solve it, after I first get the canonical Hartree Fock, there are two important pieces that are still left, how do I canonicalize, I am now assumed it is canonicalize and then how do I actually solve it, because F depends on the chi's, so the important point here is that because F depends on chi's obvious way to solve, one obvious way to solve is an iterative solution, that is assume some chi's, define the Fock operator, then solve the Eigen value equation, assume it is canonical, get the new set of chi's, again redefine Fock operator and continue to do till it converges, all mathematics you know how to do that iterative solution, when it converges what it means is that the field what is the field, the field is of chi's, chi's is the field of the electron 2, that field which is defining the Fock operator is the same which comes out after solution of the Fock operator, because after solution I am getting the spin orbitals, the field which defines the input comes at the same field as output, so my field is chi's, so that defines my Fock operator that is a box and then I solve some HF equation, I get the same chi's, if that happens that is what happens at the convergence point, normally know when that happens I am very happy, but I have actually solved the equation and this is the reason it says that the field which defines the Fock operator is the same field that comes out of the Fock operator and we call it self consistent field or symbol SCF, you might have heard the word SCF many times, SCF has been synonymously used as a HF actually, but let me again repeat HF is the method, SCF is only a procedure to solve HF many people forget, because somebody can come up with another way to solve this equation, SCF is a procedure where I solve HF equation iteratively till the field which defines the Fock operator becomes the output of the Fock operator, so that is the reason many times HF is called self consistent field, but remember SCF is a procedure, many people confuse actually, because we normally now say SCF method, we even forget that the HF did it, but HF actually did have the method, SCF is merely a procedure I can use some other procedure, of course many people write HF SCF method just to make sure that they are right, HF and then same consistent, you have seen all kinds of nomenclature in the programs and textbooks, so I am just trying to make sure that you understand that this is Hartree form, the procedure to solve this equation is SCF, even the canonical part cannot be solved easily, because of the fact the dependence of F on the chi and that is where SCF comes, I will again repeat this point of course and write down tomorrow 1, 2, 3 after everything is done, but one very important piece is canonical, that I have not touched yet, so that will take little more time, once I do that this is a small thing, but is very important, because we usually use, however I do not have to teach you canonical Hartree form, because this is still right, everything I will show you everything is same, except the orbitals, orbitals will change, energy will be same, but unless you have a canonical Hartree form equation, you will have a problem in chemistry, because you do not know what is orbital energy, the eigenvalues of the Fock operator, here there is no eigenvalues, so you cannot do ionization potential, you cannot do electron affinity, all that HOMO, LUMO that you have seen, I cannot do it, the only thing that I have achieved is the Hartree Fock energy, Hartree Fock energy will remain the same, the final energy, so I have to bring in canonical equation, I can assume that you can do it and write it, but that is not a good way of doing it, I will show how to do it, bring it formally and then you will have all those interpretations, HOMO energy, LUMO energy, HOMO minus 1, LUMO plus 1, all kinds of things that you will talk, it will all come after that, that interpretation of orbital energy, Kupman's approximation will not come unless I go to canonical equation, so this is a very important piece, just for the time being to make you understand I have assumed it is canonical, but this is a very important point, even if it is so, the Fock operator is depends on the spin orbital, so it has to be solved by SCF procedure, self consistent field procedure, so if I ask you to write what is self consistent field, you should be able to write properly, okay.