 So that is what we will go discuss today, we have derived this in spin orbital form, so we will try to see how to get integrated spin integration and get an equation only in terms of spatial orbital just as we have done. So remember our Hartree-Fock equation was an Eigen value equation of the Fock operator, so this is a form that we derived and this is called the canonical Hartree-Fock equation. And the ionized potential is minus an epsilon j if an electron is ejected from chi j. Again when I say this is only a classical way of saying it means that in the n minus 1 electron determinant chi j is absent. And if you add an electron an orbital r then the electron affinity is minus epsilon r. I told you yesterday, I told you that if there are 2-3 approximations one is that the entire thing is single determinant, yeah I am saying Cookman's in terms of exact single determinant that is one. Second is that within single determinant I have assumed that the rest of the n minus on spin orbitals or the rest of the n spin orbitals are not changing. Remember I wrote down and this is valid only up to the first order. So obviously it is not exact, it is an approximation even within single determinant. So today I was supposed to show how does it move with the exact because something is missing here and of course both of them are missing correlation because of single determinant how they match up, okay. So this is our equations where the f of 1 is h of 1. I hope all of you can now by now write it. So integral chi j star 2 1 by r 1 2 1 minus p 1 2 chi j 2 d tau, okay. I do not want to write this as direct notation because it is incomplete integration. So any direct notation will not actually speak correctly. On the other hand if I take epsilon i that is chi i f chi i in which case it is a complete integration because the coordinate 1 is also integrated and that could be written as so one integration over d tau 2 comes because of the definition one because of chi i f chi i, right. So this is the anti-symmetrized direct notation, yes. No in the sense that this is only over d tau 2 it is a function of 1. So I cannot write a direct notation. So that is the reason I am writing in this notation. I mean I have no notation for this chi j 1 by r 1 2 chi j or whatever only over d tau 2. I do not have a notation for this that is what I am doing. I can invent a notation but there is no standard notation but as soon as I do this then it is a number. So of course this is integrated over coordinate 1. So I get this in the first term and the second term already has integration over d tau 2 in the fork operator and when I take chi i 1 integration then the rest of them come. So you get complete integration over d tau 1, d tau 2 and because of 1-p12 I am writing together as anti-symmetrized element and we showed that the e-heart refock is not sum of the orbital energy. In fact it is equal to sum of the orbital energy minus half of the sum over j of chi i chi j sorry when I wrote e i there is a sum over j here but not sum over i that is a specific index. So when I write e heart refock then of course it is sum of the orbital energy minus this and here you have a double summation. One summation coming here and another summation because sum of the orbital energy okay. So we showed this yesterday that the actual expression has this plus half so that makes it that heart refock energy is not the sum of the orbital energy but you have to subtract and physically it means you are subtracting the double countered interactions. So with this we now want to move over to how to spin integrate for closed hand systems. Once again I have defined closed hand systems. So having obtained this now our closed hands essentially means chi 1 is phi 1 alpha, chi 2 is phi 1 beta and so on and so on. So in particular chi okay. So if I write generally up to n-1, n-1 is phi n alpha, n is phi n beta, n by 2 alpha, n by 2 beta. So then we want to look at this expression and spin integrate. So let us write first let us assume that chi i is sum phi i alpha. Note that this is a specific index that will come as a result of diagonalization and let me call it phi i alpha. So I have f of 1 phi i 1 alpha 1 is epsilon 1, epsilon i phi i 1 alpha 1. So this chi i is a specific alpha spin. Without any loss of generality you can make it beta spin as well. So I am not going to discuss the beta spin. Then note our expression for Fock operator. So what we are going to do is to now multiply this by alpha star 1 on the left. Note again alpha star 1 essentially means omega 1 because these are spin functions. They are coordinates are different from these coordinates. They are r theta phi and this is the spin coordinate and then perform the integration over p omega 1 only because we want only to integrate this spin. So just integrate this spin. So let us see what we will get. So we will get an equation which is alpha star omega 1, writing omega 1 specifically now, f of r 1 omega 1 alpha omega 1 d omega 1 into phi i of r 1. But r 1 is a vector equal to epsilon i phi i i. Please make sure that you understand what I have done. This part is trivial because this is function of r 1. So this and this together is 1. So right hand side is trivial. The left hand side remember the f depends on spin orbitals. So I cannot take this out and integrate. So the integration must involve f because f depends on spin orbitals. However phi i of r 1 can be taken outside the integration. So I have new operator acting on phi i r 1 gives you epsilon i phi i r. So that is the only change that in the left hand side I have to take care because this operator depends on spin. So when I integrate the spin it must involve the operator itself. So I have a new eigenvalue equation. This is an operator in R because omega 1 is integrated which acts on phi i of r 1 a number times phi i of r. So I have a new eigenvalue equation which will replace that old equation. We will have to see what is the structure of this operator. So we will analyze this operator. So let us analyze this in parenthesis. So let us write the first term. I hope it is clear. I have written every term in full. So when I said Kij's term 2 I have written R 2 omega 2. D tau 2 has been expanded in full and of course you have alpha omega 1 d omega 1. The first term is clear. The second term I have only exchanged. So this has become alpha omega 2. This is r 1 omega 1. This is and of course this remains d r 2 d omega 2 and d omega 1. But this remains R 2 omega 2. Is it clear? This j is over 12 because these are still spin orbital indices. But remember your Kij's R 2 omega 2 are either phi j R 2 alpha omega 2 or phi j R 2 beta omega 2. All these Kij's are either phi j R 2 alpha omega 2 or beta omega 2 because they are all spin orbitals. Whereas the Kij for which I am writing note is only alpha omega 1 that is where I started. But when I am summing over j it is over all spin orbitals. So the first term is trivial. So the first term can be trivial in return. So let us call this operator some f of r 1. So now we have f of r 1 equal to h of r 1. The first term is trivial because you can always integrate alpha star alpha 1. This first term. Now let us look at the second term. I am going to write the Kij's in terms of phi j alpha or phi j beta. Which means now my summation over j will go from 1 to n by 2. That is the first change. And I am going to explicitly write this for both alpha and beta. Is it clear? So I will write this alpha omega 2 and beta omega 2 twice. So sum over j 1 to n by 2 integral alpha star omega 1, phi j star r 2, alpha star omega 2, 1 by r 1 2. I use exactly the same here phi j r 2, alpha omega 2, alpha omega 1, dr 2, d omega 2, d omega 1. It is very easy to write. Don't worry. You just have to think. See I have nothing apart from typo errors. It is very easy to write. I am just writing monotonously. I have only ensured that now my Kij's are alpha spin. The Kij's are now alpha spin. I will also add the beta spin of course. I am going to do that for the first term. So there are two terms here. There are two terms here. When Kij's alpha, Kij's beta for both coulomb and the exchange. So I am just analyzing one of those four terms. Clearly when this is alpha, now I can integrate this spin d omega to d omega and very easily. This is 1. This is 1. So everything will survive. No problem. Plus, let me take the another term. j equal to 1 to n by 2. The same term I am now writing alpha star omega 1. However, now Kij is a beta. Beta star omega 2, 1 by r 1, 2. Again, phi j r 2, beta omega 2, alpha omega 1, dr 2, d omega 2, d omega 1. Once I can analyze, although this has become beta, there is no problem because beta star beta will still give you 1. Alpha star alpha will still give you 1. So the term will survive. So this term in terms of space orbital will survive twice. They have exactly the same value. You can quickly check. Let us do the same thing for the exchange term. So I am writing the exchange term on the top or maybe here. So minus j equal to 1 to n by 2. Now come here. Integral alpha star omega 1, phi j star r 2, alpha star omega 2, 1 by r 1, 2. Now here is an exchange. So this becomes alpha omega 2 and this becomes phi j r 1, alpha omega 1, dr 2, d omega 1, d omega 2. Once again, if I analyze this term, this and this will become 1. This will become 1. No problem. So it is going to survive. And now the final term. So let me write it down somewhere, maybe here. The last term minus, what is the last term? Same exchange but phi j is a beta spin. So alpha star omega 1, phi j star r 2, beta star omega 2, 1 by r 1, 2. Now here is again the exchange. So it becomes alpha omega 2 as usual and this becomes now phi j r 1, beta omega 1, dr 2, d omega 2, d omega 1. Now look at this integration. It is 0 because you have alpha star omega 1 and here you have got beta omega 1. Similarly, beta star omega 2, you have got alpha omega 2. When I integrate over omega 1, omega 2, alpha beta, orthogonal, it will make it 0. So for exchange, this term will survive twice. Sorry, coulomb, this term will survive twice. For the exchange, it will survive only once. So that is the moral of the story and actually I could have predicted from here right away. You can see that because it is coulomb, these two chi j's can be either alpha or beta. I do not care because they have nothing to do with this. On the other hand, because of the exchange, now chi j becomes r 1 omega 1. So this must follow the spin of the first one, the chi i. And chi i is alpha spin. So only alpha spin orbitals can contribute to the exchange. For the exchange, this must be alpha because original chi i was alpha. Once again, the same thing that we have repeatedly been telling that if chi i is alpha, exchange takes place only with another alpha spin and not with the beta spin. This is something that we have been telling in various different parlour. So again, you see the same result. Since my original chi i was alpha spin, so this was, if you remember, this is just alpha omega 1 because original was alpha spin and I have integrated with alpha star omega, these chi j's cannot be beta because when I am interchanging, it will become 0. And that is the reason the exchange comes once. If my original electron is up spin, it will have only exchange with up spin when I construct the fork operator. Whereas for the coulomb, it will have contribution from both up and down spin. And the same thing will go if my original was beta. It does not matter, chi i could have been beta spin, so only beta spin would have contributed. So with this, I can now write down the fork operator in spin integrated form. So f of r 1 equal to, first of all, H of r 1, that is trivial. Now spins are all gone. So here spins are all gone. So sum over j 1 to n by 2. I am not going to write this spin because they are all integrated out. So it starts from phi j star r 2, 1 by r 1 2, phi j r 2, d r 2, this term. Everything else goes out. So what remains is this, this and this. However, this comes twice. When chi j is beta spin, exactly same thing happens. This remains. So this will come twice. So I can say again sum over j equal to 1 to n by 2, integral phi j star r 2, 1 by r 1 2, phi j r 2, minus integral phi j star r 2, 1 by r 1 2. And now here only one term survives when this is alpha. So that case, this becomes phi j r 1. And the chi i, the special part of chi i will become r 2 now because there is a exchange p 1 2, right? So again to write it in the same form, I will bring p 1 2 and phi j r 2, that completes f of r 1, exactly in the same form. If you actually notice the f of r 1 in terms of space and spin orbitals, they are almost identical. Remember in the spin orbital, we have integral chi j star r 2, 1 by r 1 2, chi j 2, d tau 2. I have instead that phi j star r 2, 1 by r 1 2, phi j r 2. Sum is only over n by 2. The coulomb comes twice. That is the only story. And the exchange is exactly same, comes once. And sum is so, actually it is identical, looks identical except that this term comes twice. So I can write it just as I wrote there h of r 1 plus sum over j 1 to n by 2, integral phi j star r 2, sorry r 2. Now instead of 1 by r 1 2, I will write 2 by r 1 2 minus p 1 2. That is it, over. Is this everybody understands? It is just algebra. So the way I was writing 2 by r 1 2, I should have written p 1 2 by r 1 2. It does not matter. So 1 by r 1 2 can be factored out. And it becomes, so that is the only difference. The form is exactly same again. In the spin orbital, what was the thing? Everything was spin orbital. This was j equal to 1 to n. And here it was 1 because spin orbitals, I do not have a sense whether it is alpha or beta. I do not care. The form was very general. Moment I am integrating, I have taken care of the fact that the alpha spin orbitals are different from beta spin orbitals in their relation to the alpha spin orbital, electron in the alpha spin orbital. Because chi i, chi i is originally phi i alpha. I told you that is how I have started. Since I am writing it for phi alpha spin, it makes a difference. The spin of the other makes a difference whether it is alpha or beta. And accordingly this factor 2 comes. But other than that, it is a fairly straightforward relation. And I think this factor 2 just re-emphasizes what we have discussed before that the every pair of electron has a coulomb, but only parallel pairs have exchange. So when I have alpha spin to start with, this must only be alpha spin for exchange. That is why there is only factor 1. Note actual expression is far little bit more complicated because of p12. But with p12, I have brought back sanity that this remains phi jr2. Otherwise it would actually become phi jr1, which is what I showed here as you can see here. It becomes phi jr1. And then your Hartree-Fock equation becomes, the right hand side I already told is very simple. So your right hand equation now becomes this Fock operator, new Fock operator, phi i of r1 equal to epsilon i. Note that the spin integration that we did for the F of r1, this is actually for closed shells. So depending on your determinant, whether it is a closed shell or something else, this will change because I am now assumed that the alpha rn by 2, betas are also n by 2. So that will change. The summation everything will change. But the physics will remain the same. If this is alpha spin, I will have coulomb with all exchange only with alpha. So if I have more of alpha spins, then this summation will be more. For the coulomb and the exchange, the beta 1 will give a separate term for coulomb. So I cannot write it exactly in this manner in that case. I have to separate out. The number of terms contributing to coulomb and exchange would not be just half. So it is not going to just be a factor of 2 minus p1. So that is all. But the basic physics will remain the same that the parallel spins will give both coulomb and exchange. Anti-parallel spins will only give coulomb. Anti-parallel will give only coulomb, not exchange, sorry, not correct. Parallel spins as I said correctly will give both coulomb and exchange. Anti-parallels will only give coulomb. The other way of saying is that for every pair of interaction there is coulomb, only for parallel pairs there is exchange. So whichever way you want to write it. So this particular form, remember this factor 2 minus p1, 2 is only for closed cells because if I have another determinant in which let's say there are n spin orbitals or let's say 10 spin orbitals, just give you an example, but 6 of them are alpha spin, 4 of them are beta spin. The summation indices will be different. This factor 2 won't come, but that is trivial. You can always spin integrate. I mean once you understand how to do it, so we specifically did for closed cell because this is more often used. So you have to be very cautious in applying this form of F of r, but whatever is the F of r, the final equation will be just this in terms of special effects. And this is again now canonical equation because you have started from canonical Hartree-Fock equation. And now we can directly get the space orbitals, what we normally call orbitals. And the spin part is gone now. So you don't have to bother about spin. And we can directly solve this, but the F must have this form for closed cell or whatever form depending on whatever is the determinant. So this is the only part that you have to derive depending on unrestricted Hartree-Fock or restricted, why I am not going to those things. The closed cell incidentally, I must also mention is closed cell Hartree-Fock is often called the restricted Hartree-Fock determinant, RHS. I hope it is very clear why it is called restricted because we are restricting the occupation of the space orbitals. And every space orbital must have 2 electrons, 1 alpha, 1 beta. So that is the reason we are calling it restricted Hartree-Fock. Very often this is called RHS. We also have similarly unrestricted Hartree-Fock, UHF. Restricted open cell Hartree-Fock, those are little bit more complex. I will tell that later called ROHF. And RHF you will see is a special case of ROHF. UHF is really different. So we will discuss all these different forms of determinants, but right now let us try to understand only RHF, which is the closed cell case, even number of electrons, ground state, very nice doubly occupied. So many of the cases can be handled by this. I hope all of you can derive derivation is actually very easy. Those who are who can physically see this could have actually seen from here. You do not have to go through all that rigmarole. Spin is going to be integrated. Here it is trivial. Here the integration is could not be done alpha star alpha only because of these guys. But then if this is alpha, strip them either alpha or beta, they are separately integrated. So here it does not matter. Two times they will occur. In this case there is exchange. If this was alpha, kaiji has to be alpha. And then forget about spin. Write everything in terms of space orbitals. Write the coulomb twice, exchange once. Make the summation over j from 1 to n by 2. You just have the expression. You know, you really do not have to spin integrate. Do all these things. I mean if you are clever you can see it here right away. Just as I did for energy integration. So many of these spin integrations I am doing in a detail. But later on you can actually see it. On the fly you can do it. Just by seeing you should be able to write it with a little practice. In fact, if you do this you may be more confused because this is a long derivation. Somewhere you will make a mistake. Here it is much easy to see because you know quite obviously this is integrated by itself. So I do not care whether it is alpha or beta. Both will survive. This is not integrated with itself. It gets mixed up with omega 1. So if this is alpha then it has to be alpha. That is all the logic. So it is actually very simple to do in a physically understand the spin integration. Is it okay? All right. So I can now proceed. So now we have gone to an equation Hartree-Fock equation for closed shell or a restricted Hartree-Fock case. Which is very useful. And we will like to solve this equation of course. Note that the variation method was applied in a very general form in spin orbit. So there is no problem with that. And then eventually we took down the closed shell part. Now I will not go into tremendous details about solving this equation except to mention. Of course let me also mention that the Kupman's approximation everything remains the same. Interpretation of epsilon i remains the same. The negative orbital energy everything remains the same. This was solved for atoms almost exactly. Almost in a numerical manner. Something that is actually so this is called the numerical solution and they are very very exact. In fact in physics this is taught quite well because physicists are interested in atoms. So they can be exactly solved in a numerical manner. So numerical solution essentially means you get a give a guess for the orbitals at every point value convert that into value and then construct the f of r1 at every point. Then try to find a phi i at that r1 which satisfies this equation for all r1. Eigen value equation. And then again change. So it is a very detailed calculation but they are all playing with numbers. Numerical solution essentially means that you start with the values of phi of at all r2 because remember first I have to get f of r1 from phi j of r2. So all these r2 values you take guess. This is basically all self consistent field procedure. Any solution you have to have a guess. Then construct f of r1 at all r1. You have a map. Then try to find out that is a numerical Eigen value procedure how to get these Eigen values at each point r1. So such that when I do this multiplication and division this should give a constant. Epsilon i is constant. So there is a trick to do that. And then you get f phi again at different points all the phi's. From there you reconstruct f of r again at different point and keep doing it. So there is a lot of computations but they are all numbers. Now of course numerical computations can be also in error depending on how many points you are taking. So that is why they call grid. In mathematics they use what is called the numerical grid. If you have a mesh which is very very tightly, grids are very very small. These are the good number and you can make it as exact as possible by making a tight grid, heavy computer, fast computer. So all that has been done. In fact the two people who did very good work are Clementi and Roeti. Many of you may actually refer to the Clementi Roeti table for atomic heart reform. They have a table of heart reform. For example lithium or let us say beryllium. They will actually give you phi 1 of r1, phi 2 of r2 or phi 2 of r1 whatever at different r1. That is all the table will give you at the end because for beryllium atom you have only two special orbitals. You call it 1s, 2s whatever. All you require is the values of those functions at different r1. So that is the final is what is output. And these r1s will be as tightly as I said meshed as possible. So Clementi and Roeti have a table. So to understand beryllium atom you have to read through the table at what value, at what point what is the value of 1s, what is the value of 2s. So this is numerical. I mean it is boring actually after a point where you have just too much data. But I mean this can be done very nicely. Unfortunately the point that I am trying to say is that when people came to doing the same solution for molecules the error became too large because for molecule in the AF operator there is a H which includes the electron nuclear exchange, electron nuclear attraction. And this is now a multi nuclear unlike an atom. So when it becomes multi nuclear for molecules then the errors in the numerical procedure become much larger. So actually people never use this procedure for calculation in molecules. So you will never find for molecule a numerical solution. Now this is a little bit technical thing I will not bother. What we will do is what is actually practice for molecules. How do I solve for molecules? I cannot solve it numerically because of this term the multi nuclear attraction of electron with several nucleus not a question of number of electrons. The problem is because they have number of nuclei. As long as this one nuclei this the potential is ferrically symmetric. So it is much easier to handle but if it is not ferrically symmetric the results the numerical integration becomes very very difficult. Because there is a lot of integration which are eventually involved you have to integrate. When I construct for cooperate I remember integral dr2. So they are numerical integration. So that becomes very cumbersome and if you are interested I can refer you to the text to read you know why for molecule it is more difficult than in atoms to do that. But I think at the end what I am trying to say that this cannot be solved numerically and hence it cannot be solved exactly. I have to find some other way of solving this. So again I note that the Hartree-Fock is an approximation that is that you should not forget. Some of you are forgetting in the very first place and I will remind you when I tell why Koopman's works better. When I say works better it is compared to the exact not compared to Hartree-Fock. So Hartree-Fock is an approximation. Now we have a Hartree-Fock equation which will be solved for molecules and when you solve this we will bring in another level of approximation and I will just tell you what this because today time oh maybe you have some more time. So I bring in some more approximations and this is what is called expansion in basis set. Many of you have heard what basis set and so I will tell you what is the basis set and why there is an approximation. So for molecules even the Hartree-Fock I am not solving exactly that is a very important part to understand. So there is nothing called unique Hartree-Fock. So you have to tell me Hartree-Fock under what approximation what is that approximation that is the basis set. I will tell you what the basis set comes what is done in molecule.