 E Hartree-Fog is known that is 2H11 plus J11, correct, that all of you agree, it is a very simple thing because you have 1, 1 bar anti-symmetrized 1, 1 bar so exchange part will become 0, so you will have only the Coulomb part, okay. This can be of course written in terms of epsilon 1 as well, so this can be written also in terms of twice epsilon 1 minus J11, okay. So there are different ways of writing this. So let me write this down that the epsilon 1 is of course H11 plus J11, okay. So epsilon 1 is nothing but chi 1 F chi 1, so chi 1 can be either 1 or 1 bar, you can take any one of them, so let us say 1, so 1 alpha F 1 alpha, alpha alpha is integrated then expand F, so if you expand F you will get 1H1 plus the other one which is 1, 1 bar anti-symmetrized 1, 1 bar. I hope you remember the epsilon formula, so epsilon i is chi i F chi i, so I want these are the things in the Hartree-Fog we have already done then expand F, so if you expand F you have chi i H chi i plus sum over all the occupied orbitals let us say B, chi i chi B anti-symmetrized chi i chi i. I hope you remember this formula for orbital energies, so it has interaction with all other occupied orbitals. So when i is one of the occupied orbitals then of course if it B is equal to i it is 0, so you do not have to worry about it. So the only other occupied orbital is 1 bar for any one of them, if I take 1 alpha then 1 beta I have to take care. So this is basically nothing but H11 plus J11, so that is your epsilon 1 and E Hartree-Fog is 2H11 plus J11, so that can be also written of 2 times epsilon 1 which is 2H11 plus 2 J11 minus J11, so either of these formulas are right, so this is correct and you can then write either this or this. I hope it is clear, so if you write in terms of 2H11 you have a plus J11, if you write in terms of 2 epsilon 1 you will have minus J11 for the Hartree-Fog energy. This is for a simple case of a 1-1 bar where Hartree-Fog is a 2 electron problem, so on this note let us now write epsilon 2. However epsilon 2 is an unoccupied orbital, so first term will be H22 as usual. Note here we are writing the energy of one particular orbital, so there is no summation over orbitals and all that, one particular spin orbital. Then I have to sum over all the occupied orbitals, but now the electron is in this spin orbital 2, so the other occupied orbitals are 1 and 1 bar, so there is an extra term. So I will write this as 2 1 anti-symmetrize 2 1 plus 2 1 bar anti-symmetrize 2 1 bar. For epsilon 2, so this is now your H22. Now if I do this expansion you will get J12, but you will also have one term which is K12, so what is 2 1 2 1? 2 1 2 1 is nothing but 2 1 2 1 minus 2 1 1 2, both will survive because they are both alpha spins now. So I have assumed that the 2 is alpha spin, so you have to first assume whether it is alpha spin or beta spin, so either way it is okay. So then this is my J12 and this is my K12, I hope you can see this, this kind of symmetries you should see quickly. So 2 1 1 2 is nothing but 1 2 2 1 which is nothing but 1 2 1 2, then last time we also use 1 2 1 2 as K1, all these are identical. So remember your 2 1 2 1, 2 1 1 2 is nothing but 1 2 2 1 and then I can interchange these 2 and 1, I can say 1 1 2 2. So these are for real orbitals and in our questions these are all real orbitals. So for real orbitals lots of symmetries are there, so they are all equal and either of them is called K12 and J12 is 1 1 1 2 1 2. So this is minus K12, so I wrote this, however if I do this you have only J12 because now when I exchange the beta and alpha orthogonality we will make it 0. So my epsilon 2 is H22 plus 2 J12 minus K12. So I will just keep these 2 results here, so 1 is epsilon 1, so then epsilon 1 is H11 plus J11 and E Hartree-Fock is 2 epsilon 1 minus J11. Now what I have to find out is the rest of the term which is 2 2 bar H22 bar because you remember my expression was 2 2 bar H minus E Hartree-Fock 2 2 bar, I am going to subtract this after I calculate this. So again if you put this expansion by Slater rules you will get this as 2 times H22 because you have H2, H term is very simple, 2 times H22 and then you have the anti-symmetrized 2 2 bar which is nothing but 2 H22 plus J22. This is anti-symmetrized 2 2 bar 2 2 bar, I do not have to write half and all that because you know only one pair counts, finally they will be identical, actual Slater rule has half and sum over all, sum over both 2 and 2 bar. So I am just writing only one term because they will all be equal. So it is 2 H22 plus J22. So now what I have to do, I have to subtract this and this because what was my quantity that I wanted 2 2 bar H minus E Hartree-Fock, this is what I wanted. So of course when this is Hartree, E Hartree-Fock, this is a number which comes out, so that is a normalized determinant, so that is one. So I have to basically subtract this from this. So what I am first going to do, write H22 in terms of epsilon 2. So now you see you look at epsilon 2, so you write H22 as epsilon 2 minus 2 J12 plus K12 and then put this here. So your 2 2 bar, H22 bar in terms of epsilon 2 will then become 2 times epsilon 2 minus 4 J12 plus 2 K12 plus J2 and now you have to just subtract. So if you subtract then you will get this quantity, so subtract these two, this and this. So you have 2 epsilon 2 minus 2 epsilon 1, so 2 into epsilon 2 minus epsilon 1 and then write plus J22 plus J11 and then just write this empty, minus 4 J12 plus 2 J12. It is a little bit of a manipulation, but it is a little difficult, I agree, but I thought I will just show you how to manipulate these quantities. So once I can reach this term, I can easily write. Here your E-Hutry-Fock which is 11 bar H11 bar is 2 epsilon minus J11. I will warn you, if you write 2 2 bar H22 bar exactly like this, there is a tendency to write this also as 2 epsilon 2 minus J22. That would be wrong because this came because your Hutry-Fock potential had one 1 bar in the definition of Hutry-Fock potential but not 2. See if you remember your Hutry-Fock potential, Fock operator H plus V Hutry-Fock. The V Hutry-Fock was defined in terms of 1 and 1 bar. So the orbital 1 and 1 bar and orbital 2 and 2 bar are not identical. So you cannot simply say that this is 2 epsilon minus J11 then write this as 2 epsilon 2 minus J22. The other way to look at it is that this is your two electrons, this is your orbital 2. What is epsilon 2? If I put one electron here, that is epsilon 2. What would be the energy that this electron will face? So that is H, that is of course its own interaction H22, one electron interaction but plus Coulomb and exchange with these two electrons and clearly Coulomb is there with parallel as well as anti-parallels. They have two J12 but parallel spins only can give exchange so minus K12. So that is what you got here. So even graphically you can see whereas for epsilon 1 it is different. I am taking out an electron here so the interaction is only with one electron J11 and that has to be, you know, cannot be parallel. Whichever electron I am taking the other one has to be opposite. So there is only J11. So it is only H11 plus J11. So I hope, so that was one, some people have done this mistake and after that of course you go into a spin, you cannot get it. So the key thing is to write this properly and then also write this properly. This is easy actually. If you write it in terms of H and J then you do not make a mistake. If you write in terms of epsilon there is a tendency of making mistake because epsilon is defined in terms of Fock operator which includes only occupied orbit whereas this expression is a generic expression. So if you write in terms of H there is no error, your E Hartree Fock is also 2H11 plus J11, this is 2H22 plus J22 so that is symmetric. But in terms of, so we are asking you to re-express in terms of epsilon 1 and epsilon 2. So that is why all this complication is coming, is it clear? So I think in the last class yesterday that was we actually ended by saying that E dci is proportional to square root of n, that was one of the main thing that we did and hence it is not size extensive. Note again that we made a distinction between size extensive and size consistent. In this case, the fact that it is not size extensive is also related to the fact that it is not size consistent but I told you that size consistency is something more. So you require size extensive but also separation of the reference space itself so which is Hartree Fock at this point it is Hartree Fock that the correct dissociation or the correct size consistency behavior of the Hartree Fock itself. So in the case of H4 dimer this is automatically satisfied as it goes to H2 plus H2. So the only issue is that since this is not size extensive, hence it is not size consistent so try to understand this. But there may be a situation where a theory is size extensive but because the reference is not separating correctly, it is still not size consistent. So in some way it is understood that the size consistency is a bigger requirement, is a more a stricter requirement than size extensive. So size extensivity can be defined even in interacting regime that even when they are interacting, the energy of this energy of the system should be proportional to n so that is size extensivity. So if it is not proportional to n just as dci is not square root n it is not size extensive and that we saw in the last class that your dci energy for n H2 molecule was always becoming delta minus delta square plus n times k12 square to the power half. So that was the form that you are getting so please understand I have done it elaborately you should be able to get this thing from the correlation energy expression by diagonalizing iterative diagonalization and then solving the quadratic equation. And we have also taken the lowest root you know there is a question somebody asked so this is of course the lowest root in the quadratic equation there are two roots so this is the lowest root because we are looking at the ground state. But I think that apart the point is the dependence of n in a wrong way that it is comes under square root and although we notice that the actual generic formulas looks like n times including the formula for e correlation n times but when you actually do this iterative solution the final result has square root n stuck in here and this is essentially the problem of dci and all truncated ci for large systems so whether if you do dqci for large system dq is also not good enough because we try to understand why this happens because when I take a product of the wave function at a particular level the dimer always has to have a higher level of excitation. So if I have quadruples quadruple into quadruple will give you octopole. So no ci if you do at a consistent level for dimer and monomer will give you size consistence so that means you have to do something for monomer something for dimer and even then we realize that that coefficient has to be a quadratic not linear. So there are two points that we highlighted so the ci can never solve this problem of product separation unless you do full ci if you do full ci and this is something that we do not want to show in a very elaborate mathematics there are lots of cancellation and it gives a correct separation simply because it is exact theories are of course correct. So we are discussing the theoretical model chemistry for approximate theories when you develop approximate theories how will you do this.