 Let me go back to the perturbation theory that we discussed. There are several issues which I have to repeat again once more. First of all, we said that to start the perturbation theory, we wrote H as H0 plus V. That is important. We made sure that the psi-hat-ree-fock is a ground state eigenstate of H0 because that was our idea of starting the perturbation theory. So we said that the H0 is a n-particle operator which is sum of the Fock operator. So this is how we discussed H0. Please understand this is a n-particle operator. So these i's are not spin orbitals. I just want to make sure because people get confused. These are the coordinates. So whenever I am writing f of something, that is a coordinate. So I have r1, x1, x2, xn basically. Fock operator in terms of spin orbitals. So r1, omega1. So these i's are not spin orbitals because the Hamiltonian is a function of electron coordinates. So I am writing the Hamiltonian. So this must be function of electron coordinates. So this is a sum of the Fock operator. Each of them, I have f of 1. Similarly, I have f of 2 and so on. Although as far as the eigenvalue equation is concerned, any one of them is good. So I have the eigenvalue equation of chi i1 as epsilon i chi i1. And please note that I am explicitly working with the canonical Hartree-Fock just for simplification. So here these are coordinates. Chi i is a spin orbital. So whether I take f1 or f2, the eigenvalue system will be same. So that is the reason when I do eigenvalue, I take only one f of 1 with eigenvalue system will be same. But when I reconstruct the zerethodder Hamiltonian, there must be some of f of 1, f of 2, f of 3 up to f of n. Just want to clarify this. There should be coordinates of the electrons, spin and space both. And we also know f of 1, f of 1 which is f of x1 is h of 1 plus sum over j integral chi j star 2, 1 by r1 2, 1 minus b1 2, chi j 2 d tau. Please note that everything in the parenthesis is a function of coordinates. So this is a coordinate, this is a coordinate which is integrated. So that is why it is a dummy. These are also coordinates. Everything is coordinate. So explicitly what is written is this spin orbital chi j. And this acts on a chi i to give me epsilon i chi i. But in the definition of f operator, only one chi j appears. I want to mention this. Again some people have confusion. I saw in the Midsam paper, they wrote chi i, chi j and that got totally confused. Because obviously this is a integration of only one coordinate of electron, not both the coordinates of electron. Otherwise I will not get the fork operator. When I go to the energy, I have a two electron integral which will bring in i and j. But here chi i1 is not there. chi i1 will be here when it acts on this. But the definition of fork operator just includes this part. So this is something that you have to first understand properly. So this actually is a 1 by r1 2 integrated over the other electron, one of the electrons in any spin orbital chi j. So this summation is actually 1 to n. But this summation j is spin orbital. This is a summation of a spin orbital. This is coordinates of electron. So they can be any spin orbital and all of them are accounted for. And then I take the coulomb and the exchange part. So the potential due to the second electron or the third electron, that is not important now. Because by summing over spin orbitals, I am taking care of all the electrons. So that is what I have said repeatedly. So I define the single particle fork operator. Its sum over all the n electron coordinates give me my n particle operator. So H0 is a n particle operator whereas f of 1 or any f of i is a one particle operator. But this is a non-interacting problem now because H0 is sum upon electron operator. Because it is non-interacting, its eigenfunctions are nothing but products or anti-symmetrize products of the eigenfunctions of the f which is basically orbital because f is a spin orbital which is basically spin orbital because f is a one particle operator. So I have already told you any one particle function is an orbital or spin orbital depending on whether it is space, coordinate or space people. So these are now spin orbitals. So I take an anti-symmetrize product of this. I will get the eigenfunction of H0. So the ground state eigenfunction happens to be of course H0 psi hard trifog which is now, so psi hard trifog you already know is an anti-symmetrize product. So chi 1, chi 2 to chi n and this gives me the zeroth order energy which is sum of epsilon i's 1 to n psi hard trifog. So this is the eigenvalue equation of H0 because it is a non-interacting problem. So if I take an anti-symmetrize product of this spin orbital which is psi hard trifog I know the result is an eigenfunction and the eigenvalues are simply sum of the orbital energies. Is it clear? So remember we had an approximate wave function psi hard trifog approximate to the exact Hamiltonian. However I now discover that this is an exact eigenfunction of another Hamiltonian H0, an approximate Hamiltonian H0. So the eigenfunction which was approximate eigenfunction of H is however an exact eigenfunction of the approximate Hamiltonian. So I have just changed the picture. Now I have made this approximate state exact state the Hamiltonian is changed from H to H0. Further with an m basis problem I could generate many determinants. So I categorize them as psi a r psi a b r s and so on and we find that for the same reason since all those spin orbitals are eigenfunctions of this Fock operator each of these determinant is also an eigenfunction of the H0. Is it clear? For the same reason because they are anti-symmetrize products of the spin orbitals. Some of them are virtual orbitals but they are also included in this solution. So let's assume that I have a solution which goes up to capital N which is much larger than N. Although I am explicitly using only capital N for the definition of the Fock operator. My solutions are capital N which is of course done in a Hartree-Fog-Ruthan basis that's a different matter. I will not worry right now how do I get the solutions. So with those unoccupied orbitals if I construct these excited determinants of H0 then each of them is an eigenstate of H0 and with an eigenvalue which can be written in reference to this as so let us call this sum over epsilon i i equal to 1 to N as E0 0 by the notation of the perturbation theory since it's a ground state eigenfunction for the zeroth order Hamiltonian let's call it E0 0. So I can say with respect to E0 0 I have one extra orbital energy and one less and so on. So I can now write down all the eigenvalue equations of H0 so E0 0 plus epsilon R. So in fact entire n-to-plie excited determinants can be formed as an eigenfunction of H0 is it clear? Note that this is the entire thing is now an n particle space so these are all n particle determinants which are formed out of one particle function or spin orbitals and they are in particular the solutions of the Fock operator and that is why it is called Hartree-Fock perturbation theory. So this sets this stage because now I have a H0 which presumably is a dominant part of H and that's a question that you can still ask but I presume it's a dominant part of H simply because its ground state eigenfunction is a very good approximation to the exact eigenfunction because it recovers 95 to 97 percent of total energy. So I presume that H0 is a dominant part of H and of course the solutions all solutions of H0 are known in this basis. So that's not a problem I can keep on increasing m I will get more and more solutions but in principle in that basis all solutions are known and please remember in quantum chemistry everything that we now do is in a basis. So even I have repeated I repeat again I had told before that there is nothing called exact Hartree-Fock. Even the Hartree-Fock is in a basis okay so exact energy is also in a basis. So whatever we are going to discuss np2ci couple cluster everything will be defined in a basis there is nothing called exact because we don't have an exact Hartree-Fock but we know when m will go in the limit m tends to infinity everything will reach its exact limit okay. So whatever we are talking is of course an m basis problem. So in that basis my entire solutions of H0 are known which are mcn in number okay mcn is the number of determinants that I can form. So mcn number of solutions are already known this is a dominant part of H so I can start my perturbation theory with that okay. So we did the perturbation theory and I again we will just write the final result of E01. E01 is something first order perturbation I derived on the day and that was the 0th order wave function which is psi Hartree-Fock ground state V psi Hartree-Fock. So it was the expectation value of the perturbation operator with respect to the psi 00 which is the ground state Eigen function of H0 because I am looking at a ground state correction to the energy. So the first order correction merely comes with this and then if you see E0 0 if you look at E0 0 since this is the Eigen function of H0 I can write E0 0 as psi Hartree-Fock H0 psi Hartree-Fock correct by definition because psi Hartree-Fock is already an Eigen function of H0. So if I take an average value exactly like this then this is obviously E0 0 which is nothing but sum of the orbital energy anyway but this is alternative way of writing because this is an Eigen function of H0 with E0 0 as the Eigen value so this will become E0 0 psi Hartree-Fock psi Hartree-Fock psi Hartree-Fock is 1 so I get back E0 0 is it clear to everybody? So this I can always write any Eigen value equation Eigen value can be written as an average value with respect to its own Eigen function okay. So if I add these two then of course I see that E0 0 plus E0 1 now that everything is same I can only add the operators okay. So I have psi Hartree-Fock H psi Hartree-Fock which is nothing but E Hartree-Fock. So we had already noticed that the psi Hartree-Fock is an Eigen function of H0 but the Eigen value unfortunately was not E Hartree-Fock it was sum of the orbital energy. Now we see that if I add the first order correction to the ground state then together I recover the E Hartree-Fock okay and we can do lots of other things that we will now talk about but the important thing to realize that within this perturbation theory any improvement to Hartree-Fock energy will only occur at second order because up to first order we are recovering the Hartree-Fock energy okay. So that is the reason the genesis that the correlation theory starts from MP2 okay. So MP2 is the first correlation energy because correlation energy is going beyond Hartree-Fock I told you already it is a difference between exact energy minus Hartree-Fock. So unless and unless you have improved from the Hartree-Fock you do not have any correlation energy. So it starts only at the second order which we will do later but I just want to make some remarks on this point that at the first order you do not have any improvement. So because I cannot write H0 psi Hartree-Fock or E Hartree-Fock psi Hartree but with the first order correction I can write the question that was asked was very interesting that can I have an H0. So can I have an H0 let us call it H0 prime some other H0 where psi Hartree-Fock remains it is a 0th order Hamiltonian remains the Eigen state but the Eigen value now becomes E Hartree-Fock. So can I have this question mark you understand the problem can I have such a H0 H0 prime I am not saying this H0 some other 0th order Hamiltonian such that I can write such a nice equation this is very nice because there I can interpret the Hartree-Fock is very nicely everything as an Eigen function and Eigen value of a 0th order Hamiltonian if it is so then of course you would say that the correlation energy should start from first order we will see what happens which means if I write H with some other Hamiltonian H0 prime plus V prime of course if H0 becomes H0 prime V will also change. So let me call this V prime then is this possible that is a question we will come back to this and how to make it possible and then do you have a correlation energy from first order that is a question so I have not answered the question I will come back. Can I erase this I will just give this on the board if you realize little bit from these equations that what is the difference between E Hartree-Fock and E00 only one number E01 right this is a number remember so there is only one number which is differing from E00 to E Hartree-Fock so it is very easy to bring the E00 to E Hartree-Fock by simply writing a new partitioning where I write H as H0 prime plus V prime and make H0 prime H0 plus this number which was E01 before my previous H0 gave me E01 that is a number I can simply add that number do you understand what I am saying so I redo I rescale my Hamiltonian by adding a number which means my H0 is nothing but prime is nothing but old H0 plus Psi Hartree-Fock old V Psi Hartree-Fock if I do this then it is very easy to see that this is exactly what I will get because what happens Psi Hartree-Fock remains the eigenstate of H0 prime because it is only scaled by a number eigenvalue will be scaled by the same number right so the eigenvalue will become E Hartree-Fock so in fact the question that I am asking is actually a trivial question because this is only a number I can always scale and all my eigen solutions of new H0 prime will also be scaled by this number right so there is no problem they will all remain eigenstates is it clear please ask question if it is not clear I know these are not done so I am trying to again repeat so I am now reconstructing a new Hamiltonian H0 prime new 0th order Hamiltonian by simply adding to my old H0 this quantity which is not a problem which I can calculate I have V I can calculate this I will do the calculation more explicitly little later and of course my V prime will also change old V minus this correct I will come back to this V prime little bit later but I just want to tell you that of course my H must be same so if I have added a number in H0 same number must be subtracted to have the V prime now remember my perturbation should be now V prime I cannot have H0 prime as 0th order and V as the perturbation that is wrong so my perturbation operator is also changed to V prime which is simply the old V minus psi heart rate of V psi heart rate if I do this then it is very clear that I can answer this question so I have I have indeed I can take this box but then the question is what happens to MP1 so when I do the MP1 now remember how do I get the first order correction perturbation and I average value with respect to psi heart refock so what is my E0 prime so let me call it E0 prime 1 so what is my E0 0 prime that is E heart refock right that is the new so whenever I am saying prime it is a new new partitioning so what will be E0 prime now psi heart refock right that is still the eigen function V prime psi heart correct so what is the result replace V prime psi heart refock V psi heart refock and this is a number this is a number so the number will simply come out because psi heart refock is normalized to 1 so it will become psi heart refock V psi heart refock and that is still 0 so this was a number but the same number has been subtracted in the V prime so your V prime average value with respect to psi heart refock it is 0 so although it is nice that I could get E heart refock as the as the eigen value of H0 prime I still do not have a correlation at MP1 that is 0 so in a way as far as correlation is concerned either H0 or H0 prime I do not care both are good because they do not they do not give me any extra solution at first order as far as correlation energy is concerned the improvement only starts from second order at the second order of course things will start to change which I will come little late much later when I do the second order perturbation theory but I just want you to get convinced that at the first order by simply adding a number it does not help any question there are many many things that you can do this is not the only way to get H0 prime I can keep doing lots of things in fact there is to be very good papers which are simply titled what is a good H0 some of the papers are titled what is a good H0 or partitioning of the Hamiltonian how do I partition the Hamiltonian and lots of interesting paper in the 1960s have been done by Diner, Cleverie and Malryu they have written a lot of interesting paper in the 60s Malryu is who came in this Goa conference yeah the same man so he did a lot of work on what kind of partitioning would be good and you can see that at the first and up to first order it is trivial but at the higher order second order onward things will change depending on how I partition okay so second order onward results will keep changing so what is a good H0 is a very interesting title of a paper you know how do I get a good H0 and can I play with it and of course I will get a different perturbation theory I will get a different perturbation series from the second order onward depending on how I change my H0 is it clear of course second order I have not done but trust me that the from second order results will start to change depending on what is my H0 prime I want to convince you that by just adding a number like E01 previous E01 I do not get any change only thing is that I can be happy that the Hartree-Fock is a eigen value of H0 prime that is my only happiness but then E0 prime is 0 first order correction is 0 so eventually what is important is 0th plus first order is E Hartree-Fock that part is not changing either in unprime or prime partition okay before I go forward and look at different partitioning let me now analyze what is V just want to write down V explicitly note that your total Hamiltonian is sum over H of I plus 1 by R I again these are coordinates of electrons so whenever I am writing Hamiltonian they are coordinates of electrons okay I's are all coordinates and then I have a Fock operator which if you remember was H of I plus sum over kaij star 2 1 by modulus R I minus R 2 sorry into 1 minus P12 by Pi 2 kaij 2 d tau please note that you have to be you know since I have started writing I I am managing I so what did I do everywhere coordinate 1 I have written I it does not matter okay so it is just that you should be very comfortable in writing this so this should not be P12 this will be Pi 2 interchanges I and 2 so it does not matter so then I am saying my H naught is sum over Fock operator right so what would be my V now so if you look at this part if I sum over Fock operator I first get this part but then I get an additional part okay which must be subtracted from this so let me call this part this part sum over j this whole part let me call this V Hartree Fock so let me write F of I plus H of I plus V Hartree Fock where V Hartree Fock I is this in the parenthesis so that is exactly that extra Coulomb and exchange operator okay and your H naught contains sum of all such V Hartree Fock I okay remember again I again repeat that this j is a spin orbital this I is a coordinate okay because j is a spin orbital okay normally I would have written F of 1 but then when I want going to do any F of I a dummy variable I must I must not worry so then what happens to V so V now becomes V Hartree Fock V now becomes V Hartree Fock subtracted from 1 by r i j so it becomes 1 by r i j minus sum over I V Hartree Fock correct I am just talking of this partition of course if your H naught becomes H naught prime then V prime will have a further subtraction okay but this is the V you can now see my H naught is sum of the Fock operator so if I do H naught plus V I get back my Hamiltonian please make sure because when I do H naught I have a sum of Fock operator so sum of H of I sum of V Hartree Fock I and V has 1 by r i j minus V Hartree Fock I so that V Hartree Fock I sum gets cancelled and I am with H of I sum plus 1 by r i j sum correct so I get back my Hamiltonian so please make sure so basically this part has been added to H of I to give me a Fock the same part the sum is subtracted to give me V so many people may think that V is just 1 by r i j that is not true since I am starting from Hartree Fock perturbation I am starting from F as my original operator one particle operator to start with if I would have started with H that is if my H naught was just sum of H of I then of course V would have been r i j but that I have discarded long back because I know that is not a good one particle solution that was this thing that I discussed that would have been the simplest to have a non interacting picture and this would have been perturbation but that I have discarded since my solution now includes an additional operator I have to make sure that that operator is subtracted when I define the perturbation operator and then you can see what is psi 0 or psi Hartree Fock V psi Hartree Fock this is my E01 note this is my E01 I hope you can do this I hope you can do this this is an average value you can use later rule to tell me now so I have two terms psi Hartree Fock 1 by r i j psi Hartree Fock minus psi Hartree Fock sum over one particle operator V Hartree Fock I psi Hartree remember again this is a coordinate of electron so can you write this later rule remember this later rule is the first one that we do it type A average value so tell me this first part half of half of A B I am calling it instead of i j A B chi A chi B anti symmetrize chi A chi correct do this part this is the one electron part like sum over H of I you did how did I do sum over chi I H chi I correct so write it sum over A chi A V Hartree Fock chi A that is the first thing I will write because V Hartree Fock is my one part is it agreed everybody agrees so this is a one particle operator so I will take its eigen function its average value sorry its average value with respect to each of the spin orbitals A equal to 1 to n I am explicitly writing A and B just not to confuse this I and J okay so I write chi A V Hartree Fock chi A and then the coordinate can be anything 1 1 1 whatever does not matter then what will I do I will I will expand V Hartree Fock from here okay now if you expand V Hartree Fock you already have let me let me write this in greater details in actual integral form so what is chi A V Hartree Fock chi A so this is equal to integral chi A star 1 V Hartree Fock 1 chi A 1 data 1 agreed now I will expand chi A V Hartree Fock by this expansion this is my V Hartree Fock okay I will bring in another sum of the orbitals so this is equal to integral chi A star 1 V Hartree Fock has one more integration over d tau 2 so that will anyway come but before that another chi J which I now call chi B sum over B please remember that only J I am replacing by B to be more consistent so B equal to 1 to N spin orbital chi B 2 1 by R R I is now R 1 1 by R 1 2 1 minus P 1 2 that I should be replaced by 1 now and then chi B 2 so I have this chi J 2 chi B 2 and then this is this is a this is a V Hartree Fock so V Hartree Fock started from here correct and then I have chi A 1 then d tau 2 is a part of V Hartree Fock and of course d tau 1 is something that I have to do okay sorry yeah chi B star yeah I can shift it around and I can write this as sum over B equal to 1 to N chi A chi B anti-symmetry is chi A chi correct by just shifting this around that is okay because this is not exchange this is just that I am writing one first and two first it does not matter whichever I write right so this becomes chi A V Hartree Fock chi A now what I have to do I have to sum over A remember my this quantity E naught 1 this minus this right I am doing by a letter rule so your final E naught 1 is half of sum over A B chi A chi B anti-symmetry chi A chi B minus now there is a further summation A comes and of course summation B is already here so A B chi A chi B anti-symmetry it is exactly the same thing except the half factor right so you get finally minus half sum over A B chi A chi B anti-symmetry chi A so that is your E naught 1 so in a very long form E naught 1 is that same repulsion energy with the minus half minus so now what is happening remember in the orbital energy when I did that I had this part double counted that is why it was different from Hartree Fock energy sum of the orbital energy and that double counting is precisely going to be subtracted when I add E naught 1 and hence I am getting E Hartree Fock you remember my E naught 0 plus E naught 1 is E Hartree Fock E naught 0 was sum of the orbital energy which is actually E Hartree Fock double and this part double counted this part is actually includes half so now I have subtracted this to restore sanity so I get back E Hartree Fock okay so I just want to complete this again instead of I j I will write A B so it makes it easy so epsilon A if you remember was chi A H chi A and this is something that I have given in this papers also plus sum over B chi A chi B anti-symmetry chi A only sum over B here many people have got confused with this epsilon A also orbital energy is a number so it must be two electron integral but no sum over A because I am writing for a specific spin orbital this is actually obtained this is nothing but chi A F chi A correct if I do chi A F chi A you will easily get this because B is the incomplete integration but this completes the integration over dita 1 dita 2 so I have the orbital energy so if I do sum over E A E A which is my E 0 0 you can see this is sum over A chi A H chi A plus so this sum over A is here sum over A B chi A chi B correct which is different from E Hartree-Fock whether the half factor is missing here now when I add E 0 plus E 0 1 this minus half will come okay and you will get back sum over A chi A H chi A plus half A B chi A chi A this 1 minus half would give me the plus half and that is exactly the Hartree-Fock energy okay so this is a very long way to derive remember I initially told you very quickly because you just add H 0 and V you get H but if somebody is not convinced show me by really is later rule integration you can show it no problem okay you can actually integrate and show that the result that we got is exactly stands true and of course if I I have done just for a F 0 H 0 plus V do H 0 prime plus V prime does not matter we just scale add something and subtract something anyway that is much easier to see okay so it is a long algebra if you do it this way because I am first calculating this calculating some of the orbital energy then I am adding then I find that what was missing here from E Hartree-Fock this half factor that is comes back because this contain minus half this comes because of the following reason that you have first the 1 by R 1 2 part which is which actually has half then when I subtract the V Hartree-Fock it does not have the half so you get a full minus 1 so you get a minus half so you remember the same term is coming in various ways this chi A chi B anti-symmetrize originally I had E Hartree-Fock as half of this E 0 1 is half of this minus 1 of this to gives me minus half add that to E 0 0 I get back half it is just really playing with 1 and half okay plus half minus half so it is nothing it is a very simple algebra but make sure that you are convinced and I would put it the way I put it that is E 0 0 is size Hartree-Fock H 0 psi Hartree-Fock E 0 1 psi Hartree-Fock V psi Hartree-Fock add that you will get Hartree-Fock energy okay so that is easiest way to understand but I think this long expression is important because you learn that whatever slatter rule you have learned you know you can apply and get the same results okay.