 So, we will start on perturbation theory. Those who have already done 4 to 5 must have gone through a detailed perturbation. We will have a short review of the perturbation theory. But let me first tell that this perturbation theory can be used in quantum mechanics, in physics, even mechanics, engineering problems. It is a very general mathematical theory. It can be used in various fields. In quantum chemistry also, we can use it even to solve one particular problem. For example, if I give you a harmonic oscillator which is perturbed by an anharmonic force field, let us say x cube, then that x cube can be taken as a perturbation and you can calculate energy in what is called the orders of perturbation. So, it is not that perturbation theory is used for the improvement of Hartree form. It can be used in several problems. Right now, what we are going to do is to use perturbation theory to improve Hartree form. So, that will be our target. Can you look at perturbation theory as an improvement to Hartree form? And of course, when you say Hartree form, we have incidentally, you wanted to give the paper, did you give yesterday? So, can you give it to today with your doubts? To improve Hartree form for the many electron problem. So, I just want to revise the perturbation theory very quickly. I will not go through the order analysis right now, but let me tell you how the perturbation theory starts. It starts with a separation of the Hamiltonian into what is called a zeroth order Hamiltonian and the rest. Now, the idea of a zeroth order Hamiltonian comes from the following that it says that the zeroth order Hamiltonian must be such whose entire solutions, so let us say I am calling it psi i 0, must be known. Entire eigenvalue spectrum must be known. Of course, the underlying approximation assumes that I do not know the eigenfunction of the Hamiltonian. That is very clear because otherwise I will not do anything. So, I do not know the exact eigenfunction of the Hamiltonian, but I want to split the Hamiltonian into two parts. One part is a significant part H0 whose entire spectrum must be known. That is one of the preconditions of using perturbation theory. So, if you remember and then the rest is perturbation. So, V is perturbation and H0 is the dominant part of H. So, H0, I will lose the word it right now is the dominant part of the H, which means these solutions of H0 at least for the ground state is a dominant contribution to the ground state solution of H. So, let us assume that H has the solution which is psi i E i psi i and I order them as i equal to 0, 1, 2 etc. Such that E0 is less than equal to E1 etc. Similar ordering I can do here saying that E0 0 is less than E1 0 less than E2 0 and so on. Then what we are saying that since H0 is the dominant part of the Hamiltonian E0 0 is a good approximation to E0, psi 0 0 is a good approximation to psi 0 and we are only talking of ground state perturbation theory. So, we will not bother about the other approximations. So, that is the meaning of the dominant part of the Hamiltonian that the solutions of H0 are pretty good as far as the ground state is concerned, at least for the ground state is concerned. So, E0 0 and psi 0 0 are good approximations to E0 psi 0 respectively. Since we are only talking of ground state perturbation theory, we are not bothered whether correspondingly E1 0 is a good approximation to E1 or not. I mean you may also argue E1 0 is a good approximation to E1 and so on but at this point we are not bothered because we are looking at only ground state perturbation theory. I can only comment at this stage that the E1 0 is an approximation to E1 but it will not be as good an approximation to E1 as E0 0 is to E0. So, that is the only comment I will make. Beside that I will not worry about it. So, we must be able to write the Hamiltonian in this form that is our first job and then we will see the energy corrections and all that. I will actually since some people have not done perturbation theory, I will revise it again very quickly but that is when we require to do it. So, right now my first task is to have a Hamiltonian split into H0 plus V. Now, when you do that we notice that we have a wave function which is a pretty good approximation to the exact wave function. We already have a wave function. What is that? That is a Hartree Fock wave function. We have already done Hartree Fock. So, can I make Hartree Fock as a ground state wave function of some Hamiltonian? So, can Hartree Fock that we have done determinant be a Psi 0 0. So, I am now asking the reverse question instead of first finding H0. I already know that there is a good approximation to Psi 0 which exists. So, I already know that the Psi Hartree Fock can become my Psi 0 0 because it is a good approximation to the wave function although single determinant is not exact etc. It is a good approximation and from our experience we know that the E0 0 is 95 to 97 percent of the total E0. So, they are pretty good. They are higher than E0 but they are almost good. So, I can write Psi Hartree Fock as Psi 0 0 and somehow I should be able to get then H0. So, now can I rethink? Can I do a reverse manner that Psi 0 0 which is Psi Hartree Fock is not an Eigen function of H? I think all of you agree. Otherwise, we will not be here to discuss anymore. The quantum chemistry will be over. So, it is not an Eigen function of H but then we have to find out whose Eigen function it is because that will be my H0. Is it clear? If I know, if I can identify Hamiltonian whose Eigen function is Psi Hartree Fock, then I can think of that as a H0 and then I will see whether the entire solutions of that H0 is available or not because that is another condition. So, can you guess an H0 whose Eigen function would be Psi Hartree Fock, ground state Eigen function? Let us see. Who can guess? You have already done the Hartree Fock problem. Please re-look the Hartree Fock. Psi Hartree Fock is a slater determinant of spin orbitals. Those spin orbitals are Eigen functions of the Fock operator. So, let us say I am taking canonical Hartree Fock. Now, can you tell me what H0, what many particle H0 will give Psi Hartree Fock as a ground state Eigen function? Who can tell? Think in the reverse manner. We had a non-interacting theorem. Remember, we have a non-interacting theorem in which it says if an operator is a sum of some operators whose solutions are known, one particle operator, then the total Eigen function of that operator is a product of those one particle operators, one particle functions. So, now do a reverse analysis. I have Psi Hartree Fock which is a product of some spin orbitals, chi 1, chi 2, chi n. I know a one particle operator f of 1 whose Eigen functions are these chi i's. Now, can you tell me the operator H0 whose Eigen function will be Psi Hartree Fock? It is a very simple question. Sum of chi i, sum of the Fock operator. I hope all of you are able to think properly. Is it agreed? Because now we have a sum of the Fock operator. They have the Eigen functions of chi i. Hartree Fock is a product of that. So, of course, the Hartree Fock will be an Eigen function of H0. So, I can identify at least one H0. There may be others whose ground state solution is Psi Hartree Fock and what will be the Eigen value? E sum over E i. Is it clear to everybody? So, basically I am using the non-interacting theorem that H0 is sum of the Fock operator. Solution of the Fock operator is known. So, the Eigen function of this operator would be a product of this spin orbital. Of course, instead of product I am taking anti-semitic product, does not matter and the Eigen value will be sum of the orbital energies. Sum of the energies. Is it clear to everybody? So, now what I am suggesting is that we have got a Psi Hartree Fock which is not an Eigen function of the actual Hamiltonian. But it is an Eigen function of some other Hamiltonian H0. So, it is interesting that I have got a model Hamiltonian H0 whose Eigen function is Psi Hartree Fock with the Eigen value sum of epsilon i. Of course, this is a problem that we are not interested, but we have found a H0 to which Psi Hartree Fock would be an Eigen function. And we feel that since Psi Hartree Fock is a good approximation to Psi 0, exact Psi 0, the H0 is a good starting point for perturbation theorem. This H0 is a good starting point to do perturbation theorem. Is it clear the argument? So, why am I choosing this H0? Because this H0 has an Eigen function, ground state Eigen function, which is Psi Hartree Fock, which is a good approximation to exact Psi 0. Hence, we believe that this H0 can be a starting point to do a perturbation theorem. However, note one very important point here that the Eigen value of this problem is however not the Hartree Fock energy. And I want to repeat that because Hartree Fock energy and I have said this many times, the Hartree Fock energy has been defined as Psi Hartree Fock, 8th Psi Hartree Fock and which I have said many times not the sum of the orbital energy. In fact, it is sum of the orbital energy minus that over counted terms, half of whatever, Ig Ig at the symmetries. So, this is something that you must note, but I have no choice. You may wonder, can I get a H0? And I will come to this point whose Eigen function is Psi Hartree Fock and whose Eigen value is the Hartree Fock. That would be even better. So, I will come to that point if it is possible or not because that would be even nicer. At this point, I have something which is not the Hartree Fock energy, but let me start living with it first. So, what will I call this quantity? Sum of the epsilon i, i equal to 1 to n is by my definition here would be E0 0 because my definition of E0 0 is the ground state Eigen value of H0. So, what is my ground state Eigen value of H0? This is sum of the epsilon i which is actually E0 0. So, E0 0 is actually not the Hartree Fock energy unfortunately. So, that would have been a very nice idea that Psi 0 0 is Hartree Fock, E0 0 is Hartree Fock energy that would have been nice, but as I see so far it is not the case. E0 0 is actually just the sum of the orbital energy which is somewhat different from the Hartree Fock energy, but let us start living with this and see where we get. So, this H0 becomes a very nice starting point to do a perturbation theory. I have to of course remind you that there are another very important condition for doing a Hartree Fock that is I must have a complete solution of H0 not just a ground state solution. The ground state solution should be a good approximation to the exact ground state, but in addition to that and all solutions of H0 must be known. So, I have to now go back and see do I know all the solutions of H0. So, if you look at it very quickly the F has many solutions, in fact let us only talk of canonical Hartree Fock, F has many solutions N of them which are lowest energy are actually used in Hartree Fock to construct ground state, but there are other unoccupied orbitals through which I can construct others later determinants which I had discussed that all MCN determinants. If you now analyze this since all those spin orbitals are eigenfunctions of the Fock operator by very definition all those determinants are also eigenfunctions of H0, because all determinants are anti-symmetrious product of some set of spin orbitals occupied unoccupied altogether. So, let me write that down. So, I have MCN determinants, please note that the Hartree Fock itself I have not solved exactly, I have solved in some basis, m number of spin orbitals I have got whatever it is, actual Hartree Fock is infinity, infinite solutions. So, in fact everything I must also mention everything that we are talking is actually in a basis yesterday I mentioned this correlation energy as E0 minus E Hartree Fock or E Hartree Fock minus E0 depending on your sign. The point to note that this is exact energy this is Hartree Fock energy, but what is Hartree Fock energy? There is we are we cannot get exact Hartree Fock energy, because exact Hartree Fock energy will be obtained if I solve this exactly if I solve this exactly which I cannot solve, because for molecule I am introducing a basis. So, even the Hartree Fock has been approximated, even the Hartree Fock has been solved in an approximate manner. So, everything finally that we discussed is in a basis. So, this is in a given basis let us say m basis, this is also in a given basis, this is also in a given basis. So, when you say correlation energy that is also not a sacrosan number. In for a same molecule with a 20 basis whatever will be the correlation energy will be different from a 100 basis calculation, because all the numbers will change E0 will change, because when you do full CI your number of basis will number of determinants will increase if m increases. So, everything will change, but if you want exact values then all this m in the limit m tends to infinity everything will be exact. So, you have to go to the limit m tends to infinity in that basis then everything will become exact, then you will have exact energy exact Hartree Fock energy exact correlation energy. So, that is something that we do not do. So, it is important to realize that everything that we have been talking is in a finite basis. So, that is something that we have to leave you. So, with this finite basis we have only finite number of solutions of H0 which are all MCN. All MCN determinants are solutions, because they contain orbitals or spin orbital which are Eigen function of the Fock operator and my H0 is sum of the Fock operator. As long as H0 is sum of the Fock operators any product which is formed or any anti-symmetric product which is formed out of those Eigen functions of F are exact functions of H0. So, let us analyze these MCN determinants. One of the determinants is psi Hartree Fock. We have already said this. One of the determinants is psi Hartree Fock. What is the other class of determinants? Other class of determinants can be constructed in the following manner. Excite one electron from n occupied orbital to m-n virtual orbitals. So, I have n occupied orbitals, m-n virtual orbitals. I will give a diagram to show this and I call that psi AR where A is occupied orbitals, R is a virtual orbital. So, all possible psi AR, not one of them. Similarly, I can have two spin orbitals differing, A B to RS. I hope it is clear from the nomenclature and then you can have A, B, C, RS, T and so on, triples up to all n. How many I can excite? I can excite n. When all n have been excited that is the final thing. I cannot do any more. So, all MCN determinants I can categorize as first Hartree Fock with reference to Hartree Fock, singly excited, doubly excited, triply excited and so on up to n-toupley excited. So, let me write down those MCN determinants. So the MCN determinants, so let us assume that the Hartree Fock provides m solutions of chi i. That is the first assumption. Finite number, m solutions of chi i such that f of 1 chi i 1 is epsilon i chi i 1. So, that includes my occupied as well as virtual orbital. These m solutions, so m number of spin orbitals are now grouped into a subset of what I now call occupied n spin orbitals. Remember, these are one particle function. So, out of n spin orbitals I can actually have only one determinant for an n particle problem. So, those occupied n spin orbitals and the rest which I call m-n unoccupied or sometimes they are called virtual spin orbitals. So, I have grouped them into two subset. One of them what I call occupied. Occupied by definition are the ones which constitute the Hartree Fock determinant. So, my Hartree Fock let us say provides m solution. m can be as large as possible depending on your basis whatever. So, then I have got n of them. So, let us say this is my n which constitute my Hartree Fock. So, together this is chi Hartree Fock. Note that chi Hartree Fock is not a orbital. It is a product of orbitals, a spin orbitals. So, all together is chi Hartree Fock. Then I have got m-n. So, rest of them are all virtual. So, this is a line. It is an imaginary line with a boundary between occupied and virtual. That is all. This is not an orbital line. So, many in solid state actually this is called the Fermi level. It is called the Fermi level in the solid. This is called the gap between valence and conduction band. So, this is the Fermi level. So, anyway in the quantum chemistry we do not call it Fermi level, but this is somewhere in between. Let us see. It does not matter. So, we then we are saying that all since there are m-spin orbitals I can construct m-c-n determinants. One of them is Hartree Fock and the rest can be generated by singly excited, doubly excited etc. So, lifting one electron from here anywhere, two electrons from anywhere here and so on. So, that is one way to look at it and maximum you will have all n lifted here. So, let us calculate the number. If you look at all m-c-n determinants they must be categorized only into these parts. With reference to Hartree Fock all occupied, one of them missing going to up, two of them going up, three of them going up, four of them going up and so on because there is nothing else left. So, if you calculate this number is of course one, how many psi ar is there? How many psi ar is there? n into m-n. Then I have psi abrs. So, psi abrs will be how many number? n-c-2 into m-n-c-2. Is it clear to everybody? Please those who are not understanding if you are in the back I want to hear your voices. So, how many ways I can lift two out of n? n-c-2 because I cannot lift from the same spin orbital. Obviously, two electrons are not there, Fermi principle. How many ways I can put them into m-n? This is m-n in number. This number is m-n, m-n-c-2. So, total number of determinants that I can find out is n-c-2 into m-n-c-2 and so on. So, triples will be n-c-3, m-n-c-3 and so on up to n-2 fully excited determinants. So, if I write this whole thing, there is a very nice mathematical combinatorial sum. The first number one can be written as n-c-0, m-n-c-0. So, it means by the same language or mathematics, I am not lifting anything and you can see this is 1, this is 1. So, 1 into 1 is 1, plus all singly excited, n-c-1 into m-n-c-1. I am writing everything in terms of combinator. Then I have plus n-c-2, m-n-c-2 plus n-c-3 and so on, m-n-c-3. This is number of triples and eventually I will go up to n-c-n, m-n-c-n and I will stop here. This is all the numbers that I have and the sum total of this number is how much? m-c-n. This must be because all the m-c-n numbers, I have grouped them in Hartree-Fock, which is just 1, plus all singly excited, all doubly excited, all triply excited and so on up to all n-toupley excited. All n-toupley excited, this is again 1, it is just m-n-c-n. In fact, there was a combination problems if somebody has done permutation combination, remember in high junior school, high school, wherever you have done it. This can be very nicely written as sum over r, 0 to n, n-c-r, m-n-c-r equal to m-c-n. They used to ask proof, I do not know if you remember, somewhere down the line you might have done. They used to ask proof this. Actually today you realize that the CI is just this we are doing actually. I am categorizing this in all singly excited, doubly excited. This was a math that was there. You can see the simple way to prove actually the way we proved it. Take this m, divide into two subsets and say that the entire combination is m-c-n and this is the different grouping. That is one way to prove this actually, this particular problem, but I think you have done it in many different ways. I do not know if you remember, at least I remember our time we have done it. Somewhere in the high school it was there, some such question. You have done more difficult things than CI. That is all I want to tell you. You have already done much more difficult things than what we will do. What we are now trying to say that CI is nothing but taking a linear combination of all such determinants. That is your full CI. You get what is called exact function, exact energy in this m basis. Of course we are never getting anything exact, remember. Whenever I am calling exact, it is in a basis, finite basis. Limit goes to m tends to infinity and that will go towards exact, only in that limit. Otherwise nothing is exact. This number of determinants can be categorized in this manner. Now our task is to come back to the problem. Do I know all solutions of H0? So I come back to the problem. Which means each of the determinants that I have written here must be an eigenfunction of the H0. That same H0. And quite clearly the answer is yes because they are constructed out of the same set of spin orbitals which are linear combination of the Fock operator. So obviously they will all be eigenfunction of this thing. So let me now write down what is the value. So let us say take one function H0 psi ar. This is an eigenfunction of H0, psi ar. Now can you tell me what is the eigenvalue? So how will you quickly write it? Some reference you have to use. So it is sum of all these energies of all these spin orbitals. That is clear. In the Hartree-Fock I had n of them. One of them is missing. A is missing. R is added. So the best way to write is the E00 which is sum of the orbital energy minus epsilon A plus epsilon R. This is nothing but sum of the orbital energies. Please remember this is nothing but sum of the orbital energies between i equal to 1 to n which are occupied in this occupied set. And then from there one of the A is always occupied. So again I remind you our notation at this stage. I am going to use A, B, C, etc. as occupied and then may be P, Q, R, S, T, etc. as virtual because now the notation has to be clear. So this is one of those n orbitals from where I have lifted the electron that is subtracted and to one which is added a new one virtual orbital that is added. So this becomes my Eigen value of this for this determinant. Eigen value of H0. And I can find out so on. Psi, A, B, R, S, all of them I can find out with reference to this. This is a very convenient reference because I do not want to write everything again. I will get lost. So this I know has all the Hartree-Fock orbitals energy summed up then subtract whichever has to be whichever are in the down. So if it is A, B, R, S, subtract epsilon A, epsilon B, add epsilon R, epsilon S. So that will become the energy of the WXRA determinant whose energy for H0 differs from psi Hartree-Fock energy by these numbers. By these numbers only the change of the orbital energies because for the ground state psi Hartree-Fock E00 is already sum of the orbital energies. Note that again please remember that E00 is not the Hartree-Fock energy however. So where is Hartree-Fock energy? We will see later. Everything is in talked in terms of orbital energies or sum of the orbital energies. There is nothing called Hartree-Fock energy yet. So please remember that. We will come back to that point little later. So I am now convinced that each of those MCN determinants is an Eigen function of H0 whose Eigen solutions are known completely. So H0 can be a very good starting point to do perturbation theory. So that is the first thing that I this H0 is a good starting point to do perturbation theory. Is that something that I want to first convince you? Then of course we will have to set the perturbation theory itself. How do I do perturbation theory? So first I will make a quick overview how the perturbation theory starts. So the perturbation theory starts with the following imaginary Hamiltonian H of lambda as H0 plus lambda. It is always good for those who have done it anyway. Remember good to have a small revision. Yes in that M basis of course. M basis you have only MCN number of determinants. Each of them can be written in some form like this and all of them Eigen solutions are known. Because there is nothing else. MCN determinants complete the full space. There is no other. So all of these determinants are Eigen solutions that it is not. And I am now if you are arguing that all these solutions yes, no it is only in this M basis. Everything is in the M basis and if M becomes larger then more solutions will come out. But then I will be able to find out because there will be also Eigen solutions of the Fock operator. So the perturbation theory starts with an imaginary Hamiltonian of this kind. When I have been able to separate a dominant part of H0 then you can very easily see that if lambda is 0 this particular revision I am doing little bit quickly because many of you have already done it. At lambda equal to 0 this fictitious Hamiltonian becomes the same as the 0th order Hamiltonian. At lambda equal to 1 this becomes a full H and of course at lambda different values they keep changing between H0 and V. So I can have lambda 0.5, 0.1, 0.2. So I can switch on the perturbation from lambda equal to 0 to lambda equal to 1. So lambda equal to 0 there is no perturbation and lambda equal to 1 I have got a full perturbation and I got the full Hamiltonian H. So in some way if you start from this switch from lambda equal to lambda equal to 0 to lambda equal to 1 you get the effect of the full perturbation taking place. Then what we do is to analyze the solutions of this H lambda. So if you look at the solutions of H lambda I can write in a very similar manner it will be psi i lambda E i lambda psi i lambda and I am particularly interested only in ground state. Again everything is ordered from i equal to 0, 1, 2, 3, etc. So I am particularly interested in H of lambda psi 0 lambda E 0 lambda psi 0 lambda and note when lambda equal to 0 the H of lambda becomes H0 and the entire solutions are known. Ground state as well as other solutions are known for lambda equal to 0 only but not for other values of lambda. So we will exploit these to find out what we can do. For the first order energy and that is what I will right now do it is quite easy to analyze this. So let me do the following. Write psi 0 lambda in general as a power series in terms of lambda. So at lambda equal to 0 its eigen solution is psi 0 0 because the operator becomes H0 its eigen function is psi 0 0. For all those I am actually addressing to them but for all those if you have doubt please raise hands you have done it already. Then I do a power series expansion. So lambda times some function which I call it psi 0 1. So this is basically a first order correction to this wave function and then I have lambda square psi 0 2. There are many ways of writing this. You can write a Taylor series expansion so that you put 1 by 2 factorial and so on. There is a mathematical series expansion. Each of these are called order corrections. So it is a first order correction, first order with respect to lambda. This is second order with respect to lambda and so on. Of course eventually my final solution is lambda equal to 1. So all these corrections once known the lambda does not appear. Only the coefficients will appear you know half one third one third three factorial and so on. Even you need not write this. You can absorb this in terms of psi 0 2. If you write only power series lambda square psi 0 2. This is a Taylor series expansion but does not matter. I mean that is sometimes even easier to understand. In the same way I can write E0 lambda as a power or Taylor series. So I can write E0 0 which is the 0th order for H0. Lambda E0 1 plus half lambda square E0 2 and so on. What is important to realize is that each of these are also derivatives of psi 0. Second derivative first derivative and so on at lambda equal to 0. I think you can do this expansion easily. If you take a second derivative of this what will you get? 2 lambda into 1 by 2. So that will become lambda times just this. So this psi 0 2 will then this lambda will also go in the next derivative. So you have 2 times psi 0 2. So that 2 will cancel out with 1 by 2 factorial. So you have just psi 0 2. So you have psi 0 2 as second derivative E0 2 as second derivative of E0 and so on. So each of these can be related to the derivatives of psi or energy with respect to lambda. So in some sense the order corrections are basically nothing but derivative corrections. More often we actually like not to put these factors half. We just like to write as a power series because they are just unnecessary terms. They can be absorbed in the definition. Then it becomes much easier to analyze this. Up to first order it does not matter. When you come to second order we will try to remove this. But right now we do not have to bother. We put this into this Strodinger equation. The main Strodinger equation for lambda and expand for the ground state. So what will you get? You will have this as H0 plus lambda E. Then psi 0 lambda will become psi 0 0 plus lambda psi 0 1 plus the rest. Rest I do not require right now. I will just write the rest and while writing the rest I will have to take care. If I am doing Taylor series or power series I have to be consistent. Then the right hand side similarly E 0 0 plus lambda E 0 1 etcetera. Again you write psi 0 0 plus lambda psi 0 1 etcetera. So I write this equation. Then what I do is to equate terms on the left and the right hand side. And I will get equation in orders of lambda. So the first equation that I get is H0 psi 0 0. Remember I am now first getting equation for lambda to the power 0. So I am extracting on the left and the right hand side terms which are independent of lambda. So I have only one term H0 psi 0 0. Rest have them all lambda. And the right hand side also I have got one term which is E 0 0 psi 0 0. This is no surprise because this is exactly the eigen value equation of H0. So it is a big deal. I have not got anything actually. But it is good to know that I have got what I know from this equation. What is important is the first order equation. So that is the first order perturbation equation. So that will now be H0 into psi 0 1. Note that I am not writing the lambda H0 into psi 0 1 plus what is the other term? V into psi 0 0. V into psi 0 0, good. And then on the right hand side I have similarly E 0 0 into psi 0 1 plus E 0 1 into psi 0 0. So this becomes the first order perturbation equation which will determine potentially psi 0 1 and E 0 1. That is something that I do not know. So I will get first order correction. And eventually I can write a second order, third order and get all orders corrections. So it is not really a big problem to do this. So what we want to do is to solve this first order equation. Before I solve this equation, I want to write this in a manner which I like better. That is to write this part on the everything on the left hand side. So H0 minus E 0 0 psi 0 1 plus V minus E 0 1 psi 0 0 equal to 0. You agree? So I have taken these two terms together. I have taken these two terms together. The reason I like this better because I can write it very easily from the original Schrodinger equation which is H minus E psi equal to 0. Original Schrodinger equation is just H minus E psi equal to 0. I am writing first order equation. So how do I write first order equation? First order of this into 0th order of this. Plus 0th order of this, first order of this. My total order has to be 1. So both cannot be first order. Then product will become second. So if you see this, the 0th order of this is H0 minus E 0 0 into first order of this. First order of this is V minus E 0 1 because first order of H is H0 plus lambda V. That is V. So V minus E 0 1 into psi 0 0 equal to 0. Actually you should be able to write second order onwards. We will do that in the next class. But it is easy to write it in this manner. So I just thought I will reflect you. Of course you could have argued that I can say H psi equal to E psi and do the same thing on the left and the right hand side. It is the same thing except that somehow I like it in this manner. I do not know why. So I have to now solve this equation. So for the solution to the first order energy is actually give away. It is a very cheap solution. Very quickly you can get and that is what I will do right now. And when you go to the second order I will come back to these equations again. So the way to solve this equation is the following. First you take this equation and project these with what is called psi 0 0 star and integrate. So how do you do this? You multiply by psi 0 0 star which is already known and integrate. So when you do the integration you have a scalar product H0 minus E 0 0 into psi 0 1 plus psi 0 0 V minus E 0 1 psi 0 0. Now you can quickly see that this part is 0 because psi 0 0 is an eigenfunction of H0 with an eigenvalue E0 0. So if you take the conjugate of that equation, you have already an equation which says H0 minus E 0 0 psi 0 0 equal to 0. So if you just take conjugate of this, this becomes 0 because H0 is a Hermitian operator. Of course I must say that when you split H as H0 plus V, H0 is Hermitian. Of course that is an important condition that you should remember. So it is a Hermitian operator. So this becomes 0. So I am only stuck with this final part. This whole thing is of course equal to 0 because right hand side is 0 anyway. So what I have now is psi 0 0 V psi 0 0 equal to E 0 1 because remember E 0 1 is a number and this is our normalized unity. So I just get psi 0 0 V psi 0 0 scalar product equal to 1. I will come back and make lot of other comments on how when I go for psi 0 1, how psi 0 1 is normalized to psi 0 0 and so on and those who have already done may recall intermediate normalization. But the point is I do not require at this point just for the first order. First order correction I do not care. They need not be normalized because that kind of terms are not coming. It will come for the first order correction to the wave function not for the energy. So up to first order correction to energy I really do not require to even assume that. So you can see that if I know the perturbation V and just take the expectation value of psi 0 0 I get E 0 1 and this side is already known. V I know by definition I have split H is H 0 plus V and this I already know of course because I have started with H 0. To get E 0 1 I really do not need to do anything. So now I will do this for my Hartree-Fock perturbation. So note that what is psi 0 0 psi Hartree-Fock? V psi Hartree-Fock equal to E 0 1. Let me write it down E 0 0 once more. What is E 0 0? Psi Hartree-Fock H 0 psi Hartree-Fock. Do you agree? Note that sorry note that Psi Hartree-Fock E 0 0 was an eigen function of eigen value of H 0 with psi Hartree-Fock as eigen function. So I can also write this equation. So if you remember H 0 psi 0 0 which is psi Hartree-Fock is equal to E 0 0 which is sum of the orbital energy psi Hartree-Fock. That is a ground state solution for H 0. So if I take an average value of H 0 with respect to psi Hartree-Fock. So integrate after projection with psi Hartree-Fock star then this psi right hand side will also become psi Hartree-Fock psi Hartree-Fock which is normalized unity so it is E 0 0. So this is another way of writing E 0 0 which is a ground state solution of H 0 and I can write it in this star. So if I put this together E 0 1 becomes the expectation value of the perturbation operator and the E 0 0 was the average value of H 0 both with respect to psi Hartree-Fock correct. Is it agreed? Because that is a trivial any eigen function solution any eigen value solution will always satisfy this. So that is a trivial nature. Add these two. If I add these two what will you get? You will have the operator added right everything else is the same. So you have psi Hartree-Fock H 0 plus V which is H psi Hartree-Fock what is that? Psi Hartree-Fock star H psi Hartree-Fock integration what is that? E Hartree-Fock remember we are struggling to see where is E Hartree-Fock. If you add this my left hand side is E Hartree-Fock and that is nothing but E 0 0 plus E 0 1. Remember we said E 0 0 is not E Hartree-Fock and we are very unhappy that my 0th order wave function is Hartree-Fock wave function but 0th order energy is not a Hartree-Fock energy. But now I am contending that with that H 0 perturbation if I do the first order correction and add the first order correction then I will get the Hartree-Fock energy because if I add H 0 plus V that is H and psi Hartree-Fock H psi Hartree-Fock is E Hartree-Fock. So together I get the Hartree-Fock energy yes no no no no I have not made any such statement. Where is the first order corrected wave function required here? So the question that is asking is the question that I asked can you find a Hamiltonian for which psi Hartree-Fock is an Eigen function and E Hartree-Fock is an Eigen value? Obviously the solution that I have written for H 0 it is not it is not. So I will come back to that how to do that in the next class yes that is something that is there but even if I have not been able to do I can recover Hartree-Fock energy by adding E 0 0 plus E 0 1 which means after doing first order energy correction I have actually got Hartree-Fock. So till now I have done nothing it is just to give you a comfort level that the Hartree-Fock is there somewhere it is not lost. So any improvement to Hartree-Fock energy any correction any correlation energy calculation will now start from second order and that is the reason you find that the first correction that you talk of is MP2 you never talk of MP1. So I just thought first I will tell you because the perturbation starts from a different point of view where the 0th order energy is not the Hartree-Fock energy 0th order energy is only some of the orbital energy after I added the first order correction I get the Hartree-Fock energy and hence the correction to Hartree-Fock energy which is what I am looking for actually starts from the next order. Now the question that I asked is that can I have H 0 for which Hartree-Fock energy is a 0th order solution. In that case my correction will start from first order I will come back to that question. So you may still say why there is no such theory where something dash 1 is a correlation correction you know we will come to that later. So I hope it is clear I have not written explicitly how V looks like we will do that in the next class how does V look like remember V is not just 1 by Rij what is V? V is V is H minus H0 what is H you should be able to write H is sum over H of I plus the average operator or the Fock operator let me call it now V Hartree-Fock. So this becomes my H0 sorry I should have written it later. So sum over H of I first I am writing H plus 1 by Rij correct that is my H minus H0 which is sum over H of I minus sum over Hartree-Fock potential I call it Hartree-Fock potential and this Hartree-Fock energy or Hartree-Fock operator will include the Coulomb and the exchange operator. So integral phi for Kij star 2 1 by R1 to all that is Kij 2 d tau 2 that is all stuck here in the Hartree-Fock potential. So if you see this part will get cancelled but your V is actually not 1 by Rij sum of 1 by Rij V is sum of 1 by Rij minus this Hartree-Fock potential and that is understandable because you have already added to H to your original sum of the one electron term this Hartree-Fock term in the H0. So when I do V your 1 by Rij must be reduced by the same term otherwise you will be over counting that Hartree-Fock potential. So I will write down formally what is V and then we will try to analyze what is E01 because although I know that together it is E Hartree-Fock I must say what is the difference of these two and of course the difference of these two you must now know already because you know that E00 is sum of the orbital energy. You know the difference from orbital energy to Hartree-Fock that must be E01 but I must get it by the application of this later rule also by allowing that V to act and so that everything should match. You already know how to get it. E Hartree-Fock is E00 minus half of I g, I g anti-symmetry. So you already know how to get it and we must get the same result by expanding this and adding to E00 we must ensure that you are getting Hartree-Fock energy. To do that I have to do a detailed analysis of V so I will start from the V but at this point I just want to mention that if I do a first order correction to the energy add that to 0th order actually get Hartree-Fock energy and hence any correction to Hartree-Fock energy any correlation energy contribution starts only from second order. This particular separation of Hamiltonian into H0 plus V was actually first done by Meuler and Placer. In the next class I will give the actual reference for those who are interested can also read the Meuler and Placer's original paper and there are lots of variations to this. One of the variations is what we discussed. Can I have H0 whose Eigen function is high Hartree-Fock whose Eigen value is E Hartree-Fock and what happens then for E01? Then I should have a correlation contribution from E01. So why is it not being talked about? I will say that in fact it is possible to have such H0 but we will see what happens to E01 in that case. So there are lots of variations to Meuler plus H but this is the beginning where H0 is just some of the Fock operator V is the rest so basically V is just the sum of 1 by Rij minus the Hartree-Fock potential. This is what I call Hartree-Fock potential in the Fock operator. So basically I have written one particle Fock operator as H plus V Hartree in a very simple manner. So that V Hartree-Fock has all the terms from other case so that has to be self consistently evaluated but if I write it in this manner this becomes my capital V and I will come back to this and evaluate this psi 0 V psi Hartree-Fock V psi Hartree-Fock and see that this actually gives me Hartree-Fock energy by even application of slater rule. I can use slater rule for this and you should get the same value that you know you should get. So we will come back to this.