 So, I think I just want to quickly revise the variation method. The three important theorems that we said is one is Euler theorem that is, if you do a variation of psi tilde A psi tilde by psi tilde psi tilde for all psi tilde and this is a loaded all means entire Hilbert space, then the stationarities of E tilde are the solutions of the Schrodinger equation. So, that was the first theorem. I did not prove it. However, this proof is not very difficult. We can do at some point of time. The point is that because of this, it is actually impracticable because we cannot really scan psi tilde over entire Hilbert space 2. Then we said that given any psi tilde, it is always greater than or equal to E naught. So, let me write this E naught i equal to 0 1 etcetera, where E 0 less than equal to E 1 less than equal to E 2 and so on. So, E 0 is the lowest of them. So, for any psi tilde A psi tilde, so for any arbitrary psi tilde, it is always greater. So, even in a restricted space, so this I do not have to do all now, restricted space variation of E tilde provides an approximate solution. So, the minimum of E tilde is the best solution of under the variation. So, that is what, so for this restricted space, the minimum is the best solution. However, this best solution can keep changing because your space may keep changing. When I say restricted space, you may decide to have one restricted space, I may have another restricted space. So, within that restricted space of variation, this is the best solution. So, this does not give, so there is nothing called a variational solution. That is a misnomer. Whenever you say this is the solution that is obtained by variation method, you have to also define what subspace of variation did you do. So, that is important, otherwise it has no meaning. One of the restricted subspace that we decided, we actually discussed is a linear subspace of variation, where we have k number of basis or k number of known functions. Whenever we say basis, remember these are known functions. So, these are of course, basis means not one electron function. These are the basis for the problem, which means n electron functions. So, for example, for a n electron problem, they are slater determinants. So, slater determinants are valid functions because they are anti-symmetrized. So, they can form a basis. So, let us say I have a k number of basis, which is what is the symbol that I used last time? That is b, dd. So, d. Then if I take my psi tilde as a linear combination of this basis, c i tilde d i, then my variation subspace is a linear subspace. So, let us say i equal to 0 to k minus 1 or 1 to k depending on how you want to write. So, variation is a k dimensional subspace. That is my subspace and the parameters are c i tilde, the k c i tilde. So, again this is a linear subspace, but this can vary. There are two parameters. One is the value of k itself, another is the basis itself. So, same k dimensional subspace, I can use a different basis. So, results will be somewhat different. However, if you take an infinite dimensional basis, then it does not matter what basis you use because infinite dimensional everything is exact. I hope you understand. I can choose any d i if this k was infinity, which means it is actually spanning all subspace, then it does not matter because in infinite dimension any function as long as they are linearly independent, their combination is exact. So, their combination will provide you another basis, but not so for a finite basis. In fact, if it is infinite dimensional, this will actually go over to the theorem 1, which is the Euler variation. So, that means essentially I will get exact results. So, then we said that this solution to this is an Eigen value equation where I consider a matrix of this H over this basis. So, I wrote down this c j tilde sum over j equal to E tilde c i tilde. So, this is what you get by doing the very process of variation. So, this is the parameter c i tilde, which are obtained by solving this equation. If you analyze this equation, it is a matrix Eigen value equation. Remember, my Hamiltonian was an operator. However, when I take this integral of this Hamiltonian between this known basis d i and d j, which are within this k dimensional, then what happens is that this becomes a number which depends on k number of i, k number of j. So, I can rearrange this as a matrix of H i j, where this is a row and this is a column. This becomes another column multiplied by a number times the column. So, this becomes a typical matrix Eigen value equation. If you notice matrix Eigen value equation as a matrix times a column equal to number times a column. However, if it is a k by k matrix, then it has k number of Eigen solutions, not one. You have k number of Eigen solutions. So, each of the Eigen solutions, I can write with some superscript. Each of the Eigen solution I write with a subscript and basis comes in a superscript. So, I will get typically E i in the basis k. So, this is how I label my Eigen solutions. So, these are my ith Eigen value in the k dimensional basis. Of course, if the dimension is k, then your number of Eigen values can be only k. So, 0, 1, 2, 2, k minus 1 in a k dimensional basis. Then we know that this is a matrix noted that not only of course the rule 2 exists, which means all these Eigen values are certainly greater than or equal to E naught, but there is a further this thing exists, which says that any E i in a k dimensional basis is always greater than or E i tilde is always greater than or equal to actually E i. So, these Eigen solutions which are approximate, they are not only greater than or equal to E naught, but they are actually greater than or equal to its corresponding exact Eigen solution. So, this is a very nice theorem which gives you an upper bound for the excited states. We further noted a given E i tilde k plus 1 basis which is generated by adding one member to the original k basis. Please remember not a different k plus 1 is always less than or equal to E i in k basis. So, this is another theorem that we saw and each of them remains an upper bound. So, however, when they go down, they will never cross their own corresponding states. So, that is the meaning and this was actually was the content of the separation theorem by the McDonald, the reference to which I gave the physical review 1933 paper, which says that between every two roots of the Eigen value problem lies one exact solution. So, this was the content of the separation theorem. Actually, from there you can derive both of these, but they will remain upper bound and as you increase, actually it is as you increase the basis, it will keep on decreasing, but it will never cross this because of this theorem and that was the content of the separation theorem and this is what we said and this is a linear subspace is a very important subspace and in a way, if you remember we looked at the form of the exact wave function, that was also a linear variation, a combination over determinants. So, if you want to get those coefficients, we are actually going to use this Eigen value equation. So, this is how we will get the coefficient of expansion. Did I also do the proof from here to here? So, we just projected to a given d i and we got the solution. So, I think that completes the variation method whatever I did in overview. This is a very important method. So, I think it is you know more you talk about it, better it is, you are always going to gain. So, remember when you do Hartree-Fock, we are actually going to do a variation method, not the linear variation. Of course, linear variation is C i, configuration interaction. So, when we come back to our old problem that we know that an exact wave function can be written as a linear combination of all determinants and of course, we are talking of any electron determinants that can be formed from a complete set of one electron, which are basically spin orbitals. These functions can be spin orbitals. So, they are very often called the spin orbitals. So, this is a statement which is basically it says that it is the full configuration interaction. So, please understand the statement. I have first a one electron function, which are the basis of one electron problem. So, it can be any one electron problem because you remember the Hermitian operator Eigen functions are always complete set. So, I can take any one electron problem. Note also that I am not explicitly stating, but these functions are all orthonormal because they are they will usually come from the Hermitian operators. These basis are also orthonormal similarly. So, everything is an orthonormal basis that we are talking right now. So, what we are saying that from this one electron basis you form first for any electron all determinants. So, of course, this complete set is usually infinity again, but let us say that we have a pseudo complete set which are M in number. So, this spin orbitals are M in numbers. So, then we can generate total number of determinants. We have already seen this which are MCN. These MCN determinants are now the basis for the N electron problem. So, again please distinguish the basis that we are talking of. We have a one electron basis. This is a one electron basis and these are N electron basis. Quite obvious because this is only one electron. So, one function so N electrons cannot be put there. Here it is a determinant. So, I have N spin orbitals and I can put N electron. The Pauli principle or the anti-symmetry forbids me to put N electrons in one orbital. So, of course, I need N and that has to be a determinant. So, these determinants are now N electron function and quite clearly they are the basis for the N electron problem. So, if you want to do a CI for the N electron like this, these GI's must be determinants built out of some one electron function. It does not matter which. What I am now saying that if you take them all, it is a full CI within this M dimensional basis. Of course, if M goes to infinity, it will become exact. So, we also define full CI even if it is not exact that is full CI in a restricted space. The question that I am now asking is that this is far too complicated because you have so many determinants out of this basis. So, can I identify one single determinant out of this MCN which is very good and how do I get that? So, that is the question that we are going to ask. As I said, this is what the chemist want that a picture of an N electron in one determinant represented in one determinant. So, how do I get that determinant? So, this is the question that we asked in the Hartree-Fock problem. So, I hope the problem statement is first clear. So, what are you looking at Hartree-Fock? So, let me first write down the Hartree-Fock theory, the statement of the problem. So, we want to get one determinant now. So, our CI tilde cannot be linear combination of a determinant, but it has to be only one determinant. So, obviously, one determinant will have one N set of spin orbitals. So, let us call this chi 1 tilde, chi 2 tilde, chi 3 tilde and so on, 2 chi Nt. I am calling this tilde because right now I do not know the solutions. So, they are trial solutions and this is a shorthand notation for the N electron determinant. It is actually a determinant, N by N determinant. I hope all of you can write it. Each row has a spin orbital, each column has a coordinate of the electron or vice versa, but that is the first one is what we are following. So, now we are saying that obviously if I know the basis, given basis, this problem does not exist because if I know the basis, what am I doing? So, in fact, this problem is actually to find this basis. This one electron basis that we are talking about, it is not going to be complete of course, but at least N of them, I want to find such that this determinant is the base determinant, which essentially now means through the variation method that this will be the minimum because we know in the variation method that it is always greater than or equal to e naught. So, for the ground state at least, so this is the, let me also specifically write that we are looking at only ground state now. It is very, very tough to do excited state and heart refog incidentally. So, it is a ground state because you are not doing linear space. Remember excited state comes on a linear variation. This is no longer linear variation and hence to get excited state is much more difficult. I will come back to this point. So, it is a minimum with respect to my variation of chi 1 till day, chi 2 till day, etc. to chi n. So, if you look at this problem, this problem is actually to find this basis, one electron basis. What are the spin orbitals? So, the question that you are asking, what N spin orbitals will give you a determinant? Of course, N spin orbitals can give you only one determinant. What N spin orbitals will give you one determinant, which will produce the lowest energy within all other such N set? So, I keep on changing this N set, which N set gives you the lowest because I need one determinant only. So, I do not need anything more than N spin orbitals. Remember, but I do not know which N spin orbitals. So, I do not need the capital M here. Like here, there is a capital M. I do not need that. I need only N, but which N will give me the lowest. So, here actually the problem is to find this basis, the one electron basis from where we should start. In fact, this will become the heart refog basis from where we will start doing chi later because that is already a good basis. So, we first want to find out this basis so that this arbitrariness later will also go. We will know which basis to start with. So, what she is asking is a very good question. It is a research problem actually. Why do you do an energy variation? Can't you look at another property to be a better tool? So, let us say I am not interested in energy. I am interested in dipole moment. Can I get a wave function which is single determinant, which will give me the best dipole moment. Unfortunately, there is no variation principle with dipole moment. So, that is the problem. So, but people have still tried the problem with the constraint variation. I will come back that later, but that is what the question that she is asking. Why you are bothered about only goodness of energy? Let us say essentially the question that you are asking. Why not goodness of some other property? For one, I do not have a variation principle for some other property. Second is that energy is most important, whatever it thermodynamics. I have to say the thermo, but for a specific problem, maybe dipole moment may be important. But in general, energy is more important. That you will agree because you want equilibrium structure that is based on energy. So, you want interaction energy. That is also based on energy. You want spectroscopy. That is also energy. So, what we need to do now is to simplify this problem. First of all, let us assume as before that my N spin orbital that is chi 1 tilde, chi 2 tilde, etcetera up to chi n tilde are orthonormal. That is very important. So, this is the only condition I am putting. The condition I am putting is only to make this denominator one. If you start with an orthonormal basis and then if I represent this chi tilde as chi 1 tilde, chi 2 tilde, chi n tilde, then one can show that this chi tilde, chi tilde is equal to 1. So, if these are orthonormal and this is my determinant, then chi tilde, chi tilde equal to 1. So, this is a simplification that I am going to apply. So, my e tilde in that case will simply become chi tilde h, because that denominator goes off. However, to knock off the denominator, I have to remember that as you are changing your spin orbitals, they must remain orthonormal. Otherwise, this condition is gone. So, let me again restate the problem. So, I want to minimize e tilde subject to a condition that all my chi i tilde, chi j tilde is equal to delta i j for all i j. All i j of course within 1 to n that is clear. I am only looking at n spin orbitals right now. So, this makes sure that they are orthonormal. So, basically that becomes a more refined statement simply because I do not want to work with the denominator. Otherwise, this was a good statement. There is no problem with this statement. So, this statement is now revised that I have only a numerator which I want to minimize subject to the fact that these spin orbitals must be orthonormal. Is it okay? It is the same statement actually. So, you must be convinced that these are two same statements. It is just that with this condition, I have ensured that the denominator is 1. If you do not use, then you must have the denominator here. Nothing will change. No physics will change. But this is just for simplicity that I would like to minimize this subject to this delta i j. So, this becomes mathematically the Hartree-Fock problem. The physically the Hartree-Fock problem is to choose a set of n spin orbitals which gives you a best description of a determinant. So, to a chemist you can go with that determinant because chemist is not interested in the full CI MCN. You can at least go with one determinant. So, this is what the lithium 1s square. So, we cannot do this. So, that is the physics that why am I trying to do it? So, please try to understand. I am trying to do because a single determinant is a nice chemist description. So, I want to bring a single determinant. Who is single determinant? That is the question because single determinant has n spin orbitals. I can take any n spin orbitals. How do you know that is the best description? So, that is where the Hartree-Fock. That is the question the Hartree-Fock tried to answer. Then, the mathematically Hartree-Fock method turns out to be just this, this line. That is a mathematics. If you have understood the physics, mathematically I can say minimize this quantity where site yield is this determinant subject to this. Is it clear? So, I am going to use Lagrange variation and all that. We will do that. I will come down to that later. But even before, so once you understand that the physics and the mathematics, these two you must go together. Once you understand that, then I still have a problem. What is this? In terms of this chi-tillness because this is a monstrous thing on the left and the right. It is a determinant. Yes. So, that is the result that you will get. That is the outcome. 1 by R12 is here in Hamiltonian. So, we will see that. No, no, no, no. 1 by R12 is not going to change. The equation that we will get for chi-i will be in one electron equation where an average effect of 1 by R i will come. Again, that is an outcome of the result. I am not going to make approximations. I am just going to do this. After that, I am going to get an equation. Approximation is just here. H is full Hamiltonian. Yeah, full Hamiltonian. After the minimization, I will get a Hartree-Fog equation. So, the statement that I made was an analysis of that equation, that there is an average effect. So, that you will see later. So, once again, let me go through this. There is no average effect here. That was also respect to psi only, single determinant. This is Hamiltonian. This is an electron problem. So, Hamiltonian cannot change. What you are talking is the one electron equation that I will get. After this, I will get a one electron equation to obtain chi-i's. So, there, how does 1 by R ij come in that one electron problem is what you are discussing. So, this will come later after I solve this problem. So, there I did not solve this problem. I finally get the equation. So, I already analyzed the equation. Here, I am going to first derive the equation, then analyze. That was non-interactive. After that, when there is interacting, I said that you cannot have a single determinant. So, you have a full c i. Yet, I want a single determinant. So, how do I get it? By minimization. So, now, whatever you have learnt in 4 to 5 will come. So, just hold on. But look at the broad picture. If I have just h of h is some h of 1 plus h of 2, I do not need it because that is a trivial solution. My single determinant chi 1, chi 2 is just the solution of h. So, I have solved the problem. So, there is nothing to do. If there is a 1 by R 1 2, then of course, the exact solution is full c i always. However, I still want a single determinant. So, 2 electron single determinant. So, if you are talking of 2 electron problem, I want a chi 1 chi 2. Which chi 1 chi 2? How do I know? So, that is what I am trying to do. There, I told you the equation of chi 1 and chi 2. And in that equation, there is a 1 particle operator, Fock operator. So, that I interpret it as an average of, so that is not an approximation. That is an interpretation of this approximation. You can say that way. Approximation is this that I am trying to do single determinant. So, that will be an interpretation of the approximation. But the interpretation I can do only when I get the results. So, there of course, Hartree-Fogger covered in just one or two lectures. So, I did not go through this. So, this whatever you are telling will come quite late now because I am going to go through the derivation. This class I am going to go through the actual derivation, not just give you the results. So, that was a one particle equation. This is completely n particle. But after minimization, I will get this one particle equation, which is what you are referring to. So, that will come much later. So, before you do that, I have a small technical issue to discuss. How do I calculate this? So, how do I calculate this in terms of chi 1, chi 2 to chi n? Because my basic variables now are not side tilde. What are my basic variables? chi 1, chi 2 to chi n. So, unless I write E tilde in terms of chi 1, chi 2 to chi n, I cannot start the variation process. So, that is a little mathematics. Again, I will not do the mathematics, but I will simply... So, what I have to do is to actually find out this quantity, side tilde is side tilde in terms of chi 1.