 This was actually done for molecular Hartree-Fock, was actually done by Clemens-Ruthan, he is known as CCJ-Ruthan, we see the first C is Clemens, it is a very famous paper and with him there is a work also by Hall, these are all very very similar, seminal papers in the 1960 Ruthan Hall equation, which was done first for closed shell and then for open shell systems which are not closed shell where the equation F of R1 will be somewhat different and then they introduce the same basis set for molecules and that was called the Popol and by Popol and Nesbeth, they have referred to as Popol-Nesbeth equations, we will not worry about it, right now let us worry about only Ruthan Hall equations, it is very important to know the history, you know how Popol and Nesbeth came in 1960s and so let us go to Ruthan equation with that little this thing, so what Ruthan proposed is that since I cannot solve this, let me expand this phi i of R1 as a linear combination of basis, okay, so I take a basis, let us say mu C mu i and some basis function A mu, so let us call it R1, R1, you understand, so every one electron function can be expanded as a linear combination of a basis and I choose this as a basis, A mu, I am deliberately writing A mu because for molecule this can eventually be an atomic orbitals, LCAOMO, this is actually LCAOMO if this basis is atomic, but the basis can be anything, but then you know to write this, this basis must be complete, right, which is again a problem, just as I had a problem that I showed for Ci, I cannot do complete because that is infinity, so usually I am going to truncate it to some m dimensional basis, remember, so these are my m dimensional basis for space orbitals, so of course I am going to get how many spin orbitals, 2m, right, my space orbitals are capital M and I am going to get 2m and then if I want to do Ci, I will get exactly 2m Cm, that is a problem that I gave, but I had, I just cheated, my m was not space orbitals, spin orbitals, right, remember, so MCn is the right answer, but here I will get 2m Cm, determinants, if I want to write determinants after this and I am going forward after Hartree-Fock, I will have 2m spin orbitals, n electrons I will get 2m Cm, one of them is Hartree-Fock of course, right now I am trying to find this, so this is an approximation, so as I told you Hartree-Fock is an approximation, then the next approximation is a basis, I am going to come back to this later, what is the basis, how do I define the basis, but let us assume that I know a basis which is finite dimensional capital M, then what can I do, so this is what Routhan did, it is a very simple mathematics which I can complete today, so what Routhan did was to, yeah, I will finish in 10 minutes, Mu equal to 1 to M, Cmu I, Mu of R1, just see what I am doing is putting this expansion for phi I of R1, both on the left and the right hand side, okay, equal to epsilon I sum over Mu equal to 1 to M, Cmu I, is it okay, of course R1 can be R2, it can be R, does not matter, it is one electron dummy variable, this R1, so what I have done, this Cmu I's are my unknown coefficients, this is known, basis is known, so your problem of Hartree-Fock was originally to find phi I of R, now the problem of Hartree-Fock is to find this coefficients, correct, if I know the coefficients I know phi I of R, so my problem has changed, from finding a function, set of n functions, so n by 2 functions, I now have a set of numbers, how many numbers of course if I have m dimension, I can get also m functions, capital M functions, which means I will have 2 m spin orbitals and many of them will now become unoccupied orbitals, just as I explained before that I always also get more than n spin orbitals, I will get more than here, this m ideally should be much larger than capital M, as large as possible but ideally much larger than capital M, so that it is nearly complete, as complete as possible, okay, so with this I put this here, so my targets are now these coefficients, I have to find out this coefficient, so what do you do is the following, multiply this function by one member of the basis, let us say a nu star R1, okay, f of R1, a mu of R1 and then integrate over dr1 multiplied by c mu of I, sum over mu we can bring outside, this is numbers, so they can come out with integration sign, so the integration is over a nu star R1, f of R1, a mu R1, dr1 multiplied by c mu I, so for a given nu, the nu is fixed, so I am multiplying by a particular nu right now, do the same thing on the right hand side obviously, so epsilon I, sum over mu equal to 1 to M, again c mu I can come outside and then you have an integral a nu star R1, a mu R1, so that becomes my equation, right, now what Duttham and Hall proposed was that all we need to know is these numbers and these numbers, integrals, remember they are again definite integrals, so they will become numbers, this will also become a number, so let me call this number a nu star R1, f of R1, a mu R1, dr1, so let me call this matrix of capital F nu mu, note that how many numbers I will have as, so this is equation is for all nu, one at a time, so how many nu's are there, M, obviously because this mu is going from 1 to M, so this will become M square in number, which I can rearrange as a matrix, so I am calling it F nu mu matrix, this entire set of numbers, I will write it, I am just calling this matrix, defining the matrix, this is a definition, I am not writing the equation, okay, similarly let me define a nu star R1, a mu R1, dr1 as a matrix of S nu mu, S is basically overlap matrix, overlap between two basis functions, if you do this then your equation becomes sum over mu equal to 1 to M, F nu mu, right, so whole thing is now F nu mu matrix, C nu i equal to epsilon i sum over mu S, actually I should have written this also outside, because this is sum over mu, so C mu i should go outside, I mean inside this integration, okay, not outside, inside the integration, so you have S nu mu C mu i, right, so this equation is a matrix of F multiplied by, so it is F times C, you can see the matrix multiplication rules apply here, repeated multiplication, this side is bit more complicated, I will analyze it little later, but imagine if they were orthonormal basis, what would happen? This will become a delta, right, so then mu would be equal to nu, so in an orthonormal basis, so let us say S nu mu equal to delta nu mu, if this equation will then become F C equal to or F C of nu i element equal to epsilon i C nu i, right, because mu would be equal to nu, so it will become epsilon i C nu i and this is nothing but an eigenvalue equation of a matrix F, so this would then become an eigenvalue equation, if I can construct the matrix F, which is much easy to do because matrix I know how to solve eigenvalue equation, however it is not quite true, because in general these need not be orthonormal, in fact this is what Rutheran proposed that in general these basis functions need not be orthonormal for molecule, the reason is the following, I would like to take the basis function as derived from different atoms, so one let us say I have water molecule, so this is where chemistry comes, so I require oxygen basis, I require hydrogen basis, oxygen atomic orbitals, hydrogen atomic orbitals, they cannot be orthonormal, why, because oxygen atomic orbitals I always ask this question, they are eigenfunctions of oxygen Hamiltonian, hydrogen atom orbitals are hydrogen Hamiltonian, so they cannot be orthonormal to each other, eigenfunctions of the same operator orthonormal to each other, but not different operator, so to take advantage of this chemistry Rutheran deliberately said let us not make this orthonormal, because a physicist would normally say you know forget about this just make it orthonormal, then it becomes very easy to do this problem, but obviously the chemistry dictates that I need to take when I do LCAO picture, I need to take different sets of atomic orbitals from different atoms, so I must not make it orthonormal, but I am just telling provided it was orthonormal, this would become a simple eigenvalue equation, but not quite, because of the spoiler, so we will see how to handle this, eventually of course you have to solve this equation in a non orthonormal basis, this will be actual Hartree-Fock equation that is solved today in Gaussian, remember all this Gaussian and all that solve actually this in a basis, so you have the 61G blah blah blah all that will define this basis and we will come to that and after that this is the equation that is solved, so forget about everything that we did, okay, I have to also tell you how to construct this matrix elements, you already know what is F of R, I am going to do the integration and how do I, so there will be also numbers, after I do DR1 each of them will become a number, so you can yourself try to do that and they will become H, matrix element of H plus coulomb and exchange matrix elements all that will come up when I do the integration, so I will show that, so once I construct this, this has to be solved and you will very easily see that because of the fact that the F has spin orbitals or space orbitals which are again in terms of basis, this matrix will now depend on coefficients, so again the SCF problem will come back, this is not a matrix defined by itself, so this matrix is actually a function of C because of the fact that the F contains space orbitals, so I will come back to that how to do that, so the problem has not vanished but now there is a second level of problem not only it is a iterative solution, I do not know it is not an eigenvalue equation, so how do I handle it, so that is something that is very important, yes, coefficients will be derived self-consistent because they are my unknowns, now the basis is already known, but depending on different basis my results will be different, so that is why I say there is no unique Hartree form, so people say SCF dash 631g, dash 631g star star, 321g, so every time you do Hartree form or DFT, DFT is very similar to Hartree form, you later see, people ask what is the basis, so now you understand why, the Hartree form is unique but unfortunately we are not able to solve for molecules, so we are bringing in these levels of approximations, so the level of approximation that we brought in for molecule really starts from here and then we do this derivation and eventually come to this equation, two atoms, same atoms located at different coordinates, yes because the molecular with respect to origin everything is defined, so your Hamiltonian is different, there is a coordinate, so it is a good question that you ask me, but errors will be too much, people have tried numerical Hartree form of course, that is towards exact, but the error is too much from the exact for molecules, yes everything is exact with Hartree form, I am talking about Hartree form exact, I mean I wrote down somewhere Hartree form itself is not, no, no, Hartree form cannot be exact anyway, I mean if you are talking of exact wave function, I am again repeating, all that we are doing when moment is Hartree form, it is approximation, okay.