 The Fock matrix in atomic orbital basis, expression that I gave, F mu nu equal to h core mu nu plus blah blah blah. Can somebody write down the entire on the blackboard? We wrote nu mu does not matter, you have to be consistent. If you are consistent it is okay, h core nu mu, yeah lambda and sigmas are atomic orbital, they are going up to 1 to m, yes just write the exchange, good, yeah very good, good, good, thank you. It is fairly simple to write actually and if you understand that if this is nu mu, this should be nu mu, that is very clear, h core part. All you need to do then to bring in 2 more indices of atomic orbital, 1 index of molecular orbital. So, this is the only part that you have to be careful. So, one of them which is the second index must be molecular orbital, because remember whenever you are writing C, your second index is the molecular orbital index, because if you remember your molecular orbitals, they are expanded as a linear combination of atomic orbital. So, mu and I have given you the symbols, the mu is atomic orbital, the Greek symbol, the i, j etcetera, molecular orbital. Since you are writing column wise, the second index must be molecular orbital index, so this must be j, it is very fairly simple. Then you are left with this lambda and sigma, you write them lambda and sigma. After that you have a two electron integrals, important thing to remember that it is mu 1, nu 1, mu 1, so this and this must be same for the Coulomb part and the second one should be one of them lambda 2 to sigma 2 and obviously whatever you have put star here must come as a star here. So, if you have doubt whether this should be sigma, this should be lambda, there is no doubt that depends on what you have written here, sigma lambda can be interchange. If this is in the left then the coefficient, corresponding coefficient must be star, when you are expanding the jth orbital, fairly simple to write and then just the exchange. You should be able to write this actually just looking at it, I would write it because this is anyway going from nu to mu. So, in the atomic orbital, this must be nu to mu, lambda, sigmas are summed up that is 2, 2, 2, only issues the coefficient, if you have written this on the left, this coefficient must be star. If you write this as sigma, this as lambda, this must be star, this must be unstar, is it clear? All of you should be able to write by inspection, I know nothing is required because you remember your original thing, how did you get it, phi j star 2, phi j 2, so I expanded phi j star 2 as c lambda star a lambda star, so that is how the a lambda star has come here. Remember all these are a lambda, you know we have just abbreviated, so a lambda star has come here. The right phi j was sum over sigma c sigma j a sigma, so that is why this remains c sigma j and a sigma comes here and both of them must be coordinate 2, phi j star 2, phi j 2 and I am now doing the integration over the integral over nu and lambda, nu and nu, so they come as 1 and 1, when I do the fc equal to sc, f nu, so it is very fairly simple actually. So, I wanted, you know I am bit disappointed that more people did it in volunteer and even more disappointed that nobody could write without seeing the text and I think this is where you have to, this is where you have to understand how these things are coming, there has to be little bit more physical feeling, it is not just mathematics, it is very simple actually. So, the idea is that after knowing this we wanted to solve the pseudo Eigen value equation fc equal to s, I call it pseudo because the s is not a Kronecker delta, it is not an identity matrix, so I want to solve this. The important point now to remember in this solution is that the input f depends only on atomic orbital, explicit, however it depends on the molecular orbital through the coefficients and that cannot vanish because obviously this is where my guess will be there, coefficient guess, but otherwise explicitly my input f and input s, these are my two inputs f and input s, they are explicitly on atomic orbitals, the solution to this equation will give me this c and of course the orbital elements. So, c is a mixed thing, it has one atomic orbital, one molecular orbital because the cell CAO may expansion, so that is what I am going to get. So, the whole idea is to convert this into an equation involving atomic orbitals and that is what we are able to do, so it is a m dimensional equation, m dimensional pseudo Eigen value equation, you will get m by m matrix, but of course this m by m matrix will represent m molecular orbitals in m atomic orbital, because it is m by m matrix, so each column is a molecular orbital, so I have got m molecular orbital and not the n by 2 that I required, I had more of them, if I have more of them nothing wrong, I am going to use them later, is it clear? I can further simplify this by looking at this expression that the sum over j runs only here, the j is not there anywhere. So, I define some matrix j c lambda j star c sigma j, actually a matrix that is defined is two times this and this is commonly written as a P sigma lambda matrix, now it is obviously P sigma lambda because j is summed up which is the molecular orbital, this is summed up and what are the two indices left, the two indices left are lambda and sigma, we decide to write it in this manner P sigma lambda, if you write it in this manner, this equation explicitly will not involve any molecular orbital, because now the sum over j can be performed and I can write the f nu mu as written as h core nu mu plus sum over lambda sigma, then I write this as P 2 I can factor out, so I can write this as P sigma lambda times nu lambda nu sigma again 1 by r 1 2 I am not writing explicitly, minus half P sigma lambda nu lambda sigma mu is half because this P already contains factor 2, so I can take out of course P as common and write this minus half that is very easy, so I can do a further simplification, so if I write this further simplification then I can say sum over P lambda and I can put the bracket here and say minus of half this, so I can say sum over lambda sigma P sigma lambda nu lambda mu sigma minus half of nu lambda sigma P, again indices are very easy, please do not never get confused, if I make it mu nu, accordingly you make it, this will be mu, this will be nu and so on, this will remain as P sigma lambda, sigma lambda will then follow this and this, so first index is for the coefficient which is unstirred and the star one comes later and that I have done deliberately I will explain to you later, so if you look at this F nu mu it does not have any reference to molecular orbital explicitly because this is a matrix which also depends only on atomic orbitals. Now the entire reference to the molecular orbitals has gone into this matrix P, so P contains all the reference to the molecular orbitals in definition of F but otherwise explicitly it has no J, if somebody gives you P matrix then you can just construct it, I do not need to worry about molecular orbitals, this P has a very special name, so P sigma lambda or whatever this P matrix is actually called charge density bond order matrix, many of you have already heard this and this is a very very special matrix and it is used in chemistry many times to even evaluate bond order, many of you have talked of bond order, in fact when you come to evaluate bond order we are going to use this P matrix, that is why I am defining this P matrix, it is famously called charge density bond order matrix and the P matrix contains many many things, it will actually be utilized for calculation of population analysis, many of you have done Mulliken population analysis, Love Dean, many of these will actually the P matrix is a very very important matrix, why, if you look at the P matrix it contains all information of molecular orbitals through this sum over J, it contains all the information of molecular orbitals and what does it actually tell you, it actually tells you for each molecular orbital what is the contribution to a sigma atomic orbital and a lambda atomic orbital coming from all molecular orbitals because there is a sum, so if you look at the C matrices, let us say I take a 2 by 2 C matrix, so this is my C1 which is the first column, this is the C2 vector which is the second column, so essentially this can be written as C11, C21, C12, C22 by my notation, so this gives me the first molecular orbital, this gives me the second molecular orbital, this is the contribution of first atomic orbital in the first molecular orbital, this is the contribution of the second atomic orbital in the first molecular orbital and so on, so when I am multiplying let us say one element, let us say I am doing a multiplication of C21 star or C21 does not matter into C22, let us say I have to sum over J, so I can keep both of them same does not matter, so C21, so one of them star 2 plus 2 C22 star C22, what does it tell me, it tells me what is the contribution of the first molecular orbital to a second atomic orbital where both the indices are second atomic orbital and then add to it the contribution to the same second atomic orbital from the second molecular orbital, so what I am getting here is what I would call it P22, so P22 is a charge density bond order matrix for the atomic orbitals 2 and 2, this is nothing to molecular orbital, so it tells me after I have known the coefficients it will tell me what is the contribution from all molecular orbitals to a matrix element of 2 and 2 or matrix element of 2 and 1 for example, then what will be P21, it will be 2 C21 star C21 into C11 star plus 2 C22 into C11, C12 star, so again first, so I am just rewriting this what, these are the two coefficients I am multiplying, so I have defined my sigma and lambda as 2 and 1, so I am trying to find out what is the contribution from all molecular orbital to a matrix element of which is second atomic orbital and first atomic orbital, so I am taking its contribution from the first MO and the second MO, that is all, how do I take first MO, the second index must be 1, second MO, second index must be 2 and the first index should remain 2 1, which is atomic orbital, so remember the coefficients, first index is atomic orbital, second index is molecular orbital, I am again repeating, so this tells me what after the molecule has been formed, after the MO calculation is done, what is the contribution from all MOs to a matrix element of the atomic orbital, so that is the reason it is called charge density bond order, because when I say charge density or bond order, these are atomic properties, so when I calculate molecular orbital, molecules are delocalized, orbitals are delocalized over all the molecules, but I want to get back what is called the atomic properties from molecular orbitals, so this is the reason this picture itself is called, what is called atoms in molecule picture, AIM picture that, so density is also like this, charge is also an atomic property, you want in a methane molecule what is the charge of carbon, how do I calculate, because what I have is a molecular orbital which are delocalized, so I have to find out what is the charge, so I will tell you how to calculate charge and all that from here later, but at this point I want to introduce this matrix, which is basically a contribution to a pair of atomic orbitals, a pair of atomic orbitals from all molecular orbitals, so it is in a simple way to remember, so the P is a contribution of MOs to pair of atomic orbitals, so this is very useful, because after the molecule has been formed or the SCF calculation is done, I want to get back some information about the atomic orbitals, so this is my hope, this is the matrix which will actually give me something, so this is the reason these are called charge density bond order matrix and are used to extract atomic properties from the molecular calculation, purely atomic properties, because I have said my molecular orbital is a combination of atomic orbital, so after doing the calculation I want to get back atomic properties, so obviously this combination coefficients will be required, so that is the reason we are calculating the C's and from the C's we want to get back a contribution to any pair of atomic orbitals, so I have just written one pair which is same, two to one pair which are different, so in a way atomic orbitals are linear combination, so I want to get back from molecular orbitals atomic properties, reverse, I started from atomic character, I went to molecular character, having done that I want to extract atomic character again back, so this is my matrix which will allow me to do that, so I just thought I will introduce why this is physically also important, I will come back in the application, but at this point I just want to note that with this matrix I can write an explicit expression of F only in terms of atomic properties, because all your coefficients, the molecular orbital coefficients are actually hidden here, yes, P sigma, okay, fine, so this P sigma lambda is a contribution from all molecular orbitals, which is n by 2 of course in this case for the closed shell to a pair of atomic orbitals, sigma lambda, so that is the pair, so I am extracting the atomic orbital contribution from the molecular, just as you said it is really a reverse way and this matrix allows me to do that, yes, you had a question, two and two, atomic orbitals, two and two, so these are all atomic orbitals now, okay, so we will see then how to get this for a bond order and all that, that we will see later, how will you define this, because this has, when you sum over j, this has two indices, because it has two pairs, I multiply the two c's, you look at the expression carefully, it has c lambda j, c sigma j, yes, because I have to, mathematically I have to c, because this is a contribution of two coefficients, one of them can be lambda, one of them can be sigma, they can be identical, they can be different, we will have double summation, so double summation comes purely, because in the original expression I have phi j star 2, right, 2 into 1 by r 1 2 minus p 1 2, whatever, so 2 into 1 by r 1 2 minus r 1 2 into 2 minus p 1 2, okay, phi j is 2, d tau 2, so remember, I have expanded this and this, so in this expression a pair of atomic orbitals will come, alright, so this p matrix, if I write this at the p matrix, I will come back to the discussion of p matrix, but at this point mathematically, I just wanted to tell you that I can now write f only in terms of atomic orbitals and nothing else, because now j is gone, however j is not gone, because j is there in the p obviously, j cannot go, alright, so the problem now is that p is not a defined matrix unless I know j, so where is my scf gone, scf is hidden here, so instead of guessing c, I am now going to guess p, charge density bond order matrix and then construct the Fock matrix, then solve this equation to get c and once I get c, I can reconstruct p and I go back here, so I just have one more step of reconstructing, it is just all mathematically also nice, basically instead of doing this multiplication here, I am doing this multiplication separately and bringing it here, it does not matter, it is the same mathematics, but it is done because it is very useful, this charge density bond order matrix, so physically it is also very useful, alright, so I think with this, I hope all of you should be able to write the Fock matrix in atomic orbital basis, please remember the entire scf, apart from this guess, entire scf only involves atomic orbitals, please remember, you just get these integral, they are all atomic orbitals, two electron integrals, I have discussed the symmetry, number of integral symmetry, all that I have discussed last time and then this is also a guess of the atomic orbitals, but this is what is not known, everything is in terms of atomic orbitals, the construction of the Fock matrix, the output simply gives you the coefficients c and that is where the molecular orbitals come in, but when you write an SCF program, everything is in terms of atomic orbitals, the integrals are actually, integrals are phase from atomic orbitals, coded and everything, so that is very important, of course there is a lot of literature in how to sum this, I am not going into that details, this is a very, very difficult thing, it looks very simple, innocent, but computer program wise it is very, very complicated, simply because you cannot, I told you yesterday, you cannot do loop driven programs, what is done, because most of them are sparse matrix, so what will be given to you is only non-zero elements of this, non-zero values of this, they are indices and then you have to construct, so it is not a trivial thing, there is a lot of sorting of these integrals which go in the back, I am not going to teach you SCF programming, because SCF programming you really want to learn that is almost a half a course, I can tell you, believe me it is so complicated, just how to program this, it looks, if you are a naive programmer it is trivial, simply because you are not allowed to write like this, because most of the zero, you cannot multiply zero with something, you are wasting your time, because the result is known, so you must multiply non-zero values of these, non-zero values of these etc., so how do I do this, so there are lots of interesting sorting, one of them that is very commonly used is raffinetti sorting and in fact I do not know if any of you are interested in writing program, at some point of time I wrote the SCF program, I mean just for fun, SCF programs are available, but because we wanted to parallelize the program and that time parallel programs are not available, fortunately in Pune, SIDAC had come in 88, 89, all of you have heard SIDAC Center for Development of Advanced Computing, so Pune was the first center of parallel computing in India and it was just within 3 kilometers from my lab, so we took a lot of projects and parallelized, so parallelizing you have to run it all over again, I mean write the code, so one and a half year it took me to write SCF code, after doing lot of coding, lot of experience of coding before, it took one and a half year to get correct result, so it was a big project, in fact it was a project from SIDAC to write a parallel SCF code on a four transputer, they call it transputer that time, now it is called processors, so how much factor can you get a factor 4, you do not get a factor 4, but you can get 3, factor 3, 3.2 depending on how good you write, scale up because sometime is wasted, yes, one, look one simple choice is very simple, assume this is 0, so your F score, F mu nu is just this, that means coefficients are 0, just this and with this you do this calculation, at least you start to get a C, bad C and then you put it back, assume this is 0, so you will still get a solution and that is a bad solution, that is something I could have done by neglecting 1 by R12, it is just the one electron part, but with that you will get a solution, I do not call it SCF, you get a solution and after that you start your Hartree fault, that is always there an option with you, if you do not know how to choose, simply assume neglect this, basically neglect this stuff and after that you build up, but there are better ways of doing it of course, but that is another issue, how to converge, physically how to converge SCF solution, what are the guesses, all right good, so with lots of discussion, now we have come to the F matrix, so let me just write down, how do I start, before you start actually the most important thing is of course the Born-Oppenheimer approximation, so BO geometry, I hope it is clear what I mean by Born-Oppenheimer geometry, so you just do Born-Oppenheimer, we are doing Born-Oppenheimer approximation, so nuclear fixed, so assume the geometry first, what is the coordinates of the nuclear, because that is where I am calculating the electronic structure, once I do that geometry I choose basis, again each of them has to be expanded, how do I choose basis, that will be expanded, Gaussian all that I have not even brought in, but choose basis which are the basis of atomic orbitals, usually the basis of atomic orbitals, how do I choose, I will see that, once I have choose this basis calculate 1 and 2 electron integrals, so nu H mu and nu, nu sigma mu lambda etcetera, of course if I calculate this exchange is also included, all m to the power 4, whatever number that you will tell me on that day with symmetry, so some number you will tell me, so all these you calculate 1 and 2 electron integrals, because that is the heart of the thing, once you do that then number 4 guess P, at this point I do not know how to do that, does not matter, guess P, once I guess P construct F, I have given you the expression, how to construct F, integrals are already available, so this basis is basically this basis of mu nu lambda sigma, mu nu, that is my basis A mu, A nu, whatever I called A mu, construct F and then solve F c equal to S c, I do not know how to solve it, after I do this, number 6 I solve it, we have to expand this, 6 A 6 B all that, how do you solve it, that we will come to, come little later and then we come back to construction of P again, because now I have a C, so number 7 reconstruct F, reconstruct P, once you reconstruct P, number 8 go back, go back to step 5 and after you do this step 6, there is a convergence test here, from the earlier iteration, earlier values and present value, of course the first time there is no earlier value, but next time onwards, you do a convergence test, we do not know how to do that, we will talk about it, either on C or on P, does not matter and if it succeeds, exit, else go to the step 5 and continue, once you go to the step 5, you continue this process, so that is, if you have to write a simple program, algorithm is very simple, should be first able to write an algorithm, how do you do some of these things, we will now come, but that is about it, once you converge, you know the molecular orbital AOMO coefficients, you are done, because it means you can calculate molecular orbital, you can then calculate energy, everything you can calculate, we will see how to calculate energy, is it clear, of course the whole thing stops with the geometry first, without a geometry nothing is defined, because your basis is the basis of atomic orbitals, so basis will contain the geometry, so your basis will be like exponential minus alpha r minus something geometry, so the geometry will actually be contained in the basis and I will come to that again later, so it is very important to have a geometry, so when you calculate Hartree-Fock energy, we are calculating at a fixed geometry, of course then you can change the geometry, you can do another Hartree-Fock and eventually you add nuclear-nuclear repulsion and I have already told you, you generate what is called potential energy surface and then see at what geometry it is a minimum, that is your equilibrium geometry, so that is much further, this is only a single state SCF is what we are discussing, three atomic orbitals, three atoms or three atomic orbitals and how many atoms, because I can have a single atom and three atomic orbitals, it is a molecular calculation, so there must be more than one atom I assume, so the best you can do is hydrogen H2 and if you have H2 then 3 is bad, because if you have one hydrogen one, the other hydrogen must also have one, so it will be either 2 or 4, if you have LiH you can have 2 for Li and A, 4 for hydrogen let us say yes, you can even have 2 for hydrogen no problem, that is the simple 1S A plus 1S B, it is even better, 4 mole 4, 16 numbers, of course yes, 4 and 4 energies, it is absolute that is the Eigen value equation, what is so surprising for you, if I have an Eigen value equation AC equal to CE, forget about the fact that it is not Eigen, this is exactly what you do, if I have m by m matrix I get m Eigen vectors, each Eigen vector has m number, so I get m square plus m, so indeed you are right, so my number of elements unknowns are m square plus m, you may wonder are they are m square plus m equation or there are, please remember this is matrix equation, so you just expand, then there are normalizations that bring in another m equation, so actually there are m square plus m equations, but anyway that is the Eigen value equation is correct, you are right, so if you have 4, 16 plus 4, 20, 20 unknowns, yeah you want to draw something, Radhika no, you want to draw something, yes if geometry changes basis will also change, same basis functions but they will change, of course, of course, of course everything has to be read up, everything has to be read up, because basis will change, I will tell you what is this functions atomic orbitals, in fact you know already in hydrogen for example, you have exponential minus alpha r, but you assume the capital R to be 0, otherwise it is actually r which is the nuclear, electronic coordinate and this is the nuclear coordinate r alpha or r whatever, r A for the atom A, this is called the Slater function, typically, so this is what the coordinate comes, now if you are doing a single hydrogen atom, we never discuss this, because the geometry is 0, 0, 0, origin, you just assume it, but when you have molecule, you cannot have every all atoms in origin, so they will become a single atom, so you can choose one of them origin, but with respect to that others have to be defined, so that is the reason the molecular calculations are complex and because this is stuck in the definition of A, you have to do all these things again, so every step has to be repeated and of course if you change your basis then anyway you have to repeat, like instead of 4 if you want to take 8 and different basis functions, then we will come to that later when you actually discuss, yes any other question, because I am working in the Born-Openeur approximation, so that was my first underlying principle, I am not bringing a nuclear, I have already separated the nuclear, it is the only electronic Hamiltonian, so that is the reason, if you want to take then you have to solve both electrons in the nuclear together, that will be very complicated SCF, the structure of SCF itself will change, is it alright, can I move ahead, I think it is important to discuss this, these are, this is just to get one energy, minimization will happen after you calculate the surface for all different geometry and see where it is minimum, but there are better ways of doing it, they are called what is called analytic evaluation in which all you will do later is to look at that potential energy and you calculate its gradient, because you know where is the minimum, the first derivative must be 0, yeah and then we have to take the analytic derivative, that is another step is there, for getting the minimum, but I am not going to discuss those things right now, in fact the derivative calculation of energy derivative calculation by itself is a subject, in fact I myself have done lot of research on that, so that I will not worry right now, I am only talking of, so this course we are only talking of single point energy calculation, please do not confuse, this course we are not going to go to, how do you get the analytic gradient, so this course what we are discussing is given a born of an hammer geometry, how do you calculate energy, at different level, heart refog, post heart refog that is it, okay, but yeah indeed your questions, since you ask the question I have to answer, alright, now there are two possibilities, one is that I can finish the Koopman's discussion, before I go to how to solve fc equal to s, okay, or I can start solving fc equal to s, that is the next step, how do I solve fc equal to s, guess p I will come later, guess p is more of an intuition and so on, assume we can guess p, that is not a big problem, you have constructed a real technical point is how do you solve it, because it is not an Eigen value equation, so how do I normalize and so on, but maybe I feel that we are getting too far ahead of the Koopman's what I discussed, so I should come back to that, finish it today and then we will go to that, okay, so I think right now just remember the steps and we will have to discuss this step 6 in greater detail, how do I solve it, but if you solve it everything can be done, okay, the only other thing I must tell you is that just as I wrote the heart refog equation, I should be able to write the total energy for closed shell, I hope all of you should be able to write this, first you write in spin orbital, then you do spin integration, right, so what will be the result, can somebody tell me quickly, I have already done the spin integration, in terms of the molecular orbitals, spatial orbitals, so I equal to, can I write it, I equal to 1 to n by 2, 2 times hii, 2 times hii is basically phi i h phi i, okay, I hope all of you are in sync with the symbol, then you have a sum over ij n by 2, well now I am doing integration, 2 times jij, right, 2 times coulomb minus exchange, right, we have done this, this, this, this step of half, half was there, you are right, in spin orbitals, when I did this space integration, remember coulomb has four terms, hii, hii, hii, hii, each of them can be alpha or beta, they survive and the exchange is only parallel or anti-parallel, so twice, so I get, I get this, 2 jij minus kiij, where you know what is jiij, what is kiij, now I will ask you a simple question, can you write this down in terms of only atomic orbitals, I hope you understand my question, these are all in terms of molecular orbitals, so I have to expand wherever molecular orbitals comes in atomic orbitals, can you write this down in terms of atomic orbitals and wherever possible, use the definition of P and write down in terms of P, wherever possible, you understood, so try to avoid coefficients, C, so my question that I am posing to you is write down the Hartree-Pock energy in terms of atomic orbitals, indices, I mean integrals in terms of atomic orbitals and so on, I hope you understand the steps that I followed, we are first in terms of spin orbitals, then I did a spin integration, got everything in terms of molecular orbitals, then I am saying that molecular orbitals are linear combination of atomic orbitals, so I did another round of expansion and I got for example, the Fox matrix from the same way that I did, f ki i equal to epsilon i ki i, you remember this is spin orbital, then I went to f of r, phi i of r, epsilon i phi i of r, this is spin integration, then I came to f c equal to f c, this is by LCAO expansion, so this is the trend that I did for the Hartree-Pock equation and this is what you should remember, this is my original Hartree-Pock equation, let us not forget, it is a good time to reflect actually, I had a non canonical Hartree-Pock equation even before that, which I am not writing, by unitary transformation I have already got this, then here did I do the spin integration, then I did a LCAO expansion and wrote everything in terms of coefficients, numbers, matrix equation in which I took a special attention to write f only in terms of atomic orbitals by using p, but does not matter, that is only a mathematical thing, what I am asking you is that I have E Hartree-Pock in terms of spin orbital, which I had given you first, which is spin integrated to this expression for closed shell, again whatever discussion we are now doing is for closed shell, so this is spin integrated. Now I do an LCAO expansion, question is can you write the E Hartree-Pock now in terms of atomic orbitals, I hope you are, is the question clear, so I have already done up to this point, so wherever there is a molecular orbital you must expand in terms of the atomic orbitals and write everything in terms of, remember you have been given the integrals of the atomic orbitals, they are known, so everything you should be able to write in terms of atomic orbitals, 1 and 2 electron, is it clear, then the advantage is after I do this entire step I got my p, that is it I require, I have the integrals I should be able to write Hartree-Pock energy, only with the p you will see later and integrals, so no reference to molecular orbital as such except the p, but p contains sigma lambda atomic orbital indices, m o's are hidden, so you should be able to write these in terms of integrals, in terms of a o integrals and p matrix that is all, this is the hint I am giving, nothing else is required, when I start expanding this, you have to remember I have to expand from here, you have a phi i, h phi i, one phi i you have to expand, another phi I have to expand, you will get 2 c, they have to be combined to a p matrix, then here do it, please do it, but it is very simple, just practice, and I am telling you again how the p's will come, note here you have 4 indices, although I am writing g i j, it is phi i phi j, 1 by around 2 phi i phi j, each of the phi i phi j have to be expanded, so there will be 4 c's, 4 sets of coefficients will come, remember not 2, so write it down, write down see what happens, good, but p is in terms of atomic orbitals, that is what I said what is p, p is at contribution to a pair of atomic orbitals, contribution of all molecular orbitals to a pair of atomic orbitals, I get p, end up the SCF I get p, that is all I need, from there I can directly go here, Hartree-Pock energy, that is what I am trying to tell you, because the integrals are already available, because essentially getting p is what, getting c, physically I am getting the coefficient, that is what I require, if I get the coefficients I have the molecular orbitals, that is all I require, yes, but this has a important physical interpretation, the p is not just mathematical, that is why it is called charge density bond order matrix, which I will define later, but at this point I am asking you to practice writing all expressions involving only atomic orbitals and p, p has the hidden molecular orbitals, but explicitly there are atomic orbitals, matrix in atomic orbitals, so all expressions you should be able to write, because this is very important for programming, we are never going to bring in molecular orbitals, everything will be including the Hartree-Pock energy, okay, of course this is not the potential energy, because to this we will add the nuclear-nuclear repulsion, which is a constant term, but this is the electronic part of the Hartree-Pock energy, which should be all written in terms of the p matrix and the atomic orbitals, so that is why this practice I am asking you to do, is it okay, expand, expand, yes, absolutely, you should be able to write it, so it is a practice, p will include only two sectors, this will include four sets of c, which can be rewritten as one p into another p, so practice it, once you do it is very easy, p times p is four sets of c, no problem, okay, each p has two sets, so you should be able to do that.