 the Rutherford Hartree-Fock equation let me just quickly derive it again for continuity. So we had we started from the canonical Hartree-Fock equation which was in space orbitals. So I hope all of you can write the equation in space orbitals. So let us assume that the phi i's are the space orbitals and our chi's are the spin orbitals. So this is the canonical Hartree-Fock equation for in space orbitals and in particular if it is closed shell. So of course this form will depend on what kind of system it is. So this is a particularly for closed shell you can get a single Eigen value equation like this where f of r or f of r1 is h of r1. Please remember that now the spin is integrated. So this is only in terms of space orbitals. So this is the spin integrated equation for the closed shell. So h of r1 plus sum over j equal to 1 to n by 2 integral phi j star 2 1 by r12 2 minus p12 phi j2 d tau. So I hope all of you can write this 2 is essentially r2 just to make sure that you understand and the integration is done over d tau 2. r12 is of course mod r1 minus r2. So this 2 comes first of all the summation is over only n by 2 special orbitals. 2 comes because of the fact that the electron which is giving interaction can if this is an up spin it can be either up spin or down spin for coulomb and for exchange it can be only parallel spin. So there is only one term. The p12 will ensure the interchange of r1 and r2 when this operator will act on the coordinate of electron 1. So when it acts on the coordinate of electron 1 it will ensure that this becomes 1 and this one will become 2. So that will bring the exchange effect which cannot be classically interpreted but otherwise everything is exactly same as the Hartree-Fock equation for spin orbitals. So this is what we first did then we said that for molecules the exact solution of this is difficult so one has to adopt what is called the expansion in terms of a basis. So instead of solving this equation in the coordinate space we solve by expanding these special orbitals in terms of a basis because these are now my unknowns the special orbitals and the basis is a set of known functions. So the unknowns will become the expansion of the coefficients so these are basically numbers so that is the difference. So we can write now phi i of r or r1 it does not matter it is a dummy variable sum over mu C mu i and a basis which is the basis usually of atomic orbitals but it need not be so usually the basis is of atomic orbital the set mu is a set of atomic orbitals but note that this is only usually mathematically of course it can be any complete set but of course the chemistry dictates that if you have a set of atomic orbitals this becomes convergent. Since you are expanding in a basis of course there is a problem of basis set limit and usually again we cannot reach the limit unless we take infinite number of functions but let us assume that we have taken an m basis function. So in which case the Hartree-Fock itself was an approximation has one more level of approximation that is the expansion of the basis is it clear? So that is what we did and then we said that this problem now which was a problem of finding a function set of n by 2 functions is now reduced to finding a set of coefficients because these are known functions and those coefficients are obtained by the Hartree-Fock-Ruthan equation by plugging this into this equation and multiplying the equation by a member of the basis conjugate of a member of the basis and integrating over R. So that led us to the Ruthan equation which I just write this first and then explain sum over mu equal to epsilon i sum over mu S nu mu C mu i where F nu mu was a matrix element of phi nu star R F of R phi mu R d of R d tau. So when I do the integration multiply by one member of the basis and integrate the integration is of course a definite integral so it will result in a number that number will depend on the mu here and mu here so we call this F nu mu and this is actually has a structure of a matrix m by m matrix because there is a m in number so you have an m by m matrix. The C mu i's are very clear they are the coefficients of expansion of the ith what I now call molecular orbital in terms of the mu ith atomic orbital so in a very simple chemistry notation C mu i is the contribution of mu ith atomic orbital in ith molecular orbital so that is what this coefficient means right that I expand the ith molecular orbital a mu this is leading to this coefficient so this is the expansion of a mu in phi i so this is basically famously can be called your LCAO-MO coefficients so if you want to track a molecular orbital all you have to look is the ith column please note again the way the symbol that is written is a standard symbol where the molecular orbital is expanded in terms of a column so that is why it is not C i mu but C mu i so if i is 1 this is 1 1 2 1 3 1 4 1 up to capital M 1 so the first column will give you the expansion of first molecular orbital in terms of the M AO second column will give you expansion of second molecular orbital and so on so that is a standard notation so please remember it is used in all Hartree-Fock programs and this is the standard notation you can of course do it the other way round the row wise but this is the standard notation so C mu i is the coefficients S mu mu is similarly an overlap integral between phi mu star r phi mu r d tau and d tau and if these are not orthonormal of course the S mu mu will not be delta so provided of course they are orthonormal this will become a delta mu mu chronicle delta which is orthonormal set but in general they are not so the equation that you get is of this form and which you have derived it is very easy to do this just expand these integrate by phi mu star Mu star oh sorry sorry I think there is a mistake yeah I think you should have pointed out these are all A these are expanded in terms of the basis Nu star Mu nu star sorry so when I expand this in terms of the atomic orbital basis and multiply by Nu star then obviously you will get Nu star f of r Mu r so not the phi phi's are no longer there okay so these are all in terms of the basis okay and C mu i is the expansion of phi i in the basis of Mu okay so this is the famous Hartree-Fock-Ruthan equation okay we have to solve this equation clearly clearly if and this is something that you should recognize if S nu mu were delta nu mu which means if the basis were orthonormal okay you could see that this equation would have become f nu mu C mu i and here you would have got nu equal to mu so it would have become epsilon i C nu i for a specific nu where nu is specific you are summing over mu and this would actually have been what would I would call now Eigen value equation so all of you know that this is in the structure of an Eigen value equation so this would have reduced to an Eigen value equation provided S mu nu were delta nu mu but in general they are not so we must recognize that in gen this is not a general form this is the general form of the Routhan equation okay clearly then our next task is to find out the C mu i and to do that you have to write f nu mu in greater details by going to this equation integrating writing this in terms of the coefficients quite clearly if you notice that since f depends on the orbitals when I expand these orbitals the capital F nu mu will also depend on the coefficients and this is where the whole self consistent field structure which was there would again come back here because f will depend on the coefficients so whichever way you solve the equations you have to solve iterative so there are two problems one is of course iterative problem which is already half that is not vanishing that is there it is only that the iterative solution will now be in the coefficients but a further problem that is coming is how do you solve this equation so how to solve because it is no longer Eigen value equation so both there are apart from these there is a second level of problem that we have now encountered since it is not an Eigen value equation so how do I solve the Eigen value equation okay so we will come to that later but right now I hope all of you should be able to write f nu mu so f nu mu is if you start with this it will be integral a nu star r or r1 see please remember that whether I am writing r or r1 it does not matter because that is a integral which is integrated over the element so this is just a dummy variable I am writing r1 here specifically because I have to bring in r2 so it is easy to see 1 and 2 if I write that r then if I call that r1 then it is more difficult to write this I have to specifically write r minus r1 modulus r minus r1 so I am just avoiding doing that so whenever I require I am writing this r1 but it does not matter because that is a dummy variable so a mu r1 theta 1 so that is the first term now note that this can be trivially integrated because I know the operator h of r1 these are my known basis functions so I can integrate for whichever integration may be very complex mathematically complex but this is doable then the second term comes from here the second terms are more complicated because you have a sum over j 1 to n by 2 first and then there is an integration over d tau 2 and there is a further integration over d tau 1 please remember I am eventually doing an integration over d tau 1 right that is my f mu but in the definition of f of r1 there is another integration over d tau 2 I hope you can see that right so eventually there will be double integration so let me write it down carefully first I write a nu star r1 which is just comes here because I want to write this as a nu star f of r then instead of r now I am using r1 that is all this is dummy variable and there is a sum over j which I have just deliberately taken outside then I have a phi j star r2 so I have to write phi j star r2 again as a expansion in terms of the basis so the same expansion I am going to use now for phi j star so I hope you can expand this phi j so let us use different dummy variable all you have to do is to be careful about dummy variable so I cannot use mu and nu they are already taken so I must use something else so let us say I write phi j as sum over lambda c lambda j just I am writing it on the side line c lambda j a lambda so I just wrote it in the side line so that it becomes very easy whether it is r1 r2 that depends on what is there so I now have to do phi j star r2 so I have to write this as a lambda star r2 and of course there is a c lambda j so c lambda j star let me write it here outside the integration because that is the number so that comes outside the integration then you have 1 by r1 r2 and I will keep using the abstract form for the time being 2 minus p1 2 and then let me integrate the d tau 2 by putting in another phi j remember this is again a phi j but I cannot use the same dummy variable I must use another dummy variable let us say sigma so I put the sum over lambda of course already come and then I have a sum over sigma so c sigma j now c sigma j is already written so a sigma of r2 so I am expanding this in terms of sigma this phi j r2 so I have a c sigma j which I have written outside the integration and a sigma r2 and of course I must complete this by bringing in a mu r1 and now I have d tau 1 d tau 2 first and then d tau 1 is it clear I hope all of you can write this yes. So summation over lambda sigma summation over j c lambda j star c sigma j and this integration now if you look at this entire equation they are all numbers because the first term is the number of course the second term is also a number now because it is a total integration of d tau 2 and d tau 1 unlike here where there is a partial integration of a d tau 2 it has a both the integration and which is expected because from this partial integration I did a further integration of d tau 1 or d tau whatever so obviously this should become a number and I get a very compact form of what I call f mu out of which this is trivial but now this is where the problem will come because to define f mu mu mu you can see that I require the coefficients. So that is the reason again you have to do an iterative or self consistent field solution so instead of guessing the phi I have to guess the coefficients because these are my unknowns not the phi because these are known and then I have to construct the Fock matrix and then somehow solve this equation which I will come later how to solve so let us not worry right now let us say I solve this equation get the coefficients reconstruct the Fock matrix and keep doing till it converges so that will be my strategy now so instead of strategy of converging on phi I am going to converge on coefficients now there are several questions what do you mean by convergence when do I say it is converge all that I will discuss when we go into more details but philosophically that is going to be the strategy now so basically guess the coefficients guess the coefficients see me y matrix then construct the Fock matrix solve Routhan equation get a new set of coefficients let us say I call it so let us say I start with some n I call this n plus 1, n is number of iteration. Mathematically I am just writing n plus 1 and then go back and construct the Routhan equation again a F again and keep solving till it converges it comes out at the convergence point which I will discuss how do you convert but philosophically that would become your mathematical way of solving this so reconstruct after the coefficient the Fock matrix solve this equation Routhan equation again keep doing when the coefficients converge and what do you mean by convergence we will discuss later but there are several things to discuss before that one is to first of all look at this equation very carefully the Fock matrix equation very carefully then to see how to solve it this equation because as I said they are not orthonormal so it is not an eigenvalue equation so we will have to do piece by piece so let us look at this part little bit more carefully the part of the F nu mu which includes the two electron integration but before that let me write down the first part just as I wrote F nu mu I can now define integral a nu star R1 or R does not matter h of R1 a mu of R1 that is the easy part and that is very often called the h core sorry h core nu mu it is again a matrix element right depending on the basis nu and mu and this is very often called the h core now it is actually capital H but please remember this is not a Hamiltonian full Hamiltonian it is just the one electron part and it is very often written as capital now and with a core on the superscript the core essentially means that it is the one electron part the kinetic energy in the electron nuclear attraction so that is all the core has no other meaning so basically it means a one electron part somehow it is traditionally written with the capital H I could have written a small h just to continue that but that that should not bother you okay so this is what I define the d tau d tau 1 of course this is the first part of the F mu mu okay then the second part before I come to the second part I must remember that there is a P12 here of course so there is a Coulomb and exchange part so that of course I should not forget so the Coulomb part and exchange part both involves integration of this kind a nu star R1 so let me write those integral first a lambda star R2 so one of the typical integral don't worry about the 2 minus P12 but what is important is 1 by R12 so let me write down those integral a sigma R2 and a nu mu R1 d tau 2 d tau 1 this is the typical form of the integral that I must first understand because after that it is only a question of multiplication of 2 and when I exchange 1 2 it just exchanges these two but that doesn't harm and that that is that is only index change in this case so this is what is called the two electron integrals and the h core just as just as I written this these are called one electron integrals of course it is very easy to understand why they are one electron and they are two electron one electron integrals of course involves the kinetic energy plus the electron nuclear attraction the two electron is just the two electron integrals and that is very important to understand these integrals of course during the SCF process have to be calculated only once because you can imagine that the basis sets are known functions are known so I just calculate once the iteration comes because they will be multiplied by these coefficients all right but these integrals per se has to be evaluated only once okay so these integrals are very important they are they have a nu and lambda on one side sigma and mu on the another side so just as we have been doing the Dirac notation I can write this integral in Dirac notation as nu lambda right 1 by r12 sigma mu sorry mu sigma okay by the Dirac notation it is a 1 2 1 2 notation so so I just want to tell you again just as we did for the for the ijkl in spin orbital space orbital I am going to use a simple notation like these two write the full integral so I am not going to write the full notation okay so then I should be able to write this part of the fork matrix very quickly so then I can write this fork matrix now f nu mu as h core nu mu plus lambda sigma sum over j c lambda j star c sigma j okay so this part I bring in and now I write the integral so the integral will be first integral will be nu lambda 1 by r12 mu sigma however multiplied by 2 okay because you have two times this okay minus so let me write this over with the bracket minus the the term which is comes with only one but there is a p12 so it will only exchange the right side so now it will look like nu lambda 1 by r12 sigma mu that is it so your fork matrix is now written correct and you can actually now construct the entire fork matrix because these are now numbers everything in numbers so that is why we say in high performance computing despite the fact that everything was a function of r dependent finally everything becomes a number it just manipulation of numbers matrices and numbers it becomes very easy but big big numbers because remember these are no longer depends on number of electrons these dimensions depends on the capital M right you remember my phi i's for phi j's were expanded in a basis and the basis I said should be as complete as possible even though my number of electrons may be 2 4 10 whatever my basis has to be very large so the capital M was much larger than capital N so these are basically m to the power 4 in number 1 m 2 m 3 m 4 m so m into m into m into m 4 m to the power 4 and you can imagine per se these integrals are very large of course these are not any separate integrals remember these integrals and these integrals are same it just subset is different instead of so it's time to the power 4 I don't have m to the power 4 is a really very large number you can see very quickly typically today if you use a basis of 100 you can imagine what is 100 to the power 4 you're talking of you know 100 million and and later on I will mention this is the problem of high performance computing so a lot of good computer science people can help us they know how to do really so there's a those who are doing programming coding they can be actually they are they take a lot of interest because they have interest in computer science you know many of them are just good coders and and good algorithms matrix eigenvalue equation involving large numbers eventually you are forming a m by a matrix remember using data which are m to the power 4 so some summations are there okay and there is a summation over j of course which is 1 to n by 2 let me mention this and this is these are m 1 to m just to understand because these are only over the molecular orbitals so they are n by 2 and that is where the scf part will come the coefficients okay so they are molecular orbital but these numbers are actually called m to the power 4 and and something that can be quite you know difficult to handle if you don't do it properly fortunately and I will come later on the computation part many of these integrals are many times 0 fortunately because of either symmetry in the molecule depending on how you choose your basis if you cleverly choose your basis you can make many of them 0 and that is where group theory is very helpful you know those who have done group theory course would actually appreciate that this is also basically these bases have to be some kind of symmetry functions of 1 by r 1 2 and then it can be 0 so although they are m to the power 4 in number they are sparse I hope all of you heard the word sparse sparse matrix means it's a matrix but most of them are 0 there are very few non-zero so this is called sparse matrix so maybe typically for a very symmetric molecule only 3 to 4 percent may be non-zero it can be that sparse okay so you have to be you have to be clever in not unnecessarily playing with zeros because you can you can keep on multiplying you will see that you have to multiply these times these so what is the point in multiplying 0 and wasting computer time so instead you have to be clever in picking up only non-zero elements and then doing this summation now that is another issue that we will not discuss today how to do the summation because if you if you drive by the loop lambda sigma then quite clearly many zeros will come unnecessarily and you keep keep multiplying on the other hand what is done is what is called integral driven programs where you take only non-zero integrals and then do the multiplication but then the question is how do you multiply because you don't have loops here so you look at what are the values they also stored not only non-zero values but they also store these indices only for those non-zero values once I fetch the indices I know where it will go so you know it's a like a table table it's very easy to do if you are of course if you have ever done programming you realize that it's a you know it's something that the computer science people if you just tell them they will be excited so it's very exciting actually the programming part itself is very exciting but remember you can't do I have a loop driven program that is wasting the time because too much too many things are 0 so they do what is called integral driven programs non-zero integral driven programs so you only say you only store non-zero integrals and drive the entire program only through them so that is a strategy that we will not discuss today but I just thought I will tell you