 So, we must go back and revisit the Hartree-Fock and some of the points that we want to tell, I think you look at it very carefully, some practice problems I will give so that you can also prepare for your exam. So, I think I just should go back from the beginning that you had canonical Hartree-Fock equation which gives this pin orbitals. So, that is from the beginning and you had expressions of f of r which is the one electron part plus sum over j chi j star r 2 1 by r 1 2 how do you want to write it now? Can you somebody tell me 1 minus p 1 2 good 1 minus p 1 2 chi j 2 d tau that was our first expression. Then we went to the closed shelves we just want to give a quick summary till this point where we expanded so we wrote the chi j of r as phi j alpha phi j beta. So, we have n by 2 sets now instead of n, remember this was n, now we have n by 2 sets. Then we had an expression for the Hartree-Fock equation which is very similar to this but only for the space orbitals, however since they are operator eigenvalue equations each of these eigenvalue equations gives much more than n or n by 2 once the operator is defined you get infinite number of eigenfunctions and eigenvalues that I want to mention either here or here. This f of r which is now in space orbital actually I should write this f is not f of r, f of r omega just to correct it because you have spin orbitals. So, this should be actually f of r omega this particular one is just f of r. So, that is f of r is h of r plus sum over j equal to 1 to n by 2, then phi j star r 2 1 by r 1 2 2 minus b 1 2 phi j of r 2 all these spin orbitals you should write r and omega. So, here we are writing only in terms of r. So, the spin has been integrated. So, this becomes the canonical Hartree-Fock equation for the space orbitals for the closed shell systems. After that for molecules we expanded each of phi i of r as a linear combination of the atomic orbitals or some basis which you now call a mu of r. This is a known basis of known functions by projecting by expansion of this equation and then projecting by different a mu. We wrote the equation which is f mu mu c mu i sum over mu equal to epsilon i s mu c mu i. So, that is your Hartree-Fock-Ruthan equation and then we gave you an expression for f mu mu which in terms of the matrix elements of h which you call just h core mu mu which is the matrix element of h in the basis of the a mu and a mu. So, h core mu mu plus now can somebody tell there are several expressions you can tell in terms of coefficients and density matrix yes there are several expressions that you derived, but I think you should be able to derive one from the other I can ask you to derive one anyone. So, let us okay lambda sigma let us write it in terms of coefficient first sum over i 1 to n by 2 now tell me in terms of coefficients c lambda i i okay star c sigma i see you have the f of r here basically I am taking its matrix elements over mu mu they are expanding phi j. So, that j instead of j now you have told i I have no problem that was a dummy index. So, now you have lambda so I am doing nu so it should be nu first then lambda here right yes 1 by r 1 2 which I am not writing mu sigma that is a Coulomb integral what about factors this factor 2 will remain or no what is this factor go yeah this will be a factor 2 minus the exchange is it okay just you have to just look at this expression okay instead of j you have written i that is all and you are expanding this in lambda expanding this in sigma. So, the coefficients are coming and of course you have further integration over nu and mu so now it becomes a complete integral in this was a function of r so it becomes a complete integral so that is what you have written you can write incomplete also that is you just expand this in terms of atomic orbitals write f of r that also you can do and then later on take nu mu it is one and the same thing okay so you should be able to derive anything from anything so I am just saying these are my practice problems some several such problems may come to write so you must practice properly so then we define a density matrix of P mu nu which is 2 times sum over i equal to 1 to n by 2 C mu i C nu i star do not worry about what dummy index I am using what is important is that this comes here this comes later with the conjugate in this case so that if you write in terms of a matrix then the matrix P is 2 times C C dagger you know those who are more familiar the matrix notation that will be easier because you should be able to show this from this very easily okay so this is a column matrix square matrix this is also a square matrix if you multiply C C dagger because you can see that this is C dagger i nu so it becomes P mu nu so that is one way to remember how to write this in this I mean 2 C C dagger is much easier to remember then I can write this entire matrix in terms of projection operator by summing over i the summing over i is there so the same expression now can be simplified with just lambda sigma this is gone so now this becomes P what so what did I had remember what did I had C sigma lambda good this factor 2 goes however there is a factor half here so this becomes my expression for the Fock matrix in the A basis then we said that this basis is a basis which is an orthonormal basis so we cannot actually solve this equation so we went to a basis which is non orthonormal basis sorry so the basis is now an orthonormal basis which you do by transformation Mu prime as some x nu mu nu to make this base new basis normalized you remember that I must have x dagger S x as an identity matrix that was a condition please remember each of these x dagger S x is identity matrix so one of the identity one of the matrices that we suggested was S to the power minus half which satisfies this then we said how to get S to the minus half you diagonalize S get the unitary matrix U then you take the diagonal matrix consisting SD which is which consists of the eigenvalues and simply do back transform so that's your S to the minus half so U is a matrix of eigenvector for S so diagonalize S SD consists of all the eigenvalues diagonal elements construct S D to the power minus half and back transform once you do this then this matrix this equation can be written in terms of normalized matrix a normalized basis so that becomes your F prime C prime as C prime E remember that this matrix was FC equal to SE where this E is again a diagonal matrix consisting of the eigenvalues epsilon i so you get F prime C prime equal to C prime E and F prime is x dagger F x and what are C prime? C prime was x inverse C right or C equal to x C prime so this gives you the strategy to solve the entire SCF iteratively of course iteratively but every time after you get the S every time you construct the F matrix from either the coefficients or the charge density the bond order matrix you reconstruct F prime diagonalize from C prime you get back C then reconstruct P reconstruct F from this expression go back to F prime and then again re-diagonalize and keep doing it so I hope that flow should be very clear to you how to do that and then you get eventually the convergence test is also done over the P matrix so the coefficient these are basically the convergence in the coefficients so if you take the matrix elements of P matrix element by element take the difference square if the total sum is less than certain preset value then it is converged so that is basically the entire Hartree-Fock before you go forward to analyze little bit about the charge density bond order matrix. I also mentioned that there is another orthogonalization procedure which is this is a symmetric orthogonalization the other one is called the canonical orthogonalization so it is important to note what is a canonical orthogonalization so I will just spend a few minutes on that so for canonical orthogonalization we use x as u sd to the power minus that is it so you do exactly the same you first diagonalizes construct sd construct sd to the minus half but just do a transformation with u so the difference is there here we did u dagger here you stop with just this obviously it is not symmetric as you can see because s to the minus half of symmetric because s was a symmetric matrix we have lost the symmetry that is why it is called canonical but it is sometimes convenient particularly when there are this linear dependence that we are discussing that sum of the eigen value of s becomes 0 in that case this is convenient so x equal to u sd to the power minus half I hope you can show that this x also satisfies this condition I hope you can show if I ask you to show you should be able to show right difference is that here is s to the minus half was u sd to the power minus half u dagger right x is the canonical transform canonical orthogonalization this is love dean u sd to the power minus half u dagger in love dean so now we are proposing a canonical x which is just u sd to the power minus half so I hope you will be able to show this that x dagger sx is still identity yeah of course you have to substitute it is very easy to show right is it clear to everybody but first first get convinced that this satisfies this otherwise it cannot be an orthogonalization x right so that I hope you can all see because x dagger is adjoint of this so first 0 to the power minus half then u dagger will come then s will come then u will come and u dagger s u is already sd so you have sd to the power minus half, ssg to the power minus half so this you will actually get sd to the power 0 so x dagger or sx will actually become sd to the power minus half u dagger so that is your x dagger then you have then you have x, which is u sd to the power minus half. So, this gives you sd to the power minus half, sd to the power minus half. So, that is sd to the power 0. So, that is identity. So, both of them give the same. So, this is a second orthogonalization that is very often also done and is sometimes preferred simply because of the following reason that if I have some of this sd very, very small, some of the eigenvalues, then the way to do it is that you reorder this in the steps of from higher to the lower ones and just the lowest ones you simply eliminate. So, you reconstruct a new sd to the power minus half which now has less number of elements. Let us say you have total m number of elements and small k are 0 or close to 0, then you have m minus small k number of elements and then you transform this. So, you get then a matrix which is actually a rectangular matrix but you do not use the rest of them. You have only a columns m minus k. So, you just use only that part of the matrix to transform and do the rest of the jobs. So, that is why this is easy. In this case it is complicated because you have both even Eudager. When I do that, if I would have done the same thing here, you have both even Eudager and then again it will transpose to give you the full matrix. So, that would have been little bit more complicated where here because the right Eudager is not there, it is easy to actually use this by reordering. So, that is why this is more preferred. Again it is a very technical thing. Most of the time we do not have linear dependence in basis. So, this point is irrelevant in all such cases. We can actually use any one of them and I prefer this simply because it is symmetric. It looks nice that you have both. Here you have one side Eudager, it does not have another side. But otherwise it is very similar. So, you have to do the diagonalization of S in the beginning and then construct this. I will ask you a very small proof. Do a practice problem that all the eigenvalues of S are positive. There is something I am not worried about right now. If they are not positive it is very hard to take inverse square root of Sd. I hope you appreciate because there is a negative number. How do you take the inverse square root? So, you cannot take a square root of inverse number, a negative number. So, the point is that all these Sd elements which I have not mentioned, all the elements of Sd are actually positive which means one can show that the eigenvalues of S are positive. Please do this as a practice problem. This problem is also given in Zabo and Aslund. Please note that Zabo and Aslund is our reference textbook. So, this problem is already there in Zabo and Aslund. Hint is given. How to solve? Please see the hint and practice it because this is one important piece in this whole construction that the eigenvalues of S are positive which I have eliminated but it is easy to see. So, I had already given modern quantum chemistry. This is our reference book incidentally except for the symbols. Symbols may be different. Modern quantum chemistry by Zabo and Aslund. Attila, Zabo and Neil Aslund but anyway should be able to find it. It is downloadable. I think this book is downloadable and I think many of them have already downloaded. That is my reference book. So, many of the problems are actually given there. So, this is something that is already given. Hint is given but not the solution, entire solution. So, please try to think and do it. Let me come back to the analysis of this P-metrics. Before that, I think let me also tell you. So, when you read, please remember that the Zabo and Aslund symbols are different. For example, the spin orbitals are chi but the molecular orbitals are psi instead of phi and the basis is phi instead of A. So, Mulligan notation is very simple. Interchange the second and third index. Very simple. So, you can always go back and forth. So, that apart, I mean most of it is from that textbooks.