 So now we have derived the canonical Hartree-Fock equations. So I just want to quickly tell you what are the features. So we had a original Hartree-Fock equation which was a general Hartree-Fock equation for spin orbitals which was not in the Eigen value form. Then we said that we can do an unitary transformation of the spin orbitals. So all the spin orbitals were transformed from the Hartree-Fock spin orbitals to a new set of spin orbitals chi i prime through a transformation matrix. So that is what we did. I am just graphically showing what we did. Then we said that the original Fock operator, if you remember the original equation was f chi i sum over j lambda i j chi j, that was the equation that we got one particular equation. Then we said you could transform chi i to chi i prime. The F through this transformation becomes F prime which remains invariant. So F prime is equal to F. So that is what we first showed. Then we said that we can choose a u, so choice of u such that the lambda becomes diagonal. So with such a u, then we can write the new set of equation as F chi i prime and the diagonal lambda essentially means that you have a u dagger lambda u which is some epsilon, which is a diagonal matrix. So you have epsilon i chi i prime. So this is what we said is a canonical Hartree-Fock equation. So this is what basically the few steps that were followed to derive the canonical Hartree-Fock equation after you have derived this. So this is the first derivation which is a general Hartree-Fock equation and then you transform to get a canonical Hartree-Fock equation. So note that we have several u. I can keep on transforming but there is only one u which will make lambda diagonal. So canonical Hartree-Fock is unique whereas you can have several non-canonical Hartree-Fock because I can start from this equation, make one more transformation such that it is not diagonal. So that will also become non-canonical Hartree-Fock equation. Lambda will become some lambda prime but non-diagonal. It does not matter but only one u can uniquely diagonalize lambda. Every u will not diagonalize. So that is the whole problem of the eigenvector solution of the lambda. How do I get eigenvalues and eigenvectors of lambda? So that is the reason I am saying that the canonical Hartree-Fock is a unique thing whereas from canonical I can back transform to get several non-canonical Hartree-Fock. So you can see the other way round that I have canonical and then I can keep on transforming through several u on, u2, u3. All of them will become non-canonical. You can look at it in reverse manner. Initially I got one particular non-canonical equation. All the non-canonical equations will be same because f is same as a prime by any transformation. All that will happen that this matrix will keep changing. So lambda will become some lambda prime. So u dagger lambda will become lambda prime. In one particular case, this lambda prime will become diagonal. I hope it is clear. So that particular case is called the canonical. Otherwise in general, I can keep on transforming any matrix to lambda prime. I can keep on transforming. It is not necessary that this is diagonal. Any arbitrary u does not diagonalize whereas a specific u diagonalizes. So I will have several non-canonical which will have same form because any arbitrary u if I transform chi to chi i prime, f remains invariant. But lambda will change. So in general I will get f chi i as lambda prime chi i and I will keep on having several non-canonical form. Only change will be in lambda prime. One of the u will make this lambda prime diagonal. So that is epsilon and then that is called canonical. Is it clear? So canonical Hartree-Fog equation is a unique equation and we are of course very comfortable with the canonical equation simply because it is an eigenvalue equation. It is very simple to analyze the equation. So we will not bother about such a structure where this lambda prime is non-diagonal. But I just want to tell you the chemistry. There is a lot of transformation where non-canonical equations are also very useful particularly many times the bond localized orbitals are non-canonical. They are not canonical actually. In fact canonical equations are such that the molecular orbitals are spread over all atomic orbitals. So completely delocalized. There is the non-canonical ones which are sometimes very useful for chemistry, the orbitals. But then the problem is there is no one number for each orbital. There is a lambda matrix governing the transformation of f, transformation of chi i by f, there is a matrix. So that interpretation becomes difficult. But the chi's are sometimes more significant in non-canonical case as far as chemistry is concerned, spin orbitals, the densities. So you should remember this. But as of now we are interested in f of chi i and we said at the end of the class that now that we are going to concentrate only on canonical equation we will no longer call chi prime. So we will write this as f of chi i epsilon i chi i. Please note that these chi i's are now canonical chi i's. So this is one thing that I want to repeat that while f remains invariant, chi i does not remain invariant. Chi i keeps changing but that change does not change f. So when I am writing this, this chi i is not an arbitrary chi i. This chi i is a chi i which makes this lambda diagonal. Is it clear? The interpretation of canonical equation. So that is those are the spin orbitals that I am talking of and not the original spin orbitals here. So the one which is transformed by u such that the lambda becomes diagonal. It is just that since I am going to always work on this from now on I am dropping the prime. But you must remember that these chi i's are canonical heart reform and the original chi i that I derived may be different from this. This is very useful because for every spin orbital there is only one number which is the eigenvalue. So that is why they are very important and these numbers are now called the orbital energies. So these epsilon i's are called orbital energy and such a sense could not have been given here as you can clearly see because for each i there is no one unique number. So there is a matrix which you get but here there is a unique number which actually comes from an eigenvalue equation. Of course the eigenvalue equation I have already noted is not a straight forward equation. You have to do an iterative solution because f depends on chi i's so you require what is called self consistent field. I have also discussed that but having done that, having solved this we get what is called the heart reform equation now and these are then the final orbital energies after we have solved this equation. These orbital energies are of interest because of an approximation which was developed by Koopmans. So I will just state the approximation and actually this approximation was given by Koopmans based on again a variation method. So the basic idea of the Koopmans approximation, the basic outcome of the Koopmans approximation is the following that if you take out from this n chi i's, if you take out one electron from a particular chi i, so let us say I take out an electron from chi j, so one electron is knocked out from one of the spin orbitals chi j, then in my original Hartree-Fock I will now have n minus 1 electrons. So imagine that I have now Hartree-Fock which is basically chi 1, chi 2 etc. to chi n. So one of these spin orbitals chi j, let us say chi 2, chi 3, whatever chi 1, I have knocked out an electron. Of course I do not know which one I have knocked out but one of the spin orbital is empty because electrons does not have a number. So the spin orbital is empty that is more important. Again I repeat whenever you talk in Hartree-Fock electrons it is actually we are talking in terms of spin orbitals. So that is why all the summation etc. the coordinates of the electrons are dummy because they are usually always integrated. So I have knocked out an electron from chi j which means I now have n minus 1 electron state. So what will be the state? The state will be now n minus 1 electron with some n minus 1 spin orbitals. Now the Koopman said very interestingly if you now do a Hartree-Fock for n minus 1 electron system you will not get the same spin orbitals. Why? Because your Hamiltonian has changed. For n minus 1 electron Hamiltonian has a different external potential. So with the new Hamiltonian when you do a new Hartree-Fock you will get different n minus 1 spin orbitals. So you can do the same exercise for the canonical Hartree-Fock for n minus 1. You will get different n minus 1 and then of course you can get En minus 1 in Hartree-Fock and then you can calculate ionization potential as En minus 1 minus En. I hope all of you know this is the definition of ionization potential. So this is the amount of energy required to ionize the electron. So that is what you will get. So this is known, this is nothing to do with Koopmans. What Koopmans said that the approximation says that to first order one can assume the approximation that the n minus 1 spin orbitals in psi n minus 1 are the same set of n minus, are the same set of spin orbitals in the n electron problem, except Kaij. So if I have done an n electron Hartree-Fock then Koopmans said that if I do a normal Hartree-Fock of course everything will change. But what Koopmans said that to a first order and that is important, this order is very important. One can assume that when I am doing an n minus 1 electron wave function that wave function to the first order it is correct to assume that the rest of the n minus 1 spin orbitals are remain the same. Which means you get an approximate psi n minus 1 prime which is Kai 1, Kai 2, etc. Kaij minus 1, Kaij plus 1, Kai n. Except Kaij is not there. This is missing. J is not a dummy. I decide let us say I knock out an electron from the outer valence, outer valence. So it will be the outer valence Kaij. That is very easy. That is not the issue. When I do an experiment I can knock out an electron depending on the amount of energy that I give. So if my energy is not sufficiently high then it will knock out one outer valence. If it is sufficiently high then it will knock out inner valence or from the core because they are all bound. Outermost electron is not classical. It means quantum mechanical. It means the one which has the least binding. So I have different n minus 1 electron energy, one which is closest to an electron. That will be the outermost. So when I am saying outermost it is only to give you a classical picture. Actually you have n minus 1. That is why I am giving the actual picture that you have n minus 1 electron wave function. Normally it will be different but I assume in Koopman's approximation that I generate an n minus 1 electron determinant by keeping all these bin orbitals identical. That is an approximation. So I am writing a psi n minus 1 prime in this manner. I can call it psi n minus 1 Koopman's just to show that this is an n minus 1 electron wave function derived on the Koopman's approximation. There is no harm. This is not a Hartree Fog. I am repeating. Hartree Fog and n minus 1 will be different. So this is Koopman's n minus 1 where I do a Hartree Fog for n electron and just derive an n minus 1 electron determinant with Kaizhe missing. So is that clear? That is the quantum picture in terms of wave function. That is an approximation obviously because if I do a Hartree Fog everything will change. Koopman's original paper actually shows that his approximation is very good to the first order. That is important. In fact that is what we do not normally prove but that was original theorem. Then Koopman's said the following. I can now construct en minus 1 with this. So I will construct en minus 1 which I call en minus 1 Koopman's through psi n minus 1 Koopman's. H, H of course is now n minus 1 electron Hamiltonian, h n minus 1 psi n minus 1 Koopman. Note that since I am using the same spin orbitals it remains orthonormal because these are orthonormal spin orbitals. So it remains normalized psi n minus 1 Koopman's. You have 1 by square root n minus 1 factorial of course. That is the only difference. Otherwise it will remain normalized unity. So I am not going to divide by this. So this becomes my en minus 1 in Koopman's approximation. So I am calling it k. This is not actually n minus 1 even in Hartree-Fock. What will be en minus 1 Hartree-Fock? If I actually do a Hartree-Fock calculation for psi n minus 1. So I do the n minus 1 electron Hartree-Fock and calculate this. This will become the en minus 1 Hartree-Fock. Of course I am not even talking of correlation because that I do not know how to improve beyond Hartree-Fock. But I am saying within the single determinant I can calculate n minus 1 electron energy in Hartree-Fock and n minus 1 electron energy in Koopman's. This will be vastly different from this because when I do Hartree-Fock the orbitals will change. So I do not know how it will change. I am assuming that in the Koopman's the orbitals are same. So then Koopman's approximation says if I now calculate Ip which I defined as en minus 1 Koopman's minus en Hartree-Fock. This is Hartree-Fock. This is approximation. Then this result is nothing but minus epsilon j. It is a very interesting result. Epsilon j, j is the orbital which is missing. j is a specific orbital, remember. So of course depending on j I will get different Ip and that is why there is no single ionization potential. In a system there are many many ionization potential. Each of them are actually called main p, main ionization. I can knock out one electron from chi 1, one from chi 2, one from chi 3, chi 4 and so on. Each one will lead to different Ip. It so happens that the Koopman's approximation works better if your Ip is an outer valence Ip. Which means the electron, the spin orbital, the spin orbital which is knocked out is an orbital which has the highest energy. That is called the outer most. Now you have to define what is the outer most. Outer most means one which is homo. So that is easy to knock out because its energy is low and the ionization potential is also going to be small because it is minus of epsilon j. So obviously that is going to be small because that is a negative value, large negative, large value but negative means it is a small value. Absolute value is small so this Ip will become actually small. So it works very well. I can define for all Ips depending on j. Note that the j is changing so there is no non classical picture here. j is not a spin electron, j is a spin orbital. So I am defining nu r and nu r psi n minus 1k by knocking out one of the spin orbitals. So I have now n minus 1 electrons in the rest of the spin orbitals. How they are distributed I am not bothered but one is vacant. So then depending on which one I have vacated your Ip in Kubman's approximation so I call it Ipk will be minus epsilon j. That particular vacancy. Very often this Ip is called delta SCF. I hope you understand what is delta SCF. Delta means difference, difference of two Hartree form. So it is called delta SCF. Delta, I hope all of you understand. But the Kubman's is different. Here the orbitals have all changed. In the Kubman's orbitals are same. So that is why I said this and this Kubman showed that it is correct up to the first order. Now what is important is to see Ip, this is the definition of Ip. Remember this is the definition of Ip. So I can put 3 bar within the Kubman's approximation. What is important is to show that this is equal to minus epsilon j. I hope all of you will be able to show this now because I have given you everything. I have given you this. You know what is E and Hartree form? Just subtract and show that the difference is minus epsilon j. Can you do this? So I can give this as a first home assignment without proving that assume Kubman's approximation. So Kubman's approximation essentially means that all the n minus 1 spin orbitals in psi n minus 1 are same as the n electron Hartree form except for of course chi j, bar in chi j which is knocked out. So I have knocked out an electron 1 spin orbital chi j. When I am saying one electron from chi j it means one spin orbital is now vacant. Then this will become my E n minus 1 Kubman where psi n minus 1 Kubman's has all the n minus 1 spin orbitals of Hartree form except chi j. This is vacant. Then use later rules for this for the n minus 1 problem. Use later rule for this E n Hartree form which is already known to you and just show that the residue is minus epsilon j. It is very easy to show. In fact you can already see because this has something more. So the j part is always missing. But you have to show correctly that the Coulomb exchange and the one electron part. One electron part is very easy. That is very easy but you have the Coulomb and the exchange. So you have to show this that finally it will be equal to minus epsilon j. So that is your first home assignment. Show this part. Show that under Kubman's approximation. I have already defined what is Kubman's approximation. I p is minus epsilon j if chi j is the spin orbital missing in n minus 1. This is of course under Kubman's approximation. So it means the rest of the n minus 1 spin orbitals are same. So that is your first home assignment.