 after having done this, let me now state that what is my lambda ij? So, lambda ij if you note your lambda ij is nothing but matrix element of this quantity with chi j, chi j1 f of 1 chi i1 and lambda ij prime is nothing but the same thing, but with chi j prime. So, what is the reason lambda will change to lambda prime? Because the lambda depends on chi j. It is a matrix element of Fock operator. It is very easy to see this. If you just make a multiplication because they are orthonormal only one term will survive. So, of course you can say k, kj, ij whatever there is a dummy index, there is specific index. So, the point is that this is the matrix of Fock operator in a basis. This is also a matrix of the Fock operator in the new basis which is unitary transformer. Now we know from the matrix Eigen value equation that if I choose a proper unitary transformation any matrix can be diagonalized. Now this is why I am doing unitary transformation. Now you understand immediately? So, without any further ado, I would now say that choose a matrix u because I have several unitary transformation. So, I will say choose u, this is my choice now such that lambda is diagonal. So, something like this u dagger lambda u is some epsilon where epsilon is diagonal matrix. So, this is a diagonal matrix. You can always write u, u lambda u dagger does not matter whichever way you want to write. This is always possible because this is a matrix and not only that this is a Hermitian matrix. I must also make sure that you understand this is a Hermitian matrix. You take a dagger of this, you will see that this chi i star will come here and this will become lambda j i. So, this will also flip and the Fock operator by itself is Hermitian. I should have said this before that if you look at the Fock operator the F operator is also Hermitian. I think I must mention this point. Once again I hope you are able to see this. H is Hermitian. This is my j j. j j is also Hermitian because if I take adjoint of this what will happen? This chi j will come here, this chi j star will go here. This is already a Hermitian operator. Same way exchange P12 is a Hermitian operator. So, entire Fock operator is Hermitian. This is nothing but a matrix element in an orthonormal basis of a Hermitian operator and hence this is also a Hermitian matrix. This original lambda is a Hermitian matrix. Please note that there is a difference. This is a Hermitian operator. This is a Hermitian matrix because this is a matrix. These are numbers. Whenever you have numbers they cannot be an operator. Of course number can be an operator but in this case these numbers can be arranged in the row and column. So, that is the reason I said it is convenient sometimes to look at j i because then you will see that the j is on the left, i is on the right but it does not matter. However, you look at it. The point is that this lambda is a Hermitian matrix because Fock operator is Hermitian. Since it is a Hermitian matrix I can always find a unitary transformation such that the matrix can be diagonal. So, now I am saying that let us choose a u such that lambda is diagonal and then rewrite this Hartree-Pock equation. Quite clearly now this is my lambda prime. So, lambda prime is now diagonal. So, lambda prime ij is now epsilon let us say epsilon i delta ij. That is the meaning of diagonal. So, this is a diagonal matrix. So, the new lambda prime in the new basis will become diagonal after transformation. Note that I am transforming everything to chi but the lambda transformation has to happen in this manner because lambda has one chi on the left, one chi on the right. So, both of them have to be transformed and quite clearly I can always choose a transformation such that this will become diagonal. Now quite clearly if you say that any unitary transformation I take I cannot show that it is diagonal. Any arbitrary transformation no I cannot show. So, now I am actually deviating from here I am saying this is a matrix let us choose u that is very important such that it is diagonal. So, this is my choice. The question is how do you choose u? Now that is a point where you are actually saying how do you diagonalize a matrix. Now that is something that I will not bother right now. Why do I choose? So, then there will be question because I am following variation method. My mathematics gave me a matrix. So, if you choose that is another constraint. I did not know my mathematics what it will need to. I started variation psi A psi I only had a constraint where chi should be orthonormal and lo and behold I got this equation. I did not know what equation I will get. Once I have got this equation now I am trying to analyze how can I make it diagonal and then I find that this is the Hermitian matrix. So, Hermitian matrix can be diagonalized by unitary transformation but then I have to worry about what happens to the rest of the equation through the unitary transformation. So, one thing I understood that the f remains invariant and then I can write this rest of the equation is just f. So, this becomes chi i prime. This spin orbitals are now transformed spin orbitals of course but that I do not care. So, eventually I can then write f of 1 as some new chi i prime 1 equal to epsilon y chi i prime 1. You understand? Because now your lambda prime in the new basis is diagonal by my choice and that choice will be dictated by how to diagonalize the matrix. So, that is the part I am not going to discuss here. That is the mathematics. All of you should know by now how to diagonalize the Hermitian matrix. How do you get this equation? Provided I get u, what I am now suggesting is that the new spin orbitals can be chosen such that they are eigenfunctions of the foc operator. New foc operator is same as old foc operator. So, I am not writing f prime. Is it clear? And now I have got a canonical form without any problem which is a nice eigenvalue equation which is what we wanted all the time and which is what you are used to. So, I will give the interpretation to what is epsilon i. You already know these are orbital energies. These are the orbitals and everything. But I will give the interpretation. Today what we have done is that we have started from the non canonical Hartree-Pof equation which we have derived last time. So, we have started from non canonical Hartree-Pof equation which is a general which this is the form that we derived from the variation method. So, this I am not going to worry. It is a non canonical Hartree-Pof equation. So, this is something that I have derived from the variation method. And the form of that was f of chi i is sum over j epsilon i j chi j. I am not writing f of 1 explicitly. It is an operator equation. So, it does not matter. So, I have started from this non canonical equation. Then I discussed the effect of unitary transformation on this equation. I noted the fact that the f is Hermitian. That is the first thing. And hence, sorry, not epsilon. This was lambda. It does not matter. And lambda is a Hermitian matrix. This is something that I notice and I preserve it in any basis. It does not matter what is the basis. As long as this equation is satisfied, whether you do unitary transformation or not, this is a Hermitian operator. So, this is a Hermitian operator and the lambda is a Hermitian matrix. Now, I do a unitary transformation. So, unitary transformation leaves f in variant. So, that is the next thing we showed. Unitary transformation also leaves the spin orbitals orthonormal. It retains the orthonormality of the spin orbital. In general, any basis, if you have any basis, you can do unitary transformation. So, that also we saw. Then what we said that after having known all this, we are now merely saying, choose a transformation such that lambda is diagonal. This is my choice. So, please note that by an arbitrary unitary transformation, I cannot make lambda diagonal. So, you will get arbitrary lambda prime. But now I can choose such that it is diagonal. So, my Hartree-Fock equation then will become canonical and it will become f of chi i prime in the new basis as epsilon i chi i prime. The same Hartree-Fock equation which I started now has a canonical form simply because I am able to make this matrix diagonal by unitary transformation, by suitable unitary transformation. I could have right away said that this is a Hermitian matrix. So, simply diagonalize this. But then you would have asked question, if I diagonalize this, what happens to the rest of the equations because of unitary transformation? So, that is the reason I have to show elaborately that the f operator remains invariant, the unitary, the spin orbitals remain orthonormal. So, I can do it. I have no problem. Otherwise, it was very easy to see. If I just wanted to make it diagonal, yes, that is what we know, just do unitary transformation. But it might have created havoc on the rest of the equation. Then I have to re-understand the equation itself. So, the point that I am trying to show that the equation does not change because the fork operator remains the same. That is a very important thing and I get a canonical Hartree-Fock equation. So, this is canonical. And all my interpretations become much easier in the canonical form. But note, however, and I will show later that the total energy of the system remains same whether you use non-canonical or canonical. The only thing, however, that has changed is the spin orbitals. So, spin orbitals are different, but eventually I will also show that the wave function remains the same. Energy will remain the same. I will also have to study what happens to the wave function because wave function was in the old spin orbitals. Now, I have new spin orbitals. I must understand has the spin old function changed and I will actually show that that does not change. Hence, energy also does not change. Total Hartree-Fock energy. So, everything remains same. Orbitals, however, change. These epsilon i's are called orbital energies. So, they have a meaning and this has no parallel for the non-canonical part because non-canonical it has a matrix. So, I cannot say that one number for one orbital. So, it has no parallel. So, the orbital energy is a very special term. Remember, which is used only for canonical Hartree-Fock energy. If I tell you this is the orbital energy, that means we have solved canonical Hartree-Fock energy. Otherwise, orbital energy has no meaning. And lot of interesting things will be done interpretation, physical interpretation will be done through the canonical Hartree-Fock energy. So, let us look at what happens to the wave function first. So, let us look at the wave function in the new spin orbitals now. So, if you remember the wave function, now I call it Hartree-Fock wave function. In the original basis, this was chi 1 1 whatever, chi 1 2 etcetera, etcetera, chi 1 n, chi 2 1, chi 2 2, chi 2 n and so on. So, it is a determinant chi n 1, chi n n. When I do the new Hartree-Fock, what will happen is that each of this chi i's will now become chi i prime, which is related to u by the matrix transformation of u. So, one can write this matrix itself. How will this matrix look like? Let us say this chi Hartree-Fock is determinant of u. You understand? And a is this matrix. So, I am just rewriting this. This chi Hartree-Fock is a determinant of a matrix. We have never seen it this way. Of course, there is a 1 by square root n factorial. So, let us not forget that. That is fine. This is not going to change anyway. So, I am worried about what happens to the determinant of a when the basis is changed. When the basis becomes chi i prime, it will be very easy to show that the new matrix a prime will be nothing but a times u. Please do this exercise because each of this chi i, when I write as a sum over u chi j, you can very easily separate this into the two product of two matrices a and u, which would mean that the new chi Hartree-Fock prime, please check this yourself. Chi Hartree-Fock prime, which will be determinant of a u, which is nothing but determinant of a times determinant of u. What is the determinant of u? It is a phase not 1, exponential i phi. Please do not say 1. 1 is an incomplete answer, right answer, but only an incomplete answer. I hope all of you will be able to find determinant of u. We have done that in various courses. From the very fact that u, u dagger equal to identity, you should be able to do this. Determinant of whole thing, determinant of u into determinant of u dagger, determinant of u dagger is very easy, determinant of u star. What is determinant of u dagger? It is u star. So, this determinant is u u star is equal to 1. Remember determinant is a number. So, when I take determinant on the both side, this is 1. This is determinant of u, determinant of u dagger. Determinant of u dagger is nothing but u star determinant. So, this, so it is a number, it is a number. So, let us say x, x into x star equal to what is the value of x, exponential i phi. Not because in the complex plane, so determinant of u is nothing but exponential i phi. So, that means my psi-hat-tree-fock prime is nothing but determinant of a, which is psi-hat-tree-fock times a phase factor. And all of us know that the phase factor does not change the wave function. I hope you remember the basic quantum mechanics that I can multiply my wave function by any number or any complex number like exponential i phi, no property will change. So, the energy obviously will not change. I do not have to prove you now because hat-tree-fock energy is nothing but psi-hat-h, psi-hat-f. So, psi-hat-f on the left, psi-hat-f on the right, psi-hat-f on the left will give you, psi-hat-prime on the left will give you exponential minus i phi. And on the right, it will be exponential plus i phi, it will become 1. So, it is very easy to show that the energy also remains in value. So, because of this now, e hat-tree-fock prime is psi-hat-tree-fock prime H, psi-hat-tree-fock prime and is same as psi-hat-tree-fock H, psi-hat-tree-fock because these two together, these are related by only a phase factor. So, only phase factor on the left and on the right. So, the left one will produce you exponential minus i phi, multiplies the exponential plus i phi this one. In fact, that is the reason no properties change. So, you can always multiply anything by phase factor. Of course, phase factor also leaves the wave function normalized. That is also important. That is the reason I am not dividing by anything. If I on the other hand multiply by 2, then what will happen? This will actually become 4 times, 2 into 2 because it is no longer normalized. Then I have to also divide by psi-hat-tree-fock prime nor which is again 4. So, I would have got that 4 cancelled. That is the physics. But here I do not have to do because this remains normalized and I see that energy remains invariant. So, the hat-tree-fock energy, hat-tree-fock wave function, they all remain invariant. So, the only thing that is changing however are the spinorbitals. So, spinorbitals are not sacrosanct. Canonical spinorbitals are one, non-canonical or something else. And I can actually generate several spinorbitals, set of spinorbitals. And that is the reason in the molecule, you must have already read that let us say methane molecule. I have spinorbitals which are delocalized over the orbital, over the entire molecule. They are called delocalized orbitals. Forget about spin, delocalized orbital. I can then make a unitary transformation such that the spinorbitals are only one direction of the bond, carbon-hydrogen. You must have seen directional properties. So, they become localized spinorbitals. So, delocalized versus localized spinorbitals. Actually it turns out that these delocalized ones are the canonical ones. That is a later part. So, all the canonical orbitals are actually delocalized. So, it is very hard. That is the reason when the molecular orbital theory, we start with this orbital, they are all delocalized. It becomes more and more difficult to do simple chemistry sometimes. But then this is the way to go forward. Olden days when people issued valence bond, they were all localized. We know that these are the bond, these are atomic orbital, they are all localized. But then they did not have orthogonality property. Lots of problems were there. Here everything is preserved. So, when the high performance computers came, this became the method. Even though sometimes the chemistry gets a little fuzzy because I can always get back the chemistry later, so this becomes an easy way to do the canonical Hartree-Fock equation. So, with this I think we will not worry about the non-canonical Hartree-Fock. From now on, whatever is the Hartree-Fock is the canonical Hartree-Fock. So, whatever in the future that we will discuss is only canonical Hartree-Fock equation. So, I hope all of you realize that they are identical. There is no difference. And hence, since we are only going to discuss canonical Hartree-Fock equation, I am not going to write this prime anymore. So, the prime and unprime was my creation. So, my canonical Hartree-Fock will be chi i because there is nothing called the old chi i now. Is it okay? This is not any cheating. It is just that my symbol I am changing. Why to write prime all the time? So, I will say drop the prime and we will start with f chi i equal to epsilon i chi i and we assume that these are the canonical arguments from the rest part of the discussion that we will start. So, we will actually look at the physical interpretation of canonical Hartree-Fock equation. What is the, how can I write an expression for orbital energy? What is the interpretation of orbital energy? Particularly in terms of Koopman's theorem that the negative orbital energy is ionization potential of electron affinity. All that will come. So, basically interpretation of the orbital energies and some more interesting aspects of the Hartree-Fock theory, before we go do the same equations in spin adapted form. Remember, we have written everything in terms of spin so far. You may like to know, can I do the spin integration? How to do? It is a trivial exercise. So, I will just spin integrate the same equation very quickly. For a specific system, let us say again closed shell. We have already discussed what is closed shell or W-occupied orbitals. We will actually show how to do that and then we will have a Hartree-Fock equation only in spatial orbitals. Now, it is in spin orbitals. Hartree-Fock in spatial orbitals. Then we will see how to solve this. The solution is not so easy. So, how do I solve? But before that I will do the spin integration and how do I solve it and that is where the matrix and lots of things will come. Basis said that you have heard that will all come in the solution. Right now there is nothing called basis set. The basis set will come when I try to solve the spatial orbital Hartree-Fock equation. So, that will be the way and then I will go over beyond Hartree-Fock later. I will close the lecture today.