 So, I go back to hydrogen molecule. Hydrogen molecule is something that all of you have read very well. So, hydrogen molecule is two atomic orbitals. The simplest basis that you can think of, what is the simplest basis? 1sA, 1sB. Remember, I did give this as an exam problem. I am now going back to this. You have two molecular orbitals. And this is exactly what you do, Routhan equation. You solve the Routhan equation, you get those. So, you have a plus or minus and you have 1 by square root 2, 2 into 1 plus minus s, where s is overlapped in 1sA 1sB. So, you get what is called the bonding orbital or an anti-bonding orbital. And that is very famously done in the textbooks as two 1sA functions, giving you two molecular orbitals. So, I just want you to reflect on these that whatever we are doing, it is actually in line whatever you have read. It is not so simple, but this is the Routhan equation. This is not just the Hartree fog because molecular orbital has a combination of atomic orbital. So, this is sigma g. If you note, this is sigma g. So, this is all that you have read. And you have two electrons here originally which are now paired here. And this determines the amount of binding and so on. So, my Hartree fog wave function for hydrogen molecule is sigma g square for the space part and then you have an alpha 1 beta 2 minus beta 1 alpha 2, correct? I mean that is trivial to show because you have a sigma g alpha and sigma g beta. If you expand the determinant, you have a sigma g 1, sigma g 2 and a singlet part. This is singlet. I am not going to worry about this part, but I am going to come back and worry about this part. So, let us look at sigma g square. So, it is a very simple Hartree fog, the simplest Hartree fog that you can do. So, I analyze the sigma g square. What is sigma g square? It is sigma g 1 into sigma g 2, correct? That is what is sigma g square. So, let me expand sigma g. Sigma g is barring the constant, basically some constant times 1 SA plus 1 SB, right? And again 1 SA plus 1 SA. When I am writing this, please note this is for the coordinate 1, this is for the coordinate 2, correct? But both of them are same. These are now written in terms of atomic orbitals. And let me, so let us say that is constant K. So, it is K square times 1. So, let me now expand this further. So, I have K square into 1 SA 1 into 1 SA 2, correct? Remember, this is a coordinate 2. This is coordinate 1. So, 1 SA 1 into 1 SA 2 plus 1 SA 1 into 1 SB 2, right? Plus 1 SB 1 into 1 SA 2, correct? Plus 1 SB 1 into 1 SB 2. I am sure you have done this many times before in the MSC. So, I want to go back. When you write it in this term, you get what is called, what chemists call balance bond picture, correct? Balance bond pictures are in atomic orbitals. So, I hope you can recognize that there is a balance bond picture. This is an ionic term, correct? This is an ionic term. So, these are ionic term. These are a covalent term. This is a covalent term. So, if you analyze the restricted Hartree-Fock of hydrogen molecule, you can clearly see that it is a linear combination of covalent and ionic terms, which is something that you probably would expect that it is not purely covalent. It is a little bit of ionic term should be there, but what is surprising is that the ionic terms are of the same proportion as the covalent term, right? I think you have read this in the MSC that that is why we say that in standard molecular orbital theory, the ionic terms are over represented. Many times, this comes in various exams, maybe in NET and several MSC exams. So, I am just repeating that why it why does it happen? And this has a serious outcome on a solution, particularly when I want to calculate hydrogen molecule at a large distance. So, this R goes to infinity. So, I want to do the same hydrogen. Remember, we are doing a Born-Oppenheimer calculation. So, we fix the geometry. Now, I fix the geometry at a very large r. What would you expect? You expect the wave function to be only a product of the hydrogen atom wave function. Is it clear? Because it is a non-interacting problem now and the energy should be some of the two hydrogen atom energies. Unfortunately, if you do R H F, that will not come. So, let us say that this is my actual curve. So, this is H plus H energy. This is the potential energy of H2. Remember, this is the total energy of H2, which means apart from the electronic energy, I have added the nuclear-nuclear repulsion. If you do Hartree-Fogg, then you will get something like this. So, this is your R H F. It will actually overstep this line at a large distance. So, this is your distance R H H and this is the total energy, what I call V or whatever. So, it will actually overstep. Of course, the equilibrium geometry is not bad, except that it will not be exactly equal to exact. That is something that you expect because this is an approximation and by variation theorem, you expect that this should be higher than this. So, that is also fine, but what is important is that qualitatively, it does not approach this and that is a serious problem because then it actually predicts a wrong separation. In fact, what it approaches is quite close to H plus H minus. In fact, if you do an excited state, both of them converge the same number at large distance. Now, quite clearly, this happens because of this fact because at large distance, you expect the hydrogen molecule to be purely covalent. It is just hydrogen into hydrogen. Whereas, in the intermediate distance, there is some ionic terms. As you know that Heitler and London of course did not include, but there is a Wang who actually include the, we have not done valence bond incidentally, but I hope you remember valence bond in the MSc. So, there is a little bit of ionic term which is still okay, but in the regime when hydrogen is separated, it should be purely covalent and this ionic term must vanish, which it does not. So, that is one of the problems. The second way to look at it is that when I do sigma g square, they are spin paired. So, they are spin paired, but when you go back to the 1s a and 1s b and that is what exactly happening at large distance, then the spins tend to remain paired. Whereas, in the two hydrogen atoms, this spin can be anything. I mean, I have written it up and down, but it could be down or up, anything, but here one of them remains up, one of them remains down and that means enough variational flexibility is not given. So, the energy again is higher. So, there are many ways of understanding the same problem, but essentially in fact we will come back to this problem because this is a very serious problem. Why Hartree-Fock fails? So, even single bond stretching, when I stretch the bond Hartree-Fock fails. So, that is the essential message that I am going to give and we will come back tomorrow to analyze more why it happens and how do I correct it? Because the correction is very important and you will realize the correction will now take you beyond Hartree-Fock. That means you have to go beyond single determinant. So, this is one example from where I can say why Hartree-Fock fails. Why apart from the fact that you require accuracy that I talked about, there is a qualitative failure even single bond stretching it cannot do in some cases. The reason it is happening, it is very clear is that originally system is closed shell, it is spin paired, but when it goes back these two are open shell. So, if it is open shell the spin should have enough flexibility. So, today it is realized that the RHA fails to stretch bonds for systems which are originally closed shell, but breaks up into two open shell fragments. So, this is an example where a hydrogen molecule breaks up into two hydrogen atom. Each of the two hydrogen atom is open shell. Remember open plus open can be closed because number odd number odd number even number that is not a problem. So, the total number is closed, but it breaks up into two open shell in all such cases restricted Hartree-Fock fails. In fact, it is well realized today and one of the problem is because of the spin. So, spin is a very important part to understand this failure. So, it is today understood that not only hydrogen, but if I do Li2 to Li plus Li, again there is a failure. However, if I do beryllium dimer BE plus BE, RHF is good. So, this is not good, but this is good. Same way if I want to break Li2 to Li plus plus Li minus RHF is good. Why? Because Li plus and Li minus both are closed shell they are not open shell. So, there is a lot of understanding today when it fails when it does not fail. So, whenever closed shell system breaks up into closed plus closed RHF is good. So, it depends on what reaction you are studying. In fact, I am pretty sure many people do not even bother about it. They are looking at a reaction of something fragmenting into two parts. You have to look at the spin of those two parts. If originally it was closed shell, if they are breaking into two closed plus closed then it is fine. A reaction can break up into any parts depending on which is the reaction channel, the channel of the reaction. So, in a simple way I can say Li2 can go to Li plus Li, can go to Li plus Li minus and so on. Of course, if you ask me ground state, this is the ground state. This is not the ground state, Li plus Li minus. I hope all of you know that any AB diatomic molecule and I will come back tomorrow and discuss this when I discuss this more and there is a very nice theorem and Moolikan actually said why that any diatomic molecule AB in gas phase, remember everything that we are doing is gas phase separates ground state into A plus B. So, obviously if I give choice the ground state will be Li plus Li. So, it cannot do, but if I want an excited state then maybe RHF will be able to do, but that is really not what happens. I hope all of you know this that any diatomic molecule in gas phase always separates into A plus B, remember this. Sodium chloride for example, and I ask this question many times people get confused, the immediate answer is Na plus Cl minus wrong. Even sodium chloride in gas phase it is Na plus Cl, even such an ionic compound. The reason is very simple, the ionization potential of any atom is always greater than the electron affinity of any atom in the periodic term. You can take out the periodic term. So, the most easily ionizable like cesium, even its ionization potential is larger than fluorine or chlorine. So, the energy required to ionize is always larger than the energy that you can gain by negative ion, which means A plus B minus, which means A plus B minus energy will be always higher than A plus B. This is a very simple thing, you can look at the periodic table and that is the reason in the gas phase even sodium chloride will become Na plus Cl, but what you know of course sodium chloride as Na plus Cl minus is in water, there something else is happening because of what solvation effects, because of H plus OH minus they are stabilizing the ion ion pair. So, that is an extra thing, but in gas phase where there is nothing else and that is our calculations are all in gas phase, remember it becomes A plus B. So, of course lithium 2 will go to Li plus Li and RHF will fail for the ground state. I will come back to this discussion particularly I am concerned about the hydrogen which is such a simple system that it cannot give a correct result, what we will discuss is how do I correct it and that will be the first example where I will show that qualitatively you have to go beyond single determinant, even qualitatively. Of course I have told that most of the chemistry is on difference energies, remember. So, even though Hartree-Fock gives you 95 to 98 percent of the total energy it is not good enough, because the difference energies are much less. So, you need to get numbers which are within point 1, point 2 percent accuracy, but apart from the numbers there are certain qualitative lessons and this is one of them that I want to come back and discuss tomorrow because today it is late, but at this point please understand that even the sigma g square has both covalent and ionic parts in equal proportion. So, obviously there is something wrong in the simple RHF and of course we will see how to correct them by adding determinant we can do UHF there are lots of things by which you can correct, but UHF is not good because we lose spin. So, how do I correct them by bringing by keeping them spin adopted that will be my challenge, alright. So, after this discussion and telling you the problem of Hartree-Fock will actually go beyond Hartree-Fock. How do I now correct the techniques or the methodologies in particular we will do perturbation and CI as the two major parts. Variation theorem I have already handled including linear variation please remember the Macdonald theorem and so on, but perturbation I will have to revisit although I had done it before in 4 to 5 and many of you have done, but in this class I have not done. So, I will do perturbation again thoroughly before we go to actual perturbation theory to correct Hartree-Fock which are called MP2, MP3 etc. So, we will slowly move forward. Thank you.