 In fact, if you go further, you can also have something called electron affinity and before you do that, let me first write down again epsilon i. So, please note how to write epsilon i. Epsilon i is nothing but chi i f chi i. Note that whenever I am calling orbital energy, I have always a reference to canonical Hartree-Forcky question. So, that part is clear. So, if I just take the matrix element or expectation value of f with chi i, it is trivial to show that this is called the orbital energy. And you can actually write down. You have the matrix element, you have the Fock operator. So, you can write down this. So, can you write this down? It will become chi i h chi i. So, this will help you now to clean up the home assignment very easily after this. Once I am able to write down chi i h chi i plus sum over j chi i chi j, anti-symmetrize chi i chi j. I hope all of you can show this as well. I will not write anything more because this will be a part of the assignment actually. When you get epsilon j, you should be able to write this because you have the coulomb and the exchange operator. I have combined both coulomb and exchange operator as a chi i chi j. Remember, the sum is only over j, not over i. I have already repeated because we are talking of a specific epsilon i. And if you also note, our E Hartree-Fock had a sum over all i chi i h chi i plus sum over i here, but it had a factor half. That is the difference in the Hartree-Fock energy that you have a factor half, which all means that the E Hartree-Fock cannot be sum of the orbital energy. I have discussed this before also, but let me again repeat it because if I take sum of the orbital energies, this sum will come here. Here also, sum will come but without the factor half. So, I have already repeated that if you take sum of the orbital energies, what is happening is that the interaction between the electrons are double counted. So, once you are counting, let us say you have a 2 electron, you are taking epsilon 1 plus epsilon 2. When I take epsilon 1, I take the interaction of 2 and 1. When I take epsilon 2, I take again interaction of 1 and 2, but between 1 and 2, there is only one interaction as far as total energy is concerned. So, that is why this factor half is coming here to prevent double counting and that is the reason the orbital energy sum is not Hartree-Fock energy. Again, repeating, many people have this problem. They get homo, lumo, everything, they just count as a total energy. Hartree-Fock energy is not sum of the orbital energies and that is very important to realize. In fact, you can very easily show what it is. It is actually sum of the orbital energies minus something. So, if I have n electron, minus half of sum over ij kai i kai j. That is trivial to show because if I take all the sum of the orbital energies, the half factor will go. So, I have to subtract half of these to get half of this. So, these are trivial mathematics. To show this, you can keep playing around either way. So, this is the part which is double counted. So, this is a single pair energy. A single interaction between i and j has a half factor. When I take sum of the epsilon i, the half factor goes because you are counting it twice, i on j, j on i. So, this is actually a double counting term which is to be subtracted. So, and you can see from the mathematics that if I do epsilon i, I plus 1 here, subtract this minus half of this, you get plus half of this. Is it okay? I think with this background, you should be able to do this quite easily. Remember, you have to use letter rule to write down en minus 1 kupmans. That is something I am not going to do. That is the only part that is missing. And when you do letter rule, remember the spin orbitals are same. So, same chi 1, chi 2, except the chi j would not be there. So, when you do these summations, summation will be only over n minus 1 spin orbitals barring chi j. Otherwise, the letter rule is same. And you should be able to then subtract and show this. This is only a simple arithmetic. At the end, it will be simple arithmetic. Remember the formula of epsilon i, that is very important. So, it says that it is one electron, it is one electron potential plus the sum over j shows that all interaction due to other spin orbitals, coulomb and exchange, due on i. So, that is why there is no summation over i, only summation over j. Having done this much, let me now get back to the canonical Hartree-Fogging question and note a very important point that this is after all an operator eigenvalue equation. Once an operator is defined, I am just calculating an eigenvalue equation. While doing the operator eigenvalue equation, of course, I am doing an SCF procedure. So, that is an iterative eigenvalue equation. I guess chi, then construct f, reconstruct chi and so on and so forth. But at the end, it is an operator eigenvalue equation. Given any operator in Hilbert space, we know how many eigenvalues and eigenfunctions are there for any operator. Yes, in an entire Hilbert space, infinity. It is an infinite dimensional problem. It is not a two dimensional problem or three dimensional problem, although I am only interested in n spin orbitals. So, for the Hartree-Fog, when I do this, I get actually several energies. Spin orbitals, I order them as en, then en plus 1, en plus 2 and so on. What I do for obvious reasons because Hartree-Fog is the minimum energy and that is very easy to show that you actually like to take only this part. Now, it may also happen that energy is not some of the orbital energies. So, you may argue why not this becomes lower than this. In some sense, this becomes the Hartree-Fog, then we reorder. So, that is a little complicated thing. I will not go into that. But for a very standard cases, if the energies are ordered like this, the Hartree-Fog determinant also includes the lowest n energies. But in the process of this solution, I also get this because it has infinite eigenfunctions and eigenvectors, eigenfunctions and eigenvalues. So, what are the meaning of this? So, what do I do with these spin orbitals which are beyond n whose energies are higher than n which I am not going to use obviously in Hartree-Fog. So, what do I do with those? These spin orbitals from n plus 1 are called unoccupied orbitals. So, they will have a very important part in quantum chemistry particularly when I come to the electron coordination. So, they will play a very significant role and note that together all these orbitals form a complete set of spin orbitals not just n. So, if I have chi 1, chi 2, chi n, chi n plus 1 which are whose energies are ordered like this, these together form a complete set. Don't forget this. Just the n will not form a complete set although I required only n because obviously later on I may require a complete set of expansion. So, I am just mentioning that all the eigenfunctions of a Hermitian operator form a complete set and of course f is a Hermitian operator which we have already shown. So, this together will form a complete set which will actually include what I now call unoccupied orbitals. So, this is the first time you are defining what is called unoccupied orbitals. So, through the canonical Hartree-Fock equation we have many, many orbitals and in principle their number is infinity. N of them will be used in Hartree-Fock. Of course, the rest of them will all be called unoccupied orbitals. Usually they will be higher orbital energy than the Hartree-Fock orbital energy. Hartree-Fock is a ground state. So, that is the lowest energy. So, usually this will be all higher energies. The lowest of these are what is called LUMO. The highest of these is what is called HOMO. Those who are already used to HOMO and LUMO will now appreciate that among the unoccupied orbitals this is the lowest because I have ordered accordingly. Among the occupied orbitals this is the highest. So, this is called the HOMO. This is called the LUMO and they have a very, very interesting chemistry to play because the energy difference can be very small, can be large and depending on that lots of interesting chemistry takes place. You already know that. So, the unoccupied orbitals have a very important role in quantum chemistry of electron correlation but even more important even before that we will see that they have a role even within the Hartree-Fock and what is that role that we will see. Now, what happens if these two become degenerate? Of course, those are the questions. Then there will be a lot of problems. That is basically the multi-reference problems and so on. We will see that later. Yes, we never find but in principle there. We will see how do you know which are the lowest energies. After you get the energies just find out which is lowest. That is important. I can find out for any operator Eigenvalue equation even a finite them. Not all. I can still order them. As long as they are greater than them, they will be unoccupied orbitals. As long as the number that I find out is greater than capital N, then I can order them and find out some of them unoccupied orbitals. So, if you are asking how do you solve these? Yes, there is a way to solve all these spin orbitals. In fact, numerically one can solve for atoms very easily. No, no. What I am trying to say, then numerically you can solve all the spin orbitals. What you are essentially saying is that maybe the lowest spin orbital is missing. Is that what I am saying? No. Numerically there is a procedure of finding the lowest energy, the second lowest energy. So, that is the Eigenvalue equation. And for molecule, of course, we use what is called the basic set. So, we got a matrix Eigenvalue equation in which we exactly will know how to again get the lowest energy. So, there is a way to find out. So, right now let us not bother how do you find out. Let us assume that we have the energies which you can order. And these are the final energies, whatever energies you have got, we can order them and then we can call these unoccupied orbitals and we can call these occupied orbitals. So, there are two subset of orbitals we are defining. One is called occupied orbitals. What is the number of occupied spin orbitals? Always n. Number of unoccupied orbitals will be very large in principle up to infinity, depending on how many you can find out. So, there will be a lot of approximate ways to find out, but in principle they will be very very large and in principle they will be infinity in number. But this number is always capital N. So, and this then we can now relate an important theorem again by Koopman's for electron affinity. Before I tell electron affinity can somebody tell me how to define electron affinity? What is the definition of electron affinity? Of course, you can tell energy that is gained if in a system I have an electron etcetera etcetera, but can you tell me in terms of n electron energy and n plus 1 electron energy? What is it? Just like a road for Ip. No, do not worry about Hartree-Fock. It can be any approximate. Just write En. So, what is the electron affinity? En, En minus En plus 1. Is it everybody agrees with this? If n plus 1 electron energy is less than n electron, then it is stable system and then we say that the electron affinity is positive. So, positive electron affinity means n plus 1 electron energy is actually lower than n electron energy. That can happen many times. Normally, you may think that it is higher energy. So, how it will become negative? But it can become lower because this energy made orbital energy may become negative. So, when I put an additional electron here, the energy may actually go down and it happens many times. In fact, that is what is exactly told in the Koopman's. So, this is electron affinity. So, the Koopman's approximation says that if I now exactly do the reverse of what I said here, if I put an additional electron in this system, that means I generate a n plus 1 electron thing. So, first Koopman is saying let me generate a n plus 1 electron wave function such that I have all chi 1, chi 2 to chi n and then I add an additional n chi n plus 1 or let us say chi r, where r is outside. So, r is in the n plus 1, n plus 2 set. So, one of them, r is in the set of n plus 1, n plus 2 to infinity. So, I add an electron, I add a spin orbital rather which means now I am defining an n plus 1 electron determinant. Of course, this chi r can be anywhere. It does not matter. That is not important because they are all anti-symmetrized. Then, Koopman says that the electron affinity in Koopman's approximation can now be called as minus epsilon r. Just like I said minus epsilon j, the electron affinity will be minus epsilon r. So, again it is very interesting. What I said, if this energy is negative, what is minus epsilon r? That is positive. So, electron affinity is positive. So, in a very simple way you already know this. If a Lumeau is a negative, then you know that one can put an extra electron. Electron affinity becomes positive and that is exactly what is seen in this Koopman's approximation by saying minus epsilon j or epsilon r. So, again the proof goes exactly in the same manner. If you can do that home assignment, it is exactly identical except that here there one missing here one addition. So, when you do the summation, you have to remember that you have to add an extra chi r or chi n plus 1 whatever you can take Lumeau chi n plus 1. Again this works very well. It works well if r is a Lumeau. I hope you understand what I mean now. That means r is n plus 1 just the next one. If you put it at a very high energy it does not work so well that is 1. The second is that you have two Koopman's approximation. One for IP, one for electron affinity. So, Koopman's IP is better than Koopman's A. Between these two approximation IP works better than electron affinity. If you compare with experiment, experiment actually final result. So, electron affinity does not work so well. Remember first of all there are lots of approximation that Hartree-Fock itself is approximation. Koopman's is a further approximation. So, the two levels of approximation were using and we are saying when I compare with experiment that is exact to exact. In that case IP works better than the electron affinity and I will show you by simple bar diagram why it happens. One why electron affinity gets actually worse. So, I can also define electron affinity Hartree-Fock by the same way as defining En Hartree-Fock or Scf minus En plus 1 Hartree-Fock. In which case of course everything will change. So, it will no longer be Koopman's approximation. All these spin orbitals will change if I do Hartree-Fock. So, that is a completely different ball game. Are the statements clear? The most important part of the statement is of course let us say for IP is not this. This is an algebra. The most important part of the statement is something which is not said, which has not been discussed. Many times unstated things are more important than stated things. I hope you know that. Those who know literature will appreciate what I am saying that the unstated things are all. What is unstated? That is this is an approximation which Koopman's said works very well in the first order and we are not discussing that. If I assume this, the rest is algebra. What you are going to do? Home assignment is an algebra. Koopman's this cannot be such a celebrated theorem. It is just application of later rules and doing plus minus. Anybody can do that. The original Koopman's theorem was to show that this n minus 1 spin orbitals or for electron affinity, the n spin orbitals can be assumed to be the same to the first order. So, you are not making an energy second order. Actually, wave function first order, energy second order. That is another issue that the perturbation you know that energy always goes one order higher. We will show this. So, this is something that was important part of the Koopman's approximation, which is always omitted in all textbooks. In fact, in the textbooks, they never discussed this that why the Koopman's. So, Koopman's actually used a variation method to show that if you do this, you are only making an error beyond first order. Of course, this is an error. This is an approximation, but the error is only beyond first order and that was the main content of the Koopman's approximation. The rest is history because if it is so, it is going to be so. What is so great about that? This is only an application of Slater. Koopman's cannot get credit out of this. So, this was a real Koopman's approximation of application of variation method and analyzing the error bar. That is the most important thing. Taliya Koopman, 1931, he did this approximation and the story about Koopman's that I have told in many classes that eventually applied the same variation method in economics. Economics has a lot of variation as you know, market, stock market, etc. And 1935, he got economics Nobel Prize. Actually, applying the Koopman's did not get a Nobel Prize in chemistry. In fact, he found that quantum chemistry is too crowded. There are a lot of big people. So, this was not big enough to get Nobel Prize. But he applied his knowledge and that is very important, domain knowledge into a field of economics. How does stock market vary? The same variation method remembers stock market. Euler variation, all that he applied and then he showed that to second order how much will be a loss if you invest. Always remember this, that you can do a lot of things provided you know how to use your knowledge sometimes beyond your domain because the skills are same. For example, when you do matrix eigenvalue equations, lots of interesting things have been done in quantum chemistry. Most of them are done long back by engineers or by mathematicians. Matrix problem is they have been applied and lots of new things have been done, new approximations have been made in quantum chemistry. Similarly, somebody in quantum chemist goes out and applies this economics. So, Koopman's is a very interesting example and also remember the name of the Koopman's. Again, a lot of people make mistakes, I want to make. This is the name. So, when we say Koopman's approximation, the apostrophe comes after s. I want to make sure that you don't write Koopman and then apostrophe. His name is Koopman's. So, when you say Koopman's s, there is no other s. So, you just have apostrophe. So, that you know in English. So, I just thought I will correct this. Actually, the name is Koopman's. Thalia Koopman's. Thalia Koopman's and so that's it is called Koopman's approximation. So, please remember the small errors. Because many people make these errors, I thought you should not make. Okay. I will come back to this point and I will actually show why did I say that for IP it works better with a bar diagram and then we'll move forward.