 So, let us go back to Koopman's again. I hope you remember now, Koopman's approximation that if you take out an ionization from one molecular orbital, now I am talking of molecular orbital. Do not talk of atomic orbitals now. Molecular orbital, then the energies are minus epsilon. Remember, even when here I am doing fc equal to sc, I get p plus energies I get. I should not forget as an eigenvalues. Of course, energies also I get, which are actually molecular orbitals eventually, with m or energy they are actually molecular orbitals. So, I get this, so do not forget it. Then I am saying that my ionization potential is minus epsilon i and the whole approximation was built on the fact that the molecular orbitals, now in term I can generally write molecular spin orbitals of n minus 1 electron wave function is built out of the n electron or n spin orbitals of the original problem which is n electron wave function minus with just the chi i removed. So, except for that chi i, everything else is there. So, that is my n minus 1 electron and I am saying the energy difference is only minus epsilon i. So, this works very well. There is no problem. This particular thing works very well, particularly for ionization. So, I am trying to give a physical reason why it works well with a small diagram. So let us assume that I have an electron Hartree-Pock, which I call En SCM. So, this is something that I have already done. Then I construct what I call En minus 1 electron energy under Kupman's approximation. So, I hope it is clear. Under Kupman's approximation means exactly this and what Kupman's is saying essentially that this difference is epsilon i. Please note that you always require more energy to ionize. So, this will be always higher than this. I p is always positive. So, n minus 1 electron energy will be higher than any electron energy. You require energy to knock out and that result is minus epsilon i because epsilon i is actually negative. So, with this minus it becomes positive. So, this is your Kupman's. This is constructed by freezing the rest of the n minus 1 spin orbitals where it is and only Kaia is left out. What we are saying that this is pretty close to exact. So, let us see why. What are the effects that are missing? One of course is that this is not the Hartree-Fock of n minus 1 electron. This is not the Hartree-Fock. So, essentially this Hartree-Fock has to be done. Each of these are missing what is called the correlation energy. Remember, exact energy is different. It is not Hartree-Fock. It has a correlation energy. So, let me first see what is the energy of the exact. It will be somewhere here on a bar diagram, e n exact. It will be lower than Hartree-Fock. I hope that is very clear to everybody because you have already mentioned this by variation method that whatever we are doing it will be greater than the exact energy and if you do full CI and all that that will be actually exact energy. So, this is called the correlation energy of the n electron problem. I can call it e correlation n. So, you can say that e n scf plus e correlation of n electron gives me e n exact or you can reverse it depending on what sign you want. But let us keep it like this. In this sign, in this definition, this become negative because it is going down. Let us look at the n minus 1 electron problem. First of all, this is not even Hartree-Fock because it is the Kupman's approximation. So, if I actually want to do Hartree-Fock, what will be the result? Will it be higher or lower? Is it clear to everybody? That will be lower because Hartree-Fock is the best single determinant and this is a single determinant obtained by freezing the rest of the n minus 1 spin orbitals to the n electron spin orbitals. So, if I actually do a Hartree-Fock for the h n minus 1 Hamiltonian, this will only be lower. So, somewhere here will be e n minus 1 scf. This energy is called the relaxation energy. This is just a name. Why it is called relaxation energy? Because I had frozen the orbitals, Hartree-Fock, I allow the orbitals to relax to a right environment, its own Hartree-Fock environment. So, I call it relaxation energy. So, let me write this e n minus 1 Kupman's plus this relaxation energy which is again negative gives me e n minus 1 scf. So, I get e n minus 1 scf. However, to compare with this, I must also do correlation here. So, what will correlation do? It will push it down further. So, it will push it down further somewhere here let us say e n minus 1 exact. So, I can write now e n minus 1 scf plus e correlation of n minus 1 gives me e n minus 1 exact. So, I am using a sign that each of them is negative. So, everything is negative. So, I am just pushing it down. Is it clear? Now, what turns out is a very interesting effect. This is well known and this is something that I this time not trying to prove that the e correlation n is less than because this is actually negative. So, it is less than e correlation n minus 1. If you define this as a positive, it means the correlation energy is greater. That means there is a greater lowering of n electron compared to n minus 1 electron. Now, this is very simple to understand because as n increases, overall energy increases. Correlation energy, the lowering also increases. So, everything so, this gap is much higher than this gap. So, if I say this is delta 1, this is delta 2, delta 1 is larger than delta 2 graphically you can see. So, that means delta 1 is larger than delta 2 just gap. I am not looking at sign now. However, there is a delta 3 here which is relaxation. This is greater than this to the extent that delta 1 can come pretty close to delta 2 plus delta 3. Now, of course, this can be much higher. It can upset little bit but since this is already greater than this, it is quite likely that these two together would be somewhat close to delta 1 or maybe little more than delta 1. Then, what am I comparing? I am comparing this with this. This is n minus 1 exact with this. If you look at between these two graph, this is suppressed, this is depressed by delta 1. This is depressed by delta 2 plus delta 3 and they almost would match. So, this is a very interesting thing where because of two errors, what is the two errors? One is the correlation missing in both. Another relaxation missing here, they cancel in a way that you get nearly exact values. So, of course, if it is exactly equal, then you will get exactly equal value. That does not happen but it is pretty close to the exact because this guy very much matches the sum of these two. First of all, this is anyway higher than this. So, this guy can be little lower, little bit higher but it is pretty close and delta 3 ensures that this gap is similar to this gap which means this is equal to this. I think graphically you can see that this top has been replaced here. The top has gone here. So, you can very easily see that the bottom has gone here. So, this has gone here, this has gone here. So, as long as this is equal to this, this is equal to this. So, this is IP Koopman's, this is IP exact because this is exact minus exact. I am only saying that delta 1 is greater than delta 2. That you agree. Here, there is an additional delta 3. So, hopefully these two together would match this. Then it is exact but it is not exact. It is pretty, what I said that the Koopman's IP works better. It will never be exact because already I see that delta 1 is greater than delta 2. So, all I need is a top off here which is coming in the form of relaxation. The top off may be much more than I would require, less than I require. So, to that extent there will be difference from the exact. I am not saying that it is exact. So, what I am just saying that the Koopman's IP is a good approximation to the exact IP. Since I have a delta 2 plus delta 3 which may come close to delta 1, I am saying that the Koopman's approximation is greater. It is other way round. I am now arguing why Koopman's approximation is a good approximation. So, that is not the right. See, there is a theorem which says that the dominant part of E is proportional to n. It is as simple as that, proportional. It is as simple as that. If it is size extensive, I do not want to bring in all those terms now and that is the exact theories are size extensive. So, correlation energy. So, basically E correlation also becomes proportional to n. E correlation means minus E correlation or whatever, the change. I can say modulus of E correlation that might be easier. It is always proportional to n, the value. So, that is the reason. It is very simple. But how much difference? How much is the relaxation? That is where the approximation will not be exact, but it will be pretty close. So, in fact, this argument, I am showing to tell that the IP Koopman's is a good approximation to IP exact. But it is not exact. It cannot be exactly equal. It is a good approximation to exact IP. That is another story because if it is outermost orbital, this relaxation is small. What will happen if it is a core orbital? You know, this guy will become very, very large. Then this delta 1 minus delta 2 will not have. This is the most important thing. Then again, it will become back. So, it works even better for, see the whole issue is that if you look at this, what is the value of delta 3 compared to delta 1 minus delta 2? That is what it comes down to. That is what it comes down to. If this is exactly equal to this, then it is exact. If it is much higher, then again, it goes on the other side. So, if you take out an electron from the core, what are you doing? Koopman's approximation is very bad because you are disturbing the system so much, but you are not allowing the system to relax. You are telling, no, no, no, you stay where you are. So, if you have a crowd and you want to pull out somebody from the inside, everybody naturally should be disturbed, but you do not want them to be disturbed. So, they will all go into high energy state. Yeah, naturally, because once I am taking out, normally what do you expect? Everybody will resettle, but if I take out from a crowd outside, nothing will happen to the others. So, it is a very simple thing. So, because you are disturbing the system too much, so this delta 3 will become very large and if delta 3 becomes very large, then somewhere here will be man a n minus and Koopman compared to this. Then again, this gap will become too large. So, it is a delicate balance. This balance is achieved very well for outer valence IP. So, this balance where it is a very good approximation is more true for outer valence IP. So, you have to understand the spirit of it. If you take out inner valence, already balance is destroyed. If you take out core, then there is no point, because relaxation is much more, much larger than the correlation. The correlation is unimportant. Then you better do what is called delta SCF. Just take SCF difference. You will get pretty good result because the whole effect is coming because of delta relaxation. So I have to understand the physics and accordingly have to respond. So, this is something that I just wanted to tell. I do not know if I had derived the en minus 1 minus en as epsilon i. Have I derived it? I will close it here.