 So, let us take an example. Let us say helium atom. Helium atom has two electrons in one spin space orbital. We normally call it 1s, but now I have to be careful because whatever I am calling 1s has to be obtained by Hartree form. I have already told you that part I have skipped. How do I get phi? I have skipped, but let us call it 1s whatever I get and this is a common chemist notation whatever hydrogen atom orbital they call it the same name, but this 1s is actually complete misnomer. This is 1s for the helium atom. This is different from 1s for hydrogen atom, but let us call it 1s does not matter or I can call it phi 1. I think that is safer. So, my determinant is now phi 1 alpha, phi 1 beta. Note again whenever you write determinant and this is a golden rule you must write in terms of spin orbitals. You can never write determinant with only space orbitals. I am writing the form of energy in terms of space orbitals, but determinant cannot be written in terms of space orbitals. Without spin the anti-symmetry has no meaning. So, I must specifically write phi 1 alpha, phi 1 beta. Then I am going to use this is a closed shell by my definition. You have two spin orbitals, one alpha spin orbital, one beta spin orbital and the space parts are identical, phi 1. So, it is by my definition of 1, 2, 3 it is a closed shell. Is it clear? Of course, all this you know already, but I am just putting it in the very rigorous perspective now. So, now I want to calculate psi is psi. Apply, do the spin integration directly. I have the form. To simplify, let me also call phi H phi i as H ii. So, that some of the notations will get simplified. So, then what is the result? Result is 2 times H 1 1, because phi 1 H phi 1 you have two electrons. So, it will become two spin orbitals will become two times H 1. There is no other further summation, because there is only one special orbitals. Then let me write down the rest of the parts. So, when i equal to 1, i is only 1. So, you have a J 1 1 of course and this part is silent, because you have only one orbital. So, there cannot be i not equal to j to no meaning, because there is no 1 and 2 here. In terms of space orbitals, remember these are in terms of space orbital. I have only one space orbital, that is the result, because if I would have applied here also it is same i would be equal to 1, j would be equal to 1, 2 J ii minus K ii is J ii, because J ii is equal to K ii. Remember this form. So, very easy. I have a two electrons, each of them they are one electron energy multiplied by two and one coulomb interaction between two parallel spins. Note that there is no exchange interaction here. Don't forget, there is no minus K 1 1. Actually, there is a K 1 1, but unfortunately one of these J has cancelled it. So, there is no exchange here. There is only coulomb. Finally, what survives is only one coulomb interaction and note I am going to come back to this. Let us take another determinant, phi 1 alpha, phi 1 beta, phi 2 alpha, phi 2 beta. So, this could be helium atom. This could be beryllium atom, right, four electron. This could be beryllium atom. So, phi 1 is 1s, phi 2 is so-called 2s, whatever it is. This is not again hydrogenic 1s and 2s. We will find that out. But given a phi 1, phi 2, you can now apply the same form. So, what will be the energy? 2 H 1 1, right, plus 2 H 2 2. That is the first thing. Then I apply this. I have a coulomb J 1 1 plus J 2 2. That is clear. Now, you have I not equal to J. There is a J 1 2. So, it can be 1 2 or 2 1. So, it will become 2 J 1 2 minus K 1 2. But now, I can be 2, J can be 1. I have not done that symmetry. So, I have to write it once more. Plus 2 J 2 1, which is same as 1 2 minus K 1. Because I did not use I less than J. If I would have I will use I less than J, this should have been multiplied by a factor 2, which I did not do, but you can do it here. So, you have 4 J 1 2 minus K 1 2 and so on. So, let me analyze this. How do you get this 4 and 2? That is important. And this is where a good physics will come. So, let me write down this beryllium. 2 electrons here in your notations. So, this is phi 1. This is phi 2. Again, this is so called 1s and this is 2s orbitals. So, what are we doing here? We are first taking for each electron its own kinetic and 1 electron attraction, which is H 11, H 11, H 22, H 22. So, that part is cleaned out. Then for each of the orbitals, the 2 electrons sitting has only coulomb interaction with anti-parallel spins. If you notice further, I have 4 pairs of J 1 2. How does it come? There are 4 pairs of anti-parallel spins, 1 sitting in 1, 1 sitting on 2, 4. This. So, I have got up and down and here also I have got up and down. So, total number would be 6 and there are 2 pairs of parallel spins, up up, down down. So, that gives you 2 times K 1. So, one very important physics you get here that without looking at it, you can actually say that parallel spins have only, parallel spins sorry, have exchange interactions plus coulomb interactions. So, that is how your 4 J 1 2 comes. 4 J 1 2 came because this has a coulomb interaction with this as well as this. This has a coulomb interaction with this as well as this. So, everything has a coulomb interactions. Anti-parallel spins like 1, 1 here or 1, 2 have only coulomb interactions. So, eventually the final outcome is that if I have anti-parallel spins, they have same space arbitrarily, yes. Other possibility? Yes. Correct, correct. Because J i, because J i is same as K i, if i not equal to j, if i equal to j it cannot be both of them. Well, whichever way you interfere it. That is the Pauli principle also in a way that you cannot have 2 up spins. The point that I am trying to say anti-parallel spins have no exchange. That is the important thing. So, anti-parallel spins have only coulomb. Parallel spins have both coulomb and exchange. That is important. So, if we have to count coulomb, count both parallel and anti-parallel. If you want to count exchange, count only parallel spins. So, now you look at it. There are how many pairs of electrons are there total? Four electrons? How many pairs are there? Do not have to worry about who are in which orbital. Six pairs, four C2. Two of them are here and four of them are like this. All the six, there should be six pairs of coulomb and you see there are six pairs of coulomb. One, two and four. So, the six coulomb is divided into J 11, J 22 and four J 11 because each pair has coulomb. However, only a parallel pair has exchange. So, now you have only two pairs. So, in a way the coulomb, so the other way of writing this is coulomb is for all pairs. Note that when I am saying this, this is in terms of the space orbitals J and K after the spin integration. So, coulomb is for all pair, exchange is for only parallel pair. When I say up, up it could be down, down. Parallel or anti-parallel? That is all. So, count out the number of parallel pairs, number of anti-parallel, all other pairs. So, this is parallel pair is two. Again, when I say parallel, it means either both of them alpha, both of them beta. Both, they should be same spin basically. So, that is what I mean by parallel. So, it is very easy. So, I give you any determinant. You should be able to write down in terms of coulomb and exchange. In fact, what comes out later and this is something that I derived for closed shell, that this result is fairly general result for all determinants. The spin integration I perform only for closed shell starting from this letter's rule. This letter rule is valid for any determinant, any one single determinant and if I do the spin integration, this result that I got is a fairly general result. So, let us assume that I give you some open shell model, a very high spin state. 5, 1, 5, 2, 5, 3, each of them are up. So, it is a quartet state. So, it is a high spin state. So, what will happen here? 5, 1, 5, 2, 5. Do not worry about ground or excited state. It is a determinant. So, your energy will be h11 plus h22 plus h33. What will remain now? They are all parallel. So, everything will remain. So, I will have j12, j13, j23, minus k12, minus k13, minus k23. Everything will remain because all pairs are now parallel pairs. So, Coulomb does not get anything extra. This is not a closed shell, but I said I told you that this result is a fairly general result. So, even if I did not do the spin integration for open shell, but even if I start from an open shell determinant and do the spin integration in terms of space orbitals, it turns out that you get the same result. So, I am not going to derive it. It can be easily done. I am actually showing the outcome of the result. So, let us assume that I change one of the electrons. I make this like this now. So, same three electrons put in two orbitals. So, it is a lithium atom ground state doublet. I will have now h11 twice because two electrons are in h11. h33 will vanish of course, because there is no 3 now in the occupied. 2h11 plus h22 and now I can rewrite the Coulomb in the exchange. All pairs will have Coulomb. That is the first important thing. All pairs will have Coulomb. So, I have how many pairs now? 3C2, 3 pair. That is very clear. There has to be three exchanges. So, one of them is J11 and 1, 2 is 2. 2J12, correct? And 1 parallel minus K12. That is it. Actually, once you understand this rule, simply by inspection of the determinant, you should be able to write the energy. So, that is what I am teaching you. You do not have to go through the spin integration. I did the spin integration for closed shell and then I just declare that the result that I get in terms of the Coulomb in exchange for parallel and anti-parallel spin is a general result that you get for every determinant. So, now what I am going to do is to simply apply that result and write down fairly straightforward, in a fairly straightforward manner. Is it clear? Let us take another very good example, which is what many of you use in inorganic chemistry and so on. So, let us say there are two orbitals, one and two, which are degenerate and I have two electrons. So, there are different ways of putting it. Let us put it this. I am only going to analyze two cases. One is this, another is this. I could have put two electrons also in one. It does not matter. That is easier case. What will be the energy in each case? I will just write down the energy quickly. J11 plus J12, sorry, H11 plus H12, H22, yes, H11 plus H22, then you have a J12 and that is it, right? Let us write it down here. Again, you have 1 and 2, H11 plus H22 plus J12 minus K12. Now it is very important because you not only have one pair but that is also a parallel pair. So, parallel spins have both exchange and coulomb. Anti-parallel spins have only a coulomb. Now it is very important to note that each of these exchange integrals and coulomb integrals, if I integrate actually in terms of special orbital, they are always positive. The values are always positive. Again without proof, we are stating. So, K is a positive quantity. So, which will have a lower energy? The second one and that is essentially the Hohm's rule. What does the Hohm's rule say? If you have a degenerate orbital, the electrons go parallel. That is a proof of this actually. So, I can take anti-parallel, do the spin integration. This result will come from the spin integration as I told you. It is a fairly general result. I did not do the spin integration for open shell but fairly general result. So, this energy will be greater than this energy, which is a statement of the Hohm's rule. So, remember Hohm's rule is a consequence of anti-symmetry. Because of the anti-symmetry, I am getting the coulomb and the exchange. Many people feel that the Hohm's rule is a very important rule. It is a fundamental rule. No. The fundamental rule is anti-symmetry. Once I have a slatter determinant, write this slatter rule for the energy, do the spin integration. This is the outcome and hence the Hohm's rule. I have asked many, many people why it does not happen? They will say, Hohm's rule. I remember, why is Hohm's rule? That is what I am asking you. They do not tell exchange. They do not tell this. If you understand that there is a parallel spin which is exchange and this is what is called exchange stabilisation. Many times you have heard this nutter. The exchange is stabilising. So, the Hohm's rule is actually a consequence of the slatter rules for the matrix element, which basically comes from the anti-symmetric determinant. So, it is basically a consequence of determinant or anti-symmetric. Anti-symmetry is of course a very high rule. That is the first important rule. You can do anything, energy. You cannot break anti-symmetry. Symmetry is the most fundamental rule. You cannot choose a symmetric wave function for electrons. That is not allowed. Within that, I find that the Hohm's rule applies. So, the point is that this is a very important concept, physical insight that you get from the spin integration, the energies and so on. In fact, this is also something that is done in the inorganic chemistry. If you remember, T2G and EG levels get split and you call it 10 delta, etc. If this is very, very small, it is essentially becoming like a degenerate. And if it becomes degenerate then the spins tend to align parallel. And that is the origin of the high spin complexes in inorganic. You have already done it. If this is very, very high then of course they will become anti-parallels because for example, if this is very high then obviously this is no longer good. I should do this probably 1 and 2 as well as this. Note that in the same system, I will have H11 plus H22. I will have H11 plus H22. And this of course if you do this, this will have a J minus K. This will have only J. So, this is good. But on the other hand, if I put this guy on 1 then what will happen? This will have H11 plus H22. This will have 2 H11. Each of the H11 is much less than H22. So, I have a lot of energy saving here. Although I am losing the exchange, this gap is helping me. And that is the reason Helium atom in ground state is not 1s1, 2s1 because you may ask the same question. Then why do anything become paired? Why does anything become closed shell? It is a good question to ask actually because then this term takes over. The one electron energy is very high. Then it does not matter. I can afford to lose. And that is why systems become closed shell, remember. Because closed shell by actual nature in terms of exchange should not be favored anytime. Everything should become parallel, parallel, parallel. But if this gap is very high then I am losing instead of H11 plus H22. 2 H11 is much less. Do you understand? So, that is why it becomes low spin complex. So, you are not going to chemistry we teach this. It is not that strict. Is there a class here? Oh, okay. That is another class. Is it clear? So, this actually not going to chemistry high spin, low spin complex is also consequence of this. It is not taught in the same manner. So, if I do a Helium atom 1s2s under external field where the 1s2s gap is decreased. Indeed the ground state of Helium atom will become triplet. That is how we get from singlet Helium to triplet Helium under pressure. Because all you are doing is making 1s and 2s degenerate. Otherwise there is no reason why it should be 1s2. I mean whatever is 1s. Again, I have not derived the 1s. It will never be 1s2. It should be 1s alpha of 2s alpha or 1s beta, 2s beta. Just because of the one electron energy, the 2s is much high compared to 1s. So, I can afford to lose the exchange and gain in 1 electron energy. It is very important and that is why a lot of physics can be done. So, if I have a non-degenerate system, I can bring them close and make it triplet, make it a high spin and very interesting physics can be done actually. And people are doing it. Materials under pressure, they behave differently. The spin changes and everything is because of this. So, you have to understand this thing. The problem is of course the exchange interaction is small. So, obviously in this case H22 minus H11, that gap is much larger than the K12. So, K12 stabilization does not really help. So, that is what happens and that is the genesis of a low spin and a high spin complex in the organics which is also basically the same concept. Just again please understand, only from 2 electron you can get so much insight. You know many people say 2 electron make a high, but all insides are coming from analysis of 2 electrons. If you have the physical insight, you have the 2 electrons that really gives you a lot. Having said this, we will get to the Hartree-Fock and this is something that we will use it later because when you do Hartree-Fock, we are eventually going to do right in terms of space orbital as I told you. So, this expression at least for the closed cell, please remember. We will write it down.