 I think I will leave, there are lots of examples which you can actually read from the Zaboslu but I will leave these to go to the next topic which we are discussing with the size-gun systems. In fact, we just said in the last class that if I have H4 dimer, octraphox separates correctly because when I do VCI it does not, ok. So, a better level of calculation apparently worsens the quality of the level, right. So, I will try to come to that. So, let me first do H2 molecule, can you write the correlation level to expression plus H2? So, what is the matrix for the correlation level, you can be right quickly, 0, ok. Let us also take, simplify the problem, H2 molecule in two basis, this is a question that I have asked, ok. One molecular orbital, let us say I call it one, another molecular orbital I call it one. So, what is my Hartree-Phoff? My Hartree-Phoff is when these two electrons are in the orbital one, right. And what is double C i? I just put this here, ok. So, your web function phi, let us say exact web function is 1, 1 bar, let me write it explicitly, plus C 2 2 bar, you agree. And both of them are actually spin adapted singlet, ok. If I take a single excited determinant, then of course, I have to make sure that they are singlet. So, this is doubly excited singlet. So, if I want only that, this is almost like a full C, exact web function. So, it does not matter, I am positive of d C i now. So, this is my phi d C, I hope the nomenclature is clear. So, first one is 1, 1 bar, then second one is 2. So, 1 and 2 are spatial orbitals. When I am writing 1 in the determinant, it automatically makes 1 alpha, 1 bar is 1 d, because determinant cannot be written in terms of spatial orbitals, please remember. So, when I am writing 1 inside determinant for a web function, this cannot be a spatial orbitals. So, the default notation is the spin is attached to this spatial orbital and that spin is alpha. If I write bar, it is d. I hope you will remember, this is a standard notation that is used. So, my C i web function becomes 1, 1 bar and 2, 2. So, I have this as the Hartree-Caw and then doubly excited configuration is this. So, this is my 2 bar, this is my 2 bar and of course, there is a coefficient. So, now let me write down the C i equation. So, the first term is h minus e Hartree-Caw. So, the first term by definition is 0, right, we have psi Hartree-Caw, h psi Hartree-Caw minus e Hartree-Caw that is 0. The second term is psi Hartree-Caw with psi w x n, there is only one term. So, that means I need to find out the determinant 1, 1 bar h 2, 2 bar. What is the result of this determinant? It is a better rule, you can anti-symmetrize 1, 1 bar, 2, 2 bar. So, you expand this in terms of spin orbitals and then show that only one term survives. So, that is 1, 1, 2, 2, right because when I do the spin integration, now these are all in special orbitals. When I do the spin integration, it is survived. So, what is this term? This is called the electron 1 here, electron 1 here, electron 2 here, electron 2 here. So, this is actually an exchange integral. This is not a Coulomb integral. It looks like 1, 1, 2, 2, but in Dirac notation. In Mullikan notation, it will be 1, 2, 1, 2. So, this is actually an exchange integral between the two special orbitals. What would be the Coulomb integral? 1, 2, 1, 2. But this is 1, 1, 2, 2 and the exchange term of this becomes 0 because of spin integration. I hope all of you can see that because it will become 1, 1 bar, 2, 2 bar that will give this 1, 1 bar, 2 bar, 2 minus 1, 1 bar, 2 bar, 2 will give you 0 because the spins are about it. So, I have only one integral and that is an exchange integral between two special orbitals. I call it k12. So, this value is k12. And then you have a determinant. You have a matrix element between 2, 2 bar, h minus e r, 3 bar, 2, 2 bar. I hope all of you again can use letter root. Yes. Which one? This one? Why it is called exchange integral? What is 1, 1, 2, 2? Let us see 1, 1, 2. It is phi 1 star. Let us say 1 is phi 1. Phi 1 star 1, phi 1 star 2, 1 by r12, phi 2 star, phi 2 of 1, phi 2 of beta 1, beta 2. So, is it not exchange integral? What is the coulomb integral? When there is a charge density, one electron is sitting on one orbital, another electron is sitting on another orbital and interacting. Here you see the electron 1, here is sitting on phi 1 star 1, phi 1 star, here it is on phi 2. So, that is the density. After that I get it. Exchange term is 0. Special orbital exchange means that that coulomb, what is a coulomb density? If an electron is sitting on phi i, one electron 2 is sitting on phi j, then there interaction to the density is called coulomb. Here it is exactly opposite. So, electron 1 is sitting in phi i star 1. Here it is phi 2. So, please do not confuse with that exchange. That is an exchange for this integral in terms of spin orbit. After spin integration, I get an integral only in terms of space orbital. This by definition is an exchange integral. The coulomb integral means electron 1 has to sit on orbital 1, electron 2 has to sit on orbital 2. But if you look at here, electron 1 is sitting, not on orbital 1. This side on 1, this side on 2. That is exactly what is the coulomb integral is, exchange integral is. So, if you I just want to again tell you what is exchange integral is. In the Hartley Pog itself, we have done it. So, what is coulomb integral between two special orbitals, j, a, b. Let us say a, b is the two orbitals. So, this will be phi a star 1, phi b star 2, 1 by r 1. And then this should be again phi a 1, phi b 2. This is the definition of coulomb integral. How I will write the notation is comes later. The coulomb integrals means the density of one electron is in phi a, phi a star phi a interacting with phi b star phi a. Correct? You agree, Richard? Now, question is how do I write it? That is a later part. So, when I write it, this becomes a, b, a, b. So, that is coulomb. When I write the exchange, then the same integral will become phi a star 1, phi b star 2, 1 by r 1 2 and this will interchange. So, this will now become phi a, phi b 1, phi a, beta 1. This will become exchange. Now, if I use Dirac notation, this is a, b, b, a. So, then you have the, you have the notation that this is. So, here it is like this, that you have one integral or one, one, another, you know two. So, this is why it is called exchange integral and I am calling it k, k, a, b. So, that, by the notation itself, if you wanted to have a coulomb integral, this should be 1, 2, 1, 2. Remember, 1, 2, 2, 1 is same as 1, 1, 2, 2. I hope you are not confused. See, this 1, 1, 2, 2 can again be written as 1, 2, 1, 2 provided their real special argument, which is the case for our, so that is exchange integral. Now, you can see it is an exchange integral. 1, 2, 1, 2, 1, 2, 1 star, that is your same integral, a, b, b, a. So, I wrote that as a, b, b, a and you can see that is a, b. It is identical. So, I have just written as 1, 1, 2, 2. So, please see the, please expand this and see physically what it is. So, this is k, 1, 2. So, anyway right now it is just a number. In fact, why am I calling k is what I am trying to justify, otherwise it is a number. This quantity, the 2, 2 bar, this quantity which will come here is let me call this 2 delta. I can expand this by a letter rule, which actually I gave one problem in the quiz and I defined this as 2 delta. So, this becomes 2 delta. Then you have 1 c equal to e correlation once, very simple problem, but this is something that I gave in the quiz lecture. You can set up the equation for the H2 in a very simple two basis problem. In two basis problem, there is only 1 w x are written. So, GCI can be very easily solved and this can be actually solved. Absolutely it can be solved. So, write down e correlation as k 1, 2 times c. So, e correlation becomes k 1, 2 times c. That is your correlation energy and then the next is k 1, 2 plus 2 delta c equal to e correlation c. This is something that we have done it already because very similar expression we are doing. So, then you write, so k 1, 2 into 1 plus 2 delta n to c equal to e correlation c, put the c on both sides. So, solve for c and substitute it. So, you have e correlation minus 2 delta c equal to k 1. So, then you substitute for c, very simple algebra, substitute for c equal to e correlation minus 2 delta inverse, 1 by this. Actually it is a number, all these cases is simply a number. And then substitute here, say e correlation is k 1, 2 times c which is k 1, 2 square divided by e correlation minus 2, right. I hope all of you can see this. So, it is a k 1, 2 c, c already has a k 1, 2. So, it is k 1, 2 square divided by this. And now you can see, I told you what initially that we have to use iterative method. So, you can see for DCI, you use iterative method. Remember? So, you first put e correlation equal to 0, get a new value of e correlation, put it back here and keep doing. That is the Davidson's kind of routine. So, we have the correlation energy in terms of correlation. So, of course, you can solve it iteratively. I could have completely diagonalized this matrix and got Jacobi householder, but I did not want to do it. I wanted to show you iteratively. So, this is your correlation energy. You can get a root from here. This is very trivial because you can just write a quadratic equation. Only one number, e correlation into e correlation minus 2 delta equal to k 1, 2 square and you solve it. So, quadratic equation, take the lowest root. I hope all of you will be able to solve this. Give me the actual correlation energy for the lowest root. Again, I have a two problem, two basis problem. So, I will get two roots. One ground state, another first excitation. And because it is a linear variation, each of these roots will be an upper bound. Remember, again, McDonald's theorem. I am of course interested in the lowest root. See, if I ask you to find the lowest root, you should be able to find for here. Now, let me switch to the next problem, which is dimer of H2. H2 dimer. So, the problem is the following. I have two hydrogen molecules which are separated infinitely apart. That is my problem. So, what are my orbitals to start with? Each of the hydrogen molecules is sitting on one. I call it hydrogen molecule A, hydrogen molecule B. So, I have two special orbitals. One hydrogen A, hydrogen B to start with. So, what is the Hartree-Pock of this? Two electrons here, two electrons here. Remember, these are the monomer orbitals. But when I have separated it, the orbitals remain the same because there is no interaction between these two hydrogen. And I told you that the Hartree-Pock energy is also some of these two energies. So, Hartree-Pock separates completely. Then I have two A and two B. Again, they are same as the monomer one and two. I have to write A and B separately because their coordinates are different. The two hydrogen molecules are far apart. So, now I have to construct the DCI waveform. So, for two things. One is DCI for H2. Another is DCI for H2 dimer. Let me explain the problem because I cannot finish it today. DCI for H2 dimer. Then I will have to show that the EDCI for H2 dimer is not the summation of EDCI of hydrogen. Two times. What I will try to show in the next class is that the double CI for hydrogen dimer will not be the double CI, two times double CI for hydrogen. Even when they are far apart. So, this is infinitely apart. The distance is 2 H2 molecules. So, these are basically non-interacting H2 molecules. So, two hydrogen molecules H2 dimer non-interacting. And remember the interaction, the channel in which I am breaking is H2 plus H2. So, normally what I would expect? I would expect the energy to be some of the two hydrogen molecules. But what I will show is that if I do a DCI now on this system, my energy will not be the same. What is also important is that the DCI wave function of H2 dimer of H4 will also not be a product of the DCI wave function of the hydrogen times the hydrogen. Of course, there will be anti-semit based product. But either way, this will not be a product. Because you know the non-interacting theorem that for a non-interacting case, the energy should be some, the wave function should be product. So, what I will show is that if I do a DCI for the H4, it breaks into H2 plus H2. That means two H2s are far apart. Neither the wave function will be product, nor the energy will be something. Whereas, if you do Hartree-Fock alone, the energy was some, the wave function was product. So, this particular aspect of correct separation of wave function and energy at a non-interacting regime is called size consistency. So, what is size consistency? So, size consistency essentially means that if a super system, I call it a super system, system of two H2 molecules, and I call generally super system AB fragments into A plus AB at a non-interacting. That is when whatever you call it, RAB tends to infinity. The distance between the two fragments goes to infinity. Then, the energy of AB should be some E A plus AB, and the wave function AB should be an anti-semitrize product of A. This is called the anti-semitrize product. Because the anti-semitrize product, because overall wave function must be anti-semitrize. This capital A is just an anti-semitrizer operator. So, do not worry about it. But essentially, everything has to be anti-semitrized, not simple product. It is an anti-semitrize product. So, this property is called size consistency. If this happens, size consistency, often it is called also size separation. Clearly, by this discussion, if I can show this, it would mean that the DCI is not size consistency. So, all theories which satisfy this are called size consistent theories. This is the property of size consistency. The theories which satisfy this are called size consistent theories. So, clearly, I will propose that DCI is not a size consistency. Once I am able to show this, I just want to tell you the final result. And we will see. Remember the monomer of DCI we have already done today. So, I can easily show. So, I have got the DCI for H2. Please remember this. This is the result. Of course, you can take the quadratic equation and solve it. I mean, I just give the result. I will directly go to DCI for H2 type. And similarly, I will calculate correlation energy at this distance. When R between H2 and H2 is infinity, there will be a lot of simplifications. And then I will see that the correlation energy that I get will not be twice the correlation energy. So, that is mathematically I will first show. Then I will argue physically why it happens. I mean, that is more important. Why is DCI not size consistency? What is wrong? And that is important to understand physically because then only you will be able to correct. But the next level of theory we should be able to correct this. And I will also show that the perturbation theory that we did, MP2 is actually size consistency. So, DCI is not but MP2 was good. Except that MP2 does not recover all of good amount of energy but it is good and DCI is not good. However, you remember an approximation of DCI was MP2. But then that is an approximation. So, that approximation, the 2-3 approximation that we did would be size consistency but not the actual doubles. Okay? All right. So, tomorrow I will start from here. And please follow these because unless I am sorry that some of you did not take the quiz. Actually, this was one of the problems that I asked in the quiz. I just set up the problem and asked them to solve. One of the reason is that I am going to come to this. And this is also in the line with the iterative method that we did. In a very simple 2-by-2 problem. Double CI where there is only one W-x at a time. The problem can be very easily solved mathematically. Okay? And of course, this is exactly solved. I do not have to solve iteratively. What I am trying to show you here is the exact solution because it is a quadratic equation. So, I can exactly solve it. I am not going to do iteratively but you can do iteratively taking E correlation equal to 0 then next E correlation is this minus 2 delta then put it here, keep doing it. But in this case, you do not need to do it because this is only one number. I can solve it by quadratic equation. Analytically, I can solve it. That is what I am going to use. I could have also solved same thing by actual diagonalization of 2-by-2. That is same as this. So, if you diagonalize, what do you do? You do minus lambda, minus lambda. You get a quadratic equation. Do you remember diagonalization method? Simple diagonalization method, not even Jacobi householder like Cramer's rule, you get a lambda square and that is exactly this. That is exactly this quadratic equation that you are hitting. So, anyway, either way, it would have been the same result. So, we will come back to this problem, very instructive problem and this is where much of the physics of the quantum chemistry method is actually large. Then we will go back to NP2 and then at the end we will go to couple cluster but in between I want to do little bit of second quantization and diagram. Let us see how much I can do. At least the rules, just the rules of diagram which will be simple read. All right.