 me also write the same CI equation by actually subtracting e Hartree-Pock from the beginning. Let me write this equation little bit differently. My Strodinger equation itself remember was H psi naught equal to e naught psi naught. Let me write the Strodinger equation by subtracting from the H e Hartree-Pock. This is a number. I define a new operator H minus e Hartree-Pock, right psi naught and this becomes my e correlation psi naught, correct? Because I subtract from the right hand side also e Hartree-Pock. So basically what I have done is subtracted e Hartree-Pock from both left and the right hand side. So I have got a new operator on the left hand side which is H minus e Hartree-Pock. This operator is very often used in quantum chemistry simply because its eigenvalue is directly correlation energy. I hope you appreciate why. If I take Hamiltonian, its eigenvalue is not the correlation energy. It is actually e naught. So if I define a new operator which is H minus e Hartree-Pock, then this operator eigenvalue is e correlation. Eigen function remains the same which is exact eigen function of the Hamiltonian because I am simply subtracting a number. This operator is very often called and this will be actually made much more clear when you do second quantization later but let me just write this. It is called normal ordered Hamiltonian and very often it is denoted as H n, H subscript n and the entire equation now can be written in terms of this operator. So H n psi Hartree-Pock plus C d psi d. I am just writing in a very cryptic term the same double CI in intermediate normalization. So C naught is 1 equal to e correlation psi Hartree-Pock plus C d psi d. Except that the Hamiltonian is now H minus e Hartree-Pock. I have subtracted e Hartree-Pock from the Hamiltonian. That is all. So I do this exactly the same thing now. So I have got psi Hartree-Pock, H n psi Hartree-Pock. What will be the result of this? Can somebody tell me? What is the average value of H n with respect to psi Hartree-Pock? Zero. Everybody agrees? Because H will give you e Hartree-Pock. I have subtracted e Hartree-Pock. So that is zero. So this is gone. So I can write this as zero plus psi Hartree-Pock H psi d into C d equal to e correlation. So now you see I write the same expression of e correlation. Remember I had written this expression for e naught. Since I have to subtract e Hartree-Pock, this actually went off. That is all. So I told you already this is e correlation. Remember minus b dagger C d. This is exactly my b dagger. This is my C d. So your e correlation becomes b dagger H n. Of course, now everything H is H n. I have already moved to normal order Hamiltonian. But I think the notations are important. Otherwise it will confuse you. I will not be confused but you will be confused. I agree. So I am talking everything in terms of H n. In fact, later on when we write we forget about H. In turn, anybody is an H n. So whatever is H is H n. Then you would put this in another psi d, another w excited. So you get d. I have already told you that matrix is d. The psi w excited H n. But then now this d is different from the previous d because Hartree-Pock is subtracted. So this is a d is now defined as psi double H n psi double. So what is the difference? The difference is that this d does not have e Hartree-Pock. So whatever is the old d, I am subtracting e Hartree-Pock from the diagonal. Only from the diagonal of course. The number can be only subtracted from diagonal. So I define my new d. I can call it d prime which is with H n and here it does not matter fortunately. You know why? Because you can only do from the diagonal. There is no diagonal element here. So whether it is H n or H, the value is same because there are no diagonal elements. So it does not matter. That will become 0. You write e Hartree-Pock here. H minus e Hartree-Pock. e Hartree-Pock will come out. These are orthogonal. So I hope you can see this. These are small, small things. I may ask you. Are they equal? Are they not equal? I have to think. So I did not bother about it. But here I have to bother. For the d, I have to bother. So then if I write it as d prime, then this equation then becomes d prime. I am now rewriting the same equation that I wrote there. So d prime is my, I got lost. Yeah, d prime was my, this to psi d to psi d. But before that I had b plus d prime, sorry, d prime cd equal to e correlation cd. So I am rewriting the same expression with the normal order Hamiltonian. That is all. I am really not doing anything new. So the same expression now I can rewrite. I am removing this part. Maybe I will remove this part. So now do the same thing. Get cd from here, substitute in the correlation energy. So you have d minus e correlation. Now it is no longer e naught. It is e correlation. That is the only difference. When I am using d prime, it is no longer e naught. It is e correlation. With everywhere Hartree-Fock is subtracted. That is the only thing. Times cd equal to minus d. So you write cd as minus d minus e correlation into an identity operator inverse cd. And your correlation energy equal to minus b dagger d prime, sorry b. So you get exactly the same expression that we got there, except that I am using d prime here. I am using instead of e naught e correlation there. So when I want to do the, so I have expressions only for the correlation energy. So that is the only difference. I also had a expression for correlation energy. The difference is I had exact e naught here. So how do I start the iteration? I had to write e naught equal to e Hartree-Fock. Assuming e correlation is 0, e naught would be e Hartree-Fock. And d will have e Hartree-Fock put in. I have subtracted the e Hartree-Fock right everywhere. So now I am using d prime. And now I can start the correlation by assuming e correlation equal to 0. My first iteration starts from there. You have a d, sorry you have a d inverse. You have a d inverse, d prime inverse and then you continue. So in fact, when I do the first iteration, remember in my correlation energy, this would be Hartree-Fock, e Hartree-Fock. In my correlation energy, it is just d prime inverse, nothing else because e correlation is 0. So what will be the, what will be d minus e naught inverse? Now I come back to this approximation. Do the same approximation here. So assume now d minus e correlation or rather you can simply assume that d inverse is diagonal. So do the first iteration with e correlation equal to 0. So start the iterative procedure by guessing e correlation as 0. So what will you get in the first iteration? You will get only d inverse. There is no d minus e naught or e correlation or d prime. d prime inverse is diagonal. Anyway d is also diagonal. So I have assumed d and d prime both are diagonal. So d prime is also diagonal and this diagonal will directly now come as the difference energies because now there is no e Hartree-Fock to subtract. So when I do the d prime, we will do this exercise. So the only difference is depending on whether you use h or h n, there will be e Hartree-Fock padded or not padded. So here now e Hartree-Fock is not padded. So everything is just gone. If I was doing that, I had to write d minus e Hartree-Fock inverse because your e naught cannot be 0. In the first iteration, e naught must be e Hartree-Fock. e correlation is 0. So when I subtract e Hartree-Fock, I would have got the difference of orbital energies. With a d prime, I make a diagonal approximation and I assume this diagonal has only difference of orbital energies. I have to justify how good is this approximation from a to b. So I will do a slatter rule for this d prime inverse and then justify how good it is. Of course, it is an approximation. There is some other term, but that is not bad to approximate. So eventually what we will use for this ci is actually normal order. So we do not use this anymore. So we do not write e naught here. We write their e correlation. This becomes my d prime. So everything enters ci equation is now in normal order and intermediate normalized. Is it clear? I will come back to this tomorrow. Tomorrow we have slightly longer time. I will come back to this and start again from here. Yes, it does not matter. The upper bound would be on e correlation. It is one and the same thing. e Hartree-Fock is always greater than e naught. Finally, e is also greater than equal to e naught. So it does not matter. The upper bound will remain in e correlation. So it just shifted. Remember, normal order is nothing, but it is just shifting the entire thing by e Hartree-Fock. Just subtract. So it is a mathematical. It is very simple. I will come back to this, but in the next class what I will do? I will only derive the equations in normal order. So it will become easy and then write what is d prime by slatter rule and show if it is diagonal. I assume it is diagonal and what would be the diagonal elements? So that is what I will do. Psi A B R S, H N Psi A B R S. H N essentially is H minus e Hartree-Fock. How is it different from the orbital energy sum? So what is this approximation? Then this will recover MP2. So please note that the construction with H N or H is identical. There is no difference. It is just another convenient way of looking at. There is no new physics that I am introducing. Neither is a new physics is introduced by intermediate normalization. So these are all convenient way of writing the same equations, but it is useful because later on H N would be used and intermediate normalization is also used for the perturbation theory and many other theories which will actually follow. So it is better to write C i also as a intermediate normalization in which your C naught is just one. So when you diagonalize your one term is fixed as well and learn to write in terms of H N so that the eigenvalue of the H N is not directly e correlation energy. That is the only difference, not e naught because I am no longer interested in Hartree-Fock. That is already done. So I can subtract that. So I write away, write in equations only for e correlation. So there is no new physics being introduced, but it is just the way of writing. So please read these, how to do these and I think it should not be very different. So this is where I had written the old one and these rest of the parts are H N including this, but otherwise this all the procedure is exactly same. And I think we will also do C i S D. We will tell in the next class there are some pitfalls of double C i. So what are the problems of C i doubles? We will also do C i S D and probably after that I will only make generic comments about C i why it is not a good method. Before we move forward to the second quantization and we will see when it comes. Second quantization and diagrammatic forms of perturbation theory. Again this will be done in a reasonably loose manner because these are highly rigorous. I do not have that many lectures to do diagrammatics perturbation theory. There are so many diagrams starting from Feynman, Goldstone, Hogan-Holtz. So we will just come to diagram. We will only give you the rules. The rules have been derived and at the end we will go to methods which are neither perturbative nor variational and one of them is couple cluster and that is where we will try to close the lecture.