 So, let me restate what we did in the last class yesterday. So, we wanted to write the Strodinger equation in a somewhat different form and this is something that will be more often done. So, write the Strodinger equation in this manner H minus E Hartree-Fock psi 0 is E correlation psi 0. So, this becomes our Strodinger equation. Instead of saying H psi 0 equal to E naught psi 0, we are subtracting E Hartree-Fock from both sides. So, we have the new operator which is basically the operator minus a scalar number and if it is a matrix representation then only from the diagonal part this E Hartree-Fock will be subtracted that is what it means. So, you have E correlation psi 0. Is it clear everybody? So, this is what we call normal order H or H n. So, in terms of normal order H the eigen value of this normal order Hamiltonian is itself correlation. So, because this is what we are really interested in now. So, then we presented the CI equation DCI equation by first projecting this new Strodinger equation. So, H minus E Hartree-Fock psi 0 as E correlation. Note that we are using intermediate normalization. So, psi Hartree-Fock with psi 0 integral is 1 in the intermediate normalization. So, you have only E correlation here. So, that is the equation for the E correlation, very simple equation. So, you expand of course psi naught. So, that is what we will do now in DCI. So, psi Hartree-Fock H minus E Hartree-Fock and then I expand this as psi Hartree-Fock plus all doubles. So, I call this H Hartree-Fock H minus E Hartree-Fock. In a simple block I am calling it psi double sorry H minus E Hartree-Fock psi doubles multiplied by C double. So, this is just a notation that we introduce. Essentially, this means all sum over all WXRA determinant, specific WXRA determinant and their coefficient. So, this psi D is not just one function, but it is the linear combination of all WXRA determinant. C D is also a coefficient. All the coefficients are stuck in. So, this is just a simple way to write. So, this becomes my E correlation. It is very easy to see that this first term is 0. Remember when I had E naught, this first term was E Hartree-Fock. E Hartree-Fock plus this was equal to E naught. So, basically that is what I expect for the correlation energy E naught is subtracted. So, this becomes my expression for correlation energy, only this. So, E correlation is nothing but psi Hartree-Fock H minus E Hartree-Fock sorry, all psi doubles into C doubles. So, basically all I need to find these coefficients and then I will get the correlation energy. So, to find the coefficients I project the same equation with now doubles. When I again write psi doubles, it means each of the WXRA determinants which you are calling CDTU because I have a dummy index here, ABRS. So, that is okay. So, last time I called CDTU, but blockwise the equation will look like this H minus E Hartree-Fock psi Hartree-Fock plus psi doubles H minus E Hartree-Fock psi doubles C doubles equal to, now when I put this W excited with psi zero, it is zero by intermediate normalization E correlation CD. This Hartree-Fock part will become zero, but the W excited part will survive giving the same coefficient whatever I am projecting here. So, if this is CDTU, you will get E correlation times C CDTU. So, this is my CI equation. In fact, later on you will see that making this zero will be advantageous if I can make this zero and that is exactly where I will transit from CI to couple cross term. So, that zero came involuntary. So, it is E correlation times CDTU. So, this is what the equations were and we presented that this psi d H minus E Hartree-Fock psi zero is the column B. So, this becomes my column B dagger. Note that E Hartree-Fock here is just incidental because psi Hartree-Fock psi doubles as zero. So, B and B dagger are nothing but psi Hartree-Fock H matrix element directly with Hartree-Fock and doubles. I do not need to write E Hartree-Fock. So, yesterday I presented this as Hamiltonian matrix element, but the same thing I can write for H minus E Hartree-Fock because this is anyway zero. So, it does not harm. And this quantity is what I call d or d prime in the normal order. So, here you have to be careful whether I am subtracting E Hartree-Fock or not subtracting E Hartree-Fock. So, there will be different matrix elements. So, I am calling it d prime. So, then my equation was E correlation is B dagger C or CD whatever B dagger CD which means double column of doubles. So, this is a column matrix element. This is a conjugate B dagger rho and then I have B plus d prime CD equal to E correlation CD. So, these are the two sets of equation in the matrix form, but you should be able to write these very easily in the normal form. I hope that everybody will be able to write. So, for example, if I want to write the normal form, it will be following. So, B matrix element of CDTU. What is B CDTU? Let me write this here. B CDTU is nothing but the WXRA determinant being a CDTU. So, it will become psi. So, explicitly if I write psi CDTU H or H minus E Hartree-Fock, it does not matter. H psi Hartree-Fock. Is it clear? So, one element of B would look like this matrix element. So, then I can write this as B CDTU plus sum over d prime. This will have two elements. One is CDTU, another is ABRS, sorry RS AB, ABRS where the same notation. ABRS which means one of this is CDTU and one of this is ABRS sum over all of them. So, sum over CDTU, sum over ABRS and then you have CD which now should be ABRS. I hope all of you will be able to write this because if I do a matrix multiplication, this should be D prime sum i, ij times CD of j. So, these ij's are WXRA determinants, so CDTU to ABRS and I am summing over ABRS. So, you have the sense of matrix summation. So, this you will get as E correlation times C CDTU. So, that is what you get from the equation. If you remember, if I project to the specific CDTU, this is what you will get. So, this is just written in the matrix form and you can see the CDTU is a specific index. So, that is one survives. ABRS is dummy index, that is summed up. So, this is the matrix where each index is a combination of four orbitals. So, this I can call capital I, capital J, capital J, capital I. So, let me write in a simple how I am writing in matrix form. Let us say I generate a capital I which is a combination of CDTU and a capital J here which is a combination of ABRS. Then this equation can be written as B capital I equal to D prime ij, one of them is i, one of them is j. This is i, this is j. Cj or CDj whatever you call it, CDj, sum over j equal to E correlation CDI. I have defined CCDTU as i. So, you can see this is now in the matrix form. So, B column i plus D ij, CDj equal to E correlation CDI. Just for simplicity, I can drop this capital D because you understand the C is essentially WXRA determinant. So, we are discussing only double. So, let me drop this D so that there is no confusion. So, this D I am just dropping it from everywhere because we are anyway discussing DCI. So, we do not have to write the capital D. So, then you can see that this is exactly the matrix equation that you will get. So, that is how I wrote this matrix equation. So, one should be able to transform from here. Normally, I would project this with specific CDTU. I will get an equation from there. How did I write the matrix equation? I am just showing you. So, then what we did was to substitute from here the CD. So, I told that CD is nothing but so D prime minus E correlation into identity times C. Now, I am not writing CD times C equal to minus B and then I wrote C equal to minus D prime E correlation identity inverse B. So, I pre-multiply. It is very important where you multiply. I pre-multiply by this inverse. So, this goes off. You have C equal to minus this inverse B. Again, everything is now handled in matrix. It is very easier. You can always go back to the actual indices. That is not very difficult because each of the index of this is basically a combination of four indices, four orbitals. Two occupied, two unoccupied, two virtual. I am exciting from a set of A, B, CD, a set of two A, B, or CD, two R, S, or T. Then eventually I get E correlation. So, then in my DCI, E correlation then becomes B dagger C. So, E correlation becomes minus B dagger D minus E correlation identity inverse B. So, that just I am writing now E correlation as B dagger C. So, I am just simply putting this here and I have dropped the fact that it is CD. C essentially means it is CD. So, this becomes the correlation energy which you can see cannot be solved one time without an iterative solution. The same correlation energy could be obtained by the eigenvalue equation. Here I am not writing an eigenvalue equation. So, please note that this is in another form the eigenvalue equation. So, this is an iterative way of solving the eigenvalue equation. If you remember my original problem, it is H minus E Hartree-Caw. If I take the matrix element between all states, so basically you have all the states here i and j, let us say, then you have sum over Cj, then you get E correlation times this column Cj. So, in a very long form, this essentially means there are two sets of equations, one projected with E Hartree-Fock. So, psi Hartree-Fock H minus E Hartree-Fock psi Hartree-Fock. So, this is 0. Then you have psi Hartree-Fock, another block psi Hartree-Fock H minus E Hartree-Fock which is B-dagger psi doubles, which is B-dagger. Then you have similarly psi doubles H minus E Hartree-Fock psi Hartree-Fock which is B and then you have psi doubles. So, this is the eigenvalue equation psi doubles. This is my full matrix multiplied by the C which is 1 Cd equal to E correlation 1 Cd. So, this was my original equation. If you remember, you please do the multiplication. So, this is anyway 0. So, you have 0 minus 0 B-dagger B-d multiplying by 1 Cd equal to E correlation 1 Cd. So, that is the same equation that I have written. So, you have E correlation into 1 equal to B-dagger C or Cd and then you have B plus D prime into Cd equal to E correlation Cd. So, I hope you can write the matrix equation row into column, but this is basically a block of numbers. So, I just for the for then later on I just read that since in double C I we need not write Cd, I wrote C I. So, either way you can write in a long form, short form you should all converse in the same thing. So, different ways of understanding, but basically this is my correlation energy. So, you can say that this and this should be identical because this is an eigenvalue equation. This is a algebraic equation and that is a very important understanding that the eigenvalue equation can be written in this manner and this is a clear case that I have a 2 by 2 eigenvalue problem. I am writing the row and column as the two equations this and this and from here I automatically get this. So, please I think eigenvalue equation has never been presented in this manner, but the eigenvalue equation can be present it can be anything remember. I am just doing for the correlation energy it can be any general eigenvalue equation of a 2 by 2 block or simply 2 by 2 you can always write it in that form. So, this actually gives you an iterative means of solving the eigenvalue equation because this is an iterative equation. So, that is what I want to tell you this is important because actual eigenvalue equation is very complex because it have a large number of large matrix diagonalizing this matrix by Jacobi or householder is very expensive and you do not need all the roots all the eigenvalues I need only the ground state or the lowest first or second excited states. So, the iterative solution actually gives me these now depending on the guess you will get one particular solution. So, you will get either ground state first excited second excited state the guess is very important because this is an iterative solution. So, that is how you will get. So, this is basically transforming the eigenvalue equation in this manner. This is what you will normally get from the CI equation. So, transform everything in this manner. So, what we want to do is now to simplify this correlation how do I do an iterative solution. So, first to solve this equation first I assume that the E correlation is 0 which means my exact energy is nothing but Hartree-Fock energy. So, that is my first guess. If I use this then obviously I will get the correlation energy as equal to minus B dagger B inverse B because E correlation is 0. So, this goes out. So, that is my first iterative solution. So, I start with the E correlation 0. This is my first iteration. So, this is a simple numerical methods how to do it then what you do put this value back here. Then you will get a second equation E correlation second iteration and keep doing it till the correlation energy converges and of course the coefficient also must converge at that part of time but you can directly now the coefficients are actually hidden coefficients are no longer there. I have substituted the coefficient equation and directly write an expression for correlation energy with everything that is known but the only problem is that there is a correlation energy in the right as well as in the left hand side. So, I have to iterate the coefficient I have actually substituted. So, it is not gone in some this. So, this is actually gone into the iteration. So, let us assume that I start doing it and that is what I said last class that I stop at the first iteration. So, assume E correlation is 0 and stop here then we can analyze this expression. So, this expression now I can write in a full form what is B dagger remember this is my B dagger this was my B dagger and the other one was B. So, let me write B dagger fully it is a matrix multiplication. So, one index of B dagger which is now 4 orbitals then D prime 2 indices B another index which are all 4 or 4 indices. So, if I write explicitly then it becomes minus what is B dagger psi Hartree-Fock H minus E Hartree-Fock into psi doubles. Now, I am writing psi doubles explicitly as psi ABRS it does not matter what index I use ABRS CD2 because they are dummy ABRS and then you have the D inverse which is psi ABRS H minus E Hartree-Fock then another dummy index psi CD2. So, I have to sum over also CD2 then B of this index please remember how the matrix multiplication were B dagger I D prime the inverse I j B j. So, this is my j now what happened inverse yeah inverse of course I did not write that yeah inverse of course I am not completed it. So, then I have B which is psi now CD2 H minus E Hartree-Fock psi Hartree-Fock that becomes your first iterated value of E correlation and remember again it is a number I start from psi Hartree-Fock go to 1 W x R A determinant come back from 1 W x R A to another W x R A back to Hartree-Fock. So, I have to come back to Hartree-Fock because it is a final energy. So, then the second approximation that I do that assume I told you last class assume D prime to be diagonal. So, within the CI approximation the CI solution I made two approximation one is to first start the guess of course the E correlation is 0 that is not real in approximation but the real approximation is stopped here then I can write this expression this is not correct this is not a DCI this is only first iteration of DCI remember and then within the first iteration of DCI I assume that this is diagonal which means CD2 is identical to ABRS in my summation right because this index has to be same as this index otherwise it is 0 this D matrix is 0. So, if I assume this to be diagonal then E correlation becomes sum over ABRS and note again when I am writing a sum over ABRS I have A less than B or less than S I mean that is assumed because the all doubles excited which is obviously has a A less than B or less than S I am not explicitly writing all the time but they are anyway ordered. So, then you have psi Hartree-Fock H note again that E Hartree-Fock need not be written because it is 0. So, psi Hartree-Fock H psi ABRS and then you have psi ABRS H psi Hartree-Fock divided by the psi ABRS H minus E Hartree-Fock psi ABRS. I hope it is clear now because it is diagonal the inverse can be actually put as a denominator otherwise it is a matrix. So, matrix inversion was very difficult but moment I assume this is D prime to be diagonal its inverse is very easy it is another diagonal matrix with in diagonal elements as one by the elements. So, I can actually put this under the sum. So, the whole thing is just summation over ABRS. So, that becomes my correlation energy. So, it is very easy two sets of summation is gone only one set of summation and if you now look at it this is very close to MP2. If you remember MP2 the numerator was exactly this denominator was just orbital energy difference here it is not so there is a slight difference you can actually analyze the denominator. So, a further approximation where you make the denominator as only orbital energy difference will actually give you exactly MP2 values but you can see that from DCI to MP2 I have made very large number of approximations to get MP2. So, I have lost the structure of DCI of course because very as soon as I stop here it is no longer DCI then I have made further approximations. So, it is not DCI I am just showing how can you recover MP2 from DCI but it is no longer variational no longer upper bound will be there because I am not doing DCI upper bound will be there only when I converged it correctly when I solve the DCI equation exactly yes H minus oh this is 0 ERTFOG need not be written here these are orthogonal I have been repeatedly telling you that B I can write as H minus ERTFOG or H they are identical because the WXRA determinants are orthogonal to psi heart report. So, I this this is just cosmetic I did not write so that is the reason when I wrote H the full energy equation I also use B when I am writing correlation energy also I am using B I am not putting B prime but D I have to distinguish D and D prime. So, this is basically like an MP2 or a perturbation result second order perturbation result where denominator is just little bit different and you can actually calculate this denominator this denominator can be calculated by slater rule that is very easy to calculate remember again ERTFOG is just a number which will come out because both sides are ABRS they are orthonormal. So, ERTFOG will simply come out so really all you need to do is to use this slater rule type 1 for psi ABRS H psi ABR I hope all of you will be able to do that. So, if I ask you to calculate psi ABRS that can be a question by itself so psi ABRS H psi ABRS. So, what we have is the following we need to calculate this. So, our denominator here is just this I am now simplifying the denominator. So, I have a Hamiltonian matrix element minus ERTFOG. So, that can be taken out later at any time that you want I hope you will be able to write this I leave it to you I will not do this exercise you should be able to use slater rule. So, there will be some H etcetera you should be actually able to write this also in terms of orbital energies. So, there are several expressions of this slater rule you can write the H as H0 plus V where H0 becomes your one particle operator which is sum of the Fock operator that will give you orbital energy as H of Ii right normally you write H as sum over H of I plus 1 by Rij correct. But learn to write like this which is sum of the Fock operator. So, instead of H you have a Fock operator and then V has 1 by Rij minus the V Hartree form you note that V Hartree form I have defined in the perturbation theory. If you write it in this form your first term is actually sum of the orbital energies then something will be there please try to write it. The point that I am trying to say is that if you forget about this then your entire term here is just the sum of the orbital energies right. And if you then subtract E Hartree Fock where E Hartree Fock is also approximated as sum of the orbital energies then what you will get here as an approximation you can get this as epsilon R plus epsilon S minus epsilon A minus epsilon right. Please note that let us make an approximation that I will write Hartree Fock energy also as sum of the orbital energies I forget about this term at all. So, I am only using H as H naught plus V even in Hartree Fock energy. So, my Hartree Fock energy is nothing but psi Hartree Fock H naught psi Hartree Fock. So, I would have got only sum of all the occupied orbital energies right this is gone when I construct this this is also gone. So, again sum of the orbital energies except that AB is not there instead RS is there. So, if I subtract from Hartree Fock everything will get subtracted here there will be RS here there are AB. So, epsilon R plus epsilon S minus epsilon A minus epsilon B have you understood. So, if I approximate H. So, under the approximation that H is nothing but the 0th order Hamiltonian H is equal to H naught the full thing including a Hartree Fock your denominator D is nothing but epsilon R plus epsilon S minus epsilon A minus epsilon B alright. But please remember if I give you if I ask you to write fully you should be able to do this best later room please practice this here I am just showing that under the approximation that H is only H naught I get this denominator as epsilon R plus epsilon there is a minus which I forgot epsilon R plus epsilon S minus epsilon A minus epsilon B if I now substitute this here it is MP2. So, now you can understand this exactly MP2 because this is my AB RS anti-symmetrize this is RS AB anti-symmetrize divided by the orbital energies right. So, that would be exactly the second order perturbation. But you can see there are several steps. So, one is of course stop at the first iteration assume D prime diagonal then assume the diagonal elements as epsilon R plus epsilon S minus epsilon A minus epsilon. So, there are three essential approximation by which DCI can be shown to be equal to MP2 and then I can write under that approximation as this as this will become AB anti-symmetrize RS I hope all of you can write this later rule this will be RS anti-symmetrize AB divided by the epsilon R plus epsilon S minus epsilon A minus epsilon B with a negative sign here. So, actually you can write this as A plus B minus R plus R minus which is again always negative for the ground state energy is always negative I have told that that is that is not necessarily true for DCI energy. So, only the approximation of DCI energy that I am discussing. So, you can actually write that correlation energy and eventually you can release this A less than B then you get a 1 by 4 factor you have done all these. So, if I write all A all B all R all S you have a 1 by 4 factor you can spin integrate we have done this as an exercise spin integrate and write it as a closed shell for a closed shell element you can write a spin integrate reform exactly the same thing will happen except that actual CI is different because actual CI I have to keep iterating. So, I will get completely different result. So, I am going to discuss the actual DCI here I wanted to show as a side show that starting from DCI how can I get MP2. So, in a way it is a side show, but do not forget that actual DCI is this, but I think it is a good understanding that how DCI can lead to MP2 under the what are the approximations that you require. So, that DCI can actually lead to MP2. So, in that sense DCI and MP2 has a lot of similarity one is that of course MP2 has only WXRA determinants that I have already stressed DCI also has only WXRA determinants by very definition because I am doing configuration interaction only with Hartree-Fock plus WXRAT. So, it was important for me to know what is the similarities and differences. So, that is the reason I am trying to look at DCI and MP2, but DCI is something more more or less I do not know but something different and because you have to keep iterating when you iterate lot of new things will come because very first iteration remember I have a B and a B. So, there is already a second term and then I take whatever approximation I take I will have another B times B. So, you can see the number of V elements are increasing. So, in terms of perturbation order you will get higher and higher order. So, that is something that you can clearly see that when I do DCI actual DCI the connection with the orders of perturbation is lost. Everything comes from WXRAT, but lot more terms will come in terms of perturbation order. Is it good that you do not know because there is no variational bound to the perturbation order. So, we do not know, but finally I know that if I convert DCI that is at least an upper bound for a completely different reason because I have used variational method, but at every single iteration I have no idea about the bound. So, there may not exist bound because I am not solving a DCI equation. Is it clear? Yes. So, that is the convergence. I am putting the convergence on correlation energy. They should actually translate to the same. Well see the convergence is exact. Yes. That means between the two iterations if it is 0 it will be at the same point. So, depending on how your criteria results will change little bit. So, these are all basically numerical problems. I am not going to discuss numerical problems, numerical algorithms. Depending on what you put a convergence criteria that is what he is asking, I may get slightly different results because there is no way that I can translate what delta of C is what delta of E correlation. It is very hard for me to tell. I only know if the delta is 0 here that delta will be also 0, but we never can make it 0. So, that point remains. But I think it is convenient to put correlation energy as a criteria and usually you use at least of the order of 10 to the power minus 7 or minus 8 atomic unit as a delta. So, it will it results will be that much accurate 7 to 8 digits in atomic unit. So, atomic unit it is important. So, you know that we have a larger amount of accuracy. So, that is what typically we will use. In fact, this is also the starting point of what is called Davidson's diagonalization routine. I will not mention this. In fact, diagonalization routine should be actually covered in some other course, numerical methods or maybe computers in chemistry that came 481 I do not know. So, we will see. But this is actually a starting point of Davidson's diagonalization routine. It is a very famous routine that is used for CI. Please remember to get some states. The only problem with iterative methods are that you are always starting with a guess. So, you will reach only one result. So, you cannot get all the states. If you want to get all the states, you have to go back to full diagonalization which is Jacobi household or whatever. We have sidestep from there. But in a way, this equation actually represents all provided I can converge to all the roots. So, basically it is a highly non-linear equation. I hope those who are again those who have not done little mathematics I want to tell you. If you have a quadratic equation, how many roots you have? Two. If you have a cubic equation, you have three. Am I right or not? You do not seem to be convinced. If you have a linear equation, you have one root. If you have a quadratic, you have two roots. If you have a cubic, you have three roots. So, polynomial. What is the degree of polynomial here? This equation, the exact equation. It is also a polynomial equation in correlation energy. What is the degree? Oh, why quadratic? You have inverse. If I inverse expansion, it is infinite expansion actually. So, in principle, I can have many, many solutions. Because it is a matrix, eventually that is going to limit my number of roots and the number of roots will actually become the number of the entire set of equations, entire set of the thing. Otherwise, any inverse expansion, even if it is a matrix inversion, has many, many roots. So, this is the highly nonlinear equation. So, this itself should be able to recover all the necessary roots. The problem is how? So, if you do an iterative solution, it is very hard to get all the roots and that is essentially a numerical problem. So, because you have to guess. How do I know the guess for all the roots? Guess for the ground state is easy. Why? Because I know that the ground state energy is dominated by Hartree-Fock. Remember, Hartree-Fock is not a good method for excited states. We have done this only for ground state and I have repeated that. So, for ground state, I can assume that the correlation energy is 0. But if I use Hartree-Fock and try to describe the another excited state, it may be very bad, whether Hartree-Fock is very bad for their excited state. So, here correlation energy may be very large. So, depending on what you start with, you may get some results. So, I am just trying to convince you that what I have got here is ground state because I have started in the correlation as 0 and no wonder with all these approximation, I can eventually recover the MP2 ground state energy with approximations, of course, which is completely different from DCI to the different structure. But I think it is important to realize this.