 I think started with first dci, so let us write down the equations first. So now that we know that the method of projection gives the same as the variation method, we are going to use the method of projection to write the equation. We have done that, it is a matrix Eigen value equation, but I want to rewrite this once more. So h then C0 psi Hartree-Fock plus sum over a less than b. So we are writing only doubles, so no singles, because doubles are the first important terms. So it is what is called truncated ci, which means I have truncated the ci expansion only to doubles, it is not a full ci, it is not exact even in the m dimensional basis. So then what we do is to project, so first we project with this psi Hartree-Fock. So we will get psi Hartree-Fock 8 psi Hartree-Fock C0, so that is the number and then we will get here plus a less than b or less than a psi Hartree-Fock h psi a b r a. So this will survive. Remember when you did only singles, this was the block which was 0, because of Bravais theorem, but here this will survive with these c a b r s plus equal to E0. Now it is very easy to see here, the psi Hartree-Fock psi Hartree-Fock is 1, so you have E0 times C0 and these are orthogonal, so you do not have to write. The right hand side when I project with psi Hartree-Fock, you have only coefficient C0 because this is equal to 1, so coefficient comes out, psi Hartree-Fock is orthogonal to psi a b r s, so you do not have to write. So this is the reason you get a Eigen value type of equation. Then you have the next set of equation which is one of the specific W excited determinant, let me call it now psi CDTU, I do not want to use the dummy variable a b r s, so it is one of the W excited determinants where a b is equal to CD, r s is same as TU. So psi CDTU is the specific index now, CDTU are specific index, so I do the same exercise, you have h psi Hartree-Fock C0 plus sum over a less than b r less than a psi CDTU h psi a b r s C a b r s. So this is basically the matrix element of the Hamiltonian between doubles and doubles, as we talked in the last class the doubles and doubles h dd, so this is that part into the coefficient, again you have an energy, this part is very easy because when I do psi CDTU this is orthogonal to psi Hartree-Fock, even this is orthogonal to all psi a b r s unless a b is equal to CD, r s equal to TU, so that time only c CDTU will survive. So you can see whatever is the configuration with which I am projecting that corresponding coefficient comes in the right hand side, that is all, the right hand side is very simple and that is the reason we call this matrix Eigen value equation. So this is h times c, you can see h times c equal to e times c, one of the c's, you have actually the set of equation, so if I write this as a matrix problem then you have the first part, so there is a matrix of h where the first part is e Hartree-Fock because this is nothing but e Hartree-Fock, please remember and then you have a set of numbers which are psi Hartree-Fock, h psi a b r s and then on the column side you have again psi CDTU h psi Hartree-Fock and then you have, maybe I push it little bit further where it looks nicer, so that is my matrix. So you can see this is for all CDTU, this is written for specific CDTU but you have to keep on changing, so as I change I have all the numbers here, I will have all the numbers here and this, so this is basically one number, this is a row, this is exactly a conjugate column and this is a square matrix, square matrix between doubles to doubles and then you have c0 and c a b r s, the doubles coefficient equal to e0 the same thing c0 c a b r s or c CDTU whatever. So you can see that whatever I am multiplying I will have exactly the same number coming up, these a b r s are just a generic thing, they are all doubles. So in a very simple way I can write this as e Hartree-Fock, psi Hartree-Fock, h psi doubles, so psi d I am introducing psi d just to say it is a block of all doubles and then you have similarly psi d h psi Hartree-Fock and then you have a psi d h psi d is c0 c d equal to e0 c0 c d, of course c d is not a one number, c d is a column, so c d is a column, so it is a let us say n number of w x r a determinant, so it is total n plus one dimension in that case, so this is actually a set of number c d. So similarly this is a block, this is a row and this is a corresponding column, so it is a 2 by 2 block, so 2 by 2 block, so that is the reason when you have only singles then these two blocks become 0 because of Brillouin's theorem and the matrix becomes a block diagonal, so you just have this one value e0, so this is the reason we are calling it matrix eigenvalue equation, so I just want to rewrite. So if you now add singles, doubles, triples whatever I can keep on writing very easily by the similar block which I will explain little later, so this becomes your double CI equations. So I have to calculate the coefficient and then put it in this equation I will get the energy, how do I calculate coefficient from this equation, so the second set of equation will give me the coefficient, if I solve this equation say linear equation I can put it back here that is one way to look at it or you can say I simply construct the Hamiltonian and diagonalize that is one and the same thing, you construct the Hamiltonian and diagonalize then you will get the e0 and all the coefficient, of course if you construct the Hamiltonian and diagonalize we know that you are not going to get only one e0, you will get the entire dimension, so whatever is the total number of elements you will get that many e0, so this is what leads you to what are called excited states. So all those higher energies will now be actually excited states and they also follow the theorem that I told McDonald separation theorem, they are the upper bounds to exact excited states not just for the ground state, so all those numbers will come, so there are many ways of solving it, so one is to simplest way to think is to construct the matrix and then diagonalize, however very often this is a very tedious task because the number of WXRA determinants if it is very large then the matrix becomes very large and diagonalization of large matrix is very often time consuming, further the advantage that I have in diagonalizing is that I get lots of energies but very often I am not interested in those energies because let us say I am only interested in getting ground state, so I am doing a lot of work for no reason, so is there a simple way to solve this Eigen value equation without actually diagonalizing the full matrix, so you have to understand that this matrix size is very very large and we do not want, if it is 1 million by 1 million matrix, you do not want 1 million Eigen values and what will you do with 1 millionth excited state, this is not interesting chemistry, we often want ground state and maybe first or second excited states, so there are better ways of doing it and those are actually called iterative solutions, iterative solutions to this Eigen value problem which is much better, so something that I will mention in during the course of the CI problem if not today maybe tomorrow, so there is a iterative solution in which you can at least get 1 or 2 roots ground state and first excited state quite quickly and at a reasonable accuracy but you do not get all the states, so you do not have to actually diagonalize this full matrices, so let me just try to again understand whether you are all in sync with me, so essentially you have to construct a matrix of the Hamiltonian in the basis of the determinants which are included in the CI that is the first thing, so first determinant is psi Hartree-Fock, so that gives you a number e Hartree-Fock, then there is a set of numbers everything is a number of course, there is a set of numbers which is between psi Hartree-Fock and all psi A, B, R, S, similarly I have all W, X are determinant project this with psi Hartree-Fock, there is a set of I call them a column and then between doubles and doubles, so I have the full matrix which I am looking at as a two block, one is the Hartree-Fock block which is one by one and the rest is all doubles, so if my doubles number is n then what is the dimension of the matrix n plus 1, is it clear? Because Hartree-Fock you have to add, so my total number is Hartree-Fock plus all doubles, so if I give you some small problem you should be able to calculate at least the dimension of the matrix, I give you a small basis, two basis problem, you should be able to calculate how many W excited, if in a DCI you should be able to find the dimension of this problem and should be able to write it as a doubles, block of doubles, is it clear? Before we go forward let me also mention that we introduce something called intermediate normalization in the perturbation theory, so you may like to know is there intermediate normalization here, can I do this? Clearly the wave function that I have proposed to you is not intermediately normalized, I hope you can see this, the proposition of this is that psi Hartree-Fock, if I do psi Hartree-Fock and exact psi naught that should be equal to 1, if you remember intermediate normalization, if I do this here the psi 0 which I now call psi 0 DCI as I have proposed, what is the value? Can somebody told me? It is not 1, what is the value? As I have proposed the DCI wave function, what is the value of that? C0 plus anything is there, can there be anything? Because this is orthogonal, so only C0, only C0 and I have no control over C0 because C0 will come out of the solution, I have no control over C0 except that I know that C0 must be very large, very large because in the exact wave function Hartree-Fock is a dominating term, C0 must be close to 1 and of course C0 square plus all coefficient a less than b or less than s C a b r s square or mod square equal to 1, you can call this mod square, that is the normal normalization. But obviously this is the most dominating term and these are very very small, typically when you do a CI, this C0 may be 0.95, 96, 97, so you can imagine that large, so it is very very large, the Hartree-Fock itself is very large, the probability of Hartree-Fock would be easily about 90 percent or more and then all the rest contribute to the last 10 percent. So you have to understand that these values are very very small because you have already said Hartree-Fock is a very good wave function except that we cannot be satisfied with that because of chemical accuracy that we want for the properties, for the difference energies. So we still have to take those small small quantities as accurately as possible. So this is the thing that we get and clearly this is not an intermediate normalization because for intermediate normalization you must have C0 equal to 1. So the question is can I introduce intermediate normalization in CI? So if you look at the problem then you will realize that this can be trivially done, simply set C0 equal to 1. So I set C0 equal to 1, so if I write DCI in intermediate normalization then what will happen is that this wave function will not have C0 because that is one, the rest will remain as it is, so the C0 will go, this will go, this will go, this will go. So it is absolutely trivial to write it, everything is same and your equation here will be exactly same except that this will be 1. So everywhere I can put C0 equal to 1 and immediately you can see that your wave function is now intermediately normalized because obviously you know by previous discussion that if that is so, if C0 equal to 1 then of course C0 C0 cannot be 1 because C0 C0 is nothing but mod C0 square plus this quantity, this is already 1. So obviously this will be greater than 1 because these coefficients have to survive, otherwise of course you are doing nothing. So they are non-zero, survive so this will be obviously greater than 1 and that is something that we have to accept and I can always renormalize later to get back C0. If I renormalize then I will get back 1 by square root n, that will be my C0. So this is very easy to do, I hope you realize that this is really not a problem. Often we want to write it in this manner only because the perturbation theory we have done with the intermediate normalizations, we like to compare with the perturbation theory. So in this case it is easy to put the DCI equation in the, in the intermediately normalized form, any question? You want to ask anything? Yes, you do not need to do as I said, you do not need to do but it is convenient, the convenience is the following that you will immediately see because I want to calculate the difference of E0 and E Hartree-Fock as a correlation energy that comes out easily otherwise there is a C0 which gets stuck here with E Hartree-Fock as I showed you. It becomes E Hartree-Fock times C0, I will come to that so that becomes very easy to look at the correlation energy but you do not need to do, you are right. I can still divide by C0 and do it, it is not a big problem and comparison with the MP2 and all that that I have done becomes more easy but so I just want to tell you the intermediate normalization is not a new theory of course it is the same DCI, I have forced this equal to 1 and obviously this coefficients will change because I have forced C0 equal to 1 so everything will get normalized, everything will get renormalized. So let us look at the equation in intermediate normalize, so I have finally this block. So let me go back, let us see how the iterative solution will look like in intermediate normalize form, so essentially I am going to look at this. So the first term is E Hartree-Fock, so what I am going to do, I am going to multiply the first row times the first column, so E Hartree-Fock, this is a set of numbers which is a row, which is a row, let me call it bidagger, we will see why I am calling it dagger, bidagger multiplied by CD, so this becomes bidagger C or CD, please remember I am now using a matrix notation, bidagger is a row, so B is a column, so this will be my B, this is bidagger, it is a bidagger times this column, is it clear to everybody, I mean I am just using a matrix, so this is a matrix, this is not a number, this is a matrix, obviously this has to be a set of numbers, so it is a row, this is a column, it is a one dimensional matrix actually, bidagger CD equal to E naught, so first row into first column equal to E naught and you can quite clearly now, why I use E naught, why I put E naught equal to 1, because you can clearly identify the correlation energy as this, I do not have to divide by C naught later, so that is easy, so whatever is the bidagger and CD, CD is the set of coefficients, lovely excited coefficient, bidagger is this matrix element, where Hartree-Fock is on the left, Psi D on the right and obviously it is a set of numbers, set of Psi D, so all this together becomes a column index, column index, row index, so this is basically, this is bidagger let us say, this is some bidagger and this is some column, CD, so this is my bidagger, this is my CD, so you just multiply this, you get one number, if you multiply row times column, you get one number, so that is what you will get, because I need a number here, these are all numbers, by dimension you can see that, then I go to the second equation, the second equation I have CD, so sorry I have B, note that if this is bidagger, this must be B, so B into 1, so B, again this is a column, it is not just a number, B into 1, plus you have now a set of block which is the matrix now, D, let me call it D, I put 2 bar to denote that this is a 2 dimensional array, it is a matrix, why it is 2 dimensional, because you have a set of numbers here, you have a set of numbers here, so it is a 2D matrix, is it clear, one side is A, B, R, S, another side some CDT, so all doubles with all doubles, so one row will be one doubles, so basically I am looking at this, these elements, all elements, so one row will be defined by CDT, one column will be defined by B, so I am calling it a 2D matrix, it is really a matrix, square matrix, of course it is a square matrix, this should be multiplied by what? CD, it is a column, this is a column, now you see look at the dimension of the problem, it is exactly same, this is a column, this matrix times column, column is also a column and you will get a column on the right side, because it is E0 times CD, so everything is beautiful, that means this column, this is a column thing, so it is actually a set of equations, if I want to write it is a set of equations, I have written in one equation, it is actually a set of equation, Bi plus Dij CDj equal to E0 CDj, it is a set of equation, but I am writing only one equation here, in a matrix form, I hope all of you are sophisticated enough to write things in matrix form now, by now you are sophisticated, mathematically sophisticated, so and here of course everything is a number, so this is a one dimensional equation, this is the dimension of the equation is number of W excited amplitudes here, so I need to calculate this quantity, remember what is unknown, what is known, Bidagar is known, because that is the matrix element, that is later rule, I can apply this later rule, I can find out, what is not known is the CD, CD is in the intermediate normalize form, the coefficients are all that is unknown in the CI, the wave function is otherwise known, so I need to get CD, how do I get CD, I look at the second equation, from the second equation B is again known, so from the second equation I will try to get a form of CD and substitute this here, so matrix form, so all of you should be able to do this, so how do I get, so I will write this as B equal to E0 times CD minus D times CD, or I can write the reverse manner, I can write this as D times CD minus E0 times CD equal to minus B, again it is a column equation, this is the matrix times a column minus a number times a column equal to a column, is it okay, note again this is known, I am repeating this is my B which is just the matrix element of the Hamiltonian between W excited determinant and Hartree-Foggan, I will do that actually, I can use Slater rule and find it out, I have not done it yet, but that Slater rule can be trivially applied, I need to find out CD, so what will I do, I will write this as D minus E0, but note that now I have a problem, this is a matrix, this is a number, so how do I multiply, I have to multiply this by a identity matrix, correct, this is basically subtracting from diagonal elements of D E0, so how do you do this, you write this as E0 times identity matrix times a column CD equal to minus B, so I have a matrix times a column equal to another column which is minus B, so now I should be able to get CD from here by inversion, formally I can write an inversion of this matrix, multiply from the left, this will cancel, I will get CD, so let me write this now, so all W excited, although I am writing for a CD, remember I am writing for all the configure, all the amplitudes of W excited, so CD equal to minus D E0 inverse, so CD is just negative of D minus E0 inverse, we do not need, we do not need, yes you are right, we will see, yes that is where the iteration will come in, absolutely, you have already got the point, we do not know E0 also, so but let me write CD formally in this manner and then write the correlation energy, so correlation energy is nothing but E0 minus E hard to pop, let us define it in this manner which is nothing but B dagger CD, so I put this value of CD, so I have very nicely got minus B dagger D minus E0 into identity inverse, just want to make sure that I put 1 bar and 2 bar right, just for the time being it does not matter as long as you understand or for the time being I want to be correct formally, so this is what you get, so your correlation energy is in some way dependent on this total energy, you can actually subtract the E hard to pop from both sides, I can write this in terms of correlation energy itself but or I can write the E0 whatever, whichever way you want to do, so there is a there is a simple way to do is that when I write the E hard to pop itself, you can write as E hard to pop minus E hard to pop, so you can make it 0 and everywhere you will get correlation energy, so that is again a trivial thing but let me not worry about that right now, it depends on how I define B E C dagger etcetera, so I can write this quantity in this manner, now note this quantity very carefully, the D is a matrix element between the Hamiltonian of the Hamiltonian between two sets of W X are a determinants, minus some E0 inverse and then you have one set of elements which is doubly excited and then B dagger or B, B dagger and this is B, psi C d t right, so you have a B dagger and B, compare with the MP2 energy, if you look at MP2 energy there are exactly two such sets B dagger and B, if you look at MP2 energy what was MP2 energy psi hard to pop ABRS, ABRS into psi hard to pop right, so that was really B dagger B divided by some orbital energy, if that orbital energy happens to be this, some approximation of this then I can get MP2 from here obviously it is not, so only some approximation of this will actually give me orbital energy, so one of the approximations is basically to write D minus E0, remember D is this whole Hamiltonian inverse as simply the difference of the orbital energies of the one doubly excited to another doubly excited, so this is where approximation has to be made, so we will come back to this, so otherwise of course you will not get it, if you see it normally there is a iterative procedure required because I do not know I can take out E hard to pop, I can write this in terms of E correlation and then I have to do an iterative procedure, so first what you do you start with E correlation equal to 0, calculate this quantity, find out again E correlation, put back and keep doing, keep doing you will get final value, so that is the trivial way to do this that you can actually iterate and get a value of E0 depending on where we start from, so you can start from E correlation as 0 and then you can keep iterating, of course during the iteration it is not very easy because every time you have to do this matrix inversion which is also not a very easy thing to do, a matrix inversion of this kind where this is a Hamiltonian, so this is where approximation have to be done, but formally that is how this step will go, you have to just invert this matrix D matrix D minus E0 matrix, so basically in this matrix simply on the diagonal term whenever there is ABRS, ABRS subtract E0, so one simple way to approximate this is assume D to be diagonal that is easy, what do you mean by D to be diagonal? It means any time ABRS and CDT were different the elements are 0, so all I have is a diagonal matrix where diagonal elements are psi ABRS H psi ABRS and you subtract your E0, subtract the total elements, so assume D to be diagonal under the diagonal approximation, I hope all of you can easily do this, you can actually invert this because it is a diagonal matrix, so all you have to find out are the matrix elements between doubly excited determinant psi ABRS H psi ABRS, so you can use later rule to calculate this, so you have one particle term of all Hartree-Fock spin orbital except AB instead of that RS will be there and then there will be two particle term where again AB is replaced by RS, so you can actually do this calculation quite easily and this is not very difficult, so with this approximation it is very easy to diagonalize this matrix or iteratively diagonalizes matrix and this is something that was initially attempted as a first approximation of a DCI method that let us do a trivial diagonalize, no there is no bound except that your E0 finally E0 has an upper bound to the exact amount, this is of course I should call it not really E0 but E0 tilde because this is not the exact E0 remember, this is an approximate E0 where I have used DCI, so this has a bound, so only to that extent whatever results you get it will not go but at every iteration there is no bound, the bound is only when you converge, so you cannot say that the first iteration it will be greater than exact E0, second iteration also it is greater than exact E0 that cannot be showed, so essentially this is my exact E0 and these are my iterations, I start with iteration 1, let us iteration 2, I cannot guarantee that it will converge like this, you understand, these are these points are iterative values, it may so happen that it may go like this, like this, like this and eventually reach here which is higher than E0 exact, so this is let us say 0th iteration value, this is first iteration, I am just plotting E0 basically, so this is my first iteration, this is second iteration, third iteration, let us assume that it is converged at fourth iteration, when it is converged it must be greater than exact E0 because of the upper bound theorem, but I cannot guarantee that every iteration this will be greater than E0 and iteratively it will actually monotonically approach, that is not that I cannot guarantee, so if you make a diagonal approximation then of course this can be easily calculated D-E0 inverse and you can then show further that if this D-E0 inverse is simply the difference of the orbital energies of diagonal orbital energies, so basically energies of A, B, R, S, then this will actually give you MP2, I hope that you can see very easily because B dagger B is already MP2 numerator and then denominator this will is a diagonal approximation, so it will simply come as the denominator epsilon A plus epsilon B minus epsilon R minus epsilon S will come, actually this will become epsilon R plus epsilon S minus epsilon A minus epsilon B, but this negative sign will change it to that, so you can actually get an MP2 and I will show when I write this full form next time, I will actually write the full form by a slider rule, I hope all of you can expand this by a slider rule, then I will show what is the approximation which will give to MP2, so what I want to tell you is that while there is an exact solution of BCI possible, we can make two approximations, one is a diagonal approximation of D minus E0 inverse, then further approximation is to assume the value of the diagonal as only the orbital energy difference, then what I get is an MP2 energy, so MP2 comes from BCI following two approximations, if I do an iterative solution, remember iterative solution, so first iteration itself, so I have iterative solution, so at the very first iteration, I make an approximation that D minus E0, one inverse, one is diagonal, so that is the first approximation, A and then you assume this diagonal as the difference of the orbital energies of this A, B, R, S, now they are only diagonal, so no CDTU is there, so it is only A, B, R, S on both sides and you can clearly see that with this you will get the, of course when I say orbital energy difference, it is D minus E0, not just D, D minus E0, you will have orbital energy difference and then you can clearly see that I will get MP2, so from the DCI, the process of DCI, very first iteration under some approximation gives MP2, which is very interesting to see that there is a connection between CI and perturbation, but quite clearly that value will have no upper bound, because I made already, first of all it is only first iteration, so it is a first iteration, I am stopping at first iteration, second is that I am making two approximations, the D minus E0 is diagonal, I am also assuming the diagonal as the difference of the orbital energies, so obviously it is no longer CI, so there is no upper bound in any case, as I explained to you is only when you have completely converged the DCI solution, so this will actually give you an MP2, so if you have an iterative solution, then you can recover MP2, MP2 energy with the following approximations and this is important to understand the difference of DCI and MP2, of course if I do SDCI you will never get MP2, because MP2 does not have singly excited configurations, MP2 only has also doubly excited configurations, so only with DCI you may be able to get, but not the exact DCI, an approximate DCI can give you MP2 energy and I think it is important to understand next, I think next class what we will do, we will actually try to expand this to see what approximations will give you that as the difference of the orbital energies, so I am not stopping just to show you MP2, of course if you go ahead further then the relation with MP2 will I am not stopping, I am saying under what approximations can I recover MP2, it is actually an iterative solution, so to recover MP2 I have to stop at the first iteration itself, because you can see if I go ahead then of course this will become much more complicated, you will have a further term, so you continue to build many more times this coefficient, so I have to actually substitute that coefficient here, so whatever coefficient I have got I have to go and substitute there again, so that will actually give you third order fourth order, in fact you can also make a analysis, higher order terms will start to get generated, but only from doubles, because MP3 also has only doubles, MP4 has singles I have not done this, so these are something that I will like to do when I introduce diagramatics, so please remember that perturbation theory is not over, I will come back to that when I introduce diagramatics, but right now without introducing diagramatics I am just saying that if I take a DCI then you have a very simple way to recover MP2 energy, under some approximation at the very first iteration you can recover MP2, if I do full again value equation solution then of course you lose it, you do not really see where the MP2 came and went, only when you iteratively solve you can see the MP2 value, so the details can be worked out, but I think the philosophically you should understand what we are trying to do, because essentially this is the crux of the equation, this is the crux of the equation and this inversion is actually quite complicated, matrix inversion, so we have assumed a diagonal approximation and then we have assumed the diagonal values as the difference of energies and then only you can recover MP2. Before I come back to MP2, the further details on the DCI,