 is for ground state, the first set of determinants that we must include is actually W x r e. Because there is no Brillouin's theorem between Hartree-Fock and W x r e determinant. So now this block will not be 0 and obviously ground state will start to improve by interaction with the W x r e determinants and you can do ground state W x r e and then singly excited. You will see that the singly excited contribute. So I will just look at the structure now. Now that you have understood the structure very quickly I will write down. So let us first do take Ci doubles. Ci with only double so this is no longer there. So I have Hartree-Fock and W x r e determinant. So what will be the structure of the matrix now? Now I can directly write again I do not have to derive this. So I can do the method of projection and keep deriving similarly. So first structure is E Hartree-Fock one number. Then I will have a structure of elements between Ci Hartree-Fock H Ci A B R S correct exactly in the same manner and this will be the conjugate. So this will become Ci CDTU H Ci Hartree-Fock. Please note that I am again using just like here I had used Ci BS for projection I am using Ci CDTU a specific determinant which is part of A B R S. So and then you have this block which is Ci CDTU H Ci A B R S. I hope all of you can just derive this routinely just starting with the Schrodinger equation do method of projection for Ci Hartree-Fock then write Ci CDTU project to Ci CDTU and then you will get this equation C0 and all C A B R S or C CDTU whatever equal to E times C CDTU. If I am projecting with Ci CDTU I will get E times C CDTU but actually it is a column from that column only CDTU will survive okay. So that is an eigenvalue equation in general. Now you see this is no longer 0 because these are there is no Brillouin's theorem for this. In fact these integrals are nothing but A B R S A B anti-symmetrized R S and this is R S CD the CD anti-symmetrized or TU anti-symmetrized CDU. So these values will all be there okay. So this is a completely coupled matrix and you have to now diagonalize this. So for the first time the ground state will no longer be E Hartree-Fock. Ground state will now change because of the coupling terms here here and so on okay and you will have a separate result and now again you have to use letter rule for each of them. So this is of course quite clear you will be using only letter rule C but this block this small block which I am now calling HDD just in this terminology of HSS is the doubles into doubles. So let us look at this block little bit more carefully. So I have H doubles which is psi CDTU. So you have H doubles into doubles that is psi CDTU. So one of the double XRA determinants H with another double XRA determinants. Of course eventually there will be some and so on to get an eigenvalue equation but I am just looking at this matrix element. Now you can see even by letter rule lot of integrals will vanish even by letter rule and lots of integrals will remain of course by letter rule. So for example so many possibilities are there let us say A equal to C, B equal to D, R equal to T, S equal to U. So I have two double XRA determinants with respect to Hartree-Fock but they are same. Which rule you will apply? First one the rule A that is the rule A. I can keep on changing maybe A is not equal to C but B equal to D, R equal to T, S equal to U. Which rule you will apply? B, A is not equal to C but B is equal to T. Everything else is equal so rule B only one occupancy difference. So if I give you such problems will you be able to identify which rule to apply first that is very important. Then write the rule is second level of problem. So these are all technical problems important. Of course if you have you can have two occupancy difference rule C but then there are many integrals which will be 0. You can see there will be C4 up to 4 difference when everything is different what is the occupancy difference? A is not equal to so A B A is not equal to C, B is not equal to D. So A is not equal to C, B is not equal to D, R is not equal to S, T is not equal to U. You understand? So the ones which are projected here CD to TU they are different. So different places T and U have problem and the ones which are replaced here are also different different places R and S have problem. They can be different places can be always interchange that is a negative sign that is not a problem where it goes but eventually if you look at the differences they have differences of R, S and TU in different different places. So here you have RS and CD, here you have TU and AB. So let us take a 4 electron determinant. This will have RS, CD because it originally started with AB, CD. Let us say 4 electron. I am now replace AB with RS. So this determinant is nothing but RS, CD and this determinant would be nothing but TU, CD sorry AB, TU. So you can clearly see that if I compare the RS, CD and AB, TU all are different. So it is a 4 occupation difference. I hope you can see. So I am taking a simple 4 electron problem where my psi heart reform is AB, CD. These are the spin orbitals. Then I am writing what is psi CD to U? Psi CD to U is CD replaced by TU. You should be able to write this. So it is AB to U. And then I have psi AB, RS. They are all different. So it is a different block. So AB is now replaced by RS. So you have RS, CD. Since they are different everything is changed now. Of course if some of them are equal then it would be different case. So a maximum of 4 occupation difference which are anyway 0. There are 3 occupation difference which are also 0. Only up to 2 they will remain. The point that I am trying to say that this block will have lots of 0s coming from later rule and then again further because of spatial symmetry of course. But even rule A is applicable here. Just like rule A was also applicable here for single, single when AR is equal to BS. So between there will be diagonal terms. So basically they are the diagonal terms. So wherever I have a diagonal terms here you are applying rule A because diagonal term is nothing but same determinant of both sides. It has to be rule A. And the diagonal terms can be rule B, rule C and 0 or 0 depending on what they are. So this requires little bit of practice and I will ask everybody to do it. Take a 4 electron problem, 5 electron problem, 6 electron. Do this. Make it different. Just see what you are getting so that you will know what rules to apply for the CI problem. So essentially this is how we analyze and then you diagonalize because of the fact that there is a coupling term. Now the energy will change. So what will be the result of the energy? We will discuss. So in fact the CI problem will require at least 2 to 3 classes. We will see to discuss this. So we will first do CI doubles because this is very important. Then we will see how singles in the presence of doubles starts to contribute. We have already noticed that singles alone does not change the ground state energy. But we will see if I have taken doubles and then singles, then singles has a role even for ground state energy despite Brillouin's theorem because doubles is already there. So that coupling will actually help. So we will see this but first we will more thoroughly analyze double CI and we will place this CI doubles in perspective to perturbation theory. For example MP2 energy. How these energies are different? They are by itself very interesting study before we go to CI SD and after we tell CI SD the rest is really a technology. You know triples quadruples. I am not going to bother. What we are going to do after that is to look at the deficiencies of approximate CI. Full CI is of course no problem but we can never do full CI. As I told you MCN is too large. So what are the deficiencies of approximate CI is something that we will see why CI is not good. Why perturbation is better, CI is not good and why couple cluster is definitely the best. So that will come later part of the course when you understand what are the important chemistry that we need to ensure that our theories must have. So next class is we will continue here.