 What we will do now is to go back to the DCI and look at the structure of the DCI equation a little bit more and understand. However, before I do that, let me also tell you that it is trivial to add singles and triples etcetera to the CI equation, something that I would not like to do, but in details but at least SDCI is important to do. So let me just tell you the structure of the SDCI. So all you need, forget about now approximations, go back to the actual equation of CI. So all you need is the matrix of the Hamiltonian. So I am again writing in terms of normal order. So what is the first term? 0. H minus E Hartree-Fock, psi Hartree-Fock is 0, just as it was there before. And then I have, I am now writing in a more cryptic form H0S, I hope you will be able to understand now. This is the matrix of the Hamiltonian between Hartree-Fock and Singly excited. Remember it should be H minus E Hartree-Fock, but it is not necessary to write E Hartree-Fock for the first term. Then I have H0D, this was essentially what I called Bidager in the previous exercise. So H0D, so it is a three block. So what I am now writing is a CISD equation. So CISD has three blocks, Hartree-Fock which is only one by one block. Then you have a block of singles, Hartree-Fock then you have a block of doubles. Of course this should be larger than this, quite clear, number of doubles should be larger than the number of doubles. Then you have another block which is exactly opposite, HS0, then another block, column block which will be HD0, this is what I called B, just trying to put you in the perspective. This is the new block that is coming in, it is a block matrix. What would be this block? This block would be HSS, between one set of singly excited to another set of singly excited and this block will be HSD, this is matrix, these are all matrix, however this is a rectangular matrix because number of singles is different from number of doubles. This is a square matrix, so please understand from the block itself you will know and then you have similarly a reverse of this which is HDS and eventually another square matrix which is HDD. You can try to write as nicely as possible, those who can draw well they should be able to do that. This is my matrix of the Hamiltonian and then I have one CSCD, so I have this block of CS, I have the block of CD equal to E correlation and exactly same, 1 CS and CD and so on. This becomes your CI, it is very simple. What you do is block wise, first block times the first column is into E correlation into 1, that is your E correlation. I had written down that Vdagar C will just change as more terms, however we know that we have a Brillouin's theorem, so now I am going to apply the Brillouin's theorem for the Hartree-Fock, so this block is also 0 and this block is also 0. Interestingly, your correlation energy remains as Vdagar CD or Vdagar C, because this is 0, so 0 times this is 0, so correlation energy does not change, is it clear? Because of the Brillouin's theorem, so the question that you should ask then why should I do CI doubles, why should I do CI SD, sorry I should be happy with CI doubles, the answer is no, because when I calculate C then you will see the changes will take place. My generic expression is Vdagar C but what is the value of C, if that is different then correlation energy will be different, so that you will see for the next set of equations. This is 0 of course, so this is second row times the column, now HSS will act on CS plus HSD will act on CD to give you E correlation CS, so I have a CS equation, do I require it no, because my E correlation is Vdagar C, so I require only CD, so I do not care about CS equation but come to CD, now you have a problem V into 1 plus HDS into CS plus HDD into CD equal to your E correlation CD, so your CD now depends on CS, I hope you can understand, it is a coupled equation, I will write them down, I am just trying to explain, so to solve this I need CS, if I need CS that means my CD changes from the DCI and hence CI's the result will be different, so why is it changing, it is changing because of this block, this block is not 0, because singles by itself does not contribute because Hartree-Fogart is 0 but singles contributes with doubles and doubles mixes with Hartree-Fogart, so in an indirect way singly excited determinant is changing the energy, not directly but through doubles and we know that the doubles interact with Hartree-Fogart, so if I can change the doubles amplitude then my result will change and that is done because doubles amplitude is affected by singles, so let me write down in the same block for the CI HD equations, before I go back to more discussion about DCI which is simple, note again that I can go back to normal order, so these are all that I am writing is normal order, so whatever you call it HN, since everything is HN these are all please whatever I am writing now is normal order Hamilton H minus E Hartree-Fogart, so I do not have to write this I hope, you will understand what it means, so let me write down the expressions, so I have the first expression is E correlation equal to Bidager CD, so that is of course given first expression is Bidager CD because of Brelois theorem this does not contribute but then you have a second equation which is 0 into 1 but you have now HSS, again please remember assume by default normal Hamiltonian, so I am not going to write E HN all the time assume by default normal order Hamilton, so the H that I am writing is normal order now, so you do not have to write the symbols each time, so HSS into CS plus HSD into CD equal to E correlation into CS, is it all right, then the last set of equation is HDO which is my B, B into 1, so B plus HDS into CS plus HDD into CD equal to E correlation into CD, note that if this term was not there then my CD equation would have been exactly like double CI and your correlation energy is exactly like double CI, so I do not care about this equation, so what is HSS I am least bothered, the problem is this guy the DS or SD, this guy is not 0 and that is the reason CISD gives a different result, so CISD provides different result from DCI or many people of course to be consistent call this also CID, I mean the literature you will find all kind of nomenclature, this is called CISD, this is called SDCI, so whatever nomenclature you want to use you use no problem, so this provide different result from DCI because of the block HDS or HSD that is identical block, again by default it is a normal order Hamiltonian, so that is the reason CISD gives a different result because if this was 0 then of course your CD equation would have been also exactly identical and correlation energy anyway does not depend on singles, so result would have been exactly identical, but now just because of this block I have to solve this in a coupled equation, so I it requires CIS this is known this is not 0, so I require CIS, so I have to solve these two equations in a coupled manner to get both CIS and CD and then put this here and the result will be, of course we know that the result will not significantly vary simply because this is coming at a slightly higher order this coupling term, so this is a coupling term which changes the DCI result only to some extent, so I have some numbers which I wanted to show and that is the reason I have brought this book today, how much the double CI changes from CISD, I think it is good to know before we go back to the discussion of the DCI, so E correlation is always same, what is asking if there are triples, so there would be one more block, but this is also 0 because Hartree-Fock with triples is also 0, so the correlation energy expression is always doubles, nothing else, but then now doubles should again depend on triples, so every time you are adding newer and newer excitations doubles is being modified, if doubles would not have contributed to triples then there would be no discussion, so what is important is doubly accelerated, whom they are interacting with, whomever they are interacting with will affect the correlation energy, is it clear? So triples will also change for the same reason, but we will not discuss CIISD triples, we will just go to CIISD, so