 So, we will continue with the CI and we wanted to show some of the numbers just to tell you what is the contribution of the doubles and singles and so on and it was very clear that the doubles was the most important part of the correlation energy compared to the singles mainly because of the Brillouin's theorem and of course the two particle nature of the electron correlation and we have discussed that two particle nature that the two particles are more likely to come close together and they will then repel and plus your Hamiltonian is also one and two particle interactions so they can bring two particles closer. So, in some sense these are two important facets and because of that WXRA determinants are also very important. Singles are important only because they affect the doubles actual energy please remember our actual energy depends only on doubles I think I showed you that that whatever is the full CI even if you have full CI the energy correlation energy depends only on WXRA amplitudes correct but however WXRA amplitudes themselves depend on singles, triples, quadruples and so on. So, it is very easy to see the structure of the CI matrix it will do a full CI because you know because of Slater rule any excited determinant cannot have interaction with another determinant which is of more than two excitations that will become zero so you can see the structure in the following manner that you have of course we are talking only of correlation energy the first one is always zero and then we have H0S, H0D, H0T and so on in a full CI this is zero because of Brillouin's theorem so this is an additional simplification that we have and similarly in the column we have singles with Hartree-Fock, doubles with Hartree-Fock, triples with Hartree-Fock and so on I hope these symbols are clear to everybody okay then you have the diagonal block of singles singles doubles singles triples and so on note here on the first row everything is zero even this is zero because of Slater rule so the only term that actually contributes is Hartree-Fock with doubles and everything else is zero from here onwards everything is zero so this is zero zero and so on these blocks and then you have Hs D, Hs sorry Hs T you will have more term Hs Q and so on similarly doubles Hartree-Fock, double singles, doubles doubles, doubles triples doubles quadruples and so on continue I hope all of you can write this now so you have triple singles so it is just filling the matrix okay so this is your diagonal triples triples and so on so depending on how many you take your matrix can be written in a block form these are all blocks now this is the only block which is one dimension which is Hartree-Fock Hartree-Fock to Hartree-Fock this is one dimension of rest of either rectangular or square so this is your matrix and then you have one, Cs, Cd, Ct, Cq etc whatever is there equal to same thing E correlation one, Cs, Cd, you can clearly see when I multiply the first row times the first column since everything is zero the only thing that multiplies is H double O double Cd to give you E correlation into one I hope it is clear it is a first row into first column gives you the first number here correct so your E correlation depends only on Cd because all this is zero so HOT will multiply with Ct but Ct does not matter what is Ct this block is zero remember these are not number multiplications it is a block wise so this is a rectangular one block of a row multiplying by column Ct is a column number of Ct so it is a number by number multiplications so each of them there is a summation for the point that I am trying to say that the correlation energy always depends on Cd only one that really affects the correlation directly however when I do this next level of equation I start to get equations for Cs, Cd, Ct and that is where you can see when I look at the Davos equation it is Hdo into one plus Hgs into Cs, Hgd into Cd, Hdt into Ct and so on, Hdq into Ct all this will affect the E correlation times Cd right when you do multiply second row times the column so everything will affect okay so only thing that will not affect is what are zero by sweater rule so for example in single row I know from here it is zero right from the double row up to here it will actually survive then the zeroes will start and triple throw it will go much further and so on however in the triple throw this itself is zero by sweater rule so there is a lot of structure in which zeroes will come so there are some simplifications but essentially as you build everything will get affected so Cd will be affected by up to quadruples then quadruples are affected up to Hdq so in a way it is a chain reaction so that is why the full Ci if you do full Ci results are going to be different although correlation energy depends only on Cd the Cd will be affected by others so that minor differences will be so that is the reason Cisd and DCi or Cid are giving different results yeah because the doubles amplitude changes it is not that the correlation energy directly depends on single double amplitude changes because of the effect of Cd is it clear okay that was the numbers that I was showing let me go back to some more numbers you also saw the effect of the full Ci at the basic set that is the basic set things change one of the important things is to look at ionization potential so let's say I am doing a Ci calculation and then trying to calculate the ionization potential what is ionization potential I calculate an n electron and n minus 1 electron okay I do a Hartree-Fock so there are many ways of calculating ionization potential one would be to do Hartree-Fock for n and n minus 1 and supply one is to use Pupman's approximate remember these two are different so when I do Hartree-Fock I am taking relaxation effect in Pupman's I just take the orbitals of the n electron system we have explained that and in fact that works out very good because of the cancellation between relaxation and correlation okay so sometimes so Pupman's gives a better result than doing Hartree-Fock separately you can do Ci for both so when I am presenting the results they are results which are exactly you know done done in the same manner so we are looking at the ionization potential nitrogen ionization potential has been upgraded to S so we are looking at the lowest ionization potentials so two of them so one is three sigma g one is one five they are simply called three sigma g one five u because these are the orbitals Hartree-Fock orbitals from where I have taken three electrons so that is the meaning of three sigma g and one five I hope all of you remember n two bond or molecular orbital diagram the last two orbitals homo and homo and homo minus one are three sigma g and one five in fact there is a there is a problem in ordering them which is higher which is lower depends on the Hartree-Fock what basis you use so that is another issue that you will see here so three sigma g and one five u so if I do Cook months if I do Cook months I just write k that means just look at the orbital energy of the Hartree-Fock you see this is 0.635 again everything is an atomic unit and this is 0.615 so it is easier to ionize was the one five u let me also look at the telling the basis set the basis is a very large basis 6s 4p 3d slatter type orbital is a very large basis set that was okay so these are the slatter and each of them can be expanded in Gaussian but they are nearly exactly so if I use a very large basis set my Cook months approximation within the Cook months approximation at least does not have basis set error because a very large basis set and you can see that one five u is actually because your your ionization huh 2f here yeah nothing wrong why I told you star basis right when one valence polarizes that is d but you can use also f g h I have 2f you mean not 2f 2 number of f type function sorry this means just like here it is not 6s you did you did not question this right 6s means 6 number of s type functions 4 number of p type functions in the basis set of nitrogen see number of d type function and 2 number of f type okay sorry yeah that is a simple basis set normal place okay because principle quantum number is not of any relevance so we never talk of principle quantum don't confuse with that okay so this is Cook months and then you have sdci 0.58 this is an important result 0.61 and you have experiment 0.57 all are in affirmation 0.62 now this is a very interesting result right I hope you can see the reflect on the results if I do Cook months which is a good approximate actually within the half new form the ordering of the sigma g and one five u is different from what you get from experiment because experiment shows actually it shows that this this should be the easiest to ionize that means essentially your lowest n minus 1 electron state will be doublet sigma why doublet sigma I hope you can you know at least some amount of symmetry because I am taking out electron from sigma g so the final n electron state will have doublet and it will be a capital sigma okay that will be the lowest state of not the doublet phi but what Cook months shows is different that doublet phi is the lowest state of n minus 1 or n 2 plus so I do n 2 plus and you can see that the sdci gives reasonably correct value not exactly correct but the order is at least so this is in fact shown as one of the failures of Hartree-Fogg that Hartree-Fogg when I do Cook months approximate it is supposed to be better than delta sdc anyway difference of Hartree-Fogg because of the cancellation is still not able to give the correct train even the qualitative forget about actual numbers but if I do a correlation calculation sdci or dci I get reasonably right results so this shows that the importance of correlation that the Hartree-Fogg itself is not a good method there is a another result of lowest IP of water whenever I am saying lowest ionization potential means you are taking out electron from the homework I hope that is clear because that is easiest to take out because ionization potential is exactly reverse and negative of the orbital okay so the lowest IP of water again some numbers this is actually again a very large basis at 39 st o basis it's very large basis so we have we have calculated h2o and h2o plus to ci and takes a take a difference so this is Cook month this is only the lowest one 0.507 again everything is anatomically this is sdci which is 0.452 and experiment 0.463 so again just shows that what is the level of correction this was kind of error that you had when you do Cook month now sdci almost brings it close to that and in the case of nitrogen it even gives the right order 39 st o basis not really primitive so basically these these these are contracted Gaussian so they are a lot yeah that that is that details are there I am not giving the details to the basis set but it means there are 39 see each letter type orbital is expanded in terms of primitive Gaussian right contracted Gaussian in a in a language 39 contracted okay so this is a very large basis as you can see then each each of them is expanded I mean the details are there it's unimportant but the point that I am trying to show is that this is a very very extensively group basis set and the scf is still not good quite away when you do sdci the IP comes out quite well then there are effects of singles and doubles and this is a this is a very important table let me also write this table so it actually comes out as c minus opus is actual practice so let us see what it gives so co-dipole moment this is scf will also give you the energy and dipole I mean energy is not very interesting but so if you do scf minus 112.788 again it's an atomic unit everything is an atomic unit and dipole moment is minus 0.10 so very interesting result I will just show you then you do scf plus 138 doubles I just want to clarify what does mean this is again a 30 this is again a basis set some extended basis set it doesn't matter the point that we are trying to say that even when you choose doubles you can choose them selectively w xr a country how tell me how how can you choose selectively w xr a country if you look at the wave function from the perturbation theory the wave function of the first order content w xr a detail right what are the coefficients of the determinant it is the matrix element divided by the orbital energy right so for example each of the coefficients are psi hardly for each psi doubles whatever is the doubles let's say a b r s divided by epsilon a plus epsilon b minus epsilon r something like this some denominator and this of course for the correlation energy a mod square of this divided by so each of this number tells you what is the contribution of each a b r s into the correlation error this value divided by this one if this is very small then you can actually neglect this so you can assume that that particular a b r s is not very important which means which can happen in two ways either this value is very small or this value is very large which means there is a large gap between these two occupied and two virtual and that that that means very high line virtual orbitals you need not worry because the denominator itself will kill it okay so that that is a way to select doubles in ci so many times we do double ci but we don't take all doubles so you do a very quick calculation of this number and then decide this number again can be calculated as a b anti-symmetrize r s by a stator rule and then quickly calculate what is the value and then decide the double so that is the calculation that has been done so let us do this 138 doubles the value of energy becomes minus 113.016 dipole moment remains minus 0.06 let me let me tell you the positive dipole moment is c minus open just to give you a reference so you can see that this is actually c plus or minus okay this is also c plus open which is wrong because the positive that this is how the table is going to positive means it is c minus o plus so scf I do a dci the dipole moment is still wrong energy of course improves marginally but energy is not dc I want to you have to focus on this dipole moment still comes out wrong c minus o plus okay small c minus open this is much larger c minus o plus it is reduced but it is still wrong correct then we do scf plus 200 doubles this is interesting so I take more number of doubles some more the result is minus 113.034 and this still does not improve minus 0.07 so it tells you that by just including doubles it still remains c plus open qualitatively it still remains wrong c o but what I now do is the following scf plus some doubles 138 doubles but instead of adding the 62 extra double they have added 62 singles so number of configuration is still 200 but we have added some more singles instead of doubles so it is a cisd but a truncated cisd within the cisd also have selected and here the result is minus 113.018 again energy you see almost saturated it does not matter one way or the other but now look at this number plus 0.02 and experiment I mean the energy is not important experiment dipole moment is plus 0.04 I think it is a very interesting table so I will try to reflect on this so what does this show tell me first what did you learn from this I give you some numbers now what did you learn much more than energy that is important dipole moments are affected by singles much more than the energy that is first very important whereas for energy I told you doubles is more important but now what we see is that when you come to dipole moment actually more than doubles what is important is in moment I take singles I start to get the right result so one important lesson if you do DCI for doubles for dipole moment it is a bad calculator if you are going to calculate dipole moment you must accept except that we never do cis because energy does not improve so the ideal thing to do is CISD it is very important to do CISD to get the dipole moment corrected most cases because the real correlation contribution to dipole moment comes from this and we will I will tell you why why why that is so but let us reflect on the result and as soon as you took some singles you can see that the dipole moment went in the right direction qualitative in fact the CO dipole moment and ionization potential of nitrogen these two were told as a very significant results showing that the SCF is bad first of all SCF is bad but if you do not take singles even correlation calculations are bad in many cases for dipole moment and this is also another example CO is a very brilliant example because dipole moment CO is very small see plus 0.044 see it is a C minus O plus but it is almost neutral it is only plus 0.044 it is almost neutral carbon and oxygen electronegativity difference very very small but unlike what you would expect it is not C plus O minus it is very small C minus O plus and that is why it is a test of a theory so if you have to if you have to approach a 0 which is not by symmetry it is a much bigger test of a theory if it is by symmetry you will automatically come even if you make some small small errors but if it is not by symmetry by cancellation of many many effects then all effects must be taken correctly to have that cancellation and this is the example where to reproduce these one has to be very very careful okay so you can see that with reasonably large calculation I am almost closed but the singles are important if you just keep on improving bubbles nothing will happen this will remain negative so the result will be qualitatively still wrong and that is the reason the CO is a very difficult molecule to calculate in fact if you do Hartree-Farr first of all you will get wrong in a good basis also first of all you will get a completely wrong so you do a charge analysis of CO with Hartree-Farr I mean it is totally wrong it will show C plus O minus okay and then everything all your all your chemistry will go wrong so this is one good example why electron correlation is important and why not only electron correlation the correct way of taking electron correlation is so that is that is a very important part to analyze now why does dipole moment get affected by singles more than the energy that somebody tell me that I hope all of you know what is dipole moment operator it is a one electron operator okay dipole yeah r r dot f yes we know as is of course marker right right so you so if you have a dipole operator mu and all of you know it's basically u charge times r r i sum over that so it's a one electron operator so these quantities like psi Hartree-Farr psi singles this is very very important it's also one electron operator in fact if you do this calculation psi Hartree-Farr mu psi doubles what is the result of this is actually 0 so because by slater rule this is a one electron operator so this is actually 0 so if I do double ci the dipole moment will not directly connect to the Hartree-Farr of course you will still say why did double ci change the value to change the value because I had other terms like psi doubles mu psi double because when I do double ci this side is also ci Hartree-Farr plus doubles this side is also Hartree-Farr plus doubles I am taking an average value of dipole moment the web function where left and right hand side so this is all I will do for for dipole moment because I don't have any variation of matter so I calculated the ci web function I calculate the dipole moment as an average value so that means psi Hartree-Farr plus c d psi doubles on the left and c c s psi s then again mu and the same team under that Hartree-Farr c s right so this is the ci s level of calculation so this is how I calculate and of course the normalization denominator the denominator is important because it is intermediate normalization so remember the whole web function is not normalized unity so I have to take the denominator but if I now calculate this first one gives me Hartree-Farr example it does not change then it gives me psi d mu psi h s which is 0 but then you have psi s mu psi h s which is very significant and similarly the converse terms so I have doubles to doubles and singles to singles these are all that survive but quite clearly when I have singles I have a very large term coming from this and that actually starts to affect the dipole moment more than this term this term only shifts little bit this is the more dominant term because it is a direct interaction with the Hartree-Farr example and this survives whereas for doubles this is 0 so that is the reason it is important to have singles when you calculate a dipole moment so never do a one electron property not only dipole moment any one electron property calculation without single because you will of course change the result but a major part is now not taken which is exactly opposite to what you said for the energy for energy we said singles does not really affect because for Hamiltonian there is a Brillouin's theorem for dipole moment there is no Brillouin's theorem okay so these are small small things but it is important to remember okay