 So let us continue with the double C i. So yesterday we did for the monomer hydrogen molecule. So I hope you remember that we derived the form for the E correlation as E correlation in terms of E correlation itself. That was important to remember. So we had E correlation equal to k12 square. Is that what we derived? E correlation minus 2 delta. So if you solve this equation, it is a quadratic equation. I hope some of you have solved this equation. So E correlation is equal to, has anybody solved it? Quadratic equation, very simple solution. It is delta minus delta square, plus k12 square to the power half. So that is the result of the solution of the quadratic equation. But 2 delta, remember 2 delta was the, what was 2 delta? I remind you, 2 delta was the diagonal element of the 2 2 bar H minus E Hartree form, which is basically the W excited amplitude, W excited configuration, sorry, on this side and W excited configuration on this side. That was your 2 delta. So delta is half of this, what is used here? And if you remember, we first obtained E correlation as C times k12, where C was the coefficient for the W excited return. So our, so it was 1 1 bar plus C 2 2 bar. Note again, what we are discussing hydrogen molecule in minimal basis. So this is what is called hydrogen molecule in minimal basis. Minimal basis means there are only two orbitals. That is exactly what you have learned, sigma G and sigma U. Remember when you do MO theory, you have 1 SA and 1 SB, two atomic orbitals, LCAO, it gives you sigma G and sigma U. So you have two atomic orbitals to start with, you do a Hartree Fock, you get two molecular orbitals. So this is what we are calling 1, this is what we are calling 2. So 1 is the bonding molecular orbital sigma G and 2 is the anti-bonding sigma U. So your Hartree Fock is this. So this is your Hartree Fock which is 1 1 bar and your DCI determinant will be 2 2 bar. So that is exactly, so it is a very simple problem that we are doing beyond Hartree Fock when you do DCI. So our wave function then is 1 1 bar plus a constant times 2 2 bar. So this is my Hartree Fock. So 2 2 bar is the other configuration which I am doing. So now we have to calculate C and put it back, E correlation is C times K12, put it back here. So this is the result that you get. So that is what we showed last time. I did a solution for C, remember the double CI equation. So what we want to show, what happens to the two hydrogen molecules now? So now this is for a single hydrogen molecule, two hydrogen molecules which are far apart. So we are discussing the case of a non-interacting dimer. So we are again doing a DCI, DCI in the minimal basis. So for each hydrogen molecule I have two molecular orbitals, 1 and 2 which I am calling as sigma G. So now what I will do is to rewrite this diagram saying that this is hydrogen 1A, hydrogen this is 2A. So this is for the H2 molecule which I am calling molecule A. Actually they are same hydrogen molecule but just to distinguish the coordinates will be different. So I am calling molecule A and then similarly for H2B I will have 1B and 2B. At a non-interacting regime their energies will be exactly identical. So they are degenerate at a non-interacting regime. Now I am doing a Hartree Fock. First at everything that I am doing is for the non-interacting regime. I am not discussing what happens when these two hydrogen molecules come apart. Of course then there will be interaction between these and they will be shift. They will no longer remain degenerate. So I will have two molecular orbitals. They will have different results. But at the non-interacting regime the Hartree Fock is very simple. Two electrons here, two electrons here. You agree? It is H4 system. So four electrons in 1A and 1B. So this will be the Hartree Fock for the dimer. So psi Hartree Fock will be 1A, 1A bar, 1B, 1B bar. Remember of course everything is a determinant. I am just writing the diagram. So again I have already explained that whenever I am writing without bar it means alpha spin orbital is attached. If I am writing with bar beta spin orbital is attached. 1A, 1B, 2A, 2B are spatial orbitals per se. But when I am using them inside the determinant they have an alpha and beta attached because determinants cannot be written in terms of spatial orbitals. It has to be always in terms of spin orbitals. So that is my psi Hartree Fock. Now I have to discuss what is the doubly excited configuration. How many doubly excited configurations are possible? So now start thinking. So these two electrons can of course be excited here. That is one or these two electrons can be excited here. These two electrons can be excited here. These two electrons can be excited here or one can go there and etc. But first let us understand that since they are non-interacting any interaction between a dimer, between a monomer A and B is actually going to become 0 because if you calculate the integral, the integral will become 0. So physically one can actually rigorously show but I have physically you understand that I cannot push an electron from here to there because they are infinitely apart. So the only possibilities are actually the interaction here and interaction here. So what is now my total wave function phi 0? 1a, 1a bar, 1b, 1b bar. Again I am writing in the intermediate normalization. So this is the Hartree Fock. I told you 4 electron Hartree Fock plus this gets excited here but this remains here because it is only doubly excited. So then this will become 1, 2a, 2a bar, 1b, 1b bar coefficient C. We can say C1 plus C2. Then I push this here, this remain here. So 1a, 1a bar, 2b, 2b bar period. These are the only two doubly excited configurations that are possible because anything else will include excitations from here and there. There is possibility of one here and one here but we are talking of a wave function which is totally singlet. So that is the reason we are eliminating that and we are doing DCI anyway. So singlet, singlet. So that is the reason we are not considering this. That will count as DCI but not DCI for the monomer. Not DCI for the monomer but DCI for the dimer. So if you are considering only the singlet function this is all that you will have. So this is a special case but it is alright. For all discussion this is okay because that is actually going to complicate the matter even more if I take a singles here, singles here. Yes, it does not matter but finally yeah but finally if you look at H4 it is doubly excited. I am looking at H4. And the reference to 1a, 1a bar, 1b, 1b bar. So we restrict up to here because the discussion is very clean here. Even before I actually do the calculation first I can recognize one important fact that the C1 must be equal to C2 by symmetry. Point that we are trying to say that size consistency means when the systems are very far apart the energy should be some. Whether what is the linear combination? Yeah that is okay. I am not exciting between A and B. Excitation here that is okay. That is not a problem. Overall system exists Hamiltonian exists. The point that I am trying to say if I solve a DCI for that Hamiltonian it should be twice the DCI energy of the monomer which is not the case. That is what I will show. So that is the reason DCI is not a size consistent method. Any size I yesterday I told any size consistent method must give this result that if you have A and B if I calculate the energy when A and B are far apart the result should be energy of A plus energy of B. That is physically intuitive and correct but I will show that the DCI does not give this whereas Hartree-Fock for this system will give actually that is very easy to see. So let me go ahead. See without even doing the calculation you can actually show that C1 is equal to C2 because of symmetry because both the hydrogen molecules are identical both of them are hydrogen. So there is no reason to distinguish between this determinant and this determinant. It is just either on this hydrogen or this hydrogen. There is no interaction between them anyway. So actually C1 would be equal to C2 and then I can set up the CI matrix just as we have done it. So first one is 0. Then I will have Hartree-Fock with one of the WXR8 configures in this row if you remember. So what will be that? So I have 1A, 1A bar, 2A, 2A bar, H then any one of the determinant you take. Let us say 2A, 2A bar, sorry 1B, 1B bar. Then I say 2A, 2A bar, 1B, 1B bar. So I am exciting this. So this will be one of the elements here. You remember the elements how to write Hartree-Fock with all WXR8 configuration in row and column. What is the matrix element for this? Remember this is again one determinant with another determinant where two of them are different. So only those two will come. So it will become 1A, 1A bar, anti-symmetrize, 2A, 2A bar. If you now remember we exactly encountered the same one for the monomer except that time we did not call A, just call 1, 1 bar, 2, 2 bar, anti-symmetrize. We did spin integration. One of these survived, one did not survive. The one that survived was 1, 1, 2, 2 in special orbitals which is just 1, 1, 2, 2. Either A or B does not matter now and that was called K1, 2. So that is exactly what is going to come here, 0, K1, 2 and this one is automatically K1, 2. I do not have to do it again. When I do the other one, 1A, 1A bar is identical. Again it is 2B, 2B bar. So I will get exactly the same integral just as the same integral as the monomer. So I can fill the column, first column K1, 2, K1, 2 and then you have 2 delta, 2 delta. That is also very easy to see because if I calculate these integral between the two of the determinants which are actually identical determinants like this, like I did here, 2, 2 bar H minus E Hartree-Fock. I am doing exactly the similar calculation here. So I will get exactly the same result as 2 delta and then you will have some numbers. This is Hartree-Fock. This is one of the doubles. So doubles with itself and doubles with the other doubles. So that means now I have to take this and this. Quite clearly this is 0. I hope you can see this because 2A, 2A bar, 1B, 1B bar, 1A, 1A bar, 2B to all four spin orbitals are different between these two determinants. So between one of the WXRA determinants to the another WXRA determinants, the matrix element is 0 because I have already told that if you have more than two occupation difference, Slater rule says it is 0. Is it clear to all? So this is the one of the WXRA row. So the first one is K1, 2 with Hartree-Fock. One with itself is 2 delta and then matrix element with the other one. So it is a 3 by 3 matrix. Now this is going to become 0. This is also going to become 0, same way. So this is my Hamiltonian matrix for this and then I will have 1 CC. Actually you can write C1, C2 and show that C1 is equal to C2 because you will get identical equation. In fact I have assumed it but please remember if I ask you to show you should be able to show that assume C1 and C2 are different, you will get exactly identical equation for C1 and C2. So they have to be equal but anyway I know this. So I am just kind of doing a shortcut. So E correlation, 1 CC. Remember now there are three determinants. So there will be two 1 CC. So these are the matrix and these are the coefficients 1, C1, C2 which are equal. So I have written 1 CC. If you do not assume equal anyway they will come equal, no problem. So if I now do the same multiplication, exactly the same way instead of diagonalizing by Cramer's rule let me write it down. So again 0 into 1, K1 into C, K1 into 2 into C. So I have got 2. So what is my E correlation? E correlation is 2 times K1 into C. Beautiful. If you remember here for the monomer, let me highlight for the monomer it was E correlation C times K1 into 2 and here the E correlation is 2 times K1 into C. This looks beautiful actually. So just from here you would argue that this is alright, this is size consistent except that we do not know the value of C yet. If the value of C is identical to here then of course it is size consistent. But this will be a different C because I am solving a different problem. I only assume that the C1 is equal to C2 but I cannot assume that the C is same as this C. So I have to actually calculate the C. So we will go to the next equation to calculate the C. So second equation, any one of the equation I will take K1 into 1 plus 2 delta into C, K1 into plus 2 delta into C equal to E correlation times C. So K1 into 1 plus 2 delta into C, 0 into C equal to E correlation times C and of course the other one will be exactly identical 2 delta C equal to E correlation C. So I can actually calculate from here again the value of C and put it back there. So your C would be K1 into 2 divided by E correlation minus 2 delta. So your equation here would become then E correlation equal to 2 times K1 into square divided by E correlation minus 2 delta just as I did here. It is exactly same except that it has been multiplied by 2. So you think everything is still fine except that now if you iterate and convert the equation, quadratic equation solutions will be different. Solutions will not give you 2 times this. See this E correlation looks 2 times K1 into square by E correlation minus 2 delta. Here also you have E correlation equal to K1 into square divided by E correlation minus 2 delta but unfortunately the result of E correlation will be different. Just solve this quadratic equation then you will see what you will get. So let me solve this quadratic equation and the solution of this quadratic equation will give you just like there E correlation equal to delta not here delta minus delta square plus 2 K1 into square to the power half. Unfortunately these 2 will actually come here inside the square root that means this is no longer equal to 2 times this. 2 has come inside the square root instead of multiplying by the whole thing. Although the iterative equation looks that it is still 2 times. The coefficient is identical, E correlation is exactly 2 times but however the result when I put it back because C is dependent on E correlation that is the problem here. So to say that this C is identical to that C is actually wrong. The form is only identical. The actual value is no longer identical because the E correlation is different. I do not know this. I know this only after I substitute here and then solve the quadratic equation I will get a different equation. So I will just kind of now rewrite all the equations in one place. So this is the monomer. So what we have got now is the for the dimer the result is the following that I get still first result is E correlation is equal to 2 times K1 to C and C equal to exactly the same result that we got here K1 to divided by E correlation minus 2. Note this equation and this equation with the equation here and I have not written the C here but C is exactly same here. So I can write the C here K1 to by E correlation minus 2. Note that everything looks identical at this point. Everything looks identical that it is just 2 times because C looks identical to this. However when I substitute this I get E correlation equal to now instead of K1 to square C I get 2 times K1 to square C that is the only difference divided by E correlation minus 2. Note that if this E correlation was not here something else was here which is the constant this should have been actually 2 times but because of the fact that the E correlation depends on E correlation I have to make an iterative solution and the result of the iterative solution gives me completely so if you solve the quadratic equation and this will give you E correlation equal to delta minus delta square plus 2 K1 to square to be power half. Please do this quadratic equation solution yourself. You will see that the result is not 2 times this. However everything looks 2 times this monomer result. If you look at the monomer and the dimmer everything looks 2 times still now except when I convert this I get a different result and that is essentially the problem of DCI that the 2 comes in but it does not come in as a multiplying factor here but it actually comes in inside the square root and hence the correlation energy is not 2 times the correlation energy of the monomer. Obviously you agree that that value is not 2 times this in fact it is close to square root 2 times if at all if I expand this properly because of the square root. So this is a very subtle point because in the beginning people did not realize because the forms looked exactly identical even the coefficient is same provided this correlation energy is identical. So you put it back here looks alright still it looks alright till you actually iterate and this is the real problem of all the linear expansion and the mathematics is very interesting actually what you get and till you actually converge you cannot put any bet with everything depends on this guy. When I converge I get this different which means C is actually different coefficient is actually different everything is different. So that is the message that the correlation energy is not size consistent. So essentially I can say that for DCI correlation energy is not size consistent I hope all of you have got this if you are not convinced you can do C1 and C2 you will see you will get identical equation for C. So the exactly same equation you will get for the C. So that you can anyway do it yes I am going to make that comment so true. So the problem has come because of linear expansion unless it is a full CI in fact the same problem I can now solve in full CI which means for monomer I take only doubles for dimer I take up to quadruples let us say only doubles and quadruples forget about the rest of the singles and triples which will actually generate more than the singlet functions then you will see identical results. So only for full CI energy will exactly in fact there is another exercise to show that they do a DQCI for this and show that now the energy is exactly twice the monomer energy the reason is that the full CI is exact. So very simply to understand all exact theories must be size consistent because that is what it is otherwise there is no point doing theory if my exact theory also is not size consistent then the theory is wrong I am not even I am not even theorizing correctly right my theory itself is wrong structure is wrong but here it is simply coming because it is not an exact theory for monomer it may be exact but for dimer it is not exact and I will show you what is missing now why it is not exact yes yeah but it does not matter it will make more complication that will essentially mean that there will be more terms here. So that will not help I mean I have I can I mean here I am not taking singles anyway and here you are going to take two singles amplitude that will even make more complication there will be another set of coefficients so there is no way gold energy will become too tight. So essentially the result that we are saying is that so the result that we have reached is that DCI is not size consistent that is important. So I have these two results in front of you this this and finally this or this so as you can see that this is not two times this presumably when the energy is not two times the monomer the wave function will also not be a product obviously so now I am going to physically show why the wave function is not product I have not discussed the wave function directly except that I wrote DCI wave function so again I go back to the same diagram that I did 1A 2A 1B 2B I write the monomer phi 0 monomer let me call it 1 this 1 is just to indicate that it is for the monomer 1 one mark so what is the C I 1A 1A bar plus C 2A 2A bar correct this is for the monomer A I write this for the monomer B what will be the monomer B separately 1B 1B bar plus C 2B 2B bar you agree I have written the C I functions DCI for each of the monomer separately I have to show that when I write the DCI function for the whole in the non-interacting regime it is not a product of this it will not be a product obviously so I am going to show this and it is very easy to see this actually and that is the part of the problem. So when I now write DCI for the monomer DCI for the dimer sorry DCI for the dimer let me call this 2 phi 0 2 means there are 2 N mark 2 mark so it is a dimer of the so 2 phi 0 so what is the result it is 1A 1A bar 1B 1B bar plus some coefficient C or K this may be different let us say C 1 1A 1A bar 2B 2B bar I have excited this plus some C 2 which is same as C 1 now 2A 2A bar 1B 1B bar so that is exactly the C I wrote I just wanted to distinguish between this C and this C that is all because they are actually not same I know this. So now what I have to show is that this wave function is not the product of this wave function and that is the reason it is not size consistent because energy is should be some wave function should be product so we will take a product of these 2 functions. So let me now do 1 phi 0A into 1 phi 0B now remember when I am taking the product it should be anti-symmetrize product which means the entire product should be anti-symmetrize so that is a simple thing you do not have to worry what is that A so all I have to take is this into this in an anti-symmetrize manner so what is the first term this into this what will that give me that will give me the 4 electron determinant 1A 1A bar right and 1B 1B bar so essentially I am taking the product of the 2 dimer heart reform so that will be generate me a 4 electron heart reform after anti-symmetrize then I take a product of this into this so I will get 1A 1A bar with a coefficient 1A 1A bar and then 2B 2B bar which is actually one of my WXRA determinant and then I will get another C of the other one 1B 1B bar 2A 2A bar which is the other WXRA determinant at least the determinants have come the coefficients are different but then very importantly there is a fourth term which I cannot avoid which is C square remember this 2A 2A bar 2B 2B bar so if you now compare this with the dimer you can quite clearly see why it is no longer can no longer be a product even if C1 is equal to C you will never generate this term this term comes out of a quadruple CI you have to actually it is a 4 electron excitations right you have to excite 2 electrons from here as well as from here and that cannot be done by DCI of dimer because dimer DCI can never generate quadruple excitations so that is a problem so how do I generate this to make a product I must do a quadruple CI here but then again quadruple CI is not good enough because my coefficient has to be square what is the CI? CI is a linear expansion alright just for this particular problem because quadruple's double CI is a full CI you will eventually get exact results if you do a DQCI here but if you take let us say 10 hydrogen molecules then DQCI is also approximate so at no point of time if you have a n number of H2 non-interacting molecules any approximate CI will be size inconsistent because of this fact that you can never generate the C square term and whatever level you truncate here and whatever level you truncate here cannot be same because product term always contains higher excitations so there are two problems one is that the degree of excitations is different second is that the coefficients come non-linearly so there are two essentially two different problems so we actually see that the wave function of the dimer is not an anti-symmetrized product for this hydrogen molecule so two important reason that we understand here is that so let me write specifically DCI so what is the two reasons one is that product of the monomer contains a quadruple excited configuration remember monomers are done at DCI don't forget the monomers are DCI but when I take their product there is a quadruple excited configuration that's the problem and then the whereas the dimer because I am doing DCI for both so dimer DCI does not have this configuration so basically you if you want to do a CI at the same level of excitations it is always going to have a problem because product will always contains higher level like doubles into doubles is quadruple right so product will always contain higher level so this will become a serious problem further note that the product contains quadruple excited configuration in a non-linear manner I will just make this statement in non-linear manner so you understand what is non-linear because this is c square so the whole spirit of CI anyway will not work if you want the dimer to have the product term of course when I say product term it is a product of the monomer I mean that is understood in the context so this product term contains this quadruple excited configurations and its amplitude is actually c square so if it is c square it cannot even be described by CI because CI describes only linear excitations configurations which are mixed in a linear manner is this clear yes any question if there is any question I mean I mean this is the difficult point but I think you have to understand so the question there later we will have to ask what kind of wave function will achieve this very interestingly perturbation theory achieves this I am not going to make a statement here but mp2 achieves this in a very different way with mp2 you do not first construct wave function you construct the energy and perturbation order and it somehow achieves this that is a beauty about the perturbation but if I think in terms of wave function just like I am thinking in CI that I write wave function as a combination of determinants and then I want to do this it is not working so what kind of strategy will actually lead me to this particular property of size consistency and that is where I will mention that the couple cluster when I come that the couple cluster actually achieves this in a very normal manner because couple cluster is not a linear expansion it is actually an exponential expansion so that is the real reason that it happens and I think I can emphasize this here if I have a linear expansion like 1 plus CA this is CA is very generic monomer A into 1 plus CB that is what is happening 1 is Hartree-Fock C is the WXRA determinant you will get 1 plus CA plus CB plus CA into CB so when I do dimer I am also doing only up to here because it is a linear expansion when I am doing monomer it is a linear expansion but its product has this quadratic term which I am not able to take in the linear expansion so if you think mathematically anything like 1 plus X or 1 plus C will not work what would of course work is the following if I had exponential CA instead of 1 plus CA and an exponential CB instead of 1 plus CB then you can see the product the product is nothing but exponential CA plus CB so if I take an exponential function for this instead of linear an exponential for this and exponential for this where my C is CA plus CB that is no problem but exponential of that then the product rule satisfy that is mathematics so exponential X into exponential Y provided the commute is exponential X plus Y they are numbers no problem but if you have 1 plus X into 1 plus Y that is not 1 plus X plus Y so let us say this is my function in the computer so this would be my function for the monomer dimer X plus Y exponences or if it is 1 plus X is a function this will become 1 plus X plus Y you can quite clearly see that this is not equal because you have additional term which is X into Y so this is actually mathematics so mathematically I know that if I can actually go instead of a linear expansion an exponential expansion exponential is product separable because e to the X e to the Y is e to the X plus Y and this is actually the genesis of the couple cluster method I will come back much later right now I am just making a point here why it does not work for the CI so you can at least appreciate why an exponential will work that is all I wanted to tell you why 1 plus 1 plus C will never work okay so if your dimer has the coefficient which is this plus this then that is the only function that will work but be that as it may right now we come back to CI and just make this point that there are two important points to discuss one is that the product of monomer contains a quadruple excited whereas dimer does not have note that this product term even if I had taken if I would have taken as a linear expansion it would still not have worked because the product term must come with C square so that is why the second noting is also important because the first if you just note this it is passing the buck for the time being okay because even if you had taken in a CI it would still not have worked so that is why I am saying point A is just passing the buck for the time being you are happy why it is but the that but there is a deeper problem all right however let me again mention if for H2 dimer I do DQCI just singlet it is exact and hence for a completely different reason the result will be size consistent okay that point this C square will then be shown to be equal to something because it is a full CI but if I do DQCI for a large number of N H2 molecules then you can see that DQ is also not good enough because if I do DQ for D for all D into D into D into D you will have much more than Q right and then they will come also the nonlinear term so it will never be exact I mean you can do for hydrogen if it is more than hydrogen then it is even worse let's I take two benzene molecule then also the same issue will come up right so either way it becomes very bad and in fact the problem become worse as a two increases to N so if I go from 2 to N when N is very large the problem will actually get worse as you can immediately imagine because there will be many many terms in the product which will be missing not just quadruple now there will be H2 Puli X Tritade, Octopuli X Tritade all will be missing dimer will only have DCI so everything will be missing unless you do again a full CI which is impossible for large N because full CI we can't do so the problem actually gets only worse