 A very good example is, of course, hydrogen. Hydrogen is always a very good example. So, this is basically a calculation of hydrogen molecule H2 molecule at 1.4 atomic unit. So, this is of course, the experimental bond distance as you know. So, these are numbers, I am actually quoting the numbers. So, that is the reason I have brought this book. Zip was able to remember numbers. So, basis set is good to know. STO 3G and let me analyze this STO 3G. We have already done STO 3G, remember in the heart before. So, what is the basis set? It is just one valence orbitals and each of them expanded in terms of three gaussians. So, STO 3G. We had 431G. I hope you remember what is 431G. Please remember, exams are coming. So, remember this basis set. So, you have core orbital which is expanded in terms of four gaussians. You have a split valence. The first one expanded in terms of three gaussians. The second in terms of one gaussian. In number of contracted gaussians, however, it will depend not on these numbers, but how many entries are there. I can remind you. 431G. Then you have. So, 431G and 631G would actually give same number of contracted basis, but except that the base core is better described here. So, slight difference, but I am going to give you 631G star. Actually star star is not important because it is hydrogen and then a very large basis, 10S, 5P, 1D and then exact. Now, what is exact? For hydrogen molecule, almost near exact results have been obtained by somebody called Colos with only weights. Many people might know or know their name. So, they are very important result. So, Colos is a Polish. Both of them are Polish, Colos and only weights. It is a very famous paper which appeared in JCP, Jungle of Chemical Physics, 49. So, those who are interested can read 1968. It is a very famous paper by Colos and only weights where hydrogen molecule is almost beaten to death. So, you cannot do much better than. So, they had a very, very large calculation, etcetera, etcetera, almost near exact. So, it is beaten to death. So, you do not have much of a chance in hydrogen molecule to work. So, this is, I am just quoting this is exact result. This is much larger than this basis. This is exact results, very large. So, now, I have, I am quoting results from CI calculations in each basis. So, let us see DCI or CI doubles and SDCI. So, these two numbers, again remember what I am quoting is only correlation energy, not the Hartree-Fock. So, if you calculate total energy, then you have to add Hartree-Fock for each basis. So, remember there is a little difference. So, the difference in Hartree-Fock will not be exact difference in total energy will not be exactly same as the difference in Hartree-Fock correlation energy, because Hartree-Fock is different in each basis. So, please remember this. So, what we are presenting is only correlation energy. So, let me try to write carefully. So, DCI minus 0.02056. Again, these are all in atomic units. So, I hope all of you know what is an atomic unit. What is an atomic unit? Hartree. So, what is it? How many electron volts? Hey, come on. 27.2116. I hope I have not forgotten. Half a Hartree hydrogen atom, 13.65 electron volt. It is actually 23 point something electron. 23 point exactly decimals I have forgotten, but one electron volt is 23 point something kilo calorie. So, one atomic unit is 600 and something. Yeah, 627. That is how it is come. 27 point something into 23 point something. And then one kilo calorie is 4 point something kilo joules. 4.16 etc. kilo joules per. So, you can imagine what is an atomic unit. One atomic unit is about 2400, 2500 kilo joules. So, it is very dangerous to make a mistake in atomic unit because eventually your chemistry is taking place in kilo joules. And that is the first thing I said why correlation energy is important. So, you may say what is this number you know? 1, 0, 2 atomic unit is a very large number in terms of kilo joules. You are already talking of 50 kilo joules. If you have 2500 in order. So, 0.01 is 25. You are talking of 50 kilo joules error. No chemistry can be done. So, I am first trying to convince you why was it important and that is also important thing to realize because we do not understand the difference in atomic unit. Hydrogen ionization and I repeat that hydrogen ionization is almost 1400 something kilo joules. 13.65 into 23, half heart rate. So, about 300 then 4.12. So, 2500 whatever is divided by 2. So, about 1200 are kilo joules. It is very high. So, if I want to I told you before if I want to deprotonate the system in gas phase it is very difficult because not only have to let us say carbon hydrogen I want to deprotonate. It is not breaking C radical to H radical. C S 3 minus H plus. C S 3 minus electron affinities are always very very small. So, I break the bond and then I want to ionize hydrogen and that itself is kills me 1200 odd kilo joules. It has already killed me. Then ionization is very important. So, H plus getting H plus is very very difficult in gas phase of course. Even solution it is a completely different because they will get solvated. So, getting H plus gives me some energy. So, again if you do a gas phase calculation do a gas phase calculation those who have Gaussian etcetera of a molecule and do a calculation in terms of ionic solvent. You will immediately see the deprotonation energy how it drops down. It is very important good lesson to understand because what you are talking is in gas phase of course. There is nothing else to solve it deproton. So, let me go back to my book for once. Normally I do not like to I had to see books while teaching but sometimes I will do. So, minus 0.02056. So, you can see up to this decimal in STO 3G there is no difference up to this decimal of course there is a difference beyond maybe 7th or 8th digit. What does it show? The DCI and S DCI really does not distinguish the energy in STO 3G basis. So, that is why I am showing the basis effects. So, now let us see 431G. So, at 431G this is minus 0.02487 minus 0.02494. So, you can see the contribution of singles is still very very low at the 5th digit 7 everything else is 0. You go to 631G star minus 0.033 sorry 03373 minus 0.3387. So, it is reasonably large now and of course with 10S 5P1D it is minus 0.03954 minus 0.03969. It is almost similar and the exact actually interestingly is still minus 0409. So, you can still see that these are still higher than the exact of course exact is a far higher basis. Now, remember for hydrogen molecule this is exact. S DCI is exact in that basis because it is hydrogen molecule you cannot have triply excited determinant. You have only 1 and 2 electrons. So, that is exact. So, this exact is only because of the basis set. I told you there are two kinds of problem one is truncation in the ACI another is basis set. So, the basis set itself is not large enough. So, that is why this number and this number are not same. But otherwise these numbers are exact numbers within this basis set. For the given basis set these are exact numbers where S DCI is exact for hydrogen molecule. Is it clear? So, this also shows that as the basis set increases the effects of the singles on the correlation energy actually increases. Of course, the Hartree Fog also is changing. So, in that sense total energy change will be even more. The Hartree Fog itself is going down further. So, if you look at the change in the total energy this will be more than just the change of the correlation energy because Hartree Fog is also changing. Here we are only discussing correlation energy. So, for each of these numbers Hartree Fog in that basis is subtracted. And what is subtracted from one basis to another basis is a different number with Hartree Fog in that basis. So, Hartree Fog itself changes. So, you are not subtracting the same number. So, there is a little bit of a difference. So, if you calculate total energy the difference will be somewhat more because of the fact that basis to basis it is important. But I think it is important to note this first number that if you use a bad basis set really S DCI does not improve. So, the improvement starts to come when you have larger basis set. Of course, this will become much more pronounced. The effects of singles will become much more pronounced when you go to larger molecules. It is very clear that here the effect is very less because molecule itself is very small. So, DCI and S DCI it is really not going to make too much difference. If you go to water methane we have results for that also and numbers will be very large and the numbers will be significantly different. In fact, for all those who are following the text book of Zabu and Aslund should see you know in the chapter four or five whatever chapter four I think they have a 4.3 some illustrative calculations. So, please see that illustrative calculations you will learn from the table. Just read and analyze the table. They have made an analysis. I do not want to go into the details of the tables, but I will just make some general remarks. Yes, you wanted to ask a question. Yes, of course, correlation effect itself is very very small. So, you recover almost the full correlation in DCI because if the basis is larger then of course, you have a larger summation over R S. So, the correlation energy C I is variation. C I is variation in terms of the determinants in terms of basis also. So, it is simply because you have more number of determinants. So, when I am doing in a larger basis said I have more number of determinants. So, everything is larger. So, that is the reason I expect. I mean if this does not happen I would have been surprised. So, I expect that results should actually further go down. So, the correlation energy should improve further. Note again that in the correlation energy I have a negative sign everywhere simply because of my definition of correlation energy E naught minus E Hartree fog. If you define as E Hartree fog minus E naught then they will become positive, but you know that is a very minor part. All your equation there will be sign change. So, that is the very minor point, but we are sticking to the definition. So, all my correlation energy should have a negative sign. Otherwise, you have not done, you cannot have a positive sign. So, I think you can look at many more results which are actually given in the illustrative calculations. I just wanted to highlight one or two points that if you now look at the potential energy curve you know if I remember the Hartree fog I have a discussion of the potential energy curve that the RHF fails to describe the potential energy curve correctly. I hope you remember the discussion that the hydrogen molecule you cannot dissociate. Do you remember this? That if I do the hydrogen molecule then this is hydrogen plus hydrogen and the gas phase once again in the gas phase. So, this is what it should happen. At r tends to infinity this is the potential energy E or V whatever you call it and this is r sorry. So, at r tends to infinity it should actually go to H plus H. However, if you remember the RHF curve it was actually lower than this and then eventually goes here which I said it is closer to H plus H minus because of the spin pairing. So, if you now do CISD in particular it is exact. So, what will happen is that this number well will go down and eventually this will also go correctly. So, this is Sdci qualitatively for hydrogen molecule because it is exact. However, Sdci will be always larger than the lower than the Hartree fog RHF. This can be some other number. Let us say this is dci. It does not matter. For hydrogen molecule of course as I told here dci and Sdci really does not change anything. So, Sdci is more or less exact number. So, if I call this exact then I need not write this dotted line. I will just call this Sdci. So, this is Sdci because I do not want to write dci and this is RHF. So, dci is almost same as I said for Sdci. Most of the base is set. Now, of course this base is set to base is set this results will change. But important thing is qualitatively this is approaching h plus h because it is an exact method. Note also that when I do this calculations you have to be little bit careful. If you calculate hydrogen so the question is what is this number? This number is E hydrogen molecule minus 2 times E hydrogen atom. The E hydrogen molecule I am calculating in a given basis. Let us say STO 3G, 431G, 631G star etc. Then the question is what is E hydrogen atom? That also must be computed in the same basis. You cannot take exact hydrogen atom energy 13.65 whatever you mean because this has a basis set effect. That number does not have. So, I must recompute the hydrogen atom in the same basis. Then only you will see that this difference is going to 0 at r tends to infinity. So, somebody is interested should be able to do this calculation. So, let us say hydrogen molecule in 631G whatever star basis or star star basis and this is same identical. Then you do a hydrogen atom calculation also at 631G star. This is possible remember although it is atom you can still do a basis set calculation for atom. It is not necessary but I am going to force the calculation in Gaussian. You can do that and take for hydrogen molecule r equal to let us say 10000 angstrom. I mean which is nearly infinity. Please remember that it has to be very large distance. Then you calculate this quantity. So, this should go to almost 0 at that distance because I have been able to break dissociate the bond. 10000 angstrom is almost like infinite distance. So, I should be able to get 0. Please repeat this calculation for CISD and do that for DCI. You will have a small difference because singles in a large basis do not do it at STO 3G. Do it at a larger basis DCI and SDCI. You will see that DCI will not be able to break it correctly. It will not go exactly same but it is almost same. What is more interesting is and that is where I will discuss DCI only now is if you do a larger system and try to dissociate. Instead of hydrogen dissociation, let me dissociate hydrogen dimer and this is of this EH4. In that case, EH4 at this r tending to infinity would be equal to 2 times e hydrogen. Just as I said that the e hydrogen molecule should be equal to 2 times e hydrogen atom in exact case. Is it clear? It depends on where I could have of course broken here. I could have made H3 plus H. Then it is a different story but let us say I am doing H2 plus H2. So, a linear H4 molecule I just dissociate. I should get 2H2. It does not matter. If it is not linear also, I will get the same thing at r equal to infinity but I am just giving a linear as a simple model. Now, if I do RHF, note that H4 is a closed shell system. So, I can do RHF. What will be the result at this distance for RHF? Will it be able to dissociate or not? Answer should yes because I have already discussed if it is closed shell going to close plus close no problem. So, RHF will give the correct dissociation. So, I call it correct dissociation. If you now do DCI, of course in this case it is a 4 electron system. So, even SDCI is not exact. Remember you have to do SDTQCI up to quadruples. So, let us forget SDCI. I am just doing DCI. What would you expect now? Hartree-Fock has already dissociated correctly. I am now trying to improve it. I am trying to improve EH4 as well as EH2 and then I am trying to remember I am calculating the difference. It is clear that each of them result will go down first because I am doing compared to Hartree-Fock. I am doing CI. So, EH4 will also go down. EH2 will also go down. The question that I am asking what happens to EH4 minus 2 EH2 for this dissociation. Unfortunately, this is not 0 when I do DCI. Whereas, Hartree-Fock that is 0 and DCI it is not 0 and this is a very, very disturbing fact. Note and I will come to that why. Note that RHF which is a lower level of theory could at least predict the dissociation correctly. Of course, RHF could not do for H2 to H plus H, whereas DCI could do, but that is only because the system is so small. DCI is almost exact DCI or SDCI. But when I go to larger system DCI does not give. There is a small difference and this is going to build up as the molecule become larger and larger. So, if I take very large molecules and dissociate into close plus close DCI is not going to give. So, what I had in RHF is apparently lost in trying to improve the theory and this is a serious issue because I have been always talking of beyond Hartree-Fock. So, beyond Hartree-Fock one of the important point is to improve the results. Am I only going to improve the total energies? No, the answer is no because that is not good enough. I have already told you chemistry is about difference energies. So, I must get the difference energies correctly and I am not getting that. So, if I want to dissociate H4 I would rather trust RHF than DCI. This has been a very, very disturbing fact that will lead me to the back to the discussion of perturbation and eventually to the couple cluster. If I do MP2 for the same system, EH4 that is equal to 0. That is interesting. If you do MP2 it dissociates. DCI does not. I told you in the beginning that several approximation of DCI can give MP2 but if you do a full DCI it is different and it actually bad. So, this is bad. This is good. This is of course good also good. So, this leads to a very important discussion of what is called theoretical model chemistry. What is a good theory? Is it good enough to lower the energy? Answer is no. Obviously the answer is no because eventually I am going to describe chemistry. If my chemistry comes out better at RHF and MP2 and not at DCI I have to worry. So, this is a very major worry for the CI community. The community of people who believe in CI. Of course later one of the persons who believed in CI is very famous Arne Davidson whose algorithm I told you Davidson's algorithm for CI. So, Davidson himself came up with the corrections to DCI. So, what he does? He does a DCI and then adds some correction ad hoc correction which is no longer CI. So, it is no longer variational and then he shows this is going to correctly dissociate. Now that is of course a different matter. Many people use what is called Davidson's corrected CI. Then it is okay. It is reasonably okay. But note that Davidson's corrected CI is no longer CI. It is not fully variational. So, what Davidson does is to change an ad hoc corrections and makes it dissociate because otherwise Davidson is a CI person. So, he was very, very disturbed. This was actually kind of found out more in the 70s as that is that then Davidson was very active that time. So, obviously he came up with the Davidson's correction to the CI to make it dissociate correct even for such cases. All cases. What we will discuss next time is why doesn't it dissociate correct? This is very important to understand from the looking at the DCI wave function. Why doesn't it dissociate correctly and how does MP2 do it well? And of course later on I am going to destroy a couple cluster which also will do it well for all such cases. So, important thing to realize that a good RHF does not guarantee the correct dissociation when I correlate. So, this is a very disturbing point and many people don't understand this. This property of correct dissociation is known as size consistency. So, if a theory is size consistent it means it should be able to dissociate correctly. So, DCI is not a size consistent theory. Of course, full CI is always going to be size consistent because full CI is exact. So, hydrogen molecule you will not realize but H4 you will realize even though RHF dissociated the so the DCI is not DCI does not dissociate. So, one of the important properties for a model chemistry is not only to include correlation but to include correlation in a way that it should be size consistent. I am going to come down and slowly write at the end of 2 or 3 more classes I will write down what are the model chemistry and this was first pointed out by John Popple and Rod Bartlett in 1976 international journal of quantum chemistry symposium issue in a back to back papers if I remember. So, those who are again interested should read you know it is very important to read articles and know the history. 1976 as let us that they actually coined this term called model theoretical chemistry. So, people thought just improve the energy but that is not good enough. There are many more things which are important and I will come down that later what we will discuss in the next class is why is DCI not size consistent what is a physical effects and then how to include those effects because then only we will be able to improve the theory and how does MP2 do that. And that is why perturbation theory is good in terms of size consistency perturbation theory other problem because it is not variation. In fact it is very interesting when you do couple cluster it is also not variation. So, as soon as you have variational you have a problem with size consistency. In fact I have thought a lot about philosophically that variation is good but is it really good you understand my question it is good because it is upper bound but is it really good. In fact I am beginning to find beginning to discover this as you grow old this is wisdom which I didn't know when I was a student. So, there is something called wisdom then you ask lots and lot more questions is variational good now after doing everything I am questioning variation method is it always good because when I do perturbation it is not variation it is good when I do couple cluster it is not variation it is good when I do CI it is not good or you re-ask the question when is a variational method good. That may be a further good question to ask and we will ask that question and that is something that I asked in my PhD thesis can I do couple cluster which is good in a way that it is variational and yet it remains good you understand the point. So, that was what we call variational couple cluster that was the first time and in fact my entire fame in the beginning of my career was because I was the pioneer of variational couple cluster this term does not exist this couple cluster was known to be non-variational I will show you later. So, I was the first person to think and I thought we are thinking out of the box it is good but it is not variational can I marry both you understand can I marry variation with couple cluster. So, I think we will come back here.