 continue with the basis sets today. So, first as we said in the last class yesterday that we have to choose first the basis that we want to choose for the atomic orbiters. So, one of the basis that is very common is valence only basis which essentially means for the first row atoms like carbon we have 1s, 2s and 2p, for the hydrogen it is just 1s. So, that is called the valence only basis and these are typically slater orbitals alright which will be expanded in terms of Gaussian. So, we will that is a general philosophy for all atomic orbitals. So, first we will only describe the atomic orbital set then we said valence double zeta please note that when you say valence only basis we are still including carbon 1s because core cannot be completely neglected what I mean by valence only is that polarization function would not be there we can simply call it valence basis. So, that is basically it means that up to valence it will be there. So, then you have a valence double zeta which means carbon would have just 1s and the actual valence 2s, 2p will be doubled as 2s prime, 2p prime. So, 2s 2p, 2s prime, 2p prime and for hydrogen this will be also 1s, 1s prime because hydrogen 1s itself is valence. So, I am first describing what are the atomic orbitals the slater type of functions slater functions what are the slater functions that are there. So, this will be the hydrogen 1s, 1s prime and up to this point we said we further said however we do not use any slater orbitals each slater function is expanded as a linear combination of Gaussian functions and we decide to call this primitive Gaussian. So, a primitive Gaussian which we call let us say Pgf primitive Gaussian function a primitive Gaussian for s type is typically some constant k into e to the minus alpha r minus r a mod square which is centered on atom A where a slater has only r mod r minus r a. So, then it means that is slater function. So, I have a slater function of s type which would be now on centered A which would be now written as a linear combination of various Gaussians. So, Pgf which are characterized by the exponent alpha which I just call alpha i. So, what is required is of course going to be s type that is for clear that is s type Gaussian because I am expanding s type atomic orbitals. So, what we require in our basis expansion is to know for each slater function let us say l number of Gaussians. So, what is the length of the Gaussian and what are these coefficients and the exponent as we said that each slater cannot be written as one primitive Gaussian. So, it has to be a linear combination of primitive Gaussian. The two important difference between slater and primitive Gaussians we highlighted that at r equal to 0 its derivative is 0 whereas in the case of the slater it is non-zero. So, that behavior is different at r equal to infinity these decays much more rapidly than this later because of the fact that you have a square. I hope all of you realize why at r equal to 0 it is 0 if you take a derivative of r square r will come out and r will make it 0 whereas in the case of just exponential minus r it will not become 0 it will remain non-zero. So, I think that is a very simple algebra that you should be able to do I just told these two differences and because of these two differences as well as the differences in the intermediate region s slater cannot be mapped to one primitive. So, it has to be mapped to a combination of primitive and this is usually preset. This combination is usually preset from an atomic calculation, from an earlier atomic calculation. So, this is something these coefficients are not something which are part of the SCF these are known using this we actually calculate the atomic orbitals, atomic orbital integrals for the valence only. So, before I go forward to further basis where I include more than valence let us spend some time and look at two particular basis sets which are of use double zeta valence and this is actually a name given by Dunning and Huzinaga Dunning. Another is 6, 3, 1g, 4, 3, 1g type 3, 2, 1g etc type which are which are the popl's basis. So, I will just try to explain these two examples of course there may be many, many basis sets just to tell you what this means. So, first the Huzinaga Dunning basis in the Huzinaga Dunning basis the core remains single zeta. So, that means core is not expanded core remains as one function of course core will be expanded in terms of primitive Gaussians, but core is not expanded in terms of 2. So, what will happen is that you have an expansion for 1s function. So, you have an expansion of 1s function in terms of pgs. So, we will worry about how many functions and all that expansion in terms of pgs. So, some number will be expanded and then you have 2s, 2s prime, 2p, 2p prime these are expanded again in terms of primitive Gaussians with some exponent, some exponent and some coefficients. Usually what is done is that these exponents are kept same for both s and p, but not for the prime and unprime there that is different, but whatever you use for 2s will be same for 2p, 2s prime will be same for 2p prime. So, one of the basis that was very commonly used by Huzinaga Dunning was called 9s 5p stroke 3s 2p for example. I will explain this why this is double zeta valence. The first in the bracket are number of primitive Gaussians which are used. So, total number of primitive Gaussians used are 9s type 5p type. From this 9s type and 5p type we use, we construct a contraction this is called the contraction. So, these are often these are also called contracted Gaussian functions cgf just like these are called pgf. So, now actually we do not work with later function we work with contracted Gaussian. So, we contract this 9s type to 3s type functions and 5p type to 2p type. What are those 3s type functions that is very clear you have 1s, 2s, 2s prime and these are 2p, 2p prime. Please note that these 3 and 2 are not the principal quantum numbers these are just the total number. So, these are 3 total number of s types this is total number of p type we judge that in this case it happens to be 2p, 2p prime. So, do not confuse and you have a total number of pgf which are 9 5. So, out of the 9s type few of them will be used for 1s and a few for 2s, few for 2s prime. So, typically what is used is for 1s we use 6 core a 6 primitive Gaussian out of the 9 for 2s we use 3, one of them become common and 2s prime is usually uncontracted. So, 1 so 3pgf this is 1pgf. So, it appears that 6 plus 3 plus 1, 10 you do not have 10 here but one of them is actually common Gaussian and each for each of these expansion you have coefficients those are actually least that is given to you it is not important but just to remember this prime remains uncontracted this is actually called uncontracted because I am using only one primitive. Now, you may argue with this but then this is only an additional basis to 2s, 2s prime. So, it does not matter one is sufficient because actual 2s is taking care. So, it has 3 number of 2s. So, similarly you have p functions which are usually make 4 and 2p prime they make 1 this was usual Huzinaga Dunning convention. So, this is just one typical basis. So, 5p is contracted 4 plus 1 5. So, this is a typical basis that I am just giving you but you can have n number of basis. In fact, there are further basis in which this can be 13s, this can be 7p contracted to 3s 2p only. So, that means for each of them I am using more number of primitives and always more the merrier please understand more here, more here everywhere more is better in basis set. So, I can keep on expanding my basis set question is that if you have more number somewhere rather computational time will increase. So, that is where you have to play the balancing game. So, this is typically the Huzinaga Dunning basis set. I will explain now the Poples basis set because this is more commonly used today. In fact, it is interesting that the Huzinaga Dunning basis set is almost outdated today. Most people use Poples basis set after the Gaussian has come. So, that is of course marketing because Poples started the Gaussian. So, Poples of course does not use Huzinaga Dunning basis set in Gaussian. So, eventually people have forgotten Huzinaga Dunning basis. That is unfortunate. I may also say that the Huzinaga gave the primitive Gaussian coefficients and Dunning gave the contractions. That is Dunning gave these coefficients, Huzinaga devised this alpha i's. What are the primitive Gaussian? So, that is their contribution, Huzinaga and Dunning. But it is almost forgotten today. Everybody uses the 631g, 431g etc. So, I will now come to explain these. These are actually much easier. So, let us say I start with 631g which is a very common basis set. So, 631g essentially means this dash is very important. This dash separates core from the balance. So, this 6 is the number of primitive Gaussians used to expand 1s. These three are the number of primitive Gaussian that is used to expand 2s and this is uncontracted for 2s. So, the 631 are actually not the number of atomic orbitals. They are the number of primitive Gaussians for each atomic orbital. So, when I write 631g, first you have to understand that is this. So, this is basically double sheet of balance of Huzinaga and Dunning. Coefficients are different, exponents are different. That is all the difference. So, number of orbitals, number of atomic orbitals used in this basis set and Huzinaga and Dunning basis set is identical. Number of atomic orbitals, that is number of contracted Gaussians which is used. Finally, these later function, they are identical. That number is 1s, 2s, 2s prime, 2p, 2p prime. So, 631g essentially says this and then for 2p, 2p prime, we use exactly the same contract, same exponents 3 and 1. So, just like the Huzinaga and Dunning basis, Huzinaga Dunning has also used similar those 5p, 4 and 1. That is a little different and their exponents are little bit different from the s. Here, whatever they are using for s type, the same exponents they use for p type. So, this is very typical of the Popol's contraction. The exponents are actually identical here, same exponent they share. Only the contraction coefficients will be different. So, this is the Popol's basis set. So, you can see that it is very pretty much the same. So, you have 1s later function which is expanded in terms of 6p, pgf and then your 2s and 2s prime are expanded in 3 and 1. So, it is pretty much what the Huzinaga and Dunning did, values are little bit different and the same goes for 2p and 2p prime. What is interesting in Popol's basis is that they share the common exponents. That is a very special feature which is not there in Huzinaga Dunning and in fact, 3 and 1. Huzinaga Dunning, they are not even used 3 and 1, they used 4 and 1, but that is in a minor difference, 3 and 4 are minor differences. So, this is basically 631g. If you now take another basis set, so for example, 431g, now you can yourself tell what is 431g. What is 431g now? Same number except the 1s is 4, now no difference. Yeah, first row, carbon, oxygen, nitrogen, yeah, first row atoms. So, it is exactly the same, yeah, 4. So, if I now use 321g, it is exactly the same. So, you will then have 3, this will be 2, this will be 2. That is it, but as long as I am using of before dash 1 entry, after dash 2 entries, it means it is this basis. That is important. Of course, if I do for hydrogen, I should also tell for the hydrogen, then this is immaterial. You just look at 1s, 1s prime, both are valid. So, you use accordingly 631g, 431g. How many contractions are there for hydrogen 1s? This is no longer important. That is all. So, if you look at these basis sets, they are pretty much same Dunning and Huzinaga and Popol, but yeah, exponents are different, total number may be different and you can keep changing. Of course, these numbers can be made as big as possible. 631g is typically good, but people use more number, as I said, more the merrier. So, the point that we are trying to say that these 6 or 321 has nothing to do with the number of contracted functions. The number of contracted functions are decided by number of entries. So, these are only the number of contractions used for contracted function and this is preset from an atomic calculation. So, they do not really come in the heart reform. So, initial calculation, of course, becomes expensive. The calculation of the 1 and 2 electron integrals become more expensive and I will come to that because you have to expand that in terms of primitive Gaussian, eventually to calculate as I discussed in the last class, only once you have to do, but once you have done that, the rest of the heart reform calculation does not depend on these numbers per say 321, but what it depends is the number of entries because they define my number of atomic orbitals, contracted atomic orbitals which become my basis set M. That capital M that I was talking will be decided by this number. So, let us look at that number now. Either Huzinaga Dunning double zeta valence or the Popol's basis, the number is same quite clearly because each of them uses 1s, 2s, 2s prime, 2p, 2p prime. So, number is actually same. So, your total number M in SCF calculation remains the same except the quality of the integrals will be different because the quality of integral will depend on expansion in terms of Gaussians. So, can you calculate the number? Let us say I am doing a calculation of methane. What will be the total number of atomic orbitals, contracted atomic orbitals which will generate total number of molecular orbitals? Let us say there is no linear dependence. So, your total number of MOs that will be generated, your Fock matrix, M by M, Fock matrix. What is that M? Can you tell me now? How do you calculate? Remember, P is 3-fold degenerate. So, whatever we are doing for P, I should also tell you we do exactly same for Px, Py, Pz. So, now you tell me what is the number? 9. 9. Total methane. I have carbon, I have hydrogen. No, what is the total number of contracted Gaussians? What is the total number for this basis here? This basis is same as this basis or this is double zeta valence or who is in Nagar Dhani because number of contracted functions are identical. 1s, 2s, 2s prime, 2p, 2p prime. Hydrogen, 1s, 1s prime. So, who said 17? You are right. You have 1, 2, 3, 6, 9. 10, 11, 12, 13, 14, 15, 16, 17. What is so difficult about it? You have 4 hydrogen, each of them has 2 orbitals. So, 8, 8 already. For carbon, you have 3 plus 6, 9. So, if you do a calculation of methane, and this is very important, in double zeta valence of who is in Nagar Dhani, you will get total 17 atomic orbitals. Your SCF will be 17 by 17. Your number of molecular orbitals will be 17, out of which, of course, these are special orbitals. Remember, so you have actually 34 spin orbitals, out of which actually five of them will be used for occupied orbitals, and the rest of the 12 will remain unoccupied. So, just want to explain to you because we did that Focke matrix, Fc equal to Sce etc. So, you get that many eigenvalues, eigenvectors. So, you get so many coefficients, sets of coefficients. The set of coefficients that you will get will be equivalent to 17 by 17 matrix. So, C will be 17 by 17 matrix. You will get 17 MOs, out of which only five will be used because this is a closed shell system. Five will be used for Hartree form. The rest of the 12 will be unoccupied. So, the lowest energy among them unoccupied will be called Lumo. The highest of the occupied will be called Homo. So, if you do double zeta valence calculation with methane, and if you do not get 17 orbitals, there is some problem with input. You should be able to find out something that you should go back and check what happened. Maybe there is a linear dependence, something was removed as I told you. That could happen and reducing the number, but it cannot increase anyway. Now, if you do 321 g calculation for methane, what is the total number of molecular orbitals that you get? I am just saying what is the total number of MOs you will get, so that there is no confusion. That must be that m. If you do 321 g, good, very good. I just wanted to test it. If you do 431 g, it is same. 631 g, it is same. Many people get confused because these numbers are increasing 3, 4, 6, but it is same because you are still describing only this many number of atomic orbitals. What is happening is that the quality of the atomic orbitals in terms of the primitive Gaussian is changing. That is the only thing that is changing. So, when I go from 431 g to 631 g, the valence anywhere remains same. Only the 1s function is getting better because for the 1s function, now I have used 6 instead of 4. So, obviously 1s function will be better than the 431 g 1s function. That is all that is happening. So, I have a better quality 1s function, but the number is same. Quality is determined by number of primitive Gaussian because I am anyway not using slater. So, everything is an approximation. So, how good is this slater or what I call contracted Gaussian? That depends on the number L and also details of Di and alpha i, assuming that they are almost similar. This will be certainly better than this and because the core is better, you should also expect Hartree-Fock results to be different. Although your basis is same, do not expect the Hartree-Fock results to be same because the integrals will be different. The integrals involving the core orbital will be different because the core is different and hence the results will be different and actually results will be better here. By the variation theorem, I am always more the better. So, it will be lower, lower energy will be lower and you can do that calculation and see. Although it is a very minor difference, only the 1s core carbon is changing, nothing else is changing. So, the difference will be very, very minor in very few last few digits will change. 3 to 1g will be significantly worse because your valence, not only core is 3, your valence is also got worse. One of the valence instead of 3, you are using 2 orbiters. So, it will get worse but total dimension of the problem will be same. So, there are two different things. So, total number of MOs that I will get will be that will be determined by the dimension of the Fock matrix. Core function usually requires, one of the reason is that we are only using one core orbiter. So, that is why we try to do that as good as possible. So, that is why we usually require a larger number of cores. We do not have a double zeta. In fact, I did not tell you, Huzinaga and Danin when they first started, they actually said only double zeta, just d z. So, just d z if you do, that is not d z v. That means core is also expanded. 1s, 1s, 1s. If you do that for the methane calculation, you now have 18 orbiters. If somebody has a old program run d z v and d z, you will see suddenly one more MO has occurred. So, just because that carbon is also expanded, that was original Huzinaga, Danin, those who read earlier papers. In fact, when we used to do calculation, it was double zeta. Double zeta valence I learned much later and that is actually to be consistent with the popol. Popol was always using double zeta valence because for popol to use double zeta, he has to have two more entries here, two entries here because this is a separator between core and valence. So, this is the number of valence. This is the number of cores. So, you have to have different two entries. So, let us say the 6 is split into 5 1, then 5 1 dash 3 1 g. Popol never invented this basis set. If you have such a basis set, it would mean core is also double zeta, but popol never invented this basis set. So, popol never used double zeta actually. Popol always used double zeta valence and then somewhere down the line, Huzinaga Danin double zeta was also forgotten. People used Huzinaga Danin double zeta valence because popol showed actually, you will have one extra atomic orbital and it really does not help the results and there are a lot of studies which are made to show that for the core 1 is sufficient and then you need more number of contractions. You do it only once. So, that is why all these are punched to just 6 instead of splitting them as 5 and 1. So, there are lots of literature and basis set and they are all trial and error. Many of them have come out of intuition. So, many say that the basis set is more of an art than science. It is probably little bit of dramatic statement, but it is probably true that it is more of an art than science because you just do trial and error. You have some intuition and there is lots of literature. So, I will not go into those details, but let me now say how do you calculate the integrals before I go over to the other basis sets, some of the other basis sets which are also there. So, how do you calculate the integrals? So, for example, I take a simple integrals of mu H nu which is a one electron integral or mu nu. So, each of these are now contracted gaussian. So, they will be expanded depending on 1s, 2s, 2s prime, 2p, 2p prime whatever, it will be expanded. So, let us say mu is expanded in terms of i equal to 1 to k, some d mu i, then I have a gaussian function, I call it just mu i H and then mu is expanded similarly as sum over j equal to let us say 1 to l dj nu and gaussian function nu j, is it clear? This is primitive gaussian pgf. So, just as I used to do normal basis set expansion, I am expanded mu in terms of di mu pgf mu i. So, this just shows that it is a mu h basis, i is the number of the pgf k number, nu has l number, everything can be different. So, you have to keep track dj nu pgf nu g and of course, so the basic integration that you have to do is this. This is the basic integration that you have to do with the primitive gaussians and fortunately these integrals can even be calculated in an analytic manner, there are lot of not numerical. So, analytic integrations are possible with the primitive gaussian function and then all you do is to multiply with these coefficients which are now known. Remember, these are already preset from an atomic calculation. So, they do not come in the hard reform. I had given you a step, calculate 1 and 2 electron integrals. So, it will only come there because these numbers are already known. So, you just calculate this. Similarly, I do mu nu lambda sigma 1 by r 1 to lambda sigma. So, you will have now four summations i j k l. So, this can be i equal to 1 to k 1, j equal to 1 to k 2, l equal to a k equal to 1 to k 3, l equal to 1 to k 4. So, all different expansions for I keep writing like this d i mu d j nu d k lambda d k d l sigma and then you have this expansion in terms of i j p g f i mu p g f j nu 1 by r 1 to p g f k lambda p g f l sigma. So, it is a routine thing. I will just keep track of your symbol. That is all. I mean, it is very simple. So, all you need to do is to evaluate this integral. Again, I am writing everything in Dirac notation. You can convert it into a Boolean notation. That is not important. So, here the real simplification is here as I explained to you because now there are gaussians. So, they can be actually written in terms of only two gaussians because this p g f i mu and p g f k lambda can be combined 1 into 1. They can be combined into one gaussian on a different center and here again this and this can be combined into another center. Of course, they have centers also now. So, you have to keep track which mu. So, mu has the center a, b, c, but you can do that calculation. It is not very difficult. So, the real advantage is in the calculation of this integral and because of that the gaussians are actually done. So, this is a complicated thing that is actually done. It is not so simple to calculate a two electron integral, but these are evaluated after this simplification of two gaussians being converted to one gaussian. They can be now evaluated completely analytically. So, there are function formulas. So, these formulas are directly programmed which is very fast. So, you do all analytic integrations. So, people have done this integration is very long thing, but for computer programmer you do not care. You take the last line equal to final line. You just program that. That is the formula programming. So, that is very quick. So, please this is a lot of literature that is done. In fact, there are papers by I think SACs, somebody called SACs in general of mathematical physics. Lots of paper in the 1960s I think when the gaussians were coming up. JMP general of mathematical physics. So, those who are mathematically oriented can read those papers. Here is many papers. How to evaluate these integrals? Now, long papers. So, each of them may be 4, 5, S type, P type, D type all different depending on that because obviously if it is P type, remember primitive gaussian function will be different. It will not be just exponential minus alpha because there will be r to the power something. P type has also x, y, z. So, r to the power 1 depending on what is the quantum number. So, they are actually done for everything and very, very long papers. A lot of mathematics you know I tried to read through once. I found that probably is not useful to read through because I mean they have done it. So, everybody trust their mathematics is right. What is important is the final formula for the programmer and they just take the final formula, but those of you are actually interested in integral evaluation in terms of gaussian can go back. All the literature is available and read those papers. We will go forward. This is not the only basis set that you can use. As I said, we have so once you understand the integral evaluation in terms of primitive gaussians.