 So let me again summarize what we said in the last class before we move to the next topic that is configuration interaction is that we saw in the perturbation theory that up to second order we find the importance of, so from the MP2 description when we say MP2 for energy remember whenever we are talking of MP2 it is for energy but the wave function actually starts at MP1. I hope that is clear to everybody because the first order correction is at this first wave function corrects at the first order and hence the energy corrects at the second order because we have this formula that E0 2 is Psi 0 0 V Psi 0 1. So these two when we analyze we see that the contribution comes from only doubly excited determinants. So no singles due to Brillouin's theorem and of course no triples I want to summarize no triples and etc. etc quadruples and so on because of Slater rules. So Slater rules does not allow the V to connect from with the Hartree-Fock to triply excited because we have only one matrix element Hartree-Fock V excited determinants and back to Hartree-Fock. So only doubly excited determinants contribute so I think we just want to emphasize the importance of doubly excited determinants in the lowest order correction. So what we mean by lowest order correction is of course that this is the most important correction and this is why doubly excited determinants play an important role. Later this was expanded or around the same time by Synanoglu. I think the review I had already mentioned but very important highlights was that Synanoglu later described that the correlation takes place mostly in pairs. So what we call pair correlation theory and here there was a simple physical insight that I explained last time is that essentially for the exact wave function no two electron coordinates can be identical because if they are same the wave function must go to 0 because the Hamiltonian becomes singular so the only possible solution is that the psi must go to 0 which means there is a vanishing probability of any Ri becoming equal to Rg. Vanishing probability of course everything is in terms of probability. So Synanoglu argued that that essentially means that whenever two particles come together they would like to repel each other. They would like to go away and the only way they go away is to get excited to the virtual orbitals so that they have a larger space of a larger degree of freedom. Hartree-Fock does not give you the degree of freedom because your orbitals are only n. So if you have large m orbitals then they can go away. So this is why of course the virtual orbitals are important but what was important is to notice that only two particles can come close together. This was Synanoglu's argument. Why? Because if you look at Hartree-Fock so only two particles or two electrons have a large probability or two electrons have a large probability of coming together in Hartree-Fock of course in Hartree-Fock. So this was a very hand waving argument. So what is the reason? Because Hartree-Fock is a single determinant. So it already has the anti-symmetry built-in. It has the Pauli-Extrusion principle built-in which means no three electrons can come in the same special orbitals. So a special orbital can have at the most two electrons and because two electrons are in special orbital there is a probability that they can come together. So those two electrons must have to be correlated by exciting them to the virtual orbitals. So that is the genesis of the pair correlation theory that more than two electrons is difficult because of the Fermi principle and single electron excitation is simply not possible because Hartree-Fock has the billows condition. So then of course Synanoglu argued that there can be four particle excitations but that would be because of two pairs getting correlated simultaneously. So two pairs correlated which means in a Hartree-Fock two particles or two electrons get independently correlated giving rise to a four particle correlation. So you can have now three pairs which will give rise to six particle determinant, six-fold excited determinant. So you can break down the probability of those determinants in terms of pairs, product of pairs correlation. So we will come to that when you do CI. So essentially the coefficients for those excited determinants can be written in terms of products of the coefficients of W-excited determinants because mainly the correlation takes place in pairs. So this was the argument, the second argument he gave was that the Hamiltonian itself is one and two particle Hamiltonian. So it is not possible for the Hamiltonian to correlate more than two particles. So essentially the idea is that what Synanoglu is saying that more than two particles cannot be correlated or chances of correlating more than two particles are less. So the pair correlation is the most dominant term. So this is what the Synanoglu's theory was. Essentially every time we are talking of correlation, please understand that means the probability of avoiding. When would they avoid if they come close together? So what we are saying is that in the Hartree-Fock which is an uncorrelated description where there is no correlation, there is a chance of two particles coming close together because Hartree-Fock allows at least anti-parallel spins to be paired together and those two particles have to be correlated when you go beyond Hartree-Fock. So this was essentially Synanoglu's theory, we will expand that when we come to CI and the perturbation theory essentially tells the same thing that you have a preponderance or importance of W-excited determinants and you do not have singles and triples. But of course the pair theory goes much beyond simple perturbation theory because it also says that two pairs can be correlated or three pairs can be correlated which is absent in the perturbation theory unless you go to higher order perturbations. So you can do MP3, MP4, then of course you will see those effects. At MP2 you do not see those effects. So I think I just wanted to mention that we are not going to do rigorously Synanoglu's theory but at least in a hand waving thing we have to understand because this is a quite a difficult theory. This eventually leads to what is called the today couple cluster theory which we will do much later. So we will come back to that later but at this point I think we will now start on our next topic and we will try to relate to this the perturbation theory is the configuration interaction. I want to mention that the perturbation and configuration interaction are the two major electron correlated theories to start with. So it is the interaction of the configurations which leads to improvement of the Hartree-Fock. So this is something that we already discussed that if I have m virtual orbitals I can have MCn determinants and your full wave function is a linear combination of all the MCn determinants. I hope that is clear to everybody and we have taken an example to show by taking a two particle function let us say psi12. So what we did was to allow one of the coordinates to be used as a parameter that is for a given value of one coordinate this is only let us say coordinate 1, this is only a function of coordinate 2, the spin and of course the spin and the space. So then we can write this as Ci into the determinants of form from any chi i chi j where this, sorry, ci1 into chi i2. So this is basically a linear combination of chi i2 but I am writing ci as a function of 1 because as one changes the coefficient will only change. As we mentioned this before they are parametrically the different. Then we expanded ci1 exactly the same manner and we get some dji chi j chi i. Of course this cannot be a simple product so it has to be anti-semitrize product and this is basically the determinants. So if I have a complete set of orbitals or spin orbitals then if you form all two particle determinants out of this complete set and make a linear expansion this is in principle an exact wave function. So this is an exact wave function. So for our many particle problems an exact wave function psi 0 can be written as a linear combination. The first determinant is Hartree-Fock. Once again note that one of the determinants we want to keep Hartree-Fock which means the basis of orbitals that we use are the Hartree-Fock orbitals plus the virtual orbitals. So what is our basis? So basis that we use normally for ci is the Hartree-Fock basis. So Hartree-Fock orbital let us say n of them. When I mean orbital I mean actually spin orbitals to start with plus the virtual m minus n orbitals, virtual orbitals. Now the quality of the virtual orbital of course determines how good is your wave function but note that these virtual orbitals are always orthogonal to this n's Hartree-Fock orbitals. So these are commonly called occupied orbitals these are called virtual orbitals. But what I am trying to say that I can get some other virtual also. So I can get just the Hartree-Fock orbital another set of virtual but the property of that must be that they should be orthogonal to this. That is the only important thing. Otherwise the entire physics will become complicated. If you solve from the Routhan equation or from the Hartree-Fock canonical equation you automatically get orthogonal. You automatically get orthogonal. You can get infinity in principle but the point is that you cannot actually solve this. So for all molecules we use an expansion in terms of basis. So as I explained to you that we actually do not solve f chi i equal to E i chi i. We use the Routhan equation. So we first do a spin integration and then the spatial orbitals we expand in terms of m basis. So that is where the finiteness comes in. So in principle yes m should go to infinity. So if you solve f chi i equal to epsilon i chi i then automatically they are orthonormal. Orthonormal it is automatically preserved but if not as far as our discussion goes they should be orthogonal to this n orbitals. So I have a set of m spin orbitals. So out of each n of them are Hartree-Fock spin orbitals. So they are called occupied. So these are occupied orbitals. They are just giving a kind of a box. So their n spin orbitals are there and this determinant which is formed from here is one determinant which is Hartree-Fock. Then you have m minus n number from here and this number should be as large as possible depending on what is the value of capital M. And to be exact it should be infinity for a real complete basis set but of course we cannot get infinity so we assume that it is as large as possible and it is near complete. So that is our assumption. These are my virtual orbitals. Many times they are called unoccupied. They are unoccupied because they are unoccupied in the Hartree-Fock. So starting from the Hartree-Fock reference they are called unoccupied orbitals. So this is the description and then I can generate many, many determinants by exciting electrons 1, 2, whatever up to the virtual orbitals. So I can have two excitations, single excitations, double excitations, triple excitations and so on. So your actual wave function will then be linear combination and I explain to you that all the m cn determinants can be categorized as a Hartree-Fock plus all singly excited plus all doubly excited and so on. So I can write this as sum over a r, c a r, psi a r. Again please note the symbols that a is occupied orbital so we are using a, b, c etc. for this set and we are using r, s, t etc. for the virtual orbitals. So the symbols are very clear. So this is a determinant which is generated from Hartree-Fock by changing the spinor beta a to r. So that is part of the m cn determinant and then you have similarly a less than b, r less than s, c a b r s, psi a b r s. So once again this is a doubly excited determinant which is generated from the Hartree-Fock by two electron change from a b to r s. So this is doubly excited and then you can continue to make triply excited and so on, r s, t and so on and you can keep writing in the same manner. I am not writing so if you keep on writing up to all what is the final thing all n triply excited. The m cn determinant what is the maximum number of degree of excitation when you have excited all the n electrons here. So that is your n triply excited so that is your final number. But of course the number of determinants in each case can be infinity if your m is infinity because of this fact that there is r s. Number of degree of excitation is finite. Please understand this difference because degree of excitation depends on your number of particles. So we are talking about finite particle system so you can only excite n number of electrons but you can excite two orbitals which can go as high as possible number of orbitals. So that is in the each category the number is infinity. In principle it can be infinity and all together if you add this as well as this the number was m cn. So that is something that I showed the other day we mentioned. So this becomes an exact wave function and this is what is called full C i. So C i is basically configuration interaction so this is called C i in very common term so this is called full C i. Since our m is not infinity we have to say that we do full C i in a finite basis so this finite basis is capital M. So this is not really a still an exact wave function because exact wave function would be when m becomes infinity because otherwise my initial argument that this is i equal this is a complete set itself does not hold good if m is not infinity. So although we normally say full C i is exact but please understand the meaning of the exact is in this finite basis. So in fact everything that we talk starting from Hartree-Fock for molecule is in a basis I already mentioned to you even the Hartree-Fock we have not solved in exactly. So there is something called exact so I have exact Hartree-Fock. I have Hartree-Fock in a finite basis let us say m basis I have exact energy and I have any calculation mp2 C i full C i everything is in m basis so everything is in m basis. What is exact energy? Exact energy is nothing but full C i energy for m going to infinity I hope it is clear. So if I do a full C i that means my form of the wave function is complete. Form is complete but when you give exact energy when my basis will also become infinity. So there are two different points because here I am first expanding a basis to come to this form but this basis has to be infinity. So if I do a full C i in a limit m tends to infinity that will actually be exact calculation physically exact calculation which we cannot do. So please understand very loosely we say full C i is exact but this is actually exact in this finite basis. So everything that we talk of is finite basis similarly exact Hartree-Fock. What is exact Hartree-Fock? It is also Hartree-Fock in m basis when m tends to infinity. So we are not even doing exact Hartree-Fock. So please understand and we must understand the ordering of the energy. If we increase the basis the Hartree-Fock will improve because Hartree-Fock is variational method so I have more number of functions it will improve. So if I do a Hartree-Fock in an m basis your exact Hartree-Fock will be always lower because as m increases the Hartree-Fock goes to the exact Hartree-Fock and then you have correlation energy whatever I am doing C i or whatever in m basis and then within that you have exact correlation. If this is variational then again one can show that the exact correlation will be smaller than the correlation of m basis if it is variational. Hartree-Fock is variational so I can definitely say this but for the correlation I have to ensure how am I doing the correlation? I have not mentioned that the correlation energy is always obtained variational in fact you just now saw that we did perturbative correlation that is not variational mp to energy. However normally it is always true that if you do a correlated in a larger basis the result goes down further. It is very rare you can argue that I do a very bad Hartree-Fock and my exact Hartree-Fock is actually lower than the correlation in this bad basis because exact Hartree-Fock would improve this but m is so bad that your correlation of m may even go above exact Hartree-Fock but it is very rare that such a thing happens because then your m basis is very very poor. What I mean to say that this ordering is not clear that in a poor basis correlation energy means essentially Hartree-Fock plus correlation energy total energy will always be greater than exact Hartree-Fock or less than exact Hartree-Fock. It should be less than exact Hartree-Fock but it is not clear because your exact Hartree-Fock can be much lower. So usually m is sufficient in large that this does not happen that correlated energy is also lower than the m. Please note that when I call correlation m this means Hartree-Fock m plus correlation energy. So basically the correction in m basis. So I am adding the correction to the m basis then I am getting what I call here correlation m, correlated energy in m basis. So if m is very poor Hartree-Fock itself exact Hartree-Fock itself still may be lower than this you understand because when m is very poor because there is no variation to suggest that what will happen but usually does not happen. Usually this is the order that we always get. Nature of the function as well yes of course. Well like for example H2O3G would be a poor basis because the nature is not only that you have to have only one H2O2P but they are very poorly described by the only three Gaussians. So the nature is also very important of course otherwise most of the time we take LCAO but how am I describing the contracted Gaussian or the atomic orbit and is also an issue. You are always talking about f chi i equal to epsilon chi i. If I solve f chi i equal to epsilon chi i then it is exact Hartree-Fock that is exact Hartree-Fock. Atoms as well as molecule if I can solve the problem is you cannot solve for molecule. So we always do a m basis Routhan equation. So we expand this in m basis and we get a matrix equation. So that value that energy will be always higher than this energy. So the Hartree-Fock that we have presented to you for molecule is actually Routhan-Hartree-Fock. So you have a charge density bond order just one minute one and two electron integrals everything is in m basis. Our results they will be usually higher than the exact Hartree-Fock. Because exact Hartree-Fock is actually the same calculation Routhan calculation when m goes to infinity because then this and that becomes identical. Now my point is that when I take only an m basis I do correlation calculation the correlated energy should be by this definition should be lower because correlation energy is negative let us say it should be lower but how much lower? Does it actually cross the exact Hartree-Fock? That is a question which I cannot prove but usually our m is sufficiently good that this number goes below exact Hartree-Fock because correlation energy contribution is much more important than just the basis set contribution for saturation for Hartree-Fock to exact Hartree-Fock. So that is the point or beyond a point Hartree-Fock does not change much. So the real change comes because you are adding mp2. So then that is the reason even in a finite basis this goes down because this if m is sufficiently large this delta is very very small whereas this is large. So it will usually go down. I think this is a question of you know understanding how they really order that is a big question. How do what is exact energy? Yes I mean it is experiment but again experiment done under these conditions of theory which is zero Kelvin yes no vibronic corrections and so it is not it is still not no relativistic corrections here. So yeah I mean exact really is very difficult to say but let us say that I can do an experiment for a non relativistic Hamiltonian without any vibronic corrections I can get the electronic energy experimentally that is my exact and they are they are available. So people have vibronic correction they simply subtract it so yeah and people can do gas phase experiment to get zero Kelvin limit. More importantly this number is used as a benchmark number for all other theories in that same finite basis because at least in the m basis I can definitely say that this is an exact number in that basis you cannot get better than this number in the m basis. So if I do another theory like mp2 mp3 couple cluster some other approximate ci everything should be benchmarked against this full ci basis is same all the calculation basis is m so basis is taken out of equation. So all that you have now calculated is defining is the method so then this is the benchmark method so very often the full ci is used as a benchmark so your exact correlation energy so full ci in an m basis is used as a benchmark calculation for other approximate methods because otherwise it is exact approximate methods in the same basis that is important. So note that full ci is actually exact apart from this basis problem in so there are two different things one is number of basis another is the method itself as far as the method itself is concerned that is a form of the wave function it is exact there is nothing more to do all you need to do is to push the basis set. So if I do any other method like mp2 it is only a second order perturbation I have a neglected third order fourth order and so on so that is an approximate method if I do that in that same basis then I can compare that result against full ci and say how good is my method or later on you will see maybe even couple cluster maybe DFT density functional theory so for everything there is a benchmark required how good it is so this serves a very good benchmark full ci so full ci is often a very important method and early days of in quantum chemistry full ci used to be seen as a very important method to benchmark other approximate methods even though you will realize that full ci is not really doable because mcn is still a very large number of determinants is mcn it is very large because you have m factorial, factorial is very very very quickly it increases so very often even these many determinants are difficult to handle. So