 So when this becomes the form of the wave function, I now have to tell you that these determinants are fixed in C i, I am not changing the orbitals, the orbitals are defined. So what are my parameters here? The parameters are these coefficients C naught, C a r, C a b r, etc., which have to be determined. So these are the determined and these determination of the coefficients, so this is important to note in full C i that these determination of these coefficients are done by variational methods, variational methods. Quite clearly, if you look at the nature of the variation, this is basically nothing but the linear variation, because I am only getting the linear, the coefficients are only linear. So this is called linear variation because these coefficients are only linear in nature and we actually did a linear variation problem that if I expand a wave function as a linear combination of a basis, these are now my basis, these are my basis for the n electrons. Then you have a McDonald theorem, I hope you remember the McDonald theorem that you said not only ground state but all excited states, you have an upper bound and if I increase the basis results keep on improving, but it never goes beyond the lower states. I hope you remember the McDonald theorem. So all that will actually apply here. So the basis for the entire wave function, if you look at the basis now, what is the basis for the n electron function? We are talking of a one electron basis, but the basis for n electron wave function is now the set of determinants. So you have psi Hartreefa, then the set psi ar and so on, psi abrs and so on, all determinants. So all of them together is my set, basis set for this expansion. So please note, there is a basis set for orbitals which are one electron function. From this basis we are forming n electron determinants, how many n electron determinants are there? mcn, if I have a m one electron function. Those mcn determinants now become the basis for the n electron problem, just as one electron basis is the basis for one electron problem. We always expand a single electron function in a basis, but now these determinants form the basis. So this determinant, mcn determinants are now the basis, my basis for the n electron wave function. So that is important to note that now if we apply linear variation, remember when we said linear variation, we said exact wave function is a linear combination of a known set of functions which are basis. So what are these known set of functions, these determinants. So do not get confused with this one electron basis and the basis for the n electron problem. This basis is of course obtained from this n electron, one electron function. But the determinants are the basis for the n electron problem, spin orbitals are the basis for the one electron problem. Because obviously for n electron problem, the basis must also contain n electrons, it cannot be one electron. So these determinants have n electrons, each of them. So they become the basis for this expansion. So in that sense, this problem is exactly identical to the linear variation problem. So they are my basis and this wave function is a linear combination of this basis, now n electron basis and this coefficients if I obtain by variation principle, it becomes a linear variation. So psi hat refog is 1, then you have singly excited, that is m times m minus n times n plus w excited, you have m minus c 2 plus into n c 2 and so on up to m minus n c n into n c n which is 1. This number is m c n. So I had given this identity once, you have probably forgotten but long back I had already told this identity. So if you do this, remember these are m minus n. So if I do this, basically sum over, so there is a simpler arithmetic, sum over m minus n c r, n c r, r equal to 0 to n is equal to m c n. So that is the mathematical representation of this. So your first r equal to 0 is hat refog, 1 into 1, then the singly excited r equal to 1. So r contains your degree of excitations, so hat refog means no excitations. So what is m minus n c 0, 1, what is n c 0, 1, so that is your 1. When r becomes 1, you have m minus c 1 into n c 1, so that is your number of singly excited determinants and so on, till r equal to n. So when r equal to n, you have number of n tupley excited determinants which is again n c n is 1 but m minus n c n is your total number, there are how many ways I can excite n to m minus n. All this sum together is actually m c n, so that is the break up that you actually see. So basically all the m c n determinants have been categorized with a reference determinant hat refog, a singly excited, doubly excited, etc. So whenever we talk now of a benchmark calculation, please remember that the benchmark calculations are in a finite basis. So do not immediately jump into experiment, because the experiment is little bit more different. First of all you have to large basis, then experiments of other effects like relativistic effects, etc. that has to be taken care. But as far as theory is concerned, it is an exact theory in that basis. So this can be a benchmark for other theories. The second point to note is that although I had said Brillouin's theorem, please remember the Brillouin's theorem that psi hat refog Hamiltonian psi ar is 0 and because of this we showed, this is my Brillouin's theorem that this results to the fact that the MP2 energy does not contain, does not contain any singly excited determinant. The contribution with energy cannot be determinant, I should write does not contain contributions from any psi ar. Simply because my MP2 form had a hat refog Hamiltonian excited determinant, that's all the form is. However, when I am writing Ci, note that my exact wave function will contain singly excited determinant. There is no reason right now to eliminate that. Brillouin's theorem simply says this, Brillouin's theorem does not say that the wave function will not contain singly excited determinant. It only says that the matrix element is 0 and because of the fact and I again repeat because of the fact that the MP2 energy or correlation energy, MP2 correlation energy is sum over psi hat refog V psi excited or psi whatever determinant psi star let me call it mod square divided by the denominator which is the difference of these two, all star all excited determinant. Because of this reason, the psi star cannot be singly excited because of Brillouin's theorem because essentially H is same you can replace this H by V also, it doesn't matter as I told you H0 anyway it is 0. So that is the reason MP2 energy does not contain but because MP2 energy does not contain and there is a Brillouin's theorem, I can't jump into a conclusion that the psi 0 will not contain. That we will see when I actually solve the hat refog equation ACI equation. So as of now, of course my wave function must contain the singly excited determinant. The only thing you can argue is that because your first order wave function does not have singly excited contribution, you can argue that even in the exact wave function the contribution of the singly excited determinant will be make small, most likely it is likely to be small. What do you contribution means? The value of Car is likely to be small, of course the determinant is fixed. The value of the coefficient is likely to be small that is something that I can argue because I know that the MP2 energy is a very good cut for the correlation energy and there is no singly excited determinant. So most likely when I do a CI, the contribution of the singly excited determinant will be also less but it need not be 0. So that is something that I want to mention that do not confuse this with Brillouin's theorem. Brillouin's theorem simply says that the Hamiltonian matrix element is 0 and we will see how that affects when you actually write down the CI equations. In fact we had written down the CI equation for linear variation problem. If you remember linear variation we wrote down the equation that it is a matrix of the Hamiltonian. You have to construct the Hamiltonian matrix in the basis and diagonalize. Now the same thing we will do, my basis is only this determinants because my Hamiltonian is an electron problem, basis is determinants but what I will do I will not assume that I will re-derive the coefficients in a systematic manner. So what we essentially will do is first do a variation problem and then try to see what these coefficients are. So there is a very important theorem that is if I have and this is something I mentioned I think, if I have a wave function as a linear combination of some basis that is a phi of i which is a basis, this basis is a very generic term here. It can be a one electron problem, it can be an electron problem. If it is one electron problem these are orbitals. If it is an electron problem they are determinants but whatever is the basis, the coefficients obtain variation. This is a very important theorem, obtained variationally are the same as obtained by what I called and I had defined this before by a method of projection on to the Schrodinger equation. So this is a very important theorem and if you recall when we did the linear variation solution we actually use this theorem. We did not actually variationally obtain coefficients, I do not know if you remember this and before I go to CI I may again remind you what we did. So let us say my psi is sum over i CI phi. So what we did is the following and this will actually tell you what is the method of projection. So we write the Schrodinger equation CI phi i. So HI equal to E times CI phi i. Remember this is what we did, we just wrote the Schrodinger equation HI equal to E psi applied psi here. We are going to do that for CI later in the next class but let me just quickly revise you what we have done before, phi i's are my basis. So what are these phi i's in CI? These phi i's will be psi Hartree-Fock, psi AR, psi AVRS and so on but right now they are just basis. So then what we did was to first project this with a member of the function. So let us say any phi j, I hope you remember this phi i CI sum over i so I will now go outside. So the same thing will happen here E sum over i phi j phi i CI. So what I have done is the following, I have multiplied this Schrodinger equation very one line multiplication by a given phi j star which is a member of the basis for all j I am doing it by one member of the basis and integrate. I think we have done this many times, this is the matrix formulation of quantum mechanics, Routhan equation also we did the same thing. Then we say that this is now the matrix elements of the Hamiltonian in this basis. So I call it matrix of H, so we wrote this as sum over i, call this Hji, this is CI equal to E, since this is orthogonal, this is delta ij sum over j, sum over i so it becomes Cj, the right hand side simply becomes E times Cj, this i vanishes, sum over i vanishes and this is my eigenvalue equation. In fact if you remember this is my matrix eigenvalue equation and this gives me the coefficients by method of projection. What I am showing an important theorem, we will prove it that had I done completely variational, what would be variational calculation that is do actually psi H psi correct by psi psi and then we minimize this with respect to the coefficients, then the result that I would have got is exactly same as this. This is easier to derive, this is more complicated because you have a numerator, you have a denominator and then you have to do delta, the first derivative to be 0, it is much more complicated, this is much simpler to derive. So this is an important theorem, we have assumed last time that this theorem is true and we have only done this part. I want to first now convince you that doing this and doing actual variation gives the same result before I go ahead because when I go ahead I am not going to use this variation because it is very complicated, I am going to simply use method of projection. I hope you understand the two difference, one is that you get an expectation value then minimize this quantity with respect to coefficients, minimization which means first derivative becomes 0 and so on. Here I am not doing anything like this, I am simply writing whatever is there in the Strodinger equation and multiplying by a member of the basis star conjugate of that and integrating and lo and behold I get an equation which is a matrix equation and I call this same as this. Now note that the important theorem holds only for linear variation, this is very important. So do not do this for any other wave function, if the wave function form is a linear form like here then only this theorem holds good otherwise actual variation would mean doing this, this is truly variation but I can say for linear variation that these two are identical. So I will first show this in the next class before I actually derive the CI equation by method of projection. Please remember this method is called method of projection, it is a very important method in quantum chemistry, a very simple method, write the Strodinger equation whatever you have got just project it by a member of the basis and you will immediately get an equation and what the theorem says that if it is a linear variation then these equations are identical to what you will get by actual variation, you have any problem with the equation? No. So I am just writing phi j star phi j h phi i, CI I have taken outside this becomes my matrix element h ji is now a matrix in a basis finite basis and then you diagonalize. So that is how I get the energies. Of course you will have m roots and there will be m column vector so it will become a matrix and all that all that technology is there and that is coefficient matrix says if I put them in a matrix that is unitary matrix all that you know because it is a Hermitian matrix finally h so it is I have discussed this matrix Eigen value problem so all that will apply but important thing is to first show this because somebody will say this is not variation. So why I am going to apply variation results linear variation results, McDonald's theorem is for variation so first I will show that this is identical to this then I am going to derive the CI equation using a method of projection there I will explicitly write the basis functions as Hartree-Fock psi r and then I will apply Brillouin's theorem and a lot of things will come up after that.