 Let us start with concepts that we have been proposing. In the last class yesterday, we said that the form of an exact wave function can be written in terms of slater determinants formed out of a basis. First, let me again explain that my basis are the basis of spin orbitals which I call chi i. So these are my basis of spin orbitals. So these are basically one particle functions, that is the first thing. And spin orbitals, so they are functions of spin and space part and I deliberately choose this as orthonormalized. So that is of course the choice that is very convenient which essentially means that any two chi i and chi j, if I integrate with chi i star chi j, if I integrate the result is a chronicle delta ij. So that is the meaning of orthonormalization. So I start with a basis of spin orbital which are basically one electron functions. Let us say that I have a basis, my basis is m dimensional. So let us not worry about completeness of the basis because we have already discussed yesterday that completeness requires infinite dimension and it is very often not doable. So let us assume that it is m dimensional where m is sufficiently large as large as possible and this is where computer will play very important role. So when you actually write this quantum chemistry program, you know that your basis, different basis you use and the larger the basis usually you should get a better result. So the reason is very clear. Obviously m should be much larger than n. Typically m should be much larger than n where n is the number of electrons. So n is the number of electron, m is the basis. So these are one particle basis. Then we construct later determinants which are one particle n dimensional. So later determinants of n particle or n electron using this m basis. These later determinants, they act as a basis for the n particle wave function or n electron wave function. Just as your spin orbitals are basis for one particle wave function, any one particle function can be expanded in terms of this basis. These determinants now become the basis for the n electron wave function. What are the determinants that you can compute? Total number that you can calculate is mcn and then what I can do is to write my n electron wave function, let us say psi 0 for the n electron eigenstate or whatever as a linear combination of all these mcn determinants. That is what I mentioned in the previous thing that you can do like chi i, chi j, chi k etc., where i less than j, less than k, less than l and you can have the coefficient d ij, kl etc. Each of these functions go up to m because these are the one particle functions. If I order, I showed in the last class, if I order them in this manner, you will generate mcn determinants. So I am only taking the combinations. This is now an exact form, full form of the wave function in this m basis. So provided m is exact, this will become exact. Provided m is large enough, not exact but large enough to be accepted as exact, this is sufficiently good to be accepted as an exact wave function. This is now limited only by how good is the m dimensional basis as far as exactness is concerned. This function if you include all is what I call full ci. Now I will specifically write full ci in m dimensional basis. So then I make my point very clear that it is not necessarily exact but it is a full ci function in m dimensional basis and as m goes to infinity, this full ci function psi naught goes to exact wave function. Is it clear? This by itself is a full ci. You need not call it exact, you can just say full configuration interaction which means full essentially means that I have taken all mcn determinants within this basis. So I call it m dimensional basis and then of course I can keep on pushing this value of m as large as possible. I am making this full ci function as exact as possible towards exactness and eventually if this reaches near completion then it is an exact function. So this is what they say is a CBS limit, complete basis set limit. Essentially it means that m cannot be infinity but m is sufficiently large that it is almost complete and if I now take all the ci functions then of course it is exact function. So we will of course not worry about this. Of course we have to see how to get the coefficients, what is the energy. So all that we will come back later when you discuss the full ci wave function or ci wave method. At this point I just wanted to mention that even if the Hamiltonian is interacting there is a form of an exact wave function which is possible. Of course if Hamiltonian is non-interacting a single determinant any of these determinants should be an exact function in fact provided they are eigenfunctions of the one particle operator. So by clever choice of basis I can make all the determinants to be exact function for a non-interacting Hamilton. The only clever choice should be that these basis must be an eigenfunction of the one particle operator. Is it clear? Then it is automatic. One of them would be ground state, others will be excited states in that case. But the point that I am saying that even for a interacting Hamiltonian using these determinants built out of these basis I can make an exact wave function and here again depending on the energy one of these will be ground excited states etc. So how will a ground state be different from an excited state? It will be exactly similar form these coefficients will be different. So a ground state wave function will have one set of coefficients and another excited state will have another set of coefficients and so on. So these coefficients will distinguish from the ground to excited states in this case. So I think this is very important. Then we ask the question and that was the question that we will actually go into and that is the part of the Hartree-Fock theory is that can I identify among them one determinant which is the best? Now when I ask this question I have to also bring the consideration of the basis immediately because if you give me a basis and then say within this basis find out which determinant is the best that is one type of question. The second question can be change the basis and find out which determinant is the best? Which single determinant? That is exactly the question that I am asking the latter question. So that means given in some basis can I find one determinant which is the best in terms of energy. So that is the question that we will ask. So we will have full liberty to change the basis sets such that we identify one determinant within the new basis which will give the best energy. The best energy would be defined by what is called the variational method. So I just want to digress here on to variational method and then we will come back to the Hartree-Fock method. There are lot of other technical things that we have to do as I said but right now we will go ahead with the variation method. Before I do this I will quickly go through the variation method since most of you have done it I will not spend a lot of time but I will try to do it because in the last class I think in the 4 to 5 we have taken about 3 lectures on the various 3 to even more I think 3 to 4 lectures and variational method alone. There are many parts to the variation method. I mean again those who want to read variation method I must I refer it before I must refer again the book by Epstein, the variation method in quantum chemistry. There is a fantastic book by Salty Epstein I think this book is downloadable I am not sure but I have not found any better textbook on the variation method. If you can go through one or two chapters first couple of chapters that may be enough but this book is quite heavy. So I refer because this is one of the textbooks which puts variation method in a very systematic manner. So the first part of the variation method is the following that if I construct an expectation value using some set of trial function. So I call this E tilde where all this tilde is stand for trial. So my again function is trial function energy is of course trial and then if I vary if this psi tilde is varied over the entire space note that this entire space should not be confused with the m dimensional basis or completeness of basis because this is a n particle function I am writing variation method in general. So over the entire n dimensional space whatever the n dimensional space mean this is very often called the Hilbert space I am not going to use this terminology. Then these stationary points if you keep on varying then you will have certain points where E tilde is stationary. So the stationery is of E tilde correspond to the Eigen solutions of H. This is actually a very powerful method again I am stating without proof that if I keep on varying this psi tilde so this is a variation obviously means I keep on changing am I E tilde will change in psi tilde and there are certain points where there will be a stationery what the stationery means the first derivative will become 0 first derivative will become 0 it can be maxima it can be minima remember all this stationery it is correspond to an Eigen solution of H which means the E tilde will be Eigen value at that point whatever is E tilde whatever is psi tilde at that point will become Eigen function. So all of them will become Eigen solutions if you look at this theorem the theorem actually completely solves the problem because that is what I want to do I want to solve H psi equal to H psi and the theorem says how to do it correct. So it is a very powerful theorem which was actually which is very often called the Euler variation method in fact Euler is a mathematician. So for mathematics mathematicians Eigen value problem is a very seriously very important problem and Euler came up with how to solve the Eigen value equation in fact this is one recipe for solving the Eigen value equation of anything except that this is extremely difficult to implement because there is a very important part here is entire n dimensional Hilbert space. So you have to span a complete Hilbert space then only you can check this stationery if you do not come do not span then this stationery it is need not correspond to the Eigen solutions. So if you only restrict your variation within a subspace then whatever stationery it is you get then need not be Eigen solutions. So the proof actually requires this entire thing and again it may be mathematically very nice but physically almost impossible to implement just like the completeness of basis. So although this theorem is very elegant and I repeat that the Euler variation is an extremely elegant theorem which is never taught in chemistry or even physics I do not know it is not practically feasible and that is probably the reason it is never taught but I believe this is a very nice theorem actually in 4 to 5 we had proved it the entire theorem I am not going to do this but I am just going to make a statement here. There is a converse to the theorem and the converse is also interesting if S psi is equal to E psi that is if I know the solution of this Eigen value problem then the psi corresponds to not only psi then the psi and E correspond to the stationery it is of this function of psi A psi by psi psi. So this is a converse theorem then not only the stationery points of these correspond to an Eigen solution all Eigen solutions also correspond to the stationery. So it is both ways so that is very interesting that means you do not miss anything. So both can be proved very easily by using this one can prove that by any change of first order change in psi the first order change in energy will become 0 or conversely if the first order energy change is 0 you get E psi equal to E psi. So that is a converse I am not going to make the proof here in this class I had done that in 4 to 5 but I think for this class this statement is important. So I hope you understand the meaning of this that I keep on changing psi I look at energy and then let us say that I have a very simple plot of energy versus psi so E tilde versus psi tilde then there would be some kind of curve let us say I am very simple a very simplistic thing I am discussing then these are the stationery points so this is my psi tilde this is my E tilde. So each of them is an Eigen state of the Hamiltonian and of course the lowest of them is the ground state that is obvious. So it is just a curve which is just simply done so one two dimensional but this is what we mean by stationarity you can keep on changing psi tilde you will get different E tilde and see where are the stationery points but if you can do that over the entire space then only this theorem is valid and that essentially kills the practical utility of the theorem however elegant the theorem is then we come to the next which is the content of the all practical applications of variation method and that is no that was developed by Rayleigh and Ritz it is called the Rayleigh Ritz variation method it is developed by Rayleigh Ritz it is called the Rayleigh Ritz variation method this says that if you take any psi tilde arbitrary psi tilde and then calculate this expectation value here I am not varying the statement says for any arbitrary psi tilde if I calculate the expectation value this is always greater than or equal to the exact ground state energy this E naught is exact ground state energy. So for example here let me write this as H psi i equal to E i psi i and E 0 less than equal to E 1 less than equal to so I think it is very clear that E 0 is the ground state. So it says that whatever for any arbitrary psi tilde including this whatever I did it is always greater than equal to E naught there is no confusion with this because there is an equal to. So one of those points which is stationary will be actually E naught we now know this if I do entire search over the Hilbert space then one of those points would be E naught but this theorem is much more practical because it says that even if you do not search over the entire Hilbert space you just take any arbitrary function this is always equal to at least greater or equal to the E naught but only E naught no reference to other states. This theorem makes it much more easy to apply because now I do not have to search where is the concept of variation here the variation now comes in a different way for all arbitrary function this is greater than equal to E naught. So if I keep on varying over whatever space I want to vary the minimum of them would be the closest to E naught. So if there is a sense of variation minimum of this will be the closest to E naught quite clearly because everything that I do is greater than E naught. So the among whatever variation I have done the minimum of them would be closest to E naught how close of course I cannot tell that depends on how good you are in varying if your variation space is pretty bad still within that space minimum would be the closest but you are pretty off if you do a good variation your space will come down. So for example if this is your E naught I can have a space of variation which is bound by this so this is the best I can have another space which is bound by this then this is the best in each case of course E naught lies lower than whatever I do and in principle if I do a complete variation then of course there will be a there will be one minimum which will touch this E naught touch if I do a complete variation as I said so there is no difference from the Euler but even if I cannot do I know at least whatever I have done within that I can pick up the best terms of ground state. So this is called the upper bound property of E naught. So E naught is always bounded or E naught is always lower than that expectation value lower or equal. So this is a very very powerful method and now I can actually do a variation on whatever space I am doing and then pick up the best one which is basically the minimum. So there is a minimization this theorem again this is can be proved. So what essential is telling that if I have a Hamiltonian and just in case this psi and this psi conflicts I am going to use a different symbol for the eigenfunction or exact eigenfunction. So let us call it h phi i equal to E i phi i. So these phi i's are eigenfunctions of the Hamiltonian obviously they form a complete basis all the eigenfunctions although I do not know them but in principle they exist so a complete basis exist and they are also orthonormal because they are eigenfunctions of Hermitian operator correct. So now I have a situation where any of the psi tilde can be expanded in terms of this basis. So i equal to whatever 0 so 0 is part of the number so 0 1 2 3 etcetera okay the entire basis. Note that the basis that we are talking chi was a basis of spin orbitals this is one particle basis these are actually n particle basis. So this can be our determinants I told you in the full form of C i that the determinants form a basis for the n particle so they can be like determinant this phi i's okay but actually they would not be determinant simply because the determinant cannot be an eigenfunction of h but like that an n particle function may be a linear combination of determinant whatever I do not care what they are I do not need to know them these are the functions which are eigenfunctions of the Hamiltonian and my any psi tilde that I construct can be written as a linear combination of the eigenfunction please note that I am specifically writing from i equal to 0 so that just a symbol change. So then I calculate psi tilde h psi tilde let me also assume that and this is without any loss of generality I can do that that the psi tilde is normalized what would that mean is that the sum over i modulus C i square would be equal to 1 I hope all of you can easily show this right because psi tilde psi tilde is 1 so if you just expand these are the orthonormal sets you can see that C i psi C j C i star C j phi i phi j integral which is delta i j so it will become C i star C i equal to 1 so it is very easy to show this mod C i square please make sure these are very elementary algebra which you can do it in which case I do not need to write the denominator that is the only reason I am assuming that this is normalized unity but I will remember that this coefficient square sum is equal to 1 I have to remember this that is the condition so what would be this now you expand psi tilde on both left and the right hand side with this expansion so let us use for the left one a dummy variable j for the right one a dummy variable i and expand these as phi j h phi i C j star C i correct is very easy to do this C j star phi j star h C i phi i C i phi i okay so I am again using the direct notation to do the integral this integral can be trivially calculated why because all these phi i's and phi j's are eigen function of the Hamiltonian with eigen value E i so it is very easy for example h acting on phi i gives you E i phi i so E i comes out phi i phi j is delta i j so I can easily write this as sum over i j E i C j star C i into delta i j is it clear E i will come because of this and then phi j phi i will give me this delta i j C j star C i remains so now that delta i j is there I can write it as a sum over i E i mod C i square because now j is equal to i so j must be equal to i so C i star C i is mod C i square E i delta i j is 1 and i not equal to j it is 0 so I do not have to bother note again that this is an expansion over a complete basis okay I do not know E i I do not know phi i but this is a in principle expansion so I do not need to know okay just to prove so then we come to this situation where I now know that the psi tilde i psi tilde is sum over i E i C i square now I do the same thing for E naught so I am basically comparing this part with this part okay now I compare with E naught and I write E naught as E naught times E naught times sum over i mod C i square please note that this quantity is 1 so I have a right to multiply E naught by 1 so I just did E naught times mod C i square and then I expand this E naught times mod C 0 square plus E naught times mod C 1 square and so on plus E naught times mod C 2 square and so on is it clear I just expand this and then compare with this part if you expand this part similarly the first term will be E naught times C naught square plus E 1 times mod C 1 square and so on the first term in the expansion is same as this term the second term however is different because here every time E naught second term third term onwards here it changes E 1, E 2 and so on so that is the different so let us compare this term with this term if you compare this term with this term which is larger this term E 1 why because E 1 is greater than E naught and mod C 1 square is positive remember that positive is very important otherwise the argument is not clear correct if I multiply a larger number by positive number then it remains larger by the same positive number of course so each of this term if I compare this becomes larger than this and hence I can say that the E naught is less than or equal to psi tilde A psi is it clear the proof or vice versa whatever the proof is very trivial we have done this but again it is very interesting proof and the proof does not require the normalization to be taken but we have just assume the normalization because of the simplification otherwise this whole thing would have actually gone in the denominator that is all so nothing would have happened that denominator could have been multiplied with this E naught because if this is there then I can always say this is greater than or equal to E naught times psi tilde psi tilde so that is what I am proving so I hope you understand it does not matter then it would not have been one it would be actually psi tilde psi tilde which I can still multiply so I can show that this is greater than or equal to E naught times psi tilde psi tilde which is again the same because this quantity is always positive okay if I can ask you to prove without assumption of the normalization alright you should be able to prove that just a small quiz question that same proof you repeat without normalization then your proof will be this is greater than or equal to E naught mod Ci square which is what I have shown you are allowed to do that simply because this is always positive I hope you know that the norm of a function must always be semi either 0 or positive it can be negative okay if it is negative of course the sign would have changed I hope all of you know this would have become less than equal to so it is positive so I have proved the same thing so the proof exactly goes in the same way so the normalization by itself does not is not really a big deal it is just that is easy to show okay so this actually gives you a very nice handle to apply the variation method and this variation method is applied in many many cases you know in fact people have done particular even single particle problems which are exactly solvable you can do variation method particular in a box hydrogen atom harmonic oscillator hydrogen atom there are famous problems of variation method where you assume the wave function psi tilde to be of the type e to the minus r but just say alpha r and vary alpha tilde see what is the value of alpha you get you know the value of alpha so you will get you will see what the value you get correct so you can you can keep doing it you can do for harmonic oscillator you can take any part of the solution particular in a box in fact many times you used to keep particular in a box as a variation method you all know the solution of course but we can take a variation method like 0 to l I can give you many such functions where which is 0 at x equal to 0 0 at x equal to l and then find out the the the result so so one of them would be like x minus x into x minus l x into x square into x minus l whole square whatever all powers so a particular in a box variation method would be psi tilde as a m mn a mn x to the power m x minus l to the power n it's a good solution yeah I have given this problem I think yeah exam but it's a good problem tough problem length the problem not tough conceptually very easy but yeah length the problem and you you can find out a m so there is no sign here just different way and depending on how many terms in the expansion you take you will your results will be as good as possible so you can do it even for single particle problems which are exactly solvable you know just to test the variation method you don't have to do it but just to test the variation but but there are situations where many particle problems like helium atom there is no option in fact we did also the helium atom variation method you have to apply the variation so that is what we will do for all cases where variation is required and in particular here we will apply the variation method in deriving the Hartree-Fock equation so when I come to the Hartree-Fock I will state that so please remember the variation method so