 Here was a specific variation that we did which was called the linear variation method and I will also remind you of that because that was very interesting. So on the subject of variation this may not be required immediately but it will be required at some point of time. So the three is linear variation. So I will first tell what is linear variation. So linear variation is the following kind that I actually choose my site tilde it is no longer arbitrary. I choose my site tilde as a linear combination of some basis functions. So let us call some Ci, that Ci do not confuse with this Ci obviously, some basis function Ci times chi i, do not confuse this chi i with spin orbitals, they are the basis for n particle, n electron problems. So these are n electron problems. So I write my site tilde as a linear combination of this chi i where chi i's are known basis functions. So what are my trial parameters Ci tilde? So basically what I am doing I am changing this Ci tilde in a known basis and generating difference Ci tilde and then seeing what happens. If my basis is complete which means it is an infinite dimensional basis then I go back to the theorem one Euler variation which means I actually get exact solutions. Remember the Euler variation, the exact solution. I have already discussed such a case when these are determinants, full Ci. So if I have an infinite basis and take all determinants then I will get exact solutions which I already told you full Ci which will give you exact function and here I will get depending on the coefficients all exact solutions. Just that you do not get confused with this chi probably I should use a different determinant from Di just to show that they are determinant. So these Di's are nothing but the determinants generated out of chi. This can be one example generated out of the one particle basis which is my chi's. It can be this is an example but it need not be it can be any basis but I am just trying to tell you that in case this is complete then quite obviously I do not have to go through whatever I am telling you you can directly go to the theorem one Euler variation because then I am scanning Ci tilde over the entire Hilbert space because my Di's form the entire space and then quite obviously all stationeries of Ci tilde H Ci tilde will be one of the Eigen solutions of H which I actually mentioned without proof for the theorem one. What is interesting of course is not that what is interesting is that what happens when I do not vary over the entire Hilbert space just like I did here a proof what happens clearly theorem 2 would exist then that whatever happens this quantity would certainly be equal to E naught greater than or equal to E naught again assume Ci tilde Ci tilde is one in many of the cases for proof we will always assume the normalization so do not worry. So theorem 2 will obviously hold good that is greater than 1 theorem 1 will hold good only if it is complete theorem 2 will always hold good in case of complete it will become equal to otherwise of course it will become greater than the exact ground state energy. So then this is certainly holds good why am I talking of linear variation then if it is just to tell you highlight you theorem 1 and 2 no there are certain more interesting results that we may get when you have a variation involving such basis such linear combination of basis and that is what I will first state again without proof I will first state. If you do this calculation of Ci tilde H Ci tilde assuming that again it is normalized and do a stationary calculation so you analyze the stationary t's of Ci tilde H Ci tilde if you do that then if this is a m dimensional basis or let us say k dimensional basis then first of all what I will get is that I will get k solutions you will actually get an equation which I will show you later is a matrix equation in k dimensional space. In fact right away you can now see that the configuration interaction that I discussed can be easily done using this for a incomplete set of determinants that is if I do not take all mcn I take a few of them then exactly this is going to hold good for configuration interaction because this incomplete set. So if I have k number of determinants or k number of basis again when I am saying determinants is an example it can be any basis then this will give me k solutions that is very important first to know how it will come to k solution I will show later k solutions and let me order those k solutions as E 0 0 tilde E 1 tilde E 2 tilde etc. where this is less than equal to this these are as far as the solutions are concerned remember they are not exact solutions now because I have a k dimensional subspace so they cannot be exact solution correct. So they are they are they are not of course exact solution yet each of them must be greater than or equal to E 0 that is also clear from the Rayleigh Ritz is it clear because theorem 2 of course holds in this case. So I order the solutions that I get out of the linear variation in this manner what is interesting to note in this case is that obviously this E 0 tilde which is the lowest among the solutions is going to be greater than E 0 and and all others are also going to be greater than equal to E 0 which is needless to say but what is more interesting is that the second solution E 1 tilde is also going to be greater than equal to E 1 and in fact if I go forward E 2 tilde will be greater than equal to E 2 and so on. So all the finally the E k tilde would be greater than equal to E k this is remarkable because now we are not talking of just greater than equal to E 0 of course if something is greater than equal to E 1 or E 2 or E k it must be greater than equal to E 0 that is needless to say because E 0 is the lowest but now we are saying how much greater that the second solution must actually be so much greater that it must be greater than the first excited state energy the third one must be greater than E 2 this is remarkable because now for the first time this bound that I said for E 0 can now be changed to saying that I have an upper bound property for each of these E 0 E 1 E 2 E 3 because many times in chemistry we are interested in excited states. So far my bound property was only for ground state and now remarkably in the linear variation I have an upper bound property for excited states so this is a very very interesting problem again this problem was I had given that paper I do not know how many of you have read this paper of McDonald sorry physical review 1933 I do not remember the volume and pace number but you should be able to find out it is a very very nice paper the McDonald and actually this was expanded later by Heileras and Undheim and what I am going to talk is now called Heileras I hope all of you know Heileras's name Heileras-Undheim-McDonald theorem that is the theorem that I am going to state this is the part of the theorem there are some parts of the theorem. So the first part is that each of the solution is greater than its corresponding exact solution exact solution these are exact solution the till days are variational solutions. So I am using till day for trial function as well as variational optimized function this is after variation so from here I do a stationery that is important only after stationery whatever I get they must all be greater than or equal to its corresponding exact solutions is it clear this part so this is of course much more powerful than just the Heileras's statement because Heileras's statement only talks about being greater than or it does not say what solutions will be greater than equal to E1 or E2. So here I can get the solutions I will let you know how to get the solution this case solutions this is basically some matrix Eigen value problem but if you solve this then whatever you get by stationarity stationarity essentially means you expand this high till day in this basis and make this C I till days stationery as the first order change of this becomes 0. So that is always the meaning of stationarity first order change is 0 do that calculus you will get some results and this is what you will get yes that is that part is there that I have not said no well this part is correct that I am going to come next because this part does not say this part just says that they are each of them is greater than equal to its corresponding state. So that is the next statement that depends on how many states you have taken and so on the part that he already mentioned is something that I am going to now talk about let me start from this k dimensional basis and then add one more basis. So let me change the basis to k to k plus 1 I hope you understand the problem. So I my my trial function now becomes I equal to whatever 0 to actually yeah the solutions I have mentioned 0 I have made one but I think you understand what it means. So if I have to write then this should be actually k minus 1 to be more correct because this is a k dimensional. So please make that correction it cannot have e 0 to e k that will become k plus 1 okay I hope you understand that that is fine. So I have now added one more basis so my functions now become k plus 1 dimensional and I will now get e 0 till day e 1 till day etc up to e k till day now truly I will get up to e k till day that is why k plus 1 again I ordered them in the same manner that this is less than equal to this etc e 2 till day and so on. What is interesting in this part of the theorem is before I do that let me just for sake of clarity say that I must distinguish this e 0 till day with this e 0 till day because obviously they are different. So let me call this e 0 till day k plus 1 e 1 till day k plus 1 and so on. Then the theorem this part of the theorem first says that the e 0 till day k plus 1 is always less than or equal to e 0 till day k. So in a k plus 1 dimensional basis the corresponding solution that I am going to get after stationarity will be always less than or equal to e 0 till day with the previous basis. So adding a basis only lowers the energy and you already know from the theorem 2 Rayleigh's variation principle lowering the energy is very good because it is bringing me closer and closer to the ground state. Now I know from here lowering the energy is very good for other states also because each of them is an upper bound so each of them is coming very nicely closer and closer to its own excited states to its own state energies and so on. So I hope this part is clear but please remember when I compare this this is an addition of one function. If you compare with a k dimensional basis another arbitrary k plus 1 dimensional basis this is no longer true. This means that I must take this basis and only add one more I cannot change that basis then only the theorem is true. So if I add one more function then all the numbers will become will decrease all the numbers will decrease compared to the previous number but interestingly they will all be an upper bound so they will never cross. So what does this together mean? So this is my e 0 let me just do a simple calculation this is my e 1 this is my e 2 and this is my e 3. Let us say initially I do a three dimensional calculation. So I will get my e 0 till day somewhere here. I will have to get a e 1 till day which must be higher than this somewhere here. So this is my three dimensional basis I have total four solutions and then I will get e 2 till day which will be higher than here but not necessarily here. Now that I do not have a fourth solution this can go anywhere this can go anywhere but when it stops here I do not care whether e 2 3 is here in between e 2 e 3 or here it has to be only greater than e 2. However let me let me add one more basis to this. Let me add one more basis to this then I will have my e 0 till day 4 here which must lower this must also be lower and of course it cannot cross this also must this whatever wherever this e 2 till day 3 will now also lower will actually come down and this will come in between what is interesting is it will come in between e 2 and e 3 and somewhere here we will go to this. So this is 4 or this is 3 of course this is e 2 till day 4 and this will become e 3 till day 4 again it can go anywhere. This structure is ensured by a very interesting theorem that McDonald proposed in that paper and that is called the separation theorem. It says that between every two roots roots means solutions to this problem, variational problem lies one and only one exact solution. This statement is beautiful actually it now means that when I do 4 e 2 till day 4 cannot be here initialized e 2 till day 3 can be anywhere it cannot be here because between these two root there must lie one root. So once I do that the last one can be anywhere but moment I put another solution that one will come down. So between these two roots e 0 till day e 1 till day there is one root here between these there is one root one exact solution between these there must be an exact solution. So this must not cross e 3 I hope you understand when I do a 4 dimensional basis this cannot cross e 3 exactly 3 e 2 till day 4. So this must be below because in between there must be a root must be an exact solution the root is of course the solution up to the variational problem. So lies one and only one exact solution sorry only one this is actually the content of the theorem which essentially says that nothing can cross its own root because if it crosses then also this theorem will be violated because then between the two roots there will be no root. So essentially the variation method that I stated is actually consequence of the separation theorem. What McDonald did was actually to prove this theorem many times this is called inter living theorem many places you might have inter living means essentially you understand what is inter living there are two leaves and in between there is one exact solution. So it is an inter living theorem or separation theorem and this actually comes directly from here. So for example I go to a simple two root problem so in a two basis so I have e 0 I have e 1 so just to show so this is of course e 0 till day 2 two basis problem so this will be greater it will be somewhere and then there will be e 1 till day 2 the point is there must be one root between them. So that is the reason if this is between e 1 and e 0 that is all you have to ensure then of course the e 1 must come here so then this automatically becomes an upper bound e 1 and then if I increase the basis the next one will be automatically upper bound e 2. So the upper bound property that I actually explained is a consequence of the separation theorem and that is that is very most important thing and obviously when you add one more solution they will also go down there is no other option. So if you look at both these theorems were actually put together in this content as a content of this inter living theorem or a separation theorem and this is what McDonald has proved this basically a mathematical proof because essentially this solution come as I said of an eigenvalue equation of a matrix k by k matrix so he only analyzed mathematically how this eigenvalues of a k by k matrix that is nothing to do with quantum mechanics or quantum chemistry remember this is mathematics eventually this will be an eigenvalue problem which I will show so it is only analysis of the eigenvalues of an eigenvalue problem that if I keep on increasing the basis of an eigenvalue problem how does the eigenvalue how do the eigenvalues change so it is actually can be applied to mathematics physics engineering biology in fact engineering people have applied a long long long back but it is very interesting that this application immediately after McDonald's thing came almost immediately came to quantum chemistry very fast because the quantum chemistry was developing and in particular of course you can imagine for this problem of linear variation which is basically configuration interaction I already told you at that if I make a linear combination of a determinant then I have a configuration interaction so to the CI problem this will be actually used and I have not told you of course how to get the CI tilde neither in the context of CI nor here but that will be the matrix eigenvalue problem which I just promise I will come to that very quickly so that essentially is the content of the linear variation method and as you can see it is a very powerful method because if I do an arbitrary variation of Rayleigh-Ritz I really do not know where I stand but here I have a lot lot more idea of where I stand I also know that this solution cannot exceed this it must be because there has to be separation interleaving one root so that is a very one one exact solution so that is why the linear variation is a very powerful method in quantum chemistry again lot of problems linear variation is used for example the particle in a box just now I gave an example C Mn x to the power m x minus l to the power n that is a linear variation problem please understand except except that the basis is not normalized please remember my basis is normalized I have to again mention here so it will not be an exact eigenvalue equation there there will be a some kind of a change but basically it is not very difficult to renormalize you can always go to a renormalized basis which is orthonormalized from there and then do the problem so if they are normalized then the solutions come out very nicely and this will be the content of how to get the solutions in any of the linear variation and I will I will just show I have 5 or 10 minutes I will actually show that how this this case solutions turn out to be an eigenvalue equation so that I complete this part of the variation method okay I think I am not going to discuss too much about variation method because next class I want to go into the actual Hartree-Fock method okay Hartree-Fock theory alright so I come back to this linear variation very quickly and tell you now how to get the solutions and because I have been talking of the matrix matrix matrix so we must know so what we do a simple way so basically what we want to do is to find out such find out the the average value of psi tilde A psi tilde the first order change should be equal to 0 so that is basically the content of the variation method so I must put this here on psi tilde on this side psi tilde on this side and make them equal to 0 to do that of course I must ensure that the dis are normalized okay so if I do this let us write down the Lagrangian which I am subject to this condition so what will be the actual statement that this minus sum ij lambda ij d i dj minus delta ij right so this is what is called the Lagrange variation I hope all of you know this is the variation under constraint so I am varying this under the condition that d i dj minus delta j should be 0 I think I have done this many times so this is what I am actually going to vary so I am going to vary this c i tilde s sorry keep it like this I am going to vary this c i tilde s such that this quantity becomes 0 the first order change is so let us do it very quickly how the first order change becomes 0 so let us expand the psi tilde A psi tilde I had already done this expansion once in proving the Rayleigh Ray's variation principle so it is the same thing sum over ij c i tilde star or whatever c j d i h d j note that that time I had expanded in terms of exact Eigen functions for proving the Rayleigh Ray's principle here it is not an exact Eigen function they are basically the basis so this is what I will get here I have expanded this as i this as j so I get d i h d j minus sum over ij lambda ij d i d j actually this is not right actually this is not right d i d j are anyway orthonormal it is a c i square sorry mod c i square minus 1 that is equal to 0 so it should be c i star c i minus 1 into lambda i because obviously I told you that if these are orthonormal essentially the condition is that this must be square of this must be equal to 1 so c i star c i alright so now what I do I vary c i star first order variation and then find out what is the change or I can actually differentiate this expression with c i tilde star either way it is okay so if I do this differentiation again of course if I put this then there is only one condition I do not need to write lambda ij I can simply write this sorry there is a mistake here very first place so I can simply write as a lambda because there is only one condition mod c i square equal to 1 so there is only one Lagrange parameter okay so now if I vary with respect to c i star what will you get you will get c j times d i h d j note what I am doing is this is my Lagrangian I am varying this Lagrangian with respect to delta c i star either delta c i star or delta c i does not matter only one of them so you have a sum over j c j d i h d j minus lambda times this is sum over i okay this is sum over i lambda times c i equal to 0 this is very easy to do this because lambda time 1 will not contribute anything when I do c i star c i one of the summation is c i so I differentiate only c i will remain here there is a double summation so when I differentiate c i star c j will remain with a single summation so the entire thing will now become d i h d j times c j sum over j equal to lambda times c i for all i because I have to differentiate for all i so I am sorry for this little mistake here on the constraint so that is why little confusion there is a sum over i c i star c i minus 1 so that is and there is only one condition so I need only one parameter this obviously as all of you can recognize is a matrix eigenvalue equation you construct this element matrix element and call these are numbers right so call this as a matrix so you arrange them in the row i column j so it is a two dimensional matrix because when I integrate this basis Hamiltonian with this basis d i star d j I will get number for each i j so eventually this is a matrix equation so it is a d let us say h i j or whatever I call it c j equal to lambda c i so let us call these matrix as a h i j so this will then become sum over j h i j c j equal to lambda c i and all of you know that this is basically matrix eigenvalue equation so that is a matrix this is the column of c's which I am trying to find out this is the column of c's that I am trying to find out equal to the eigenvalue times one number and this is for all i so obviously I am going to generate the full set and then you have to solve this equation so this is how you actually get this k solutions obviously if I have a k by k matrix how many eigenvalues are there k eigenvalues again all of you know from the mathematics so if I have a k by k solution I get k number of lambdas I have written this equation for one lambda but you have actually k number of lambdas and eventually these coefficients can become a square matrix and I can write a different form and I had shown this also how to write matrix eigenvalue so matrix eigenvalue equation is very important please read it revise it so one of the revision point those who are revising must be matrix eigenvalue properly we have done this very thoroughly and if you do not understand matrix eigenvalue equation is very hard it will keep coming in quantum chemistry in different forms okay so this ci problem that we have been discussing which is basically linear variation is nothing but solution finally at the matrix eigenvalue equation all you need to do is to calculate this matrix elements the matrix element of the physical Hamiltonian in the basis whatever is my basis okay and then diagonalize that matrix okay so that is basic recipe to get all the solutions and this particular way of writing the way I wrote here expanding Cytilday and Cytilday left right like this is something that I showed for Allerith's principle but they are of course these were exact eigens solutions that was a formal basis formal expansion to show that the upper bound theorem with respect to E0 here it is a basis so of course these numbers have to be computed these are not the eigenfunctions of the Hamiltonian please remember eigen repeat many people get confused these are any basis so obviously these are not eigenfunctions if these are eigenfunctions in the Hamiltonian I would not do this my these are already eigenfunctions right so I expanded that time Cytilday in terms of eigenfunctions only to show the upper bound principle of course I do not know the eigenfunctions so these are some basis in which you calculate this matrix element and then solve an eigenvalue first and then you will get all the k solutions of the linear variation problem and they have this property that I mentioned the separation theorem each of them is an eigenfunction is I upper bound to its own exact solution corresponding exact solution and of course if I increase they all change values they all become lower and lower the entire content is what is basically given by the separation theorem I think with that we will leave the variation method now we are going to apply the variation method to the to the derivation of the Hartree-Faugh okay Hartree-Faugh equation alright so I think we will close the class today.