 So, my essential problem is the following change the variables kai-aitin-day to the first order and then calculate this lagrangian change at the first order and make it equal to 0. So, all of you agree now. So, my essential way of doing it would be change all kai-aitin-day from the original kai-aitin-day plus delta kai-aitin-day. So, this is my new value let us say kai-aitin-day ends and then I calculate delta l-tint day. So, delta l-tint day is nothing but first order change. So, this is first order change in the lagrangian and put it equal to 0. So, you all agree that this is the first order change. So, this is basically doing the derivative equal to 0. So, I do not have to consider second order change and then divide by delta kai-aitin-day and then limit. So, all that I do not need to take. It is a very simple thing. So, just make each of the kai-aitin-day change to first order and then calculate the first order change in delta, the change from the original and make it equal to 0. Is it clear to everybody? So, that will be basically doing the variation and of course, now I will have to go through the algebra what the first order change means, so that will go slowly. Note again there are two kinds of kai-aitin-day. One on the left, one on the right, everywhere. So I have basically kai-aitin-days as well as kai-aitin-days, correct? Now the question is do I need to change both of them? The answer is no because they are complex conjugates. I can only change one of them, let us say complex conjugate of the kai-aitin, so left one and then find out what is delta-intin-day and make it equal to 0. Because the other one will be simply complex conjugate of it, so I do not need to do it twice. So I can only change the kai-aitin-day left side. So basically I am changing kai-aitin-day star, so that is easy to chop off on the left side. There is no particular reason to choose left side, I can make right side. So eventually I am going to make each of the left kai-aitin-day star from kai-aitin-day star to delta kai-aitin-day star. First order change and see what happens to the delta kai-aitin. So I think you can now do this, of course when I am changing kai-aitin-day star remember the kai-aitin-day star is also changing obviously because I cannot change only the complex conjugate of the left kai-aitin-day star because they are conjugate of each other. So both of them are changing to the first order. So let us now write down what happens as a result of the change. So let me first do the first term, first term is very easy, fairly easy. So let us do the first term, remember I am changing each of the kai-aitin-day star. So this is only the first term in that expression. So what would be the result? It will be sum over i, remember I am writing delta. So I am going to subtract this term for each term. So I will now take only the first order. So first order will be delta kai-aitin-day on the left, is there any more term? Yes, plus kai-aitin-day h delta kai-aitin-day, that is all, do you agree? Of course I will have another term for both of them are delta kai-aitin-day, but that is second order, so I will not consider. So delta l-tin-day because of this change will have one side delta kai-aitin-day, another side just kai-aitin-day and kai-aitin-day is delta kai-aitin-day. Because when I am changing either one of them, the other one will also change. So the first order change of kai-aitin-day delta l-tin-day for this term is can we are taking orders of only delta, first order in delta, whatever is the delta we define, so Hamiltonian please remember this small h does not have chi, so when I am changing chi the Hamiltonian does not change, the physical Hamiltonian does not have chi, so this small h is a one particle operator, that will be constant, so that is why there is no change here and of course this h will act on that, actual integral simple it is a one particle integral, so you do not have to bother, this is a function, these are function integrals, of course this function is not equal to this one, these are change, very small function and there is an operator, so it is a one particle, so for example this one is nothing but chi star whatever is the dummy variable or h of r delta chi, so that is this integral, so these integrals can be trivially done, actually I have not defined what is delta common, remember and it is not important to define, you will see the later, but it is just a first order change, assume whatever is the first order change, so note also when I did this problem remember this first order change is for any delta change, so it is an arbitrary delta change, so this change that we are defining here is for an arbitrary delta change, I am not defining even delta change, any arbitrary delta change, so that will anyway cancel out, so I did this, now let me go back to the second term, the second term is fairly complex as you can see, first of all the second term has two terms, you know not one term, it is a Coulomb and the exchange term, so let me write down delta L in A, let me call this term 2A, so 2A is essentially the Coulomb term for the second term, so if I write this you will get half of sum over ij, so you have the Coulomb term and I have to take the first order of the Coulomb term, so remember there is a 1 by r1 to here, so I am going to now write it explicitly, so I have a delta kai i tilde, this should now remain kai i tilde, 1 by r1 to kai i tilde, kai i tilde, this is the first term, then I can change kai i tilde, remember this is for all kai i, dummy variable, so then there will be another term which will be kai i tilde, delta kai i tilde, 1 by r1 tilde, kai i tilde, kai i tilde, then there will be a third term which is kai i tilde, kai i tilde, 1 by r1 tilde, delta kai i tilde and there is a fourth term which is kai i tilde, kai i tilde, 1 by r1, kai i tilde, I think all of you can write it very easily, so for original integral which is kai i tilde, kai i tilde, 1 by r1 to kai i tilde, you are just changing each of these four one at a time, because everything is first order, so I cannot have any two changes plugged in, so I will have only one change at a time, is it clear? That is the coolant part, then comes the exchange part which is more difficult to write, simply because I have to switch this, eventually I have to switch this, so kai i tilde, 1 by r1 to kai i tilde, so I can write this delta l tilde to b, so I have to be just careful, so this will be equal to half of delta kai i tilde, kai i tilde, so same terms I am writing, rewriting it again, 1 by r1 to, except that the right hand side will now all change, kai i tilde, kai i tilde, plus delta kai i tilde, kai i tilde, delta kai i tilde, 1 by r1 to kai i tilde kai i tilde and so on, you can write this all these terms, I am just writing it but it is very easy, and clear point that it is quite boring, and hence the fourth term, I should just write the minus sign, whole thing is minus, so for everything I have just exchanged the right side and since this count the negative sign, I have to make minus for everything, so this up here it changes, actually let me put this outside, should be minus half, half also multiplies it, so minus half of all this, do you agree? So you have the delta l1, delta l2a, delta l2b and finally of course this time, last time, delta l3, so let me write this down as well, so delta l tilde 3 equal to minus, I put a minus sum over ij and then I add a lambda ij, which is just a number times delta kai i tilde, kai j tilde, plus kai i tilde, of course the first order change to this quantity, lambda ij into delta ij is 0, so I am not writing, I am writing the change because that is a constant, chronicle delta is a constant, this is also a constant, so I am not writing this minus delta ij, because we are writing only the change, so that will anyway get cancelled, whatever is constant, is it okay? So now I have got everything in the in place, so I can write down the entire delta l, except that there are too many terms, that is the only problem specifically because of 2a and 2b and so we will try to look at this term little bit more carefully and try to analyze this, the first set of terms and the third set of terms are fairly straight forward, you have this, this, this and this, out of which it will notice this term and this term are complex conjugate of each other, I hope you can see that, see if I take conjugate of this term, what will happen, how do I take conjugate, this one will become the left vector, h is a Hermitian operator, it will remain same and this left vector will now go to right, I hope all of you remember how to take matrix element conjugate, see if I have an operator a and if I have a pi and psi, what is the conjugate, conjugate is nothing but psi, a dagger phi in general and for Hermitian operator okay, so just remember this, how to take matrix element conjugate, just take the conjugate from right to left and put it down for Hermitian operator of course, you have a special simplicity, so you can notice that this part and this part are conjugate, complex conjugates of each other, if you do the same exercise, you can clearly, here it is more difficult to see, because here there are some over i in both, I hope, when I do it here, it is little bit more difficult, because this is delta chi i chi j, this is chi i delta chi j, so by itself, this is not a conjugate of this, yes, however there is a summation over all i j dummy index, if I take the entire summation, then the sum of these and some of these are conjugate feature, because I can interchange the dummy index, I hope all of you understand, if not I will cause a minute to get that point very clear, so essentially what we are doing is the following, that I have a set which is sum over i j delta chi i t in the chi j t in the and then I have another term which is sum over i j, lambda i j chi i t in the delta chi j, so I have to show that this is conjugate of this, before I do that, let me also mention a very important part, that the Lagrangian by which I am starting, the original Lagrangian just like the energy is of course hermitian, it has to be hermitian that is one of the important constraints that I will have, because the energy has to be hermitian, eventually I am going to calculate energy, so my Lagrangian has to be hermitian, so one of these constraints is, because of this you have to see that the lambda is a hermitian matrix, of course if it is a real then it is very easy, it is a symmetric matrix, otherwise lambda has to be hermitian, I hope that is very clear, because you look at the Lagrangian, what is the Lagrangian, this part, this part was i j, lambda i j, chi i t in the chi j t in the minus delta, so that was my, if this has to be hermitian, so you can see that I can always interchange this part for a dummy variable, no problem, but when I interchange i j this lambda will become lambda j, dagger right, so unless this is hermitian the whole thing will not be hermitian, so right away we have a situation that to start with my lambda is a hermitian matrix, if lambda is a hermitian matrix it is easy to show that this is conjugate of this, now normally lambda is just a real matrix it is actually symmetric, it is not even, hermitian is really pushing it too far, because these are all real numbers in actual cases, so they are actually real symmetric matrices, so you can see what happens now, if I take a conjugate of i j lambda i j delta chi i t in the chi j t star, see if I take a star of this what will happen, I am taking a star of the entire summation not each star, so what will happen this will give me sum over i j lambda i j star delta sorry chi j star delta correct, just as it occurs chi j sorry chi j delta, chi j star is not necessary, then I interchange the dummy variable i j j 2 i, so this remains sum over i j and if I interchange the dummy variable this becomes lambda j i which is actually same as lambda i j if the real number otherwise is hermitian matrix, so I get the same thing lambda i j and this then becomes chi i delta chi i which is exactly the same as this, so the conjugate of this star after summing is equal to this star after summing, is it clear, because of the fact that the lambda is a hermitian matrix or a real matrix, so if I if somebody ask you to prove that these are conjugated you should be able to prove very very clear, so first part of the prove is that since l is hermitian, l is hermitian this lambda is hermitian matrix, so this is where it starts and then it is very easy to show that this two terms are conjugated. So I have a situation in the delta l a, delta l 1 and delta l 3 is that I have two sets of terms which are conjugated so that is important to remember that I do not have to write, I can simply say that this plus compression, similarly this plus compression cc, so cc is essentially short form of compression I do not have to write, then we come down to the analysis of 2a and 2b which actually has a large set of terms, so we have to analyze this, now if you look at this set of terms we have to see whether there is a conjugate which exists here, if you take the conjugate of this star what will you get, you will get chi j chi i correct on the left, 1 by r 1 2 chi j delta chi i which is actually same as one of the terms, either of these does not matter because again there is a dummy variable i j, same exercise I can do and I can show that these two terms and these two terms are conjugated to each other, I hope all of you will be able to show this exactly the same manner because of dummy variable interchange i and j they would be conjugate to each other and in the same manner the 2b also has two sets of terms that these set is a conjugate of this term, so what is interesting is that the entire thing can be written as a set of half of the terms plus compression, is it clear, so let me now try to write down the delta l in this manner, so I will only write the half set of terms, so I will not write the entire set of terms and for the sake of consistency the half set will always contain the change on the left side, so the first set is delta chi i h chi i sum over i, I am going to put the complex conjugate later for all terms, so I am not writing this, then I go back to 2a and I write this as a half of i j delta chi i tilde chi j tilde 1 by r 1 2 chi i tilde chi j tilde and then the second term sorry the second term I am only going to write the 2 terms, however I want to write I will write, the second term is chi i tilde delta chi j tilde 1 by r 1 2 chi i tilde chi j tilde, I do not choose to write it like this, what I will do I will make a further dummy variable interchange here, I will make this i I will make this j deliberately, so then this will then become plus delta chi i tilde chi j tilde 1 by r 1 2, so these 2 sets of terms and then I will have the exchange similarly, so you can write down the exchange similarly as delta chi i tilde, so just do the exchange here chi j tilde 1 by r 1 2 chi j tilde and then minus half write each of them separately because again I want to make sure that this interchange is possible only over to entire sum not for each i j, so that is the reason I am writing it separately for the time being delta chi i tilde chi j tilde 1 by r 1 2 minus sum over i j lambda i j delta chi i tilde chi j tilde that is it and then I just say that whatever it is plus complex function all the terms it is a constant is it okay, with the sum over i I can actually I am actually allowed to write this that is important when I do a sum over, now there is of course I am going to make it 0 equal to 0 what was the reason I did this, you know why did I make such a tweaking that I did not want delta chi i the reason is if I have this I can apply a very simple rule which I will tell you now, so no if I do this sum it will be identical so only for not for each i j, so please do that sum the reason I want to write in the following let us say I have an equation which is delta chi i star and blah blah everything is there something is there integral is equal to 0 something is here you do not have to bother sum function of whatever if I integrate over the all variable it is equal to 0 for all delta chi star sum over i then the first thing that I will do is a sufficiency condition to make it equal to 0 I will say that for each i it is 0 for each i if it is 0 then I do not need to write the sum I do not need to write the sum so I will say this is 0 for each all arbitrary delta chi i star then what does it mean this means that this will become equal to 0 I hope all of you know the integral that if I have x star y is 0 for all x then y is 0 and that is a that is a strict condition for the 0 so if I have a very simple condition that a b is 0 for all a then b is equal to 0 this is a very strict equality that is what I am going to apply eventually delta chi star 1 and something is there which I do not care something is there so that something must be 0 on the right of the delta chi so that is the reason I am doing this now you may you may have problem because you may say this is only a 1 b 1 1 electron integral here you have 2 electron so that is really not the point I will show you that is not the point by taking 1 such integral so can I erase this now I will come back only to that so let us say I take 1 1 term here any 1 term so the first term so half is also not important sum is also not important but let me write this delta chi i star 1 chi j star 2 1 by r 1 2 chi i 1 chi j 2 this is my term this is the first term that I am writing is it okay so except then everything has to be that is not important how do I how do I write it in this form that is what I am going to show you remember you have an integral over delta 1 delta 2 I first performed the integration over delta 2 because this is a function of 1 so I integrate this quantity with delta 2 and allow to do that right if I integrate this quantity over delta 2 this will be a function of not delta 1 function of 1 delta 1 is not right delta 1 is a value so this entire thing will now become function of 1 some function I do not care what it is so I can write this as an integral delta chi i star 1 function of 1 and then you have a delta 1 which is exactly in this form so each of the terms I can write it in this form after integrating over the coordinate 2 I have a situation where something will be a function of 1 which multiplies with delta chi i star 1 everywhere and gives you 0 because my delta l has to be 0 finally and then I can integrate that whatever this something I am going to collect of course plus plus plus all that together is going to be is it clear so from every term I have I will have this so this f of 1 is going to be collected over these over these over these over these everything and then I can say that entire thing is equal to I can do that collection so let me write it down sum over i then I have some term which I just call it now x of 1 delta 1 that is what I am going to get finally I am going to get this form that x of 1 contains everything this one integrated by 2 2, 2, here, here here of course there is no integration over 2 it is simply h of 1 here it is a very simple thing so this is what I am going to get except that the x I am going to write in full later sum over i equal to 0 for each delta collection please note that this x is nothing to do with type this is x it could be y whatever if you do not like x I can write y just to make sure that do not get confused write x, y, say whatever you want so this is the this is the generic form of the equation that I will have except that the y of 1 has to be now expanded from the list of terms on the left is it clear so how do I make it equal to 0 two things I am going to do one is a sufficiency condition the sufficiency condition tells me that under summation over i each term is individually equal to 0 so I am going to use a sufficiency condition and this is usually of course you can argue with that that is this the only condition so I am going to say for the sufficiency condition that each delta chi i star 1 integral of course y of 1 beta 1 is 0 for of course each i but this I am not going to sum for each delta collection please correct me because when I have a sum over i how can it be for each i that is nonsense what I mean is that for all arbitrary delta that means I can change chi i arbitrarily to delta chi i see what I was doing I was changing x to delta x remember my x square problem delta x can be any arbitrary change similarly for each chi i the fact that we are changing for each i that is obvious so that is what I mean so if I but what I am now doing is sufficiency condition is that I am not putting the sum over i I am saying under the sum each term is 0 if under the sum each term is 0 of course the sum is also equal to 0 so that is the reason it is called sufficiency condition but not necessary because you can argue that for one i can be negative another i can be positive and that we are not going so we assume the sufficiency condition then I say that for each delta chi i and of course each i now both it is equal to 0 and then I use this a b equal to 0 to say that y of 1 is equal to 0 since it is for arbitrary delta chi i for y of 1 is equal to 0 for whatever i will come so we will have to only identify this y of 1 interestingly this is a one electron function the equation that I am getting from a very complicated n particle variation has now been reduced to a single particle equation I want to tell you that I have got a one particle equation this y obviously has one term which is h chi i I think the first term is quite obvious to see it starts with h of i 1 chi i of 1 right that you can easily see because the first term h of 1 chi i and then of course the rest are all there which also includes chi so this quantity this function of 1 is of course a function of chi i this spin orbitals but it is a single particle equation that is important and we will show that this is precisely the equation which is called the for spin orbitals of course since it is a single particle equation if I can get this equation in the form of an Eigen value problem it will become orbitals or spin orbitals if I can get it as an Eigen value problem we will see how to do that but essentially this is the one particle particle equation that we talk of and many of you already know that the pop operator acting on something something equal to something so all that will come as a y of 1 I mean something minus something equal to 0 but this is the spirit of the Hartree form variation okay now the point is it is a sufficiency condition so if this is 0 conjugate is also 0 so that is the reason sorry I probably probably I should have mentioned that in the sufficiency it is not only that for each i it is 0 but I am also assuming for delta L to be 0 first that this is 0 so I have another sufficiency condition that this is 0 this itself is 0 is the first sufficiency condition because if this is 0 sufficiency condition is there under the summation over i each term is 0 under the summation over i each term is 0 but remember now if you look at the second term onwards there is a summation over i called delta chi i but there is a summation over j what does the j go that j will actually go in y of 1 that is hidden now because I am only talking of sum over i I wrote the entire equation as sum over i delta chi i star 1 which I told you is an integration of the whole thing over delta 2 so that y of 1 will now actually include function of j sum over j so that does not vanish the two summations so I already said this yeah right here so I had a sum over i sum over j so the sum over j will be any a perform so your f is now sum over j this quantity so the point here I am trying to say that every entire equation I am trying to write as a delta chi i star sum over i only sum over i so basically I am trying to write it in this form this quantity will now contain the second electron just as here the second electron integration plus all the summation over j wherever required for example the first term into chi i 1 wherever required so everything will come including the Lagrange multiplier this summation over j everything will come in y of 1 so right now it is just just a function abstract function important thing to realize that because these integrations are over all delta 2 wherever 1 and 2 coordinates are there that integration is performed to get y of 1 because you have performed integration over delta 2 I have only a function of 1 is it clear so y of 1 is a very complex function it will include all the the terms like h of 1 chi i 1 all summation over j all integration over delta 2 wherever required okay so I think this is something that we should remember or when I will do this y of 1 so the y of 1 and actually you already know how I am going to do it it is very easy so it is basically the first term of y of 1 for example the y of 1 the first term itself h of 1 chi i 1 that is clear this is very trivial because then when I integrate here I get delta chi i star 1 h of 1 chi i 1 so that is the first term the second term in start I can generically show d tau 2 multiplied by chi i 1 note that this chi i 1 must remain integration is over chi j 2 but this chi i 1 you cannot vanish I want to write everything with only one delta chi i star on the left everything else should remain so this chi i 1 will be part of the y of 1 and I can keep writing like this I have just written this chi i 1 later because I am not integrating I am first integrating over d tau 2 so I am allowed to bring this here and then write chi j 1 and then go off in fact we will do this thing in the next class I think he has to also go you have to go now right he has another video recording so I think we will also stop here today so this is how it will happen as far as y of 1 is concerned and then all this together will become equal to 0 which will give you the single particle heart rate okay is it clear so 2, 3 very important thing is that the Lagrange multiplier is not Lagrangian L Lagrangian is Hermitian Lagrange multiplier is also a Hermitian matrix that is a matrix remember this is a scalar remember this is a scalar this is a matrix Hermitian matrix okay and then we are using a sufficiency condition that the for each summation over i within this summation over i the entire all the terms for each i is 0 and what is the other sufficiency condition that we said so for all i it is 0 and the complex conjugate also is we are not going to consider so first is that all the terms are going to be equal to 0 and then within this term for each i each term is 0 so these are the two sufficiency conditions I am going to use so that I do not have to bring in the concept of complex conjugate then the second part is under the summation over i I am going to write only summation over i summation over j will all be hidden in y of i so under the summation of i for each i the terms which survive are also equal to 0 the two sufficiency conditions I am going to use in deriving the happy combination alright so I will stop here