 I hope all of you by now should be able to write the Hartree-Fock equation in non-canonical form right, canonicalization will do but non-canonical form can you all write okay good. So what is the first term H of 1 right and chi I of 1 then what is the next term sum over j integral chi I star will it be chi I star chi j star okay and the coordinate 2 then 1 by r 1 2 if you want we can put the exchange together right this is a nice way of doing actually just put 1 minus 2 your problem is over as far as equation is concerned okay into yeah into chi I 1 so let me write this as okay sum over j lambda ij chi j as I said this can be made lambda ji or ij it does not matter because that depends on how you start to do. So this is your Hartree-Fock equation in a non-canonical form so anybody asked you should be able to write this very simple. So you have a Coulomb term which comes from the density of the electron 2 in chi j and you are summing over j which means you are taking the interaction over all the electrons so when I am integrating over d tau 2 chi j star chi j 1 by r 1 2 you get effective one particle interaction on the electron sitting in chi I orbital that is the interpretation of the classical term of course the exchange term it is not easy to interpret because when I actually allow p 1 to add this goes out of the integration and chi I comes inside the integration so that is the reason we cannot now call it an exchange and we saw yesterday in the DFT also that similar things are there that the exchange is difficult to integrate okay. So in fact DFT class is very much on the same thing those who are attending anyway so we are we call this density sum over j chi j star 2 chi j 2 is actually density density of the electron 2 right. So this is the density of all the electrons interacting by 1 by r 1 2 so density is nothing but probability so this is the weighted average so this is why it is called weighted average because weighted average is probability into the actual value so weighted average into 1 by r 1 that is how this electrostatic Coulomb term is interpreted the exchange term of course cannot be interpreted like that but chi I will become 2 chi j will come outside 1 and this part is of course a non-canonical form so I do not yet have an eigenvalue equation okay and this is basically the HF equation for all these spin orbitals okay. So I will get if I solve this one particular equation I will get chi I and then I can construct my wave function using this chi I as chi 1 chi 2 etcetera 2 chi n so the question is of course which n spin orbitals you choose so these are my Hartree-Fort determinant. So initially if you remember we are using tilde because it is a trial function now I want to tell you that we will drop the tilde because now it is no longer a variational or trial spin orbital these are the spin orbitals which come out of the result of the variation so now we will say that this we will drop the tilde we will say these are the Hartree-Fort spin orbitals. The one important thing we noted is that this operator Hartree-Fort operator which you called F the entire operator includes this Coulomb and exchange term which is basically dependent on the spin orbital themselves so to solve this equation you have to use what is called the self consistent field procedure okay we discussed that last time so you have to solve in an iterative manner and this is the origin of this term SCF because it is solved in the self consistent manner this procedure is also called SCF procedure okay so whether non canonical or canonical that is exactly the same. So this is basically the Hartree-Fort equation for one particle spin orbitals note that this entire term that has come has really come out of the variation method that I wanted to ensure that the energy E Hartree-Fort which is psi Hartree-Fort H psi Hartree-Fort is now minimum okay is now a minimum so we have ensured that this is minimum for all single determinants right we have ensured that this is a minimum for all single determinants if I choose all possible single determinant this particular spin are determinant which satisfies this equation is the minimum energy so that is what we wanted okay so we will go to the further part of how to canonicalize and then do lots of interpretations before we canonicalize let me talk about an effect of unitary transformation on this equation all of you know what is unitary matrices right so effect of unitary transformation on this matrix so let us assume that I transform the spin orbitals by unitary matrix so let us say that I get a new I get a new set of spin orbitals which I call now prime chi i prime after I reach the Hartree-Fort equation okay I transform them through some chi j of i sorry some u j of i chi j so what I am doing essentially is that I am making a transformation I hope all of you understand what is the transformation it means that the new set of spin orbitals are linear combination of the old set of course I can always do that so I am choosing the new orbitals which are function of the old set linear combination of the old set the combination coefficients are what I am calling as a matrix which is now unitary so this u is a unitary matrix alright this coefficients I can put them as a matrix and this is an unitary matrix of course you know what is the unitary matrix if I take adjoint u dagger or u dagger u that is equal to identity so let us assume that we transform this spin orbitals by unitary matrix so what will happen to the first thing what will happen to the orthonormality of the spin orbitals okay can somebody tell me the original spin orbitals were orthonormal because that was the condition if you remember we put while solving the Hartree-Fort equation now I am doing a unitary transformation so will the new spin orbitals remain orthonormal or will not remain orthonormal right all of you can prove this if I ask so can you show by this equation that chi i prime chi j prime is also delta i okay so how will you prove what is the strategy you will expand this side you will expand this side as a conjugate of this expand the right side as a as it is and then show by the fact that u dagger is identity these results into delta i I hope all of you can do this very small home exercise so basically you have to do the summation but it actually teaches you little bit of algebra how to manipulate the summations so for example when I write this what do I get so remember when I do expansion of chi i prime I have to bring in another dummy variable it cannot be i or j right because they are specific index so what is the dummy variable I will bring let us say k okay and what will be now chi i star it will become chi j chi k sorry it will become chi k so now I am expanding chi i prime with the dummy variable k so this will become chi k and you will have u k i right star so I am just going to piece it together and then you have chi j prime which I again expanded by another dummy variable let us say l so this will now become chi l right and this will now become u l j remember do not get confused with this because now this is j this is a dummy variable l so everything will change accordingly so this will become u l j chi l correct so this chi l has come here and this is the combination coefficient so for each given i and j I can write chi i prime chi j prime by summation like this now can we manipulate this you have a u k i star u l j right you change one of them to be adjoint so for example if I change this to be adjoint you become u dagger ikth element right and this remains as ljth element so what how do you how do you do that now how do you bring u dagger okay sum over kl so this is already delta kl so you want to say this is u dagger ik good then you have delta kl good then you have lj so I just shifted this here yes so because of delta kl that is okay so either way if you write it is a product of u dagger identity u matrix which is still u dagger u so it is an identity matrix correct so identity matrix ijth element so it is a delta ij this can be seen in many many ways or you can simply say k is equal to l so you can see it from that way u ik u k j sum over k that is delta ij so anyway this actually turns out to be nothing but delta okay so that is what we wanted to show if you do not bring the delta kl it may be complicated because you have u ik dagger and u j l and then you cannot do anything so this part is very important to recognize that this part I can bring it inside or at least make recognize this and make k equal to l then it does not matter then you can directly do it there so either way the physics is the same but if you just want to put mathematics you can simply put delta here delta kl is nothing but identity kl so you can say u dagger then now you see it is a repeated summation k is here k is here l is here l is here so it is u dagger identity u which is same as u dagger u which is delta ij so either way you should be able to do this is it clear is it clear to everybody this manipulations are very important if you get confused manipulation you will not get the result matrix manipulations you must learn alright so this is a very important thing we learn that if I do a unitary transformation on the spin orbitals the resulting spin orbitals remain orthonormal so that is a good comfort because when you derive this equation you remember we made we wanted that the orthonormality should be preserved otherwise the determinant will not be normalized unity remember when I write E Hartree fog I have not written the denominator that is only because these spin orbitals are orthonormal so if I do a unitary transformation I want to write the Hartree fog in the new basis I must ensure that the new basis of spin orbitals remain orthonormal so that is in fact this is a very general property of unitary transformation that if I have a set of vectors and I transform this set of vectors by unitary transformation the new set of vectors if you start from an orthonormal set the new set also remains orthonormal so two orthonormal vectors can always be connected by unitary transformation so remember there are no unique orthonormal vectors I have one set of orthonormal vectors I can go to another set of orthonormal vector another set of orthonormal vectors and so on each of which is connected by unitary transformation and product of two transformation unitary transformation is also unitary transformation I hope you you know that, if I have a two transformation u1, u2, right, so I transform one set to one original set to a set 1 or 0 to 1, 1 to 2, then of course I can go from 0 to 2 also by another transformation directly and that would be simply product of u1, u2. So if each of this is unitary, the product is also unitary. I think all this you should be able to prove in matrix, okay. So the unitary transformation is very, very important in physics because it preserves the orthonormal, that is a very important part of a vector space and this has been used in physics many times, okay. So that is one thing I just showed this but this is very general proof, it has nothing to do with Hartree form, this part. Take any set of vectors, do this transformation, it remains orthonormal. So now let us see what happens if I do a transformation here. If I do the transformation here, the entire equation will change because everywhere chi will become chi prime, correct. Wherever there is a chi, it will become chi prime. So I would like to know what happens to this equation. Will the operator which is the Fock operator that I have defined, will it change? Remember in the Fock operator there are three terms, I am again repeating. One is the one electron part which is the kinetic and one electron, whatever external potential, electron nuclei, any other external potential. The second is a Coulomb part and the third is the exchange part. So there are three parts of the operator. Each of them is an operator and I think it is good to write this as an operator form. So I have already defined the operator, so let me define the operator again, F, I defined F in the original basis. So what is my F? I am writing it F as H, so F of 1 as H of 1 plus sum over j and I call it a Coulomb operator, jj of 1 minus sum over j kj of 1. So let me say that I, let me write it in this manner. Can you identify jj and kj? So what is jj now? This is a Coulomb operator. It is an operator now. Coulomb integrally I have already written that is for the entire energy but here only d tau 2 is integrated. So this part is only an operator, function of 1. So I am only talking of this part which will actually act on kj 1. So what will be jj of 1, jj of 2 or jj of 1, sorry. So this is integral kj star 2 1 by r 1 2 kj 2 d tau. Note that this is now an operator because it is an incomplete integration. It is a function of electron coordinate. The same way I can write kj of 1 as integral kj star 2 1 by r 1 2 p 1 2 kj 2 because otherwise it is very complicated to write because if you actually try to write then this will become kj i 2 it is very complicated because you are writing the exchange due to this pin orbital kj and suddenly some i will come. So formal way to write is just put p 1 2 that is all that is all. So just change this two electron part for the exchange. It is no longer 1 by r 1 2 assume it is 1 by r 1 2 p 1 2. So the exchange part itself has a permutation. So then you can write jj minus kj and f can be then trivially written like this and your equation Hartree-Fock equation then becomes f of 1 chi i of 1 is sum over j epsilon ij chi j. Again I repeat this can be epsilon ij or epsilon ji and according to convenience you can choose that. So this is your Hartree-Fock equation if you remember. So I am rewriting this f operator by introducing what I call the Coulomb operator. So this is called the Coulomb operator and this is the exchange operator. Just like we have Coulomb energy and exchange energy. So we said now it is a Coulomb operator and exchange operator in the Hartree-Fock operator. So now when I doing the unitary transformation I have to see what happens to the operator. It is very clear that the h will not change. Operator will remain the same. Of course chi i will become chi i but as far as the operator is concerned it will not change though h does not depend on spin orbitals. I am only changing the spin orbitals. What will change is the jj and the kj. So we have to see how does this part not each jj but the sum over jj and sum over kj change. If I change the chi j to chi j prime so that is what we will do now. So let us take the sum over j the Coulomb operators for each of these spin orbitals. And now again we do exactly the same. So sum over j jj of 1. So that is what we are doing. So you have the jj there. So what will be the result? jj of 1 prime. Prime because now I am writing it in the new basis. So what is the result? It will become chi j prime star prime 2. 1 by r1 2 chi j prime 2 d tau. So everything I am just writing in terms of the new spin orbitals. That is all. There is no difference. Then what I will do? I will apply the fact that the new spin orbitals are unitary transformation of the old spin orbitals. So exactly the same manner I will do. So I have to write chi j star prime 2. So now we can write this. One to write this what do I do? I will bring in the new dummy index. So you can remember chi i was sum over j in ji chi j. So now what I am going to write is chi j star 2. So let us bring in a dummy index k. So there is a sum over j. So sum over j is already here. Now I bring in a dummy index k to expand this. So when I bring in the dummy index k, this becomes k. So it becomes uk and this is now j. So it becomes uk j chi k. So it is uk j star chi k. I hope all of you can see this. How to do the manipulation of the dummy indices? Then you have 1 by r 1 2. So now you have 1 by r 1 2 which will come in the integral which will not change. Then the next is chi j prime 2. Although this is also j, I have to re-expand this. So to re-expand this I have to bring in a sum over l and now I am going to write chi j prime with a sum over l. So it will become ulj chi l. So ulj and this will become chi l. Or I can write in the same manner integral. Let me write in the same manner that we are used to chi i star 2 1 by r 1 2 chi l t tau. Please make sure that you understand this. So what I have done for j on the left, I have brought in a dummy index k. For the j on the right, I have brought in a dummy index l. And I am writing this as the old spinos. Note that this is the old operator. If I sum properly when these are equal, that is important. Right now I cannot associate because this k and l are different. If you look at the operator jj, it has both of them identical. So let us see how to manipulate now. So now again we do the same strategy. First we bring k and l outside, j inside. So basically first I am summing over j. When I sum over j, you note that the j occurs only here. j does not occur here now. So I am going to sum this up over j first but while summing I will again do the same strategy of changing the u dagger. So it will become now u dagger or rather u lj u dagger jk. So I just push this on the right of this because they are all commutative. They are all numbers. So I am now writing this as first u lj then u dagger jk. Then the rest, k k star 2, 1 by r 1 2, k l 2. Note that these are the old spin orbitals. This was the new spin orbitals. I have expanded them. Now moment you write it in this form, you can see the sum over j can be trivially performed, repeated index. So this becomes u u dagger lk which is nothing but delta lk. So now I write this. So then this becomes sum over k l delta k l because the sum over j is performed over this and you have phi k star 2. Now you can already see the result that you want. Delta k l phi k star 2 1 by r 1 2 phi l 2 d tau. Sum over k l. Of course since delta k l is there, k is equal to l. So this is nothing but sum over k jk because now k and l are equal. So since k and l are equal, this becomes phi k star 2 1 by r 1 2 phi k 2. I mean I can write this l. So eventually l and k are equal. So this becomes jk 2, jk 1 sum over k. So I can write that this is same as this. So what I have proved? If I start with the new spin orbitals and sum over all the Coulomb operators, all the Coulomb operators with the new spin orbitals, that is same as the sum of the all Coulomb operators in the old spin orbitals. Is it clear? That means in the Fock operator the term, this term also does not change. It remains invariant. This term anyway does not change. This term remains invariant because the summation over this in the old is nothing but summation over this in the new. Don't worry about this j and that k because that is a dummy variable. But basically the dummy variable means that you are taking all Coulomb operators, all spin orbitals. In the new basis, you are taking sum over all Coulomb operators in the old basis. So they are identical. That means this part of the Fock operator is identical. Is it okay? I can go run through it again. So how did I prove this? I started from the sum of the Coulomb operator in the new basis. So I plug in the expression of the Coulomb operator except that the chi is became new basis, chi jf star 2 etc. So that is the sum over j. Then I did each of the new basis expanded in the old basis and I brought in the unitary matrix. For chi j star 2 here, chi j prime star 2 on the left, chi j prime 2 on the right and introduce two dummy indices knl. So then I get this expression. The rest of the expression will remain same. So 1 by r1, 2 only here chi k2 and chi l2 will come. Then I said let us sum over j first. So I just interchange these two summation, summed over j first and that is nothing but delta kl or delta lk into this. So that is what I got. j is now over. So it is only sum over kl and then since k is equal to l, this is nothing but sum over kj or sum over ljl. It does not matter, k is equal to l. So all Coulomb operators, not each of them, remember each of them, each of these jj is not same as each of the jj prime. I did not prove that. I proved that the sum is equal. So each of them will be different but total sum will be equal and that is all I want in the Fock operator because the Fock operator is a sum of the Coulomb operator. Now quite trivially you should be able to show that this also remains same. I will not do that exercise because this is nothing but this except that there is a P12. So you should be able to prove this exactly the same manner that the sum over all the Coulomb operators is identical. Exchange operator is also identical. So the first point to establish that if I do unitary transformation, the spin orbitals remain orthonormal. The Fock operator remains invariant. That is very important. So I can immediately write the same Hartree-Fock equation by exactly the same way, F of 1, only this becomes chi i prime 1. So this becomes prime now equal to sum over j lambda ij chi j prime 1. So exactly the same equation. I can write. You can say that this is also some lambda prime. It does not matter. Some lambda prime which you will see what it is in the new basis because lambda will depend on chi. You will very quickly see. So I am calling it lambda ij prime. So this will be the Hartree-Fock equation in the new basis. I am not writing F prime. Remember because I have just now showed that F prime is equal to F. So I have just now showed that F prime is equal to F. So if I transform the spin orbitals by unitary transformation, the Fock operator remains invariant. Only the orbitals will change. So this one, this change I have already taken care. So I am only now doing the change of the chi i. So new chi i, I will get a new chi i. Now you will very easily see why did I choose lambda as a lambda prime and what will happen to this. And this is where the canonicalization has a meaning. So let me know.