 Let me remind you again what we are doing. So, we are doing ground state perturbation theory. So, it essentially means that this is the ground state of the physical Hamiltonian. So, we are looking at exact ground state and we are trying to find out how the exact ground state energy can be approximated by perturbation theory. So, that is what is called the ground state perturbation theory. So, we had very quickly, I will just give you the formula. We had partitioned H as H0 plus V, where Eigen function solutions of H0 were known. So, wherever I am writing superscript 0, it means they are the solutions of H0 and I is the different states. So, of course, I is 0, 1, 2, 3, etc. So, the E00 is the lowest energy. So, I can write E00, sorry, less than or equal to E10, etc. So, this is an order. Then what we said is that the correction to the ground state at the first order. So, every time I am writing the superscript, remember superscript is the order and the subscript is the ground or excited state. So, when I am writing size 01, so that means there is a first order correction to the ground state. E01 is the first order correction to the ground state, etc. So, my actual E0 will be E00 plus E01 plus E02, etc. And the same thing goes for size 0. So, this is of course known. So, these are the things that we have to find out. So, we already know that E01, what is the formula? You take size 00, which is the ground state eigenfunction of H0 from here and then take the average value of V with respect to the size 00. So, that is your E01. So, that is the formula which you did quite sometime back. And then in the last class, we also wrote down the equation for the second order and from there we found that in the same manner, you can calculate E02 as size 00 V size 01. Second order correction we cannot still get because we need to know what is size 01. So far we have not done told you how to calculate the wave function corrections. So, size 01 is something that we have to evaluate. Size 00 is known. Remember size 01 is not known because the eigenfunctions of H0, size 00, etc. Hartree fog and determinants and so on. So, size 01 is something that we have to find out. To do that, I also remind you the two equations that we wrote. One is first order perturbation equation. So, this into size 01. So, this is the thing that we have to find out plus V minus E01 size 00 equal to 0. Remember, this is the first order equation. I hope all of you can write it. It is the first order perturb Strodinger equation H minus E psi equal to 0. So, and the second order equation was similarly H0 minus E00, size 02 plus V minus E01, size 01 minus E02, size 00 equal to 0. So, these are the two equations that we have written in the last time. And we told you that each of this equation, if I project with size 00 star and integrate, then from this equation, I get this. From this equation, I will get this because this term becomes 0. So, you directly get E01 equal to size 00 V size 00. Similarly, here this term is 0. Then you have size 00 V size 01 and E01 into size 00 size 01 that is 0 because of intermediate normalization. So, that is equal to E02. So, each of these equations have been obtained by projection. So, you project these equations to size 00 star and integrate, multiply and integrate. So, you get these two equations. We are also explicitly working with intermediate normalization. That means all ground state corrections are orthogonal to the 0th order state for all k, 1, 2, 3 etc. Corrections, only corrections of course. So, that is why this term size 00, size 01 becomes 0. So, you have only size 00 V size 01 which is equal to E02 when I multiply and integrate. So, this is something that we have done so far in the last class. So, I want to remind you. Now, our task will be to evaluate size 01. So, we have written an expression of E02, but this expression is not useful unless I know this. So far I do not know how to get the first order correction to the web function. We are only talking of correction to the energy. However, I have an equation which involves size 01. So, I have to use this equation to obtain the actually size 01. This was not used before because I only multiplied with size 00 star and if I multiply with size 00 star, this becomes 0. So, regardless of this, the integral is 0. So, I did not need to know size 01 to get E01. So, now this is what we will do today. So, I will leave those equations here. Note that my eigenstates of H0 are completely known. So, that is something that I want to remind you that H0 is a Hamiltonian whose eigenfunctions are completely known. Please be comfortable with the notations. So, whenever I am used to writing superscript 0, I remind you they are eigenstates of H0. They are completely known and these i's are different states. So, i equal to 0, 1, 2, 3, etc. In our case, when i equal to 0, what is size 00? That is Hartree-Fock web function. I just remind you when H0 is some of the Fock operator, size 00 was Hartree-Fock, but we have to now improve the Hartree-Fock. All determinants, MCM determinants are eigenstates of H0. So, they belong to this space of psi i0. I am writing the equation generally, but I am just throwing you back to Hartree-Fock perturbations that all singly excited, doubly excited, etc. determinants were eigenstates of H0. So, this is something that I fully know. I know complete eigenstates of H0 in a basis, but let us not worry about that right now. I have the entire spectrum, eigenvalue spectrum that is completely known. I am now going to evaluate size 01. When I evaluate size 01, which is the ground state correction, I first note the intermediate normalization. The intermediate normalization says that when K equal to 1, the size 01 are orthogonal to size 00. That means if I expand size 01 in terms of these states, which are complete set of functions, I must ensure that the size 01 does not include size 00 because it is orthogonal. I hope you realize what is orthogonal. The coefficient comes as an overlap. Since it is orthogonal, the coefficient must be 0. Obviously, if I expand size 01 and this is something that we will do, we will expand size 01 as a linear combination of all size 0K with a linear combination coefficient CK. I will call this CK1. Since all these CK0s are known, I can expand size 01 as a linear combination of all CK0 except K0 equal to 0. Why K0 equal to 0? Because I have to enforce the intermediate normalization. So because size 01 is orthogonal to size 00, I can expand except K0 equal to 0. Is it clear? So for example, when I do Hartree-Fock perturbation, my first order correction to the web function will not include Hartree-Fock itself but it will include all other excited determinants. So I am writing in a general way because it is supposed to be orthogonal to Hartree-Fock. That is the intermediate normalization means. So I first write this expansion in a general form. Again I am writing the perturbation first in a general form. Then we will put the Hartree-Fock perturbation theory. So what I have to evaluate now is this coefficient. So provided I know this coefficient, I know the first order correction because it is a linear combination of these determinants. So what are the coefficients? These coefficients, if you note, can be written formally as CK0, size 01. I hope all of you agree because these are all orthonormal states. So the way to get the coefficients is to project size 01 with one of the Psi K0. So if I project with Psi K0, I make an expansion. They will all become orthonormal except when the expansion includes K. Only for that term it will survive. So just to show you again, so if I have Psi K0, Psi 01, I expand the right hand side by a summation over L Psi L0 CL1. Note that this one means it is a first order correction. This K is of course this K. So I am just using instead of K a dummy variable L. So L not equal to 0. So it is very clear that these are orthonormal. There is a delta KL. So this result is only when L becomes equal to K, this survive, this becomes 1 and this becomes CK1. So this is again usual quantum mechanics that the CK1 can be given as Psi K0, Psi 01. So if I can find this quantity, then I know the coefficient of expansion and I know the entire Psi 01. K will be all the states of H0 except K not equal to 0. I am taking these are my states which are coming here. These are the eigenstates of H0. So all eigenstates of H0 have superscript 0 and subscript is K. So K0 means it is the ground state of H0. So I am using all states of H0 except the ground state of H0. So how did you want to write this? Psi K here that is Kth order correction which I do not know. What are the things that I know? I know these. What are available with me are all eigenstates of H0. All eigenstates of H0 are written in this manner. That is what we have decided and that is the simple. That is an assumption, intermediate normalization. That is an assumption. That is an assumption. I already said that the normalization is my choice and I explained in last class that if I do intermediate normalization, the full function will not be normalized but that I can renormalize later. Right now it is my choice. Is it clear? Any other question? So we want to evaluate this. I am going to go back to the first order equation and try to evaluate this. Remember initially I projected to Psi 0 0. The ground state of H0 only which is the Hartree form. Now what I am going to do to evaluate this, I am going to project to all other Psi K0. So for example I take this equation. I multiply by Psi K0 star and integrate. So what I will get? Psi K0 H0-E0 Psi 0 1 plus Psi K0. So V-E0 1 Psi 0 0 equal to 0. So I have taken this equation, first order perturbed equation and multiplied instead of Psi 0 0 star with Psi K0 star. So all other eigenstates of H0 which are known. So one by one I am going to do for each k. So remember when I did Psi 0 0 this guy became 0. But now if you see carefully it will not become 0 because Psi K0 is an eigenfunction of H0 but the eigenvalue is EK0 not E0 0. So this will not become 0. So can you find out what this will become? This will now become EK0-E0 0 yes E0 0 into Psi K0 Psi 0 1. I hope you can see this because this is an eigenfunction of H0 with an eigenvalue EK0 which is a real number. So when I take complex conjugate this Psi K0 becomes conjugate no doubt and this number is real so it just becomes EK0 Psi K0-E0 0 so that comes out and Psi K0 Psi 0 1 remains here. So I hope this small thing all of you are able to do and as you rightly said this is what this is my target quantity. This is Ck1. Ck1 is exactly the projection of the first order wave function to the kth state of H0. So that is my target quantity. So we will try to find this out but let me complete this equation. So then you have Psi K0 V Psi 0 0 I am breaking this up minus E0 1 Psi K0 Psi 0 0. Note that this I will do for all k not equal to 0 because I am not interested in k equal to 0 because my Psi 0 1 does not contain Psi 0 0. So I am only doing intermediate normalization. So this becomes my equation. Let me keep this here. So let me rewrite this equation again if you are not able to see. So first time projecting the first order equation is Psi K0 1 by 1. So Psi K0 which is k not equal to 0 then H0-E0 0 Psi 0 1 plus Psi K0 V-E0 1 Psi 0 0 0. So that is my projected equation. So I just rewritten that equation whatever I wrote here those were not able to see from this side. Then I said this is nothing but EK0-E0 0 Psi K0 Psi 0 1 plus Psi K0 V Psi 0 0 minus E0 1. So that becomes my equation. Now my target quantity is this. This quantity is 0 because remember k not equal to 0 they are orthonormal because all the Psi K0's are eigenstates of H0 and there since k is not equal to 0 they will be all orthogonal automatically to Psi 0 0. So this term really cancels out. So I can now find out what is CK1. Remember I said CK1 is only this quantity. So what is CK1 now? So CK1 will be Psi K0 V Psi 0 0 divided by EK0-E0. So let me analyze this term now because this into CK1 is equal to this quantity. So what we have essentially done is by projecting to the first order Schrodinger equation first with the Psi 0 0 star which is the ground state of H0 we got an expression of E0 1 and same thing we did for here we got E0 2. But now to calculate Psi 0 1 we know that Psi 0 1 can be written as a linear combination of all H0 states barring the ground state of H0 because we want Psi 0 1 to be orthogonal to that as a part of the intermediate normalization that Psi 0 0 should not be included in Psi 0 1. So we have expanded the Psi 0 1 in terms of all states of H0 except k not equal to 0. The reason we are doing is that these are the states which are available to me. You can of course argue that I could have expanded in any other complete set but these are convenient set which is available to me so I am expanding in this set. Then I notice that the CK1 is nothing but the overlap of this with Psi K0. So if I can formally get this then I am done. So to do that I went back to this equation and projected Psi K0 and showed this equation or this equation from where I actually obtained CK1 as Psi K0 0 V Psi 0 0 divided by this. this what we have now got. So CK1 equal to Psi K0 V Psi 0 0 divided by E 0 0 minus. So each of these coefficient for Psi K0 can be obtained in a very simple manner. So what is the prescription? Everything is superscript 0. If you look at the formula everything is known. So if I want to calculate what is the contribution of Psi K0 in the first order correction to the web function. First what I will do I will take this perturbation matrix element between H0 eigenstate ground state eigenstate with the kth eigenstate of H0. The kth one that I am trying to find out and divide by the corresponding energy denominator which is E 0 0 minus E K0 keeping E 0 0 as 1 from where it will be subtracted. So that is important to note. So this can be very easily calculated. Remember when I did E0 1 I had a similar matrix element but both are Psi 0 0 which are just an average value of E. Here I have actually a matrix element because the right and the left are different and in fact you know that if this is Hartree Fock these are determinants which are excited determinants like Psi A R Psi A B R S and all that will come and we can of course evaluate this by Slater rules. So that is something that we will do and divide by this energy difference all these energies are also known remember. We have already done this. This is of course some of the orbital energies these are some of the orbital energies plus minus the orbitals which are occupied in the kth state. So anyway that is immaterial for a general case anyway these are known because of the denominator. So what is asking is that this also tells you a very interesting perturbation theory that you already know. For example if I have a Hartree Fock then you expect the one determined one state whose energy is closest to the ground state of H 0 to contribute more because if this is less then this becomes more. So that is one of the reason they say the homo-lumo gap if it is large then there is a larger excitation to lumo. So that is a very approximate idea it just means that there is more contribution to that from that state. However there is a Brillouin's theorem that you have to remember. So certain things will not happen because of Brillouin's theorem. See I had to I could eliminate this. So it just simplifies. So remember this term this particular term. So anyway I mean these terms will further come when you go for second order third order and so on. There will be more and more such terms will come. That is one of the reasons that I because already Psi 0 0 is there I need not include it. But please note that normalization is anyway a choice I have been repeating. Finally my overall Psi 0 is a linear combination of all states because Hartree Fock is already included in Psi 0 0. So I need not include them in the correction term. So that is the physical reason why it is not necessary. But there is a mathematical reason which brings in simplicity. So that is what we are discussing. Physically anyway it is not required because I am only looking at correction. My overall web function includes Hartree Fock anyway. And I can always renormalize at the end. So this is also not a big problem. I can always renormalize. So the prescription to do a perturbation theory again I repeat that you first know all the H0 eigenstates. So you can do it for simple case even one particular case harmonic oscillator everywhere. So if you have harmonic oscillator, harmonic oscillator and which is perturbed quadratic function which is perturbed by cubic function let us say x cube term in the potential. You can use that as a perturbation. So how will you do it? Then all these Psi 0 0 and Psi 0 0 will be harmonic oscillator eigenfunctions. One of them will be ground state eigenfunction. These are excited states of harmonic oscillator. Take the V as x cube. I put up by x cube. So calculate the matrix element of x cube between the different states between the ground state of harmonic oscillator and all the excited states divide by the energy difference that will give you the coefficient for the first order correction to the energy. So if I give you a problem that there is a harmonic oscillator which is completely known one particle. So you have half kx square which is completely known and now my V of x is half kx square plus some lambda x cube near about the equilibrium. So let us say this is a very small perturbation. Then what you will do is that up to V of x and kinetic energy your entire solutions are known. So that becomes your H0 and this will become your V. So then what you can do is that you can apply this perturbation theory to first get the E0 1. For E0 1 it is very simple. Your harmonic oscillator ground state eigenfunction and take lambda x cube simply take the matrix element average value. For writing the first order correction they are going to be combination of all harmonic oscillator eigenfunctions just as I have done it here and each of these coefficients will be obtained by the matrix element of lambda x cube between ground state harmonic oscillator and excited states of harmonic oscillator. The harmonic oscillator means normal harmonic oscillator H0 that is what I have defined as H0 divided by the energy difference. So this is very simple. So every state you simply calculate matrix element and take the energy difference. So pictorially you can see the following way. This is my E0 0. This is Psi 0 0. Then I have got E1 0, Psi 1 0. These are known. Remember all 0s are known and so on. E2 0, Psi 2 0. So pictorially now I am trying to calculate Psi 0 1. I know that the Psi 0 1 should not contain these because this is already there in my Psi 0 0. So Psi 0 1 should contain by the intermediate normalization these two states. Let us say I have only three basis just to give an example. So how do I calculate the coefficient of each of the states in expansion? So this is important. So it means I just take the V whatever is the V between these two states first which is Psi 0 0, Psi 1 0. So that is your matrix element. Then divide by the energy denominator whatever is the denominator here. Then you take V between these two states divide by the denominator. If there are more states V from these three states and so on. Always from this state you go to one state up. Take the matrix element of V divide by the energy denominator. An energy denominator must come in this form that it is E 0 0 minus E k 0. So you will get the combination coefficient and once you get the coefficient you should be able to write Psi 0 1. So that is what we will do now. We will write Psi 0 1. So let me write Psi 0 1 now as a linear combination of all eigenstates of H 0 except Psi 0 0. So first you write sum over k0 equal to 0 and then write C k 1. So I now have the C k 1 which is Psi k 0 V Psi 0 0 divided by E 0 0 minus E k 0 into Psi k 0. So that is my coefficient multiplied by the H 0 eigenstates. So that becomes my Psi 0 1. Is it clear? So this is the number because this is the matrix element of V divided by the energy difference. Everything is known here so you have to calculate one by one and then multiply of course by Psi k 0. So that becomes your combination coefficients. So basically the whole purpose is I am trying to construct the first order perturbation correction to the wave function as a linear combination of what is already known. What is already known is the eigenstates of H 0. So I am trying to construct this function as a combination of the eigenstates of H 0 and I discovered that the combination coefficient for each of the k can be written in this form. So there I know everything about Psi 0 1. Of course it is a long expression because it is a linear combination but in principle I can calculate. In principle of course this summation over k will be limited by my knowledge of the eigenstates of H 0 because I told you if I have m basis then my knowledge is limited because I can get only mc and determinants. So of course that is there but other than that this is an exact expression. We are not yet doing Hartree-Fock perturbation here. Of course now I will do Hartree-Fock H 0 as Hartree-Fock and then see what are these Psi k's we will do that so there is still a lot to do. Having done this now I can go back to E 0 2. Remember my E 0 2 I had written down the expression Psi 0 0 V Psi 0 1. Do you remember? E 0 1 was Psi 0 0 V Psi 0 0. Any E 0 n I wrote Psi 0 0 V Psi 0 n minus 1. So for E 0 2 it is Psi 0 0 V Psi 0 1. So Psi 0 0 V Psi 0 1. All I need to do now is a trivial task is to expand Psi 0 1 as I have done here and push it here. So let us try to do that. Psi 0 1 has an expansion k so that I must write before. Again I hope all of you are comfortable in doing this expansions. Then I have Psi 0 0 V. Now I have to write Psi 0 1. So how will you write? You can write these coefficients later and bring this here first because the integration is going to take place with Psi k 0. This is a number C k 1. So I will first write that Psi k 0 and complete the integration and simply multiply by this number. Remember this is just a number. So I can multiply by this number or I can write the full thing Psi k 0 V Psi 0 0 divided by E 0 0 minus equation. It is just that I have written this in a different manner. This first, this later. I have switched it around because that is what that is immaterial. That is a number. This is a number I can go take it anywhere. And this looks very nice expression. I will just tell you why it looks so nice because you can see in between there is this Psi k 0 Psi k 0 sum over k. Note that if this sum over k was complete, this by itself would have been an identity operator. However, it cannot be written like this because you have E k 0 here. So I cannot take this out separately and in any case this is k not equal to 0. So do not interpret that as identity operator. That is the first thing I want to tell because that will be a mistake because the k is also here. I cannot sum just this. But it will be something which we will do. I think we have done that in 4 to 5 before. If you remember recall, so I am just repeating. So first to note however that this quantity is a conjugate of this quantity. It is a complex because V is a Hermitian operator. So if I take complex conjugate of this, the Psi k 0 will come on the left. V will remain as it is. Psi 0 0 will come on the right. So this is a conjugate of this which means I can further write this expression as sum over k not equal to 0. Psi 0 0 V Psi k 0 mod square divided by E 0 0 minus 0. This becomes my E 0 2. Do not forget E 0 1 however. E 0 1 was just Psi 0 0 V Psi 0 0. So this is my E 0 2 I am doing already. E 0 1 was very simple. Psi 0 0 V Psi 0 0. Instead of 1 it was 0. All of you agree here? I mean those who have done 4 to 5 is just a revision but I think it is worth doing because now we will draw the actual Hartree-Fog perturbation but it is worth doing. Note a very important point about this formulation. This denominator is always positive because it is mod square. So it does not matter what is the value of this. This into its complex conjugate is always positive. I hope all of you know any number times its conjugate is always positive. So a into s star it is always positive and the denominator is always negative because this is the lowest eigenstate of H 0. These are the higher eigenstates of H 0. k not equal to 0. And of course if k equal to 0 would have been there it would have been disaster. This would have been a singular infinity. So anyway I have removed that. So now you know why intermediate normalization is required further. So I have removed it for much simplification. So it is only k not equal to 0. So this quantity for all k each k is negative. Numerator is positive, denominator is negative. So it is always negative. So the entire quantity is always negative. It is a very interesting result that we get qualitatively. So what we are saying that I have no idea what is first order energy. First order correction to the ground state energy is psi 0 0 v psi 0 0. I cannot say if it is positive or negative. In the context of Hartree-Fock perturbation I had already told that I am least bothered because orbital energy some plus this was Hartree-Fock energy. Anyway so I am not bothered whether this is positive or negative. But the important thing is that after I do the correlation when I am going to do the correlation this will be the first correction. I will write what is psi k 0 in terms of determinants later but regardless of what it is it is always negative. So what does it mean? It means when I do a second order m p 2 correction the Hartree-Fock energy goes down which means the energy has improved or it has gone worse it has improved because originally Hartree-Fock was always greater than or equal to 0. However there is a question how much negative it might have gone down. See I mean that is the question that we are not bothered. So E 0 0 plus E 0 1 is of course Hartree-Fock energy and that is greater than equal to exact E 0 for sure. When I am doing when I am adding E 0 2 again this I am talking in terms of Hartree-Fock perturbation E 0 2 this is certainly less than or equal to E Hartree-Fock. So you may think that it is going towards E 0 which is generally right but it might be over corrected. You understand what is over correction. So it may actually go down further. So if this is E Hartree-Fock and this is E 0 then after I correct with E 0 2 it may even go down. It only says it is negative but there is no upper bound on this that this should be greater than or equal to E 0. There is no upper bound. There is an upper bound however on this. I hope you know why now. The reasons are different. This is Hartree-Fock energy. So up to this point I can write it as a Hamiltonian expectation value with respect to psi 0 0 because H 0 when I put H 0 here it is E 0 0 when I put V here it is E 0 1. At this 2 it can be written as a matrix element. It is a Hamiltonian expectation value with respect to some state psi 0 0. So that is always greater than or equal to E 0 2. But when I am adding E 0 2 I cannot write like this. So there is no Rayleigh-Ritz variation principle. So this result of upper bound I am getting because not because of the perturbation theory but because of my original Rayleigh-Ritz variation method because this can be written as an expectation value of the Hamiltonian. When I do the E 0 2 I have lost that variation method because it is no longer can be written as an expectation value of Hamiltonian because E 0 2 has very complicated terms. It is not expectation value. So I cannot write this as an expectation value. There is no Rayleigh-Ritz principle. I can only say that the E 0 2 is an upper bound theory but there is no upper bound theory. I hope it is clear so it can go down. The fact that E 0 1 had an upper bound property was not because of any reasons of perturbation. I have no idea of what is E 0 1 positive negative but because of the fact that added to this it generates me a matrix expectation value of Hamiltonian and that is always bound because of Rayleigh-Ritz principle. So that is really a variation principle. Let me erase and write down the formula again. So this is something that I know. So I am writing down everything now where E 0 0 is lower than E 1 0 etc. So they are ordered in this manner. This is for H 0. Then we want to calculate E 0 1 but before I do that let me write E 0 0 that is trivially psi 0 0 H 0 psi 0 0. So that is the first formula which trivially comes up. Then we write down the ground state first order correction to the energy. So that is psi 0 0 V psi 0 0 such that E 0 0 plus E 0 1 is psi 0 0 H psi 0 0 which when I do Hartree-Fock perturbation is my Hartree-Fock energy because this is Hartree-Fock by definition and by Rayleigh-Ritz principle this is greater than or equal to 0. Then the third formula that we just generated was E 0 2 is psi 0 0 sum over k not equal to 0 psi 0 0 V psi k 0 mod square divided by E 0 0 minus x and just E 0 2 itself is always negative just E 0 2 itself. So it will certainly depress the value from E 0 0 plus E 0 1 but how much I do not know normally of course when I do MP 2 I will come to that because I have to analyze what are psi k 0's then. Then we see that it does not go beyond exact energy but actually recovers about 60 to 70 percent of the entire correlation energy. So the difference 60 to 70 percent of the difference it actually recovers. So it does not go below it does not actually over correct so that is a good thing. So this will eventually become our formula for MP 2 and this is the first correction to the correlation energy. Of course to write MP 2 I have to now go back to Hartree-Fock perturbation where H 0 is some of the Fock operator write them down in terms of determinants that we will do tomorrow and write the final form of the Hartree-Fock perturbation. This is a more general perturbation theory that I have done. I would also mention what I did in the 4 to 5 plus that there is something called resolvent. I cannot take this as a projection operator but there is something called resolvent or not which is defined as sum over k not equal to 0 psi k 0 psi k 0. So this is like a projection operator divided by E 0 0 minus E k 0. So this sum over k everything summed up because k comes only here. If you allow this if you understand this resolvent then the entire E 0 2 can be written without the sum over k not 0 psi 0 0 V r not V psi 0 0 very simple. Note that our E 0 1 was psi 0 0 V psi 0 0. Now I have written E 0 2 also with psi 0 0 psi 0 0 except that instead of V it has become V r not V. So in a way you can see the similarity with the E 0 1 formula. Obviously there has to be 2 V because it is second order perturbation. So that also you should notice that first order I have 1 V second order I must have 2 V for the wave function also the same thing first order correction to the wave function had only 1 V in the numerator. It has 2 V but they do not come together. The 2 Vs are spaced by the resolvent that is the whole idea. So essentially what is happening pictorial is the following that I go from psi 0 0 through the V I go to 1 higher state of H 0 then again through V I come back to psi 0 0 divide by the energy denominator each time. So if you look at the picture the picture that I drew E 0 E 1 0 etc. I gave you a picture for psi 0 1 first order correction. Now I am giving a picture for E 0 2 what you do is that you go from this is psi 0 0 go from psi 0 0 through V to psi 1 0 then come back to psi 0 0 again. If you interpret these functions let us say K equal to 1 what you do psi 0 0 V psi psi 1 0 back to psi 0 0. So you are going to psi 1 0 come back to psi 0 0 each time you do that take energy denominator then you do the next one go here come back here take the energy denominator. So this is the pictorial way of looking at it while going here there is a matrix element well coming back there is a matrix element is just the complex conjugate when I come back it is like a scattering process. So I go here come back here and each time I take energy denominator keep doing for all states. So this is the way to remember so each time you go there is a matrix element each time you come back there is a matrix element. So you are always going from psi 0 0 coming back to psi 0 0 when you do E 0 1 you did not do anything like going back you started from psi 0 0 immediately was in psi 0 0 with V. So basically if I write E 0 1 in the same process all these were not required I started from here and come back without going anywhere. Here I am going up and then coming it is a ladder I am going up and coming back but eventually of course for all E 0 you have to start from psi 0 0 and come back to psi 0 0 remember that is important when I did psi 0 1 I told you it is a combination of various states that is for psi 0 1 but if I put that in E 0 2 remember this is a combination but this combination is such that when I expand this I come back to psi 0 0. So that is exactly what is happening because each of the coefficients has this number. So eventually I start from here come back to one of the psi k 0 and back to psi 0 0. So all energy correction in fact when I do E 0 3 E 0 4 everything will have the same structure and we will not do that but it will become more and more complicated because I have only one two scattering here quite clearly you can see if I have third order energy there has to be three scattering how to do that you may wonder how to do that I will show you one process where I am not going to do third order correction how do I do three tell me I go from one to here then I will go from here to here there is another V then come back straight because I have odd number now in fact that will be reflected in my expression. So all these expressions are much easily understood by diagrams in fact I am not immediately going to diagrammatic picture but I just want to convince you that what Feynman said the diagrams are extremely good pictorial representations of the algebra. The algebra is actually unnecessarily you get lost diagrams are very easy if I tell you this diagram you will be actually able to write the matrix element the first one will be psi 1 0 V psi 0 0 multiplied by psi 2 0 0 V psi 1 0 then psi 2 0 back to psi 0 0 or psi 0 0 V psi 2 0 I am coming back because there has to be only three for E 0 3 so actually all such processes together will give you a contribution to E 0 2 E 0 3 E 0 4. So pictorially the diagrammatically the perturbation theory can be very nicely understood in fact all these things that are done will be actually converted to diagrammatics and that became what is called diagrammatic perturbation theory or MBPT many body perturbation theory we will not do too much about it but I want to convince you that actually diagrams are much more easy way to pictorially fathom what is happening in the perturbation theory and understand the process okay if you understand that eventually for all energy I have to start from here and come back here by this scattering process and that is exactly what has happened through the resolvent I have a much easier way of writing I do not have to write the denominator the denominator is involved in the resolvent resolvent has resolvent does not have a property a projection operator I am again repeating so you may wonder what are the properties of resolvent we will discuss that if we do r naught square what will happen and so on yeah we can discuss those question and some of these we discussed in the previous class so that is an independently it is a good algebra alright I think today I will stop here and now actually put H naught as some of the fork operator and find out what are the Psi k 0's what will be the Psi k 0's and then calculate this when I do this I will require the final slatter rule that I told you that is the matrix element of Psi 0 0 Hartree-Fock with doubly excited determinant that will that will come in so tomorrow's class I will start with that I will start with that final slatter rule again like the first two slatter rules I will only simply write the expression we will understand the expression and then go back to this and try to complete it.