 So, I think let us once again revise the diagram rules. So, we were specifically talking about the Huygen-Holtz diagrams. So, the rule number 1, I will let me rewrite the rules little bit more clearly. So, we have each dot contributes to a matrix element, either 1 or 2 particle. So, basically 2 particle means 2 electron integral and if it is 2 particle then it is anti-symmetrized. So, for each of these either F or V, there is a dot. V essentially is of course again 2 particle minus 1 particle. So, 1 particle does not contribute. So, basically 2 particle. Then the second is that all these dots have to be joined. So, the question is that how many dots are there? So, the perturbation order, I can also write here that with 1 or 2 pairs of, so each pair consists of one incoming and one outgoing line. So, I have either 1 pair for 1 particle or 2 pair for 2 particle dots. And then energy diagrams are represented by closed diagrams involving these dots. So, I just join all the dots in a manner that they are completely closed. Then an imaginary line between the dots, let me also specify right here that this matrix element is a numerator. So, that these are constructed with sum over e of holes minus sum over e particles as the denominator. But the sum is the number of hole lines that this intersects, the imaginary line that intersects, the number of particles that this intersects. Is it clear? Then number 5 is if you have in the entire closed diagram, if you have a number of holes is h, number of loops is l, loops can very often be seen from the diagram or from the algebra. Then you attach a factor minus 1 to the power h l, h plus l. So, multiply by this factor. Then number 6 is 2 to the power minus k, where k is the number of equivalent pairs. So, what is an equivalent pair? It is either a pair of holes or a pair of particle starting from the same vertex and ending at the same vertex. So, starting from a vertex and ending at the another vertex, which is basically have to be all same. And of course, at the end it is important to write that sum over all holes and particles. When you write the algebra, we have to ensure that all internal holes and particles are summed up. So, we did some practice problem. So, one was the MP2 itself. So, we said this is MP2, sorry, this is MP2 in the Hugen holes, which is 1 by 4. This is A, this is B, B and Q, 1 by 4, PQ AB. So, the matrix element that this contributes is basically the pair of incoming on one side and outgoing in the other side. Or since they are all real, it does not matter. You can have outgoing here and incoming here. So, this is this. Then we did the third order diagrams. So, we saw there are three third order diagrams. So, these are three third order diagrams. One is a hole hole particle, particle ladder. This is a hole hole ladder. This is a hole particle ladder. So, when you draw, of course, you can write the expressions. If I ask you to write the algebraic expression for this, I hope all of you can write it. So, just I am asking you a question. How many, what will be the factor? 2 to the power minus, what is the factor here? 1 by 8. So, here also 1 by 8. Here there is no factor. There is no equivalent line. So, just plus, just 1 for the hole particle ladder. And then you have minus 1 to the power H plus O. What will be the sign here? Number of loops. The number of loops is the only thing that it will count properly. And when you write the numerator, actually from there also you can calculate the loops. As soon as you write the numerator, you can calculate the loops without looking at diagrams, whichever you are comfortable. So, how many loops are there? Depends on how you are calculating. Depends on how you write the numerator. Let's try to write one of them. That's easier. So, let's write this A. Yesterday I did not write the expression. A, B, P, Q, RS. The first one I am writing. So, these four are particle lines. We will put this loop rule little bit later. First let me write down the numerator. So, this is sum over A, B, P, Q, RS. 1 by 8. Then you have RS. Now it is a question of how you write it. RS anti-symmetry A, B or RS anti-symmetry B, A. You have two choices. But write it in some way. And accordingly loops will change. So, RS anti-symmetry A, B. So, you have to now remember A is going to R from the diagram. And then you have RS anti-symmetry P, Q. And P, Q, just one bit. P, Q anti-symmetry A, B. You can write any way you want. Now you divide these two imaginary lines. Remember, when I wrote that rule, please modify an imaginary line between two vertices, every pair of vertices in succession. Not every pair means not between 1 and 3, every dots in succession, every adjacent vertices. So, then you have epsilon A plus epsilon B minus epsilon R minus epsilon S divided by epsilon A plus epsilon B minus epsilon P minus epsilon Q. Up to this point, now we will count the loops. So, let us start from somewhere. Let us say A. I have to come back to A, A to R, R to P, P to A. So, one loop. Then I go from B, B to S, S to Q, Q to B. So, I have two loops. See, A to R is R to A I can do. That is unimportant. That does not matter. So, you can say A to P if you are saying A to P, P to R, R to A. So, let me change it to a BRS. Because it does not matter. It is symmetric. There is no convention as long as they are real orbitals anyway. It does not matter. They are identical. So, we are all talking of real orbitals here. You have to keep track. As long as the convention is same. So, I think that will now cause A to R, R to P, P to A. So, it does not matter. You should not get confused by this. So, the point is here you have two loops. So, the loops are very clear. That is one is this loop. So, if you look at diagram actually, one is this loop, goes here, comes back. So, that is one loop. The second is the other loop, goes from here, goes here, comes back. So, there are two distinct loops. So, this is second one. I can label with the blue line. So, exactly the same way I can do the whole-hole. So, this is the first loop, this, this one comes back here, back here. And the second loop is follow this blue line, come here and back here. So, again there are two loops. And of course, there are two whole lines. So, this is the plus sign. And I can draw exactly the same way the whole-hole, except the symbols will now change. I will have A, B, C, D and two particular lines. Let us try to have another diagram. So, you can draw this. Let us try to write the third one. The third one, I can make three loops. For the third one, for example, this can be a loop by itself. This can be a loop by itself. This can be a loop by itself. So, there are three loops. Each of these vertices, you can have a pair of lines can be a loop. But again, it depends on how you want to write it. If you interchange, then loops can be two. I can interchange one line. But if I follow this and that is the more ready convention, so A, B, C, P, Q, R. So, let me label this again. So, let us say this is R, this is A, no, sorry, this is P, this is A, this is Q, this is B, this is R, this is C. That is easy. So, I have three particular lines, three whole lines. So, look at the first vertex. Outgoing is C, A, anti-symmetrize, then R, P, then you have B, R, anti-symmetrize, Q, A, no, sorry, this is Q. The second one is B, R, then A, A, C, sorry. So, B is going out, R is going out, and then A is going in C, A, C. Then the last one is A is going out, P is going out, not A, Q, Q, Q, Q. Q is going out, P is going out, and then B comes in A comes in A. So, you can see one loop is A to P, P to A, and R to C, C to R, Q to B, B to Q, in the same way. This is Q, sorry. So, that second one should be also Q. So, I am drawing this diagram again here so that everybody can see. So, the labeling that I have done is PA, outer line, this is P, this is A, then you have Q and B. So, this is B, this is Q, this is R, this is C. So, that is the label we have done, and this algebra corresponds to this. So, now you have to see the sign. How many internal holes are there? How many loops are there? So, how many loops are there? Three. The way I have written, and of course, the internal hole lines are also three. So, it remains plus, and no 1 by 2 factor. There is no equivalent pair of lines. So, that becomes the algebra. So, if I give you a diagram, perturbation diagram, you should be able to write the algebra using anti-symmetry matrix. Any problem? Oh, numerator. I am always forgetting the numerator. Numerator is easy. Of course, you just write it down. So, you have epsilon A plus epsilon C, A and C, P and R. See, numerator, you do not have to bother. There is no order or anything. Just see the lines. It is very easy. And then you have epsilon A plus epsilon B minus epsilon P minus epsilon. You just have to keep track of your symbols. Yes, total energy is some. So, MP3 is this plus this plus this. So, that is why I have written plus, plus, plus. So, up to third order, the diagrams, if you look at the third order diagram, they are also very similar to this diagram. And that is why they are called ladder diagrams. And there is no other possibility of writing any more lines. In fact, it can be seen that such diagrams actually come because of W-excited determinants that MP3 also contains only W-excited determinants. I said that even in the first quantized, if you look at the third order energy, it is Psi 0 0 V Psi 0 K, then Psi 0 K, sorry Psi 0 K, sorry Psi K 0, Psi K 0 V Psi L 0 Psi L 0 V Psi 0 0. So, this is basically one of the major, and then of course, you have a denominator. So, here you can see that you have only Hartree-Fock to doubles. And of course, this K must also be doubles because I have to come back to Hartree-Fock. So, you have doubles to doubles and back to Hartree-Fock. So, you have no other terms. Now, you should be actually able to see the first quantized MP3 diagrams. They are not the algebraic expression. You will realize that the actual expression is not so simple. The actual expression is a combination of two expressions out of which a part of first expression cancels the second expression. And we have only these particular terms which in diagrammatic notations are basically just the unlinked diagrams. And that was the basic idea of what we said is that the link cluster expansion. All parts of the diagram must be closed and linked. So, if we refine that H 1, all diagrams, one of the rules that we wrote, all diagrams are not only closed, all energy diagrams, but also they must be linked. So, this is something that I thought I would not bother so much, but I can at least show algebraically where the problem comes up. So, you can go back to the algebraic expression of perturbation theory and we can write E 0 3. So, let us start from E 0 3. I hope you remember how to write E 0 3. Psi 0 0 V Psi 0 2. So, if you remember E 0 3 expression, it was Psi 0 0 V Psi 0 2, second order in the intermediate normalization, second order. So, we have to now get what is Psi 0 2. So, how do I get Psi 0 2? I go back to my second order equation H 0 minus E 0 0 Psi 0 2 plus V minus E 0 1 Psi 0 1 minus E 0 2 Psi 0 0 equal to 0. What we need to do is to find out what are the expansion of Psi 0 2 in terms of Psi 0 k's. Because remember, exactly like first order, we will write Psi 0 2 as a linear combination of all k not equal to 0, which I now call C k 2 Psi 0 k. The same expression that I use for the first order except that the coefficients will now be called C k 2. So, I have to find out the C k 2. I have to find out the C k 2. So, a specific C n 2 is nothing but Psi 0 n Psi 0 2. So, this is the matrix element that I have to find out. If I know this matrix element, then I know the entire expansion of Psi 0 2. I will put it here and I will start to get E 0 3. So, this is also Psi n 0. So, let us now find out this what is the value. So, what do I do? I start with this expansion and project with a Psi n 0. So, if I project with Psi n 0, I will get E n 0 minus E 0 0. So, I will get first E n 0 minus E 0 0 into Psi n 0 Psi 0 2. This is what I want to find out. So, that is what I get from the first expression. I am projecting with Psi n 0 from the left. Then I have Psi n 0 v. I want to write this as two terms Psi n 0 v Psi 0 1 minus E 0 1 Psi n 0 Psi 0 1. This is not 0 because this is the first order. So, this is the expansion that I get. So, I write this as this is bracket. So, I have Psi n 0 Psi 0 2 equal to, I will have two terms. One is Psi n 0 v Psi 0 1 minus E 0 1 Psi n 0 Psi 0 1 divided by the denominator changed. Instead of changing sign here, I have just changed the denominator sign. Is it alright? So, now you can see that when I put this back here, I actually have two terms. One coming from here, one coming from here. So, if I explicitly write the E 0 3, I will first get Psi 0 0 v. And now I can expand this by putting an n here or this is basically the R, the solvent kind of thing. So, I can expand this by writing this as sum over n Psi n 0 and then C n 2, which is basically this term that I wrote. So, this would be now Psi n 0 v Psi 0 1 minus E 0 1 Psi n 0 Psi 0 1 divided by, it looks big but it is actually not big. What I have done here only, because I have to write Psi 0 2, I have written Psi 0 2 as sum over n not equal to 0, Psi n 0 and this coefficient. This coefficient is just here. So, I have just put this whole thing here including the denominator. So, it is really not big. So, just the denominator I have written here. And then you can see that I have two expressions. One is this, another is this. Each of the expressions have Psi 0 1, which I now know how to expand and I will put that Psi 0 1 expansion in each case and then I will complete the algebra. So, let us write the Psi 0 1. So, what is Psi 0 1? This I already know. So, Psi 0 1 is sum over some m. I am using a different dummy index from n, sum over m against C m 1 but now I know what is C m 1. So, can somebody tell me again Psi m 0 v Psi 0 0 Psi m 0. I can write Psi m 0 here divided by the denominator E 0 0 minus E 0 0. So, all I have to do is in this expression of Psi 0 1, I have to substitute that. So, you can see now there will be two double summation and because Psi 0 1 also has a denominator, there will be two denominators. So, if you are just little bit careful with the algebra, you should be able to write the entire expression now. So, then your E 0 3 becomes, let us do it carefully, sum over m, sum over m both not equal to 0. Sometimes you just write prime here to indicate that 0 is excluded. You have Psi 0 0 v Psi m 0. That is the first term. Now I am multiplying this Psi m 0 v and now I have to bring Psi 0 1. So, I do an expansion of m and then I write Psi m 0, close this. It has first a ket vector so with which this will be closed and then another factor will come Psi m 0 v Psi 0 0 divided by E 0 0 minus E n 0 which is already there and one more which is coming from here E 0 0 minus E n 0. So, you have a double summation, but the story is not complete because I have a third term, I have a second term which is E 0 1 which just comes out, which is E 0 1 which can come out and then product of this with this. So, E 0 1 times again there is a summation, remember E 0 1 is a number so I can simply take outside. E 0 1 is just a number so I have a sum over m and then I have Psi 0 0 v Psi m 0 which is the first term multiplied by this into Psi m 0 and now I write Psi 0 1 which is Psi m 0 Psi m 0 v Psi 0 0 divided by again these two denominators. So, it is very similar expression except that v is not here, it is directly Psi m 0 Psi 0 1, yes, we will come to that orthogonality, I am coming to that now, so these are the two expressions. You can see here that the expression becomes simplified for the second term because this is now delta n m. So, n must be equal to m, so although there is a double summation it actually becomes single summation, remember here it did not happen because there is a v, otherwise the expression exactly identical is very easy to write, take out the v from here, instead of that an E 0 1 has come. So, if you look at the dimension it is same, you have 3 v here, 1, 2, 3, third order, here you have 1 v and 2 v but the third is this E 0 1. So, if you look at dimension in terms of perturbation order it is still 3, instead of v coming here it has gone out. So, this is of course this can be taken out so n is now equal to m, so I can put this as Psi n 0 and the summation over m can be knocked out and I can call this also E 0 square, right, okay, you write, now it is very easy just write it, just put n equal to m, Psi 0 0 v Psi n 0, Psi n 0 v Psi 0, it is basically square of this just like second order energy but denominator is not second order, like second order denominator is a square, otherwise what is multiplied with E 0 1 looks like exactly E 0 2 except for this denominator, otherwise E 0 2 was exactly same, Psi 0 0 v Psi n 0 or Psi k 0 back to Psi 0 0, if you see this Psi 0 0 v Psi n 0 back to Psi 0 0 divide by the denominator but now you have a square denominator that is only difference, otherwise I would have written as E 0 1 into E 0 2, it would have been even simplified which is also dimensionally third order but actually it is not, this remains.