 Without further discussion, I will tell the rules of the perturbation order. If you want to write MP3, there is a very important theorem called the linked cluster expansion. Again, without proof, I will just state the theorem that all MPn energy diagrams must contain n vertices of V perturbation V. Of course, in this case, we know that the V Hartree-Fock does not take. So, I must join and join such that this form linked and closed diagrams, I think it is important to understand these two terms linked and closed. It is easy to see that this is closed. Closed means there should not be any line, any hands not shaken. So, if I have four people, there must be another four people to hold their hands. Now, if somebody odd person out whose hands are not shaken, nobody has held the hand, that is not a closed diagram. That is an open diagram. One hand is open. You understand? Here, of course, it is closed. It is linked. Link means no part of the diagram should be such that it is not connected to the rest of the diagram by at least one line. Here you can see it is not the case. There is no part of the diagram which is left alone, but such a case will be there. In fact, I can show you one case, a very simple case in MP4. Let us say I am trying to draw a diagram of MP4. I am directly going to fourth order just to show this. So I will have four V lines and let me just consider right now two particle operators. Let us not worry about one particle operator. So I can draw a diagram like this. I can draw another diagram like this. Remember, there are four Vs, V, V, V, V. This is closed diagram, all hands shaken. Yet it is not linked. Why? Because this part of the diagram is nothing to do with the rest of the diagram. So such a diagram is not allowed by the link cluster expansion. So if I want to draw MP4 diagram, such a diagram I cannot draw. There must be at least one line connecting here with here. This is an unlinked diagram, unlinked but closed. The theorem, the link cluster expansion theorem that Goldstone first proved is that this must contain n vertices and then should be joined such that they form linked and closed diagram, the energy diagrams. And then once you form these diagrams, apply the rules. All the rules that I said minus 1 to the power h plus l, 1 by 2 to the power number of loops, all that you, equivalent lines, all that you apply after you draw the diagram. So if you see the MP2, there is only one diagram possible. So we can now quickly go to MP3. I have three vertices. Now start to think diagrammatically. Think of joining. If I ask you to join the MP3, can you do this? How many ways you can do this? That is the question. Remember you cannot do this because this guy is in trouble. This is unlinked open. This is terrible, not even closed because these four hands are not shaken. So you cannot do this. So let us think again. This is really diagram drawing at this stage. Draw diagram such that they are linked and closed. Yes, good, good. This, this, this and remember the rules. Every pair of vertices must have an income, two incoming and two outgoing for the one by Rij. And actually I will just show later, maybe tomorrow that a one particle operator will not exist even here, just like in MP2, it will not exist even here. That I will show algebraically, maybe tomorrow. So incidentally tomorrow there is a class at 5 o'clock. So this is one diagram but let us level it. Now while leveling, there are many possibilities. For example, both these lines can be holds from here. If these two are outgoing from this vertex, then these two must be incoming. I have no choice. Now I have no choice, correct, because this must have two incoming and two outgoing. Now look at this vertex. This has already got two outgoing. This must be two incoming. I have no choice. And now the third vertex is automatically satisfied, is beautifully satisfied. If you are not satisfied, something is wrong. So this is A, this is B, this is P, this is Q, this is R, this is S. There are four particle lines, two whole lines. I can draw something else, just the opposite, yes, absolutely, why not? So this is particle, these two are holds. So this is B, this is Q, this is A, this is B, this is C, this is D. I am not through yet, there is one more diagram, yes, where this can be whole, this can be particle. And if you do this, automatically jigsaw puzzle will set it, no problem, the rest will automatically fall in place. You do not have to do, you do not have much option. Now it does not matter whether this is outgoing or this is incoming, that will take care in the sign, in the loops. It might not be the same, overall creation annihilation is same, not whole particle, I never said whole particle is the same, no, no, no, not overall diagram, each one. This is two annihilation, two creation, this is two annihilation, this is two annihilation, two creation, every vertex has two creation, two annihilation, number of creation annihilation operator must be identical, not whole chain particle, no, no, no, that is not the rule. The rule is number of creation and number of annihilation operators must be same for every diagram. See, whole chain particle is because what is creation and what is annihilation, whole chain particles may change, your terminology. So I am talking of A, in terms of A, A dagger, number is same, you look at the count the number of creation, number of annihilation, it is identical. So these are the three diagrams which I can draw, three possibilities, that is nothing else. Can you draw any other skeleton, let me call this a skeleton, the original thing is a skeleton, within the skeleton are three, any other skeleton is possible? I mean you could think for example, you have to be, so maybe three lines here, maybe one line here, but then I am in trouble, I am in trouble, I am in trouble because this guy is orphan, whatever, correct, so this is not possible because every vertex has to have four lines, two particle vertex, if this was a one particle operator it is possible. Now remember I have to be very careful because my original perturbation v has both two particle and one particle, tomorrow I will show you that just like mp2, mp3 does not have v heart refog, that is v is only 1 by r i j, but later on that v heart refog may come, then you have to be careful, right. It can be 1 by r, because originally your v is 1 by r i j minus v heart refog, this is the one particle operator, see this could have been v heart reform, but in mp3 such is not possible, so right now for mp3 we have only these three diagrams and you can actually write the algebra, till we add, after this I will just sum, sum, sum, so for every diagram I will write the rule and I will just sum them, my final energy is sum of these three, so another diagram will come because of one particle, I will add it, all these diagrams will be added, for each diagram I gave you the rule, for second order I have only one diagram, so there is no add, so here I will write a rule for this, rule for this, rule for this, my final energy will be sum of this plus this plus this, okay, no addition I cannot combine, these are three separate diagrams, so this particular diagram is often called whole-whole ladder, there is a nice name in physics, I hope you can understand that these diagrams are called ladder diagrams, so you start from here step in step and this is a whole-whole ladder because the ladders are all whole, these four lines are very important, this is called the particle-particle ladder, right, PP ladder and this is called whole-particle ladder, HP ladder because the ladder itself has one-whole-one particle, so this can be now A, this can be R, this can be B, this can be S, this can be C, this can be T, so I have to bring in three-whole and three-particle line and I can now apply my rules to write the expression for each of them and sum, that will become your MP3 energy, you know if you have to do by algebra I have to take long time I can tell you, first of all apply the Goldstone's lean cluster expansion, that theorem and then simply sum, unfortunately I do not have time to prove the theorem and so on, so I am just stating how to quickly do it, you can do MP4, MP4, MP5, whatever but at least diagrammatically is much easier, actually after some time you should be able to think in terms of diagram, thinking in terms of algebra is not correct, you have to think in terms of diagram right away, so can you write the algebra for this, at least one diagram, can you try, if I give you one, one diagram I hope you will write the algebra, rules, very quickly write the expressions, so sum over everything, whatever is there I will sum, what is this vertex, RS anti-symmetrized AB, sorry AB anti-symmetrized RS, then second one RS anti-symmetrized PQ and the final one PQ anti-symmetrized AB, that is the first thing, just write the very easy, now look at, now all you have to do is the loop rule and number of equivalent pairs, loop rule is very easy, do not even go through this, just look at the algebra what you have written, A goes to R, I have to come back to A, R goes to P, P goes to A, one loop is over, every line can be traversed only once, B goes to S, S goes to S, goes to Q, Q goes to B, another loop is over, so I am only two loops, all lines are now consumed, is it clear, I have two loops, it is basically like this, one loop is like this, this, this, another loop is like this, this, this, diagrammatically if you see, I have two loops, so I have minus 1 to the power 2 for the loop and number of internal holes, 2, so what is my formula plus, minus 1 to the power 2 plus 2, remember my holes lines are minus 1 to the power h plus l, number of internal holes plus number of loops, so I have minus 1 to the power 2 plus 2, so plus sign, as it stands, if you change it, your loops will change, automatically anti-symmetry will be taken care, so you do not have to worry, then equivalent pair line, how many equivalent pairs are there? Three, this is one equivalent pair, AB, this is another equivalent pair, this is another equivalent pair, so you have plus 1 by 8, so diagram is over, now you go to the next diagram which is also exactly same, 1 by 8, only the matrix element will change, go here, finish, your MP3 energy is over, is that easy actually and denominator of course, you have to write, denominator is very easy, so now denominator you write here and here, so one is epsilon A plus epsilon B minus epsilon R minus epsilon S, there are two denominators, another here which is again epsilon A plus epsilon B minus epsilon P minus epsilon, is very easy, two denominators just some product, but I think it is important to know the algebra because that is what the theory is coming, diagrams are later transcription of the algebra, in a simple way I am translating the algebra actually, so I think it is better to first learn the algebra and the theory, so this is the diagram, this is the algebra for this particular diagram and then when you write program, you can program this, once you have algebra and I just wanted to give you a flavor of diagrams, I mean this course is of course, it is completely you know hopelessly incomplete, but I just thought I will tell you that how beautiful they are and they actually depict the physics very nicely, what is happening they depict the physics very nicely, in fact if you look at the first quantize MP3 that is what we will do later, you will see why there is no V-Hertree fork in this V and what diagrams are there, because I am going from Hartree fork through one V to a W excited, this is the W excited, from one W excited to another W excited back to Hartree fork, so if you look at MP3 actual expression psi 0 0 V psi doubles, psi doubles to another psi doubles back to psi Hartree fork, now you see there cannot be a singles, because both sides there are Hartree fork to something and I will write this expression and because of Brillouin's theorem, they must go to only doubles, so the doubles must be connected to doubles in the middle, so if I write E03, you will realize now the beauty of this diagram, what it is actually showing, so if you look at E03, in fact we can have a very nice discussion on E03 itself, that how the link cluster expansion comes and all that, unfortunately it is a good discussion, I just have run out of time unfortunately, when I think the course should start, I have run out of time, but that is what happens always, but anyway, so the point is finally what you get is the following psi 0 0 V psi k 0 psi k 0 V psi l 0 psi l 0 V psi 0 0, I have 3 V's divided by whatever is the denominator, I will not bother right now. Now you see the beauty, since I am starting from Hartree fork, this must be doubles, correct, this side also I am starting from Hartree fork, so this must also be doubles, remember sum over kl and of course kl not equal to 0, so that is the intermediate normalization, so both are doubles, so inside also must be doubles to doubles, this expression can actually be derived from the original perturbation expansion, there is a term of this which cancels another term and something that I do not want to show, that is the whole idea of size consistency and the link cluster theorem, so what survives is only these diagrams and you can see that I do not have a possibility of going to singles, because of Brillouin's theorem, because both the sides are locked by doubles, then I can only go from doubles to doubles and this is what the diagram actually shows, that this is a double, this is a double, this is a, I hope you can figuratively think it is a doubles, because you have four lines, so it is actually doubles, it is going to PQAB, ABCD, CDPQ, so I can actually write down the doubles also for each diagram, but there are three possibilities, so you have to sum this plus this plus this, which will change when I come to E04, E04 what will happen, this will be doubles, but there will be additional term, so now doubles can go to singles or triples, come back to doubles and then back to Hartree-Fock, it is a scattering process, so if I start from Hartree-Fock, I can go to doubles, through doubles here and come back to Hartree-Fock here through doubles, this is third order, but if I have fourth order, I can have a singles here, from singles I can go to doubles, okay not singles here, doubles here, from doubles I can go to singles, back to doubles, back to Hartree-Fock, that is possible, that chain is possible, this is psi naught to doubles, from doubles I can go to a singles, this is singles, let us say this is singles, this is doubles, back to Hartree-Fock, this is easy to mark, so from Hartree-Fock I am going to doubles, so no violation of Brillouin's theorem, doubles to singles is allowed, singles to doubles is allowed, back to Hartree-Fock, no violation, so at the fourth order onwards singles will start to come, singly excited determined, that also I can see from diagrams, even triply excited determinant, they will also come, because I can go from doubles to triples, back to doubles, even quadruples will start to come, because from doubles I can go to quadruples, it is a two occupation difference, come back to doubles, back to Hartree-Fock, of course I cannot have a pen to pull, because of this later rule, doubles cannot be connected to pen to pull, so at the fourth order I have in terms of CI language, I have contribution from singles, triples, quadruples, singly excited, triply excited, quadruple excited, apart from doubly excited, and that is from the algebra that you can see, you will see MP4 has singles, triples, quadruples, there is a meaning of that, in fact there is MP4 calculation in which they will show results, MP4 only doubles, then MP4 doubles plus singles, triples, quadruples, how much is the difference, that is very important to note, because I know doubles is most important, if I do the correlation as a contribution only with doubles, which is like this ladder diagrams, on singles, triples, quadruples, what is the difference, this is important to know, because single, triples, quadruples are very small, in fact you can without any, you know looking at the table, you can say that this difference will be very, very small, that is clear, because they are negligible, and your doubles is the most important. So there is a contribution of course of doubles here, so this can be doubles also, as well as singles, what I mean is that doubles to doubles of course is allowed, as well as single, triples, quadruples, yes, psi, a, b, r, psi, a, r with, one, yes, in the same way, the same way, the two particle terms, they will all come, well, these lines can be same, when I am summing over b and c, b can be equal to c, see, this is summed over all b, all virtual, all occupied, this c is also summed over all occupied, okay, so there it will be same line, one line, one line, so there will be b, when you draw the diagram in c, only one line will come, because there is a one particle and there is a two particle, so we will try to draw, so the point that I am trying to show is that, if I have mp4 from only doubles, it is very easy to draw, that is Hartree Fog going to doubles, doubles going to doubles, doubles, Hartree Fog, that is always a possibility, at all order that is a possibility, right, in addition you have doubles going to single staples quadruples, so if you want to do mp4 with only doubles, which are basically called the ladder diagrams, then exactly do like this, exactly extend it and all three diagrams like this, so these are called ladder diagrams, at all orders of perturbation, so ladder diagrams come, so you have n things and keep on extending one by one and two-two line in between, so this is basically Hartree Fog to doubles, doubles, Hartree Fog, so you can always write these at all order, mp5 and you have whole, whole, particle, particle, whole particle, exactly identical, okay, and it is very interesting because this series itself, if you sum, you get a very interesting result and that will be actually a couple cluster doubles linearized version, it is called LCCD, you know, I will not bother about it, but these ladder diagram themselves are very important, mp4 you should be able to draw this, I hope you can draw this, three, three lines you can draw like this, now can you complete it, one line here, so here itself, so this is one line, now you can see this is, this is a very asymmetric diagram, this is basically all four lines, anyway all four lines, but from the very nature you can see that this is not like the ladder diagrams, so I have three lines here, one line here, but every vertex has four lines, two incoming, two outgoing, so this will come from the singly excited, there are triply excited, all those things we will actually bring in, okay, because everything that I construct will be eventually a two-particle integral, if I adjust v, okay, and so on, you can actually start constructing other diagram, but this is actually easiest diagram to think of, three line here, three line here, one line here, okay, in between, and now you see when I do a line here, the denominator will have only two, not four lines, here, here there will be four lines, so this is a clear thing that is coming from a singly excited determinant, okay, so if you write the diagram, anyway we'll do that.