 If you look at this term and if you want to diagrammatically represent this term, let's forget about this. Just look at this term diagrammatically, then how this diagram would have looked like. Note that every energy diagrams are closed. So E01, I am looking at only the second term, E01 would have been a closed diagram. So somewhere it starts there and ends here, whatever it is, it has only going from what is E01? E01 is just size 00, V size 00. So I start from a dot and come back to the dot. I don't go anywhere. How many lines is there? Unimportant now. I just come back to the dot. So E01 is a closed line like this and then I have the rest of the diagram which is something like a second order term. Something like a second order term. If you expand this, again I will have to write this as A, B, R, S, R, S, A, B, all that is there. It's also W excited. N is W excited but then my rule will not allow because extra denominator is there. So I have to refine the rule. But this rule refinement is not necessary because this diagram does not survive. This particular part does not survive. And this is the famous content of the link cluster theorem. So what it says is that the part of this diagram, so if you expand this algebraically, this has also some terms which are unlinked just like this. So let me first expand this term. The second term is actually an unlinked term. I hope it's clear that this is unlinked because if you look at the algebra, whatever I write the algebra, this part of the diagram which is E naught 1 is disconnected, not linked with this part. So this is an unlinked diagram. So the second term is purely unlinked. The first term, if I analyze the first term which is this, the first term if I expand has also an unlinked term. In fact, this is the only part that I am not going to do, a statement I am making. So if I do an algebra, it will also have an unlinked term or which is basically now known as a term which does not scale with n proportional to n because all link terms scale with proportional to n. So unlinked term does not proportional to n. So this has a wrong n dependence plus a link term, I am just calling a link term. But let me define all unlinked term as a wrong n dependence, all link terms are right n dependence. So basically in terms of n dependence, I have a wrong n dependence here. If I expand this, I will have a term which is partly, a part of which is wrong n dependence and that part will completely cancel this part. So what can be shown is that these two unlinked terms cancel each other. So what this mean and this is the content of the Goldstone-Linck cluster theorem, what this means is that what survives is only the link part of this. So this has a link part, this is an unlinked part. The unlinked part comes with a plus sign which cancels this unlinked term. I am not going to show this, I am just making a statement and I have no derivation. This is the unlinked term because you can see that there is a E0 1 which is just multiplied by another quantity which goes from size 0 0 to size 0 0. So this part and this part are not joined if I draw a diagram. So that is why this concept of unlinked term but actually it means it has a wrong n dependence. Denominatory is anyway wrong. So I had to devise a new diagram rule for this if I had to but fortunately it does not survive. It finally does not survive so I do not have to devise. So the diagram rules that Hugenholz described are only for link diagrams. Actually if there is an unlinked diagram, the diagram rules have to change. So the content of this is that the first term has, first term itself has an unlinked term and a link term and the unlinked terms of the first term cancel this second term which is purely unlinked which has no link term leading to E0 3 which is only the link part of this. So your final E0 3 would become, the final E0 3 is just the link part of the first term. So for nm not equal to 0, psi 0 0 V psi 0 n psi n 0 psi n 0 V psi m 0 psi m 0 V psi 0 0 plus the denominator E0 0 minus Em0 into E0 0 minus Em0. I will now say that what survives is only the link part L. So this is not the full, what survives finally is not the full first term but only the link term of the first part. So what we diagrammatically represent is not actually the full term, diagrammatically we are representing only the link part because the unlinked part of this I need not worry because it cancels with this. It has, so that is something that I am not deriving, that is the part. Unlinked term essentially means I told you all energy, correlation energy must be proportional to n, like DCI has wrong n dependence, square root n. So all unlinked terms are basically wrong n dependence. So it is not exactly n to the power 1. So that is why they are called unlinked terms. So this is the diagrammatically I am calling it unlinked term but actually in terms of physics there is a part of this which is proportional to n, there is a part of which is not proportional to n, that part which has a wrong n dependence is exactly same as this part and they cancel each other. So leaving you only with the link part or the part which has a right n dependence. So whenever these diagrams have right n dependence, we call this linked, this is what I am calling. So the diagrams which have a right n dependence are actually represented by link diagrams. So all diagrams which have right n dependence. So these are all wrong n dependence term and there is a wrong n dependent part from here. Remember here it is an algebra. So I am not telling you it is linked or unlinked, it is an algebra. So if the algebra when I expand I will have an unlinked term. Here it is very easy to see the unlinked term because you have some number multiplied by something. So that is very easy to see, there it is not very easy to see. So if I expand the algebra I will get this and this unlinked term will cancel this leaving with me exactly same thing but I am now specifically writing capital L just to make sure that the entire first term is not programmed or diagrammed. So what I am diagrammatically representing is just the link part of this. No, no, no, capital L is linked, capital N, this is not basis or number of electrons, so as the size of the system increases. Yes, it is not only MP3, all MP at all orders are size extensive. So mylaplacid perturbation is a size extensive thing, DCI is not, it can be possible. MP2 is only negative, that is another issue. So very often it has oscillating behavior like this and then eventually it will convert somewhere. So there is no bound. So that is another issue. But all perturbation theory is size extensive because it has a right end dependence finally. You cannot see but eventually it cancels. Now this can be shown at third order by algebraic expression. The beauty of the Goldstone's theorem is that Goldstone showed this cancellation at every order. So basically this such a cancellation takes place at all nth order, MPn. So thus you can say that the perturbation theory is size extensive, which was not true for DCI. DCI was not size extensive, SDCI is also not size extensive, in fact we had elaborate discussion that any approximate CI is not size extensive, but perturbation theory at each order is size extensive, so that is a good point. So I just wanted to highlight that what we are programming is actually the linked part of the first term. The second part we do not even program, they just cancel. So in a way if you like the fourth order, fifth order, it is very easy to write. Keep writing like this, only the linked part, second term. So this analysis I have actually done in the text books. You will see that the unlinked term of this first term arises from n equal to m, yeah. I mean psi n 0 b psi n, I mean psi 0 0 b psi n. Psi 0 0, yes, that is an addition. n and m are? No, they cannot be 0, they cannot be 0. When n equal to m, see here when I summing, there is a sum over n, sum over n. When n equal to m, there is a part of it which cancels out with this. So this is, no, this is of course unless you do the algebra, it is difficult but it is there in Jabba Osloom text book and many other text book. In fact, Goldstone Shoha, no, middle term will not, it will survive, it will survive. But what I am trying to say, the rest of the terms will look like this, when m equal to n, except in the middle term. A part of n equal to m cancel, not all, part of n equal to m will cancel exactly with this term. And so that is why this is little bit more complicated, so we will not attempt to do this in this class, whether how already time is over. So I think we are just making a statement that let us not worry about how the algebra is derived but this part, so do not try to derive it, do not try to see it. So we just make a statement that the first part has an unlinked term, a wrong n dependent part which cancels the wrong n dependent part here and leaving you with only the link part of this. So diagrammatically what we have programmed is only the link part. And I will leave you with a fourth order diagram, so please practice. If I give you a fourth order diagram, you should be able to write the algebra. So this is a fourth order diagram, very nice diagram, it has also its symmetry. This is a fourth order diagram, so this is one v, another v, another v, another v, so again link diagram, so you can of course label it in many ways. So let us try to label it in some way and so let us say that I label this with the whole line and these two are particle lines. Then the rest can be labeled very easily, this must be whole line because it is going out and then again I have an option here, whole and particle and then this is going in, this is going out and then this must be going up, must be a whole line, yes. So this is one diagram, you can label it in many ways but let us say I give you this diagram, I hope you will be able to write the algebra now, so fourth order diagram. So count out the number of holes, number of loops, number of equivalent pairs and write down the algebra. So I can just tell you a diagram, third order diagrams are completely done, there is nothing left, so I am testing you with the fourth order diagram, so I give you a diagram. So you have to first, the first trick is to label them, so for example this is A, follow a label, labeling system so this is A, this is B, these are P, these are Q, many lines are there, then this is another hole C, this is R, this is D, this is E, remember they are not same number, so you have particle lines are only A, B, C, particle lines are P, Q, R, hole lines are 5, there are 5 hole lines, 3 particle lines, this particular diagram, it is an odd diagram, so then you have to sum over all the lines and then write down the rest of the factors, I hope you will be able to do that, it is a good practice problem, again it is there in textbooks, so there should not be any problem, I think it flows