 We conclude our course today with the discussion of a question that we had asked earlier that take ethanol we say that there is a certain kind of splitting of CH3 proton resonances because of the CH2 protons coupling with CH2 protons. Why is it that there is no splitting of CH2 resonance because of coupling of the CH2 protons? Is there no coupling? That is what we generally think right when we talk about NMR, I think it is even casually said that protons with same chemical shape do not couple with each other perhaps that statement has been made in some course or the other and that statement is not quite accurate. There is no reason why protons with same chemical shift should not couple with each other. After all the basis of coupling is that they are magnets. So why will they not couple with each other? So today let us see how we can perform a time independent perturbation theoretical treatment of coupling of protons with same chemical shift and whether that leads to any kind of splitting in the NMR spectrum of such systems. So what we are discussing today really is no splitting for A2 systems. To start our discussion as usual we are going to write the Hamiltonians and we are using perturbation theory. So our job is to first write the unperturbed Hamiltonian and then write the perturbation correction to the Hamiltonian. So can you help me with this? What is the unperturbed Hamiltonian going to be in case of A2 system? Let us remind ourselves what it was for the AX system. Yesterday we have discussed the AX system. What was the Hamiltonian for AX system? Minus gamma B0 1 minus sigma A IZA right and then what was the next term? Minus gamma B0 1 minus sigma X multiplied by IZX operator. In this case what is the difference from AX system? There we have two kinds of shielding constants right sigma A and sigma X here since it's A2 in both the terms we have to use sigma A, there is no sigma X right, if that is the case what I'll do is instead of writing IZA and IZX there are still two nuclei right. So I'll write IZ1 and IZ2 and I hope you don't have a problem if I write it like this. Do we agree that this is the unperturbed Hamiltonian? And by the way in case I have not said it earlier this entire discussion is from Macquarie and Simon's book. It is also there in Grebiel's book but I find Macquarie and Simon to be very easy to understand in this part so you can follow that book. Are we all okay with the unperturbed Hamiltonian right? H0 is minus gamma B0 into 1 minus sigma A multiplied by IZ1 plus IZ2 where 1 and 2 denote the two nuclei both of which are of A kind. What will be the perturbation term? First order correction to the Hamiltonian it will be like earlier HJ, this time I will not write JAX, I'll write HJAA divided by h cross square multiplied by what? I1.I2 is that right? So I might as well expand since we discussed it yesterday I'll write like this, IX1.IX2 plus IY1.IY2 plus IZ1.IZ2. This here is my first order correction to the Hamiltonian okay and of course now what is the next step? We only found out structure for a note. Now here these look like this right? So I am not going to go right to the point of the problem but the clothes are going to be like the dunwalli, for fun, right? So earbuds and everything. So my question here is what's left over is both l0 and l3 because it's already different detail and if you understand the equation I'm going, I think you guys unperturbed wave functions, yesterday I wrote Psi, so today let me write Phi, just to be in line with the aquarium Simonscope, there is no other reason, Phi 1 will be as usual, now I will write Alpha 1, Alpha 2, yesterday we had written Alpha A, Alpha X, right, so here I will write Alpha 1, Alpha 2, and let me write Phi 4 as Beta 1, Beta 2, what is Phi 2 and what is Phi 3, what was Phi 2 yesterday, for AX system what was Phi 2, Beta 1, Beta A Alpha X, and Phi 3, Alpha A Beta X, can I keep the same wave functions just substituting A and X by 1 and 2, is there a yes or no, yes, no, why, why do we take a linear combination, so we come back to something that we learned earlier, remember indistinguishability, yesterday we were talking about AX system, there was a difference between the two nuclears, right, they had, they experienced different kinds of fields, so it was, I could distinguish between A and X, now I cannot, and since I cannot, then we have to go back to our old friends the linear combinations of the wave function, okay, same thing that we did in electronic spectroscopy, it's just that there we were talking about electronic spins, here we are talking about nuclear spins, that's the only difference, so this is 1 by root 2, Alpha 1, Beta 2, minus Beta 1, Alpha 2, this one is 1 by root 2, Alpha 1, Beta 2, plus Beta 1, Alpha 2, all right, what is my job now, my job is to find the energies of the four levels denoted by the wave functions 5, 1, 5, 2, 5, 3 and 5, 4 and work out the transitions and when we do that, as you'll see something new will crop up that we did not discuss, that we did not need to discuss yesterday, so these two wave functions 5, 1 and 5, 4 are exactly the same as the wave functions that we handled yesterday, so there is no reason why their energies should be of any different form compared to what we had yesterday, agreed? Yesterday also, Psi 1 was Alpha 1, Alpha 2, right, Psi 4 was Beta 1, Beta 2, so whatever expression we got for their energies, we should get similar expressions here, you can do the math, it'll be the same thing, so maybe I'll just write that, where do I write, maybe I'll write here, I'll keep that part available, what will be the energy of Phi 1? What was the expression yesterday? Yes, minus H cross gamma B0 multiplied by, yes, 1 minus, what was that? Alpha A plus Alpha X divided by 2, isn't it? Sorry, why am I seeing Alpha, sorry, Sigma, sorry, my mistake, Sigma A minus Sigma X divided by 2, was that right? Today is there any difference between Sigma A and Sigma X, so what is Sigma A plus Sigma X divided by 2, Sigma A? So now I'll get 1 minus Sigma A, that's it, what else did we do? We wrote this as nu 0, actually H nu 0, when nu 0 is the Larmor frequency of a Bayer proton, so if you want to write it in terms of frequency, then this is going to correspond to frequency of A nu, yes, all right, what was the correction term? What was the first order correction? H, J, there it was AX, here it is AA divided by 4, okay, I'm not doing the math, please do it yourself, you'll come to the same result, the math is exactly similar to what we did for AX yesterday, with one less complicating factor there is no Sigma X, there is no JX, everything is A, there is no X, okay, so E1, I hope you can agree on that, what is E4? Maybe like yesterday we'll write the same thing, E4 will be just this thing with a plus sign, what about this, will this become minus? No, so both alpha alpha and beta beta states actually get destabilized as a result of coupling, so the second term we still have a plus sign, the only difference that we'll have as we saw yesterday is that in case of level 1 we'll have a minus sign here, there'll be a stabilization, in case of level 4 there will be destabilization and we'll have plus 1. Now let us see what is going to be the scenario for E2 and E3, E2 and E3 of course should be different from what we obtained for the AX case, right, because the wave functions are different, let's see what it is, so this is where I'll write the energies and this is where I'll work it out, so let's work out E2, 0, how do I get E2, 0? I make this Hamiltonian operate on phi 2, right, so what do you get then? Minus gamma B0, 1 minus sigma A, IZ1 plus IZ2, this whole thing operates on 1 by root 2 alpha 1 beta 2 minus beta 1 alpha 2, what do I get? What do I get? How many terms do I get? IZ1 operates on alpha 1 beta 2 and on beta 1 alpha 2, so 2 terms there, IZ2 operates on alpha 1 beta 2 and beta 1 alpha 2, 2 terms there, but one thing I hope you can understand is that since I'm using IZ it'll always be eigenvalue, isn't it? It'll always be an eigenvalue equation and eigenvalues are going to be either plus H cross by 2 or minus H cross by 2, so what I'll do is I'll write these eigenvalues here and I'll write the wave function here, 1 by root 2 is common, let's not bother about that, that will come anyway, okay. So what happens when IZ1 operates on alpha 1 beta 2, how much do I get? I get back alpha 1 beta 2 first of all, what is the eigenvalue? Yes, little louder, IZ1 operating on alpha 1 beta 2, so beta 2 is a constant as far as IZ1 is concerned and what do you get when you operate IZ1 on alpha 1? H cross by 2 alpha 1, right, so that's why I've written alpha 1 beta 2 there and the eigenvalue is H cross by 2. What happens when IZ1 operates on beta 1 alpha 2, minus H cross by 2 and I write this one here minus beta 1 alpha 2, is that okay? Next is IZ2 has to operate on alpha 1 beta 2 and minus beta 1 alpha 2, IZ2 operating on alpha 1 beta 2 what do you get? What is the function? I get back alpha 1 beta 2 and what is the eigenvalue? IZ2 operating on alpha 2 and beta 2, minus H cross by 2, excellent. Now when IZ2 operates on minus beta 1 alpha 2, I have minus beta 1 alpha 2, this I can close, what is the eigenvalue? IZ2 operating on beta 1 alpha 2 I've taken the minus there already, so plus H cross by 2, so what do I have here? I have 2 into the wave function, isn't it? So it is an eigenvalue equation, but what is the eigenvalue? 0, is that right? Eigenvalue is 0, is that surprising? First time I was surprised, why is it not surprising? I do so much of math and then get eigenvalue 0, why is it not surprising? I should be surprised. See earlier what did we have? You had alpha and beta, right? So stabilization of one, destabilization of one was not the same, so you had some net energy. In this case, one is alpha, one is beta, both are equivalent. So stabilization of one is exactly offset by the destabilization of the other, energy is 0, okay? So what we learn is that your energy of, well at least E0, let me write, of level 2 that is 0, okay? Akansha, I'm sorry, can you say that again? How did I get this? Let's do one term then, open which bracket? No, we did everything. So what do we have? If you open the bracket, you have I, I'll just write I1 and let me call this, I don't know, chi1 and chi2. So I'm going to have, this is I1, this is I2, I have I1 chi1 minus I1 chi2 plus I2 chi1 minus I2 chi2, is that right? And then let us look at one of these terms. What is I1 chi1? That is your, you can write like this, I'm not writing 1 by root 2 here, beta2, IZ1 alpha1, isn't it? Because IZ1 will not operate on beta2, beta2 will be a constant as far as it's concerned. What do I get from here? IZ1 alpha1 is h cross by 2 alpha1, so you get back h cross by 2 alpha1 beta2. See h cross by 2 alpha1 beta2, that is how we evaluated every term. Yes, each term is h cross by 2, but then it's not always h cross by 2. If it is beta wave function, it will be minus. So that's why in two cases it is plus h cross by 2, in two cases it is minus h cross by 2, that is why it cancels off. Are you satisfied? You understood, very good. She doesn't look very convinced. Which multiplication? So that is what I just explained. This part you're okay, right? So I have to do it term by term, isn't it? What is the first term? First term is IZ1 operating on alpha1 beta2. So let me explain in a different way now. So what I have is IZ1 operating on alpha1 beta2. That is the first term, you agree? What will be the second term? IZ1 operating on minus beta1 alpha2. So minus IZ1 operating on beta1 alpha2. What is the third term? IZ2 operating on alpha1 beta2. What is the last term? Minus IZ2 operating on beta1 alpha2. Now what I am saying is IZ1 operates on alpha1, but not on beta2. So beta2 is a constant as far as this operator is concerned. So you can take beta2 out. Here also, you can take alpha2 out. In this case, take alpha1 out. And in this case, take beta1 out. What is IZ1 alpha1? Yes. So that's where we get it. You're right. So let me, I think I made a mistake, fine. So let me write like this. You have to take into account 2. Then it falls in place, right? It will work out. Fine. So we get 0. Work it out for phi3. You will get 0 once again. So now, what is the picture that we get for the uncoupled scenario? This is your phi1. This is phi2 and phi3. This is phi4. What is the energy gap here? This is 0. This is your minus nu0 multiplied by 1 minus sigma a. What is that? It is 0. Well, then h multiplied by that. So essentially this is h nua, isn't it? If I just write in terms of frequencies, this is nua. What about this? This is 0. This is plus nu0 multiplied by 1 minus sigma a. That is once again, if you write in terms of frequency, this is nua. So in the uncoupled scenario, you expect to get one line at the resonance of a. Now, what is the correction term? Now, to do the correction term, again, we have to remember what we get. So tell me ix operating on alpha. What do I get? ix operating on alpha? Yes, h cross by 2 multiplied by beta. It's not an eigenvalue equation. Iy operating on alpha, what do I get? Yes, Iy operating on alpha. This is what we discussed yesterday. Ih cross by 2 multiplied by beta. Iz operating on alpha, that we know it is h cross by 2 alpha. That is the only eigenvalue equation. What happens when ix operates on beta? What do you get? Iy, this is something we discussed yesterday without derivation. This is what comes from the spin matrices. Since we do not really talk about, we do not have the scope to get into relativistic quantum mechanics and all that, we have to take this action. Iy operating on beta gives us, what is this one? Minus. So this is the thing to remember. Minus ih cross by 2 alpha. Iz operating on alpha, that of course we know is minus h cross by 2, sorry, why am I writing alpha? Iz operating on beta, that of course is minus h cross by 2 beta, alright. Now, so essentially we have to see what happens when this operator operates on things like alpha 1 beta 2 minus beta 1 alpha 2 or alpha 1 beta 2 plus beta 1 alpha 2. So to make our job easier at the end, let us do something. Let us see what is ix 1, ix 2 operating on alpha 1 beta 2. What is Iy 1, Iy 2 operating on alpha 1 beta 2 and so on and so forth. We will just write down those terms and then we will take it from there. So ix 1, ix 2 operating on alpha 1 beta 2, that will be ix 1 alpha 1 operated, multiplied by ix 2 alpha, you know, beta 2. What do I get? ix 1 alpha 1 is what? h cross by 2 beta 1 and ix 2 beta 2 is h cross by 2 alpha 1 or 2, alpha 2. So I get h cross square by 4 beta 2, alpha 2. Is that right? So this is what is going to happen throughout. As long as you are using x or y, 1 and 2 will get interchanged. As long as you use z, they will not get interchanged and it will always be h cross by 4, whether there is a plus sign or minus sign that we have to see. ix 1, ix 2, alpha 1 beta 2, that is done. What is your ix 1, ix 2, beta 1 alpha 2, what will it be? Now I think we can work it out mentally. ix 1, ix 2, beta 1 alpha 2, what will it be? h cross square by 4 alpha 1 beta 2. Is that right? Then iy 1, iy 2, alpha 1 beta 2, what will that be? Iy 1, Iy 2, alpha 1 beta 2. So first of all it will become beta 2, alpha 1. Then you have h cross square by 4 also. What is the sign? Plus, i into i is minus 1 and there is a minus 1 already. Oh, sorry, sorry, sorry. I changed the sequence as well as the indices, beta 1, alpha 2. What is iy 1, iy 2, alpha 1, alpha 2 operating on alpha 1, alpha 1 beta 2 have already done. So beta 1, alpha 2. What will it be? h cross square by 4 alpha 1 beta 2, plus or minus? Plus, iz 1, iz 2, alpha 1 beta 2. What is that? Minus h cross and this time there is no exchange, alpha 1 beta 2, h cross square by 4 alpha 1 beta 2. And what is iz 1, iz 2 operating on beta 1 alpha 2? iz 1, iz 2 operating on beta 1, alpha 2. What do I get here? Did I write the minus sign here? Minus h cross square by 4 beta 1 alpha 2. This is what alpha 1, alpha 2, alpha 1 beta 2. What will we need in order to work out the correction? Now, what is the expression? What is the expression for the first order correction to energy? First order correction is psi 0, first order correction to Hamiltonian, psi 0 integrated over all space. This is what we have learned from perturbation theory. So essentially what we will try to do is we will try to evaluate this h1 psi 0 and then let's multiply by psi 0 integrate over all space. What is h1 psi 0? Let us see. We are talking about phi 2. What is h first order phi 2? That will be hjAA. Now, what we have seen there is that no matter which term we use, we will always get either plus or minus h cross square by 4. So I can take h cross square by 4 outside the bracket. So that h cross square and this h cross square will cancel. And you are left with hjAA divided by 4. I'm trying to evaluate this. What happens when this first order correction to Hamiltonian operates on phi 2? Phi 2 is alpha 1 beta 2 minus beta 1 alpha 2. Now, take any of these operators, operate on any of these terms, you always get either plus or minus h cross square by 4. So I hope it's not very difficult to see that h cross square by 4 will be common outside the bracket. And the moment we do that, this h cross square, that h cross square will cancel. You will be left with hjAA divided by 4. Now that will be multiplied by whatever. So once again, let us try to do it like this. Here I'll write the numbers. Here I'll write the wave function. So ix1, ix2 operating on phi 2. 1 by root 2 is common anyway. So I'll write it here. ix1, ix2 operating on alpha 1 beta 2. What do I get? Alpha 1 beta 2. Why did I erase it? The first thing I erased. Is that right? So h cross square by 4 I have taken out. So here I get 1. So if I write it that way, again I'll land in the same problem, isn't it? So let's not try to do that. I've understood where the problem was. The problem was in my writing. 1 by root 2, okay. But I can take h cross square by 4 outside the bracket. There's no problem with that, isn't it? Sorry? Okay, let's see. We'll just do it. 1 by root 2 you can take. That's not a problem. It'll be 1 by 2. Fine. What are you saying? Right now I'm only operating. I'm not integrating yet. It is 1 by root 2. It'll become 1 by 2 when you write multiply also. Okay. So first I get beta 1 alpha 2. Then what do I get after that? Beta 1 alpha 2 I have written. Next is ix1, ix2 operates on minus beta 1 alpha 2. Wait, say which one you are operating. Iy1, Iy2 operates on alpha 1 beta 2. What do I get? You have plus or minus. We need agreement on that, plus or minus. Don't forget we have taken h cross square by 4 outside the bracket. Okay. ix1, ix2 operating on alpha 1 beta 2 has given us beta 1 alpha 2 multiplied by h cross square by 4. Then ix1, ix2 operating on beta 1 alpha 2 also gives us h cross square by 4 beta 1 alpha 2 but there's a minus sign here already. Right? That's why we have written minus alpha 1 beta 2. Next is iy1, iy2 operating on alpha 1 beta 2. What do I get? h cross square by 4 beta 1 alpha 2, plus or minus, plus, h cross square by 4 has gone out. So beta 1 alpha 2. Next one iy1, iy2 operates on beta 1 alpha 2. What do I get? Alpha 1 beta 2 plus sign or minus sign, minus. What is it? Beta 1 alpha 2? Alpha 1 beta 2. Then what is the next one? Iz1, Iz2. Right? Operates on alpha 1 beta 2. What do I get? Iz1, Iz2 operates on alpha 1 beta 2. Minus alpha 1 beta 2. Last, Iz1, Iz2 operates on beta 1 alpha 2. Minus but then there's another minus already. Right? Plus beta 1 alpha 2. Right? What do I have then? What do I have? Beta 1 alpha 2 plus beta 1 alpha 2 plus beta 1 alpha 2 minus alpha 1 beta 2 minus alpha 1 beta 2 minus alpha 1 beta 2. Right? What was the wave function? Alpha 1 beta 2 minus beta 1 alpha 2. Right? So this is how I'll write it. Minus phi h j a a divided by 4 into 1 by root 2 alpha 1 beta 2 minus beta 1 alpha 2. Is that right? And what is that wave function? 1 by root 2 alpha 1 beta 2 minus beta 1 alpha 2. It is phi 2. So I can write phi 2. So then, what is the first order correction to E2? You have to let it be the first order left multiply by phi 2. Integrate over all space. What do you get? Integral phi 2 star phi 2. What is it? That is 1. So minus phi h j a a by 4. So what we get is E2 is then 0 minus phi h j a a divided by 4. And that is a different result. From what we got yesterday for the second energy level, is it not? In that energy, yesterday what we saw is alpha alpha and beta beta were destabilized by h j x divided by 4 and alpha beta and beta alpha were stabilized by the same amount. That is not the case here. Alpha alpha and beta beta are still destabilized by h j a a by 4 this time. But what we see is alpha beta minus beta alpha is stabilized by phi h j a a by 4. So in fact from there itself, perhaps we can figure out what will be the first order correction to energy of phi 3. The total change in energy should be 0, isn't it? Stabilization has to be equal to destabilization. So when do you get 0? What will be the amount of stabilization of E3 is 0? It will be h j a by 4, isn't it? Because 2 into h j a by 4 plus h j a by 4 that is equal to 3 h j a by 4 that added to minus 3 h j a by 4 is equal to 0. But let us see if we get that. So this is the same exercise but this time I want to use phi 3 instead of phi 2. Can you tell me what is h first order operating on phi 3? Phi 3 is, well 1 by root 2 is there of course, alpha 1 beta 2 plus beta 1 alpha 2 this time. What will it be? I have written, whatever it is, it will be h, it will be minus 3. Suddenly I got distracted, h is okay. What will the first term be? Ix1, Ix1 is same thing, actually I should not have erased, I should have just changed the signs. But anyway I have erased now, cannot unerase, so just bear with me and tell me what the answers will be. Ix1, Ix1 operating on alpha 1 beta 2, yeah, alpha 1 beta 2 you will get beta 1 alpha 2, right? Plus or minus, beta 1 alpha 2. Then Ix1, Ix2 operating on beta 1 alpha 2, what will you get? Plus alpha 1 beta 2, right? Make a mistake, no, right? Next one what is it? Plus alpha 1 beta 2 or beta 1 alpha 2? Beta 1 alpha 2. Then Iy1, Iy2 operating on beta 1 alpha 2, what is it? Alpha 1 beta 2. Then let us do the last two, Iy1, Iy2 operating on alpha 1 beta 2, minus alpha 1 beta 2 and Iy1, Iy2 operating on beta 1 alpha 2, minus beta 1 alpha 2. What do you have now? What do you have now? So first of all this is plus, this is HJAA divided by 4, right? That is phi 3, right? So first order correction to the third level is HJAA divided by 4. So this is the energy, HJAA divided by 4, simple math but it is just that you have to do it. Now let me erase everything and draw the picture. What happened to this one? This got destabilized by an amount HJAA divided by 4, right? What happened to this? It was destabilized by the same amount HJAA divided by 4. What happened here? Phi 2 and phi 3 were degenerate, right? If you did not consider coupling, the energies were zero. Now phi 2 is stabilized and phi 3 is destabilized. Destabilization is by same amount, HJAA divided by 4. Stabilization is by oh man. Now I have parallaxed. Stabilization here is by HJAA divided by 4. So these are the four levels. Looking at this, we might think that we should still have splitting but let us not forget something. Alpha alpha, beta beta, alpha beta plus beta alpha, right? And this one is alpha beta minus beta alpha. Have we encountered these wave functions earlier? In electronic spectroscopy, yeah? So alpha beta minus beta alpha is unique, right? And the other three form a group. What is that group of alpha alpha, beta beta and alpha beta plus beta alpha? What is that called? Triplet. And what about alpha beta minus beta alpha singlet? And we have studied spin selection rule. In fact, we derived part of it. What is spin selection rule? You cannot have a singlet to triplet, triplet to singlet transition. So this transition is forbidden, okay? Phi 1 to phi 2 or phi 2 to phi 3 or phi 3, not phi 3, phi 2 to phi 4, these are all spin forbidden transitions, okay? And this is what we have learned. What about phi 1 to phi 4? Is phi 1 to phi 4 allowed? Phi 1 is alpha alpha, phi 4 is beta beta. Both are triplets. Are they allowed? Is that allowed? Why not? One photon rule. One photon rule, right? Both have to be flipped. That is not allowed. This is also forbidden by one photon rule. What are we left with? Which are the transitions that are allowed? 1 to 2 is not allowed. 1 to 3 is allowed. I have drawn this wrongly, right? This is also destabilized, sorry. So 1 to 3 is allowed? 2 to 3 allowed? No. 3 to 4 is allowed? What is the energy of this transition? You have this, right? This is E1 where this is minus. I am saying E1 to E2 is allowed. Sorry, E1 to E3 is allowed, right? What is the energy difference between E1 and E3? We have all the correct way, yeah? Sorry? What is the energy difference between E1 and E3? Sorry? H nu dot? H nu 0? Multiplied by 1 minus sigma A, and that's it. There's no signature of coupling, right? So H nu 0 into 1 minus sigma A is the energy gap that you expect in the unperturbed system, right? So what will be the frequency of transition here? Nu A. And what about 3 to 4? What about 3 to 4? What is E4 minus E3? Same thing, HJA by 4, HJA by 4 will cancel, and you will be left with H nu 0 multiplied by 1 minus sigma A. If I write it in terms of frequency of transition, again it will be nu A which is nu 0 into 1 minus sigma A. So what we learn is that it's not as if there is no coupling. Coupling is very much there. It is just that this additional spin forbiddenness comes in for A2 systems because of the indistinguishability of the protons, and that is why the number of allowable transitions goes down. The only transitions that are allowed are between two levels both of which have been destabilized to an equal extent, right? From here destabilization is by HJA by 4, from here also destabilization is by HJA by 4. So it is as if the two levels have just moved together, okay? Same is true for this and this, phi 3 and phi 4. The two levels have been destabilized by an equal amount. That is why you do not see the signature of coupling in the spectrum. It does not mean that coupling is not there, coupling is there. Of course you might say that you are making it up. You have just done some mathematical wizardry and you are saying coupling is there. The situation is equally explainable in terms of this diagram. Even if I do not invoke coupling, I get the right picture, right? So how do I know? How do I know that coupling is even required in this scenario? Well, we know it because we have only discussed the two extremes, right? Yesterday we discussed AX. AX means sigma A and sigma X are very far apart from each other. Today we have discussed A2. That means the sigma's are exactly the same. What perturbation theory cannot do is discuss situations of AB where sigma A and sigma B are not very far apart from each other. For that you have to use the variation method which you are not going to do in this course this time. But when you do a variation treatment, you see that when you go to AB, the situation is somewhat like this. So we are saying that for AX, this is a situation, right? Two lines for A, two lines for X. For A2, situation is this. You have one resonance. For AB, the spectrum that you get is somewhat like this. These two doublets move closer together and the ones inside gain intensity. The ones outside lose intensity. These are called second-order spectra. When sigma A minus sigma X is sufficiently large, much larger than the coupling constant J, then you get what you got called first-order spectra. We have discussed first-order spectra so far. We call it first-order spectra because you can explain those spectra using first-order perturbation theory. You call this second-order spectra because you have to go to higher-order perturbation theory if not a different level of theory altogether if you want to explain this kind of spectra. So you can imagine like this in the hypothetical situation where let us say you have a control over the sigmas. You start with the situation where the two sigmas are very different. You get doublets. Then by some magic, if you keep on decreasing the difference between the sigmas, the spectra start moving closer together, the inner ones become stronger, outer ones become weaker and finally the merge, the outer ones are vanished, the inner ones merge together to give you one line. Of course this is only a very qualitative hand-waving way of explaining the scenario. The correct way of understanding it is by using variation method which can take you all the way from A2 to AX. Variation method is also discussed in Macquarie and Simon. It is not in the syllabus this year but you are welcome to read it. It's not very difficult. You can understand if you read it yourself.