 So, we have been discussing about the product operator formalism of the density operator which we wrote like this summation Bs of T Bs and then we looked at the various basis operators these ones for one spin we had a set of basis operators there were 4 for 2 spins the basis operators number were 16 for 3 spins then we have a number is 64 this is 4 to the power whatever that is so this is 4 to the power 1, 4 to the power 2, 4 to the power 3 and we also had said that these are all products of angular momentum operators and we also saw what these individual operators represent in the density matrix where do they represent we have various kinds of single operators and we also have operators of like this which are products of 2 angular momentum operators and we will also have products of 3 angular momentum operators ILX, ILZ, IMZ and so on so forth. So, these are some illustrations and we saw exactly what these individual some of these individual operators represent for the one spin 2 spin cases basically we looked at those which are representing in phase magnetization of single spin and also for the 2 spins we looked at in phase magnetization in all the 3 cases of 1 spin 2 spin 3 spin cases. We also looked at the products of this type that K, I, KX, ILY how they represent a mixture of double quantum and zero quantum coherences and by combination of more than one of such operators we can generate pure double quantum coherence or pure zero quantum coherence. So, and we these ones therefore they are present in the density operator looking at the density operator and looking at these ones we can make an interpretation of what is the system how the system is evolving through the pulse sequence what kind of coherence is generated through the pulse sequence this is the physical meaning of those calculation of the time evolution we will come to it very soon. So, to continue with this discussion what sort of spectra we expect in the for the anti-phase magnetization in the 3 spin case that is what we will we are going to see. So, now we consider the 3 spin example. So, the 3 spin example we represented as KLM all the 3 are weakly coupled spins and we saw earlier in the 3 spin case also for operators like IKX, ILX, IMX what sort of transitions are occurring how do we represent these transitions in the 3 spin system. Let us do a recap on the energy level diagram of the 3 spin case once more we have 8 energy levels here and these are represented as 1, 2, 3, 4, 5, 6, 7, 8 essentially it is a duplication of the 2 spin case and in one of these 4 energy levels you have the the 3rd spin as M spin as in the alpha state here and see you see here this is all M spin is in the alpha state and here in these 4 energy levels the M spin is in the beta state. So, that is how you generate the 8 energy levels and the transitions accordingly are as K transitions and L transitions and M transitions the ones which connect from this group to this group or all the M transitions the ones which are in this group they are basically the K and L transitions and that is indicated here. So, they are indicated by different colors here so and therefore we can understand what these ones represent the arrows indicate what sort of spectra you will get when you actually measure the spectra of the 3 spin system and we have seen therefore in the earlier cases that we have in phase magnetization of the K spin if one is trying to see then you will have 4 lines which are in phase and if we have similarly for the L spin and similarly for the M spin. So, now we are going to see if I have an a term of this type in the basis operator for the 3 spin case remember for the 3 spin case also we have 64 operators 64 operators will consist of single angular momentum operators the products of 2 angular momentum operators and products of 3 angular momentum operators. Now, we are going to consider a 2 spin product for a 3 spin system and this is the operator like 2 ikx ilz in a 3 spin system what does this represent? So, it will be very similar to what we had in the 2 spin case itself that we will have these 2 transitions there is one transition this is the K transition alpha K is going to beta K this is the K transition and correspondingly this is the K transition alpha K to beta K but now you see the arrows in this case the arrow is like this and in this case the arrow is like this. So, this represents a an anti phase term these 2 terms will have opposite signs and that is indicated by again the representation here if 1 3 is positive then the 2 4 transition is negative and that will be that will be in this block the same thing will happen in this block as well 5 7 and 6 8 will also have opposite signs if this is 5 7 is positive 6 8 will be negative. So, therefore we have here 5 7 transition and 6 8 transition the separation between these 2 will be the coupling constant between the spins K and L the same thing is here of course the separation between these 2 or these 2 will be the coupling constant of spin K with spin M. So, you have therefore a doublet of a doublet and you can see here the anti phase nature is for the coupling of K and L whereas if you look at these 2 peaks which arises as a result of the coupling between K and M they are in phase. So, therefore with respect to the M spin these 2 are in phase this whole doublet is has the same sign as this doublet this is this is negative positive and this is negative the same sign pattern here for this one and this one. So, therefore we call this as the K magnetization which is anti phase with respect to L but in phase with respect to M. Let me repeat that again here. So, we have here 4 lines the 4 lines come as a result of coupling between K to L and K to M and we can represent here the same in a slightly different notation I can write that here itself. So, let us say I have one line for the K spin in the absence of any coupling then I will have here 2 lines as a result of coupling J K L and let me represent this as plus and minus because I get because of the K in this system we will have the 1, 3 transition and the 2, 4 transitions or have the opposite signs. Now, there is a K M coupling subsequently. So, K M coupling results to splitting like this then we will have situation like this then we will have this one we call it as plus plus here and minus minus here and this is K M coupling. So, in this case I have assumed that the K L coupling is larger than K M coupling but you can also have a situation where K M coupling is larger than the K L coupling. So, in that case what one could also do is we have this here then we have these 2 as the result of the K L coupling and then we will have splitting again due to the K M coupling then we will have this sort of a situation. So, this then will be plus minus this will be plus minus because this is plus and minus this splits into 2 when then it will give to the minus minus here then this plus splits into 2 then we get plus plus here. So, this is the kind of a situation what you are seeing in this diagram. So, this is what you are seeing in this diagram we have these 2 transitions are shown here as plus and minus that is this situation and this is plus minus and plus whatever that is you may call this plus this minus or vice versa does not matter. So, you get different kinds of patterns depending upon the relative magnitudes of the coupling constants. So, here we have assumed the K M coupling is larger than the K L coupling. The K L coupling is resulting in this kind of anti-phase peaks which are positive and negative here and this is anti-phase peaks and these peaks are in phase because this is due to the K M coupling if this is positive positive or negative negative and this is negative negative or positive positive correspondingly. So, therefore, this is the way you get for K X L Z in a three spin system if you were to take K Y L Z in a similar situation you will get the same sort of pattern in both these cases but they will have dispersive line shape because of the K Y term if I take a K Y term it will have a dispersive line shape. So, similarly for if I were to take L Y M Z or M Y L Z. So, accordingly one will have such kind of patterns in the individual multiplets of the L and the M spins. So, this is an illustration to show how the different transitions appear in your spectrum. Now, let us consider a further extension of this we will consider three spin product. A particular example I have taken here is the three spin product I K X I M Z I L Z for I K X I L Z I M Z notice here the transverse is only for the K whereas the L and the M spins are along the Z axis. Therefore, these ones this term obviously represents the K magnetization because whatever is in the transverse plane is what we observe and we call that it represents that magnetization. So, this is the K magnetization now we can intuitively we can think that this is something which is anti-phase with respect to both the L spin as well as the M spin. So, what do you expect here? So, in this situation earlier this group was had the same sign as this group if this is negative positive this is also negative positive that is because we had the in phase character with respect to the M spin. Now, the K magnetization in this situation will be anti-phase with respect to M as well. Therefore, the first anti-phase term appears because of this K L splitting and secondly this whole term is inverted because of the anti-phase nature with respect to the M spin. Therefore, this now which was negative here now becomes positive and this one becomes negative. Therefore, you get a pattern here this is negative positive positive negative the coupling constant here is of course then this is the K L coupling and from here to here that remains the K M coupling. Now, you see if you look at the splitting due to the K M it is one line here another line here now these two are having opposite signs with respect to each other and therefore this is anti-phase with respect to the K M splitting as well. So, it is anti-phase with respect to the K L splitting this is negative positive K L splitting it is also anti-phase with respect to the K M splitting if this is negative this is positive similarly this and this have opposite sign. Therefore, we call this as anti-phase magnetization of K with respect to both L and the M spins. So, in the diagram here you can also see if I take this as positive direction here this as a positive negative direction here and then correspondingly I will have this as negative direction here and this has a positive direction here the same thing is represented schematically this is the schematic representation of how this peaks will appear in your spectrum when we record when your density operator contains this sort of a term this term sort of a term is a basis operator present in your density operator then if you were to measure this then you will get the spectrum which will produce a spectrum like this, this particular term of course various other terms will produce different kinds of spectrum. So, we are interpreting how what are the contributions of the individual basis operators in your NMR spectrum. So, therefore so much for the physical meanings of the individual basis operators what do they represent in your NMR spectrum but now we come to the next objective that is we must be able to calculate the evolution of the density operator consisting of all of these basis operators through the pulse sequence. I again repeat this slide here this was this slide you already seen before this is a multipulse experiment with so many pulses and the system is evolving with the time starts with the time 0 we have this pulse here and H1 tau 1 pulse here then evolution and dust pre-precision for the Hamiltonian H2 tau 2 then P3 H3 tau 3 P4 and so on so forth. So, for free evolution we have to calculate the evolution under the Hamiltonian this is for these ones and then for the pulses we have this evolution under pulses P rho P minus 1 and rho here can be a summation of all the basis operators and this is the same equation what we had earlier start with rho 0 keep applying the pulses and the evolution under the various Hamiltonians as we have discussed before. So, now we explicitly put this basis operators here for the free evolution I call this as rho dash the rho dash is summation Bs e to the minus i by h cross ht Hamiltonian t then you have this basis operator Bs e to the i by h cross ht and for the pulses we call it as rho double dash then this is Bs P Bs P minus 1 this is the constant therefore this will be this can be taken out and we have this pulse here P Bs and P minus 1 the basis operators set for single spin we have this expression when all the multiple spins are also constructed from these individual angular momentum operators as I mentioned before. Now, let us go further and try and calculate the evolution under the Hamiltonian and evolution under the pulses what happens when you apply the pulses what happens for when the density operator evolves under the Hamiltonians. Now, let us write the Hamiltonian once more here explicitly we will write the isotropic Hamiltonian for weakly coupled spin systems in liquids consist of two terms and this is the Zeeman term where it is due to the interaction between the nuclear magnetic moments and the magnetic field and this is represented as omega k i z k summation over k the k is the same index which goes over the various spins. So, we can write as k z dot z k does not matter earlier we used k z here but does not matter it means the same thing we are referring to the same thing this is the processional frequency of the k spin and this is the coupling constant term where we have jkl is the coupling constant and the operator term for the interaction is i z k i z l or you may call it as i k z i l z. So, it is the same thing just the indices are changed here. So, we could have written it as i k z i l z as well same thing and in briefly we write it as h z plus h j this is the coupling Hamiltonian this is the Zeeman Hamiltonian the first term represents the chemical shifts and the second term represents the scalar couplings. For evolution of a basis operator B s obviously we can write this B s prime this for the row prime I write it as B s prime now because we have considered in the particular operator B s here. So, B s prime I write it as e to the minus i h t B s e to the i h t is the very simple extension of the previous equation. For h now I write explicitly e to the i h z plus h j t B s e to the i h z plus h j t. So, basically putting for h here summation sum of the Zeeman and the coupling terms. Now, for the weakly coupled spin system you remember what the Hamiltonian consisted of. So, Hamiltonian consisted of i k z term here and k z l z here. So, these are all z components the z components therefore this part of the Hamiltonian commutes with this Hamiltonian because k z operators the z operators they all commute. So, and they all belong to the same spin. So, therefore these two will come out. So, when we do that then of course we can actually calculate the evolutions of these operators separately without affecting the result in any manner. So, if I were to write explicitly this I can write this h z plus h j t as h j plus h z t does not matter because this only means there is an order of the evolution order of the calculation is going to be interchanged. So, here I keep the same h z plus h j this is h z plus h j and this part I turn it in the reverse manner I write it as h z plus h z t. What is the consequence of this? For the B s prime I will have evolution under the h z t I can calculate this therefore this will be e to the minus i h z t B s i to the i h z t and multiplied by e to the minus i h j t on this side and on this side e to the i h j t explicitly. The central term will represent chemical shift evolution. So, let me write that explicitly here. So, I write it as B s e to the minus i h z t e to the i h z t and then here it is e to the minus i h j t t and e to the i h j t. So, this particular term here this is the chemical shift evolution and from here to here this after that this will be the j evolution. So, we will first calculate what is the chemical shift evolution of a particular basis operator B s. So, for this we take an example of i k x let us say B s is equal to i k x and we will calculate the chemical shift evolution that is under the influence of Hamiltonian h z. So, the B s prime will be e to the minus i omega k i z k i k x e to the omega k i z k t. Of course, you be careful that somewhere I have used this z k and k z they get mixed up here. So, here I have used k z here and here it is z k. So, essentially I am meaning the same thing. So, this will be e to the minus omega k i k z t i k x e to the i omega k i k z t. We have previously derived this formula e to the i minus i beta i x is given by this expression cosine beta by 2 minus 2 i i x sin beta by 2. This expression we have derived earlier by explicitly expanding this exponential function as a series and then we figured out that this can be represented a simple manner like this. So, we will use the same trick here. So, we will write explicitly this one in the form of this kind of an equation. So, B s prime will be cosine omega k t by 2 minus 2 i sin omega k t by 2 i k z here that is this and then I have this i k x that is this and then I have this for the other side I have cosine omega k t by 2 minus 2 i sin omega k t by 2 i k z. So, let us explicitly calculate this we get here cosine square omega k t by 2 i k x this one this one will give me cosine square this multiplication will give me cosine square omega k t by 2 and i k x will stay here and then this product notice here I should have here a plus sign because this is e to the i omega k z k s. So, I should have here a plus sign. So, and then I will have here this product. So, this with this term gives me 4 sin square omega k t by 2. So, this is minus 2 i and plus 2 i gives me minus 4 i square that is plus 4 and sin square omega k t by 2 but these operators I cannot move around I have to keep them here i k z i k x and i k z. So, these ones will stay here these do not commute and therefore I cannot move them around and then I will have here 2 terms of the cross term one due to this and the other one due to this this gives me minus i sin omega k t a commutator of i k z comma i k x the 2 terms clubbed together can be represented in this manner. So, I have a commutator here i k z i k x now here we have to calculate what this 3 product term implies. So, we have here i k z i k x i k z what does this imply. So, let us try and calculate that using this matrices once more I have put here i k z is half 1 minus 1 and i k x is half 0 1 1 0 and i k z again is another half 1 0 0 minus 1. So, you put this 3 matrices because this is the k spin I can simply take the product notice all of these belong to the k spin therefore I can simply take the product here and this gives me minus 1 by 4 i k x. So, therefore that 4 cancels then I will get here cosine square omega k t by 2 i k x minus sin square omega k t by 2 i k x plus the last term this one remains this one is there and this gives me k z k x this gives me i k y. So, this commutator gives me i k y. So, therefore what I get I get here i k y and because of the i i minus i square. So, I get plus therefore I get here a plus sin minus omega k t k y. Therefore what I got finally cosine omega k t i k x plus sin omega k t i k y. So, after all this calculation what we have got is i k x as a result of evolution under the i k z Hamiltonian I get these 2 terms see if I went to write here a simple manner in a symbolic form i k x under the influence of the Hamiltonian h z gives me it simply means a rotation gives me i k x and so I write to represent this in a pictorial manner like this. Let us say we draw the 3 x's the 3 x's we label them as i x i y and i z of course it can be is a very general thing here it can be k spin l spin whatever it is. So, I just represent it as x y and z and this is this represents the Hamiltonian part what the circle here implies this is the Hamiltonian and I have the transverse magnet this is the basis operator here k x here the basis operator and under the influence of the Hamiltonian this is the time evolution that is going on this rotates in this manner right. So, after a certain time t if your rotation is here then I have i k x cosine omega k t plus i k y sine omega k t right. So, that is so this represents a rotation in this manner. So, you can extend this argument further to say that if I were to take a i k y here suppose it were i k y the same rules will apply. So, this represents the full rotation of the entire magnetization components. So, if I were to start from i k y here then after a certain time tau if the magnetization is rotated like this then what I would get out i k y cosine omega k t i k y because this will be cosine this if this it has rotated like this. So, therefore I will have this component cosine omega k t i k y and minus i k x sine omega k t in that case. So, if I were to write for the i k y let me also write that here. So, if I were to start with i k y then this will go i k y cosine omega k t minus i k x sine omega k t right. So, that represents a rotation of the y magnetization from here to here right. So, therefore we have derived this principle to how this evolution can occur. So, we stop here.