 We have been discussing spin echo as an example and trying to calculate explicitly the density operator revolution through the pulse sequence. So we did that last time for a particular case of two spins with coupling and without coupling, without coupling first and then with coupling later. And we are considered the spin echo sequence which was 90 degree X pulse followed by time tau then a 180 degree X pulse followed by time tau and then the evolution was calculated until this point. So we are now going to make a slight variation in this in this sequence to see what difference does it make, whether the phase of the RF pulse does it make a difference and if it does in what way does it make a difference. So that is the question we are going to ask and therefore we will simply change the phase of this pulse instead of a 180 degree X pulse we will apply the pulse along the Y axis and see what happens. So as before we will first consider two spins without coupling which means the Hamiltonian is simply Hz which is the Ziemann Hamiltonian. The evolution happens in the same manner here under the influence of the Ziemann Hamiltonian. Now we have calculated explicitly the density operator in the previous case and the difference here is only at this point. So up till here the calculation is the same therefore we will not repeat that calculation here but simply take what was the density operator from the previous class at this point and that is rho 3, rho 3 is minus iky minus inside bracket iky cosine omega k tau minus ikx sine omega k tau and then we have minus ily cosine omega il tau minus ilx sine omega il tau. This was the result of chemical shift evolution from rho 2 to rho 3 and then the individual frequencies are here omega k and omega l for the two spins. Now the difference comes we are now applying a pulse along the Y axis and not along the X axis. So what does it do? So if it is along the Y axis this term does not change this is invariant so this remains as iky whereas this one goes from ikx to minus ikx and therefore this becomes plus here. So this becomes plus ikx sine omega k tau similarly minus this ily cosine omega l tau and this becomes plus so plus ilx sine omega l tau. Now put this again explicitly you have it minus iky cosine omega k tau and minus ikx sine omega k tau and similarly here minus ily cosine omega l tau minus ilx sine omega l tau this is at time point 4. Next we have to evolve under chemical shift once more for the next tau period. Now there are four terms and each one of these has to be explicitly evolved. So this is the first term iky so this gives me here minus cosine omega k tau and iky evolves as under the chemical shift once more like this iky cosine omega k tau minus ikx sine omega k tau and the second term which involved ikx will give me this minus sine omega k tau in inside bracket you have ikx cosine omega k tau plus iky sine omega k tau. Similarly for the l spin you have the same two type of terms minus cosine omega l tau multiplying ily cosine omega l tau minus ilx sine omega l tau for the first term and the second one here is minus sine omega l tau ilx cosine omega l tau plus ily sine omega l tau. Now we notice what happens here so rearrange the terms this is minus cosine omega cosine square omega k tau and then we have the other iky term is here and this is minus sine square omega k tau iky if you put iky take iky take minus iky inside bracket you will have cosine square omega k tau plus sine square omega k tau and so therefore what is in the bracket is one therefore you get a minus iky and what happens to this term so this is plus ikx sine omega k tau cosine omega k tau and here it is minus sine omega k tau cosine omega k tau this will cancel. So from these two we retain only this this part and now similarly if you see here we retain this part these two terms will be remaining this is cosine square omega l tau ily and this is sine square omega l tau into ily and therefore these two terms will remain and we get ily here and these two terms will cancel and we therefore in the end for rho phi I get minus inside bracket iky plus ily this is exactly equal to rho 2. So therefore what has happened if you recall this let us see what rho 2 in the previous class minus iky plus ily this is what we started off with. Now in the previous class at rho phi we had here plus sine it became rho phi is equal to plus iky plus ily so the effect of changing the phase of the 180 degree pulse is that this sine has changed and it has remained as rho 2 which is minus iky plus ily the chemical shift is completely refocused that effect is the same there is no change in that one except that we have a change the sign is now become minus instead of the plus in that case. This also we had seen in a vectorial picture when we discussed the spin echo earlier. So this provides a mathematical rationale as to what is happening in the sequence. Now let us consider the next case where you have two spins with coupling Jkl now we have Hz plus Hj this is the Hamiltonian now. Now we have seen before that the chemical shift is completely refocused at time 0.5. So therefore we need not to calculate the chemical shift evolution once more here because you remember we said evolution under Hz and Hj can be calculated independently. It does not matter which one you calculate first which one you calculate second it does not matter at all. Now we assume that we have already calculated the chemical shift evolution under the chemical shift the influence of the Gma Hamiltonian therefore we do not need to calculate that once more we straight away take that result here the rho phi was minus iky plus minus inside bracket iky plus ily therefore we start from there and now we calculate the evolution under the coupling Hamiltonian. So this is the first term there gives me minus inside bracket iky cosine 2 pi Jkl tau why do I take 2 pi Jkl tau now because I am evolving for the whole time period 2 tau. So I did not evolve under the coupling before therefore I am now evolving for the whole period 2 tau therefore I get here cosine 2 pi Jkl tau minus 2 ikx ilz sin 2 pi Jkl tau similarly the L frequency gives me ily cosine 2 pi Jkl tau minus 2 ilx ikz sin 2 pi Jkl tau here notice the difference is here you have ikx ilz here you have ilx ikz. So therefore that was the simplest case we could calculate the evolution under the coupling in a simplified manner we do not need to calculate the coupling the chemical shift evolution once more we can assume that we have already calculated and you simply calculate the evolution under the coupling the 180 pulse is applied to both the spins. Now we consider the next case which is case 2 here the 180 degree pulse is applied only to the L spin notice the pulse sequence here we have the L spin channel here and here is the K spin well this is also this is the K spin however this is the hard pulse which may be applied to both the channels but to the L spin we apply only the 180 degree pulse though if you consider this as applying only to the K spin then of course we have this but we are applying the first pulse to both the spins then we will of course also we will have a 90 degree pulse here when we say 180 pulse is applied only exclusively to this. So we could in principle say that in this calculation we have calculated it assuming that we have applied a one the initial 90 degree pulse to both the spins. So we can add that here that so the first 90 degree pulse is applied to both the spins therefore both are excited and then we have the evolution going on under the influence of the coupling Hamiltonian we could have done this experiment this way as well in which case of course we apply the pulse only to the K spin we need not calculate the evolution of the L spin magnetization here because L spin magnetization does not exist here. So if you have this pulse also then we have generated both then we can have the evolution of the L spin as well going on. So therefore both are correct in you can choose whichever way you want to do it. So here we have considered both the spins and therefore we said the rho 2 is minus iky plus ily the first 90 degree pulse is applied to both the spins on K and L therefore we have this one now. Now what is rho 3? The rho 3 is minus the evolution now under the chemical shift. So minus iky cosine omega K tau minus ikx sin omega K tau this is the K spin evolution and the L spin evolution gives me minus ily cosine omega L tau minus ilx sin omega L tau. So put it expand it and rearrange it properly and we have rho 3 is minus iky cosine omega K tau plus ikx sin omega K tau and likewise minus ily cosine omega L tau plus ilx sin omega L tau. So then now you see here there are 4 terms this rho 3 has 4 terms now each one of them we will have to evolve under the coupling. So therefore rho 3 prime will now evolve under the coupling to rho 3 prime use the result after the coupling evolution. So the first term here gives you this coefficient remains the same. So we have here iky cosine pi j Kl tau minus 2 ikx ilz sin pi j Kl tau. The second term gives me this coefficient is the same sin omega K tau we have ikx cosine pi j Kl tau plus 2 iky ilz sin pi j Kl tau. The third term gives you minus cosine omega L tau plus ily cosine pi j Kl tau minus 2 ilx ikz sin pi j Kl tau and the fourth term gives you plus sin omega L tau ilx cosine pi j Kl tau plus 2 ily ikz sin pi j Kl tau. So therefore the result is now we have a total of 8 terms here. Let us rewrite this in a making some rearrangements. So the rho 4 is now after a 180 degree expulse now that is after the on the L spin only okay. So rho 3 prime was after the coupling evolution 8 terms are there and now we apply 180 degree expulse only on L. So therefore what happens there is no pulse applied here so this remains the same this term remains the same and here again ikx nothing happens to K spin but the L spin 180 degree pulse is applied. So therefore this ilz goes to minus ilz and therefore this becomes plus here. So this gives you 2 ikx ilz sin pi j Kl tau that remains this we will put it in a different color for that reason. Now once again here nothing happens to this term and then on this term ilz goes to minus ilz and therefore this gives me a minus sign here minus 2 iky ilz. The third term now this one ily goes to minus ily because this is the expulse therefore ily goes to minus ily. So this is again in a different color and whereas this one it is invariant lx is invariant therefore nothing happens to this term this remains as minus 2 ilx ikz 0.1jkl tau and in the fourth term this is ilx here since this is the expulse this will be invariant nothing happens to this whereas this one the ly goes to minus ily and therefore you get a minus sign here you get minus 2 ily ikz sin pi jkl tau. Now therefore you rewrite this as 8 different terms right we have now 8 terms here okay each one of them you expand it you have 8 terms. Now what we do each one of those 8 terms you have to evolve under the j coupling because the j coupling is present during the next tau period. So each one of those we now evolve so this is the first term from there you get cosine omega k tau cosine pi jkl tau and iky evolves as iky cosine pi jkl tau minus 2 ikx ilz sin pi jkl tau and the second term gives me minus sin pi jkl tau cosine omega k tau and this is the operator which we are evolving and that gives me 2 ikx ilz cosine pi jkl tau plus iky sin pi jkl tau in these cases you should remember that summary of the evolutions which are given earlier because it is all what I am using here those summary equations which are there for the evolution under the coupling and evolution and the chemical shift is what is being used. The third term gives you sin omega k tau cosine pi jkl tau and it is a kx part which is evolving ikx cosine pi jkl tau plus 2 iky ilz sin pi jkl tau and the fourth term likewise gives me minus sin pi jkl tau sin omega k tau and inside here is 2 iky ilz cosine pi jkl tau and you get a minus sign here minus ikx sin pi jkl tau. So similarly the fifth term is cosine omega k tau cosine pi jkl tau ily cosine pi jkl tau minus 2 ilx ikz sin pi jkl tau the sixth term gives me sin pi jkl tau cosine omega k tau then inside bracket you have 2 ilx ikz cosine pi jkl tau plus ily sin pi jkl tau the seventh term gives me sin omega l tau cosine pi jkl tau and inside the bracket you have now ilx cosine pi jkl tau plus 2 ily ikz sin pi jkl tau and finally the last term gives you minus sin pi jkl tau sin omega l tau and inside bracket you have 2 ily ikz cosine pi jkl tau minus ilx sin pi jkl tau this is at the end of j evolution from the after the 180 degree pulse. So this is adjust before the spin echo at the time of the spin echo. Now let us rearrange this and look at these terms which actually tend to cancel. Now we notice here that earlier which I put it all in sin color and now I put the same in the red color and say these terms cancel. This cancels with this so we have here plus and here it is a minus and therefore this term cancels this remains and similarly here this term cancels with this term 2 iky ilz term cancels with this 2 iky ilz term the coefficients are the area of the plus sin omega k tau cosine pi jkl tau and sin pi jkl tau. So it is the same coefficients here with a minus sign and therefore this cancels with this. Similarly this term cancels with this term and this term cancels with this term. So now therefore half of the terms are cancelled. So let us pull together the other terms and you have now the rho phi is equal to minus cosine omega k tau cosine pi jkl tau iky cosine pi jkl tau and minus sin pi jkl tau cosine omega k tau iky sin pi jkl tau plus sin omega k tau cosine pi jkl tau ikx pi cosine pi jkl tau and minus sin pi jkl tau sin omega k tau minus i k x sin pi j k l tau plus cosine omega l tau cosine pi j k l tau i l y cosine pi j k l tau plus sin pi j k l tau cosine omega l tau and here i l y sin pi j k l tau plus sin omega l tau cosine pi j k l tau i l x cosine pi j k l tau and finally minus sin pi jkl tau sin omega l tau and inside bracket minus ilx sin pi jkl tau. So, therefore, now you see we have the operator terms are either iky ikx or ily ilx and then we have the various coefficients for those. Now, we pull together all those which belong to particular operators and then you get this. So, here what we have is the coefficients of iky. So, iky is this operator which was there and we pulled together those terms which have iky. So, here we have then minus cosine omega k tau iky and inside bracket we get cosine square pi jkl tau plus sin square pi jkl tau. Similarly, for ikx here you have the coefficient sin omega k tau and inside the bracket here cosine square pi jkl tau plus sin square pi jkl tau. Similarly, for ily here you have cosine omega l tau and inside bracket here you have cosine square pi jkl tau plus sin square pi jkl tau and then for ilx you have the sin omega l tau and inside bracket here you have cosine square pi jkl tau plus sin square pi jkl tau. Now, notice everything that is present inside the bracket is all 1. Therefore, it simply is simply equal to 1. Therefore, I only have these terms remaining. So, cosine omega k tau iky plus sin omega k tau ikx plus cosine omega l tau ily plus sin omega l tau ilx. Now, where is the coupling? Therefore, the coupling is all vanished. All the terms which contain the coupling information have vanished. So, therefore we say the j coupling evolution is refocused in other words this is called as spin decoupling. So, if I apply a 180 degree pulse only on one of the spins then it results in spin decoupling. So, we can do this calculation for evolution for with the initial 90 degree pulse applied to one spin only then also we can do the same calculation. But we have since we have included both it is actually more general. Now, we notice one thing in the previous case we after the row 4 that is after this point we calculated the j evolution. So, we switch to the order of the evolution. Earlier we had we are calculating the chemical shift evolution first and then the j evolution. Now, we have done the j evolution and then we are going to do chemical shift evolution now. So, after this term we will now do chemical shift evolution because here I am demonstrating to you that you can actually switch the order of chemical shift evolution and coupling evolution and that is what we have done here. So, now after the row 5 I now calculate the chemical shift evolution. Chemical shift evolution gives me therefore cosine omega k tau here Iky cosine omega k tau minus I kx sin omega k tau. The second term gives me sin omega k tau I kx cosine omega k tau plus Iky sin omega k tau cosine omega l tau Ily cosine omega k tau minus I lx sin omega k tau and finally sin omega l tau and here inside I lx cosine omega k tau plus Ily sin omega k tau. So, this is for the last top period this is after the 180 degree pulse we evolve for the next top period under the influence of the chemical shift evolution. So now you see what happens you look at the terms once more rho phi prime this part we read it as it is but notice here there is a cancellation. So this term can this is cosine omega L tau sine omega L tau I L x this is plus sign here and minus sign here and this is sine omega L tau cosine omega L tau this is the plus sign here and these two terms will cancel. Because this is the chemical shift evolution of the L spin therefore the L spin must have the omega L as the evolution okay and then we can see that these terms will cancel and what will happen to this this is cosine omega L square omega square L tau. So what do you get I k y minus I k y cosine square omega k tau that is this and this minus sign is multiplying everywhere right. So I k y cosine omega square tau minus I k x sine omega plus sine omega k tau I k x cosine omega k tau plus I k y sine omega k tau therefore this is the plus sign this is the minus sign here I pull this together I take minus I k y inside bracket I cosine square omega k tau minus sine square omega k tau and then you have this one here this I k x sine omega k tau okay and cosine omega k tau and this is sine omega k tau cosine omega k tau these both have the same sign this minus minus plus and this is also plus so therefore this gives me I k x sine 2 omega k tau okay. So similarly here the what we have here cosine I l y cosine cosine square omega L tau and and this is I l y sine square omega L tau okay because this is omega L tau here and now inside bracket therefore you will get cosine square omega L tau plus sine square omega L tau that is 1 and therefore I simply written this as I l y. So therefore your rho phi prime is minus I k y cosine square omega k tau minus sine square omega k tau plus I k x sine omega k sine 2 omega k tau plus I l y. So this is substantially simplified now what we have here this is cosine 2 omega k tau therefore I have here minus I k y cosine 2 omega k tau plus I k x sine 2 omega k tau. So the k spin has evolved under the chemical shift whereas the L spin chemical shift also is refocused you do not see any term remaining here which contains omega L. Therefore when I apply 180 degree pulse only on the L spin the L spin chemical shift is refocused but the k spin chemical shift continues to evolve for the whole period 2 tau therefore I have here starting from my I k y it was minus I k y therefore I have inside minus I k y cosine 2 omega k tau plus I k x sine 2 omega k tau. So L chemical shift is refocused whereas k spin chemical shift has evolved. So therefore in summary we have seen the different ways of dealing with the spin echo we have done explicit calculation under different conditions when though in the spin echo the 180 degree pulse is applied along the x or the y axis what happens when there is only chemical shift evolution when what happens when the 180 pulse is applied on only one spin that leads to spin decoupling. This is the common technique which is used in all multiples experiments for decoupling during the course of the pulse sequence particularly in along the indirect detection periods in multidimensional inverse spectra. This we will discuss in greater detail but we have demonstrated here the general principle as to how these things are working and explicitly shown by density operator calculation the various terms that you hold under the influence of the different Hamiltonian.