 So far we have discussed extensively the basis operators, the evolution of the basis operators which forms the basis for calculating the evolution of the density operator through a given pulse sequence. Now we will just take a summary of all of these time evolutions and I will listed here a few illustrative evolutions and by and large we will be using these only most of the time we will be using these many equations for calculating the evolution of the basis operators. Here I kx is supposed to represent the in phase magnetization of the k spin and it evolves under the influence of the chemical shift Hamiltonian H z for a time t and then it gives you I kx cosine omega kt plus I ky sin omega kt. I ky under the same Hamiltonian for the time t gives you I ky cosine omega kt minus I kx sin omega kt. Notice here the change in the sign depending upon what your basis operator is you have to be careful with regard to what sign you will get here. These I have written for k spin but it can be for any other spin, L spin, M spin whatever. So in your density operator expression you may have combination of all of these and then each one of them will evolve with its own characteristic frequencies as are indicated here. Now I kx again under the influence of the coupling Hamiltonian for a time t gives you I kx cosine pi jkl t where jkl is a coupling constant plus 2 I ky I lz sin pi jkl t. Here we consider the coupling between the spins k and L and therefore you have the coupling constant kL here. If it were a coupling of some other spin k to M then it will have a coupling constant here as jkm and here it will be accordingly I mz. So therefore depending upon what coupling one is looking at then you will have such kind of expressions. If you have multiple couplings then you will have to evolve these step wise first under one coupling and second time under the second coupling but the same equations will be applicable. Each term will have to be evaluated again for another coupling and so on. So I ky under the influence of the coupling jkl that gives you I ky cosine pi jkl t minus 2 I kx I lz sin pi jkl t. Once again node is the change in the sign here if we are starting with kx you get plus here if you start in the ky you get a minus here. This is basically due to the convention you have used of a particular kinds of rotation the sense of rotation in your the group. Now if I had for example all these were in phase magnetizations of the k spin. We also treated anti phase magnetization of the k spin we had the basis operator such as this 2 I kx I lz under the influence of the coupling Hamiltonian it gives you 2 I kx I lz cosine pi jkl t plus I ky sin pi jkl t. We do not need to consider a chemical shift evolution because it will be simply similar to what we have here where there is I kx evolving you write put this equation the other things remain the same. So we do not need to explicitly write the chemical shift evolution of such a term important thing is the coupling evolution of this. Likewise 2 I ky I lz under the influence of the Hj Hamiltonian for time t gives you 2 I ky I lz cosine pi jkl t minus I kx sin pi jkl t once again notice the change in the sign. So in summary it is essentially it is these set of equations by and large we will be using for calculating the evolution of density operator and the corresponding basis operators in any given pulse sequence. Therefore it simplifies your calculation quite substantially and it is important to remember these equations pretty well so that we can easily write the evolution of the density operator through a given pulse sequence. Now let us move forward the next thing to see is when you actually calculated the evolution of the density operator obviously you will be getting a mixture of various basis operators adding together in the summation. Now we have also seen earlier in the density operator depending upon what sort of elements that are present there when you actually make a measurement some of those ones appear and some of those do not appear. So therefore when you have such a kind of a situation we have to ask a question as to among the various basis operators which will form the density operator which of these are observable and which of them are non-observable. So what we measure is a transverse magnetization. So which is represented as mx plus imy that is m plus and this is called as m plus this is also called as raising operator or ladder operator whatever. So this is what we measure transverse magnetization is what we measure this is the x component of the magnetization this is the y component of the magnetization and this is the complex magnetization we measure and we have seen earlier that we have to calculate the expectation value of the corresponding operator for such a kind of an observable. If this is our observable the corresponding operator is i plus which is ix plus iiy. So then what we have to do we have to calculate the trace of this operator with the density operator that you might get at the time of detection. Just before detection what is the density operator we have to use the trace of that one. So what we have to calculate for measurement is trace i plus rho and rho is the density operator just at the time of the detection of your signal. Now since rho is a summation of various basis operators here. So we will have to calculate trace of the individual basis operators to see what basis operators contribute to observable signal and which ones do not. So that is what is indicated here. So you calculate trace BSI plus and notice that this actually is quite symmetric BSI plus is the same as trace i plus BS it does not matter which order you put them in and then this should be non-zero for BS to be observable. So this is the condition we have derived earlier the expectation value has to be non-zero for it to be observable. So this is the condition. So we are going to therefore calculate the traces of this individual basis operators with this i plus operators. Let us take a few examples here. So let us first consider BS is equal to ikx this is the in-phase basis operator whose evolution we saw earlier and trace of ik plus ikx. Now I have taken it for k magnetization here and if it were l magnetization you will put l here and so on so forth. So if it is a sum of various spins you will have to take the sum of all of those things here and then calculate the trace. But this is enough to demonstrate what sort of things we are going to get for all the individual spins. So ik plus if I write here this is ikx plus iiky that is ik plus and then have the ikx here. So therefore if you make multiplication here trace of ikx square that is this and this the first term gives you ikx square and the second term gives you plus i trace of iky ikx. So trace of iky ikx. Now how do we calculate this? Since both are operators of the same spin so what we will do is I will take the individual spin operators here and take a multiplication. So the trace ikx you know is half 0 1 1 0 and therefore I take a square of this I get 1 by 4 here 0 1 1 0 0 1 1 0 plus i trace of 1 by 4 this is iky half 0 minus i i 0 is iky and ikx is again half 0 1 1 0. So therefore trace 1 by 4 0 minus i i 0 0 1 1 0. So what is this one the trace of this what is the trace? Trace is the sum of the diagonal elements. So therefore this gives me half this is 1 plus 1 2 therefore this the first term gives me half and the second term gives me 0. Therefore this is observable because this is non-zero. Therefore this term this basis operator is observable. Let us look at the second one iky so I do the same thing here ik plus with iky so ikx plus i iky iky and therefore this is i trace iky square this one here iky square plus trace ikx iky this one here ikx iky. So use the same matrices again here for iky I put here 0 minus i i 0 0 minus i i 0 and for this 1 by 4 0 1 1 0 0 minus i i 0 and we take a multiplication of this. So here I get i is here and this gives me 1 by 4 iky square obviously is 1. So therefore this is 1 0 0 1 plus trace 1 by 4 i 0 0 minus i. So this again gives me half but there is a factor i here therefore use me i by 2 and this one gives me 0. Nonetheless it is non-zero and therefore this is observable. Now let us look at this. This is once again K magnetization but now it is anti-phase with respect to the L spin. So I take this Bs is equal to 2 ikx i Lz this is anti-phase basis operator. Now doing the same trick so ik plus here I have 2 ikx i Lz. So therefore I get here 2 trace ikx plus i iky and ikx i Lz is here and the 2 I have taken it out. So then what I get? So this gives me ikx square i Lz this remains as it is and this gives me 2 i iky ikx i Lz. These 2 are of the same spin and this is of a different spin. Notice here these 2 are of the same spin and this is of a different spin. Therefore what I have to do to calculate this I actually have to take the matrix representation of the individual operators here and then take a direct product. So therefore to evaluate this so I have here 2 trace and this gives me 1 by 8 because there are 3 terms here half half half gives me 1 by 8. So then ikx square is 1 0 0 1 and therefore I have here 1 0 0 1 and then i Lz is 1 0 0 minus 1 this is for a different spin. Notice here so let me write that so that it becomes clear to you. So this is for the k spin and this is of the L spin. So therefore this is again for the k spin and this is of the L spin. So when I do that I take a direct product of this and direct product here. So I get here 2 trace 1 by 8 1 minus 1 1 minus 1 all other terms are 0 and here I have 2 i trace this gives me minus i i i minus i and all other terms are 0. So therefore what we get if I take the trace this is the sum of all the diagonal elements this is 0 and so also it is 0 here and there the whole thing is 0. So therefore notice this anti-phase magnetization here is not observable 2 i kx i Lz is not observable. Similarly if I take 2 i k y i Lz which is again anti-phase basis operator of the k spin anti-phase with respect to L. So I substitute here i k y i Lz it is the same calculation and you have i k y i Lz here so use 2 i trace i k y square i Lz plus true says i k x i k y i Lz. So once again we do the same thing here and this is for the k spin and this is for the L spin and here it is again for the k spin and this is for the L spin and then we take a direct product take a direct product of these 2 matrices so I get here 1 minus 1 1 minus 1 and here again I get y minus i minus i size and you see the trace of this is 0 and the trace of this is also 0 therefore this is non-observable. So therefore I have illustrated here how to calculate whether a particular basis operator is an observable or not when you have sum of various kinds of operator terms in your density operator not all of them contribute to your observation when you make a measurement some terms will make contribute to the measurement and some will not and those which do not contribute to the measurement you can simply ignore them so far as your signal is concerned. Now if you do a similar exercise for the various operators representing the double quantum and zero quantum then you will find that they are also non-observable. So however anti-phase basis operators evolve under J coupling into observable terms. Now if you are doing an experiment where at the beginning of detection so you have the anti-phase magnetization and the anti-phase magnetization evolves under coupling as you have seen evolves under coupling into an in-phase magnetization here for example if you took 2 i k x i L z it evolves into i k y term here and similarly 2 i k y i L z evolves into an i k x term here. So after that evolution is done if you consider evolution during the detection period and during the signal collection and therefore this will these terms will then contribute to observable magnetization these themselves are not observable but these ones evolve under the coupling into observable magnetization as i k y or i k x. So therefore anti-phase operators evolve under J coupling into observable terms. Therefore if you are at the beginning of detection if you have this anti-phase term and during detection if you do a decoupling of the two spins then this terms will not be observable but if you do not decouple you allow them to evolve during the detection period under the influence of the J coupling then they will contribute to observable magnetization. So we are now prepared the ground for calculating the evolution of the density operator through any given pulse sequence. What we are now going to do is you consider a simple example the first example this is the spin echo this is considered to be one of the most crucial and fundamental pulse sequence which is used in many of the multiple experiments and we will now explicitly calculate the evolution of the density operator for this particular pulse sequence the namely the spin echo. Now you recall the previous lectures what is the spin echo sequence it starts with a 90 degree pulse and then there is a tau period then you have a 180 pulse here and again the same tau period and we said that the magnetization which was decaying here now it rebuilds and then causes an echo at this point. So and all the field and homogenities will be refocused chemical shifts will be refocused this is what we said and we are now actually going to explicitly calculate that using the density operator formalism. Now notice here these phases are two chosen arbitrarily they can be anything you can use x or y or combination of those and we will take some illustrations of those as well we keep for simplicity for illustration we keep the first pulse as 90x and we have the various time points indicated here this is time 0.12345. So this is the beginning of the detection period we will do our calculation until this point we are not going to do a further evolution here this will be the normal FID and the data collection happens here. So we will see what is the effect of this pulse sequence until this point therefore we evolve the density operator through these steps and reach up to this point. So now we consider two spins k and l and without coupling assume to the first case that there is no coupling between the two spins. So at time 0.1 this is a row is a row 1 what is the density operator it is just the sum of the two i z operators. So because you remember the equilibrium density operator was the i z operator which is the sum of the z magnetizations of the two spins. So therefore this is i k z plus i l z. Now if I apply the 90 degree x pulse then I get the density operator at time 0.2 this gives me minus i k y plus i l y so this is the 90x pulse. Now during the next period tau this evolves under the influence of the chemical shift because there is no coupling so we need to consider only chemical shift evolution. So I consider the evolution of the individual these operators here so i k y gives me i k y cosine omega k tau minus i k x sine omega k tau. For ready reference I also put that here those of you who can see this is the small picture but this I hope some of you can see it with good eyes. So we have the x here the y here minus x here minus y here and this is z. So there is a rotation around the z which is the chemical shift evolution. So the i y goes to cosine omega k tau here this is this component and this component will be minus i k x sine omega k tau. Similarly for the l spin so this will be minus i l y cosine omega l tau minus i l x sine omega l tau. So this will be the density operator at the time 0.3. Now what I do at this time point I apply 180 degree x pulse to both the spins. So 180 degree x pulse what does it do? This means the same it shifts i k y to minus i k y rotates by 180 degrees so it is a rotation from here to here. So is minus i k y cosine omega k t this remains the same and it has no effect on i k x and therefore this remains as i k x sine omega k tau. Similarly for this i l y goes to minus i l y minus i l y cosine omega l tau minus i l x sine omega l tau. So this is the result of 180 degree pulse there we now got the density operator at time 0.4 in the pulse sequence. So rearranging those terms remove the signs I rewrite this order of 4 as i k y cosine omega k tau plus i k x sine omega k tau plus i l y sine omega l tau plus i l x sine omega l tau. So during the next tau period I will have to evolve under the chemical shift once more again there is no coupling therefore evolve under the chemical shifts only. Now there will be how many terms each one of these produces one term here so it evolves under the chemical shift and so therefore I will have total of 4 terms here also there are 4 terms each one I will have to evolve under the chemical shift. So therefore cosine omega k tau this remains here now for this i k y I write here i k y cosine omega k tau minus i k x sine omega k tau plus for this term now sine omega k tau and i k x term here i k x cosine omega k tau plus i k y sine omega k tau notice the change in the sign here. So we have to be careful with that these are very important. Similarly now for the l spin this is cosine omega l tau devolved i l y cosine omega l tau minus i l x sine omega l tau and i l x term gives you sine omega l tau into i l x cosine omega l tau plus i l y sine omega l tau. These ones are put in different colors the reason is these ones will cancel out see here i k x what is this product here cosine omega k tau sine omega k tau and here it is sine omega k tau cosine omega k tau and the same i k x term therefore this will cancel this term. So maybe I can just indicate that to you by explicitly writing that there. So this will cancel this similarly for the i l y so this will cancel this one. So then what we are left with so we are left with i k y cosine square omega k tau and this will give me sine square i k y sine square omega k tau that is this here and similarly for i l y cosine square omega l tau plus sine square omega l tau and obviously this is one and this is also one therefore I got here rho phi this is the density operator at time 0.5 this is i k y plus i l y so everything is vanished all these time dependencies is vanished the frequency dependence is vanished we started with the density operator rho 1 as i k z i l z plus i k z and after the first 90 degree pulse I had minus i k y plus i l y and I got back i k y plus i l y. So the chemical shift evolution has completely been refocused except for the change in the sign here and that of course will depend upon what are the sign of the 180 degree pulse and accordingly we could have had a different minus sign here as well we will also see that later. So therefore the chemical shift information is completely refocused we got back the basic operator as we had in the beginning. Now let us consider the situation when we have the coupling as well now the coupling means when you have the coupling your total Hamiltonian will be h g plus h j this is the chemical shift Hamiltonian and this is the coupling Hamiltonian. So we will consider case 2 cases here case 1 180 x pulse is applied to both k and l spins in the spin echo sequence. So therefore evolution under h z and h j can be considered independently and now as before your rho 1 is i k z plus i l z and rho 2 is minus i k y plus i l y 90 x pulse since evolution under chemical shift is refocused at the end of the spin echo we will calculate only evolution under h j through the spin echo because we have already shown that chemical shift is refocused. So we do not need to calculate under the chemical shift again therefore we will only calculate evolution under the coupling constant. So therefore j evolution I put here picture once more here 2 i k y i l z and those of you can see it clearly then you have that indication here as to what terms you will get. So rho 3 prime is the one this is the result of j evolution and I get here I had minus i k y plus i l y right. So the first term gives me i k y cosine pi j k l tau minus 2 i k x i l z sin pi j k l tau and you have the minus sign this is for the k spin and this is for the l spin minus i l y cosine pi j k l tau minus 2 i l x i k z sin pi j k l tau this is only influence of the coupling j k l j evolution. So now I apply a 180 x pulse on both k and l so this sign remains the same. So 180 x pulse takes me i k y to minus i k y so this remains this and what happens here i k x does not change l spin is applied it takes i l z to minus i l z therefore this gives me plus here so there are plus 2 i k x i l z sin k l tau and once again here minus i l y cosine pi j k l tau plus 2 i l x i k z sin pi j k l tau when I apply the pulse on both spins and this is what I will get. So simplifying this rho 4 prime is i k y this minus minus plus i k y cosine pi j k l tau minus 2 i k x i l z sin pi j k l tau and this again i l y cosine pi j k l tau minus 2 i l x i k z sin pi j k l tau. So now you will have to evolve under the j evolution once more after the 180 degree pulse. So now each one each of these term will have to be evolved so there are four terms here there will be four evolutions therefore. So therefore the first term I take away the cosine pi j k l tau part out and i k y term has to be evolved this gives me i k y cosine pi j k l tau minus 2 i k x i l z sin pi j k l tau and the second term is minus sin pi j k l tau now I am evolving this anti phase term you remember the summary 2 i k x i l z gives me 2 i k x i l z cosine pi j k l tau plus i k y sin pi j k l tau similarly for the other term cosine pi j k l tau i l y cosine pi j k l tau minus 2 i l x i k z sin pi j k l tau and the last one minus sin pi j k l tau 2 i l x i k z cosine pi j k l tau plus i l y sin pi j k l tau. So now we regroup these terms so what I get here i k y cosine cosine use my cosine square that is the first term here i and then where else is i k y i k y is present here and this gives me sin and sin this is sin square pi j k l tau so therefore i k y cosine square pi j k l tau minus sin square pi j k l tau and now these 2 terms are identical 2 i k x i l z so sin cos and here it is also sin cos so this one is minus 2 sin i k x i l z so 2 cosine pi j k l tau and that is essentially sin 2 pi j k l tau and similarly for these terms we get here i l y cosine square pi j k l tau minus sin square pi j k l tau minus 2 i l x i k z sin 2 pi j k l tau. Now what is this here inside the bracket that is cosine 2 pi j k l tau i k y cosine 2 pi j k l tau minus 2 i k x l z sin pi j k l tau plus i l y cosine 2 pi j k l tau minus 2 i l x i k z sin pi j k l tau notice therefore each of the individual spins have the tau dependence in the in the end and the two and magnetization components would have separated out in phase by this amount the 2 pi j k l tau depending upon what the value of tau is the phase separation between the 2 components of the k spin will be given by this and it also generates here an anti-phase magnetization okay so this we see and this we may not this is not observable if you of course it holds under the coupling later into observable magnetization but if you are looking at the end of the spin echo it is just this term which is observable and this is term which is observable so scalar coupling is convolution is not refocused at the time of spin echo. So we saw that chemical shift evolution is refocused and the coupling evolution is not refocused this is what we had also derived by a vector picture earlier in our previous classes very early in the in the course we had actually derived this okay so we stop here and continue with this discussion in the next class.