 In the last class we discussed the details of the two-dimensional Cauchy experiment and we will just have a recap of that one here. So two-dimensional Cauchy experiment looks like this. So for a two-spin system, so you will have a so-called diagonal peak for each of the spins and this will appear with a plus, plus, plus, plus signs all the components. So there are two signals of this spin, two signals of this spin in the one-dimensional case. So here also you have two lines here and here two lines here for the one-dimensional experiment and this produces the two diagonal peaks. And then here you will have plus minus, minus plus and then you have plus minus, minus plus. So this is the structure of the two-dimensional Cauchy experiment. We call this one as the cross peak, this is the diagonal peak, this is the diagonal, this is the cross and this is the diagonal and this is the cross and the cross peaks has all absorptive line shapes and the diagonal peaks have all dispersive line shapes because of this it has lower resolution in the spectrum. So we discussed the consequences of this. We also calculated the spectra for three spins, different kinds of three spins. In every case the same situation appears, the diagonal will have dispersive line shapes with the fine structure of the individual spins and the cross peaks will have this plus, minus, minus, plus characters with absorptive line shapes. Now the diagonal having the dispersive line shape is a disadvantage here and therefore to circumvent this problem a new idea emerged and that is what we are going to discuss today. So and that is called two-dimensional double quantum filtered Cauchy experiment which is given in this manner. It is a new pulse sequence, this first of all it overcomes the difficulties of the Cauchy experiment which we will see soon. It also introduces a new concept called as phase cycling which is much more general than what is indicated here for this particular experiment. However, this will illustrate the use of phase cycling for obtaining a kind of information you may want to have in your two-dimensional spectrum. The pulse sequence for the double quantum filtered Cauchy is given here. So you have here a 90 degree pulse. Now we have written here for the phase as phi. In the case of Cauchy we had written this as X and this also as X. Cauchy was the two pulse experiment. It has 90, T1, 90 and then the data was collected immediately after. Here we introduce a small time period here. This is extremely small time period. So this is about the after order of few microseconds. So this time period is extremely small. This is few microseconds and that is given to allow for changes in the phases as we will see soon. And then following that you apply a final 90 degree pulse and then you have the data collected as a function of time T2. Now here this phi is a phase which is changed from one experiment to another experiment. You see here it is indicated scan number 1, 2, 3, 4. For these 4 experiments they are all collected with the same value of T1. So for every value of T1 this is what you do. You collect 4 experiments and that together you make one FID. Well each one is an FID, this of course is collected in the receiver as it is plus sign means it is added. And the minus sign means the FID that comes here is subtracted. And the third experiment with this phases for the pulses this is added and here for the fourth experiment with this phase for the pulse it is subtracted. So collection of all of these is for one value of T1. So when we increment the T1 again you do this 4 experiments. This strategy is known as phase cycling. This allows us to filter out certain signals and select certain signals and that is called as selection of coherence transfer pathways. This is illustrated very well and we will see how it works. And so therefore let me repeat here for each value of T1 you record 4 FIDs with the phase of these 90 degree pulses changing as indicated here. For the first one it is x then it is y then it is minus x then it is minus y. And the data in the receiver is added here, the subtracted here, added here and subtracted here. So this whole strategy is called as phase cycling. So you may have multiples of this 4 does not mean that you can only have 4. Suppose you decide to have 8 scans this is called as a scan number. So suppose you want to have 8 scans to improve a signal to noise then what you do is you repeat this 4 once more. So you have 1, 2, 3, 4 with this then you have the 5, 6, 7, 8 again with this and the data is repeated in this manner. So you can blocks of 4, you can collect the data in this particular manner. You can use 4 scans, 8 scans, 12 scans, 16 scans and so on and so forth depending upon how much signal averaging you want to do and that is the characteristic of this particular experiment. After you do all of this you have a small time period here which is a few microseconds and this allows us to convert something what we have created here into an observation by application of this last 90 degree pulse which is done with a fixed phase. This is done with X or it can be done with Y also it does not matter and then the data is collected here. As the data is collected here remember whatever comes out is stored in the receiver in this manner. So this strategy is called as phase cycling which is a very general strategy which can be used for various other purposes as well and we will demonstrate the particular use in this particular double quantum filtered COSY experiment. Now let us try and understand this how this works and what is the consequence of this doing 4 different scans and adding and subtracting them as we indicated here. If you remember the first 2 pulses are similar as in the case of the COSY. So you have 2 90 degree pulses but the important thing is the 2 90 degree pulses have the same phase of course in the COSY also we considered 90X, 90X. So in this case also we do this 90X, 90X and this calculation for the density operator remains the same as in this case. So what we do here is we directly write the value of the density operator at this point starting from the K-spin magnetization here. Here the initial magnetization is it will be Z magnetization of K-spin and L-spin for a 2-spin system KL system and we illustrate the calculation starting from the K-spin and with the K-spin here you have the KZ magnetization you apply this 90 degree pulse and continue through the evolution as we did in the case of the COSY and arrive at this point. When you arrive at this point what is the density operator? So this is phi is equal to X rho 4 is minus ikz cosine pi jkl t1 minus 2 ikx ily sine pi jkl t1 cosine omega k t1 plus ikx cosine pi jkl t1 minus 2 ikz ily sine pi jkl t1 sine omega k t1. In the case of COSY what we did was we said this portion of the density operator is not observable so therefore we discard it. We did not want to consider this any further because afterwards when you collect the data immediately this does not contribute to observable magnetization therefore we discarded it and started continuing with the calculation further using only these 2 terms. And this we said this KX terms leads to the diagonal term and this is the 2 ikz ily and this is the Y magnetization of the L-spin and this leads to the cross peak. This leads to the diagonal peak, this leads to the cross peak because in the t1 dimension we have the frequency omega k. So omega k and this will evolve with omega k in t2 therefore this will give the diagonal peak and this will evolve with omega L in t2 therefore this will produce me the cross peak. And notice here is the Y magnetization for the cross peak at this point and here it is X magnetization of the K-spin at this point. So this is an important difference which you must remember because we are going to come back to it at a later stage. Now so when we repeat this experiment with phase of the pulse incremented by 90 degrees instead of phi is equal to X now we put phi is equal to Y. So what happens then you do the same calculation once more then what you arrive is given here. So individual steps I am not going to give you here again because you have done that in great detail when we did the Kozy experiment. So therefore we can simply take this result when phi is equal to Y the rho4 is given by minus ikz cosine pi jkl t1 plus 2 iky ilx sine pi jkl t1 cosine omega k t1 plus iky cosine pi jkl t1 plus 2 ikz ilx sine pi jkl t1 sine omega k t1. So according to the changes in the phase here you can notice the changes in the various density operator components. Here you had kx ly and here you have ky lx here you had kx now you have ky. So here at kz ly now you have kz lx. So these are the consequences of changing the phase phi. So now we are not discarding anything at this stage we are keeping all of this. Now let us look at the third experiment. The third experiment is phi is equal to minus x. So phi is equal to minus x it gives me minus ikz cosine pi jkl t1 minus 2 ikx ily sine pi jkl t1 whole multiplied by cosine omega k t1 plus and here I get minus ikx cosine pi jkl t1 plus 2 ikz ily sine pi jkl t1 sine omega k t1. I have numbered this equation says 1, 2, 3, 4 this is the equation number 3 and similarly when the two pulses are applied along minus y axis that is now phi is equal to minus y then I have here minus ikz cosine pi jkl t1 plus 2 iky ilx sine pi jkl t1 cosine omega k t1 and the other one is minus iky cosine pi jkl t1 minus 2 ikz ilx sine pi jkl t1 into sine omega k t1. So these are the 4 equations I get and these represent the density operator at the beginning at the end of the cosy sequence. So now since the last pulse is a constant phase pulse so it is not going to make any difference with respect to whether we consider the addition or subtraction of the results of all of these before or after the evolution after the 90 degree pulse. But what we do here is let us now try and see what is the consequence of the phase cycling that means if I consider the operations of these 4 FIDs and say I take here the plus sign and I subtract this second result add the third result and subtract the fourth result. If I do all of these because anyway this is the phase cycling so in the data is added or subtracted I can add those results here itself and see what is the final density operator at row 4. So the results of the 4 experiments which we have done here if I did a addition, subtraction, subtraction if I take these 4 density operators and do this operation what do I get and this algebra of course one can do it very easily and then you will find here that I get minus 4 iky ilx plus ikx ily sin pi jkl t1 cosine omega k t1. So you see this is very interesting and we have here these iky ilx ikx ily terms these ones were neglected in the cozy we said these ones are not observable therefore we are not going to consider that here what we have done is we are actually selecting only those we have eliminated all the other terms we eliminated the ikz term we eliminated the ikx term and also the ikx or ikz ily or ikz ilx whatever that was we have eliminated all the single quantum terms. Now this if you recall from the product operator descriptions this is a pure double quantum operator therefore by doing this operation at the end of 4 experiments the row 4 contains a pure double quantum operator the t1 dependent term of course the remains this sin pi jkl t1 cosine omega k t1 this is single quantum and this here is at this point of course we do not decide this is not this does not decide whether it is single quantum double quantum these are the coefficients the nature of this coherence is decided by these operators. So this is pure double quantum operator now the double quantum operator is not observable right. So therefore in order to observe this what we do is we apply the third 90 degree pulse the final 90 degree pulse transforms the double quantum operator to single quantum operator because it is only the single quantum which we can detect which leads to observable magnetization in the detection period. So therefore when we apply the last 90 degree pulse then I write the density operator row 5 as minus 4 ikz ilx plus ikx ilz and this coefficient remains the same sin pi jkl t1 cosine omega k t1. Now you see this is an important part in the earlier case when we are considering just before the detection the diagonal peak had one particular kind of a phase the cross peak had a different kind of a phase. Now you see both this one leads to the cross peak this leads to the diagonal peak why do I say this because here you have cosine omega k t1 and this term evolves now with k spin frequencies and this term evolves with the l spin frequencies and therefore this is the diagonal term and this is the cross term. Now we notice both of them have the same kind of a pattern this is also kx and this is lx it is anti phase with respect to l for the diagonal anti phase with respect to k for the cross peak and therefore the patterns of these are similar. So therefore you expect in the end that the patterns of the peaks in the diagonal and the cross piece would be similar along the t1 axis with these coefficients and this will result in the frequency modulation according to the evolution here. Now this now consists of anti phase magnetizations of both k and l spins with the same phase. So they will evolve during the t2 period into in phase magnetization of k and l spins whereby it comes observable as such these terms are not observable. I remember earlier also when these terms were presented after the evolution of the t2 we had ignored this because these ones do not lead to observable magnetization as they are unless they evolve further into ily and iky when they evolve into iky and ly then they lead to in phase magnetization which has a non-zero trace with the observable magnetization operators ikx or ilx. So therefore this will evolve in the t2 period to lead to observable magnetization which is in phase magnetization of k and l spins whereby it becomes observable. Therefore we are now going to consider the evolution of this. Let us rewrite this rho phi operator as in the simplified notation we have same this part and for the this coefficient I have written as f t1, f t1 is sin pi jkl t1 cosine omega k t1. Now let us consider this evolution of rho phi in the t2 period. Now the t2 period both the Hamiltonians are operative the j coupling Hamiltonian as well as the shift evolution Hamiltonian they are both operative. Now we can consider the evolution independent of the sequence. Let us first consider the j evolution. If I do the j evolution I take one 2 factor inside so I remain keep the second one here minus 2 and then this gives me 2 ikz ilx cosine pi jkl t2 plus ily sin pi jkl t2. The second term gives me 2ikx ilz cosine pi jkl t2 plus iky sin pi jkl t2. Now this is after the t2 evolution. So after the t2 evolution in j this term as the interface this term has remained as it is and we are generated in phase terms here with the coefficient sin pi jkl t2 but there is a residual this antiphase terms with the coefficients cosine pi jkl t2. These ones will not be observable because if I take the trace of these operators with ilx or ikx this will be 0 and therefore these terms we need not consider and we will consider only these terms which will be observable magnetizations. So next we consider therefore the evolution of this under the shift Hamiltonian. So this gives me minus 2 this first term gives me ily cosine omega l t2 minus ilx sin omega l t2 and multiplied by the sin pi jkl t2. The second term gives plus iky cosine omega k t2 minus ikx sin omega k t2 and multiplied by this and sin pi jkl t2. And we have this f t1 here and this f t1 multiplies the whole bracket. So now we consider y detection as has been the convention we have always assumed this and therefore let us continue to assume that can also detect magnetization of course but that does not matter. So without loss of generality we will assume this particular detection here and then we have the following signal. So at that point we will ignore this x terms. So we will only have this ily cosine omega l t2 sin pi jkl t2 and this one will be iky cosine omega k t2 sin pi jkl t2 and the whole thing is now multiplied by this f t1 period which is f t1 coefficient and this is sin pi jkl t1 and cosine omega k t1. Therefore what I have got here this part this gives me the diagonal peak after 2 dimensional Fourier transformation because this is the cosine omega k here and cosine omega k here. This term gives me the cross peak because I have cosine omega l here and cosine omega k here. Therefore along therefore this leads to the cross peak and this leads to the diagonal peak. Let us look at this in more explicit forms. So the signal that I am going to collect will be of this step because the operator part has been removed because you already taken the trace with ily and iky. So therefore I have only written the coefficients which are the ones which are responsible for the frequencies. The cross peak is cosine omega l t2 sin pi jkl t2 and the diagonal peak is cosine omega k t2 sin pi jkl t2 and multiplied by the sin pi jkl t1 cosine omega k t1. Let us analyze the cross peak. For the cross peak I take this and this product and that is written here cosine omega l t2 sin pi jkl t2 sin pi jkl t1 cosine omega k t1. So therefore the t1 part can be split into 2 as is indicated here. So this is the sum of 2 sin terms. Similarly this t2 dependent part is also sum of 2 sin terms here and this is when both are sin terms this is sin omega k plus pi jkl t1 minus sin omega k minus pi jkl t1 and likewise the t2 part also gives me 2 sin terms sin omega l plus pi jkl t2 minus sin omega l minus pi jkl t2. So those ones are written explicitly here the 4 terms are separated out from this product. So what this gives me? Now if you see this first term the first term has 1 t1 part and 1 t2 part. So after 2 dimensional Fourier transformation this will give me a frequency at omega k plus pi jkl and this along the F1 dimension this gives me a peak at omega l plus pi jkl t2 and along the F2 dimension. This is one component of the cross peak. This has a plus sign. The second component is sin omega k plus pi jkl t1 as the same as this and sin omega l minus pi jkl t2 and this gives me a frequency at omega k plus pi jkl along F1 omega l minus pi jkl and F2 and this has a minus sign. So similarly there will be another minus sign minus peak here negative peak with omega k at omega k minus pi jkl t1 and omega l plus pi jkl t2 pi jkl this is along the F2 this is along F1. So the fourth one will be with the positive sign omega at frequency omega k minus pi jkl along F1 and omega l minus pi jkl along F2. So therefore these cross peaks have 4 components of the cross peak have plus minus minus plus structure. So this is what is indicated here. Now notice also that all of them were sin terms and we said earlier that this sin terms will actually lead to a dispersive line shape. Each one of them was a sin term and therefore it would lead to a dispersive line shape. But we want an absorptive line shape. An absorptive line shape how do you get? It is possible to get an absorptive line shape by doing 90 degree phase shift on the signal. A 90 degree phase shift on the signal we can do in which case we will generate a absorptive line shape. Of course one could have argued that we could have done the same thing in the cosy as well. Yes indeed but then of course when you do that you see the cosy the diagonal peak and the cross peaks had different line shapes and therefore if you convert one of them into one shape the other one will be changed to another shape. Therefore if you make the diagonal peak dispersive and the cross peak absorptive as it was but if you convert that cosy diagonal peak into absorptive line shape then it will have a dispersive line shape in the cross peak which is not desirable. So now we want to have a absorptive line shape in the cross peak and therefore we did not do anything with regard to cross peak in the cosy spectrum. But here we have a dispersive line shape as it is coming therefore we convert this into absorptive line shape by doing a 90 degree phase shift. What is the consequence of this in the diagonal we will see in the next few minutes. So when I do this 90 degree phase shift the sign does not change. So I get a plus absorptive minus absorptive minus absorptive and plus absorptive. Therefore the 4 components will have the plus minus minus plus structure with absorptive line shapes. Now let us look at the diagonal peak. The diagonal peak has cosine omega k t2 sine pi j k l t2 sine pi j k l t1 cosine omega k t1 notice it is the same kind of a structure as it was in the cross peak except that we have omega k frequency here. Now you write split them out in the same manner as before then I get the same 4 terms here at frequencies omega k plus pi j k l along f1 omega k plus pi j k l along f2 and omega k plus pi j k l along f1 omega k minus pi j k l along f2 omega k minus pi j k l along f1 omega k plus pi j k l along f2. Notice these are all diagonal components therefore I have omega k here in all of them and omega k minus pi j k l along f1 omega k minus pi j k l along f2. Now once again all of them are sine all of them are sine terms. So therefore, I get here in the diagonal peak the 4 peaks at the following coordinates as indicated here nu k plus jkl by 2, nu k plus jkl by 2, nu k plus jkl by 2, nu k minus jkl by 2 and nu minus jkl by 2, nu k plus jkl by 2, nu k minus jkl by 2, nu k minus jkl by 2 and all of them have these kinds of a signs. This is positive, negative, negative, positive and this also has dispersive line shapes as they are sine terms. But now therefore I can do a phase shift. So that is the important difference. So in either case we had dispersive line shapes in both the cross peak and the diagonal peak. Therefore, if I apply 90 degrees phase shift here, here also I will get absorptive line shapes. Therefore, I get plus absorptive, minus absorptive, minus absorptive, plus absorptive. This is the important difference with respect to the Cause that both the diagonal and the cross peak have the same line shapes and also a positive, negative component structure and therefore if I do a phase correction on the entire spectrum then I get absorptive line shapes in both the diagonal and the cross peaks at the same time and additionally I will have plus minus structure in the diagonal as well. So what is the consequence? The consequence is this. So this was the Cause and this is the double quantum filtered Cause. See here I have the dispersive line shapes in the Cause and absorptive line shapes in the Cause in the cross peak and the diagonal has dispersive line shape. If I were to convert this into absorptive line shapes, this would become dispersive line shapes and vice versa and that is why we do not want to do that. We want to have a good design shape here because these are the information carriers. We do not want these ones to be masked and resolution to be lost. So it is not possible for us to do that and we keep the dispersive line shape in the diagonal peak and absorptive line shape in this. This of course was a disadvantage. Now in the double quantum filtered Cause we get over this problem. I get absorptive line shapes in the diagonal as well as in the cross peak and I also have plus minus character here. The result of this will be any tails which are present on this line shapes will get cancelled and therefore I will have much higher resolution in the diagonal area as well. The cross peak dispersion remains the same as in this case. The cross peak, the fine structure and therefore the resolution in this remains the same as it is here but the gain is in the diagonal peak you have the diagonals are absorptive line shapes and plus minus dispersion. The same thing happens for the L-spin. We started the calculation with the K-spin. So we went through the double quantum filtered and came to the observable single quantum magnetization. So it resulted in this structure here. If I were to start from the L-spin then I will also get diagonal peak here and the cross peak here for with the similar kind of a structure. So this is the big advantage. You see the experimental spectrum. The experimental spectrum you see there is a comparison here for a simple small molecule this is taken from this book here and as it is shown at 14 you can see the phenomenal improvement in the resolution around the diagonal. So you see all those things which are very close to the diagonal you cannot see them here at all and those ones are very clear in this spectrum and they are very well resolved. The consequence of that is because of the cancellations on the tails this cross peak is very poorly seen here and this is seen very clearly in this and therefore the double quantum filtered cosy saves a lot of trouble and the resolution improvement allows you to determine the structures much better. Notice also here that these ones are not simple two spin systems in the experimental spectra. These also are further fine structures you have so called in phase splitting as well and this is a consequence of multi spin system which we discussed in the COSY experiment earlier. So same thing is happening here there is a further splitting here in this in phases splitting plus plus minus minus and the same thing is retained here as well. So therefore you have for all the multiple spin systems you have an improvement in the diagonal peak and the cross peak fine structure remains more or less the same and this is a big advantage in your two dimensional spectrum. And here is another example of a much larger system that was a smaller system this is actually on a protein system. See one of the regions from the protein a phenylalanine residue here and this is actually completely merged. So it is impossible to do anything in the COSY spectrum here. You cannot identify any cross peak here in this COSY spectrum whereas those ones have become very clear here in this as well as in this area. So these cross peaks which are coming so close to the diagonal these are very well resolved in this double quantum filtered COSY. So this was a big boost for resolution enhancement in the correlation spectra. So therefore the double quantum filtered COSY on the one hand improved the resolution and allowed you to analyze complex spectra and on the other hand it also demonstrated a strategy how to select particular pathways of magnetization. So we chose here a double quantum signal and converted the double quantum signal into observable single quantum by applying a third pulse and therefore your pulse sequence was slightly modified. So this is a very general phenomenon. One can do this for various kind of filters and also use this strategy for selecting different kinds of pathways. For example if you want to select the Z magnetization then also it is possible. You have to use the different kind of a phase cycling. The five pulses which I had mentioned there are used a particular kind of a phase cycling for addition subtraction and changing of the phase xy minus x minus y and addition subtraction etc. But you can use the different combination of this phase and the receiver phases is addition subtraction etc to choose another kind of magnetization. For example the Z magnetization in which case you generate a different kind of correlations in your spectrum. So therefore that also will be used and we will see therefore this is a good illustration of the strategy of selecting the coherence transfer pathways in your spectrum. So with that I think I will stop here and we will continue with further methods of improving the spectra in the next class.