 We are now going to go over to a second category of two-dimensional experiments namely two-dimensional correlation experiments. These have been by far the most useful experiments for studying a variety of systems, large systems, small systems etc. and has been extremely useful in characterizing molecules. Then the fine structures in the molecules understanding the connectivities of the individual carbons in molecules and so on and so forth. And the typically started with homonuclear proton correlation experiments but today we also various kinds of correlation experiments which deal with proton or carbon 13 and proton with nitrogen 15 and so on and so forth. So first we will look at the proton-proton correlation experiments and this because in the initial years it was all homonuclear proton experiments only which were designed and they were used for variety of purposes. So this experiment is called as the COSY, correlated spectroscopy. So the abbreviation is COSY and it is very commonly used for all organic chemistry and biochemistry and biological chemistry and what not. So this experiment consists of simple pulse sequence it is experiment with the two pulses you start with a 90 degree pulse can be X or Y or whatever but conventionally we are simply sticking to X all the time and then you followed by a T1 evolution period then you have another 90 degree X pulse and followed by the detection during the period T2. So in this period of course you may have this what we call as the preparation this includes the 90 X pulse in the preparation then you have the evolution this pulse 90 X pulse here itself acts as the mixing pulse in the generalized context of 2D spectroscopy and this is the mixing pulse and then you have the detection here of the signal during the T2 period. Now we will see that this kind of an experiment produces a spectrum of this time this is a schematic experiment. So if I call this as F2 axis which is where we actually detect the signal and this is the indirect detection axis we call it as F1 and that comes as a result of Fourier transformation along the T1 dimension. And now let us see what is the information here if you generate peaks like this so I will have a so called diagonal here the diagonal will have the same frequency along the F2 and the F1 axis. So therefore this all these peaks which are here they have the same frequencies whether it is along the F1 or the F2 axis. Now then it displays correlations here it displays these are called cross peaks and the cross peak displays a correlation between this particular spin and this particular spin when you try to call this as protons it is this proton and this proton there is a correlation between these two where does this correlation arise from? This arises from the coupling from the J coupling in this experiment the correlations arise as a result of J coupling these two protons are coupled to each other and therefore it produces a cross peak here and this is called as the cross peak or the off diagonal peak. And now you notice that this peak is also attached to another proton and that is this proton is also coupled to this proton therefore we generate a cross peak here as well between these two protons. So therefore this forms a network. So you have a network of spins here these three protons are coupled to each other this is coupled to this and this is coupled to this and we produce peaks on both sides this will be symmetrical spectrum here and no matter which side of the diagonal we use the information is the same. Now if there is a proton in your molecule which is not coupled to anything it will produce what is called as a singlet and it will not have any correlations along any of the axis there. Therefore it will be easy to identify which of your protons or singlets which are not coupled and which of your protons are coupled and what sort of coupling pattern spin systems exist in your molecule. So this is an extremely useful information for characterizing your molecular structures. Let us try and understand this using our standard method of product operator calculation. So this is the cosy of two spins once again we choose two spins k and l. So at the time point 1 so I have written here time points 1, 2, 3 and 4 will explicitly calculate the density operator at these individual time points so that we know what is the information content here and how the information is flowing through the pulse sequence. So at time point 1 so I have the equilibrium agronization so the density operator is i k z plus i l z considering the two spins. Now individually we can calculate the evolutions of this through the pulse sequence but let us for demonstration we will consider only the k spin and whatever results we generate similar calculations can be done for the l spin as well and therefore we do not want to repeat that for the two things. So we consider the calculation for the k spin only in the further discussion. Now so if I apply 90 expulse to the k agronization I generate minus i k y as before. Now this evolves under the Zeeman Hamiltonian which is omega k i k z for a period t 1 yielding the density operator rho 3 at time point 3 in the pulse sequence. So now considering the chemical shift evolution the i k y terms are also in this manner the minus sign is kept out here so I have here i k y cosine omega k t 1 minus i k x sin omega k t 1. Now we have to consider this further for evolution under the coupling. Next considering evolution under the j coupling Hamiltonian which is 2 pi j k l i k z i l z. So density operator will be let us say rho 3 prime so the rho 3 prime will be given by keep this minus sign as before and now we individual terms these operators we have to evolve under the coupling. So i k y evolution gives you this i k y cosine pi j k l t 1 minus 2 i k x i l z sin pi j k l t 1 and you have this cosine omega k t 1 from here. And this term gives you minus i k x cosine pi j k l t 1 plus 2 i k y i l z sin pi j k l t 1 and the sin omega k t 1 comes from here. So this is the density operator at the end of the t 1 period. Now what we are doing at the end of the t 1 period we are upto 90 degree pulse again. So here we notice that after we apply the 90 degree pulse these terms will get transformed into the particular manner. We get here at rho 4 the first bracket gives me this i k z cosine pi j k l t 1 because I apply 90 x pulse this was k y here and k y goes to k z and here it was k x l z l z goes to l y therefore I get here 2 i k x i l y sin pi j k l t 1 and cosine omega k t 1. The second term remains like this the k x is not affected so it remains as k x cosine pi j k l t 1 but here you see there is a change. So I get here minus 2 i k z i l y sin pi j k l t 1 sin omega k t 1. Let us look at this individual terms. So you see this one is z magnetization of the case k spin and now this is a mixture of the double quantum and zero quantum coherences and this is the x magnetization of the k spin. Now this one here is the l magnetization this is the y magnetization of the l spin which is anti phase with respect to k therefore you see the 90 x pulse the second 90 degree x pulse has caused a coherence transfer from the k spin to the l spin. So from the k magnetization we have generated l magnetization here so this represents a coherence transfer. Both these are single quantum terms although this is anti phase this is in phase whereas this one is z magnetization and this is multiple quantum transitions here you notice that these entire these operators do not lead to observable magnetization during the T2 period because the next what we are going to have is evolution during the T2 period whatever is observable there we are going to retain and what is not observable we will ignore because it is not going to lead to us any signal. Therefore these two terms here we can ignore because though these ones do not lead to observable magnetization and these ones actually lead to observable magnetization both these are observable terms. Now the first term which represents x magnetization of the k spin evolves during the T2 period with frequencies characteristic of k spin. So during the T2 period this evolves with the frequency omega k in the T1 also we have omega k in T2 also it will be omega k therefore this will produce the so called a diagonal peak which I mentioned to you earlier the F1 is equal to F2 is equal to omega k in the final 2D spectrum. The second term which represents y magnetization of L spin evolves under the T2 during the T2 period with frequencies characteristic of L spin. Therefore this will have T2 during the T2 period this will have frequencies of L spin but T1 it had frequency of k spin. So therefore this is this will be F1 is equal to omega k and F2 is equal to omega L therefore this produces what is called as the cross peak in the final 2D spectrum. Both these peaks will have fine structures which contain the coupling information and this we will we can calculate we can see that immediately from here that the Jkl term is appearing here and therefore we will have coupling information in the in this individual peaks. Let us first consider the diagonal peak term. The diagonal peak is arising from I kx the first term in this observable part. So a chemical shift evolution leads to the density operator rho 5D this is during the T2 period. So I call this as rho 5D and this is I kx cosine omega k T2 plus I ky sine omega k T2 and the entire modulation of the T1 evolution I simply write it as FD T1 and FD T1 is cosine pi Jkl T1 sine omega k T1. So you recall back and that is just this cosine pi Jkl T1 and sine omega k T1. So I call this as FD T1 and now we consider evolution of these terms under the coupling evolution under coupling generates the density operator rho 5D prime given by this formula here rho 5D prime is now coming from the evolutions of the individual terms here. The first term gives me here I kx cosine pi Jkl T2 plus 2 I ky I Lz sine pi Jkl T2 and cosine omega k T2 remains from here. The second one gives me I ky cosine pi Jkl T2 minus 2 I kx I Lz sine pi Jkl T2 and this sine omega k T2 is from here and then all of it is multiplied by FD T1. Now among these depends upon what signal we are going to measure and we will have to see what are the components which you have to retain. Let us assume that we detect y magnetization we will have to take the trace of the density operator with I ky. So when we do that we will only have this one here this will represent our signal cosine pi Jkl T2 into sine omega k T2 FD T1. Now put back this FD T1 you have here cosine pi Jkl T2 sine omega k T2 multiplied by cosine pi Jkl T1 sine omega k T1. What it is the pattern of this evolutions of the terms in the T2 and the T1 are the same here is again a sine cosine multiplication and again here it is sine cosine multiplication. So if you want to expand this further so this will give me so this is the same thing which is written here 1 by 4 sine omega k plus pi Jkl T2 it will generate 2 terms each right. So this will generate 2 terms each plus sine omega k minus pi Jkl T2 this is from the T2 from this one here and multiply the whole thing by this sine omega k plus pi Jkl T1 and sine omega k minus pi Jkl T1. So what does this tell you that already indicates you there are going to be 2 frequencies along omega 2 that is omega k plus pi Jkl or if you want to take out the 2 in terms of the hertz if you want to write frequency then it will be nu k plus J by 2 and this will be nu k minus Jkl Jkl by 2 because if you take away the 2 pi term here so this will be nu k minus Jkl by 2 nu k plus Jkl by 2. Similarly here along the F1 dimension also I will have 2 terms the nu k plus Jkl by 2 and nu k minus Jkl by 2. So therefore this entire diagonal peak will have 4 components 2 into 2 therefore this will produce 4 components and where this is the signal that we detect where do these ones appear now I write in terms of the frequencies removing the pi part so I will write in terms of the frequencies. Four peaks with a dispersive line shape why do I say dispersive line shape because these all have sine dependence and we have seen earlier that if there is a time domain signal which is sine dependent it will produce me dispersive signals. So we will have a frequency peak at nu k plus Jkl by 2 nu k plus Jkl by 2 along the F1 F2 dimension then I will have here nu k plus Jkl by 2 and nu k minus Jkl by 2 this is along the F1 this is along the F2 and this is again nu k minus Jkl by 2 and nu k plus Jkl by 2 and nu k minus Jkl by 2 and nu k minus Jkl by 2. So we will have 4 peaks these are centered around the nu k frequency nu k is my frequency of the k spin. So and all of these are positive because you see the previous thing they all have positive components here these are plus plus plus all of them are plus this is plus this is plus this is plus so all are plus therefore I will have all positive peaks and they all have dispersive line shapes. So how does the spectrum look like? So you see this is the way the spectrum will look like I have drawn the diagonal here so which is running through this. So the similar calculation for the L spin will produce me these 4 peaks there are 4 peaks here for the k spin I have shown you the calculation for the k spin. So all of these are dispersive line shapes and they all have the same sign and it will produce me a peak of this type and likewise if you did for the L spin you will produce a spectrum of this type and now this spectrum has been taken from one of the books and chemotechnics in organic chemistry and particular things have been dropped and this will come in the next class when I actually discuss the cross peaks as well here and those peaks are also present here at this point those will be the cross peaks. So I think we have discussed the diagonal peak here and in the next class we will look at the cross peaks how they appear. So you can look at this things once more at where this peaks appear and we have the peaks appearing at nu k plus jkl by 2 nu k plus jkl by 2 nu k plus jkl by 2 nu k minus jkl by 2 nu k minus jkl by 2 nu k plus jkl by 2 and here is nu k minus jkl by 2 and nu k minus jkl by 2. So therefore this produces a spectrum which is like this and we will see later that this sort of a dispersive line shape is not a very desirable thing and obviously one has to do something different there to get better line shapes in this point otherwise this will mask the signals which are close to the diagonal and that has been one of the problems of the normal Kozy experiment and developments which have happened to remove those and we will see of course in the next class how the peaks appear in the correlations which is the cross peak and we are considered one part of the density operator now we will have to consider the second part of the density operator at time rho 4 and which will then produces the signals which produce the cross peak. So we will stop here and continue with the cross peak calculation in the next class.