 So, let us continue the discussion of the evolution of the basis operators when you will consider various kinds of pulse sequences. So, we have to know how the individual basis operators evolve under the influence of different Hamiltonians. Basically, there are two components of the Hamiltonian we are dealing with and there will be a Ziemann Hamiltonian and the coupling Hamiltonian. And last class we looked at the evolution of a particular base operators base operator IKX under the influence of the chemical shift Hamiltonian. Today, we will discuss the evolution of that same operator IKX under the influence of the scalar coupling Hamiltonian. So, to remind you as to what is the scalar coupling Hamiltonian, let me just write here the scalar coupling Hamiltonian is written as Hj is equal to 2 pi Jkl, Jkl is the coupling constant between two spins K and L and we have these operators IKZ, ILZ. So, this is the for the two spin case this is a coupling Hamiltonian. So, for the evolution what we need to calculate is we have to calculate e to the minus iHjt and we using the same basis operators this will be e to the iHjt. And if I explicitly write this here so I have this Bs double prime this is the operator that we get as a result of the evolution here you have e to the minus i 2 pi Jkl IKZ ILZ t IKX e to the i 2 pi Jkl IKZ ILZ t. This individual operator can be expanded or written in a simpler form like this e to the minus i 2 pi Jkl IKZ ILZ is equal to cosine pi Jkl t by 2 minus 4i sine pi Jkl t by 2 IKZ ILZ. This can be proved along the same lines as we derived the expressions for e to the i BX for example. When we took the expression for e to the let me write here e to the minus i beta IKZ so we wrote in a particular manner here cosine beta by 2 minus i sine beta by 2 and things like that. So, this was X here if you are applying it for the pulse this was X here if beta is a flip angle then we wrote is equal to cosine beta by 2 minus 2i sine beta by 2 IX so that was how it was written. So, in the same manner we write this here except we make note of the fact that IKZ ILZ can be written as in terms of the poly spin matrices 1 by 4 sigma KZ sigma LZ. So this will help us in deriving this equation and poly spin matrices satisfy this condition this sigma Z square is equal to sigma LZ square is equal to 1. So, this is the property which we should remember and we can prove this equation. So, therefore I am not going to explicitly prove this, this can be a kind of an exercise for the students who can practice this and prove this equation. So, we will actually use this equation now as it is and calculate the evolution of the KX operator under the influence of this. So, in this place we will have to put this and on this place what we should do IKX and this will be a plus sign when you write for I2 pi i Jkl IKZ ILZT then this the same as this except we will have a plus sign here for I sine and then you do the multiplication of this part IKX and the other part which is with the plus sign then you arrive at this equation IKX cosine pi JklT plus 2 Iky ILZ sine pi JklT. So, this is the kind of equation that will come. You notice here that IKX has evolved into a in phase IKX term and it has generated what we may call as your anti phase term it is the K magnetization which is anti phase with respect to the spin L. You remember earlier we have discussed what these individual basis operators represent therefore here we have said under the influence of the coupling Hamiltonian the IKX operator will evolve into this IKX cosine pi JklT plus 2 Iky ILZ sine pi JklT. So this is the anti phase term. So such the evolutions will keep happening and we will look at them as we go along. Now similarly you can calculate for the basis operator Iky evolution under the Zeeman Hamiltonian this is the chemical shift Hamiltonian and the coupling Hamiltonian Hj and they can be represented in this manner we will not prove this basically they can be derived in the same manner and we will just write it down here and we can use them later. So the BS prime will be cosine omega KT Iky this is for the K spin K for the single spin and we are minus sine omega KT IKX. So because here we are consider the evolution of the KX operator so the KY operator the KY operator evolves with the Iky here and minus sine omega KT IKX. Similarly the coupling Hamiltonian generates this sort of the evolution the BS double prime Iky cosine pi JklT minus 2 IKX ILZ sine pi JklT. I want to point out here one thing that when you are actually calculating IKX remember we had IKX here in this term but plus 2 IKY ILZ. Now here you see if you are starting with IKY this gives you IKY cosine pi JklT minus 2 IKX ILZ. Similarly in this case also when you had IKX evolution plus IKX cosine omega KT plus IKY sine omega KT if you are considering IKY then it is IKY cosine omega KT minus sine omega KT IKX. Therefore this indicates in some way a kind of a rotation in the transverse plane and you can pictorially represent them very soon I already indicated this to you in the chemical shift evolution earlier that such kind of a representation can be used for calculating the evolution of any of these operator whether it is IKY or IKX or minus IKY minus IKX and things like that. Now similarly we can do this exercise for the basis operator 2 IKX ILZ evolution under HZ and HJ. Now HZ works on individual spins therefore if I consider 2 IKX ILZ though which is the transverse term here the transverse term is IKX. Therefore chemical shift evolution will happen only for the IKX part ILZ part will not contribute to the chemical shift evolution. Therefore if I consider the BS prime here so this will be simply 2 and IKX term evolves as in the case as indicated here earlier IKX cosine omega KT plus IKY sine omega KT and ILZ remains as it is there is no change. So even if it is in this form the 2 product operator the chemical shift evolution works on the individual operator because we have the ILZ here and we have the ILZ coming here IKX evolves in the usual way as it is for the individual single spin. Now the coupling Hamiltonian the causes this sort of a transformation here it gives you cosine pi JKLT 2 IKX ILZ. So the anti-phase term which remains it has is amplitude modulated by this expression cosine pi JKLT and on you generate here IKY plus sine pi JKLT. Notice here that when you consider IKY rotation you got IKY cosine pi JKLT minus 2 IKX ILZ sine pi JKLT. Now we are considering this you are generating the same terms once more so it is this appears in the positive sign then you have plus sine pi JKLT IKY. So this indicates you that these kind of terms some form some sort of a group they transform within themselves the Hamiltonian here is IKZ ILZ and the operator is IKX ILZ the basis operator is 2 IKX ILZ and we have also another basis operator 2 IKY. So 2 IKZ ILZ 2 IKX ILZ and IKY they somehow form a kind of a group and similarly for the IKX as well this can be represented in a pictorial manner like this. So similarly I can write the same kind of expressions for 2 IKY ILZ under the influence of HZ and HA so BS prime here gives me the same here evolution of IKY as individual chemical shift evolution and ILZ remains here and BS double prime which is the result of evolution under the coupling yet cosine pi JKLT 2 IKY ILZ same here and it generates the IKX term. So as I said these indicate some sort of rotations in a particular plane pictorially we can represent it in like this. So we put the Hamiltonian this is for the coupling similar thing we wrote earlier for the chemical shift so the coupling the Hamiltonian is written along this axis here so IKZ ILZ this is the coupling Hamiltonian and if I take the particular base operator IKX here then it under the influence of this rotation evolution under this this circle indicates the evolution under this Hamiltonian it generates rotates in this manner so you get you get when it rotates like this you generate components of IKX and 2 IKY ILZ suppose it is rotated up till here then you generate IKX cosine pi JKLT and assign IKYLZ sine pi JKLT and if you were to continue this rotation under the influence of the same Hamiltonian like this so what you will get though this vector has moved over here so you get 2 IKY ILZ cosine pi JKLT as it is here suppose it has moved until this I want to draw a vector as well here if I wish to draw a vector here so then if this rotation has happened for a particular angle this angle and this will be the pi JKLT and you will write the cosine pi JKLT so if this is pi JKLT here then you will have 2 IKY ILZ cosine pi JKLT and this component will be minus IKX sine pi JKLT so and similarly if you continue this if you were to start here you will move here if you were to start here you will move here and things like that therefore these actually transform within themselves this IKX 2 IKY ILZ and IKZ ILZ this form a group with transform within themselves go from here to here you get this from here to here you get this and this you can go from here to here again you get this and this under the influence of the same Hamiltonian so if I were to write here in the similar manner here 2 IKX ILZ so I get here IKX going to IKY and IKY going to minus 2 IKX ILZ here and so and this continues to generate me this portion here and the minus IKY here so like that so this whole set of operators which are present here 2 IKX ILZ IKY and this coupling operator here this 2 IKZ ILZ is for kind of a rotation group when I say rotation group they transform within themselves so under the influence of one thing the one operator transforms into the other so that is the meaning of the rotation group. So this is the very these are very useful expressions for calculating the evolution of any density operator as we get along okay now that is so much for the evolution of the basis operators so though the illustrative examples which I gave you but there will be we have to do this exercise for every other rotation basis operator this will be this can be easily done because you already indicated what sort of the principles are involved what sort of rotation groups are formed and therefore it is easier to write down the expressions for all of those. Now we turn to the evolution under the influence of pulses what do the pulses do and we have seen earlier the particular basis operator gets transformed by this sort of a transformation if this represents a rotation this represents a pulse here and this is the basis operator this is the gain a pulse PPS P-1 and the pulse can be applied along the X or the Y axis and therefore I write here RQ the rotation operator RQ and RQ-1 so this represent the pulses Q can be X or Y. Now let us actually calculate the influence of this on some basis operators we take as an illustration BS is equal to IZ this is the simplest all of all of those and for a 90 degree X pulse the transformation will be we have to write the individual matrices for this pulses so we have here R a suppose we apply it along the X axis we have said 90 X pulse therefore this is Rx pi by 2 IZ Rx minus 1 pi by 2 and we have derived earlier what this Rx pi by 2 is in terms of a matrix so this will be 1 over root 2 1 minus i minus i 1 and then for the IZ we have this representation 1 by 2 1 0 0 minus 1 this is the matrix because single spin we are considering a single spin here and Rx minus 1 pi by 2 will be the inverse of this inverse of this matrix so this will be 1 by root 2 and that is 1 i i 1 so one can calculate the inverse of this if you may take a product of this matrix and this matrix you will get 1 so that indicates that this is the inverse of this. Now when you do all these multiplications you will get 1 by 2 0 i minus i 0 and this is equal to minus i y so what this has told us that 90 degree X pulse rotates the IZ into minus IY so this is also what we had said earlier in the discussion with the derivation of the pulses and that was we can write is so this is the X, Y, Z and if I have a vector here IZ operator this gets transformed into minus Y so it goes here X, Y, Z goes into minus Y gets rotated here to minus IY IZ goes to minus IY if I apply a pulse along the X axis so that is the calculation what we got from this 90 degree X pulse so IZ operator 90 X this is indicated here in the figure so actually so this figure is already there so if I write here rotation around X and this is Y and this is the Z so rotation operator is along this IX and Z magnetization is rotated into the minus IY so from the same you could also see what if your magnetization was along the minus Z axis it will go to Y if it were along the Y axis it will go to plus Z so if you are considering the this initial operator here as IY then you can imagine that this IY will go to IZ we said IZ will come to minus IY this is what we had in the explicit calculation now if you were to start with IY the same rotation will take IY to IZ okay and IX will of course be invariant nothing will happen will explicitly show that and if you were to have minus IY then minus IY will go to minus IZ under the influence of this rotation okay now that is explicitly calculated here if BS is equal to IY then what I have to say here RX pi by 2 IY RX minus 1 pi by 2 now I put the same matrices here 1 by root 2 1 minus I minus I1 this is half half this is the IY operator half 0 minus I I0 and this is 1 by root 2 1 I I1 this is the inverse of this matrix and if you calculate this you get half 1 0 0 minus 1 that is IZ so this is the same which told you before that under the same rotation the IY is going to IZ so therefore these again for a rotation groups X Y and Z these four rotation groups they transform within themselves if there is a change in sign of course that is not considered different it is the same rotation group just except with the coefficient will be different so this rotates Y into Z rotates Z into minus Y and Z rotates minus Y into minus Z and so on so this forms a rotation group you remember if I had my the rotation around the Z axis rotation around the Z axis was what that is the chemical shift evolution when I had the chemical shift evolution my operator was going from here to here IX was going to part of it is was rotating in this plane X to Y Y to minus X minus X to minus Y and so on so forth therefore these three operators for a rotation group and one of them can be the basis operator other one can be a Hamiltonian in this case the operator is the IX rotation is around this axis therefore and we have the basis operator is IZ or IY in the chemical shift case this was the Hamiltonian this was the operator and the basis operators were here and therefore one could go from IX to IY to minus IX and so on so forth so this is the way the we can look at all of this in a comprehensive manner so this IY goes to IZ for a Y pulse earlier I wrote as X pulse so what does the Y pulse do so the Y pulse take this rotation here we can do the same calculation so it goes IZ goes to IX notice we have a left-handed rotation here IZ goes to IX and IX will go to minus IZ and minus IZ will go to minus IX and so on so this is the way the we describe the rotation groups rotation under the influence of the various pulses now what about IX itself suppose your base operator is IX and I apply 90 degree X pulse intuitively you would think that okay nothing should happen to it but let us also prove it so we put RX pi by 2 IX RX minus 1 pi by 2 so I put the same matrices here 1 by root 2 1 minus I minus I 1 half 0 1 1 0 1 by root 2 1 I I 1 and this will give me half 0 1 1 0 and this is IX so therefore the IX is invariant under RX pulse so that is intuitively one would have imagined that you have put the magnetization along this same axis and you are applying a rotation around that axis and you should not move in principle now let us consider multi spin basis operators so far we looked at the individual spin operators and we now turn to multi spin operators the effects of pulses can be applied in individual spins now the pulses can be applied on all the spins or on individual spins so the effects will be different now here we consider a case we consider a basis operator 2 IKX ILZ and we have seen this operator represents the K magnetization which is anti-phase with respect to L and this is a single quantum coherence because this is KX KXLZ is a single quantum coherence X magnetization of K anti-phase with respect to L now we apply a pulse on both the spins 90XK plus 90XL so this is called as a non-selective pulse it is applied to both the spins K and L so now what happens when I apply KX pulse nothing happens to KX right because we said this is invariant X magnetization is invariant under the X pulse therefore nothing happens to the KX operator and what happens to LZ now LZ goes to minus Ly this is we calculated Z goes to minus Y right so therefore X pulse on L spin take this ILZ to minus ILY therefore I get here minus 2 IKX ILY so what is this now here we got a mixture of double quantum and zero quantum coherence this we have explicitly calculated earlier that this represents a mixture of double quantum and zero quantum coherence so what we have done here is by applying a non-selective 90 degree pulse on anti-phase magnetization of one spin we have converted that into a mixture of double quantum and zero quantum coherences so these are called as coherence transfers and various kinds of experiments will make use of such kind of transformations. Suppose I take the same operator here now apply a pulse not 90X pulse but apply 90Y pulse on both the spins so if I take X what happens to X so you have to recall when I apply a Y pulse what happens to X so and 90Y on LZ will take me to 90LX and this will take me to minus 2 IKZ so you have to go back and look at what happens when I apply a Y pulse on X magnetization let us go back and look at this so this will be easier to follow if you do that so looking at here so what happens when I apply Y pulse on X magnetization it will go to minus Z and Y pulse on Z will take me to X so therefore this apply them here so I get this will go to minus Z and LZ goes to LX therefore I get here minus 2 IKZ ILX now notice here just the phase change of the pulse has made a huge difference in the transformation this was anti-phase magnetization of the K spin with respect to spin L and now what we have here this is X magnetization of L spin anti-phase with respect to the K spin now this is a single quantum coherence this was single quantum this was single this is also single quantum when we applied a 90X pulse we had got double quantum plus zero quantum here from this basis operator now when you apply a Y pulse to the same basis operator I get a single quantum coherence of the L spin so this is a different kind of a coherence transfer single quantum to single quantum from one spin to another spin which are coupled from K spin to L spin I get a transfer and this is a single quantum to single quantum coherence just by changing the phase of these 90 degree pulse so in the earlier case when the 90 degree pulse was along the Y axis apply to both the spins so I got a single quantum coherence into a mixture of double quantum and zero quantum coherence and so this has important implications so this represents conversion of anti-phase X magnetization of K spin into anti-phase X magnetization of L spin that is the single quantum coherence transfer now suppose we apply the again once again with the same operator base operator here but apply a pulse only on the K spin for example I will not apply along the L spin at all that means I have applied selective pulse a selective 90 degree pulse on one spin only I apply to only the K spin so therefore the L spin remains unaffected because I have not applied any pulse on to that and K X takes me to the minus K Z therefore this will go to I K X I L Z takes me to minus 2 I K Z I L Z so is a completely different transformation once more now this anti-phase magnetization has now got transformed into Z Z order Z Z order has to do with the order of the of the populations in the two states on the top alpha alpha state and the beta beta states as we had discussed earlier so this anti-phase magnetization of the K spin is getting transformed into Z Z order simply by changing what kind of pulses we apply we get different kinds of transformations of the basis operators so now if I apply on the other hand a the same basis operator once more here but I apply a pulse selective pulse only on the L spin not on the on the K spin earlier when I applied along the K spin I got Z Z order so now apply it only on the Y spin so what I get here K X remains K X because no pulse applied and L Z takes me to L X 2 I K X I L X now this is a mixture of double quantum and zero quantum coherences so therefore you see once again how a single quantum is converted into double quantum plus zero quantum coherence by changing the kinds of pulses we apply so you notice therefore that a combination of different kinds of pulses with the different cases can really be used to create a whole lot of transformations in your density operator and all of these one we will use when we actually calculate various pulse sequences and that will become an important but this will be easy to calculate in a simple operator form product operator form so similarly here 2 I K Y I L Z so far I was looking at 2 I K X I L Z and we can calculate similarly for 2 I K Y I L Z if I start with 2 I K Y I L Z apply a non selective pulse on X both K and L spins along the X axis then I get this into 2 I K Z I L Y and this is now the L magnetization interferes with respect to K this was K magnetization interferes with respect to L and this is L magnetization interferes with respect to P so single quantum to single quantum coherence transfer so these are important important results which will play a very major role in how we calculate the evolution of the density operator through the pulse sequences so this is referred to as coherent transfer from spin K to spin L in general it is seen that application of RF pulses to interface magnetization in multi spin systems causes coherent transfer among the spins is from the basis of many multiple experiments in homo and heteronuclear multi spin systems so here is a summary of all of these evolutions what I have indicated here once again as a recap so because this will be extremely useful and we have to remember this very well we have to remember this by heart which operator turns which basis operator into what so which Hamiltonian converts what basis operator into what basis operator so we will have to see all of this so if this is chemical shift evolution your Hamiltonian has the i z operator here therefore we wrote writing this manner and your basis operators are i x and i y therefore i x rotates into the i y and i y rotates into minus i x minus i x rotates into minus i y and so on so forth so all transformations happen here these form a rotation group all the these three operators form the rotation group so scalar coupling evolution once again the same thing is repeated for the benefit of consolidation so you have this the Hamiltonian is 2 i k z i l z we put it along the z axis this is my axis this is my axis so this axis is now 2 i k z i l z and i k x is rotated into mixture of i k x and 2 i k y i l z and the same Hamiltonian rotates the basis operator 2 i k y i l z into mixture of 2 i k y i l z and minus i k x so if I were to start from minus i k x I will get minus i k x plus the cosine component will be minus i k x and I will get here minus 2 i k y i l z now if I want to put here the anti phase magnetization here it was a starting with the in phase magnetization this is how they transform if I start with 2 i k x i l z this anti phase magnetization and this operator remains the same 2 i k z i l z this rotates into i k y so this will be the cosine component will be 2 i k x i l z the sine component will be i k y so if I consider the rotation further i k y rotates into minus 2 i k x i l z partly it remains as i k y the cosine pi j k l tau will be i k y and the sine will be minus 2 i k x i l z minus 2 i k x i l sine pi j k l tau so this is the way rotation happens I mean when you wrote here this is the rotation is happening depending upon the value of tau your vector will be somewhere here or here or here wherever so if it is equal to pi by 2 if tau is such that this whole rotation is pi by 2 then you will go entirely from here to here if it is not pi by 2 if it is the whole thing is equal to 30 degrees or 45 degrees or something then your vector will be somewhere here so you will have a cosine component and a sine component similarly here if you are rotating by an angle pi by 2 that depends upon your value of tau then this goes completely into this but if it is less than that the vector can be somewhere here then you will have a cosine component here and a sine component so with regard to this so this is the way we actually calculate the evolutions under the influences of various Hamiltonians so to about the rotation by pulses so if I have x pulse the i z rotates into minus i where notice here here we consider a 90 degree pulse which goes into this but you can extend this argument to say well I do not want to apply 90 degree pulse suppose I apply a 45 degree pulse what happens in fact very easy to imagine here now we understood the principles how the rotations are happening if I were to rotate by 45 degrees then the rotation will take the vector from here to here so this angle will be 45 degrees so I will have a component i z and a component i y minus i y so a any angle can be chosen so if I have a 90 degree pulse it takes me to minus i y but if I have a beta pulse then I will have cosine beta component for the i z and the sine beta component for the minus i y so similarly if I apply this pulse to the i y magnetization goes for a 90 degree pulse it will go here but suppose I were to apply a pulse which is not 90 degrees somewhere 45 degrees or something or 30 degrees then I would rotate it only up till here then we will have a cosine component here and a sine component here similarly if I were to take a rotation around the y axis a pulse along the y axis the rotations will happen like this i z goes to i x i x goes to minus i z minus i z takes you to minus i x and again minus i x will take you to i z once again the same arguments apply with regard to the flip angle instead of a 90 degree flip angle if you use a different flip angle you will get different components along the y and the z axis so with that we will stop here and we will continue the discussions in the next class.