 So, we are going to continue the discussion of the density matrix for the description of the NMR experiments. As a recap I have put here the generalized form of the density matrix for an n level system. So, we have this n, this small n there is n level system and if you recall all the diagonal elements in the density matrix represent the populations of the individual levels. So, these ones I have represented as P1, P2, P3, P4 and Pn like that. The off-diagonal elements represent the phase coherences between the spins in the individual states. For example, if I take this the energy levels are labeled 1 to n, this term represents a phase coherence between the spins in the states 1 and 2. The extent of phase coherence is represented by this coefficient here C12 and this shows the time dependence of this phase coherence and that goes with this frequency omega12. So, this is an oscillating function, this is phase coherence, this is the degree of phase coherence and that is oscillating with the frequency omega12. And this will be the phase coherence between the states, in the spins in the states 1 and 3 and this is the extent of phase coherence of the magnitude amplitude and this is the phase. How does the phase change between the two spins? So, it is oscillating with time and like that for every pair of states you have a phase coherence between the individual states and so therefore, in principle you can have phase coherences between any two states and therefore, here nothing is assumed to be 0. However, in reality of course, all of these terms will not be non-zero, there will be some terms which will be 0, which some terms which will not be 0 and that we will explicitly consider when we take specific examples. Now, having seen what the density operator is, the density matrix is and we must also see what is a measurement. In fact, we have discussed this earlier also. When we measure a signal in NMR experiment, what we are actually measuring is the so called expectation value of the relevant operator, Mx or My or M plus minus, M plus minus these are called as raising and lowering operators. This is called Mx plus My. But remember here, earlier I used symbol, here I used the symbol Ix to measure x magnetization or Iy for measuring y magnetization and you can also measure in between the two and that is called as Ix plus minus I Iy and these ones are referred to as I plus or I minus. So, for convenience I have used here Mx to say it is magnetization, but these earlier I have used the nomenclature Ix and Iy, these are the operators Ix and Iy which represent the transverse magnetization components, x and the y components. The expectation value is given by this which is the trace of the density matrix with the Mx of matrix and notice here it is a symmetry, the trace of rho Mx is the same as the trace of Mx rho. So, it does not matter here which way you which order you multiply it, but the trace is invariant to that. So, this is a kind of a principle which is generally valid and one can easily verify by taking different kinds of rows and Mx's and whatever then you will find that this equation is actually valid. Now, let us take an example. We take an example for a one spin system I is equal to half. So, it has two energy levels, I is equal to half has two energy levels therefore the rho of T will be a 2 by 2 matrix. So, we have two populations here and there is a coherence, one coherence between the two energy levels. So, it is i omega 1 2 and here it is e to the minus i omega 1 2 because typically omega 2 1 is equal to minus omega 1 2 therefore we have here e to the minus i omega 1 2 essentially looks like a complex conjugate of this, this was the same as in the generalized case. If we have such a kind of a one spin system which is clearly not the equilibrium density matrix. In the equilibrium density matrix, we had the populations non-zero and these ones were 0. So, therefore this is clearly a non-equilibrium state created some somehow, so we will not go into that one now, we will come to that later when we discuss the multiples experiments. So, we will assume that we have somehow created this coherence between the two energy levels 1 and 2 and we also assume that the amplitudes here are the same. So, this is a particular amplitude here and of course the same amplitude is present here, we assume it is the same. So, this is my initial density operator to start with, we do not ask how I created this at this point and we will see what is the consequence of this, how we created it comes later and that will that also we will see in due course. So, let us exactly calculate it now, earlier we remember we calculated the matrix representations of the I X operator, I X operator was half 0 1 1 0 for a one spin system this is half 0 1 1 0 and the density operator is here, this is p 1 p 2 e to the i omega 1 to t e to the minus i omega 1 to t. So, you multiply this now, what you get here? The first element 0 into p 1 is 0, 1 into the e to the minus omega 1 to t, so therefore this element will be e to the minus i omega 1 to t and this element is second one is 0 into this is 0 and 1 into p 2 and therefore you have a p 2 here. Likewise for this multiplication 1 into p 1 plus 0 into this and therefore this will be equal to p 1 and the fourth one will be 1 into e to the i omega 1 to t and 0 into p 2 therefore we have this. So, now what is the trace of this? Remember the definition of the trace, the trace is the sum of the diagonal elements of the matrix. So, if I add these terms and you have multiplication by half here, so e to the i omega 1 to t plus e to the minus i omega 1 to t whole divided by 2 and that is the cosine function. So, the cosine omega 1 to t. Now, what I measure is therefore this coherence. I measure the time dependence of this coherence here and so including, now if I include the transverse relaxation which has not been included here, in any of these we have only included the time dependence and the frequency dependence of the phase coherence is here and the relaxation is not included. So, if I include the relaxation I have to multiply this whole, this is the transverse magnetization. So, the transverse magnetization has to be multiplied by e to the minus t by t2 where t2 is the transverse relaxation time. So, if I plot it, what do I get? I get this function which is basically my FID. So, therefore I am actually if I create the density operator which has off diagonal element which is non-zero that reflects itself in the measurement of the magnetization and what we measure is the FID as a function of time what we are measuring is the free induction decay this is the FID. Let us extend to 2 spins that was one spin. Now, Mx magnetization is now the sum of the magnetization of the 2 spins M1x plus M2x or otherwise one could have taken I1x plus I2x but let us keep the same nomenclature M1x plus M2x. Now for the 2 spins both are spin half systems the 2 spins both are spin half. So, then I will have for the 2 spin system I will have 4 energy levels here where and these ones are represented as alpha A alpha X this A and X represent the spins one spin is called A other one is called as X this is capital X not to be confused with this coordinates X alpha A alpha X and then I have the energy level 2 alpha A beta X energy level 3 beta A alpha X and energy level 4 is beta A beta X. So, these are called as product functions these are the Eigen functions of the Hamiltonian and for the weakly coupled situation as we have seen in the lectures on analysis of NMR spectra. So, for weakly coupled Hamiltonian these the spin system these ones are Eigen functions of the Hamiltonian and they represent the energy levels in this order. So, we call it alpha A alpha X alpha A beta X as number 2 beta A alpha X as number 3 beta A beta X as number 4 and we will maintain this same convention in defining the energy levels and their spin states. Now, so here for this spin state what is the azimuthal quantum number this is plus 1 N here it is 0 here it is 0 here it is minus 1 because alpha is plus half and beta is minus half and it is true for both therefore this adds to plus 1 this adds to 0 this also adds to 0 this adds to minus 1. Simply dropping these A and X subscripts here we can simply write these states as kets here as alpha alpha alpha beta beta alpha and beta beta. Now, once again I will assume a non equilibrium density operator of this form. Notice here this was again from the generalized statement of the density operator of the density matrix. So, I have here the 4 populations of the 4 states here P1, P2, P3, P4 and I have the coherence is between the states 1 and 2 spins in the states 1 and 2 spins in the states 1 and 3 and this represents the spins in the states 1 and 4 and this represents the spins in the states 2 and 3 and this represents 2 and 4 likewise. So, I will also have 3, 4 here and these are their complex conjugates complex conjugates in the sense these are actually supposed to be omega 2 1 omega 3 1 omega 4 1 but omega 2 1 is minus omega 1 2 and therefore these 1 simply become 1 2 here minus 1 2 minus 1 3 minus 1 4 similarly minus omega 2 4 minus omega 2 3 e to the minus omega 3 4 t. Now notice here the spins in the states 1 and 2 so what is this phase coherence what are what is this coherence here what is this frequency this frequency is basically the transition frequency of the spin X. So, it is this transition alpha to beta therefore this represents this single quantum coherence here we call it as the single quantum coherence because what is the spin state I mean the azimuthal quantum number change this is plus 1 here and 0 therefore this is 1 transition or the coherence single is called a single quantum coherence this is the corresponds to the spin X likewise this also represents the spin X in the single quantum coherence alpha X goes to beta X here the A spin is in the beta state here A spin is in the alpha state here and that is the difference because of this there could be changes in the frequencies and this we have seen earlier. The interesting thing is here omega 1 4 if there is a coherence between the spins states here and here and this is double quantum coherence right because this is plus 1 here and this is minus 1 here therefore the separation between them is 2 and therefore we call this as a double quantum coherence these 4 are single quantum coherences ok 1 2 1 3 2 4 3 4 are single quantum coherences and 2 3 what is 2 3 2 3 is from here to here right and this is 0 quantum coherence because there is no change in azimuthal quantum number here. So, it is 0 here and 0 there in both the states it is 0 and therefore it is 0 quantum coherence. So, let us assume that we have a density operator which looks like this or the density matrix which looks like this and what are we going to get if we calculate the trace of Mx with such a kind of a density operator. Now, once again you recall back what was the Ix operator or the Mx operator here for a 2 spin case 2 spin case this Mx operator was 0 1 1 0 1 0 0 1 1 0 0 1 0 1 0 and if recall here these ones represent the single quantum coherences these represent the single quantum coherences 1 2 1 3 2 4 3 4 and these are the their respective symmetrical elements ok. So, now we multiply this with this density operator ok I have created a density of matrix like this with all these elements as known as 0 and I have the measurement by the Mx operator ok. And therefore if I do this calculation here you see this row multiplies this column. So, that gives me only these 2 terms first row I mean first element here multiplication this to this gives me once again these 2 terms we have to see only these 2 are non-zero only these 2 will survive in this second element and the third element in the multiproduct will contain these 2 and the fourth will contain these 2 ok. So, likewise you can do for the second row as well and that one will contain elements in the first position and the first position. So, if I multiply here this will contain p 1 and 1 4 and this will contain this one and this one and this one contain this one and this one likewise this one and so you can calculate these and after that you regroup those elements which are present in the total matrix. And then you take the trace when you take the trace you have to add the elements of the density the diagonal elements of the density matrix just as earlier you got only cosine omega 1 2 t here now you will get cosine omega 1 2 t plus cosine omega 1 3 t cosine omega 2 4 t cosine omega 3 4 t. Notice none of the populations appear here in the trace the populations are not appearing nor are the double quantum and the 0 quantum coherence is appearing here that is interesting. Therefore, when you measure the Mx you are only measuring the single quantum coherence because that is what is the measurable magnetization. So, the populations are not measurable magnetization because the Mx is the magnetization in the transverse plane this is x component of the magnetization and the populations therefore cannot be contributing to the measurable magnetization and interesting thing is the double quantum and the 0 quantum coherence is even though they are present in the density operator are not detected these constitute non-observable magnetization. So, if you have to observe this you will have to convert this into observable magnetization by some tricks and this of course one will do all the time when we actually do multiple experiments. So, this is the take home message here is the when you want to measure x component or the y component of the magnetization you will only collect the single quantum coherences you would not collect the 0 quantum coherences or double quantum coherences or even the populations. Now, let us see how the various conditions are created of the density operator. So, obviously one has to do some perturbations and in a multiple experiment we have lots of RF pulses we must also know we must also know how to calculate the effect of the pulses on the density operator because the density operator is supposed to be containing all the information about the spin system which includes the populations and the various coherences and how these ones get transformed under the influence of the RF pulses. So, to do that we must actually go through this calculation once more and how do the density operator transform in the presence of the RF pulse. So, now we have to define the Hamiltonian appropriately. So, the Hamiltonian now consists of this original Hamiltonian which is your main interaction with the Zeman Hamiltonian plus the couplings the J couplings which are present which is time independent. Earlier we saw that it is the main Hamiltonian is time independent and this is the portion which is the time dependent part because RF pulse is the time dependent where you apply it for a short time generally one applies for extremely short term there is a general thinking that one can treat this as a delta function but nonetheless there is a time dependence here the RF is applied for a certain is 0 and then it goes up and then comes down. So, the frequency is applied for a short time tau therefore in this Hamiltonian there is a time dependence this time of course is extremely short this may be of the order of 1 or 2 microseconds and sometimes tens of microseconds but not too long. So, h 0 is a time independent part of the Hamiltonian and h 1 t which is time dependent represents the RF pulse. So, now we rewrite your Liouville equation d rho by dt is equal to i by h cross the commutator of rho and Hamiltonian h. Earlier this Hamiltonian was only h naught now our Hamiltonian will be h naught plus h 1 t. Now, so what was the solution when this was not present you recall that rho of t was given by e to the minus i h naught t rho of 0 e to the i i by h cross h naught t that was the solution in the absence of the h 1 t. So, maybe I shall write here for quick reference h 1 t was 0 then rho of t was e to the minus i by h cross h naught t rho of 0 into e to the i by h cross h naught t. So, this was the solution we had in the presence of the RF now we would like to seek a solution which looks like this instead of this rho of 0 here we put in here an entity defined as rho star t this is the same this we keep the same and we call this as rho star t. I will explain to you a little bit later and this is actually called as the interaction representation because we are now going into a frame of reference when the interaction is present between the RF and the and your spin system and this is also this is the Zeeman interaction and this represents the going into the rotating frame under conditions of resonance we will see that very soon. So, now what I do I substitute it there and do a little bit of mathematics here do the algebra differentiate this d rho by dt is equal to minus i by h naught t h naught this is minus i commutator of h naught with com with the rho. So, this was there. So, if I differentiate it here what I will have I will have two quantities here this is one commutator rho with h naught and the other commutator rho with h 1 t. So, this is i by h cross commutator of rho with h naught and another commutator rho with h 1 t. So, that is put there and this is d rho by dt you differentiate it and then you will get i by h cross h naught rho plus these e to the minus i by h cross h naught t d rho star by dt. So, this little bit of algebra is there here when we do the differentiation you will get the terms which will pull them together it will give you this and this term will stand out separately because here is the differentiation with respect to the rho star because earlier this was not there and this was rho of 0. So, now the this will appear separately because we have a time dependence here as well. So, therefore and this must be equal to this i by h cross the two commutators rho with h naught and plus h 1. So, put it separately out the two commutators are put out separately here. So, therefore this right hand sides of the two have to be the same this must be equal to this since this and this are the same therefore this will become equal to this. So, now what we do I want to get an equation for d rho star by dt. So, d rho star by dt I put it on one side and then you multiply on this side you take this multiply on this side by e to the i by h cross h naught t that is this and from the from this side you multiply by e to the minus i by h cross h naught t and that is this one here and what you get here will be simply d rho star by dt you take this term let me repeat it here you take this term and multiply from the left by e to the i by h cross h naught t. So, this will become 1 and multiply from the right e to the minus i by h cross h naught t. So, this term will become 1 and what is left here is d rho star by dt and on the right hand side what will happen this ones will appear as they are as we did the multiplication i by h cross will be there and the left hand multiplication will give me e to the by h cross h naught t and the right hand side multiplication will give me here e to the minus i by h cross h naught t. Therefore, you get an equation for d rho star by dt is equal to i by h cross you can put this whole thing in a bracket here. So, then we take this calculation further and d rho star by dt we expand this commutator which was rho comma h1 as rho h1 minus h1 rho this portion means the same once again there is a bracket here this bracket i by h cross multiplies the whole thing anyway. So, that is since this is all inside the bracket here it does not matter. So, what they do now i by h cross e to the i by h cross h naught t rho now here I introduce this at this point I introduce this. So, in between rho and h1 I put in this e to the minus i by and I do the same thing here as well at this point also I introduce this what is that that is basically one only know. So, here is a rho this is one. So, therefore, rho h1 is not perturbed. So, I put this in between and then this h1 and this e to the minus i by h cross h naught t is here and similarly for this I have this e to the i by h cross h naught t that comes from here this term then h1 then again in the middle I introduce this one a taken as a product of these two quantities which is actually one and rho e to the minus i by h cross h naught t now we notice something very interesting happening. So, this portion this portion is similar to is my rho star e to the i by h cross h naught t rho e to the minus i by h cross h naught t this is my rho star. Now, this one from here to here is a similar transformation with h1 therefore, I call this as h1 star and likewise here from here to here from here to here this is h1 star and then from here to here it is rho star. Therefore, what I have got I got here rho star h1 star minus h1 star rho star. So, that is basically my commutator rho star h1 star therefore, d rho star by dt is i by h cross rho star comma h1 star where h1 star is e to the i by h cross h naught t h1 e to the minus i by h cross h naught t. Now, this is called as interaction representation this represents the interaction between the Rf and your spin system and in the rotating frame this becomes equivalent to the rotating frame under resonance condition under resonance condition this one is actually is 0 and we will see what is the consequence of that. Let us try and calculate the matrix element of this h1 star Hamiltonian k between the two levels k and m take the states k and m. So, this is the branch state k and I explicitly put here for e h1 star e to the i by h cross h naught t h1 e to the minus i by h cross h naught t and on this side m and what does this give me this give me i by h cross e k because this is an eigen function of this of the Hamiltonian h naught. So, therefore h naught operating on the state k gives me e k therefore, this being in the exponential. So, I have here e to the i by h cross e k and this gives me e m this operating on the state m. So, this gives me the energy e m and I have here therefore and then the matrix element of k h1 m. So, this being numbers this can be taken out therefore, I have here e to the by h cross e k minus e m t and to k h1 m. So, this is the matrix element and this is the time dependence here the time dependence has come here and the time dependence from this we will we will see how this will slowly can be removed. Now, notice this is the energy difference between the two states m and e k and m. Now, if I substitute h1 is equal to h1 0 e to the minus e to the i omega rft because this is my rf the rotating frequency if I take e to the i omega rf and this represents your rf Hamiltonian. So, therefore, if I put that here then this e to the i omega rf can be taken out and then I put it here inside here but I given h cross here because I have an h cross here. So, therefore, e to the i by h cross e k minus e m minus h cross w rf omega rft and k h1 0 this time the independent part comes here h1 0. So, time dependence is completely taken out it comes in this term and this one is time independent and this represents the amplitude of the rf at time t is equal to 0 and during the pulse also the amplitude will not vary if this term is non-zero if the term is 0. So, what happens now? Look at this e k minus e m is the energy difference between the two levels and what is the order of this? This is typically in megahertz this is also in megahertz omega rf therefore, this difference is typically of the order of your spectral range which is in kilohertz range and therefore, this is an extreme and if your time is in in microseconds t the duration of the pulse if it is in microsecond this will be extremely small number therefore, the time independent term will be extremely slowly varying during the pulse and can be effectively considered to be constant. In fact, under resonance condition this is ideally identically 0. So, this is the resonance condition when you have the resonance condition this is by identically 0 then there is no time dependence at all. In all hard pulse experiments you notice our h1 is so large the amplitude of h1 is so large that resonance condition is actually satisfied for all the spins at the same time. Because the effective field is along the h1 field the field along the z axis the 0 field is 0 and therefore, the resonance condition is satisfied for all of them at the same time. Therefore, typically for a hard pulse which is of the order of one microsecond or even sometimes you can effectively treat as a delta function that it is 0 then this term will be completely independent of time. In that case the matrix element this can be assumed to be independent of time in other words during the time of the pulse h1 star can be assumed to be time independent and is equal to the amplitude of h1 and what will be the amplitude of h1? This is not varying therefore, it must be equal to h1 0. So, what was in the beginning that is what is going to be there. Therefore, when if it is time independent then I can write the solution of this equation as rho star t is equal to minus e to the minus i by h cross h1 star t rho star 0 e to the by h cross h1 star t. Now I can replace this h1 star by h1 because I say that it is independent of time and then it is equal to h1, amplitude is equal to h1 and now rho star 0 is equal to rho of 0 and therefore I can also replace this as rho of 0 and then you will also see that even the rho star since this is under the resonance condition this evolution also extremely small then my rho of t is equal to rho star of t. Therefore, I can write the whole thing as rho of t is equal to e to the minus i by h cross h1 t, rho of 0 e to the i by h cross h1 t. This becomes a very simple calculation now and we can calculate the evolution of the density operator under the effect of the RF pulse in the. So, notice this is valid during the pulse we should remember this. This is the thing which is during the pulse and we assume we are in the resonance condition and this density operator and the interaction representation becomes equal to what is in the normal case. This density operator transformation by the RF pulse can be described in this manner. So, now we are ready to calculate explicitly the form of the Hamiltonian h1 including the interaction between the magnetic moment and the h1. The effective field is now along the RF axis is equal to the h1. So, h1 Hamiltonian is mu dot h1 which is gamma h cross h1 this is the amplitude of h1 ix if the h1 RF is assumed to be applied along the ix axis then the transformation operator is e to the minus i by h cross h1 t. Now, what is this interaction you put here for the h1 you put gamma h cross h1. So, the gamma stays there h cross cancels with this h cross and the gamma h1 is omega 1 omega 1 is the RF you convert the RF amplitude into a frequency that is omega 1 gamma h1 gamma h1 times t and gamma h1 times t you have seen earlier that it is a flip angle is a particular angle this is the frequency and this is time t and therefore this is a flip angle. So, the processional frequency gamma h1 of the processional frequency in the rotating frame in the under the resonance condition and therefore the how much the magnetization is flipped will depend upon this time t. So, we have seen that earlier this is you can create a 90 degree rotation or 180 rotation and things like that depending upon what the this time what we will have here. So, this represents the rotation of the magnetization about the x axis by angle beta and thus depending upon the length of the pulse different rotation angles can be obtained. Now, we will also explicitly show that putting this equation for the pulse we will demonstrate that it actually results in rotation of the magnetization from the z axis or wherever. So, similarly for the RF applied along the y axis the rotation by the pulse is represented by e to the minus i beta iy. So, this can be extended to other kinds of situations wherever you want to apply if you want to apply somewhere in between then one has to put in ix plus iiy and things like that. So, those are called as phase shifted pulses and so we will not discuss about those things right now we will say we will take the simpler cases where it is clearly along the x or the y axis and this will be sufficient to understand the principles that are happening and those involved things can be calculated separately. I think we will stop here and we will continue in the next class with further description of the pulses and their effects on the magnetizations.