 Welcome, let us continue the discussion of the effects of the RF pulses. How does the density operator transform in the presence of RF? Because we need to know all of such things. How does the density operator transform under the influence of different kinds of Hamiltonians? So the RF constitutes also a Hamiltonian and we actually saw last time how to calculate the evolution of the density operator in the presence of the RF. So we had this solution here. We started up with the Liouville equation and we represented the solution like this, rho of t is equal to e to the minus i by h cross h naught t, rho star t e to the i by h cross h naught t. H naught was the main Hamiltonian and Rf Hamiltonian was represented by h1 and this represented the interaction representation and we also show after detailed calculation that under the resonance condition, we have the rho of t is equal to the e to the minus i by h cross h1 t, this Hamiltonian t, rho of 0 e to the i by h cross h1 t and this h1 represents the Hamiltonian of the RF. And we also demonstrated that e to the minus i by h cross h1 t essentially represents a kind of a rotation operator e to the minus i beta iq, we will explicitly demonstrate that further, how this will represent a rotation and therefore having said this, we will say rho of t is equal to e to the minus i beta iq rho of 0 e to the i beta iq. Notice here q can be x, y or z, generally z is only rotation around the z axis therefore it is not kind of an effect of the RF pulse but general rotation along the z axis can be represented in the same manner if the rotation is about a particular angle beta. And we define now this as the rotation operator call it as Rq beta is equal to e to the minus i beta iq and q can be x, y or z. So therefore my rho is now Rq beta rho of 0 Rq minus 1 beta and that is the same equation here we now calculate explicitly the effect of the pulse on the density operator. So to begin with let us take the equilibrium density operator i z, rho of 0 i z of course if the rho of 0 has this other entities but the operator part is the i z only therefore it will pick up only the i z part of the density operator, equilibrium density operator and put that here. So the rho of t for rho of 0 I put here i z Rq beta i z Rq minus 1 beta. So in order to do this I need to get the matrix representations of these operators these were in the form of exponential operator whereas i z by nu is in the matrix operator therefore I have to put this also in the form of a matrices. So for this we will calculate the matrix representation of the RF pulses we have to get explicit matrix representations for these terms. So how do we do it? So let us write here again Rq beta e to the minus i beta iq q can be x, y or z. Now we have earlier seen matrix representations of the operators i x, i y and i z and these ones we have calculated matrix representations of the operators i x, i y and i z and those ones were simply are given in this except for the factor half. So therefore this can be represented as half sigma q where sigma are the Pauli spin matrices basically i z is half into 1 0 0 minus 1, i x is half into 0 1 1 0, i y is 0 minus i i 0. Now we take away the half you have this matrices 1 0 0 minus 1 and this is represented as sigma z and likewise sigma x is given as 0 1 1 0, sigma y is given as 0 minus i i 0. These are called as spin matrices for a single spin half these are called as Pauli spin matrices. You can easily verify that the Pauli matrices satisfy this condition sigma z square is equal to sigma y square is equal to sigma x square is equal to 1 but one notice here this 1 is a unit matrix it is not just number 1 this is unit matrix 1 0 0 1. So this will be useful for us to calculate the matrix representations of the individual operators. So here I expand this exponential function as a series this is a typical expansion of the exponential function e to the minus i beta i x I write it as e to the minus i beta by 2 sigma x i x is half sigma x and you write it as a series 1 minus i beta to sigma x notice here there is a unit matrix here this is unit matrix here then you have the sigma x matrix and here what we will have sigma x square i by 2 whole square and sigma x square and sigma x square is a unit matrix because we have seen that earlier right and this is this is minus 1 by 3 factorial i beta by 2 to the power 3 and this will be sigma x cube but sigma x cube is sigma x square into sigma x and sigma x square is unity there although the explicit unit matrix is not written here but it is there remember this and likewise 1 by 4 factorial i beta by 2 to the power 4 and this is sigma x is 4 sigma x 4 is again 1 therefore you will get this infinite series there are terms which depend on sigma x and there are terms which do not depend on sigma x now we regroup these terms and we say put all these terms which do not depend upon sigma x as 1 minus 1 by 2 factorial beta by 2 to the power 2 notice unit matrix is present here and 1 by 4 factorial beta by 2 to the power 4 again unit matrix is present here and likewise and this bracket closes here and the other terms are minus i beta by 2 minus 1 by 3 factorial beta by 2 to the power 3 and all of these then have multiplication with sigma x so therefore this will be what is this this first bracket this first bracket is actually an expansion of the cosine beta by 2 this is the series expansion of the term of the function cosine beta by 2 and the terms in the second bracket are the represent the series expansion of the of the function sin beta by 2 therefore I have here cosine beta by 2 multiplied by the unit matrix and this is we can write this unit matrix here okay so and the i m minus i sigma x sin beta by 2 so what does this give me put this in the matrices matrix form explicitly cosine beta by 2 1 0 0 1 minus i sin beta by 2 and I had the sigma x sigma x is 0 1 1 0 therefore now if I put beta is equal to pi by 2 see if I put beta is equal to pi by 2 then what do I get I get cosine 45 here and sin 45 here and both of them are equal to 1 by root 2 okay so e to the therefore e to the i minus e to the minus i pi by 2 ix gives me 1 by root 2 and if we add these here 1 gives me 1 minus i 1 right so this this is 1 and this gives me minus i this gives me minus i this gives me 1 and 1 by root 2 factor comes out here so similarly for r y pi by 2 this is the pi by 2 pulse applied along the x axis rf is applied along the x axis this is again now a pi by 2 pulse applied along the y axis so this gives me 1 by root 2 1 minus 1 1 1 okay so similarly you can calculate matrices for the pi pulses if I choose beta is equal to pi then rx pi is e to the minus i pi ix and this gives me 0 minus i minus i 0 and r y pi gives me e to the minus i pi i y this gives me 0 minus 1 1 0 the effect of these pulses on the density operator can be explicitly calculated using these matrix representations. So for example for a density operator represented by iz so this is the easiest example to take so we take the equilibrium density operator represented by iz and we will calculate what is the effect of rx pi by 2 is it actually a rotation now we will actually demonstrate that this pulse does make a rotation okay how do we show that so the density operator here rho of t if this is the same equation back here rq beta iz rq minus 1 beta and therefore I put here rx pi by 2 iz and rx minus 1 pi by 2 so now we calculated what is rx pi by 2 that was this and now we I get here a 1 by root 2 from here 1 by root 2 from here and this gives me 1 by 2 in the in the matrix therefore I have a 1 by 4 and this gives me 1 minus i minus i 1 and iz is 1 0 0 minus 1 half factor is already taken away here and the inverse of this rx inverse inverse of this matrix is this 1 i i 1 how do I say this because now you see if you multiply this matrix with this matrix you will get 1 right so this product you take 1 1 here okay and this gives me 1 and minus i minus i square so minus i square is 1 so therefore 1 plus 1 gives me 2 we can actually do that let me do here the 1 minus i minus i 1 and multiply by 1 i i 1 so this gives me this is equal to 1 minus i square this is 2 and this is i minus i this is 0 and minus i plus i this is 0 this is minus i square is 1 plus 1 this gives me 2 and therefore if I take this 1 by root 2 1 by root 2 I will have 1 1 by 2 here therefore this is equal to 1 0 0 1 right so therefore what I have got therefore this is the inverse this is demonstrated that this matrix is the inverse of of this matrix so now if I multiply all these three what do I get I get half 0 i minus i 0 and this is minus i y okay so what I have got is the result of this iz has become minus i y therefore there is a rotation 90 degrees rotation right now therefore we have explicitly demonstrated that the z magnetization is rotated on to the so if I had the z magnetization here and if I apply the x this is x y z coordinate as the axis and this rotates the transverse magnetization along this so I get here magnetization rotation 2 minus i y so therefore clearly the z magnetization is rotated on to the negative y axis okay now for two spin system there was simple for one spin system for two spin system of course it is also simple but we have to do a little bit more calculation here now there are two spins both are independent right now I want to calculate the matrix representation of the pulse for two spins so both are i is equal to half so they have this one spin here and the second spin here so let me call them as spins k and l so if I call this as spin k and this spin as l right and each one of them is independent therefore if I want to represent a matrix we represent both of them I have to when I take the product the product means I can take a simple matrix multiplication here I have to take what is called as the direct product so direct product means each of these element when I take the multiplication of the direct product means if I take this and multiply the and I put the entire matrix here multiply the entire matrix with this element therefore I get here 1 minus i minus i 1 now so far as this element is concerned I multiply minus i with this entire matrix therefore I get a minus i here and minus i into minus i this gives me plus i square that is minus 1 here and likewise minus i into minus i gives me minus 1 here and minus i into 1 that gives me minus i here so similarly I get here minus i minus 1 minus 1 minus i and for this I get 1 minus i minus i 1 so now you see this is the 4 by 4 matrix so I should get the 4 by 4 matrix right that the ix operator for a 2 spin system is a 4 by 4 matrix so we get that by doing this direct product of the individual spins because the 2 spins are independent and they can be multiplied by this direct product method so therefore I get a 4 by 4 matrix and my operators ix, iy, iz they will also be 4 by 4 matrices and this is the way we calculate the matrix representations of the operator for RF pulses so similarly we calculate for Ry pi by 2 I have here 1 minus 1 1 1 this is for spin k so if I call it for spin k and this one is for spin l and I take the direct product so if I take the direct product I get here 1 minus 1 1 1 that is the same and then this is basically multiplying with minus 1 this whole thing so I get here minus 1 1 minus 1 minus 1 here and this will be 1 minus 1 1 1 and this will be 1 minus 1 1 1 because all these are ones here right so therefore these will be the same except this this term will be different so this Ry pi by 2 now consists of all real numbers whereas the Rx contain all imaginary i's also using such matrix representations for the pulses and the density operator the evolution of the density operator through a multiples experiment can be calculated so put this all in summary I have for one spin I have here 1 by root 2 1 minus i minus i 1 Ry pi by 2 1 1 by root 2 1 minus 1 1 1 Rx pi is 0 minus i minus i 0 this is for and Ry pi is 0 minus 1 1 0 and similarly for the two spins you have 4 by 4 matrices and you have calculated here Rx pi and Ry pi which will contain only anti-diagonal elements notice here of course here also they contain anti-diagonal element this is the diagonal and this is the anti-diagonal so I have only anti-diagonal elements here and here also I have only the anti-diagonal elements Rx pi and this indicates how this shows how the calculations have to be performed. So for now we are ready to calculate any kind of a matrix representation any kind of evolution of a multiple pulse experiment so if I have a multiple pulse experiment so I start with the initial density operator rho 0 start with this apply a pulse P1 here and then allow the spin system to evolve under the influence of the Hamiltonian H1 for the period tau 1 then apply a pulse P2 then apply and then allow the system to evolve under the influence of the Hamiltonian H2 for a period tau 2 and continue like this therefore any number of pulses what you might have here I can continue to calculate the evolution of this of the density operator through the pulse sequence and finally I will get the density operator here which is what I actually measure and then that contains the entire information about the evolution of the spin system through the pulse sequence okay how do we actually do it so therefore we say start with the rho 0 this is my initial density operator. So if I put the density operator here the first thing that is coming here is the P1 that is the pulse therefore when I have the pulse I have the P1 and P1 minus 1 right so this is the what I get after the first pulse that means for this part of the evolution the density operator here becomes the rho of 0 this is the new rho of 0 so for the beginning of this one this becomes my rho 0 so therefore this is my rho 0 for the next operation and that is the evolution under the Hamiltonian H1 for the period tau 1 so e to the minus i by h cross h1 tau 1 and on this side e to the i by h cross h1 tau 1 so now where I am I have reached here at the end of this now apply the pulse again see if I apply the pulse therefore once again I multiply by P2 here and P2 minus 1 here now if I then I reach here then of course I will have to multiply by e to the minus i by h cross h2 tau 2 and on this side e to the i by h cross h2 tau 2 and so on so this is the way one can actually calculate the matrix representations of the density operators and evolutions of the density operator through the multiples experiment and that is way one has to calculate it and I think we can take it forward in the next classes and with the explicit calculation and see what is the physics and what is the science that comes out of this calculations so we will stop here