 In the last class we started with a rigorous description of NMR, a quantum mechanical description which is called as density matrix description of NMR. We try to derive an expression for the density operator starting from the basic principles of quantum mechanics that how we make a measurement and what is the meaning of measurement in quantum mechanical terms that is when you measure the expectation value calculate the expectation value it represents a measurement and we also said that we have to take an ensemble average to represent the measurement. So as in the end we came up with this expression for the so called density operator rho is equal to 1 by N 1 minus this is the Hamiltonian divided by KT and this is an operator here. So rho is an operator and this represents the total number of states the Eigen states and that depends upon the spin system what we might have whether it is one spin system or two spin system or three spin systems and so on and what is the value of the I for most purposes we will be dealing with I is equal to half systems and we also said we will represent the Hamiltonian by the interactions and by and large it is represented by this simple expression here this is minus gamma h cross h naught and this h naught is the field and I z this is the z component of the angular momentum this is the I z operator. Now since the density operator is intimately connected with the angular momentum operator so we have to look at the properties of these operators in some detail. Now here I would like to mention that since this chapter is going to be quite intensive with regard to the operators and matrices, matrix representations, angular momentum and things like that it may be a good idea to take a small digression and spend some time on the properties of the spin space for the benefit of those who did not have sufficient exposure with quantum mechanics operators and so on and so forth. Therefore we will go slow here and try and calculate this individual we will introduce this concepts clearly and explain from the first principles how one can calculate these various elements. So now this is the same equation which we wrote earlier in the first class that any state is represented by wave function the wave function is written as superposition of the various states these are so called basis states they are form a basis set all the UMS are the eigen states or the form a basis set and this is the superposition of all the basis sets and you have the CMTs these are the coefficients which contribute to the superposition. In a simpler notation we simply write it as summation CMM and the summation goes over all the states of the spin system. So if as I said this goes from M goes from 1 to N the spin states satisfy this identity this identity also we use last time however let us come back to this identity and try and show how this comes and try and prove this. So we use this identity in the previous class and this M is the one of the eigen states or the element of the basis set and with this summation runs over all these M's. Let us take one of these element this one which is written here I take this here and I will show you that this is actually an operator. So this if I operate on the my wave function psi of t then I will get it like this this operator as it is and then for the psi of t I write this now it just instead of M here as the index of summation I use N it does not matter this just runs over all the states whatever index I use here it runs over the all the states. So summation CNN okay now this CNS are numbers okay so this CNS are numbers these are not operators therefore I can take this entire part outside of this operator term and then I put the summation here instead of putting it here then I say then CN comes here then I put the same thing back this M this is the ket this is the bra and this takes an element with the another ket here N. Now these are the sum in the expression gets simplified into this. Now you remember we also said that these states the elements of the basis set these are orthonormal. So what is the meaning of orthonormal that this bracket here this element is either 0 or 1 depending upon whether M is equal to N or M is not equal to N we write that as a chronicle delta. So this one is therefore is written as chronicle delta NM this delta NM is equal to 1 if N is equal to M or it is equal to 0 if N is not equal to M. Therefore this summation once again running on various N's here CNM delta NM this will simply become equal to CMM because for all other N's which are not equal to M this is 0. So therefore only one element of this survives and therefore we have here CMM. Now I take instead of this only one particular element which I took last time I take the whole summation I take the whole summation of all this. So therefore you notice you notice here that this is essentially projecting out one particular state here from the entire wave function. This was psi of t here and eventually I have got here only one state with its coefficient CMM. Therefore this operator this is now you understand this as an operator this operates on the wave function psi of t and gives me only one particular state therefore this is called as a projection operator this projects out the particular state M out of this psi of t. So now I am taking the entire summation of all these operators M and M all the M going from 1 to N. So this is the same here and this operator giving on psi of t give me CMM and this summation is basically psi of t again we have seen that this is in the definition. Therefore this equation if you see if I take out psi of t from both sides here then it simply reduces to this equation this is the identity which we actually used earlier and we have proved it explicitly here for the benefit of those who are not familiar with this principles. But these are generally used routinely in all various kinds of calculations in quantum mechanics and quantum chemistry and therefore this is very important to understand how these things come. Now let us do explicit calculations with regard to the angular momentum operators we said the density matrix is very intimately connected with angular momentum operator IZ and we will also deal with other angular momentum operators in due course and therefore it is important to describe the spin states and the representation matrix representations of the angular momentum operators. Let us start with one spin with I is equal to half in one spin how many states do we have in the presence of the magnetic field well there are anyway two states in the presence of the magnetic field they are non degenerate they have separate energy levels in the absence of the magnetic field they will be degenerate but nonetheless the two states are present and because we said the IZ is a quantized state and we have only two possibilities there therefore we have two states alpha and beta and these are the spin states and these are also the Eigen functions of the IZ operator as we see here. Now what are the properties of these states that is integrated here in this matrix element alpha beta goes to 0 whenever I say alpha beta actually in the conventional nomenclature those were used quantum mechanics the integral forms actually this ket this is called the ket this is actually the wave function psi and this side which is there which is the bra this is the complex conjugate. So therefore when I write like this it is essentially the integral psi star psi okay so therefore in the case of the spin states we simply write this as the bra here and the ket here and therefore this is orthogonal and therefore this is 0 all these basis elements of the basis set are orthogonal in nature and if I do that with the similar states alpha and alpha here this is equal to 1 beta alpha is equal to 0 beta beta is equal to 1. Therefore we say these states are orthonormal because this is normalized to unity and these individual matrix elements these are 1 here and the cross terms are 0 and therefore these are orthonormal basis set there are only 2 states right these 2 states are orthonormal. Now we also saw earlier that I z which is operating on the state alpha if it is m and this is the m value is this we defined in the very early stages of this course that this is the m value this gives me half I z operating on alpha gives me half alpha I z operating on beta gives me minus half alpha because these are the m values for beta the m value is minus half and for the alpha the m value is plus half therefore this gives me half and minus half and you see the same state is back here therefore these are the Eigen value equations and alpha is the Eigen state of the I z operator and beta is also an Eigen state of the I z operator with the Eigen values half and minus half. Now this I x alpha this we can take it for granted although this can be proved later this I x alpha is now giving you half beta. So I x alpha gives me half beta I x beta gives me half alpha notice here the alpha I x operator converts the alpha state into the beta state and it converts the beta state into the alpha state in fact this is what was used when we described the selection rules for energy absorption by the RF. The RF was applied along the x axis and the I x operator was coming as a perturbation by interaction of the RF field with the magnetic moment and then the I x or the I y operator is coming there and because of the change in the state here then the delta m is equal to plus minus 1 selection rule was derived as a result of this. So and that becomes conspicuous again when we will actually try to calculate this matrix representation of this I x and I y operator. Likewise if I do I y on alpha it gives me I by 2 beta and I y on beta gives me minus I by 2 alpha. Once again this alpha state is converted into the beta state here and the beta state is converted into the alpha state and therefore these are not Eigen value equations because the Eigen function is not retained. There is a change of state and but with a certain kind of a result here as a coefficient. So now I calculate the matrix elements of the I z operator. The matrix elements of the I z operator meaning what? So I have this sort of calculations I have to perform. I z in the middle alpha here, alpha here, alpha here, beta here, beta here, alpha there, beta here and beta here. So these four elements represent a 2 by 2 matrix of the I z operator. So this is called as the matrix representation of the I z operator in the basis set of the two states alpha and beta. You have the two states alpha and beta here, these form the ket individually and these form the bra individually once alpha and once beta and here the ket once alpha, once beta. Therefore I have a 2 by 2 matrix here. Now you notice I z alpha gives me half alpha. So the half alpha and alpha, this alpha alpha goes to 1. This is what we saw in the previous slide and therefore this results gives me half. Now I z beta gives me minus half beta as we saw in the previous slide. So if I take minus half out then I have the alpha beta here and that alpha beta has to be 0 because these are orthonormal. So I z alpha gives me half alpha. Therefore the matrix element will be half beta alpha and that will be 0. So and here I z beta gives me minus half beta. So therefore I will have minus half outside then I will have the beta beta and the beta beta is equal to 1 therefore this is equal to minus half. So if I put these four elements in the matrix then I will get here half 0 0 minus half. But I take the half out then I have half here 1 0 0 minus 1. So basically this will become the matrix representation of the I z operator for a single spin I is equal to half. Similarly we can do that for I x operator. Now I x operator if you remember I x converts alpha into the beta it gives me half beta. So the half beta means I take away half out then I have the alpha beta and that will be 0. So therefore I know if we see I x beta so I x beta gives me half alpha then alpha alpha is equal to 1 therefore I have here half. This was exactly the same thing what we used when we were actually calculating the selection rules with the perturbation magnetic moment interaction with the RF field and the operator part was I x there. So you see the difference of delta m between these two is plus minus 1. So therefore that is how we get a selection rule of plus minus 1 for RF induced transitions. Basically it comes from this sort of a calculation. Beta I x alpha if I do the I x alpha converts alpha into the beta state I get a half beta. So beta beta is equal to 1 so I get a half here. So here I x beta goes you half alpha and therefore beta alpha gives me 0. So therefore if I write it in the same manner here so alpha beta here alpha beta here these are the two kets and these are the two bras. So I x is in the middle so I will have four elements here. So I will have half here sorry 0 here half here half here and 0 here. Therefore this and if I take out the half half 0 1 1 0. So therefore this becomes the matrix representation of the I x operator. So similarly if I do for I y I y alpha I y beta and these ones now notice here I have a I coming here these are imaginary. So I have 0 minus I by 2 I by 2 0 and therefore I have here half 0 minus I I and 0. So this will be the matrix representation of the I y operator for a single spin half. Now if we eliminate this half we remove this half the matrices which are present for the I z I x and I y. So they are simply 2 by 2 matrices in I z you have 1 minus 1 0 0 for I x you have 0 1 1 0 for I y you will have 0 minus I I 0 and these 3 2 by 2 matrices are called as Pauli spin matrices for single spin and these are the ones which you actually calculated and these are called as Pauli spin matrices. Now let us turn to 2 spins. So 2 spins consider a and x I a is equal to half I x is equal to half. Now the individual spin states of these 2 spins are alpha a beta a and alpha x beta x. Now for generating the basis set for the 2 states we take the products of this individual spin states. This was done in the second chapter when analysis of spectra was described when you wrote the Hamiltonian you try to calculate NMR spectra using the quantum mechanical principles this also was done in a very generalized way. So for the 2 spins we calculate the product states for the 2 spins these will be alpha a alpha x alpha a beta x beta a alpha x and beta a beta x these are the 4 states which constitute the basis set. So 2 by 2 by 2 2 states for this and 2 states for this. So therefore I will have here 4 states. I represent these 4 states in the energy level diagram like this I have I label this alpha a alpha x a state number 1 alpha a beta x a state number 2 beta a alpha x a state number 3 beta a beta x a state number 4. And now the angular momentum operator a consolidated angular momentum operator for the 2 spin system with the I z or the I q or I x or the I y will be the sum of individual operators. So I z here for 2 spins will be I z a plus I z x I x for the 2 spins will be I x a plus I x x and I I y for the 2 spins will be I y a plus I y x. So therefore in abbreviated notation we write like this I q is equal to I q a plus I q x where q goes from x changes from x to x y and z. Now what are the properties of this spin states? So this will follow from the general properties of the alpha and the beta states if I label this now I will label the my 4 states as 1 to 4 right. So here therefore I take the products of the individual states I j then this is equal to delta I j the I j can go from 1 to 4 right. So the 4 states which are there so these are orthonormal states of the 2 spin system I j is equal to delta I j. So therefore how many matrix elements will be there for the 2 spin states since there are 4 states. So 4 ket states and 4 brass states and therefore this will be 4 by 4 matrix for the I z operator and likewise for the I x and the I y operators. So it will consist of these elements I z 1 1 I z 1 2 1 3 1 4 2 1 2 2 2 3 2 4 3 1 3 2 3 3 3 4 4 1 4 2 4 3 4 4 and these are the matrix elements we have to calculate. Let us illustrate this how these can be calculated. So let us calculate I z 1 1 I z 1 1 is 1 1 the 1 state is alpha a alpha x right. So therefore in the bra here I have the alpha a alpha x and the I z operator here is I z a plus I z x and the ket again is the 1 1 1 state and that is alpha a alpha x. Notice each of these operator operates only on its own spin state and it does not operate on the other spin state. For example I z a will operate only on the alpha a it will not operate on alpha x. So therefore if I am calculating the matrix element with respect to I z a I can take out the alpha x state. So this is what we shall do now if I have this 2 states here now alpha x I can take it out if I am calculating I z a here so I separate out these 2 into separate calculations. So in one case I have the I z a in the other case I have the I z x here. So if I have I z a here I keep this alpha a here then the I z a and keep this alpha a here and this alpha x this alpha x come out here as a separate entity alpha x alpha x. Similarly plus alpha x I z x alpha x that is if I am looking at this operator then alpha x alpha x these 2 are taken here and these alpha a alpha a come out separately here. Now because of the orthonormality of this this is equal to 1 and likewise this is also equal to 1 and what does I z a give me operating on alpha a it gives me half alpha a. Now therefore half alpha a half we take it out alpha a alpha a that gives me 1 again and therefore I have a half here coming from this term this is 1 and this gives me half. Similarly this gives me half and this gives me 1 therefore half plus half this gives me 1. Let us do for I z 1 2 I z 1 2 is alpha a alpha x here this I z remains the same and here I have the state number 2 which is alpha a beta x. So I do the same as I did here before so I have here alpha a I z a alpha a those are these states are taken alpha states and alpha x and beta x come out here that because I z a operated does not operate on the x spin states therefore these alpha x beta x come here. For the second one when I am putting I z x here I have to use the x spin states only therefore I have here I x alpha x and then I have the beta x here and the 2 alpha a alpha a come out. Now notice here because of the orthogonality this term gives me 0 here alpha x beta x is 0 whereas this gives me half on the other hand here this gives me 1 and this gives me 0 again because this is half minus half beta x and alpha x beta x gives me 0. Therefore both these terms actually become 0 and we vanish ok and likewise I z 1 3 and I z 1 4 will be 0. So let us do the same thing for I z 2 2 I z 2 2 is now we have the 2 state here and the 2 state here alpha a beta x alpha a beta x and I z a plus I z x. So this is alpha a I z a alpha a which is easily followed from here then I have this beta x beta x here beta x I z x beta x and alpha a alpha a. Now what do I get from here I z a on alpha a gives me half and therefore and this is 1 alpha a alpha a is 1 and therefore I get a half here. Now what do I get here I z x operating on beta s gives me minus half beta x and therefore minus half beta s beta x is 1 alpha a alpha a is 1. Therefore I get a minus half these two terms will then cancel I will have a 0 here. So therefore this the second term 2 2 will be 0. Now similarly you can do this calculation for all the other elements of the matrix and you will find that in the second row I z 2 1 2 3 2 4 and in the third row 3 1 3 2 3 3 3 4 that means all the four terms and in the fourth row 4 1 4 2 4 3 these will all be 0. So therefore in the first row also we said I z 1 1 was non-zero and other three terms were 0 so earlier. Now I z 4 4 also we can calculate and this is in the same manner when you calculate it this is beta a beta x beta a beta x here and I z a I z x in this terms. So this will turn out to be minus half minus half and this will be minus 1 because I z a on beta gives me minus half and I z x and beta x also gives me minus half. So therefore both will be minus half minus half and this will be minus 1. So totally therefore if I see list all of these elements here you can see that in this I z matrix I have only these two elements which are non-zero and all the other elements are 0. Therefore for the two spin state the two individual I is equal to half spins the I z matrix representation will be simply this. Now for the I x we can do the same thing I x also we will have this four elements here and now let us try and calculate for some of those. So if I can I x 1 1 this is alpha a alpha x I x a plus I x x and alpha a alpha x here we do the same trick as before. So alpha a I x a alpha a and alpha x alpha x and this is alpha x I s x alpha x alpha a alpha a. Now notice here I x operating on alpha gives me beta that is this here whereas this remains the same and I x operating on alpha gives me beta. So therefore I have alpha x and beta x and notice these ones are 0 this is 0 and therefore this will be 0. So similarly I x 1 2 if I calculate I have here alpha a alpha x for the 1 and alpha a beta x for the 2. So here I have therefore the same expansion I x a alpha a alpha x beta x I x alpha x beta x alpha a alpha a. Notice here this fellow actually gives me 0 alpha x beta x gives me 0 right and what about this I z x operating on this is equal to 1 I x operating on beta x gives me alpha a half alpha x therefore alpha x alpha x will be equal to 1 and then I get a half therefore I will get 0 plus half and that is equal to half. So similarly you can calculate for these elements I x 1 3 I x 2 1 I x 2 4 3 1 3 4 4 2 4 3 all these will be equal to half and the remaining elements will be 0. So thus your I x representation will look like this all the 4 elements will have a half here this 1 1 here 1 1 here 1 1 here and 1 1 here and all the other elements are 0. You can do the same exercise with I y and then it will turn out that you have half here it will be very similar to this except that instead of 1 you have this I elements. So you will have minus I minus I minus I minus I here in the upper triangle you will have the minus I's and in the lower triangle you have I I and I I here. So this will be the matrix representation of the I y operator for the two spin states. I think you will stop here thank you will continue with this discussions in the next class.