 In the last class we have seen how to calculate the matrix representations of the angular momentum operators in the basis of the eigenstates of the individual spin systems. We considered explicitly one spin system with i is equal to half and two spin systems both with i is equal to half two spins labeled as a and x and we calculated the matrix representations of the operators i z i x and i y. And we also talked about the general properties of the individual spin states. What does the orthonormality mean and how it is used in the calculation of the matrix representations of the angular momentum operators. We went through step by step and now we will continue this exercise and try and calculate the matrix representation of the density operator which is of crucial value for us in the description of the NMR experiments. So we had derived the expression for the density operator rho for a single spin i is equal to half we had calculated that the density operator can be represented in this manner which is rho is equal to half 1 plus gamma h cross h 0 i z by k t where h is 0 is the magnetic field. In fact this is how we actually started calculating the matrix representation of the angular momentum operators because we discovered here and we showed here that angular momentum operators are intimately connected with the density operator and that is the one which is actually going to tell us about the response of the spin system to various kinds of perturbations in your NMR experiments. So for the single spin we will write this density operator as in this manner is it will be a 2 by 2 matrix because the two spin states for a single spin are alpha and alpha and therefore I will have here rho alpha alpha rho alpha beta rho beta alpha rho beta beta. The states here are the ket states and the states here will be the brass states and I will have to calculate the matrix representations the matrix elements rho alpha alpha rho alpha beta rho beta alpha and rho beta beta. Let us do that. So rho alpha alpha explicitly we write it as alpha here rho here and alpha here and that is equal to now we put in this rho explicitly here. So we have the alpha then for rho we have this half 1 plus gamma h cross h 0 i z by k t and we have the alpha here. So now this one does not give us anything so I will have I take out the alpha completely from here and then I will have alpha 1 alpha that is simply alpha alpha and then I will have here this part the alpha I take out this constant gamma h cross h 0 by k t this constant I take it out and then I will have this matrix element coming alpha i z alpha because i z is an operator I cannot take that out. So this is the constant here so this I can take it out. So then I will have alpha i z alpha. So this gives me 1 because of the orthonormality so therefore half inside 1 here and half here and i z on alpha gives me again a half and therefore alpha alpha is equal to 1 and I will have here half gamma h cross h 0 by k t. So totally rho alpha alpha will become half plus 1 by 4 gamma h cross h 0 by k t. Now rho alpha beta so this is the same way alpha here beta here and rho here half 1 plus gamma h cross h 0 i z by k t and beta here. So this now gives me alpha beta here with alpha 1 beta that gives me alpha beta here and this will be 0 right and similarly here this gives me alpha i z beta i z beta gives me minus half beta therefore I will have alpha beta here and once again this will be 0. So therefore this total element will be 0. Similarly beta alpha if I calculate this will again give me half this is beta here and alpha there therefore this is half beta alpha will be 0 and here once again this constant being the same i z on alpha gives me half alpha therefore there will be beta alpha the beta alpha also gives me 0 therefore this will also be 0. Now beta beta will yield some non-zero value because as in the same way as the alpha alpha so beta half here and this gives me half and gamma h cross h 0 by k t beta i z beta but i z beta gives me minus half and therefore this will be minus half gamma h cross h 0 by k t and therefore this total will be 1 by 2 minus 1 by 4 gamma h cross h 0 by k t. So put it all together so the matrix representation of the density operator is explicitly here the diagonal elements these are the diagonal elements this will be non-zero and the off diagonal elements are 0. Now these diagonal elements are the same what we calculated on rho alpha alpha and rho beta beta and that is I can take this separate it into two matrices here so therefore this is 1 0 0 1 multiplied by alpha and gamma h cross h 0 4 k t by 4 k t into 1 0 0 minus 1 you recall the previous class that this was actually the i z operator for the single spin system and this is the unit operator so therefore half into 1 and gamma h cross h 0 by 4 k t into i z by 2 because if I take away the one one more two I have to take it out then it will be i z so therefore they simply once we know the matrix representations of the individual operators one can straight away this write these matrix elements in the for the total density operator as well but here we actually from the first principles we calculated and demonstrated how these expressions are obtained. So this will be in the simpler form rho is equal to half into unit matrix and gamma h cross by 2 k t into i z so therefore if I want to write explicitly in a similar form I can also write this as half unit matrix plus gamma h cross h 0 by 2 k t into i z here I am actually representing the matrices of i z and the matrix i the i this i is a unit matrix 1 0 0 1 and this is the i z of the single spin 1 0 0 minus 1. So this is the way also one can write once you know the basic matrix representations of the individual operators one can write the density operator also in that manner. Now that was for the single spin system now let us do it for a multi spin system you go a little bit more because we always have to deal with multiple spin systems. So if I have multi spin systems then I will have to consider the Hamiltonian in a more generalized form that it has the Hamiltonian is h z this is the Zeeman part of the interaction and this is the coupling interaction which is the h j so h z represents the Zeeman interaction and h j represents the j coupling interactions. However under high field approximation that is because your magnetic field is of the order of several teslas right. So therefore hundreds of kilo gauss so that is huge magnetic field and this h j this is in few hertz. So this is about coupling constant this is about 10 hertz 20 hertz and things like that and this is in megahertz several megahertz the total interaction here it is in megahertz and this is in several hertz therefore this is a very small quantity and often one can neglect this for the purpose of the calculation of the density matrix okay. So therefore explicitly for the two spin system A x we will write in this manner this is 1 by now n is the total number of states right the total number of states are 2 i plus 1 A states and 2 i plus 1 x states. So therefore the total number of states will be the product of these for example in the case of two spin i is equal to half we had two states here and two states there so therefore it became four states therefore in that case it will be it will be for the two spin system i is equal to half this will be 4 okay and i z again will be the sum of the two operators for the individual spin states. Now so let us recap that here i z A is equal to i z A plus i z x if A and x are spin half systems I will have these four states as individual basis set states alpha A alpha x alpha A beta x beta A alpha x and beta A beta x. So once again I will write them as in the form of energy level diagram here if these ones are non degenerate then I will have this four energy levels then I will have 1 2 3 4 as labels for this four individual states okay this is the convenience we have just labeled them in this manner one can label it any another manner also but one has to keep the convention and maintain the same convention all through okay. So now the generalized form of the density operator or the density matrix 4 will be in this form matrix representation of the of the density operator will be rho 11, 1, 2, 1, 3, 1, 4, 2, 1, 2, 2, 2, 3, 2, 4 and likewise so they will have a 4 by 4 matrix for the density operator. Let us try and calculate these individual elements more explicitly. So let us do for rho 11 the rho 11 is I have now I have dropped this A and x here because we have already seen that explicitly for simplicity I drop the A and x but it is understood that they are there this will be alpha A alpha x alpha A alpha x but simplicity I have just dropped this A and x we understand that okay they they are there. So now if I want to calculate this so this will be 1 by 4 because 2 I plus 1 A there are two states there and for x also two states therefore this total number of states will be 1 by 4 1 by 4 alpha alpha alpha alpha this alpha alpha 1 alpha alpha that is this one here and gamma h cross h naught by 4 kT and then I will have here alpha I z A operating on alpha okay these are the A states and then the x states are coming out here as alpha alpha plus and these are now the A states alpha A alpha A and here are the x states these are alpha x alpha x I z x is coming here. Now so this gives me 1 and what does this give me this gives me half right this gives me half this is 1 and this is 1 and this also gives me half therefore this is half plus half. So gamma h cross h naught by 4 kT to half plus half this is 1 therefore I will have a total of 1 by 4 1 plus gamma h cross h naught by kT. So this is the first diagonal element of the density matrix. Now we calculate rho 1 2 the rho 1 2 this is the second element of the first row and this is 1 by 4 here as before and then alpha alpha and here I have alpha beta. So the first number this gives me alpha alpha beta here this obviously goes to 0 because orthogonality of these individual states and this one gamma h cross h naught by 4 kT alpha this is the A state here alpha A I z A alpha A and then I have alpha beta alpha beta these are of the x state this alpha and this beta that comes here and for the when I take the x of the I z then I will have alpha for these are the A states and this is alpha I z x beta. Now you see this also is 0 and this gives me again alpha beta I z operating on beta gives me beta only therefore this A gives me beta therefore this also will be 0. So therefore this will be 1 by 4 into 0 plus gamma h cross h round by 4 kT is 0 plus 0 so the whole thing is 0. So similarly when we calculate it we will see that rho 1 3 is 0 rho 1 4 is 0 and we have this the first row except the first diagonal element all the three other three elements are 0. Now let us calculate for the second row we will calculate for the rho 2 2. So the rho 2 2 this is alpha beta and alpha beta here alpha beta 1 plus gamma h cross h naught by I z divided by kT. So this first element gives me alpha beta alpha beta and this is gamma h cross h round by 4 kT this alpha I z A alpha and I will have beta beta these are for the x states and this one is for the A states and this is the beta I z x beta. So this gives me 1 this gives me 1 this gives me half and this also gives me minus half. So therefore what happens here this part is 1 1 by 4 and gamma h cross h round by 4 kT this gives me half and this gives me minus half. So half minus half this is 0 and therefore I will have only this term remaining. So this term goes to 0. So similarly if we calculate it you will see that rho 2 1 in the second row and rho 2 3 and rho 2 4 will be 0 and you do the same exercise for 3 3. Now it is beta alpha and beta alpha here this one gives me 1 again and this gives I z A beta the minus half this is 1 1 this is 1 and this is 1 and this gives me half. So this will be minus half plus half and therefore this vanishes I will only have I am left with only 1 by 4. So other 3 elements in the third row 3 1 3 2 3 4 they will all be 0. The fourth row the 4 4 states if I do this calculation I will have a beta beta and beta beta here and this one is beta I z A beta this gives me beta beta and now for this beta beta here and I z x also beta beta. So this is this gives me minus half because I z A on beta gives me minus half beta and this is 1 this is 1 and this also gives me minus half beta from the x state x spin therefore I will have minus half and minus half here therefore this is total minus 1. So I will have therefore 1 by 4 minus gamma h cross h naught by 4 k T. The other elements rho 4 1 4 2 4 3 are equal to 0. So therefore what do I get? So I have here all the elements listed here 1 by 4 this is 1 1 1 1 and then gamma h cross h naught by k T 1 by 4 1 0 0 0 0 0 0 0 0 0 and minus 1 and what is this? This actually if you remember this was the I z operator for the 2 spins and this is the unit matrix this is the unit matrix unit matrix multiplied by 1 by 4 and this is the number of states this is basically the partition function. So in generalized manner this is the partition function number of states which is equal to the number of states and here again it is the number of states and this is a constant gamma h cross h naught by k T that is that is the constant k here and therefore I can write this in a simplified manner this is 1 by z plus k by z I z I z is your matrix for the I z operator therefore rho is simply equal to this. So if we knew this density of state or the partition function if you know this number of states here and I z operator you know what it is we can simply write down explicitly the matrix representation of the density operator. So I think we can stop here this is the good time to stop for the calculation of the density operator for the 2 spin states.