 Hi there, we are discussing the quantum mechanical treatment of the various magnetic interactions of hydrogen atom. We have seen that the total Hamiltonian of hydrogen atom can be written as this is the electron Z 1 term, this is the nuclear Z 1 term plus A i z, this is the hyper 1 interaction term. We also saw that we could write as two parts H 0 plus H prime. So, H 0 we call the unperturbed Hamiltonian or the main Hamiltonian which includes this one that is the most significant interaction is there. The perturbation is this, this can also be written as A s plus i minus plus times by 2. So, this is the main Hamiltonian, this is the perturbation Hamiltonian. We can now try to find out the Eigen states of this and corresponding energies. So, for that the Eigen state can be thought of as a product function of the electron spin and the nuclear spin. The symbolically we write them as M s, M i. And without knowing their expression form, we can use the property that when let us say S z operates on this one, it will give me the electron spin angular momentum quantum number which is M i s the electron spin component of the quantum number. Similarly, i z operating on M s, M i will give M s, M i. So, these two identifies the possible component of the angular momentum. M i takes the value of plus minus half and M s also takes the value of plus minus half. So, we have got 4 possible combination and correspondingly 4 wave functions are there and we should get 4 energy levels alright. So, they are designated by plus half plus half plus half minus half. So, we can find out the energy of this H 0 Hamiltonian with respect to this 4 wave functions. How do they look like? Using this formula very easily to find out that let us say G e beta e B 0 S z operating on plus half plus half minus half plus half will give this operating on this will give plus half your way to this. Similarly, i z operating on this will give similarly, we can find out all the other terms here now A operating on let us say plus half plus half will give a i z this operating on this give plus half this operating on this will give another plus half. So, this gives a by 4 plus half plus half. Suppose we have another one let us say S z i z plus half minus half. So, this operating on this gives plus half this operating on this gives minus half the net resultant will be minus a by 4 plus half this way we can find out all possible energies of the 4 states. So, this are given here let us call it here energy of this a by 4 a by 4 and other 2 I will write at the top of this. This is e minus half minus half this gives minus G n beta n B 0 by 2 plus a by 4. So, this 4 different wave functions give rise to 4 different energies and the how they are arranged they are arranged according to of course, the various values of this here the there is a mistake here. So, this mistake corrected. So, the major interaction is coming from the electron Z 1 term here then this will be very small contribution that will be added to that and this will be also small contribution. So, these are now shown here in this slide. So, when there is no magnetic field that all the 4 energy levels are same. So, first we apply the electron Z 1 term. So, the maximum splitting is due to that when the electron space is minus half with the lower energy plus half gives higher energy and then this is the interaction coming from here. Then we add the nuclear Z 1 term that is here G n beta n B 0 I Z. So, that splits this energy into 2 for M i equal to plus half and minus half plus half here gives negative energy. So, this lower energy similarly here minus gives higher energy. So, these are nuclear Z 1 term added here then we add this the hyperbolic interaction here assuming the A is positive quantity then I Z S Z when the product is negative that gives lowering of energy and the product is positive gives higher energy. So, that is the way the further these levels are changed. So, having got these 4 energy levels what are the allowed transitions now for if the micro magnetic field is applied along the Z direction then we will say perturbation due to the micro wave let us say micro wave in the Z direction we will have this sort of operator B 1 is the let us say the magnitude of the micro magnetic field applied along the Z direction. So, this 4 energy and the 4 wave function that we have there to have transition I must get a matrix element of this kind some M i M s M s M i and M s M i prime this has to be non-zero for transition to take place. Now, here you see that this is as good as saying that M s M i S Z M s prime M s has to be 0, but then all these 4 energy levels are against state of S Z. So, unless these are this is same as this will not be 0 that means M s should be equal to M s prime that means plus half minus half so that means no transition. So, micro magnetic field applied along the Z direction does not cause any transition that is not surprising we have already seen earlier that the micro magnetic field has to be applied along the perpendicular direction same thing is coming quantum mechanically also. So, states therefore suppose you apply the H micro wave along the X direction then this will look like G e beta i B 1 S X. So, the same argument now what I need here is M s M i S x M s prime M s this has to be non-zero. Now, I can write S X as the raising and lowering operator S plus plus S minus by 2 this one. Now this can be non-zero if this operating on this gives a function which is same as this one this operating on this gives function which is same as this one and what is more that this nuclear spin part is not involved in this integral. So, this can be written as non-zero so that means this must be i must be equal to i prime over delta i M i must be equal to M i prime or delta M i 0 and here because this can either increase it by 1 unit or decrease it by 1 unit. So, that gives this selection no delta M s will be equal to plus minus 1. So, these are the selection for that and so those are the two transition allowed which is shown in this slide. So, here this transition takes minus half to plus half for the electron spin without changing the nuclear spin. Similarly, for the other one here electron spin changes minus to plus half again without changing the nuclear spin. So, what is the energy gap for this you can calculate from this pair and that pair. So, one delta E turns out to be G e beta i B 0 plus a by 2 this is for M i equal to plus half other delta E is equal to minus a by 2 this is for M i equal to minus half. So, the two transitions are separated by a by 2 from the position which is G e beta i B 0. So, if there are no hyperfine interaction then the energy would have been this one energy of transition. So, I see the two lines which is split which are split from the original line position and what is more here see that this term the nuclear given term has disappeared completely from this. So, in this arrangement the way it appears that we do not see any interaction of the nuclear given term. Now, if we do the experiment let us say at a fixed magnetic field then what we see in fact the derivation here is based on a fixed magnetic will B 0 is kept fixed here. So, that does not change. So, we vary the frequency then B 0 is fixed. So, this will be equal to h nu let us call it for first transition this is h nu for second transition. So, as a function frequency if you plot it this will give a peak where B 0 is fixed. So, this will give a one peak here other peak there the low frequency peak will correspond to M i equal to minus half other one corresponds to M i equal to plus of higher energy. So, this center of this is exactly given by this one. So, this gap is now exactly equal to A. So, we say that in frequency unit this is exactly equal to the hyperbolic coupling constant because nu 1 will be mu 2 g in beta i B 0 this is the frequency unit. Now, if we do the experiment by keeping the frequency fixed where vary the magnetic field then h nu is kept fixed. So, for first transition this will be g e beta i let us say B 0 1 A by 2 for M i equal to plus half and this is kept fixed. So, second line will be at B 0 line 2 minus A by 2 this is for M i equal to minus half. So, here frequency kept fixed. So, if I plot this as function of magnetic field B 0 as this is fixed by B 0 comes up to be h nu minus A by 2 divided by g e beta i. So, this will appear at the second lower magnetic field. So, this is M i corresponds to plus half and here M i corresponds to minus half you see how it changes now. And again the exact layer the center the magnetic field will correspond to this one which is given by this is the B 0 then it will be and the gap here delta B will correspond to. So, we call this the hyperfine splitting constants. Now, we have seen this energy levels and the correspondingly we have found out the two transitions and got this relationship. Therefore, variable frequency experiment and variable field experiment we have neglected so far the cross term or the perturbation term which I was saying earlier which is h prime is A by 2 s plus i minus plus s minus i plus. Now, effect of this can be seen by considering the first order calculation and first order calculation the let us say e 1 calculate given as psi 0 perturbation psi 0. First order calculation here. Now, here this is the perturbation here and this 0th order wave functions are the wave functions that we have here this. So, if I write any of this one M s M i and put this here s plus i minus plus s minus i plus M s prime M i prime. So, here this suppose to be same as here. So, I cannot have prime here this will be same as that. Now, because this is increases the value of M s by 100 this decrease by 1 unit for nuclear spin. So, this will be 0 for all possible this 4 states that we have here. So, even if we do the calculation of the first order this term does not contribute to the energy level in any way nor does it change the allowed transitions there. So, what we conclude from here is that all possible energies that we have discussed here and the allowed transitions they are correct up to first order of calculation. And here the characteristic feature is that the centre of the spectrum corresponds to the exact g value that we can measure from here and it is symmetric with respect to the centre and we can ignore the nuclear given part all together. So, with this I have shown the final slide here this is the Hamiltonian which is good enough for our first order calculation. Nuclear Given splitting gives this and then hyperbolic interaction gives splitting here and we can get the allowed transitions from this with this we stop now.