 Hello, today we are going to discuss second order EPR spectrum calculation. Earlier we saw how to calculate spectrum up to first order. Now why bother with second order calculation because first order calculations gave quite good interpretation of most of the spectrum that we saw so far, but sometimes that is not sufficient. So, here the two important reasons. For example, when hyperbolic coupling constant is not small compared to the electron Zimantam or when there are several equivalent nuclei giving large splitings, in that case first order calculations are not good enough. So, when you doing the second order calculation we will take hydrogen atom as an example. And then try to make a distinction between hyperbolic coupling and hyperbolic splitting constants. Also try to make a distinction on fixed frequency experiment and fixed magnetic field experiment whether they are same in all respect or not. First order spectrum that we calculated and interpreted earlier gave the same value for hyperbolic splitting constant or hyperbolic coupling constant. There was no difference, but we will see some difference may appear when we do the calculation up to second order. So, let us recapitulate our Hamiltonian for hydrogen atom. This is the Hamiltonian, this is the electron Zimantam, this is the nuclear Zimantam and this is the isotropic hyperbolic interaction. This can be written in this fashion. So, here we treat this part electron Zimantam, nuclear Zimantam and A SZIZ as the unpart of Hamiltonian and the other part is treated as a perturbation. The 0th order wave functions are given as a product states of the electrons pin angular momentum component and nuclear spin angular momentum component. Since each of them is half, we get 4 states plus half plus half and this fashion. We did all these calculations earlier and 0th order wave functions are of this kind and the 0th order energy are also of this kind. Then the first order calculation requires evaluation of integral of this kind. Here this is the perturbation and these are the unpart of wave function. For this states here, all these elements are 0. Why? Because these are same for both of them. Now, when this acts on this one, it is going to decrease the value of M i by 1 unit and when this acts on this one, it is going to increase the value of M s by 1 unit. So, effectively this will give something let us say M s M i times M s plus 1 M i minus 1 and because they are different and they are orthogonal, so this will automatically give rise to 0. So, first order energy calculation is 0 for all the 4 states. So, these 4 states are also good wave functions of first order and energy is also good up to first order. To do the second order calculation, this is the expression for the second order correction to energy. Here the kth state, kth is one of the 4 states, this energy involves this sort of integral. Here psi k is this k and then sum over all possible other states which are there other than the k and this is the difference of energy between the k and j state. So, for these to be nonzero, these integral must be nonzero. So, here again, if you put a prime here to indicate that they are different, these will be nonzero when after raising this, this will be something like M prime s plus 1 and this will be M prime minus 1, this lower seat where this increases. So, for these to be nonzero, this must be same as this one. So, that in other words, s prime plus 1 should be equal to s, M s prime plus 1 would be equal to M s and M i prime minus 1 should be equal to M i. These conditions must be satisfied for this integral to be nonzero, which is also written here. If you have the other one s minus i plus then M s prime s must be equal to M s plus 1 and M prime i must be equal to M i minus 1. So, here not all 4 states will contribute to the second order calculation. State number 2 and 4 give rise to these sort of values. So, from this expression now, from this expression, I can calculate the energy correction up to second order and that turns out to be this. This is the second state, second order correction comes out to be s square by 4 and this factor in the denominator. This is the electron Z 1 term, this is the nuclear Z 1 term. So, that way I can calculate all the 4 energy corrections for the 4 states and you see that state 1 and 3 do not have any second order correction and state 2 and 4 have. This has got plus value, this has got minus value. So, with the second order calculation done, I can now get the total energy up to second order. Here the state 1 does not change. So, it remains as it is. Similar state 3 also remains as it is, but 2 and 4 get mixed because of this cross term that we have found out here. So, this 2 and 4, I put prime here to indicate that they are not pure state 2 or pure state 4. Some admixture is there. So, these are the energy of this 2 prime state or 4 prime state. So, having found these 4 energy levels, now we can find the transition energies. These are the transition energies where the mi does not change. Here also mi does not change and this energy is given for transition from state 1 to 4 and 2 to 3, this 2 prime to 3. Essentially, this 2 prime is very nearly equal to 2 and 4 prime is very nearly equal to 4. So, these are the energies that we get here. This corresponds to mi equal to plus half, this corresponds to mi equal to minus half. So, what we see here? This should look familiar that if the second order correction was not applied, energy was this. Similarly, energy was this. So, this is same as the first order splitting. The energy level is split equally by plus a by 2 or minus a by 2. But now, when you include the second order correction, both the transition energies have increased by this amount. So, that is something different. Now, we do the experiment. Now, let us say fixed magnetic field experiment and afterwards we will see the fixed microwave frequency experiment and see how the spectrum looks like. In this condition, the magnetic field is kept fixed. Then, the frequency required to cause the transition for the first one. This will be given as this is for the first transfer, where m s is equal to plus half and second one, similarly corresponds to this corresponds to m s equal to minus half. So, the spectrum will look like something like this way. This b zero is kept constant fixed. So, this frequency is higher than this one. So, energy is higher for this and lower for this one because it comes minus sign. So, to get the particular frequency now, I divide this whole thing by 1 by h. Similarly, divide this by 1 by h. So, this gives me, let us say nu 1 and nu 2. So, this corresponds to m i equal to plus half, this corresponds to m i equal to minus half. Now, where is the center of this? Now, if this was absent, the center would be just average of this and this. So, this would be g e beta e b zero by h. Now, the way this appears now, these lines are shifted with respect to center by this amount. So, again if this was absent, I would have got the first order spectrum and then some of the spectrum would have been dotted line here. This is also shifted somewhere. So, this way the spectrum would have appeared. Now, both the lines are shifted to the higher frequencies. So, the center of the spectrum has now moved towards higher side. So, the actual g value is here, but the way it appears center of the spectrum now appears somewhere here now. So, this is the important outcome of this analysis is that, if one blindly looks at the middle of the spectrum and then calculates the g value for this magnetic field, then the answer will be wrong. The actual g value is here, both the lines have shifted up and that can be understood from this sort of analysis. The next point is that, the difference between this and this, if we take the way it appears, there is a difference between this and this, then this of course cancels. So, this gap is still equal to here, which is the hyperfine coupling constant. So, this experiment gives the value of the hyperfine coupling constant though it will not give you the g value correctly unless you account for this correction and find out exactly the correct place of the magnetic field to calculate the g value. Now let us do a fixed frequency experiment, let us say nu 0 is kept fixed. So, here in this energy expression, I keep the frequency fixed on both sides and find out for which value of v 0, I will get a transition. So, if nu 0 is the frequency which is kept fixed, then the energy is h nu 0 and that will be equal to, here let us ignore this nuclear demand term unnecessarily, because just making the equation looking ugly, but one could return them if one requires it. Though it is safe to ignore, because this term is about 2000 times smaller than this one. So, unless one is interested in a very precise measurement, one might as well ignore this. So, for the second transition minus half. So, here this is constant. So, I have to find out the value of v 0 for this line and also v 0 for this line. So, it is a rearrange the equation and then that will look like this. So, the only difference is here plus minus. So, I am combining this together to write one single equation. This could be further rearranged to give rise to a quadratic equation in v 0. Here again the minus sign corresponds to m i equal to plus half and plus sign corresponds to m i equal to minus half. So, this is a quadratic equation in v 0. So, I can solve it and get the value of the magnetic field for a given value of nu 0. Now, of course, you know the quadratic equation gives two roots. Here only one of them roots will give physically meaningful value, realistic values. So, let me write that. Here I have taken only the positive square root of this one. Now, this could be simplified by rearranging the term and it will look like this. So, this pair corresponds to one root, this corresponds to the other root of this two possible transitions that I can get. So, this equation is exact. So, if I put the values now for a given frequency and let us say a known value of the this hyperbolic coupling constant, you can get the two course magnetic fields. So, since the values are known, let us see what sort of numbers we get. So, the a in frequency unit is 2 0 megahertz in frequency unit. So, in other words a by h is the frequency of this hyperbolic coupling constant. So, if we put a let us say some typical frequency for the EPR experiment of the X band say 9.5 gigahertz, then you can calculate the corresponding magnetic field for the two transitions. If they turn out to be B0 lower, lower line corresponds to 3 1 1 5.93 gauss and B0 higher corresponds to 3 2 5.67 gauss. So, we get a spectrum which in a sense similar to this, but let me draw it here now. This is this was done as a function of frequency. Now the experiment is a function of magnetic field. So, I got a spectrum which looks like this one is here. Now this corresponds to m i equal to plus half, this corresponds to minus half. Now where will be the g value? So, if you simply take the average of these two positions, then B average B0 is equal to average that turns out to be 3 3 8 0 gauss. But this is see this is not the resonance position for the g value, because both the lines are now moved towards the lower side. So, here knowing that for this I forgot to mention the g value for this is given as since 2 2 with this g value one can calculate the magnetic field. So, that g value will correspond to this magnetic field which is this is the value that will come out to be if you use this micro frequency and this g value that is the supposed to be. So, you see the difference now the average value is lower than the place for at which I must calculate the g value. So, here on other words here the lines are shifted towards the higher frequency. Here if the actual g value would be somewhere let us say here these lines both of them have gone downwards. So, if I take the average value of this average may be here B average. So, if we calculate the g value at this at the center of this thing I will again get high at g value the true g value somewhere here both the lines got shifted downwards. Now to see a little more about this complicated equation let us try to simplify a little bit by taking the approximate value of this square root because here these numbers come out ok, but they do not quite throw much light in what is going on inside this one. So, just simplify it such that we understand at least qualitatively what we are saying here is correct. So, this one I can to I can take this common and then write this way square root here. So, then as a there is the approximate approximation of the square root I can write it in this fashion. So, this bottom is 4 g e b e. So, and further simplification can give rise to this expression which I am writing straight away. So, here now it shows that the magnetic field that appears here with respect to this h nu 0 by g beta e that true center of the spectrum and from there the minus a and plus a that is the that is the splitting, but both of them now have gone towards the lower side by this much amount. So, that is what we are saying that the implication of this exact equation there of course, this is approximations this cannot be treated for exact calculation we must use this, but nevertheless it does show how the two lines have gone down. So, second point is that the way again this difference of the two line position is such that they are not same as the hyperfine coupling constant, hyperfine coupling constant this if you convert magnetic field in it this will turn out to be 8 6 gauze. Now, here if you take the difference of these two that difference turned out to be right side by side difference in line position that turns out to be 7 4 gauze. So, this is the observed splitting in the spectrum given by this experiment when you do the experiment here and this is the intrinsically the splitting because of the hyperfine interaction is 506 gauze what you see here is 509 gauze. So, this is where one makes the distinction between electron nuclear hyperfine coupling there is a strength of interaction in energy unit either right here or this one and this is the electron nuclear hyperfine splitting constant this one can convert this also frequency unit, but this is measured quantity that appears in the spectrum in the form of splitting of lines. So, one must really mention what it what it is that one is reporting in the first order spectrum these two things do not make a difference they are the same, but once we apply second order correction this difference is possible and one can see that here it is definitely measurable quantity. This is the second important consequence of the second order calculation first important consequence was that the g value will be absolutely wrong if you simply blindly take the center of the spectrum to be the correct place for calculating the g value the second one is this. Now, this hydrogen atom the sketch is very special because this coupling constant is really really large that is not very surprising because for hydrogen atom the electron is actually in 1 s orbital and that is the also the main and the principal requirement for forming contact interaction. So, this interaction is very strong and you get such large splitting most of the appear signal or organic molecules and typical hyperfine coupling constant that one sees usually this is not of much significant importance. So, one tends to ignore this, but there are cases other than hydrogen atom where this can become important one is that there could be more than one nuclei coupled to the electron and each of them contributes to hyperfine coupling constant such that overall the effect is large. So, if a set of equivalent nuclei they are coupled together same simultaneously to the electron and then second order corrections might be necessary. So, here is an example this EPR spectrum of the CF3 radical the fluorine has a nuclear spin of half. So, one we expect that this 3 equivalent fluorine should give a spectrum of this 1 is to 3 is to 3 is to 1, but what it gives is this that this what are supposed to be the intensity 3 is not split into 2 lines 1 is intensity of unit 1 this is to 1 and 2. So, this line is split here also splitting constant is reasonably large. So, to understand that let us take a simpler radical system something like Rh2 radical so 2 spin half system which are equivalent and coupling to the electron. So, here we will write the Hamiltonian again in this fashion where we couple the first nuclear spin states to get a total nuclear angular momentum and its component and then that will couple to the electron spin that is the idea. So, here again for simplicity we ignore the nuclear German term without serious consequence anyhow that is very small. So, the hyperfine term will now look like Azjj is the component of the total nuclear spin quantum number. So, when there are 2 nuclear spin total j can be 1 or 0. So, this is the corresponding components 1 0-1 this is for 0. So, now we have 8 possible states they are represented by the electron spin component and nuclear total angular momentum component j and mj. So, that we get 8 states plus half for electron and 1 0-1 0 for the nuclear spin and similarly again for minus half component electron I get another set of 4. Now, we follow similar type of calculation that you have done for this hydrogen atom exactly similar. So, first order energy calculation appears to be this same for all the 8 levels. So, first order spectrum is obtained by following the selection node delta mj is 0 delta mj is plus minus 1 and this gives a spectrum in this fashion 1 is to 2 is to 1 what we of course, expect for this type of radical. Now, when you do the second order calculation these are the states which gives non-zero value of the integral and the value is given here and then the second order calculation of energy also or in this fashion the state 2, 3, 5 and 6 they have non-zero value of second order corrections to energy and 2 and 3 have positive correction and 5 and 6 have negative correction. So, then using that one can find out the position of the resonance magnetic field and that is the way it will look like. So, here the first order spectrum gave 1 is to 2 is to 1 line position and then if you apply the second order corrections using this sort of energies then this 3 lines will shift towards lower magnetic field and this intensity 2 is split into 2 of 1 is to 1 and the spectrum will appear in this fashion. So, this of course depends on the value of the coupling constant and the magnetic field B0. If B0 is very large this correction is small and this will not be seen therefore, not only the coupling constant but also at what magnetic field the experiment is done. So, if one works at low frequency PR spectrometer this effect might become more prominent. The same way if one does the calculation for 3 equivalent spin half nuclei then the total nuclear spin will combine in this fashion to give rise to total J equal to 3 by 2 and total J equal to 1 by 2 it will appear 2 times each of them the MJ values are given here. So, exactly same calculation what we have done here can be calculated and the energy will be split such a way that all the lines will show shift towards the lower magnetic field and in particular this intensity of 3 units will be split into 1 intensity of 1 unit 1 and another intensity of 2 units all of them will show downward shift. So, here again the calculation of G value will require special consideration because centre of the spectrum does not correspond to the true magnetic field. So, this is exactly what was reported here CF3 radical. Now, the example that I showed about second order coupling and the splitting it is possible that one may not see those splitting all the time nevertheless they are there if the hyperbolic coupling constant is reasonably large or there are many equivalent nuclei which are coupled together to produce a large splitting. So, here is an example for this type of metal complex it is a trinuclear cobalt complex arranging this fashion that all 3 cobalt are equivalent and the appear spectrum is shown here. So, cobalt 59 I should have has I equal to 7 by 2. So, that will give total I will be 21 by 2 in other words there will be 22 lines and if one compute the first order spectrum the relative intensity among this 22 lines will look like this. So, in a sense there are 22 lines though one may not see the extreme end all the 20 lines are seen and they reasonably reproduce this relative intensities. But nevertheless look at this line shape narrow here and brought there each of them narrow towards the bottom and brought at the top why is that whatever you discussed so far is they just cannot none of them explain this sort of line shape. So, here the explanation was that the second order coupling constants or second order splitting constants are actually playing a role here it so happened that they are not resolved. So, look at that the 3 cobalt nuclei each of them have got 7 by 2 nucleus and splitting constant if you measure from the scale which is typically maybe 25 30 gauss which is not very large but because so many of them are there the effect can be noticeable. So, this calculation was done after second order and this is the way the result is. So, the first to the spectrum predicts the relative intensity of this sort of values now if you include second order calculation then many of this degenerate transitions become non-degenerate now this is how they appear here bunch of them are appearing everywhere and all of them are shifted towards lower magnetic field. So, the reason behind this unusual line shape. So, we saw that second order interactions are important if the hyperbolic coupling constant is large or there are more than one equivalent nuclei present there which are also contributing to a large overall splitting then there is second order calculation are necessary for an accurate measurement of the g values and hyperbolic coupling constant even to reproduce the very unusual line shape of a spectrum where these splitting are not resolved that is all.