 In this module we are going to learn about 2 applications of perturbation theory of degenerate states in atomic systems well atomic and molecular systems. But before that let us just remember where we had stopped in the last module we developed this first order perturbation theory for degenerate states. We understood that for degenerate states where you have n number of wave functions associated with the same unperturbed energy where the size denote complete set of orthonormal functions you cannot write the perturbed energies as perturbed wave functions as size 0th plus something. First of all you need a linear combination because you do not know which of these functions is going to contribute in which part of wave function they are all identical indistinguishable. So, well function forms may be different but indistinguishable in the sense that energy is same. So, there is no way in which we can say that one particular wave function will contribute and the others will not cannot do that. So, this is what it is we write the wave function we start with a linear combination of wave function then go to the usual definition of our corrected wave function usual definition of uncorrected energy of corrected energy and then we arrive at this matrix equation in which we see perturbation V ij operates on these eigenvectors to give the ek first order energies as the diagonal elements of this matrix capital X. This is something I think I forgot to say earlier this capital X is the matrix of first order corrections to energies and here the beauty is you can see from this expression here that all the of diagonal elements are going to be 0 delta ij remember. So, we are going to have this diagonal matrix where the diagonal elements are only going to be nonzero and they are going to give you the values of first order corrections to energies we start from here. The first system that we discuss is that of a non-rigid rotor and there the perturbation that we have to consider is centrifugal distortion we have already studied at length this unperturbed system of rigid rotors and the assumption there was that the molecule is like a rod it does not change bond length in the course of rotation or the bond length is equal to the equilibrium bond length. In a non-rigid rotor what happens is the bond is flexible it is not like a rod. So, what happens if you take a spring tie some weight at the end and rotate it the spring is going to get elongated right. So, we are still working within right now we have not said whether bondogram or approximation is valid or not we are sort of working within it we are not considering vibration and rotation simultaneously as such. But we are considering the effect of vibration that the average value of r0 square which is the all important term in the rigid rotor Hamiltonian that average value of r0 square is going to increase. So, we start with the unperturbed system rigid rotor for that the Hamiltonian as we know is l square by 2i when l square by 2i operates on size is 0 we get bj into j plus 1 I was thinking that I am making some mistake somewhere and I have actually written it a little later also sorry about that. Of course, I think you can see that this equation is not complete you must write the wave function also 0th order psi j this is what we have learned and ej 0th will be equal to then b into j into j plus 1 just this nothing so perturbation is centrifugal distortion as we said so it came and vanished isn't it from classical mechanics we know that the centrifugal distortion is associated with the potential that is proportional to the fourth order of angular momentum. So, we construct the operator v equal to k minus k l to the power 4 now please remember here we are going to use it what l to the power 4 means so l square psi is equal to say something what do I write let me write a psi so now if l square operates on this function this is a function right l square operates on l square psi what do I get l square operating on a psi a will come out and l square operating on psi will give me we had said a earlier psi get a square psi right of course we have done this in your tutorials and all left hand side will be l to the power 4 operating on psi so this is the meaning of l to the power 4 l square has to operate twice that is all but of course you know it in case there is a confusion in anybody's mind I thought I just say it once okay so this is your centrifugal distortion so centrifugal distortion when it operates on the wave function and remember we will still work with the uncorrected 0th order wave function there is no need to incorporate the correction term as we have learned a few modules ago we get minus k l to the power 4 operating on psi j 0th now remembering that l square by 2i operating on psi j 0th gives me bj into j plus 1 I can rewrite that and I can get l square operates on the uncorrected wave function to give me 2i multiplied by b multiplied by j into j plus 1 multiplied by psi j 0th now what do I have to do to get this eigen value of l to the power 4 I have to make another l square operate here so when I do that what do I get 2 i b these are all constants they come out j into j plus 1 constant come out so I am left with all these constants multiplied by l square operating on psi j 0th and I know that is going to give me well it is going to give me 2 i b j into j plus 1 multiplied by psi j 0th simple right so we make this operation and we get v operating on psi j 0th is equal to minus k l to the power 4 operating on psi j 0th gives us minus 4 k i square b square j square into j plus 1 whole square psi j 0th please do not square psi the wave function okay you need to understand how the operation is done I am sure you are clear with that okay so this is what I get this is the eigen value of this v operator okay so v operating on psi j 0th gives me in short we will write minus d into j square into j plus 1 whole square psi j 0th where d is equal to 4 k i square b square now see let us work with this wave function with this level j equal to 1 which is triply degenerate you can work with any level in fact I will encourage you to work with j equal to 2 or something and see for yourself what happens I mean if you work with j equal to 0 that will be cheating because there is no question of degeneracy there so let us work with this j equal to 1 which is triply degenerate m j for that the magnetic quantum number equal is equal to minus 1 0 plus 1 so the v matrix as we have discussed earlier is going to be something like this the matrix elements will be written v 1 1 v 1 2 v 1 3 and so on and so forth our job is to find the matrix elements well now if you remember the linear equations that we have written earlier in order to get solutions there the secular determinant has to be 0 this is something that is very well known premise rule we are going to use it and we will discuss a little more about that when we talk about variation theorem so secular determinant has to be equal to 0 that means the determinant corresponding to this matrix should be equal to 0 so did v will be equal to 0 something like that write minus x okay now see we will try to simplify these expressions for v 1 1 v 1 2 and so on and so forth to do that we will remember that the size is 0 is really an eigen function of the operator v and it has an eigen value minus d into j square into j plus 1 whole square so with that let us begin we will also remember that this v and h 0th actually commute right it is apparent here there is common set of wave functions so what is vii vii will be integral psi i 0th v psi i 0th overall space do we have vii here yeah v 1 1 v 2 2 v 3 what is the value you can write like this v operating on psi i 0th I know is minus d into j square into j plus 1 whole square all that is your all that is constants will come outside the integral inside the integral you have integral psi i 0th multiplied by psi i 0th well if it is complex and psi i 0th star multiplied by psi i 0th integrated over all space what is that we know what that is is not it we know that this is equal to 1 great so already we have simplified and we have found the expressions for vii so the expression will be minus d into j square into j plus 1 whole square so you take this and vi j what is vi j so basically we are going to replace this expression here but before that let us think a little bit about vi j vi j has two different wave functions integral psi i 0th whole square psi i 0th star multiplied by v operating on psi j 0th integrated overall space that will give me similarly d I have written j dash just to signify that this i and j are different minus d j dash square multiplied by j dash plus 1 whole square multiplied by integral psi i 0th star psi j 0th overall space and we know very well that these wave functions are orthonormal so this is going to be equal to 0 so what we are saying is that vi j is equal to 0 now when I say i and j what do I mean it is very clear what I mean by vi i when I what do I mean when I say i and j well not so difficult also I mean that we are talking about two different uh degenerate wave functions okay that is equal to 0 so now the secular determinant I just show you that once again so remember all these diagonal terms will become this minus d into j square into j plus 1 whole square minus x all of diagonal terms as we see here is going to become 0 so this is a secular determinant that is equal to 0 very simple right take the determinant this multiplied by this multiplied by this equal to 0 the expression that you get for x which is the first order correction to energy is minus d j square into j plus 1 whole square it has three roots but the three roots are identical isn't it because what we have essentially is minus d into j square into j plus 1 whole square minus x whole cube equal to 0 so the roots are identical roots are the same and we have the same amount of correction to energy for centrifugal distortion no matter what l value we have chosen no matter what m value we have chosen rather okay so this is the expression that we get e j now instead of b j into j plus 1 becomes minus d becomes b j into j plus 1 minus d j square into j plus 1 whole square so this is a case where perturbation removes none of the degeneracies whatever degeneracy was there will remain right we have started we have demonstrated this with this j equal to 1 state which is triply degenerate so what we see is that energies will go up but a term in m is not coming in okay so all the 3 m sub levels have gone up by the total energy by the by the same amount the total energy has gone up by the same amount for all the 3 sub levels this is an example of perturbation theoretical treatment of degenerate systems where degeneracy is not lifted at all but do not think there is no effect the energy is changing energy of e j level is decreased by d j square into j plus 1 whole square okay so much for our non-rejected order but now we want to discuss something in which actually degeneracy is lifted and that example is provided by stark effect stark effect means lifting of degeneracy by electric field and we are going to learn stark effect we are going to go a little quickly through this because we have actually introduced all the necessary tools I really would like you to try and do things yourself so let us talk about the principal quantum number 2 for hydrogen atom now I hope the change in gear was not very quick we are talking about non-rejected rotor so far now we are talking about stark effect in hydrogen atom okay n equal to 2 for n equal to 2 what are the orbitals the first wave function is 2 s second one is 2 p is 0 third one is 2 p minus 1 fourth one is 2 p plus 1 and once again even though it might have become a cliche please do not forget 2 p minus 1 is neither 2 p x neither not 2 p z not not 2 p y 2 p plus 1 is neither 2 p x not 2 p y but rather a linear combination of them 2 p x and 2 p y are obtained by linear combinations of this 2 p minus 1 and 2 p 1 as well 2 p z is actually equal to 2 p 0 okay when I am equal to 0 then we actually get the p z orbital we have discussed this in significant detail while talking about orbitals okay so v in this case is minus equal to r where is the electric field and we can write it as minus epsilon r cos theta because it is it can get an angle right so that we can write as minus epsilon into z where epsilon is a magnitude of the electric field now this is a secular determinant that is equal to 0 here it is important to remember what v 1 2 v 2 1 3 and all that is here we are just designating 2 s as 1 so v 1 1 would be the perturbation term involving 2 s orbital and 2 s orbital same with v 2 2 v 3 3 which v 4 4 fine what about v 1 2 it involves 2 s orbital and 2 p 0 orbital what is v 1 3 it involves 2 s orbital and your 2 p minus 1 orbital and so on and so forth alright now as usual first let us write down the wave function actually I should have written the theta phi part here as well but anyway to start with theta is fine so this is the theta part of the wave function as you know and now I do not remember exactly if I have we did talk about recursion relations but I do not remember if I talked about this if not just take it axiomatically that when you take a what kind of polynomials are these like where no Legendre so when a Legendre polynomial is multiplied by the coordinate then you get a linear sum of the polynomial before and polynomial after I think we did it yes recursion relation we actually discussed it right so here our variable is cos theta so I have written cos theta multiplied by the Legendre polynomial in cos theta and here I have written L because I am talking about hydrogen atoms sometimes my mistake I might have written J please correct it when talking about hydrogen atoms so it is a linear sum of the polynomial in L minus 1 and polynomial in L plus 2 okay so now what is your V ij V ij is equal to minus epsilon r multiplied by p i cos theta cos theta pj cos theta integrated over all functions space okay here we have our good friend cos theta multiplied by pj of cos theta so of course you can expand using this recursion relation and you get minus epsilon r I hope I have not missed brackets here minus epsilon r no there is no question integral I can write p l of cos theta multiplied by a into p l minus 1 of cos theta plus b into p l plus 1 cos theta okay so naturally we get 2 terms and this gives me 0 right let us see so let us take these 2 what do I get I get something like integral p l of cos theta well star of that if it is not real actually it is real here multiplied by p l minus 1 of cos theta and this gives you something similar this one is p l cos theta second one is p l plus 1 in cos theta so I get 2 integrals and now see once again do not forget that these are wave functions right an orthonormal set so this integral p l in cos theta multiplied by p l minus 1 cos theta integrated over all space that has to be equal to 0 yeah that must be equal to 0 because they are orthogonal to each other same here so finally after all this you get the answer to be 0 V ij is equal to 0 unless your delta l equal to plus minus 1 what will happen if delta l is equal to plus minus 1 delta l means the difference between i and j if j equal to i plus 1 or if j equal to i minus 1 then one of these integrals is going to survive right so if only if delta l equal to plus minus 1 we are going to get V ij nonzero otherwise in all other cases you are going to get V ij equal to 0 now let us think V 11 that is 0 V 22 V 33 V 44 very easily they are all equal to 0 you do not have to worry much yeah because of course delta l is not equal to plus minus 0 sorry plus minus 1 it is plus minus 0 so this V 11 V 22 V 33 V 44 are gone what about V 12 for V 12 is delta l equal to 0 let us see 1 means 2 s 2 means 2 p 0 so of course for this l equal to 0 for this l equal to is it 0 is it 1 is it 1 yes it is 1 delta l is equal to 1 m here is 0 not l okay please do not get confused even if i try to confuse you okay so so what we see is this 1 2 is going to be nonzero because delta l equal to plus minus 1 similarly V 21 is also going to be nonzero because delta l equal to plus minus 1 right in case i talk too much and confuse you over the last 2 minutes let me just say it once we have proven that V ij is equal to 0 unless delta l equal to plus minus 1 for V ii V 11 V 22 V 33 V 44 since delta l equal to 0 that is not plus minus 1 V ij V ii is equal to 0 no problem V 12 and V 21 are nonzero because delta l equal to plus minus 1 okay here l is 0 here l is 1 in fact since our time for this module is getting over I leave it to you to work out and prove that V 12 equal to V 21 is equal to 3 epsilon remember what epsilon is this is epsilon magnitude of the electric field not very difficult to work out so we get this kind of an expression we just go back here what about V 23 V 23 means 2 p 0 and 2 p minus 1 for both l equal to 1 so delta l is 0 again same with V 24 V 43 V 42 V 32 for all these delta l equal to 0 so all these are equal to 0 so I have already told you what V 12 and V 21 are 3 epsilon and I need to figure out what is V 13 V 14 V 31 V 41 for that well I have put the card before the horse and I have shown you the answer first so even these are equal to 0 why because now we need to worry about the non well we have to worry about the imaginary phi dependent part of the wave function also and it is not very difficult to see that this L z of which this capital phi is an eigenfunction and minus epsilon z that is just z they are going to have common eigenfunctions so by applying the logic that we have done already we can obtain we can establish that these are 0 and you can also figure out that x is equal to 0 0 plus 3 and minus 3 in terms of epsilon so now what has happened if x what is x x is the first order correction to energy so if that is equal to 0 that means there is no change in energy so what we see is that for these 4 orbitals s 2 s 2 p 0 2 p minus 1 2 p plus 1 two of them do not change in energy one is stabilized one is destabilized which one will be stabilized which one will be destabilized not very difficult to say that the one that is aligned with the electric field the orbital whose angular momentum is aligned with the electric field will get stabilized the one for which it is not aligned with electric field that will get destabilized so that has to be plus 1 and minus 1 okay so let me just draw a schematic diagram I request you to have a look at pillar's book for a more detailed label diagram here I am just going to draw a little sketchy sketch if I may call it that here we had these 4 orbitals 2 s 2 p 0 2 p minus 1 2 p plus 1 out of these 2 are stabilized and destabilized and I have told you that one has one of them is 2 p plus 1 one of them is 2 p minus 1 I have told you the direction of angular momentum with respect to the electric field now I would like you to it is very simple figure out whether plus 1 is stabilized or minus 1 very easy most of you would have got the answer by the time you finish the question but please do it sometimes it is better to ask oneself easy questions also now what I want to draw your attention to is that 2 orbitals do not change in energy so out of these 2 which is s and which is p 2 s and 2 p 0 they do not change energy right for them your x is equal to 0 so here can I say that this one is 2 s and that one is 2 p how will I do it because again energy is the same so I cannot do it so here I have to take linear combination so I have to take something like c 1 reverence I will call it maybe c and d and not even write 1 so you can write c psi 2 0th plus d psi 1 0 there is no reason for writing 2 before 1 but then it does not really matter so these are actually linear combinations of this 2 s and the 2 p 0 orbitals how to find the coefficient of that is easy and suppose what I am saying is not even right suppose they are s orbital and p orbital they do not combine even then this is fine because in that case one of the coefficients will be equal to 0 so this is general expression and here now for the first time we have encountered the mixing of orbitals right when we write linear combination what we are doing essentially is that we are saying that the orbitals have mixed so when orbitals mixed mix you get what are called hybrid orbitals you are familiar with hybrid orbitals we talked about we have learned earlier about hybrid orbitals in valence bond theory so here what are we doing a mixing 2 s and 2 p 0 we are getting hybrid orbitals okay and in this case the 2 hybrid orbitals are the same energy right and we have discussed a situation in which perturbation removes sum of the degeneracies not all not all not none sum this plus 1 and minus 1 get very different energies well get different energies and the 2 s and 2 p 0 these orbitals mix and the energy is not changed that leads us to the concept of hybrid orbitals this is the discussion of perturbation theory that I wanted to perform so the stage is said to go and talk about go back to our multi electron atoms if I only want to talk about perturbation theory but there is another method that we want to learn and that is variation method that is what we will do and then we will see how variation method and perturbation method both can be used to take us back to multi electron atoms and then we will discuss a little further so as you see 41 lectures I think are over it is supposed to be a 60 to 70 lecture course so about 2 third of the course is over and we are nowhere near molecules that is because the tools that we develop while learning multi electron atoms are very very useful when you want to talk about molecular systems we do want to talk about approximation methods that are typical of molecular systems as well like Huckel method but those are easy these are more fundamental more interesting and requires a little more attention that is why we are spending so much of time on learning these techniques and before ending the module let me also say this it is impossible to understand this unless you go back and work out everything by yourself also you should refer to books see I am typing all this on the slides it is human to are so it is very possible that there will be error in the slides that I might not notice so it is important that you also read the book books also have typos but well to a lesser extent perhaps or maybe mine is a lesser extent so please do read the books that we have referred to and please do write it out it is important to write out these things then only you will understand okay so I trust that you will be able to do that and next time we are going to talk about variation methods.