 Hello, welcome to the lecture number 24 of the course quantum mechanics and molecular spectroscopy ok. Now in the several classes we were looking at the transition moment integral and its relation to other quantities such as Einstein's E coefficient, B coefficient and the absorption spectrum through Pierre Lambert law. But from now onwards I am going to take a slightly different term and try to get the selection rules ok. Now if you go back quite a few classes then we came across this equation which I will write it down as probability of transition P of t if state f is given by E naught square divided by 4 h bar square omega f i by omega square sin square delta omega by 2 into t divided by delta omega by 2 square modulus of f mu dot E i square and we remember we called it as transition moment ok. Now it turns out that P f of t is proportional to your transition moment integral square. So, what are you doing here? You are going from a state i to state f ok and we also related all this to Einstein's E coefficient, Einstein's B coefficient and the extinction coefficient through Pierre Lambert. We also looked at the line shapes ok. All these experimental quantities could be related to this. Now it turns out that if I slightly rewrite so in the vicinity of omega f i ok that means what I am doing is omega f i is approximately equal to omega that means you will have a transition only when there is a resonance ok. So, in that scenario one can rewrite this equation P f of t. So, when you have omega f i is equal to omega I can ignore this term. So, I will get this will be equal to E naught square by 4 h bar square sin of delta omega by 2 divided by delta omega by 2 square mu dot epsilon ok. Now we know this function is the line shape ok it determines the line shape ok. Now what we have of course all these are non-zeros. So, the selection will be allowed or not is given by this integral ok. So, one can write transition moment integral T mi is equal to f dot epsilon i ok this is my transition moment integral and whether the transition from 1 to 2 or f 2 I will be allowed or not will depend on this integral. If this integral is 0 the transition is forbidden if this is 0 then it is forbidden this is not equal to 0 then we have allowed and 0 not equal to 0 they combine what is known as selection rules. Now what we will do in this lecture is to look at this operator mu what now let us think of a molecule AB ok. So, two atoms A and B are combined and A and B because they are not the same. So, there will be say let us say delta plus and delta minus. So, you have a dipole and if the distance equilibrium distances are not. So, for example, if you have a AB molecule one can draw a potential energy surface like this that is the zero of energy and the distance goes to infinity asymptotically ok. Now what I am trying to do is that of course if you go from here to say here there is a huge change in the geometry or the bond distance that it spans. So, one can always think that the movements are not. So, one is looking at around the equilibrium position if you look at around the equilibrium geometry one can write mu as a Taylor series expansion. So, what I will write mu is equal to mu naught plus d mu pi d r evaluated r naught into r plus d mu d square mu by d r square evaluated r naught into r square plus d mu d cube mu by d r cube evaluated at evaluated at r naught into r cube plus etcetera ok. One can write a Taylor series. Now let us look at it in a slightly more useful way. So, the dipole movement mu is a Taylor series expansion some fixed comes plus d mu by d r at r naught multiplied by r plus sorry 1 over 2 factorial d square mu by d r square r naught to r square plus 1 over 3 factorial d square mu by d r cube evaluated at r naught plus r cube plus etcetera. So, this is how I can write my mu as a Taylor series expansion around the equilibrium geometry and this mu naught is the permanent dipole. And what is d mu by d r how the dipole movement changes with respect to r ok. It is a first derivative and you have second derivative. So, one can write it as mu is equal to mu naught plus alpha dot r plus 1 over 2 beta r square plus 1 over 6 gamma r cube etcetera where alpha equals to where alpha equals to d mu by d r evaluated at r naught beta is nothing but d square mu by d r square evaluated at r naught and gamma equals to d square actually one could take even the constants into the. So, when I if I write this as half then I can delete this and if I delete this then one can write 1 over 6 d square mu by d r cube ok. So, one can write such equation. So, this is your permanent dipole there is a first derivative of mu this is second derivative coefficient and the third derivative coefficients ok. So, this is how one can write in terms of Taylor series expansion. So, therefore, your TMI around the equilibrium geometry is given by f times mu naught plus alpha dot r plus beta r square plus gamma r cube less etcetera etcetera. So, this is the transition moment that we need to evaluate ok. When I say dipole moment it is basically linear combination of permanent dipole moment and it is derivative at various levels of various degrees with respect to r ok. So, this is the transition moment integral that I need to consider. So, one can write therefore one can write TMI is equal to f mu naught plus alpha plus beta r cube plus gamma beta r square plus gamma r cube beta r square plus gamma r cube plus. So, that is the transition moment integral I am interested in ok. Now, let us look at this equations phi ok. So, what are my solutions? So, what are my solutions? I said H naught into i is equal to E i E i i and H naught into f is equal to ok. What are my H naught? H naught is the Hamiltonian for which I know the solutions. So, E and E f are the Eigen values of the functions Eigen functions i and f ok. Now, what is H naught? That is the thing that we have to figure out ok. To in this course from up till now we have never bothered what is the H naught. H naught we just assume is some Hamiltonian for which we know the solutions and once we know the solutions we can evaluate the transition moment integral and that transition moment integral can related to various experimentally measurable quantities. But right now we have not even thought about what the H naught could be. Now, if you write the entire molecular Hamiltonian which I am going to write. So, H molecule ok. So, H naught molecule is nothing but minus h bar square by 2 sum over alpha del square alpha by m alpha that is the kinetic energy of the nuclei where alpha is its index minus h bar square by 2 m e sigma over i del square i. So, i is the index on the electron. So, the kinetic energy of all the electrons minus attraction between the nucleus and all the electrons. So, this will be 1 by 4 pi epsilon naught sum over alpha sum over i z alpha e square by r alpha i. So, that is the distance between the electron and the nucleus. Then we have a nuclear-nuclear repulsion. So, that is equal to 1 over 4 pi epsilon naught sum over alpha sum over beta greater than alpha z alpha z beta e square by r alpha beta that is the nuclear repulsion because alpha and beta are 2 nucleus and you do not want to count the repulsion between the nuclei same nucleus that is why alpha is not equal to beta and then you have to have only 1 if alpha is if nucleus 1 is repelling 2 nucleus 2. So, it is same as nucleus 2 repelling a nucleus 1. So, it is. So, to make sure that you do not repeat the terms use this kind of notation plus similarly with the electron 4 pi epsilon naught sum over i sum over j greater than i e square by r i j. So, we have 5 terms. So, this is K e of nucleus K e of nuclei this is K e of electrons this is nothing but P e of electrons and nucleus that is attraction this is what P e of nucleus to nucleus that is repulsion and this is nothing but your P e of P e that is nothing but electron repulsion H naught of molecule multiplied by psi now which is a function of electrons and nucleus should be equal to E n psi which is function of electrons and nucleus. Now, it turns out that this is not a possible solution. So, what we do is what we use something called bond open Heimer this approximation says that the total wave function psi of electrons and nucleus can be separated as chi of nucleus and phi of electrons parametrically dependent on nuclei and the total Hamiltonian H naught can be written as H nuclei less H ok. Now, it turns out that when I look at H nucleus the H nucleus even though is this H nucleus the total energy is equal to H nucleus chi nucleus will give you E nucleus chi nucleus, but the total energy E is nothing but E n plus E electron. But you see the phi V. So, now, if you have H electron acting on phi electron and that is parametrically depends on nucleus what was parametrically depend on nucleus means for every different nuclei position the Hamiltonian of the electrons will be different give a different solution. So, that will nothing but E E plus phi. So, this E E also depends on nucleus because the wave function itself. So, if you take the total energy or the nuclear energy nuclear energy is just not the repulsion, but it also kind of has the electronic component into it and that is called potential energy ok. So, E so when you have P E that is nothing but you have E in electro nucleus plus E electron ok. So, what you do is you can draw make a potential surface ok that is E. So, that depends on how the electrons and the nuclei are arranged as a function of distance. Now, one of the issues that we have is the following. So, when you have total wave function which is in the total wave function psi of E comma n is nothing but chi of n into phi of E and parametrically depend on N ok. Now, this is my total solution and of course, it will have its own quantum numbers. So, what I am looking at is the following I am starting from a final function f. Now, this f function could be written as the following chi nucleus phi electron nucleus some quantum all of this having some quantum number let us say A, B, C or I will call it as A prime, B prime, C prime I will come to what these are and this multiplied by your mu dot E multiplied by function I which will I will call it as chi n phi E to N having quantum numbers A, B, C ok. So, what I am trying to do here I am going from a wave function initial wave function which will have quantum number A, B, C and this quantum number could involve the nuclear quantum numbers or quantum numbers related to the nuclear motion and the quantum numbers related to the electronic motion and with some values A, B, C and I am going from there and changing my quantum numbers to A prime, B prime, C prime ok. Now, what I am trying to do is the following I am going from one set of electronic and nuclear coordinates to a different set of electronic and nuclear coordinates and this transition is bought about by this operator. So, now we can think of it this way. So, this operator is really complicated in some sense is that chi n phi n E to N whole thing of A prime, B prime, C prime mu naught plus r plus beta r square plus gamma r cube whole pi chi n phi E N whole thing of A, B. So, I am looking at such a transient moment integral ok. So, when I take the total wave function ok. So, this is how we are going to look at this transient moment integral. But of course, the way it is written it is almost impossible to solve. So, we have to make right approximations to be able to solve this transient moment integral evaluate the transient moment integral and look at the selection rules which we will continue in the next lecture. Thank you.