 Welcome to lecture number 13 of the quantum mechanics and molecular spectroscopy course. We will quickly go through the contents of lecture number 12 before we proceed with lecture number 13. In the last class, I told that a charged particle when placed in electromagnetic radiation or the light will experience what is known as Lorentz force. And Lorentz force f L is equal to Q times E plus VB where Q is the charge of the particle. So, this is the Lorentz force where Q is the charge of the particle and E is the electric field, B is the magnetic field and U is the velocity of the particle or V is the velocity. And this in terms of classical mechanics force is given as mass into acceleration which could be written as D by dt of momentum. So, we use the analogy of this and converted the momentum P or P prime is equal to P minus QA where P is the original momentum and A is the vector field. Having done this then we cast the total Hamilton H as P square P prime square by 2m plus V of x. We consider only in the Vx direction and after doing the necessary math, we came across the equation in which H prime of D is given by ih bar Q by m A del that is your, but minus ih bar del is nothing but operator P. Then we could condense this equation to finally, H prime of D is equal to minus Q E naught by omega m cos omega t E dot P. So, this could be written as minus Q by sorry plus there was a plus omega m cos omega t and you know E naught into E P and this E naught into E written as E. So, Q by omega m cos omega naught E dot P. So, this is H prime of D. But in the case of particles which are atom or a molecule which has large number of particles, so which all of the H R's particles will interact with the light. So, this H prime D for this is for one charge and when for collection of charges will turn out to be minus sorry plus E naught by omega m cos omega t sigma over n Q n m n sorry this is not needed because m exchange going to be different for different particles m n because P is going to be momentum of the associated particle. So, this is for collection of charges. So, we will start this lecture with this equation. So, what we have at the end of the last class we had this H prime of D equals to E naught by omega cos omega t sigma over n Q n m n epsilon ok where Q n comma m n comma P n are charge mass momentum of the nth particle. So, this is what we have. Now, if you go back to the coefficient or the probability ok, your P of t that is the probability of f is equal to 1 over H bar square integral or modulus of integral 0 to t prime e to the power of minus i omega i f t integral f H prime of t i d t modulus square ok. That means, we still have to evaluate this integral f H prime of t. So, unless this integral is evaluated of course, you cannot evaluate this P of t ok. Now, to evaluate this integral now we know the Hamiltonian. So, we need to plug in this value of H prime of t ok. So, your integral f H prime of t i is given by f e naught by omega cos omega t sigma over n Q n m n epsilon P n. So, this is the integral that I need to evaluate ok. Now, I am going to slightly rewrite this equation or this integral f H prime of t i equals to ok. Now, e naught by omega cos omega t I will take it outside integral f because e naught is a constant omega is a constant cos omega is a constant as far as this integral is concerned f because f and i are the eigen function of the time independent Hamiltonian ok. f times sigma over n Q n by m n epsilon P n ok. Now, to begin with we will start with slightly some other equation. So, if I take a quantity called R H naught commutator that is the commutator of operators R and H naught right. Now, this can be evaluated at this commutator R comma H naught can be evaluated to be i H bar P by m ok. Now, I am going to take a small detour and try to evaluate this commutator, but always remember the commutors are evaluated with respect to some function. So, if there is an R operator and there is a H naught operator H naught is along the direction of R. So, this will be equal to minus H bar square by 2 m d square by d R square with some potential v of R that your H naught ok. Now, then all of this will act on a function f of R. So, let us evaluate the commutator. So, by the way this is this is also called minus H bar square by 2 m del R square plus v of R ok. So, now, what you have to evaluate you have to evaluate R H naught commutator with respect to f of R function ok. So, this will be equal to R times minus H bar square by 2 m del square f of R minus minus H bar square by 2 m del square ok. Now, take the first term. So, that is going to be minus H bar square by 2 m because that is constant R time del square f of R ok. Now, the second one is my this is minus of this minus. So, this will be plus H bar square by 2 m del square R f R. Now, if you take this I am just going to evaluate this. So, del square R f of R this is equal to, but del is nothing, but del del del square. So, that is nothing, but del del of del R f of R. So, one can first evaluate this. So, this is nothing, but del times. Now, this is a differential del is nothing, but d by dr. So, your differential function. So, first acts on R. So, we will give you 1. So, we will get f of R. Now, second this will act on del f of R ok. So, what we will get is less R times. Now, again now this del the second del acts on it or the second differential acts on. So, we will get del f of R plus. Now, we have again a function. So, this is one function and this is another function del of R is 1 ok. So, we will get del f of R because that is 1 plus del acting on del f of R give me R into del square f of R. So, now, I am going to plug this in here what you will get is. So, what I am evaluating you are evaluating R H naught commited. So, this is equal to minus h bar square by 2 m R del square f of R plus h bar square by 2 m what you have del f R del f R 2 del f R 2 del f R plus this is R square R sorry R del square f R R del square f of R. Now, if I explain this this will be minus h bar square by 2 m R square R del square f of R plus this 2 and this 2 we will get we will get h bar square by 2 m del f R plus h bar square by 2 m R del square f R. Now, we will see that this m and this 10 will get cancelled though. So, this will be equal to h bar square by 2 m del f R. So, this is nothing but this is equal to minus i h bar. So, there is no 2 here m here just m by m multiplied by i h sorry this is plus i minus i h bar by. So, minus i square minus 1 into minus plus i h bar del f of R I am sorry the finally, you do not have f of R because it is evaluated with respect to f of R. So, then you have to put f of R here as well. So, then you are R, but this is nothing but p of R. So, what you get R h naught commutator will be equal to i h bar by m. So, because now I want to use this p operator here. So, let us continue. So, what was the equation that we had this one? So, f h prime of T i equals to E naught by omega cos omega t acting on f sigma over n q n by m n p n i and we found that R h naught commutator is equal to i h bar p by m. Now, what I am going to do this? So, I am going to plug in here. So, this will be equal to. So, this now be equal to E naught by omega cos omega t f sigma over n q n into p n p n by m n. So, p n by m n is i. So, this is nothing but E naught by omega cos omega t. So, i h go on the other side that will be i will become minus i and h. So, you have h bar minus i that is what you get sigma or f sigma or n q n epsilon n R n. So, that is the new Hamiltonian when you have plugged in this value. So, this will be equal to minus i E naught by h bar omega cos omega t omega t into f. Now, let us expand this commutator. So, this will be sigma over n q n R n h naught minus q n h naught R n. I have just expanded this commutator R n h naught because that it will be R n h naught minus h naught R n. So, this will be equal to minus i E naught by h bar omega cos omega t. I will write it as 2 because there are 2 terms here I can write it as 2 separate integrals. So, f of sigma over n q n R n because R is just a now epsilon h naught acting on minus f sigma over n q n h naught i. So, now this h naught will act on i on give me E i. So, so this will now be equal to i E naught by h bar omega cos omega t this h naught acting on my give me E i i. So, i f because E i is a number f sigma over n q n R n epsilon i. Now, this h naught will act on the other side because you know h naught is a Hermitian operator and we know the turnover rule. So, if I use turnover rule this h naught can act on f and give me E f complex conjugate of it, but energies are always real. So, complex conjugate of E f star will be only E f. So, this will be minus E f sigma over n q n R n. Now, you will see the integral in both cases is the same. So, essentially your f h prime of t i can be written as minus E naught by h bar omega cos omega t E i minus E f to integral f sigma over n q n R n epsilon divided by i. Now, what is E i f i E i minus f i E f sorry E i is equal to delta E i f. So, this is nothing but h cross omega i f. So, this is nothing but this is equal to minus i E naught h cross omega cos omega t this will be h bar omega i f integral f sigma over n q n R n epsilon i. Now, this is h bar omega and this is h bar. So, it will cancel. So, what you will get is this minus i E naught cos omega t omega i f by omega. So, this is very important and f sigma over n q n R n. Now, if there are charges like this distributed over some this one. So, this is charge q 1 this is charge q 2 this is charge q 3 q 4 q 5 etcetera q 6 q 7 q 8 q 9 q 10 this is my center q 11 and then you have radius s R 1 R 2 R 3 R 4 R 6 R 11 R 10 etcetera. So, what is this if I take? So, this is a product of q n R n sum over if you have charges distributed like that that will give you the dipole moment. So, for n charges sigma over n q n R n should be equal to mu that is the dipole moment of the molecule. So, that is nothing but is equal to minus i E naught cos omega t omega i f by omega f epsilon. Now, what is mu dot epsilon? Mu epsilon simply means that the mu has to have a dot product with along to the epsilon that is the electric field vector. So, that will give me is given by this equation. Now, one could understand if you forget these prefactors this integral depends on the dipole moment of the atom or the molecule. By the way dipole moment of atom I do not know because its dipole moment its permanent dipole moment is 0, but dipole moment is not permanent dipole moment. You must understand mu is not equal to mu naught we will come to that at later sometime, but essentially the initial and final states will couple with respect to the projection of the dipole moment along the electric field vector. So, this is nothing but the projection of dipole moment electric field and the projection of the dipole moment along the electric field vector determines the selection rules and this will lead to selection rules. We will stop it here for this class and continue in the next lecture.