 Hello, welcome to the lecture number 20 of the course quantum mechanics and molecular spectroscopy. We will go on with our lecture after a brief recap of the previous lecture. In the previous lecture we looked at two things, one is the rate of absorption and as it is rate of emission. Based on this we can and we got two equations, rho radiation nu is equal to A21 divided by P12 N1 by N2 B21 minus okay and this under the circumstances A21 is written as A and P12 equals to B21 is equal to B, we write A by P N1 by N2 minus 1 okay. Now N1 by N2 is given by Boltzmann population, this is equal to e to the power of h nu by k t, it is the same as e to the power of delta e by okay. So this turns out to be A divided by P into e to the power of h nu by k t minus 1. Now rho radiation nu, if I look at in terms of Planck's blackbody radiation that will give me 8 pi h nu cube by C cube into 1 over e to the power of h nu by k t minus 1 okay. Now if I rho radiation nu and now if I equate these two equations then I can get A by B equals to 8 pi h nu cube by C cube and we also discussed that A by B is proportional to nu cube okay. And another thing that we looked at if you starting from the Fermi golden rule okay and using dipole movement along z axis and also using isotropic light, we said that W12 is equal to pi by 6 h bar integral to mu z dot e 1 modulus square rho 2 and this is nothing but dipole mode and this is density of states okay. Now I want to make use this equation and get along okay. Now there is something that is there is that if I have this also can be written as okay W12 can also be written as pi by 6 h bar to mu z to epsilon modulus square and then I will get the electric field out square of it rho 2. So this is okay this is probably not the right way to write. So this one can write as pi by 6 h bar to mu z so only the component of mu z that is aligned with respect to the electric field I 1 square square rho 2. The reason why I want to bring the electric field out because the energy of electromagnetic field is equal to epsilon naught. So this is energy and this is nothing but electric field and epsilon naught is permittivity of free space. So I can write rewrite this equation W12 equals to okay pi by 6 epsilon naught h bar modulus of 2 integral to mu z epsilon 1 square E rho 2 of electromagnetic radiation okay. Now so this is what we have. Now what I want to do is I want to slightly now this is based on Fermi's golden rule and you know Fermi's golden rule happens because there is a density of states it is going from a discrete initial state to state that is embedded okay in some density okay. But now let us consider for example you have a initial state 1 and you have final state 2 okay which is embedded between in some density but I want to look at only this transition okay. So I am looking for 1 to 2 discrete transition no density it is just the discrete transition okay. Now apart from that when you are looking at this discrete transition okay now all I want to do is I do not want to use. So this will happen let us say it happens at delta E 1 2 or 2 1 this is equal to E h nu 1 2 okay. It will happen at a very specific wavelength or a very specific frequency nu 1 2. So if I want to make a discrete transition if I want to go from state 1 to state 2 I have to give the right frequency nu 1 2 or the electromagnetic radiation with the right frequency nu 1 2. But now if you look at we were talking about the candle or some kind of lamp so of course that is not going to give you a monochromatic monochromatic energy or electromagnetic radiation. So nu 1 2 will not be very very single very very selective okay you will get all sorts of nu and out of which only nu 1 2 will cause the transition. If you use non-monochromatic so what I am trying to do is that I am going from state 1 to state 2 which has a energy difference of delta E and frequency of nu 1 2 okay but using non-monochromatic radiation and when you have non-monochromatic radiation you have let us say p radiation nu such that you have nu plus in the frequency nu plus nu. Now this is the radiation density okay so if I want to get the energy okay rho radiation nu I should get p radiation okay if it is a small integral but if you have a large integral then this should be integrated over nu. Now in such scenario okay your w 1 2 will be equal to pi by 6 epsilon not h bar z E 1 whole square okay energy E with radiation density p radiation p radian okay but we are looking at spreading the radiation so this must be integrated over d nu so when I integrate over radiation nu nu so what I will get is pi 6 epsilon not h bar 2 nu z epsilon 1 whole square integral okay by the p radiation nu d nu of course I am writing the same equation so when I integrate what I will get is this integral is given by 1 over h bar rho radiation okay therefore your w 1 2 is given by pi by 6 epsilon not h bar square integral 2 nu z along some epsilon 1 whole square into rho radiation so let me reiterate now according to Einstein's coefficients what was rho w 1 2 rate for an absorption it was nothing but b 1 2 okay n 1 okay now I can equate these two equations now so what you will get then b 1 2 n 1 equals to pi by 6 epsilon not h bar square square whole square now but there is a problem in this equation I do not know what this n 1 is n 1 we said is the population of the state but when you derive quantum mechanics roles okay we derive for one single molecule or one single quantum object so your n 1 must be actually 1 because all this derivation of this the initial state to final state is based on one quantum object so therefore n 1 we must be replaced by 1 so your b 1 2 it is nothing but your b is equal to pi by 6 epsilon not h bar square 2 mu z 1 whole square so let me look at so your b that is your Einstein's coefficient for spontaneous absorption and stimulated emission is given 1 pi by 6 epsilon not h bar square and the integral 2 mu z but we also know a by b is equal to 8 pi h now we know that h bar is equal to h by 2 pi or h is equal to 2 pi so if I write in terms of h bar then I will get 2 pi so we will get 16 pi square h bar mu cube by c cube that is a by b but now we already know b so a is equal to 16 pi square h bar mu cube by c cube into b which is nothing but pi by 6 epsilon not h bar square 2 mu z 1 whole square now I can do some arrangement so this h square and this h bar will go away this pi becomes pi cube 6 2 times so 3 8 so this will become 8 pi cube 8 pi cube by 3 h bar c cube mu cube mu z 1 whole square so your a is given by this and b is given by so what you have is a is some constant 2 z epsilon 1 whole square and b is some other set of constants 2 mu z epsilon 1 whole square and this constants here will be 8 pi cube mu cube divided by 3 h bar c cube and this constant is pi by 6 epsilon not h bar square now you can really see that the Einstein coefficients a and b are proportional to the square of the transient dipole both a, b are proportional to mu z 1 square of this so this is nothing but transient dipole ok the proportionally constants are different but essentially if they are proportional to transient dipole so if you know the transient dipole then you can get the Einstein coefficients a and b and if you know the Einstein coefficients a and b then you know the rate of absorption and rate of emission ok so essentially the rate absorption slash emission is related to transition. Let us suppose you have an excited state ground state 1 and excited state 2 and you have n 1 and n 2 now if you excite 2 from ground state excited state for a very large delta E and we know as the large delta E increases mu also increases or very large mu which means your a by b is proportional to mu cube that means if you have very large mu ok then mu cube will be much more than the b so for example let us say nu is nu is 100 ok so let us say nu is equal to 100 ok so nu cube will be 100 cube so that will be 1 6 0 so you have nu of 100 ok the nu cube will be a million ok so generally nu cube is going to be is going to dominate over nu that means a will dominate over b that means spontaneous emission will dominate over stimulated emission so which means for a very large delta U ok spontaneous emission will dominate over stimulated emission ok. Now you can think of a scenario in which you are going from a state 1 to state 2 your initial population is 1 and this is n 2 at some point of time ok after you excite and then you switch off the light then what will happen everything will emit spontaneously so minus d so decay of population from the excited state d n 2 by d t should be proportional to the n 2 times a that the rate constant and the population ok now if I integrate this minus d n 2 over 1 over n 2 is equal to a so we can take the other side d n 2 by n 2 is equal to minus a so this will be nothing but ln n 2 will be equal to minus a t so if you take plus some constant of integration and then when you rearrange what you will get is n 2 of t will be equal to n 2 of 0 that is initial exponential minus a t so this is so the decay will be given by a decay constant will be given by a so what means the decay constant when you excite to molecule to higher level the spontaneous emission will decay by rate constant of a now this decay constant a is equal to some constants epsilon 2 1 square this is nothing but your decay constant ok so by measuring the decay constant and this can be experimentally measured I will come to this in the next lecture ok ok decay constant in the excited of the excited of the excited state can be measured so it is like a rate of reaction ok then you can measure the rate constants that is a decay constant and that decay constant is proportional to the transition dipole so if you can somehow measure the decay constant experimentally then you can evaluate the transition dipole ok we will stop here and continue in the next lecture.