 Welcome to the lecture number 19 of the course quantum mechanics and molecular spectroscopy. As usual, we will have a quick recap of the previous lecture. In the previous lecture, I was talking about Einstein's coefficients. These are just given by if there are two states 1 and 2. So, the rate of absorption in the presence of photon h nu or rate of stimulated emission h nu and spontaneous emission in the absence of h nu, we have rates. So, let us call it rate w 1 2. So, going from state 1 to state 2 that is absorption, this corresponds to absorption will depend on the number of molecules in the state 1 and the radiation density at the frequency nu and this frequency is same as the h nu. This frequency and this frequency are same and the proportionality constant is called b 1 2. That is the Einstein's coefficient for absorption. Similarly, w 2 1 that is emission. Now, emission will consist of both spontaneous emission and the stimulated emission. So, this will be equal to n to the population radiation density radiation nu and this will correspond to the stimulated emission into spontaneous emission does not need the radiation. So, it does not need or it happens in the absence of h nu. Therefore, it does not need rho radiation. So, this is for the spontaneous emission that proportionality constants are b 2 1 and a 2 1. Now, under equilibrium condition the rate of forward reaction we know if a and b are in equilibrium. So, the rate of forward reaction must be equal to rate of backward reaction. So, rate forward must be equal to rate backward. So, that means, w 1 2 must be equal to w 2 1. In such scenario, so the b 1 2 n 1 rho radiation at nu must be equal to b 2 1 n 2 rho radiation nu plus a 2 1 n 2. I will rearrange this equation as rho radiation nu equals to a 2 1 n 2 divided by b 1 2 n 1 minus b 2 1 n 2 which can further be can as a 2 1 divided by b 1 2 n 1 by n 2 minus b 2 1. Now, we know that n 1 by n 2 that is the population of the ground state with respect to excited state in Boltzmann is equal to exponential delta e by k t or exponential h nu by k t where delta e is equal to e 1 sorry e 2 minus e 1. So, this is nothing but from your Boltzmann. Now, when I plug in this, so this should be equal to a 2 1 divided by n 1 by n 2 I will use this equation. So, b 1 2 e to the power of h nu by k t minus b 2 1 radiation. Now, if I go back little bit ahead and look at the transitions transition probability then I had one term. So, your p f of t was equal to some constants I do not you can go and look it up multiplied by what you had is delta mu f i plus nu or sorry omega f i plus omega plus omega f i minus omega square and f slown dot mu i. So, this we know is transition moment integral and there were some constants and there was this term which is the delta function. Now, you see we said that the omega f i plus omega delta and delta omega f i minus omega cannot be simultaneously correct. So, you cannot fulfill these two conditions simultaneously because in this case omega should be equal to minus omega f i and in this case omega must be equal to omega f i. So, simultaneously this cannot be right. So, we said this one will correspond to absorption and this one will correspond to stimulated emission. So, a molecule can either go absorption or stimulated emission at a given point of time. In general, it can happen both things can happen but at a given instance only one of the process will either it will go from the ground state to excel state or come back from the excel state of ground state it cannot happen simultaneously. So, for absorption process we only took this one. So, for the stimulated emission process it is. So, if you look at the equation if you now look at the stimulated emission everything else will remain the same except that this delta function will change. So, if only the delta function is changing that will depend whether the we are going up or coming down. So, there is no reason for b12 and b21 to be different. Therefore, b12 must be equal to b21 because essentially all the other integrals and the constants are exactly the same it is just the delta function that is going to be changing. So, if the probability is going to be same then the rate and the rate constant must be the same. So, b21 must be equal to b12. Now, if I go back and look at this equation and now I plug in radiation nu is equal to a21 divided by b12 n1 by n2 minus b21 this was equal to a21 by b12 e to the power of h nu by kt minus b21. But I said b21 is equal to b21. So, this is equal to I will call it as b. Similarly, a21 I will just call it as a because spontaneous emission will only happen from the upper state to the lower state and it cannot happen from the lower state to the upper state. So, essentially what you have is radiation nu equals to a divided by b e to the power of h nu by kt minus where a is equal to a21 and b is equal to b12 is equal to b21 under that approximation. So, if you have this then you have this. Now, according to Planck's black body radiation theory rho radiation nu is given by 8 pi h nu cube by c cube into 1 power e to the power of h nu by kt minus 1. Now, in all this case of course, k is Boltzmann constants. Both these equations give rho radiation nu. So, one can equate root of them. So, it turns out that a by b is equal to 8 pi h nu cube by c cube. Now, we know 8 pi h and c they are all constants. So, one can write it as 8 pi h by c cube into nu cube. So, a by b is proportional to nu cube because you know these are constants. So, what is a? It is the rate constant for spontaneous emission and what is b? It is a rate constant for stimulated emission. So, the ratio of rate constants actually because both rate constants both the a and b the rate constants for the stimulated emission and the rate constant for the spontaneous emission will act only on the population of the excited. So, not only rate constants, but rates as well. So, ratio of the rate constants or rates of spontaneous emission to stimulated emission is proportional to nu cube. That means, nu is nothing but nu if you multiply it by h it is the energy. So, if you have two states 1 and 2 and this is delta e this is equal to e 1 e 2 minus e 1 this is equal to h nu. As you keep increasing the energy difference between states 1 and 2 the spontaneous emission becomes more and more probable because the rate constant will increase keep increasing with nu. So, if you keep increasing the value of nu the spontaneous emission becomes more and more probable. So, which simply means that as the energy difference between ground and excited state increases the spontaneous emission becomes more probable and it scales as energy cube. So, a by b as you know is equal to 8 pi h by c cube. So, this is an important but this is only the ratio but we really we only know what is the ratio of A to B but we have to get the value of each. See for example, if I say ratio is 2 then I can have if ratio of A to B is 2 then I can say that you know A can be 10 and B can be 5 or A can be 2 and B can be 1. So, you know the ratio but you do not know the absolute values. So, we still have to find out values for this we will return to one of our older derivation that is the Fermi's golden rule. So, let us now look at the Fermi's golden rule. The Fermi's golden rule is given by W that is rate fi is equal to 2 pi h bar modulus of mu square rho f and this we call it as transition dipole and this is the rho fi is nothing but density of states around. Now, where modulus of mu square was given by E naught square by 4 h bar square into f mu dot epsilon i whole square which can also be written as 1 over 4 h bar square. If I take E naught square multiplied by epsilon that becomes you know electric field. So, modulus of f E dot mu square where E is equal to E naught. So, this is your modulus mu square that is my rate but in the Einstein coefficient we are using rate between states 1 and 2. So, I am going to rewrite this equation slightly differently W 21 equals to 2 pi h bar I will call it as mu 21 square rho 2. So, this will be and where modulus mu 21 square equals to 1 by 4 h bar square E dot mu 1 whole square. Now, if that is the case now let us look at slightly in a different way. Now, let us assume that my molecule or an or the transition dipole is along the z axis. So, let us just imagine there is a molecule AB and is aligned along z axis. So, the dipole movement effectively will be along the z axis. So, if the molecule is aligned along the z axis or the dipole movement AB molecule is along z axis. Now, your mod mu square will now then become mod mu z square because the dipole movement is along z axis. So, this will be equal to 1 by 4 h bar square modulus of sorry integral of 2. Now, mu z dot E 1 square because the essentially all the molecule is along z axis. So, you know only the z projection of the dipole movement or the dipole movement and z axis will get projected onto the electric factor E. Now, if that is the case in such scenario and when we take isotropic light. Now, what is isotropic light? Isotropic light means light that is spreading in all directions x, y, z. So, let us suppose you know you have a lamp or candle in the middle of the room then what will happen? The light will go in all sorts of directions x, y and z. So, the projection along the z axis will only be one third of the total light. In such scenario if you use isotropic light then you get there is a coefficient one third. So, your mu square will now be replaced by one third of mu z square because there are two things that are happening. First is the light is along the all the directions and the second is the dipole movement of the molecules is only along the z direction. So, basically you are wasting the one you know basically you are wasting the two thirds of the light that is along x and y projections and you are only utilizing the light that is along the z direction. Therefore, if I use all this then by mu z square will be now one twelfth of h bar square mu z dot e whole square where this will correspond into isotropic light. So, what am I doing now? I am using isotropic light to interact with molecules that are aligned along the z axis. So, essentially this tells us that the isotropic light interacting molecules whose dipole movement is along. So, in such scenario your w12 will be equal to what was this 2 pi h bar modulus mu square rho. So, that was our equation. So, in such scenario what we will have is 2 pi h bar into 1 over 12 h bar square mu z dot whole square. So, this will be nothing, but so this h bar and that h bar. So, rearranging what we will get is pi by 6 h bar mu z square rho 2. So, this is your rate. So, now we know this is nothing, but transition dipole with mu z interacting with isotropic light and this is density of states. We will stop here and continue in the next lecture. Thank you.