 Hello, welcome to the lecture number 17 of quantum mechanics and molecular spectroscopy. In the previous class we were looking at the transient probability and we arrived at the equation P f of t equals to E naught square by 4 h bar square omega f i by omega square integral 0 to t prime sin square delta omega by 2 into t divided by delta omega by 2 whole square integral f epsilon dot mu i whole square, okay and this I told you is the transition moment integral transition moment integral dmi and this is the modulating function, okay. Now when you plot this modulating function as a function of delta omega where delta omega is nothing but omega f minus omega i, okay and h bar delta omega is equal to E f minus E i in that scenario this will look like this something like that, okay. So, this is the function sin square delta omega by 2 divided by delta omega by 2 square, okay. Now if this function looks like this, this means the absorption will happen at delta omega is equal to maximum absorption will happen delta omega by L is equal to 0 that is what is not the resonance condition you should always see there will be very tiny absorptions even in the wings, okay and that wings is governed by this where in the wings it will get absorbed will governed by this equation and will depend exactly on delta omega, okay. Now continuing with the present lecture now let us go back and look at this equation little bit more carefully, okay where P f of t is equal to E naught square by 4 h bar square omega i f by omega square sin delta omega t by delta omega by 2 square epsilon dot mu whole square, okay I made a small mistake in the last slide. So, there is no this integral is this integral does not exist because sin square came over only it was as an integral, okay standard integral. So, now this is what we get, okay. Now this is the probability of transition from a initial state i a initial state i to a final state f of course as you look at this is a t dependent that means as long as the perturbation is there, okay the transition will take place, okay. So, longer the perturbation longer will be the transient probability, okay because that time comes here. So, you see the perturbation if can actually be prolonged and you can increase the transient probability, okay. Now there is one issue that we will look at in this lecture is the what if f state f is not isolated. So, in this case what is happening you are going from a initial state i which is precisely defined to a final state f which is precisely defined. So, we know that h naught i is equal to E i i and h naught f is equal to E f. But this is when f is separated out that means f is a clean state that does not have anything in its vicinity, okay but that may not be possible all the time. So, what if f lies midst of some states? So, let us look at the possibility that there is a initial state i which is ground state which is well separated but there is a final state f which is in between you know many such states. So, a state f is embedded in a density it is not cleanly separated state but f state f is embedded and where you can think of now let us think of hydrogen atom. Now hydrogen atom let us say E n we know is given by E n is for h atom is given by let us say E 1 by n square. So, just proportional to square of the principal quantum number and E 1 is the principal quantum number 1 state that is 1 s state. So, this is generally equal to if energy is this is minus 13.6 E v divided by n square, okay. Now if you go and look at that state and let us go and look at large values of n. So, let us say value of n is equal to say 500, n is equal to 501, n is equal to 502. Now if you go to such states. So, E 500 is equal to minus 13.6 divided by 500 square E 501 is equal to in E v minus 13.6 divided by 501 square in E v and E 502 is equal to minus 13.6 by 502 square in E v. Now if you plot these energies they will be very close to each other. In fact, the difference will be less than milli electron volts. In fact, it will be micro electron volts. So, essentially the energy levels 501, 502, 500, 501, 501 and 502 are lying on top of each other. So, such will be the scenario and the other thing is the what happens to the wave function. Now let us look at I will give you an example very simply as particle in a box, okay. Now particle in a box wave function will look like this. So, for n is equal to 1 is looks like this, n is equal to 2 it will look like this. Now go to very high values of n is equal to let us say 20 and 21, okay. So, 20 it will have 19 nodes many nodes are there and for 21 there will be just one extra, okay. So, what happens is that the wave function does not really change much, okay between n is equal to 20 and n is equal to 21. So, when you go very high in energy where the energy levels are densely packed, the wave function changes are also very small, okay. So, in this case let us say if I have n is okay let us redefine that. So, n is equal to 499, n is equal to 500, n is equal to 501. If you take three states one can describe the state by wave function on the average by any average wave function will look like psi 500 approximately, okay. It will not change too much with respect to 500 if you go down by 1 by 4 to 499 or 1 up by 501, okay. And energy also will remain more or less constant, okay. In such scenario if there are lot of states, okay. Now, what I want to now get is the transition probability p of t of a state f, okay. But not really of a state f but states around f, okay. Total probability will be nothing but integral over f, okay. p f of t density of states e, okay. Now, what is this density of states? So, that means for a unit energy, how many energy levels are packed? That is the rho e. So, rho e is nothing but density of states, okay. Density of states simply means number of energy levels. Now, unit energy is kind of you can define 1 joule or 1 calorie or 1 kilo joule or 1 kilo kohl calorie or centimeter inverse or e v or milli e v it is up to you, okay. What will be the interval that you want to define, okay. If I use that then my p of t will be equal to e naught square by 4 h bar square omega fi by omega whole square integral f, okay. Sin square delta omega by 2 by delta omega by 2 whole square rho f of e, okay. Now, that is my density of states, okay. Now, if I define very narrow range, okay. So, for very narrow range change wave function is minimal and can be, okay. So, in that case then your f epsilon dot mu i. So, what I am saying is that this f even though I am looking at transitions to many of these states I will take an average value and that change is not going to be very much different, okay. There is a modulus square here, okay. So, I am going to slightly rewrite this. So, this p of t is equal to e naught square by 4 h bar square omega fi by omega square modulus of f epsilon dot mu i whole square integral over f sin square delta omega by 2 t by delta omega by 2 whole square rho f of e t. So, that is the integral that I need to evaluate and this I told you is a constant because the wave function f in average sense is already defined, okay. Now, when you have that let us slightly look at. So, let us look at delta omega. Now, what is delta omega? Delta omega is nothing but omega fi minus omega i. So, this is nothing but e by h bar minus omega, okay, where e is equal to h bar omega fi. So, this e is the energy states that we have. So, this is your i that is e i but this is like e, okay, the states around which you are looking at the energy, okay. Now, if I define x as half of e by h bar minus omega, okay, into t, then your d e will be nothing but 2 h bar by t into, okay. Now, so my initial equation was, so I am just choosing something called variable transformation. I am just trying to change from d e or e to x and I define x in terms of e, okay. So, if I have what we had is p of t is equal to e naught square by 4 h bar square omega fi by omega whole square, okay, modulus of f epsilon dot mu i whole square integral sin delta omega by t 2 sin square into t delta omega by 2 whole square rho f of e, okay. Now, after this transformation, I can write p of t is equal to e naught square by 2 h bar omega fi by omega whole square integral f epsilon dot mu i square into t rho f of e integral minus infinity to plus infinity sin square x by x square t x, okay. Now, there is one thing that I want to do is the range, okay. Now, here I have defined over some range f, but here I have defined over minus infinity to plus infinity. So, there is a change in the integration limit, if I, so I have changed, so this change in the integration limit comes because you know you have redefined your e range in e. So, essentially I will tell you what it means, it means that if you have some energy that around e you are looking at, okay. So, essentially you are looking at this small range, okay defined by some average wave function f, but if you move away from this or above this, the average wave function is no longer the same. So, your wave function has changed. So, the wave functions corresponding to the energy has changed. So, this integral will not matter anymore. Therefore, you can extend the range from minus infinity to plus infinity because it is a similar to extension of adding zeros because the wave functions are going to be different. So, because of this energy range and you are fixing your energy to a small energy range, this integration can be extended to minus infinity. Now, it turns out that integral minus infinity to plus infinity sin square x by x square dx is nothing but pi, okay. So, if I do that, then p of t will be equal to pi e naught square by 2 h bar omega f i by omega square integral f epsilon dot mu square rho e at f into t. So, that is my final equation, okay. Now, if I define something called w fi, okay, this is nothing but that is the rate constant, sorry, rate of absorption. That is nothing but this is the probability into, this has probability by time, okay. So, that is nothing but p of t by t. So, that is nothing but pi epsilon naught by 2 h bar omega f i by omega square modulus epsilon naught i square, okay. Now, I can define some quantity such as w fi equals to 2 pi h bar modulus of mu square into rho, where now you can look at this equation and then decide modulus of mu square is equal to e naught square by 4 h bar square f epsilon dot mu i square omega f i by omega square, okay. Okay, there are two values, this is called the dipole transition dipole transition. By the way, that was trans-shell, this was t m i that is a trans-shell moment integral. So, this is called transition dipole and transition dipole is defined like this, okay. And omega fi that is rate, sorry, w fi that is rate constant, sorry, rate for absorption will depend on these values, okay. Now, for a given f, this is a constant and omega fi is constant and if you know what wavelength of light you are shining, this is constant. So, all these are constants, okay. Everything that is inside this is a constant, okay. So, essentially, so which means for a given transition mu square is a constant and you know omega fi, sorry, sorry, w fi that is rate, okay. 2 pi h bar also is a constant, mu square is constant. So, 2 pi h bar is some constant k into rho f. So, which means your rate constant fi, sorry, rate fi w fi is given by rho f. So, which means that from initial state i to a density of states around f, okay, this rate of transition will depend on the density of states. So, the rate of transition fi is proportional to density of states around f and this is called Fermi's Golden Rule. So, essentially the transition between a state i to a densely packed states around f, this rate rate of transition fi is equal to pi e naught square by 2 h bar omega fi by omega square epsilon dot mu i modular square to rho f, okay. This can be written as w fi is equal to 2 pi h bar modulus of mu square rho f of e, okay. And modulus of mu square is equal to e naught square by 4 h bar square omega fi by omega square modulus of f epsilon i, okay. So, this is the transition dipole which of course for a given transition is constant. So, you can think of it like this, your rate constant w fi is proportional to modulus of mu square and w fi is proportional to rho f, okay. So, the rate constant for the, sorry, the rate for the transition between i and a dense states f is given by the transition dipole, okay and density of state, okay. And these constitutes Fermi's. I will stop here. I will continue in the next lecture. Thank you.