 to the lecture number 25 of the course quantum mechanics and molecular spectroscopy. Towards the end of the last lecture, we were looking at the transition moment integral which is nothing but f mu dot E. Now, we wrote that mu as an expansion of mu naught plus alpha plus beta R square plus gamma R cube plus etcetera, ok. This is the Taylor series expansion, the equilibrium geometry. Therefore, your TMI will now become mu naught plus alpha R plus beta R square plus gamma R cube plus etcetera, dot epsilon, ok, where alpha is nothing but, alpha is nothing but d mu by dr evaluated R naught and beta is nothing but half d square mu by dr square evaluated R naught and gamma will be nothing but 1 over 6 d square mu d cube mu by dr cube evaluated R naught, ok. These are the coefficients, ok. So, this is nothing but dipole moment derivative. We will come to the usage of dipole moment derivative later in this course, ok. Now, in this case what mu naught is called permanent dipole moment which is nothing but the dipole moment at the, when the molecule is at equilibrium geometry. So, simply for example, if you have a molecule AB because of the electronegativity difference if there is a partial charge q plus on A and q minus on B and this distance is R, ok, R naught that is the equilibrium geometry then your mu naught will be nothing but equal to q times R naught, ok. Now, one of the interesting thing that you will still need in the time to evaluate in your transient moment integral when you write TMI this is equal to f epsilon i is the fact that we still have to evaluate the initial final wave functions. In the beginning of the course or throughout the course up till now what we have assumed that we have a Hamiltonian H naught we still do not know what that will exactly is into i is equal to E i i and H naught f is equal to. So, this we have assumed all while but we still do not know what is H naught precisely, ok. Now, what I am going to do is that let us start with a very simple problem that is the rotational state. So, what we look at is pure rotational spectra we look at the pure rotational spectra, ok. Now, if you have proved if you take the total if you have a molecule AB then and this is your R naught, ok. One can write Hamiltonian H is equal to minus h bar square by 2 m A del square A minus h bar square by 2 m B del square B plus U R EL AB, ok. Now, one can write such a Hamiltonian H, ok. Now, this is what this is the kinetic energy of atom A and this is nothing but kinetic energy of atom B and this is the relative or bond energy of AB, ok. So, we will go back to this same thing. So, your H is minus h bar square by 2 m A del square A minus h bar square by 2 m B del square B plus U R EL. Now, what does U R EL AB consist of? If U R ELB will consist of, ok, we will consist of three terms. One is electron nucleus attraction consist of electron repulsion nuclear nuclear repulsion and four electron kinetic, ok. Now, I want to make one approximation called rigid rotor approximation. Now, what am I saying when I say rigid rotor approximation? Very simply it means the following. If I have A and B and they are separated by distance r naught, ok, the distance is fixed. That means the nuclei are no longer moving, ok. So, the nuclei are at the fixed positions in this case. When you have such a scenario, then the sum of U R EL AB will be equal to constant, ok. So, the nuclei are fixed and in the AB molecule the nuclei are fixed, they are not vibrating, they are not ending, they are just fixed, ok. When such are fixed, when the two nuclei are fixed, then, ok, electron nucleus attraction will not change, the electron electron repulsion will not change, a nuclear nuclear repulsion will not change, electron kinetic energy will not change. All of them will be some constant numbers and when you add all the four constant, you will get one more constant. So, your U relative AB will be equal to constant. So, which means your Hamiltonian H R R, ok. R R because rigid rotor, ok. I will call rigid rotor as R R, ok. H R R will be now be equal to minus H bar square by 2 m A del square A minus H bar square by 2 m B del square B, ok, plus some constant. Now, one of the interesting things of, you know, energy measurements is that you can always measure energy with respect to some value, ok. For example, in thermodynamics, we only measure delta G's and we that is delta G's because we always think that the delta G of the standard states is 0, ok. And we measure delta G with respect to a standard state. Similarly, in this case one can think that one can measure energy with respect to this constant which is nothing but U relative AB. So, since I am measuring with respect to this constant, I can ignore it. So, my H R R will be now be equal to minus H bar square by 2 m A del square A minus H bar square by 2 m B del square B. There was a constant which I ignored because I am measuring energy with respect to that. So, this is my Hamiltonian. If you think of it, so what I have is the following, ok. I have A and A is connected to B by a stick that has length R naught and that is not changing. That is what I said rigid route, ok. So, the positions of A and B are fixed. Let us for a time being think that what all the possible motions that this particular system can undergo. Now, I told you the R naught is fixed, ok. So, when the R naught is fixed, it can only undergo two motions. One is that if you have A and B, ok, attached by R naught, then of course, it can move along x, say z direction, it can move along x direction and it can move along the y direction. So, the whole molecule can move in some direction, ok, along x, y and z, ok. Now, what else can you do? Because it A and B cannot move, then what can you do? It can rotate. A can rotate, A B can rotate like that without changing the distance or it can rotate like this, ok. So, A B can rotate and A B can translate. So, only possible motions are translation. What did I say? So, we had this Hamiltonian, which was nothing but h, ok, is equal to minus h bar square by 2 m A del square A minus h bar square by 2 m B del square del square B, ok. Now, we had two of them. So, there are two possible motions. One is the motion of A atom and other is motion of atom B. However, they are strict to each other. So, A and B cannot move independently. So, in that scenario, this equation can be written as minus h bar square by 2 capital M del square C m plus minus h bar square by 2 mu del square, ok. So, this is something called, it is called co-ordination transformation. I will not go to the mathematics of it, but I will tell you what. So, if I have a quantity A and B, ok, that is my body A and B. So, what you have is the coordinates of A and B, ok. And there is a mass of it which is m A and the mass of this is m B and there are some coordinates, ok. And what I will call it as R A and R B. So, the center of mass will lie somewhere here. So, this is center of mass, ok. So, this will have R center of mass. Similarly, you have another distance called this as naught which is nothing but R internal. So, there are two things that one can do is that instead having R A and R B, there are the vectors from the origin going to A and B. There are two vectors that are representing this system R A and R B. Instead of that, you could choose an alternate description that is R center of mass and R naught. So, it is an equivalent description of it. And one can do the math, ok. There is something called center of mass transformation. One can look it up in some textbooks, ok. And it will clean from that one can make such a transformation in the Hamiltonian without making any approximations. But there is only one thing that I need to define where your m capital M is equal to m A less m B, ok. And small mu is m A m B divided by m A less m B. And this is called reduced mass. And this is total mass. These two Hamiltonians, so h R R equals to minus h bar square by 2 m A del square A minus h bar square by 2 m B del square B is identically equal to minus h bar square by 2 m del square center of mass minus h bar square by 2 mu del square, ok. So, instead of having two coordinates representing A and B, now you have an alternate description in terms of coordinates representing center of mass and internal motion, ok. And this is of course valid only when these are valid under this. Now, there are two things that let us go to the Hamiltonian equals to minus h bar square by 2 m del square center of mass minus h bar square by 2 mu del square, ok. Now, there is one issue is the following. Is that this quantity is the center of mass motion. That means the whole object is moving, ok. So, this is nothing but your, so you can now calculate your h R R as two quantities. One is h center of mass, less another is h internal, where h center of mass is equal to minus h bar square by 2 m del square center of mass. So, this represents the motion of the of the entire AB molecule. Now, this is just a motion of the entire AB molecule in some direction x, y, z or a linear combination thereof. So, this essentially leads to what and there is no potential acting on it. So, essentially this leads to what is known as free particle, ok. And we know the free particle is unquantized, ok. So, one can think of having a this one as h cm chi cm is equal to, ok, E cm chi cm where E is equal to h bar h bar square k square by 2 m is the minus h bar square k square by 2 m that is the kinetic energy. So, we do not have to worry about the Hamiltonian corresponding to center of mass, because that just represents the motion of the entire AB, ok. And just if you want to move a particle or a mass or AB molecule, ok. Of course, it is not related to any spectroscopic transition. So, we would not worry about it anymore in this course, ok. Now, what we are left with the is the remaining part h internal which is nothing but minus h bar square by 2 mu del square. Now, if I had a particle AB, ok and the this one was at x A, y A, z A, these are the coordinates of A, x B, y B, z B as internal coordinates of or coordinates of B, then you can write x is equal to x A minus x B, y is equal to y A minus y B, z is equal to z A minus z B and this represents internal. Therefore, your del square internal will be equal to d square by dx square plus d square by dy square plus d square by d z square, ok. And this is internal coordinates, ok. There is of course, one problem, ok. Now, the problem is this is that when you want to, so I know I told you that AB is a rigid body and of course, you must remember is R naught will be equal to x square plus y square plus z square to the power of half which will be nothing but x A minus x B whole square plus y A minus y B whole square plus z A minus z B whole square to the power of half, ok. Now, there is one thing that I told you is that of course, this can move in the x y z direction, the entire molecule AB. But I told you that we are not concerned with such a motion because it does not lead to any spectroscopic transitions, ok. Now, the thing that we are interested is in the rotational motion either in the plane of the board that I am writing or a plane that is perpendicular to the plane of the board, ok. When you have such motion, then, ok, then you are talking about a motion of AB in a circular fashion. So, that means, you can think of your H internal, your H internal is equal to minus h bar square by 2 m del square internal. And this internal motion is the motion of the molecule without moving the center of mass. So, what is this? This is that this is motion that does not change the center of mass, ok. So, you have to keep your center of mass fixed and then move around, ok. So, you have to rotate the body either in the plane of the board, ok. So, AB can rotate or rotate perpendicular plane of the board, ok, while keeping its center of mass fixed because the motion if there is a motion with of the center of mass that is already gone into the other Hamiltonian of the free particle that I discussed little bit earlier, ok. So, now, what we are concerned is just the motion of this AB molecule in a such a way that its center of mass remains exactly in the same position. So, this will lead to rotation. So, your H internal, ok, this is equal to minus h bar square by 2 m del square internal, which is nothing but minus h bar square by 2 m d square by dx square plus d square by dy square plus d square by dz square. This should represent rotation without changing, ok. We will stop here and continue in the next class. Thank you.