 Welcome to lecture number 26 of the course, Quantum Mechanics and Molecular Spectroscopy. We will have a quick recap of the previous lecture before proceeding. In the last lecture, we looked at the total Hamiltonian H is divided in two parts which is H nuclear and H electronic, okay. And the H electronic will consist of terms Ke of electrons, potential energy of electrons and nuclei, potential energy of nuclear repulsion and potential energy of electron repulsion. All this put together one call it as a u-relative, okay and the energy that comes out of this and this the H nuclear will consist of Ke of nuclei, okay. Now that means your H internal that is the motion of the nuclei can be written as Hn plus u-relative. And I told you for a given nuclear configuration u-relative is equal to constant and we can always measure energies with respect to some constant that is u-relative, therefore one can treat u-relative is equal to 0 that means measurement of energies relative, okay. So what you get is H is equal to H internal. Now if you had two atoms A and B separated by a distance r naught, okay and which is fixed. So which means the A, B nuclear configuration is rigid. So A, B nuclear configuration is rigid that means A and B are not moving with respect to each other. In such scenario the kinetic energy of A, B which is nothing but minus h bar square by 2mA del square A minus h bar square by 2mB del square B can be written as minus h bar square by 2 capital M del square cm minus h bar square by 2 mu del square internal, okay. Now what is in where m is equal to mA plus mB and mu is equal to mA mB divided by mA this is the total mass, this is the reduced mass and this is called this transformation is called center of mass transformation. So one can think of it like this if you have an axis, okay B and this term is rA, this is rB, you can have center of mass somewhere here which I will call as r center of mass and then there is a r naught which is the distance between A and B. If the coordinates of A are xA, yA, zA and coordinates of B are xB, yB, zB then one can tell r naught is equal to square root of xA minus xB whole square plus yA minus yB whole square plus zA minus zB this I will also write it as x square plus y square plus z square square root of where x is equal to xA minus xB y is equal to yA minus yB z is equal to zA minus zB, okay. So what you have your Hamiltonian H will now be equal to or the nuclear Hamilton will now be equal to minus h bar square by 2 capital M del square center of mass minus h bar square by 2 mu del square internal, okay. Now this is nothing but motion of center of mass. What is the motion of center of mass? Center of motion of center of mass simply means that the whole object AB is moving in some direction and that will not contribute to the internal structure, okay. Therefore one can write minus h bar square by 2M del square cm if this multiplied by chi n is equal to En chi n, okay. This is free particle, okay because it has no potential energy term. So this is a free particle. So when you have free particle your energy is En is equal to h square k square by and we know this En is the not quantized. So the energy of the free particle of course is not quantized, okay. So that corresponds to the translation. So this essentially corresponds to translation, translation energy of the AB molecule, okay. Now one need to deal with this your H internal will now be minus h bar square by 2 mu del square internal. That will be nothing but minus h bar square by 2 mu d square by dx square plus d square by dy square plus d square by d z square. So this is the Hamiltonian that we need to solve. So the corresponding Schrodinger equation would be minus h bar square by 2 mu del square internal to some wave function I will call it as phi is equal to minus h bar square by 2 mu d square by dx square plus d square by dy square plus d square by d z square whole thing acting on phi is equal to. So this is the equation that we need to solve Schrodinger equation, okay. Now to solve the Schrodinger equation we have to look at one important point, okay that is the following. Is that if you have AB this is say called I will write it as R capital R, okay. And of course this can rotate in the plane of this board and also perpendicular to the plane of this board, okay. So essentially what we are looking at is that without changing the center of mass, center of mass remains fixed without changing the, now that means we have a rotating particle a particle that can rotate in x, y, z directions, okay. So now your del square internal was nothing but in Cartesian coordinates d square by dx square plus d square by dy square plus d square by dz square. Now but we have only the center of mass is fixed so we have only the rotating particle. So we can transform this into a spherical polar coordinates. So we do another coordinate transformation wherein the del square internal can be written as d square by dr square plus 2 times by r d by dr minus l square by r square h. So this is the transformation in spherical polar coordinates where I will come back to what l is. So your del square internal will be d square by dr square plus 2 by r d by dr minus l square by r square h bar square, okay. And where your l square operator is given by minus h bar square into 1 over sin theta d by d theta sin theta d by d theta plus 1 by sin square theta d square by d phi square, okay. Now let me tell what these terms are, okay. Now we had the Cartesian coordinates x, y, z to begin with. So let us suppose you have a Cartesian coordinates x, y, z so there is a point here which is represent by x, y and z, okay. So I will call it as a z axis and this is x and this is y. Now this distance is r, okay and this angle along z axis is theta, okay and the projection of this onto the x, y plane is the angle phi, okay. So that is how you define x, y, r, theta and phi, okay. So initially you had x, y, z, okay there are 3 independent coordinates that describe this point. Now you have similarly 3 independent points that is r, theta and phi, 3 independent variables. So essentially you are going from one coordinate to the other coordinate, okay. Now your del square internal, okay your del square internal will be nothing but, okay. If you ignore r, okay if I fix r, okay if r is fixed, r is fixed, why am I fixing r? I am fixing r because I am considering that the molecule AB has a fixed radius r naught or capital R does not matter, okay. If r is fixed then of course its derivatives will, you cannot have a derivative with respect to a constant, okay. You can only have a derivative with respect to a variable. So you can ignore these 2 terms and all you are left out with this. So that will be equal to minus L square by r square h bar square or this I will call it as a rotational Hamiltonian h, rotational under rigid rotor approximation. So my h rod now will be equal to L square by internal r square h bar square, okay into minus, so this is into this minus into minus, this is your del square so h bar square by 2 mu, okay. So when I rearrange this, this will be this h bar square, this will cancel, negative sign will also cancel. So what I am getting is L square by 2 mu r square is 1 mu h rod, okay. Now this is written as L square by 2 I where Ie is equal to mu r square and this is called moment of inertia, okay. So at the end what I have is my h rotational is equal to L square by 2 mu r e square, okay. Why am I writing r e or r because I want looking at the equilibrium distance. So a, b this distance is I will call it as r equilibrium distance between the 2 a and it is not changing, it is a rigid rotor approximation that means a and b are fixed at one place. So this is nothing but L square by 2, okay. Now what is my L square? L square is equal to minus h bar square into 1 over sin theta d by d theta sin theta d by d theta plus 1 over sin square theta d square by d phi square, okay. And one can quickly look at this is nothing but your angular h hydrogen atom or this is also nothing but particle on a sphere, okay. And the solutions of these are given by solutions of these are given by what I will call y, okay j m theta phi these are called spherical harmonics. You can look up any elementary quantum chemistry textbook or quantum mechanics textbooks to look at the spherical harmonics, okay. Now in spherical harmonics you have 2 important mission. One is the total angular momentum given by L square operator and there is a z component of angular momentum which is given by L z, okay. Now we know that your L square operator is your y l m is a Higand function of the L square of h bar square into j into j plus 1 into y j m theta and phi and your L z operator acting on y j m theta and phi will give you m h bar y j, now let me quickly write once more, okay. So essentially your y j m theta and phi are eigenfunctions of the operator L square it will give you eigenvalue of h bar square into j into j plus 1 y j m theta and phi and other is L z operator is also as eigenfunction of y y j m theta and phi which will give you m h bar y j m theta and phi. Now there is one important thing you must realize that if there are 2 operators have common eigenfunctions then the operators must commute with respect to each other. So which means the commutator L square L z must be equal to 0. So operators L square and L z commute, okay. Now let us go back to our Hamiltonian your H rotation was nothing but operator L square by 2 mu R e or same as L square by 2 I e. Now you can see 2 mu R e or this part or this part is just a constant. So if you multiply a constant okay the wave function is not going to change. So we have this Hamiltonian and because this Hamiltonian consists of the operator L square okay. So this Hamiltonian will have same eigenfunctions of L square okay except that you know now they will be multiplied by different constants. So you have H rotational acting on y j m theta and phi will give you h bar square divided by 2 mu R e j into j plus 1 y j m theta and phi and this is your eigenvalue. Now we have a that is rotation with the quantum number j will be equal to h bar square by 2 mu R e j into j plus 1 which is same as h bar square by 2 I e j into j plus 1. Now this is written as h b j into j plus 1 okay therefore if I write like this okay what I have got h bar square is h square divided by 8 pi square I e okay j into j plus 1. Now if I remove h then I am left with so instead of h square I can write h into 8 pi square. So if I take this as p e then I will get this. So your b e is equal to h by 8 pi square e rotation that will depend on value j will be equal to h bar square by 2 I e j into j plus 1 this is also equal to h b e j into j plus 1 where b e is equal to h by 8 pi square I e and has a unit of second inverse because you see energy is joules and h is joules second and j's are just quantum number. So this must be joules second so has to be multiplied by second inverse okay and so this should give me joules okay so b e has a units of second inverse. So finally what you have we have the following your h y j m theta and phi or h rotation okay is equal to e rotation j y j m theta and phi and your e rotation j is given by h b e j into j plus 1. So this is your energy and this is your wave function okay so we can now look at transition okay now they are quantum number so you will have rotation states that are j is equal to 0, j is equal to 1, j is equal to 2 etc okay and we can look at the transition between this in the next lecture. I am going to stop it here thank you.