 My name is G. Nareesh Patwari and I am at the Department of Chemistry IIT Bombay. I am going to teach quantum mechanics and molecular spectroscopy course. You can reach me at Nareesh at chem.iitb.ac.in or g.nareesh.patwari at gmail.co. Apart from me there are two teaching assistants for this course, Ms. Namita, Brigid Bizoy who can be reached at Namita, Brigid at gmail.com and Ms. Sumitra Singh who can be reached at Sumitra Singh 141 to at gmail.com. If you have any queries regarding those course please write an email to one of us. Let us get started. This course contains of 8 parts which may not be equally divided however we will start with introduction to quantum mechanics and arrive at the Schrodinger equation. Then we will look at the time dependent perturbation theory following which we will also look at properties of light in a classical manner. In the interaction Hamiltonian we will interact the classical light with the quantum mechanical molecule and that leads to a semi-classical picture following which we will evaluate the transient probability and this transient probability will be related to Einstein's A and B coefficients. Following the evaluation of Einstein A and B coefficients which will be, these will be related to the extinction coefficient which can be experimentally measured. We will also look at the line shapes and the lifetimes both of which can be evaluated experimentally and theoretically. We will look at the connection between the experiment and the theory. Finally we will quickly derive the selection rules for rotation, vibration and electronic transitions. All of this we hope to cover in 8 weeks. Today in the first lecture we will start with introduction to quantum mechanics and arrive at the Schrodinger equation. To arrive at the Schrodinger equation we will start with something very simple called classical wave equation. For example, if you have a wave that is travelling like this and this is the direction of propagation. So as both time progress, as the time progresses it, the wave progresses both in the space X and T and this wave of course has amplitude which is time dependent. So the amplitude keep going goes up, it will reach a maximum value and then keeps going down and then we will go to negative values as well. And at some instances the amplitude is 0 as in this case. So the amplitude depends on both position and the time. So if we define psi as an amplitude, so psi of X, T because it depends both on space or the coordinate and the time, the amplitude depends both on position and time. The variation of amplitude in space and time is given by classical wave equation which says that d square by dx square of psi of X, T equals to 1 over c square d square by dt square psi of X, T. Now the classical wave equation relates the second derivative of the amplitude with respect to space or coordinate to the second derivative of the amplitude with respect to time and the proportionality constant is 1 over c square and in this case c happens to be speed of propagation or velocity of propagation. Now you can see that the classical wave equation is a second order partial differential equation in T and X. So one can quickly realize that since this is second order partial differential equations, one of the possible solutions for psi is equal to k e to the power of i alpha. It is an exponential function i out because it is periodic nature it has to have this i. Now in this case alpha is given by 2 pi X pi lambda minus mu T and it is called phase factor. You should understand that a phase factor is in the exponent here. So you can see psi of X, T is equal to k e to the power of i alpha and your alpha is in the exponent. If something is an exponent it must be dimensionless. So you will see that X is length unit lambda is length unit. So X pi lambda is going to be dimensionless because lambda is the wavelength it is in length units. Now in the case of frequency mu it is 1 over T, T that is the time is T. So this will also will to mu T also will be dimensionless. So alpha is a dimensionless quantity therefore can be used as an exponent. So what we have is the classical wave equation d square by dx square psi of X, T equals to 1 over c square d square by dt square psi of X, T and psi of X, T is equal to k e to the power of i alpha where k is a constant. Now this is from very old classical mechanics. In the late 19th century and early 20th century when the quantum mechanics was developing then two major equations came into force and one of them is Planck-Einstein equation that is nothing but h nu is equal to e and the second one is De Brogue equation lambda is equal to h by. Now these two equations are epitome of wave particle duality ok. In the first case light waves behave as particles that is what Einstein said and in the case in the second case lambda is equal to h by p De Brogue said that matter can behave as waves and will have corresponding wavelength depending on its momentum. Now we know that in the classical wave equation alpha equals to 2 pi X by lambda minus nu T. So what I am going to do is that I am going to take this nu and substitute in this equation alpha and similarly I am going to take this lambda and substitute here and when I do that I will get alpha is equal to 2 pi X into p by h minus e into T by h ok where X is the coordinate p is the momentum. However you will know that the momentum is along a particular direction because three momentous along x, y, z are orthogonal to each other. Therefore this X when multiplied by p can be written as p X. Now this can be equal to because h by 2 pi this 2 pi one can take into the denominator of the over h. So one can write this as X dot p X minus e dot T by h bar where h bar equals to h by 2 pi ok. Now in some sense we have quantized the alpha, alpha which was from the classical wave equation is now written in quantum mechanics. Now we have psi, psi of X comma T is equal to k e to the power of i alpha and alpha which is now written in terms of quantum mechanics is nothing but X dot p X minus e dot T by h bar. Let us now take partial derivatives of psi with respect to T and X. So first taking D by DT of psi of XT this is equal to D by DT of k e to the power of i alpha. So this will be equal to when you take D by DT of k e to the power of i alpha. So i k e to the power of i alpha into D alpha by DT. Similarly if I take D by DX of psi of X comma T is equal to D by DX of k e to the power of i alpha this is equal to i k e to the power of i alpha T alpha by X. Now when I take D alpha by DT so if you take D alpha by DT the first part X p X will be a constant and what is it is going to derivative is E T by h bar. So essentially all you will get is i k e to the power of i alpha minus E by h bar. Similarly this will be i k e to the power of i alpha in this case X dot p X will be the function and E dot T will be a constant therefore what you will get is p X by h bar. So if I slightly rearrange this will become minus i E by h bar k e to the power of i alpha and this will become i p X by h bar k e to the power of i alpha. But k e to the power of i alpha is nothing but your psi of XT. So I am going to replace that so this will become minus i E by h bar psi of X comma T. This is nothing but minus p X by h bar psi of X comma T. So this is, so all I have done is taken the partial derivatives of psi with respect to T and X. So at the end of last slide what we had is D by DT of psi of X comma T is equal to minus i E by h bar psi of X comma T. Another equation is D by DX of psi of X comma T equals to i p X by h bar psi of X comma T. Now what we can do is slightly rearrange this equation. Now one can remember is that 1 over i equals to minus i. If I remember that and rearrange these two equations what I will get is i h bar D by DT of psi of X comma T equals to e psi of X comma T. Similarly minus i h bar D by DX of psi X comma T is equal to p X psi X comma T. Now these two equations which I will call it as 1 and 2. These two equations are eigenvalue equations. What are eigenvalue equations? If you take a mathematical operation say an operation in this case I you take a first derivative with respect to time and multiplied by i h bar. So that is some kind of operator. Let us say if you have an operator A and it acts on a function f of X then what you get is a constant multiplied by the same function. Such an equation is called eigenvalue equation and you will see that equations 1 and 2 are eigenvalue equations. How so? If I have this operator i h bar D by DT and I take that operator and act on psi of X comma T will gives me the eigenvalue e. Similarly, you have an operator minus i h bar D by DX and operate it on psi then I will get the eigenvalue p X and I will get back the same function back in both the cases. Therefore, in the first case i h bar D by DT will correspond to the operator energy because it gives the energy eigenvalue. Similarly, minus i h bar D by DX operator will corresponding to the momentum operator because it will give me momentum eigenvalue when you operate it on psi of X comma T. Now, we have used the concept of operator e and p X and these operators i h bar D by DT is equal to operator e minus i h bar D by DX is operator p X. Now, in quantum mechanics the concept of operators is very crucial. One of the postulates of quantum mechanics says all physical observables must have corresponding operator. This is in classical mechanics. This will have corresponding operators in quantum mechanics. So, in the case of energy, the quantum mechanical operator will be i h bar D by DT and in the case of momentum, the quantum mechanical operator will be minus i h bar D by DX. Now, if you take energy of a system, what is the energy of the system? Energy of the system is nothing but kinetic energy plus potential energy. So, in classical mechanics kinetic energy will be equal to half m v of course, if it is along x axis, it is v x square plus some potential capital V of X comma T. I will come back to the potential energy in a minute, but let us look at this. So, this can also be rewritten as p X square by 2 m because p X is equal to m into v X in classical mechanics plus v of X comma T. And this total energy of the system is also called Hamiltonian H. Now, in classical mechanics E and H can be interchangeably used. E is in total energy of the system and it is also called Hamiltonian. Now, if I want to convert this into operator, then I must take H convert into operator. This is equal to p X square by 2 m operator corresponding to that plus v of X T operator corresponding to that. Now, p of X operator I already know that is nothing but minus i H bar d by dx. That is operator corresponding to p X, take a square of it divided by 2 m plus potential energy operator v of X comma T. So, when I expand this, I will get minus H bar square by 2 m T square by dx square plus v of X comma T. So, your Hamiltonian operator H will be nothing but minus H bar square by 2 m T square by dx square plus v of X comma T. This is also equal to of course, total energy operator. Now, if you take equation 1, what was our equation 1? i H bar d by dt of psi of X T is equal to operator E psi of X T. This is nothing but operator H psi of X T. Now, if I write only like this i H bar T by dt of psi of X comma T is equal to H psi of X comma T. This equation is the Schrodinger equation. It is also called time dependent Schrodinger equation. And H will be equal to minus H bar square by 2 m d square by dx square plus v of X comma T. Now, I know what is the kinetic energy operator, but I still do not know what is the potential energy operator because I am keep on writing v of X comma T because the potential energy will depend on the problem that I would like to choose. For example, if I take a particle in a box, the potential is 0 inside the box. If I take a hydrogen atom, the potential energy will be 1 over square root of X square plus Y square plus Z square. And if I take harmonic oscillator, the potential energy will be 1 by 2 k X square. Therefore, the potential energy operator will depend on the problem that we will choose. However, the kinetic energy operator will remain always the same minus H bar square by 2 m d square by dx square. This kinetic energy operator of course, is only the one dimension. One can generalize Hamiltonian in three dimensions. By the way H is also called Hamiltonian. So, H in three dimensions will be minus H bar square by 2 m d square by dx square plus d square by dy square plus d square by dz square plus v of X, Y, Z, T. That is the generalization of the Hamiltonian in three dimensions, where the kinetic energy is along X, Y and Z dimensions and they are orthogonal to each other. And you take projections of kinetic energy along each dimension and the potential energy also is function of three dimensions. But in the present case, we will try to limit the Hamiltonian to one dimension. So, finally what you have is I H bar d by dt of psi of X, T equals to minus H bar square by 2 m d square by dx square plus v of X, T. So, this is the Schrodinger equation. It says that the influence of time on the psi of X, T is equal to the influence of the total energy operator on psi of X, T. So, the time dependence of any psi of X, T will be governed by the total energy operator and in such scenario, the psi of X, T is called the wave function. The name wave function was given by Schrodinger. However, the interpretation of the wave function was given by Max Bowen. We will stop here for this lecture and we will continue in the next lecture.