 Hello, I am G. Narish Patwari and I am teaching quantum mechanics and Malkar's spectroscopy course. Welcome to the lecture number 2. Before we go into lecture number 2, we will have a quick recap of what we have done in the lecture number 1. In the lecture number 1, starting with classical wave equation, which is nothing but d square by dx square of psi of x, t is equal to 1 over c square d square by dt square psi of x, t where psi of x, t is equal to k e to the power of i alpha and alpha is equal to 2 pi over 2 x by lambda minus mu t. Using this information from classical equations, combined with e equals to h nu and lambda is equal to h by p is Planck-Einstein equation and the De Bruyne equation, we arrived at two operators i h bar t by dt is equal to operator e or operator x. Operator h minus i h bar t by dx is equal to operator p x. We further developed that the psi of x, t is called wave function and is governed by the time dependent Schrodinger equation which says i h bar t by dt of psi of x, t is equal to minus h bar square by 2 m t square by dx square plus v of x, t acting on psi of x. Therefore, the time dependence of the wave function psi of x, t will depend on the Hamiltonian equation and this is called Hamiltonian operator h. It turns out that in many problems that we deal in chemistry, the potential energy operator v is of x, t can be m dash v of x because has no time dependence. For example, you take a hydrogen atom and hydrogen atom will remain as hydrogen atom in for the infinite time because the potential is time independent. Similarly, harmonic oscillator will continue to remain harmonic oscillator. So, most of the problems or the systems that we deal in quantum chemistry in chemistry in particular have no time dependence on the potential. Therefore, h which was minus h bar square by 2 m d square by dx square plus v of x, t can be written as minus h bar square by 2 m t square by dx square plus v of x. You will see that the kinetic energy operator which is nothing but minus h bar square by 2 m d square by dx square has no time dependence. Similarly, the potential energy also has no time dependence. Hence, the Hamiltonian becomes time independent. In such a scenario, the wave function psi of x t can be written as product of psi of x multiplied by phi of t. Because if you see Hamiltonian only depends on the coordinate, any partial derivative with respect to coordinate will have time as a constant. Therefore, phi of t can be bought out of the wave function. If you go back to the Schrodinger equation which is nothing but i h bar d by dt of psi of x, t is equal to minus h bar square by 2 m d square by dx square plus v of x psi of x, t. I will write it as i h bar d by dt psi of x, t is equal to h psi of x, t. But your psi of x t is now a product of psi of x into phi of t. So, the Schrodinger equation now becomes i h bar i h bar d by dt of psi of x into phi of t is equal to h times psi of x phi of t. Now, d by dt on the left hand side, d by dt when it becomes acts on psi of x, psi of x becomes a constant. Similarly, h which only depends on the coordinates acts on phi of t, phi of t becomes constant. Therefore, one can get the psi of x outside and phi of t outside. So, when I do that I can write this equation of psi of x i h bar d by dt of phi of t is equal to phi of t h psi of x. Now, what I am going to do is I am going to divide both sides with psi of x phi of t. When I do that psi of x i h bar d by dt of phi of t is equal to phi of t h psi of x phi of t divided by psi of x to phi of t divided by psi of x to phi of t. Now, this phi of t will cancel on both sides on the on the numerator and denominator. This psi of x will cancel on the both numerator and denominator. So, what you will get is 1 over phi of t i h bar d by dt of phi of t is equal to 1 over psi of x h psi of x. You will see that the left hand side is a function of t and the right hand side is a function of x. So, I am equating two functions of two different variables. For example, if I equate f of x with g of y this will be only possible if both the functions this will be possible only if both the functions are equal to a constant and that constant I will call it as w. So, you end up getting two equations because of that 1 in 1 in t and 1 in x 1 over phi of t i h bar d by dt of phi of t is equal to w 1 over psi of x psi of x is equal to w 1 over psi of x psi of x is equal to w. So, when I rewrite i h bar t by dt of phi of t is equal to w phi of t h psi of x is equal to w psi of x. Now we know Hamiltonian is nothing but the total energy operator. So, h is equal to and this is also equal to operator e that is something that we have discussed in class number 1. So, it is easy to replace w with e. When I do that what I get is i h bar d by dt of phi of t is equal to e phi of t and h psi of x is equal to e psi of x. So, I get two equations for which in both cases e is the eigen value. So, the eigen value e will depend on the time dependence of the system and eigen value e will depend on the coordinates of the system. However, in the first case it will depend on the time dependent operator t by dt and in the second case it depends on the Hamiltonian. And these two anyway you can see are equal. And the second this equation I will call this equation as 2 and this as 3 will realize that equation number 3 is nothing but time independent Schrodinger equation, which is nothing but equation number 3 will be h psi of x is equal to e psi of x, generically written as h psi equals to e psi. And this equation is nothing but your time independent Schrodinger equation. Now, in the time dependent equation what you have is i h bar d by dt of phi of t is equal to e phi of t. So, if I slightly rewrite this d by dt of phi of t equals to when I go on the other side i becomes minus i. So, minus i h bar will go to denominator e phi of t. So, essentially when I operate d by dt operator on phi of t I will get minus i by h bar e as a eigen value. Now, this means I can solve this first order differential equation very easily. And when I solve it the solution will be phi of t equal to exponential minus i by h bar e times t or simply written as e to the power of minus i e t by h bar. Now, the other equation is h psi of x is equal to e psi of x. But my h is nothing but minus h bar square by 2 m t square by t x square plus v of x solution to this is going to be much more complicated and I have not solved it. We will come to the solutions of it in a minute, but the total wave function that is psi of x comma t can now be written as psi of x to phi of t which is nothing but psi of x into e to the power of minus i e t by h bar. And in this case this is called phase factor. Total wave function psi of x comma t is equal to psi of x e to the power of minus i e t by h bar. Unfortunately, I still do not know how to solve for psi of x because I need for that I need to solve h psi of x is equal to e psi of x. Solutions to this equation which is nothing, solution to this equation or the time independent Schrodinger equation is beyond the purvey of this course. If you want to know how to solve this equation then you might want to look up other NPTEL courses such as introduction to quantum chemistry by Professor K. L. Sebastian. In this course of course, we will assume that we know the solutions typically for a time dependent. Typically for a time independent Schrodinger equation there are many solutions and or of the form h psi n is equal to e n psi n where n is the quantum number. Now you get many such solutions for this. In fact, in some cases you get infinite number of solutions and if you take a set of all these psi n's this will form what is known as complete set. This complete set simply means that any arbitrary function chi of x can be written as x is equal to e n is a summation over i c i psi i of x. So, any arbitrary function can be written as a linear combination of elements of this complete set. I will give an example. You have a Cartesian coordinates x, y and z. Any point in the Cartesian space can always be written as a linear combination of x, y and z. Therefore, the Cartesian coordinates x, y, z or the vectors corresponding to x, y, z form a complete set. In quantum mechanics for every psi n there is another set complete set which is made of complex conjugates of psi of n. That is because there are no rules on the wave function as to whether it should be real or complex. In general, a wave function can be complex. So, when you have a complex wave function then there exists a complex conjugate of it. In Dirac's bracket notation, this is written as a ket function and the bra function. In this case, we of course, assume that we all know what psi n's are. For example, if you take a hydrogen atom problem the psi n's will be all the orbitals 1s, 2s, 2p, 3s, 3p, 3d, etc. If you take harmonic oscillator functions will be equal to the v is equal to 0, v is equal to 1 or vibrational quantum number is equal to 0, the function corresponding to that second and third and etc. If you have a particular box then there is a n is equal to 1, n is equal to 2, n is equal to 3, 3 function etc. In this course, of course, we will assume that all these solutions are already known. Before I move on, there are few things that I would like to define. In quantum mechanics, any operator corresponding to a physical observable is Hermitian. Hermitian means operator is a self-adjoint of itself, which means a average value, I will come to what average value means, average value of an operator a is equal to its transpose complex conjugate, that is Hermitian. Now, I will define few integrals and integrals are definite integrals. In quantum mechanics, generally the integrals are definite integrals. So let us define the first one called overlap integral. This integral is given by integral psi m star psi n d tau. D tau is the corresponding volume element or the surface element or length element in for integration. So this is equal to in bracket notation will be psi m psi n and this is given by for a given system given by delta m n is nothing but conical delta. So delta m n is equal to 1 if m is equal to n and is equal to 0 if m is not equal to m. In such case, we get what is known as the orthonormal set. But overlap integral is delta m n only if all of psi m and psi n belong to the same system, wave functions of the same system. If they are wave functions of different systems, then of course you will have to really evaluate the overlap. Then other reason expectation value x, it is also called average value. This is given by and this will correspond to operator average value of an operator a, expectation value of an operator a for a given system. So this will come out as psi n star a psi n in terms of bracket notation. This is written as psi n a psi n. I do not know how much it is. I know that it is a number because you know it is a definite integral. But I do not know what is its value. I have to evaluate this number or this integral to get the value. It could be 0 as well, expectation value can be 0. And the other one I will call it as an action integral. Now what is this action integral? I will define this as psi m star a psi n. So what does it do? A acts on psi n and the resulting function gets projected on to the psi m. So the bracket notation it is psi m a psi n. For example, if I act light on hydrogen atom which is in the 1 s orbital whatever resulting function you can now project on to either 2 s or 2 p and in generally this will give us transitions. So these action integrals are very important in spectroscopy. The other thing that I want to tell you is about the Hermitian operators. As I told you that all operators corresponding to physical variables are Hermitian in quantum mechanics. So we just said that average is equal to A average transpose star. Now if I write in terms of integrals then this will be in the case of Hermitian operators will be integral psi star m a psi n d tau will be equal to integral a psi m whole star psi n d tau. This also will be equal to psi n star a psi m d tau. In bracket notation one can write psi m a psi n is equal to psi n a. Now this is properties of Hermitian operator. Hermitian operators give real expectation values. So this is so all the because all physical observables were correspond to some real expectation. So when you measure momentum it should be a real number. When you measure kinetic energy it should be real number. If you measure position it should be real number. So operators that correspond to momentum, position, kinetic energy or Hermitian operators. We will stop it here. We will continue in the next lecture.