 So far in this course we have talked about free particle, we have talked about particle in a box and then we have talked about tunneling. Today we move to another system which is very important from the point of view of quantum mechanics and this system is there of a harmonic oscillator. It will take 2 or 3 modules to complete the discussion. So today we are just going to introduce ourselves to the problem and we are going to learn a very important concept called ladder operators. Harmonic oscillators of course is something that all of us are familiar with from once again class 11 and 12. We know that a harmonic oscillator is an oscillator that obeys Hooke's law. What is Hooke's law? f equal to minus kx or you can write minus kx equal to m d2x dt2 where x is the displacement from mean position and we might remember the classical examples of harmonic oscillator. Pendulum of a clock is a harmonic oscillator provided the angular displacement is not more than 4 degrees and there are many such examples and oscillatory motion is very closely related to circular motion and that is why we get results that look very similar to what we get in circular motion and we get to work with things like frequency of oscillation and so on and so forth things that we have encountered in rotational motion. So as usual we are first going to remind ourselves of the classical picture of harmonic oscillator and then we will go into the quantum mechanical description. So let us get started. Why should we start why should we discuss harmonic oscillator in a quantum chemistry course anyway? The biggest reason from the point of view of a chemist is that a harmonic oscillator provides an approximate model for a vibrating diatomic molecule and then if you do that then this k that is there the force constant it basically tells us how strong a spring is if it is for a spring. So what we do is we approximate the chemical bond between two atoms in a diatomic molecule to start with as a spring with a spring constant k. So automatically k becomes bond strength. So if we can determine k quantum mechanically or spectroscopically then we get to know the bond strength which is a very important parameter one of the most fundamentally important parameters of chemistry. Now when we try to talk about molecules of course the classical description is not going to be complete and we have to discuss quantum harmonic oscillators. So whenever we go from classical to quantum regime we find that several differences come in discrete energy levels come in we will talk about wave functions and we will see we get one more very interesting concept that comes up. But before going there let us develop the treatment a little bit. First of all let us simplify this Hooke's law a little bit and let us rearrange d 2 x dt 2 is equal to minus k by m x. Now this is a differential form which we can differential equation form which we can try to solve. And the trial solution that we use is x of t is equal to a sin omega t plus b cos omega t. Once again the sin omega t cos omega t is reminiscent of what you would get in circular motion omega here is root over k by m and that as we might know is the angular frequency of oscillation. To see whether this is a valid solution or not what we will do is we will differentiate it twice and we see what we get. So let us find dx dt what is dx dt we have a sin omega t plus b cos omega t. So when we differentiate sin omega t with respect to t omega comes out and sin omega t derivative is cos omega t. And the second term plus b cos omega t once again omega comes out and cos omega t when differentiated gives you minus sin omega t. So what you get is omega multiplied by a cos omega t minus b sin omega t. And since we are talking so much about eigenvalue equations it is very obvious that this is not an eigenvalue equation. So this function that we have used is not an eigenfunction of the first derivative. What about the second derivative because second derivative is what we are interested in we want to solve this differential equation. If we differentiate once again what will we get the first term a cos omega t now gives us minus omega a sin omega t and omega multiplied by this omega that is there already gives you minus omega square. Second one gives us omega multiplied by b sin omega t and minus sin is already there. So now omega square comes out and d2x dt2 turns out to be omega square minus omega square multiplied by a sin omega t plus b cos omega t see what has happened we have got back x of t in this equation first of all it is an eigenvalue equation. Secondly if you compare this equation with this equation it is very clear that minus k by m is equal to minus omega square or omega is equal to root k by m. So this is something that comes from classical mechanics. If you want to go into the quantum world then what we need to do is we need to write the Schrodinger equation. For that we need to remember that potential energy in this case is going to be half kx square. So this is a parabolic potential and let us also remember that there is a relationship between omega the frequency of angle of frequency of oscillation and the force constant k. So instead of k we might as well write something in omega. So we will write half m omega square x square that is your potential energy. So now we take this and we plug it in Schrodinger equation the first term of the Hamiltonian remains the same it is a kinetic energy operator in place of potential energy we are going to write half m omega square x square. So here is what we get minus h cross square by 2m d 2 psi dx2 plus half m omega square x square psi. Now it might be worthwhile to say here that usually what we do is we do not work with m we work with the reduced mass because it is a two body problem when we talk about an oscillator it is a diatomic molecule HCl there are two atoms they are moving with respect to each other and it is difficult to formulate the problem that way. So what we do is we reduce it and instead of m we write mu the reduced mass and that is what we work with but for now we will just write m whenever the need arises will switch conveniently to mu and I am not going through the detail of it because that is something that we must have done earlier in physics classes. Okay so now before going further let us discuss a little bit about why a simple harmonic oscillator would be a more or less valid model for a chemical bond. We know from once again our class element well knowledge that a potential energy surface for a diatomic molecule is something like this it is not really a parabola when the two atoms are very far away from each other then their interaction energy is zero and then when they come closer then the attractive forces predominate and the energy goes down until a minimum point which is the equilibrium bond length and after which the energy increases sharply due to inter nuclear repulsion okay. So it is very clear that this curve that we have drawn the solid curve is definitely not a simple harmonic oscillator but if I try to draw a simple harmonic oscillator and superimpose with it you can see that from for small displacements from the mean position the two curves are more or less superimposable right say up from here to here the two curves are more or less the same and remember simple harmonic motion requires very small displacement from main position anyway. So if you are going to work with small displacements from main position then simple harmonic oscillator model might be a valid model so that is the premise within which we are going to work for now. Later on when you want to do spectroscopy not in this course when you want to do spectroscopy then you have to consider an harmonicity of the oscillator but let that be the story for another day whoever is interested can go through the lectures on our molecular spectroscopy course which are now freely available on YouTube right. So simple harmonic oscillator is a good approximation for small displacements. So we work with the parabolic potential we write it in Schrodinger equation as we already done in fact we have written it in a little more complicated form that we will come back later and then when we solve it we solve it with boundary condition sorry psi is not coming here this is actually psi psi is 0 at x equal to plus minus infinity. If you remember the conditions that arose from Born interpretation the system cannot exist beyond x equal to plus minus infinity right. So for continuity like in particle in a box the boundary condition is the psi equal to 0 at x equal to plus minus infinity not wherever this potential energy surface is and as we are going to see later on right now what I am doing right now is that I am sort of putting the card before the horse and giving you a brief summary of the results later on we will actually arrive at them. The reason why I want to do this is that we are going to do a little bit of mathematical manipulation I would not like us to get lost while doing that I would like everyone to know what lies there at the end right because the end result is what is of use as application in chemistry. So let me give you the end result first so that even if you get confused while we go through the next discussion it is not so much of a problem to understand the application. So when we work it out we will see that energy is quantized energy EV is equal to V plus half into h cross omega but omega once again is the angular frequency of oscillation and V is the vibrational quantum number ranges from 0, 1, 2, 3, 4 and so on and so forth. So this is what you will get we will get discrete energy levels and they are going to be equispaced that is very easy to see we just work out what EV plus 1 is going to be subtract EV from EV plus 1 you will always get h cross omega. So in a simple harmonic oscillator quantum harmonic oscillator we get equispaced energy levels and energy gap between the between two successive levels is always h cross omega that is one quantum of vibrational energy great. Now we will of course get wave functions as well we will work all this out right now I am only giving you a summary of the answers we will work out and we will see that we get wave functions that look some were similar but not exactly similar to what we get for particle in a box but in a very major departure from particle in a box you see that some wave function is there outside the potential energy surface as well and I would like you to take this as an exploratory question I would like you to find out and maybe post on the forum why you think the wave function exists outside the potential energy surface as well of course it dies down pretty fast but let that be a homework for you people great that is one thing. The second thing is that as I said vibrational quantum number ranges from 0 to infinity. So the lowest energy you can get when you put b equal to 0 is half h cross omega. So you see vibration energy for a quantum harmonic oscillator can never be 0 a quantum harmonic oscillator can never be rest even if you lower the temperature to 0 Kelvin you get a minimum energy of half h cross omega this is called the 0 point energy this is a very major departure from a classical harmonic oscillator classical harmonic oscillator can actually be a stationary quantum harmonic oscillator cannot be stationary why because if it does then its position is determined completely uncertainty in position is 0 what is the position x equal to 0 it will be in the equilibrium position and what is the uncertainty in momentum that also is 0 because it is not vibrating anymore so delta p is 0. So delta p equal to 0 and delta x equal to 0 is not allowed it is not allowed because of uncertainty principle that is why a quantum harmonic oscillator can never be at rest 0 point energy is there when we talk about rigid rotor later you will see that a rigid rotor can be at rest even a quantum rigid rotor and we will discuss why that is the case why the departure from quantum harmonic oscillator but for now the very important take home message is that for a quantum harmonic oscillator the lowest allowed energy is half h cross omega and we will work it out in the next module this is called 0 point energy as I said. Okay now let us come back to the earlier slide and let us start trying to find solutions for Schrodinger equation this is our Schrodinger equation minus h cross square by 2 m remember m will be replaced by mu in due course d 2 psi dx 2 plus half m omega square x square psi equal to e psi. Now there are two ways of finding solution of the Schrodinger equation for a harmonic oscillator the first way is a power series method sort of Brute's force method which is of use in many other applications as well I am not very sure whether we want to do it here but we will see the method that will definitely work out is the algebraic method using ladder operators this is a fantastic method and it introduces us to this very important concept of ladder operators as we will see by the way the discussion I am performing is from Griffiths quantum mechanics book I find that this book is very nice reading as well the treatment is nice and it is as if the author is talking to you so you get a good feeling if you read this book but you can study this from any standard quantum chemistry quantum mechanics book that you are comfortable with. So let us go ahead and let us try to learn the algebraic method using ladder operators for finding solutions of Schrodinger equation for a quantum harmonic oscillator. So here is your Schrodinger equation let us see if we can write it in a little bit of simpler form. So first of all if you look at the kinetic energy term first one minus h cross square by 2m d 2 side dx 2 here the operator involved is minus h cross square by 2m d 2 dx 2 that is the kinetic energy operator as we have discussed already of course here we are talking about motion in one dimension only so that is why you do not have this del 2 del x 2 del 2 del 2 del y 2 del 2 del z 2 business. Now kinetic energy is directly has direct relationship with linear momentum and as you know the linear momentum operator is h cross by i d dx or you could write minus i h cross d dx and of course kinetic energy is p square by 2m so you can write this kinetic energy operator minus h cross square by 2m d 2 dx 2 in terms of this p hat operator and it of course is going to be p square by 2m so I can write the first term on the left hand side of Schrodinger equation as p square psi by 2m for the sake of convenience I am not writing the hat anymore but please do not forget that p here is actually an operator it is not a number. So p square psi by 2m is the first term and while writing the second term what we will do is eventually we want to write it in some simple form. So look at the second term we have omega square and x square and we have m so it can very easily become m omega x whole square if I divide it by another m and the good thing is if I do that the second term will also have 2m in the denominator the first term already has a 2m in the denominator so I can take it out outside the bracket. So what I will do is I will write the second term as 1 by 2m multiplied by square of the product m omega x that operating on psi. So I have just rewritten the two terms on the left hand side of Schrodinger equation. So now taking 1 by 2m outside the bracket the equation becomes 1 by 2m p square please do not forget that p is an operator plus m omega x whole square so this together is the Hamiltonian operator that operating on psi gives us e psi. I am saying operator operator operator so many times because it is important that we do not forget that we are not dealing with numbers we are dealing with operators even x here is really the position operator it is a different matter that when position operator operates on the wave function it gives you the eigen value of position multiplied by the same wave function. So now see this is our Hamiltonian 1 by 2m multiplied by p square plus square of m omega x. Now you look at this part p square plus m omega x whole square if there is some way of factorizing it it would be nice of course if the numbers then it is very easy because it is in the u square plus v square form. So and we know that u square plus v square can easily be factorized as i u plus v multiplied by minus i u plus v yeah u square minus v square everybody knows that is that involves all real numbers if I have u square plus v square you can still factorize it by using the imaginary number iota i u plus v multiplied by minus i u plus v just check whether this is okay or not okay I will give you a second for that right fine let us go ahead. So now the issue is these are not numbers as I said these are operators and the problem with operators is that if there are numbers then u into v would be equal to v into u commutation relation would hold u v equal to v u or u v minus v u equal to 0 operators however do not necessarily commute and here we are working with the momentum operator and the position operator we are going to show you in the next module that these two operators definitely do not commute okay if the operators commute then some special property arises again let us take a change take on that but operators do not necessarily commute. So it is not easy to factorize this u square plus v square kind of expression into two products but what we will do is we will still try and see what is the form of this kind of product we will take i into p plus m omega x and we will take minus i into p plus m omega x multiply them together and see what we get and what we get is going to make our life a lot more interesting as far as your harmonic oscillator is concerned. So what we will do is we will try to evaluate i p plus m omega x multiplied by minus i p plus m omega x and to do that what we will do is we are going to write this in a little simplified form once again remember i p plus m omega x is an operator in itself p is an operator x is an operator so the linear sum is also an operator and same holds true for minus i p plus m omega x what we will do is we will construct an operator out of i p plus m omega x and we are going to call it a minus similarly we are going to construct an operator out of minus i p plus m omega x and we are going to call it not very difficult for you to guess now a plus now why this is called a minus why that is called a plus you will see but for now let me just tell you what the form of the operator is a minus is actually 1 by square root of 2 h cross m omega multiplied by i p plus m omega x where did that 1 by 2 root over 2 h cross m omega come from it came from hindsight we are not the first people working this out it has been done many times and people people who did it for the first time actually found out that the subsequent result becomes simpler and more meaningful if we use this factor of 1 by root over 2 h cross m omega as Newton had said if you see further then that is because we stand on the shoulders of giants let us take the advantage of standing on the shoulders of giants that is where this factor comes from it has not fallen from sky right if you did it for the first time you would be perfectly justified in not even writing this factor we write it because we know where we are headed all right similarly a plus is 1 by root over 2 h cross m omega multiplied by minus i p plus m omega x all right so we have defined the factors then what we want to do is we want to work out the product remember these are operators once again I am repeating many times because this is the concept that often does not sink in if I just say it once and go go ahead so if we there are two ways in which we can work out the product of a minus and a plus we can either work out a minus a plus or we can work out a plus a minus okay remember the operator that is on the right operates first on the function operator on the left operates next so what we will do is we are going to evaluate a minus a plus okay I will request you after the module to evaluate a plus a minus yourself because we will use a plus a minus as well okay so this is what we want to do now let us take a break come back and start right from here in the next module