 Welcome back to the lecture on the solutions using differential equations. We were looking at the second order differential equation in the last lecture and towards the end of the lecture I introduced the power series method. I gave you a simple example of solving the power series using the power series the differential equation d square y by dx square plus k squared y is equal to 0 ok. And if you remember we obtained the solution y of x as 2 series 1 and even n coefficient series n equal to 0 to infinity minus 1 to the n k x raised to 2 n by 2 n factorial which is identified with cosine k x and an odd n series 0 to infinity minus 1 raised to n k x to the power 2 n plus 1 divided by 2 n plus 1 factorial and a 1 by k as the coefficient which is identified with the sin term ok. So, the purpose was to introduce the series method and also to obtain the recurrence relations which are here the coefficients for the recurrence relations are here and the k to the 2 n a naught as the various coefficients a 2 n ok. Now today we shall do a fairly rigorous solution of the harmonic oscillator problem using the power series method and this might continue to the next lecture because the algebraic details are quite involved and let us go through them slowly so that the application of the power series method and also the ideas of convergence of the various terms in the power series are all clear. But before we do that we have to transform the Schrodinger equation for the harmonic oscillator one dimensional h psi is equal to e psi of x ok to a suitable form and eventually identify that form to be the same as the Hermites differential equation. So, that will be the first step and once we identify the Hermites differential equation we can use the power series solution similar to what you see here for the Hermite equation and then study the solutions for the convergence of the power series ok. So, the first step is to identify from the Hamiltonian for the harmonic oscillator through some steps two sets of I mean two problems one is because the domain for the harmonic oscillator is all of x from minus infinity to plus infinity it is important that the solutions that we proposego to 0 at the boundaries become finite, but in this case since the boundary is infinite the solution should be should approach asymptotically 0. So, first thing is to obtain an asymptotic solution for very large values of x and very large negative values of x both ok that would give us a functional form which would be the limiting form for the harmonic oscillator wave function. It will ensure that the wave function is well behaved, but having attempted the asymptotic form we should find out the solutions for the harmonic oscillator at all intervals of x not just large values large extremes of x and the second part is where we would end up using the Hermites differential equation and obtain what are known as the Hermite polynomials ok. Therefore, let us write the steps through which we would carry this process first one is the Schrodinger equation for the harmonic oscillator the kinetic energy term is minus h bar square by 2 m where m is the mass of the oscillator d square by d x square which is all of this is the kinetic energy operator and we have the harmonic potential half k x square acting on psi x giving u e times psi x ok. The force constant and the mass of the harmonic oscillator are the only two input parameters of the model h bar is a constant h by 2 pi planks constant and x is of course, the oscillator variable away from the equilibrium position which is x is equal to 0 when the potential is 0 the harmonic oscillator is at a minimum potential and therefore, this is the potential energy operator and the force constant k is given by of course, the frequency of the oscillator as the relation is the frequency is 1 by 2 pi square root of k by m ok. This is the fundamental frequency of the oscillator in fact, this is the only parameter that we need to know when we talk about the harmonic oscillator, but it is dependent on two parameters namely the mass of the oscillator it is inversely proportional to the square root of the mass and the frequency is also directly proportional to the square root of the force constant. Therefore, if the oscillator is stiff that is its frequency is very large fundamental frequency is very large it is a reflection that the bondwhichis responsible for the oscillator vibrations about equilibrium that is a very strong bond and also if the masses are very large you see that the frequency decreases. So, if you have hydrogen molecule and you are thinking of the oscillation of the hydrogen atoms about the equilibrium position versus a H D a hydrogen deuterium molecule you can see that the mass is replaced by the reduced mass of the system and in the case of H D the reduced mass is much more than in the case of hydrogen and therefore, you see that the frequency of the oscillator is less when you substitute the masses by a heavier isotopic mass ok. Those are details that we will see over as we go along, but the nu is the fundamental frequency associated with the oscillator its expression is exactly what you have in the classical harmonic oscillator case and we can use that here or we can leave it as k ok. For this exercise I would use this particular form ok. Now, all you need to do is to write this as a differential equation involving d squared by d x square psi and so all that you do is to multiply by 2 m e and you are bringing it to the side therefore, its 2 m e by h bar squared minus half k x squared psi is equal to 0. I have removed the minus sign. So, this becomes minus a half this is minus e and when you bring it to the side. So, this is what it is ok. Now, e is the parameter or the energy the Eigen value that we have to find out psi is the wave function that we want to solve this equation and obtain ok. Therefore, given k and given h bar squared we have to get both the psi and d e and the domain x is all the way from minus infinity to plus infinity theoretical ok. No oscillator is harmonic for all extensions of vibrational or amplitudes yes. 2 m e by 2 m e by h bar squared this one over here you can I am sorry yeah this also should be multiplied by k m by h bar squared x square ok. Thank you yeah ok yeah that is that is correct ok yeah I am I am sorry yeah. So, this is the yeah I should have been because anyway dimensionally you can see that both of them are 1 I mean 1 by L square ok yeah that is ok this is just a multiplication mistake thank you. Now, the extension of x is all the way from on both extremes. Therefore, what is the solution psi of x for very large x such that psi of x goes to 0 for x going to the limits plus or minus infinity. So, that is the first thing that we have to look at ok. So, let me make some substitutions as per the notes what I have here is 2 m e by h bar square is call it lambda ok. So, d square by dx square psi plus lambda minus and the k m by h bar square I have written as alpha squared x square psi is equal to 0. The dimension of lambda you can see is twice the mass energy by h bar square this will be 1 by L square like here 1 by L square psi is common. Therefore, this is dimension is 1 by length square and alpha square is 1 by length to the power 4 multiplied by x square. So, alpha has the dimension of 1 by L square and alpha is given by k m by h bar square square root ok ok. So, dimensionally the equation is 1 by L square times psi plus terms with 1 by L square times psi is equal to 0 and the solution for very large values of x is that alpha square x square is much greater than lambda. First we shall write to the differential equation as given here as 1 by alpha d square psi by dx square because I am going to transform the variables plus lambda by alpha minus alpha x square of psi is equal to 0. Then you can see that alpha x square will be a dimensionless variable. So, let us do y is equal to root alpha times x then the psi of y is psi of root alpha x ok. So, what we do is that the differential equation therefore, becomes d square psi by d y square minus y square sorry minus lambda by plus lambda by alpha minus y square psi is equal to 0 that is easy to see because all that we do is d by d y d by d x we are replacing with this 1 by root alpha and d square by d x square times 1 by alpha will become d square by d y square and here of course, alpha x square is y square the point is that y is dimensionless. Remember alpha is 1 by L square therefore, alpha x square is dimensionless ok we will do that ok. Now, let us do the asymptotic limits the asymptotic case is that lambda by alpha both being constants are parametrically dependent on the oscillator and y being very large as x goes to infinity y also goes to plus or minus infinity whatever that limit is the asymptotic solution that we will look for is d square psi by d y square minus y square psi is 0 by removing the lambda by alpha from this by assuming that the y is much much greater than lambda by alpha very large values ok. Now, so what does that give you we need to look for the solution psi of y which will satisfy the asymptotic equation d square psi by d y square minus y square psi is 0 and suppose we take the solution to be exponential plus or minus beta y square it has to be plus or minus y square or even powers because of the structure of this differential equation, but we will limit ourselves to this and you will also find out depending on the value of beta whether it is a positive constant or a negative constant we will only retain one of the terms that is if beta is positive we will keep e to the minus beta y square because that is one which will go to 0 when y is very large ok. So, we need to find out what is the value of beta and it can be obtained by substitution and the result is I would give you that the answer turns out to be beta is equal to 1 by 2 and this positive. So, the solution psi of y is exponential minus one half whole square ok. Now, that is only the asymptotic solution ok. What we need is the general solution when the lambda by alpha cannot be ignored and when y is small and therefore, all that we need to do is to propose a wave function psi which has as asymptotic character the exponential, but otherwise it is dependent on y and therefore, the proposal is psi of y is equal to some function of y times exponential minus a half whole square ok. This will take care of all the intermediate values and the small values of y and is such that that for very large values of y this will be the dominant term which will take the wave function to 0 at the boundaries which are at the extremes. So, the proposal is that psi of y be this which satisfies the differential equationthat we just wrote down namely d square psi by d y square plus lambda by alpha minus y square psi is equal to 0. There is no approximation here ok. So, our substitution is psi of y is given by h of y e to the minus y square by 2 ok. What does that become when you do the substitution? You get the simple looking differential equation d square h by d y square minus 2 y d h by d y plus lambda by alpha minus 1 h is equal to 0 ok. So, straightforward substitution I leave youleave it to you toobtain this this is called the Hermites differential equation ok. Therefore, to summarize what we did for the last 20 minutes or so, we started with the harmonic oscillator problem transformed it to a form that as a simple differential equation introduced a dimensionless variable y in terms of x which is the position the harmonic oscillator amplitude and the constant alpha which has been defined earlier as the product of the force constant times the mass of the oscillator divided by h y square the square root of itdimensionally y has no dimensions and x is 1 x is the length this is 1 by L using the dimensionless variable. We rewrote the differential equation and looked for solutions which are valid at very large values of y and which will truncate the wave function to0 as y becomes large on either side of the axis and that gave rise to an exponential form e to the minus a half y square and later what we did was to write the exact solution for the harmonic oscillator equation as a product of a function yet to be determined multiplied by an asymptotic form which ensures that the wave function behaves reasonably at the extremes and given that substituted into this equation we get the Hermes differential equation. So, this is step 1 the next step is of course, to solve the Hermes equation by using a power series method after a brief pause we shall continue that ok. So, now we shall start with the power series method or infinite series. So, the proposal is h of y is equal to sum over n equal to 0 to infinity a n y to the power n and therefore, d h by d y the first derivative is n equal to 1 to infinity n a n y to the n minus 1 and d square h by d y square is sum over n equal to 2 to infinity n into n minus 1 a n y to the n minus 2 ok. So, this we have to substitute in the differential equation and obtain relations between the coefficients which are not known and that process is the process of obtaining the recurrence relations ok. So, if we do that d square h by d y square is n equal to 2 to infinity n into n minus 1 a n y to the n minus 2 minus 2 since you are multiplying d h by d y with y obviously, this will become sum over n equal to 1 to infinity n a n raise to the power y y to the power of n ok n a n multiplied by y raise to n and then we have plus lambda by alpha minus 1 sum over n y raise to yeah y raise to n is not it. Now, n minus 1 is multiplied by this y yeah so it is correct. So, lambda by alpha minus 1 n equal to 0 to infinity a n y to the n is equal to 0 ok. So, you can now collect coefficients coefficients of x to the 0 coefficient of x raise to 1 x raise to 2 etcetera and we want them equal to 0 we want them equal to 0 and so on ok. So, let us look at to the coefficient of x raise to 0 there is one term which is 2 into 1 a 2 coming from n equal to 2 and there is no term here that is one term which comes from n equal to 0 which gives you plus lambda by alpha minus 1 a 0 is equal to 0 ok that is the sorry coefficient of y raise to 0 dot x y to the 0 and what you have is the first term which gives you a 2 is equal to minus 1 lambda by alpha minus 1 by 2 times 1 a naught this is product 2 into 1 2 times 1 ok. The next if we were to look at the term associated with y the term associated with y you will have one here which is 3 into 2 a 3 ok you will have one term here which is minus 2 n is times 1 a 1 and then you have one more which is lambda by alpha minus 1 a 1 that is equal to 0 ok. So, which gives you a 3 is equal to minus lambda by alpha minus 1 plus 2 times 1 divided by 3 times 2 a 1 ok. So, I will write one or two more terms a 4 let me write that is easier to look at a 4 comes from y raise to 2. So, it is 4 into 3 and it is connected to y squared y squared. So, it is minus 2. So, let me write the term a 4 is 4 times 3 a 4 minus 2 times 2 a 2 plus lambda by alpha minus 1 a 2 is equal to 0 or a 4 is given by 2.2 times 2 let me see lambda by minus ok let us write that lambda by alpha minus 1 the minus sign divided by 4 into 3 times a 2, but a 2 is given by minus 1 lambda by alpha minus 1 divided by 2 times 1 times a 0 ok. So, you can see that there is a recurrence relation between a naught and a 2 and a 4 and so on. So, the even number even indexed coefficients a 2 a 4 a 6 are all given by one recurring relation with an unknown constant a naught and likewise for a 3 and a 5 a 7 if we write these things down you would see that the odd numbered subscript subscripted coefficients are all connected to one undetermined or unknown coefficient a 1 and let me write therefore, the general recurrence relation for you. Let me do thatthe separate cases of odd and even a's a 2 n is given by minus 1 raise to n lambda by alpha minus 1 minus 2 times 2 n minus 2 let us see that is one term likewise there are other terms. So, times that times lambda by alpha minus 1 minus 2 times 2 n minus 4 ok times until you reach the free term lambda alpha minus 1 and this vanishes this all of these divided by 2 n factorial ok. So, you could see times a 0 ok. So, a 2 n is a series of terms with the minus 1 raise to n you can see the progression 2 n minus 2 2 n minus 4 and so on until you reachthe 0 here and this is one recurrence one relation of course, the other relation iscan also be written directly from this, but anyway I will write it as an independent one for the odd n's that is a 2 n plus 1 it is minus 1 raise to n y 2 n plus 1 factorial lambda by alpha minus 1 minus 2 times 2 n minus 1 ok times lambda by alpha minus 1 minus 2 times 2 n minus 3 times all the way 2 lambda by alpha minus 1 minus 2 into 1 ok all of this multiplied by the coefficient a 1 ok. So, this is the relation between a 2 n plus 1 and a 1. So, you can see that the odd terms that is a 1 a 3 a 5 etcetera are given by that and this is the relation between a 2 n and a 0 and this is for even the n's n equal to 2 4 6 etcetera ok. Now, the study of this take this further into solvingthe equation with the h of y which we wrote as sum over n equal to 0 to infinity a n y to the n. We need to substitute these coefficients in here and you would see that this series keeps onforever there is no limit on the value of n. Now, we need to see whether the series converges at some finite value or or converges to some finite value or or the coefficients growing independently of each other. So, if we have to look at all those properties, then one needs to pay attention to the ratios between the coefficients a n and the next a n plus 1 and you have to see that this ratio becomes the as you take the term a n plus 1 by a n a n plus 2 by a n plus 1 etcetera. The coefficients should follow a pattern they should go to 0 or they should decrease according to well defined mathematical properties for the series to converge. So, there is a whole lot of issues that need to be taken into account in looking at the series and seeing that the series converges and that part we will continue, but let me stop at this point namely that the two recurring coefficient relations are the ones with which we will start the next lecture. We will continue and substitute this into the series, obtain the conditions for truncating the series of finite values of n and then obtain what are called the series of polynomials which will depend on what value of n you truncate the series and those are called the Hermite polynomials. And these Hermite polynomials need to be obtained with the convergence of the series and the multiplication of this with the exponential minus y square in mind then those solutions will be the solutions that you would see as the solutions of harmonic oscillator. We will continue this in the next lecture.