 In the next couple of modules we are going to complete our discussion of harmonic oscillators. In these modules we are going to talk about wave functions to a little greater extent and we will discuss an interesting recursion formula that relates the wave functions with each other and very briefly we will mention why that is important. But before we begin let us recapitulate what we have studied so far in this topic. We have studied the Schrodinger equation for the quantum harmonic oscillator and starting from the elementary form we have rewritten it slightly in terms of momentum 1 by 2 m multiplied by p square plus m omega x whole square this is a Hamiltonian it operates on psi to give us e psi. Then we have introduced and we have talked at length about this over the last 2 or 3 modules we have introduced 2 operators a minus and a plus these are called ladder operators and we have studied various aspects of ladder operators first of which is that the commutator between a minus and a plus is 1. Hence we have learnt how we can rewrite the Hamiltonian this time in terms of the ladder operators. So, there are 2 ways in which we can write it because the operators do not commute with each other it is h can be written as h cross omega multiplied by a plus a minus plus half or you can write h cross omega a minus a plus minus half remember a plus a minus and a minus a plus are not 1 and the same because the commutator is 1 and not 0. Which one do we use? We use any one of these forms whichever is convenient for the particular problem we are trying to solve. So, what we have done is we have learned that these are called ladder operators for a reason if a plus operates on psi for example it produces a new wave function which is an eigen function of the same Hamiltonian operator with the interesting phenomenon that the eigen value of energy is more than the eigen value of the original wave function psi by h cross omega which is one quantum of vibrational energy. So, h operates on a plus psi to give us e plus h omega multiplied by a plus psi. Now a plus psi is not necessarily the actual wave function which is one step higher in the ladder it could be the actual wave function multiplied by a factor. So, later on when we do a little bit of treatment with this you will see we are going to write it as a 1 multiplied by the function because remember all wave functions have to be normalized anyway there is no guarantee that when ladder operator operates on a wave function it gives us a new wave function which is normalized. But the good thing is even if a wave function is not normalized Hamiltonian can still operate on it and give us the correct value of energy. Why? Because if you look at Schrodinger equation in this form any form the normalization constant would come on left side as well as right. So, it does not matter this is one important thing that we should note in quantum mechanics that it does not matter really just to find the eigen function it does not matter whether the wave function is normalized or not. Of course, it does matter when you want to work out the average quantities as we have seen earlier. So, a plus is called a raising operator because its action on psi is to take us one step higher in the vibration energy ladder and a minus takes us one step down. So, it is called the lowering operator. Now this is something we have said earlier knowing a wave function in principle using the ladder operator we should be able to work out all other wave functions by going up or down the ladder one step at a time and knowing the energy of one level we should be able to work out the other energies as well. In fact, we have noted that vibrational quantum numbers have values of 0 1 2 3 so on and so forth and the expression for energy of vibration for a harmonic oscillator is E v equal to v plus half into h cross omega which also tells us that for the lowest value of v v equal to 0 the energy is not 0 it is half h cross omega but we have said it several times let us not go into that once again and we have also learned that the lowest energy wave function is of this form m omega by pi h cross raised to the power 1 by 4 multiplied by e to the power minus m omega 2 h cross x square do we have to remember this expression not really what do we have to remember we have to remember that it is a Gaussian function e to the power minus some constant into x square if you remember that that is enough there is no need in this course or science in general to remember each and every expression that we come across right so this is what we have learned so far and the question we stopped at in the previous module is what are the wave functions for v equal to 1 2 3 and so on and so forth so we will take 2 approaches first of all we are going to use the ladder operator and work out the wave function for v equal to 1 and then we are going to do a more rigorous analytic approach using a power series solution to arrive at the general expression for the wave functions and we will find out their energies as well and that is what we will take us to the recursion formulae but first let us see how we can very conveniently go up the ladder using the ladder operator so this is what we are going to use we know psi 0 already and we have with us the step up ladder operator a plus what we will do is we will make a plus operator and psi 0 so we are going to get a psi 1 as and this is what I was saying a plus psi 0 but we should not forget the constant a 1 because there is no guarantee that a plus psi 0 is a normalized wave function so psi 1 wave function for v equal to 1 is a 1 multiplied by a plus psi 0 okay so first of all good let us get ahead now so we write like this what we do is we write the form of the a plus operator 1 by root over 2 h cross m omega multiplied by minus ip plus m omega x operating on psi 0 please remember this p here is an operator x here is an operator x is not so much of a problem because the action of x is just to multiply the wave function by the value of the position but let us not forget what the form of the momentum operator is momentum operator as we know very well by now is h cross by i ddx or minus i h cross ddx so in the next step we are going to substitute this expression for momentum operator into the expression for psi 1 right so we are going to make the momentum operator plus m multiplied by minus i plus omega x operate on psi 0 so when we do that this is what we get we have just written it out I will give you a moment to absorb this I will read it out meanwhile but please read for yourself and you will get what what I have got so the first term would be minus ip operating on psi 0 right p is h cross by i ddx so minus ip would be minus h cross ddx so minus h cross being a constant comes out and ddx operates on the wave function one fourth power of m omega by pi h multiplied by e to the power minus m omega by 2 h cross x square in case you are confused by me rattling out all these expressions please go through this yourself and convince yourself that what we have said is correct the first term here is just an expansion of minus ip psi 0 what about the second one second one is easier m omega x psi 0 m omega x is there we just write the expression for psi 0 from here in this expression so this is what we have got but of course this is only the beginning and not the end what we should do next is we should differentiate this function and that is not all that difficult also right because this m omega by pi h cross to the power one fourth is just a constant it will come out and we have to differentiate this Gaussian function with respect to x not so difficult so that is what we need to do here I have just shown every step so that in case anybody is confused you can go through the presentation later on and you can convince yourself this can be self-study material and as I have said a following by and large the textbook on quantum mechanics by Griffith at the moment so this is what we get since we differentiate e to the power minus m omega by 2 h cross 2 x we get back the same exponential function but it has to be multiplied by 2 x multiplied by m omega by 2 h cross minus of that so minus and minus becomes plus h cross in the numerator h cross in the denominator cancel each other 2 x that 2 and the 2 in the denominator cancel each other and finally we are left with m omega x m omega by pi h cross to the power 1 by 4 is from the previous line m omega x comes from differentiation of e to the power minus m omega by 2 h cross x square and you get back the same term anyway so this is what we have got a 1 multiplied by 1 by root over 2 h cross m omega multiplied by some of these 2 terms well the first time is m omega x multiplied by one fourth power of m omega by pi h cross then multiplied by e to the power minus m omega by 2 h cross x square second term is also the same just add them up 2 comes out and this is what you get a 1 multiplied by 2 by root over 2 h cross m omega multiplied by m omega multiplied by m omega by pi h cross to the power 1 by 4 multiplied by x multiplied by e to the power minus m omega 2 h cross x square okay and of course you have 2 m omega in the numerator root over 2 h cross m omega in the denominator so you are going to simplify this and you are going to get root over 2 m omega in the numerator and h cross in the denominator right that is what you will get and you have to normalize it while normalizing you are going to use a standard integral since we have talked about standard integrals already I am not going through it once again in any case you need to look at a compendium to solve it I am just telling you that when you use the standard integral a 1 turns out to be 1 so this is what you get psi 1 is this okay so essentially this is the wave function we have got this we had got already psi 0 we have got this psi 1 which often looks like a sine function but actually is not really a sine function it is x multiplied by e to the power minus m omega by h 2 h cross x square multiplied by a constant it is a product of constant vth order polynomial in x okay here I am jumping the gun a little bit but it is not very difficult for you to see that x is of course a first order polynomial in x and I am saying vth order polynomial already because if you look at psi 0 you see we have the this e to the power minus m omega by 2 h cross x square in psi 0 as well as psi 1 okay the Gaussian function is there in both the wave functions right some constant will be there constant will defend will differ from function to function but what we see here is that here there is nothing else in x which means you have x to the power 0 for psi 0 for psi 1 you have x to the power 1 and as we will see later on as you go higher up you will get terms in x to the power 2 x to the power 3 and so on and so forth you are going to get polynomials in x and order of that polynomial will be the same as the vibrational quantum number v right so this is a an example of how one can use ladder operator to go up a ladder and given a wave function help you work out the wave function that is immediately next so from psi 0 we have worked out psi 1 so what we can do is we can keep on doing this we can keep on using ladder operator and finding your psi 2 psi 3 psi 4 and so on and so forth one by one and one thing I should say is that you do not really need to normalize like this there is another way by using Hermitian conjugates by which one can do the normalization but that requires a little more of linear algebra right now we are not going to go into it maybe later on if you get time we will come back and we will expand we will get into that as well but what we have obtained is a way in which we can generate the wave functions knowing one wave function one by one but what we really want is something that is more general can we do everything at one go can we find a general expression for all the wave functions that is what the analytic method allows us to do it is a little more rigorous so since I think most of our students of this course will be chemistry students some of you might not have studied too much of math so if you find this a little bit dry or a little bit intimidating please bear with us because the real beauty of this harmonic oscillator business will not come out unless we are little persistent and unless we crack all this math that we are going to do right we will go step by step wherever we have to make assumptions we will tell you but by and large we are going to actually work out everything so please bear with me and please work out yourself you have to work this out using a pen and paper by yourself after the lecture or even during the lecture was it and work it out so that every step is clear to you okay so I hope everybody has a pen and a paper at hand let us get ahead now for using the analytic method we go back to the Schrodinger equation in which we have this kinetic energy term on the left hand side potential energy term is written as half m omega square x square remember what omega is it is angular frequency of vibration on the right hand side we have E psi okay so we are going to try and obtain direct solutions of this by using a power series method what is power series method we will see when the time comes but first let us rewrite the variable this equation is in x since we are not the first working out this x equation we know that the final result will be neater if we work this out in terms of not x but another related variable xi this letter that you see on the left hand side here this is the Greek letter psi xi sorry in English is written x i xi equal to root over m omega by h cross multiplied by x why do we take this because well somebody has worked it out already and it is it has become clear that making this substitution makes the working easier makes the answer look a little neater so here goes epsilon equal to root over m omega by h cross x that is what we are going to use so using that what we will see is we are going to transform the kinetic energy term and we are going to obtain minus h cross omega by 2 d 2 psi d psi 2 most of you might be able to work this out by yourself but as promised we will go through this step by step so to do that since we know that xi equal to root over m omega by h cross multiplied by x we have defined it that way let us make x the subject of formula so x is root over h cross divided by m omega multiplied by xi right so the first thing to do would be to differentiate x with respect to xi so dx zi is a constant square root of h cross by m omega that is fairly straightforward now what we want is we want to know what is d 2 psi d 2 d psi 2 sorry I will say that again what is d 2 psi d psi 2 why because d 2 psi d psi 2 is there in the kinetic energy term of the Hamiltonian so d 2 psi d psi 2 is equal to the derivative with respect to xi of d psi d psi so what is d psi d psi d psi d psi d psi dx multiplied by dx d psi I hope all of us are familiar with changing the variable during differentiation this is how we do it so d psi d psi is equal to d psi dx multiplied by dx d psi that is what we are going to use and then we are going to differentiate it with respect to d psi once again so let us do that we know what dx d psi is that is a constant let us bring the constant out and then we are left with d d psi of d psi dx right so that then will be again equal to root over h cross by m omega d dx of d psi dx multiplied by dx d psi same thing once again we know what dx d psi is we write it so the root sign goes and we get h cross by m omega multiplied by d 2 psi dx 2 so d 2 psi dx 2 very easily is m omega by h cross d 2 psi d psi 2 so what we do is instead of d psi dx 2 we are going to write this and then we are going to multiply by minus h cross square by 2 m so minus h cross square by 2 m multiplied by m omega by pi h cross d 2 psi d psi 2 gives us minus h cross omega by 2 d 2 psi d psi 2 right that is what we have written here so we have worked out the first term on the left hand side of Schrodinger equation in terms of not x but psi now let us work out the second term this is much more straightforward because all we have to do here is that instead of x square we have to write the term in xi square and that is very easy square of it this what will it be x square is equal to h by m omega h cross by m omega multiplied by xi square it is as simple as that this is what we will use x square equal to h cross by m omega xi square so half m omega square x square will be equal to half m omega square multiplied by h cross by m omega xi square so m and m cancel one of the omegas cancel you are left with h cross omega by 2 xi square we will write that left hand side now is written in terms of xi minus h cross omega by 2 d 2 psi d omega d psi 2 plus h cross omega by 2 xi square psi there of course will be equal to e psi now what we can do is we can take this whole thing on to the right hand side because we want to write our differential equation in a need form that is why we are doing this so we can take this second term to the right hand side and then what we will do is that entire right hand side will multiply by minus 2 by h cross omega then on the left hand side we will be left with only d 2 psi d psi 2 this is what we will get d 2 psi d psi 2 is equal to 2 by h cross omega multiplied by h cross omega by 2 xi square psi minus e psi so you can think that we have actually taken the first term to the right hand side or you can just well you can see it is just algebraic manipulation not very difficult to understand so on the right hand side we have 2 by h cross omega multiplied by h cross omega by 2 xi square psi minus e psi okay so in the right hand side we have 2 terms both of them are psi multiplied by some constant or the other so you can take psi common outside the bracket and also the nice thing is you have h cross omega in the numerator here h cross omega in the denominator here 2 in the numerator here 2 in the denominator here they are going to all cancel if you open the bracket so the first term will become xi square second term will become minus e well minus 2e divided by h cross omega right this is what we will get d 2 psi d psi 2 is equal to xi square minus 2 e by h cross omega multiplied by psi okay again this is an eigenvalue equation and this is a differential equation which we can solve without much hassle but I would like to draw your attention to this e by h cross omega once again what is this e by h cross omega is essentially is energy and h cross omega is the energy of each quantum so it is a number of quantum of energy is not it so we are going to write it as k k equal to 2e divided by h cross omega so k essentially is twice the number of quantum of energy that is there okay so depending on which vibrational level we are talking about this number will be 1 2 3 4 something like that 10 20 whatever so the differential equation we get finally is d 2 psi d psi 2 is equal to xi square minus k multiplied by psi what is the next step the next step obviously is to try to solve it but let us tidy up our board a little bit this is what we are saying we have rewritten Schrodinger equation in terms of xi square and k where k is 2e by h cross omega and xi let us we forget is x multiplied by square root of m omega divided by h cross h cross great so now let us understand something we have said that k is just a number like 1 2 3 4 so on and so forth right so what happens if xi is very large then you can neglect k with respect to xi square then d 2 psi d 2 psi 2 becomes xi square psi so what we will do is first we are going to work this out what is the solution for very large value of xi and then we will go to the general solution so first of all the general solution here would be a e to the power minus xi square by 2 plus b e to the power xi square by 2 just work it out yourself differentiate twice you will get back this same equation but one thing to remember is that mathematics is not the be all and end all for us mathematics is only the tool that helps us get to the physics so we must remember that the second term is problematic e to the power xi square by 2 so as xi increases either in positive direction or in negative direction this e to the power xi square by 2 is going to increase which means that it is not a normalizable function that means that this part of the wave function should not be there because remember we are looking for acceptable wave functions only and a wave function that is not normalizable is not really acceptable so we can simply write psi equal to a e to the power minus xi square by 2 for very large values of xi okay and that should remind us of the wave function that we got already using ladder operators that is also a multiplied by e to the power some constant multiplied by well instead of x you can write it in terms of xi square cannot you right so this that is what it is so what will the general solution be if you think of small values of xi as well we can restore generality by remembering that we will need this e to the power minus x square by 2 function exponential factor which becomes 0 at plus minus infinity that will be required whatever we have has to be multiplied by it but there is no guarantee that the pre exponential factor well should not call it pre exponential factor it is a Gaussian term pre Gaussian factor what is the guarantee that it is a constant and it is a constant for large values of xi but not necessarily for all values of xi so to keep it very general we write it as h function of xi okay will not consider that it is a constant let us say it is a variable okay something what it is we will see so what is d psi d psi let us work that out first d psi d psi as you can see very clearly is a product of two functions right one function is h of psi h of xi and the other function is e to the power minus xi square by 2. So we know what the derivative of a product is let us do that the first one what we do is we keep e to the power minus xi square by 2 as it is and we differentiate h and since we do not really know what it is just write dh d psi multiplied by e to the power minus epsilon square by 2 what will the second term be h which is a function of xi will remain intact and it will be multiplied by the derivative of e to the power minus epsilon square sorry xi square by 2 which will be minus 2 xi multiplied by e to the power minus xi square by 2. So this is what you will get minus h h of xi multiplied by xi e to the power minus epsilon square by 2 okay this is these are the two terms that we have got for d psi d psi what is the next step the next step is to first write it a little cleanly let us take e to the power minus xi square by 2 common and then we differentiate once again I am not doing this step explicitly I leave it for you to do it. So we get d 2 psi d psi 2 is equal to the Gaussian factor multiplied by d 2 d psi 2 of h minus 2 xi multiplied by first derivative of h with respect to xi plus xi square minus 1 multiplied by the function h of xi. So we have obtained this expression for d 2 psi d psi 2 which we can now replace in this expression of Schrodinger equation and then we can go further ahead that is what we will do in the next module.