 Welcome back to the lectures on the quantum mechanics and the elementary or model systems in quantum mechanics. The last lecture was on the power series method to the harmonic oscillator problem and I stopped at the point where we arrived at two recurrence relations well essentially one, but recurrence relations for two independent coefficientsfor the solution of the harmonic oscillator differential equation. So, let me recap that in the first two minutes and then continue in this lecture with the solution and the solution to the wave function of for the harmonic oscillator ok. So, differential equation that we wanted to solve was d square h by d y square minus 2 y d h by d h by d y plus lambda by alpha minus 1 is equal to 0. This was the differential equation and we proposed to the solution that h of y is equal to sum over n 0 to infinity a n y to the n and if you remember the overall solution that we were looking at for the harmonic oscillator we hadpsi as exponential minus y squared by 2 h of y. This was the equation that this was the solution that we were looking at ok it was basically beta y square, but it does not matter it is minus y square ok. The wave function must go to 0 at the boundaries and here the boundary is that y isplus r minus infinite infinity. The nature of this function therefore, should be such that the whole thing goes to 0 for very large values of y. It looks like it does, but if h of y for example, grows faster than exponential of minus y square by 2 growing slower then psi of y may actually tend to increase. Therefore, it is important for us to see what is h of y. We have obtained a relation for a n we have obtained a relation for a 2 n in terms of a naught that was the last line of the last lecture and a 2 n plus 1 in terms of a 1 ok. Therefore, we know that the series that we have h of y n equal to 0 to a to the n a n y to the n. We know what the a n's are and we need to look at a n plus 2 by a n because you see it is either a naught or a 1. So, the series alternates with the coefficients. So, if we take the ratio of this and if this has a certain pattern of convergence slower than that of this then we have to relook at the h of y and the solution is that we will truncate h of y to finite number of terms and the justification for the truncation is that h of y otherwise will not allow the psi of y to converge to 0 ok. So, let us look at a n plus 2 by a n if you recall the coefficients will be minus lambda by alpha minus 1 minus 2 n by n plus 1 times n plus 2 which goes as minus 2 by n and goes to 0 as n goes to infinity. So, coefficients are will behaved, but if you look at e to the y square for example ok which we can write as 1 plus y square plus y to the 4 by 2 factorial plus etcetera y to the 2 n by n factorial and so on. And if you look at the coefficients a y of n plus 2 y of n if you take the ratio of the two terms coefficients coefficients of the power of y to the n plus 2 and coefficients to the multiplying the term y to the n if you take that ratio that also is 1 by n by 2 plus 1 and which goes as 2 by n for n going to infinity ok. So, there is a problem here that even though h of y is such that the terms as we increase more and more coefficients as we add more and more coefficients and we write this as a larger and larger series the convergence property of this is very similar or it is the same as the convergence of e to the y square. And you have seen that if that is a case this whole thing will actually go as e to the y square by 2 for very large values of y and that is a problem with this not satisfying the boundary conditions. Therefore, it is important for us not to have an infinite series h of y needs to be truncated at some finite value some finite number n ok. And to truncate that at some finite number n what it means is that the coefficient a n is not 0, but the coefficient a n plus 2 is equal to 0 is equal to a n plus 4 is equal to etcetera. Therefore, it is important for us to choose the power series in such a way that all terms beyond a certain value of n have 0 coefficients. How do we do that? We do that by looking at the structure of the coefficients please remember a 2 n was written earlier in the last lecture as minus 1 to the power n. If I remember it is lambda by alpha minus 1 minus let me see it is 2 a 2 n was yeah 2 into 2 n minus 2 I think we had this and lambda by alpha minus 1 minus 2 times 2 n minus 4 and so on till lambda by alpha minus 1 all of this divided by 2 n factorial times a 0. Therefore, what we want to do is to have a certain value of n such that perhaps a 2 n is not 0, but the a 2 n plus 2 will be 0 that will be 0 if one of this leading terms if this leading term goes to 0 or if we choose lambda by alpha minus 1 to be equal to whatever that follows ok. Here we are talking about a 2 n for a 2 n plus 2 this will be lambda by alpha minus 1 minus 2 times 2 n that will be the leading term. Therefore, if we set lambda by alpha minus 1 is equal to 2 times 2 n then you can see that a 2 n plus 2 will be 0 and likewise all the other terms which involve a 2 n plus 2 and so every term in the power series with n greater than this coefficient will be 0. So, this is a way of truncating the power series to finite values. What value of n? n equal to 0, n equal to 1, n equal to 2 each 1 will give you 1 hermite 1 h of y ok. If we truncate h of y at n equal to 0 which means that is we say a 2 is 0 a naught is 0 then you have only 1 coefficient constant a 0 ok. I have not yet said what is a 2 n plus 1 we will do that in a minute ok in terms of a 1. Let us assume that we are considering only the even terms in here ok. We are considering only the even terms then the even terms are such that if we let the a 2 to be 0 then only one term a naught then h of y is simply a constant and that will happen you see that a 2 will be 0 when n is 0. Therefore, lambda minus by alpha minus 1 is 0 that will give you what is known as a h naught of y which is a constant a naught. If we truncate this at n equal to 1 then lambda by alpha minus 1 is equal to 2 into 2 4 which means that we will have a naught a 2 but not a 4. Therefore, what you will have is h naught y with the leading term a naught you will have 1 and then you will have plus y square times some coefficient. So, it should not write this as a naught sorry it is a naught plus a 2 y square and that is it a 4 is 0 and all the others are 0. Therefore, this is h 2 h 2 of y ok. So, if we consider only the even terms and for the moment we do not worry about the r terms we consider only even terms then we have h naught of y h 1 of y each one is a finite has a finite number of powers of y and each one is obviously a polynomial in y the first one is of course, a constant the second one is a quadratic in y the next one h 4 of y will be a quartic one in y and so on. What about the odd coefficients? If we keep a 1 as the only term and what a 3 a 5 etcetera to be 0 then we have the polynomial or we have the h function 1 of y which will be a 1 of y. If we choose the terms here I have written a 2 n. So, let me write a 2 n plus 1 here and tell you what that condition is. So, if we write a 2 n plus 1 that is minus 1 to the power n you have lambda by alpha minus 1 minus 2 into 2 n minus 1 ok times lambda by alpha minus 1 minus 2 into 2 n minus 3 times. So, on till the last term lambda by alpha minus 1 minus 2 into 1 all of this by 2 n plus 1 factorial times a 1. Therefore, if we leave a 1 not 0, but a 3 we want it to be 0 a 3 has to be chosen by having lambda by alpha minus 1 to be equal to 2 that is that is the only term it will not the others will not be there. Once a 3 is 0 all the other terms are a 5 a 7 etcetera or all 0. So, you have the same thing that lambda by alpha minus 1 minus 2 n 2 into 2 n ok ensures that a 2 n is non 0, but a 2 n plus 2 is 0. So, therefore, the condition is the same whether you are with 2 n or with 2 n plus 1 the corresponding n value should determine this factor ok. Therefore, what we have is when we do that truncation this term the the way function that we wrote down as e to the minus y square by 2 h of y ok. Now, becomes psi 0 with h 0 of y as 1 possible solution psi 1 of y with e to the minus y square by 2 h 1 of y and so on. Therefore, what we are doing is that to ensure that the way functions are do not explode, but they become 0 at the boundaries we are providing solutions which are solutions obtained by truncating the power series to finite number of terms. The polynomials that you get by doing that are known as Hermite polynomials Hermite polynomials. Just to complete this part of it we should then give the successive terms namely a 2 a 4 a 6 etcetera in terms of a 0 which we already know here in here. Therefore, if we assume a naught to be a constant some suitably chosen constant we can immediately calculate a 2 by using that condition namely it is lambda by alpha minus 1 ok times that ok. After all lambda by alpha minus 1 is truncated at some value. So, you put that value here and you put get the numerical factors this will be substituted by whatever n that we chose and then obviously, this will fold beyond a certain point to 0 ok. So, we know the explicit expressions for a 2 a 4 a 6 etcetera in terms of a naught the moment we know where to truncate the series because by doing that we have chosen the value for this in terms of an integer ok two times an integer or whatever that may be two times twice n or whatever ok. Any integer where we choose to truncate it that integer get substituted here. Therefore, we have explicitly numerical expression algebraic, but with the numbers for each of the coefficients and without doing explicitly those things I will write down the solutions that we get as the final result which you can verify and it is also one of the background quiz that you can do to convince yourself that that is the answer ok. The answer when we do the truncation is the following ok H naught of y is 1 we were assumed a naught to be 1 ok H 1 of y is 2 y we have chosen a 1 to be 1 again I guess ok I will leave it leave a question mark there H 2 of y is minus 1 you can see that a 2 minus 1 lambda by alpha minus 1 is chosen already and therefore, what you have is 1 minus 2 y square ok which istimes 2 factorial times 2 factorial ok which gives you 4 y square minus 2 H 3 of y is given by minus 1 2 times 3 factorial into y minus 2 third y cube which gives you there is of course nothing else which gives you 8 y cube minus 12 y ok. So, you can obtain a whole series of them and what is interesting is that H naught H 2 let me write H 4 and also H 5 I will stop beyond H 5 H 4 is 12 minus 48 y square plus 16 y to the power 4 and H 5 of y is 120 y minus 160 y cube plus 32 y to the power 5 ok and so on. So, what we have therefore, iseven an odd in y H naught of y even in y plus or minus y is the same H 2 of y 4 y square minus 2. So, H 2 of minus y is the same as H 2 of y it is an even function H 4 of y 12 minus 48 y square plus 16 y to the power 4 H 4 of minus y is also the same thing therefore, all the even ends naught 2 4 6 etcetera are all even functions in y all the odd ones H 1 of y is 2 y therefore, H 1 of minus y is minus 2 y therefore, H of y is minus of H of minus y which is an odd function. So, this is an odd function this is an odd function 8 y cube minus 12 y 5 120 y 160 y cube 32 y 5 it is an odd function therefore, all the odd ends contain polynomials which are odd functions of y what about exponential of minus y square that is y square by 2 that is an even function. Therefore, you see that the Hermite polynomials give rise to wave function solutions for the harmonic oscillators with odd and even parity alternating between the ends corresponding to odd n or corresponding to the even n. So, what is the final result? The final result for the harmonic oscillator wave function is that psi naught of y is e to the minus y square by 2 psi 1 of y is e to the minus y square by 2 times 2 y psi 2 of y is e to the minus y square by 2 of whatever we had written already 4 y square by 2 now you can see so on ok. So, a table of these harmonic oscillator polynomials are given in the lecture set and I would request those of you who are not certain about the solution to substitute the solutions in the differential equation and see that the answer that you get is 0 by choosing the lambda by alpha minus 1 corresponding to that n value you will know that ok. So, this is the approach here has been if you recall the last lecture looking at the Schrodinger equation we looked at what is called an asymptotic solution or the solution for very large values of the argument the variable x, x was changed to a dimensionless variable y because you remember x was the position variable that is the amplitude of the oscillation. So, we removed the physical parameters the dimensions out of that by redefining the x in terms of y and then looking at what are called very large solution very large values of y solutions which gave us the exponential minus y square by 2 as the asymptotically applicable solution and then with that we wanted an arbitrary solution that is a solution valid for all arbitrary y values y small y large etcetera. Therefore, we multiplied the asymptotic solution with a functional form called h of y which was assumed to be an infinite series and then we use the convergence argument that this function cannot forever be an infinite series or ever increase for all values of y we had to truncate that in order to have a meaningful solution and the truncation led to us a whole series of solutions depending on when we truncate what n value what integer value we truncate and therefore, you see that the harmonic oscillator in principle provides an infinite number of solutions because there is nothing here which tells me that I can I need to stop at some point if I stop at that point that is the value, but there are depending on where I truncate the n I have that many different size solutions. So, the solution of the Schrodinger equation for the harmonic oscillator problem using the mathematical properties of the differential equations and convergence immediately gives us an infinite number of solutions, but you know that a harmonic oscillator approximation is not really a valid approximation for very large extensions of or very large values of amplitudes. So, these are theoretical, but in practice when we consider very large values of extensions we will have to worry about unharmonicity and so on otherwise these are exact solutions obtained for the half k x square potential energy problem or the harmonic oscillator energy problem in one dimension ok. Now, there are a few mathematical properties that I would like to write down here, but let us pass for a few seconds and then we will come back to writing some of these properties ok. So, I shall give the value of n in a minute, but we will also now see that each n each psi n corresponds to a given energy E. Remember when we set lambda by alpha minus 1 is equal to 2 n for any given value of n 2 into 2 n we put in, but that was for the coefficient a to the 2 n and therefore, for any Hermite polynomial of order n it was truncated at that ok. So, if we do that lambda please remember recall from the previous lecture lambda is 2 m e mass of the harmonic oscillator by h bar squared and alpha we have just now written that alpha is k alpha is equal to square root of k m by h bar square or square root k m by h bar. So, if we substitute this lambda by alpha is equal to 2 n plus 1 then what you have is 2 m e by h bar square times alpha is 1 by alpha is h bar by square root of k by m and that should be equal to 2 n plus 1 ok. So, let us cancel out these things and the m becomes square root m. So, the answer goes over to e is equal to h bar into square root of k by m times n plus a half taking the 2 to cancel out and this is h times 1 by 2 pi into square root of k by m times n plus a half and you know that that is the harmonic oscillator frequency nu and therefore, the energy of the harmonic oscillator is h nu times n plus a half where n is the index associated with the wave function psi n ok. So, we have an n plus a half that is very interesting because that is going to give us the idea of what is known as the 0 point energy later on, but this has to be kept in mind. Therefore, psi n is associated with a given energy E n also indexed by the same n as you have it here. Now, what about the normalization constant? Mathematically one can work this out, but I am only going to write the final answer it is easy to check these things and verify text books give you these things over and over again. So, what you have is n is pi by alpha 1 by 4 2 to the n by 2 n factorial 2. 1 half what exactly is this? This is the normalization constant such that the psi of n of x psi of n of x d x now is equal to 1 minus infinity 2. So, we have to write this psi n of x from whatever is given here this n square and we have write polynomial we have to write n by this expression and now we can write therefore, the wave functions psi naught of x is going to be alpha by pi because it is how do I define this yeah. So, I should define this as 1 by n here ok and if I do that then what do I get psi n 1 by n squared psi n of x psi n of x d x is 1 and then we can see that that is given by the corresponding expression. So, and this n is what I have here pi over alpha and so on. So, psi naught will be alpha by pi because it is 1 over and it is to the 1 by 4 e to the minus alpha x square by 2 ok psi 1 of x is given by 1 by root 2 alpha by pi to the 1 by 4 e to the minus alpha x square by 2 h 1 of root alpha x ok psi 2 psi 2 of x normalized is given by 1 by 2 root 2 alpha by pi to the 1 by 4 e to the minus alpha x square by 2 h 2 of root alpha x and so on. What is what is important is that 1 psi is an exponential of that is a Gaussian function to indexed by the same n by the harmonic oscillator and 3 that the function the argument of the harmonic oscillator is dimensionless and 4 if you remember the integral of psi star psi d x is equal to 1 tells you that the dimension of psi has to be 1 by square root of length because d x is length and you can see that it is alpha to the 1 by 4 alpha is l square dimension 1 by l square dimension alpha is 1 by l square dimension therefore, alpha to the 1 by 4 is 1 by root l. So, these pointers should be kept in mind that alpha is in the numerator and alpha x square come as a as a product or root alpha x come as a quantity so that we do not write functions which have difficulty I mean you cannot interpret those functions. So, those are things of course, you have to worry about these constants because you require the wave functions to be such that they satisfy the probability distribution, but other than that you can see that everywhere it is alpha to the 1 4th alpha to the 1 4th alpha to the 1 4th and so on. So, you can see that it is 1 by square root of the length which is a one dimensional problem because 1 by square root length 1 by square root length times length dimensionally it is a number it is no no dimensions ok. So, these are important these are things which one has to keep in mind when one is doing this for the first time. After some time of course, physicists have a very nice way of doing these things by removing all these dimensions from the process right in the beginning and writing what are known as reduced variables or dimensionless variables or redefined constants and so on, but you know we are not physicists and what we want to do is to understand the problem of molecular parameters to be interpreted properly. So, it is important to keep track of all these things. Now, in the given a given that I have another 10 to 12 minutes left let me summarize some of the properties of the Hermione polynomials I am just going to give you this for you to verify, I do not intend deriving these properties here ok. In the same way I did not tell you how we get the normalization concern there are details of calculations which we can always omit and then there are other things which we have to remember and keep in mind ok. So, let me just write down an extremely important relation that the Hermite polynomials satisfy it is called the Hermite polynomial recurrence relations recursion sorry it is not recurrence relation it is a recursion formula that is they satisfy this formula namely any Hermite polynomial n plus 1 order of y is connected to the Hermite polynomial of order n and the Hermite polynomial of order n minus 1 y is equal to 0 ok. This is very important it basically it is also a very it is it is worth remembering this relation even though I do not remember it now I just copied it from the paper, but it is worth remembering at least in the first few years and manipulate with them to know that all these polynomials the Hermite polynomials and the others that we would study soon like the log air polynomials or the spherical harmonics or the Legendre polynomial all of them have this beautiful what are called recursion relations and also they have very fundamental properties they are all called orthogonal polynomials in that the Hermite polynomials corresponding to different orders or orthogonal to each other with a weight function which we call as the exponential function and so on. So, this is worth remembering and all that you have to remember is this relation and the fact that H naught of y is equal to 1 and H 1 of y is equal to 2 y ok. If you remember these two and this in principle you can get H 2 n plus 1 ok n is 1 from n minus 1 which is 0 H 0 and n which is H 1. Therefore, if you want to get H 2 of y you are using n is equal to 1 1 here ok then minus 2 y H 1 of y plus since it is n is 1 it is 2 H 0 of y is equal to 0 and this we already know is 1 ok. Therefore, it is 2 and we know that H 1 of y is 2 y 2 y ok and therefore, this whole thing being 0 implies that H 2 of y is 4 y squared minus 2 which was written down a few minutes ago as rising from the solution of the differential equation yes and in the same way you can immediately write H 3 y also like n equal to 2. So, if you do that H 3 y is 2 y times H 2 y ok minus 2 into 2 H 1 y because n is 2 and you can see that immediately that it is 2 y into 4 y squared minus 2 minus 4 into 2 y. So, that gives you 8 y cube minus 12 y ok. So, this is also something that we wrote down ok. So, you can construct all these polynomials just from the first and the second one using this recursion formula is one of the things which is worth remembering ok. Another way of remembering the generation of Hermite polynomials is through what is known as a generating function formula ok which let me write down ok. The H n of y can be generated by this particular form minus 1 this is n minus 1 to the power n e to the y square times d n by d y n on e to the minus y square ok this is another generating function. So, if you want H 0 of y obviously, it is 0 there is no derivative and minus 1 raise to 0 is 1 e to the y square the derivative is since it is 0 it is 1 e to the y square times e to the minus y square is 1. H 1 of y is easy to write down again H 1 of y is minus 1 e to the y square d by d y of e to the minus y square which you know is minus 2 y times e to the minus y square and this minus and this minus go away and e to the minus y square and y square cancel you get 2 y ok. So, you can generate the whole Hermite polynomials also through successive differentiation of an exponential of a Gaussian function and then multiplying by the inverse of the Gaussian function and a sin factor ok. So, this is another interesting formula a lot of these things have been studied for several at least 250 years or more because the Hermites equation is a fairly old equation studied by Hermite and orthogonal polynomials are very well known in mathematics they are there are very rich discussions on the second order differential equations and so on. So, I would not elaborate this beyond this point our objective was to convince ourselves that we know at least with some degree of confidence we can write down the Hermite polynomials by solving the actual differential equation of the Hermite equation or the originally the Schrodinger equation for the harmonic oscillator with the help of the harmonic potential. And we have seen that the wave functions are ah given by these functional forms yes one last expression that I must write down which is also very very important is that these functions are orthogonal Hermite polynomials are orthogonal ok that is integral minus infinity to plus infinity psi n of x star psi m of x the star is actually a redundant quantity here because these are all real functions ok. But in quantum mechanics we always have to remember this because very often if we forget the star it comes out at the most innocuous place as something that we have forgotten and it gives you all kinds of nasty answers. So, keep the star always as star even if you do not use it psi n star of x psi m of x dx is delta n m the normalized for normalized ah harmonic oscillator wave functions. So, that is ah orthonormal and the corresponding expression that you will have is that the n n n m the normalization constants associated with the Hermite polynomials and minus infinity to plus infinity psi n star is e to the minus alpha x square by 2 and of course, there is this also gives you that. So, it is e to the minus alpha x square h n of root alpha x h m of root alpha x dx and this integral is delta n m. So, this is the orthogonality orthogonality of the Hermite polynomials justifying the term that these are orthogonal polynomials when the n and m are the same that is going to be 1 given these normalization constants and this is known as the weight factor in numerical calculations numerical integrals ok. The weight factor weights the Hermite polynomials for different values of x according to the e to the minus alpha x square ok. Alpha is a constant associated with the problem the moment you choose a harmonic oscillator alpha is fixed the x goes from minus infinity to plus infinity the exponential alpha minus alpha x square contributes different values to different Hermite value the the function at different points of x and therefore, if you have to do this integral numerically this ah function comes in as a weight function and a numerical method which ah calculates integrals using this kind of Hermite functions is also often referred to as a ah Hermite quadrature you will not do that here, but when you do numerical work involving these integrals ah it is important to remember that that is this quantity ok. The exponential minus alpha x square acts as providing the right amount of all it is all it does is that for very large values of x it contributes so little that the Hermite polynomials you can pretty much neglect for large values of x exponential of minus alpha x square is almost 0 almost close to 0 therefore, the product is not contributed so much this kills them. So, if you have to do this integral numerically for finite values you are very happy that you have something like this ah the only thing that you want to make sure is that the alpha is reasonably big and therefore, you do not have to calculate the numerical values of these integrals for very large ranges of x which is important when you do computer calculations ok. These are some things which you have to keep in mind and in the next lecture what we would do is to look at the functional form more carefully using graphics and also ah the consequences of the Hermite polynomial and the harmonic oscillator wave function being ah everywhere that is except for some finite values of x where the polynomials go to 0 ah the Hermite polynomial there are the roots of the polynomial we call it you know after all it is a it is a power series in x finite power series and therefore, it has for each value of n it has n roots and apart from these n roots the polynomial is never 0 and therefore, the wave function is never 0 in the real physical space all the way from minus infinity to plus infinity except for a finite number of points and that leads to a very interesting consequence known as the tunneling or things that we were not very familiar with things which do not exist in classical mechanics and the half ah factor in the energy is another interesting phenomenon which leads to what is known as the 0 point energy therefore, these are consequences that as ah chemists and as spectroscopies we will have to remember and these in the next lecture we will continue until then thank you very much.