 Hello, so we continue with our discussion of Hilbert spaces. We want a basis for L2 or the real line with the usual Lebesgue measure dx. And we are going to see that the Hermite functions furnishes with a complete orthogonal system. So, what are the Hermite functions? Before we define the Hermite functions, we needed the Hermite polynomials. In the last part of the last capsule, we started out with this Hermite differential equation and we found that they have polynomial solutions and these polynomials are called Hermite polynomials. Unlike the Legendre polynomials, the Hermite polynomials do not have any standardized normalization conditions. So, there is no universally accepted convention for normalizing them. So, let us call these Hermite polynomials f lambda x. For the time being we used the f lambda x as the notation, we will change it later. The polynomials f lambda x when lambda is 0, 1, 2, 3, these are solutions of the Hermite differential equations which are polynomial solutions when lambda is a non-negative integers. They are orthogonal in the following sense. They are orthogonal with respect to the measure e to the power minus x squared dx. Of course, f lambda x is a polynomial. A polynomial is not in L2 with respect to the Lebesgue measure, but it is in L2 with respect to this measure e to the power minus x squared dx. So, orthogonality means integral over r f m x f n x e to the power minus x squared dx is 0 if m is not equal to n. Simply take the e to the power minus x squared and write it as a product e to the power minus x squared by 2 into e to the power minus x squared by 2. Give 1 e to the power minus x squared by 2 to f and the other one to the other f. So, define h n x equal to f n x into e to the power minus x squared by 2, n equal to 1, 2, 3, etcetera. These are the Hermite functions. A Hermite function is a certain polynomial called Hermite polynomial multiplied by e to the power minus x squared by 2. These Hermite functions will form an orthogonal system of functions in L2 of r. So, let us dispose of the proof of this theorem. So, first of all, we recall the differential equation. This f lambda x, the polynomial satisfies this differential equation y double prime minus 2 x y prime plus 2 lambda y equal to 0. We must first convert it into self adjoint form. I already explained to you the self adjoint form for the Legendres equation and for the Bessel's equation. For the Hermites equation to convert it into self adjoint form, simply multiply by e to the power minus x squared. So, the first two terms after multiplying by e to the power minus x squared gives you d dx of e to the power minus x squared y prime. And the other term is 2 lambda y to the power minus x squared equal to 0. And so now what you do is that, when I take lambda equal to m, I know that f m is a solution. When I take lambda equal to n, I know that f n is a solution. So, I write down those two equations, d dx of e to the power minus x squared, f m prime plus 2 m f m e to the power minus x squared is 0, d dx of e e to the power minus x squared fn prime plus 2n fn e to the power minus x squared equal to 0. Again multiply the first by fn, second by fm, integrate by parts over the whole real line and subtract. So, multiply the first by fm and integrate by parts, the boundary term 0 will be 0 because e to the power minus x squared decays very, very fast and so polynomial growth of fm or fn is not going to control the decay. So, the boundary terms drop out. So, other term that you get integrate by parts is minus integral over the real line e to the power minus x squared fm prime fn prime and this term will appear from the both the equations and they will drop out when you subtract. So, what do you get? You get that 2m integral fm fn e to the power minus x squared is 0 to n integral fm fn e to the power minus x squared is 0 when you subtract you are going to get m minus n times integral over the real numbers e to the power minus x squared into fmx fnx 0 which is exactly what is stated here. This completes the proof of orthogonality of the hermit polynomials with respect to this measure e to the power minus x squared dx. Now, we want an explicit formula for these hermit functions just as we had the Roderick's formula for the Legendre polynomials we are going to get a Roderick's type formula for the hermit function. So, now let us look at this expression 7.7 you take e to the power minus x squared you take the nth derivative dn is going to be a polynomial times e to the power minus x squared and multiply by e to the power plus x squared and so the exponential factors will cancel out. So, the right hand side of 7.7 is obviously a polynomial in x that polynomial let us call it qnx it is a polynomial of degree exactly n. Now, I am going to show that these polynomials are going to be orthogonal with respect to the measure e to the power minus x squared dx. In other words if you assume m is less than n we have to show that integral over r qnx qmx e to the power minus x squared is going to be 0 that is what we have to show. Okay, so what is this expression substitute from 7.7 qnx and put it in the integral e to the power plus x squared and e to the power minus x squared cancelled out the nth derivative of e to the power minus x squared remained and qmx remained and I am going to keep on integrating my parts and throw the derivatives on to the qm factor but every time I do integration by part there will be boundary terms but as I explained before these boundary terms will cancel out and so there is no need to worry about the boundary terms eventually all the derivatives would have been transferred to the qm factor we would have got integral nth derivative of qmx e to the power minus x squared but qm is a polynomial of what degree? Degree m and m is less than n and so differentiating qmx n times will kill this completely this is going to be 0 and the orthogonality of these polynomials qn's has been established. So now we are in the following situation v is a vector space of all polynomials what is the inner product of the vector space integral over r fx into gx into e to the power minus x squared dx. So in this vector space we have got two sets of polynomials q0, q1, q2 that we have just described and f0, f1, f2 those were the provisional notations for the Hermite polynomials. And now observe that q0, q1, q2 are polynomials of degrees 0, 1, 2 etc f0, f1, f2 are also polynomials of degrees 0, 1, 2 etc. So the linear span of q0, q1, q2 up to qn is going to be exactly set of all polynomials of degree less than or equal to n and the same is true for the linear span of f0, f1, f2 up to fn. So linear span of the first n plus 1 q's is the same as the linear span of the first n plus 1 f's and this is true for every n and so by the fundamental orthogonality lemma there exists a sequence of constants cn such that fnx equal to cn times qnx for every n. So there is equation 7.8 that I had displayed on the slide. A briefly recall the fundamental orthogonality lemma I recalled for you that if v is a vector space endowed in a product with respect to which you have two sets of orthogonal systems of non-zero vectors such that the linear span of the first k plus 1 of the v's is the same as the linear span of the first k plus 1 of the w's for every k. Then there are scalars ck such that vk is ck times wk. So now except for determining the constants cn's we are practically found the Hermite polynomials. What are the Hermite polynomials? It is a constant multiple of qn. The nth Hermite polynomial is a constant multiple of qn namely e to the power x squared nth derivative of e to the power minus x squared. So now we can select the constant cn's in such a way that after normalization I get a unit vector. So since we are in a Hilbert space we could try to normalize it to be unit vectors in the absence of any other normalization condition one condition is as good as any other condition. So if you demand that inner product of fn with itself is 1 that means fn is cn times qn so mod cn squared integral minus infinity infinity qn qn e to the power minus x squared is 1. Again put the definition of qn again put the definition of qn and we can calculate the value of cn. Remember what is qn e to the power x squared that will cancel with this e to the power minus x squared and the nth derivative of e to the power minus x squared I will transfer the nth derivative on the other qn's and I will do the job. So here is the calculation qn is this and this dn minus 1 apply to minus 2x e to the power minus x squared. Now I will apply Leibniz rule and I am interested in the leading term not the lower order terms. So when n minus 1 derivatives have to be applied to this the leading term will be obtained when all derivatives fall on e to the power minus x squared and if you keep doing it successively I will see that the leading term in qnx is minus 2x to the power n plus lower order terms L ot stands for lower order terms. So from this we infer that if I differentiate qnx n times I am going to get minus 2 to the power n into n factorial. Why are we interested in this thing because now we want to calculate the inner product qn qn e to the power minus x squared which is basically qn times the nth derivative e to the power minus x squared integrate by parts n times the boundary terms cancel out all derivatives will fall on the qn you will get minus 1 to the power n each time you integrate by parts you pick a minus sign integral minus infinity to infinity nth derivative of qn times e to the power minus x squared dx. But nth derivative of qn we just computed there was a minus 1 to the power n that goes away 2 to the power n n factorial times root pi and so this is the normalizing factor and now I would divide by this normalizing factor and I will get the normalized Hermite polynomial 7.9 namely e to the power x squared nth derivative of e to the power minus x squared upon 2 to the power n by 2 pi to the power 1 4th square root of n factorial. But this normalization is usually not used it can be used for calculations when you want to compute Fourier series and Fourier coefficients when you work with this complete orthonormal system but otherwise this normalization is usually not used. Now we call this e to the power x squared nth derivative e to the power minus x squared is qn and that is a constant multiple of fn. Now we put in this minus 1 to the power n and this is what we are going to call as hnx. So, for us the nth Hermite polynomial is going to be the expression given by 7.10. So, family of Hermite functions are you take this Hermite polynomials hnx multiplied by e to the power minus x squared by 2 this will form a complete orthonormal basis for L2 of the real line that is the next important theorem, theorem number 75. The completeness assertion will be proved later we cannot prove it in this particular capsule at least we cannot complete it in this capsule. These functions are extremely important in quantum mechanics and the study of the Schrodinger operator and the literature on these functions is very vast and you also see multi-dimensional analogues of these things all over the place. We are also indicated in chapter 4 that these are the eigen functions of the Fourier transform operator. Remember that the Fourier transform is a operator on L2 it is almost a unitary operator except for scaling. If you rescale it you will get a unitary operator and the Fourier transform has four eigen values and each eigen value has infinite multiplicity and these Hermite functions are the corresponding eigen vectors and it gives you a complete orthonormal basis. So, let us recapitulate these ideas quickly. So, what is the basic thing? The basic thing is the Hermites differential equation y double prime minus 2 x y prime plus 2 lambda y equal to 0 lambda is an integer n you are going to get the solution h n x. Now, what happens is that this differential equation transforms to u double prime minus x squared u plus 2 lambda plus 1 u equal to 0 where u and y are related by this equation y e to the power minus x squared by 2 equal to u. And this is a more convenient differential equation to work with why because this differential equation in u has no u prime term. So, the Abel level formula which you study in elementary differential equations will immediately give you that if you take two linearly independent solutions of this and the Ronskin is going to be a non-zero constant. Of course, if you take two linearly dependent solution the Ronskin will be easy of course. The other feature is that this differential equation is invariant under Fourier transform. When you take the Fourier transform of u double prime you are going to get minus chi squared u and when you take the Fourier transform of minus x squared u you are going to get the second derivative of the Fourier transform and this of course is a constant. So, you will simply get the Fourier transform u hat will fall on this. So, if you take this differential equation and you take the hat you get the same differential equation. It means that if you have a solution of this differential equation which is rapidly decreasing that is in the Schwarz class then u hat will also be a solution of the same differential equation and that will also be in the Schwarz class. Now, we know that the Hermite function h n x is a solution of this differential equation. In chapter 4 when we were discussing this we were not sure about this, but now we know completely what the solutions are that lie in the Schwarz class. So, you can think of this discussion as completing the discussion that we started out in chapter 4. So, we have an example h n x equal to e to the power minus x squared by 2 capital H n x that is a solution of the differential equation the Schwarz class and it is Fourier transform h n hat little h n hat is also a solution of the same differential equation. But now if you take two solutions which are the Schwarz class and I take the Ronskian the Ronskian first of all is going to be constant because u prime does not appear. But this constant is going to be 0 because as x tends to infinity I am going to get 0 in the limit. So, h n and h n hat little h n and little h n hat must be linearly dependent because they are both solutions of the same differential equation and both are in the Schwarz class. So, h n hat must be a multiple of h n a nonzero multiple in fact. So, h n is an eigen function of the Fourier transform with lambda n as the corresponding eigen value. What are the eigen values that is the next important question that I leave it to you to ponder over. It is not difficult to guess what are the eigen values of the Fourier transform remember that if I take the Fourier transform and if I take the Fourier transform again I am going to get 2 pi times the reflection map. So, using that you can figure out what the eigen values are is going to be root 2 pi minus root 2 pi i root 2 pi and minus i root 2 pi. So, now some exercises prove the following three term recurrence relation for the sequence of Hermite polynomials h n x namely h n x is what minus 1 to the power n e to the power x squared nth derivative e to the power minus x squared prove that h n plus 1 x minus 2 x h n x plus 2 n h n minus 1 x equal to 0. How to do this 7.11 you must go back to chapter 5 and see how did we prove the three term recursion formula for the Legendre polynomials. So, look at these three terms and this particular term 2 x h n x that is not a Hermite polynomial it is a Hermite polynomial multiplied with x. So, this middle term is not a Hermite polynomial but it is a polynomial of degree n plus 1 and because of polynomial of degree n plus 1 it is going to be a linear combination of h naught h 1 h 2 h n plus 1. So, let us write it as a linear combination summation j from 0 to n plus 1 c j h j x and now we need to determine the c j's exactly as in the Legendre's case when j is less than or equal to n minus 2. So, for 0 1 2 all the way to n minus 2 the c j's are going to be 0. So, what is going to be left over is that this minus 2 x h n x is going to be c n plus 1 h n plus 1 plus c n h n plus c n minus 1 h n minus 1 these are the only things that are going to survive. Again parity arguments h n x is odd function if n is odd h n x is an even function if n is even. So, x h n x and h n x have opposite parity. So, c n will be 0. So, minus 2 x h n x will be simply a linear combination of two terms c n plus 1 h n plus 1 plus c n minus 1 h n minus 1 you have to calculate c n plus 1 and c n minus 1 you have to multiply by h n plus 1 and integrate and finish the problem. In the case of the Legendre polynomials we had the added advantage that p n of 1 is 1 unfortunately we do not have that simplicity here. But if we go back to the previous slides you will be able to figure out what are the leading coefficient of the Hermite polynomials and you can use that to simplify your calculations. But whatever method you use for calculations you will get equations 7.1 mark. The second exercise show that the zeros of h n x are real distinct remember that we have a Roderick's type formula you can use the Roderick's type formula and there are other ways to prove this. So, usually when you have a three term recursion formula the three term recursion formula helps us to do these things. We will not discuss this part and the interlacing of zeros show that the zeros of h n x and the zeros of h n plus 1 x interlace again a characteristic feature of these orthogonal systems of polynomials. The important exercise for us is to find the generating function for the h n x or the exponential generating function as it is called. The equation 7.12 basically gives you the exponential generating function for the sequence of Hermite polynomials. Multiply by e to the power minus x squared and you get exactly the exponential generating function. I put the e to the power minus x squared here so that this equation 7.12 is going to be used very crucially for proving completeness in the next camps you. How do I prove 7.12? Instead of proving 7.12 multiply by e to the power x squared and this e to the power minus x squared goes away from the right hand side and the left hand side simply becomes e to the power 2 tx minus t squared instead of t I am using z and I am going to think of x of 2 zx minus z squared as a entire function as a holomorphic function in z and because a holomorphic function in z I am going to write it as a power series b0 plus b1 z plus b2 z squared plus da da da where the b0 b1 b2 are functions of x. Directly by expanding the exponential check that b0 of x is 1 which is the same as h0 of x check that b1 of x and h1 of x agree. So, 1 factorial b1 and 0 factorial b0 they agree with h1 and h0 respectively. Now we find a three term recursion relation for these bj's and compare it to the three term recurrence relation for the h n's and they will more or less agree and you will be able to get the relationship that you want. You want to show that bnx is hnx upon n factorial that is what we want to show. Let us do a simple calculus exercise. Let us differentiate e to the power 2xz minus z squared it is 2x minus 2z e to the power 2xz minus z squared differentiate this again n times apply d to the power n to both sides where capital D stands for differentiating with respect to z and you put in z equal to 0. So, when you take a power series and you differentiate n times and put z equal to 0 you get n factorial times the nth coefficient. So, you get the left hand side you will get dn plus 1 the n plus 1st derivative and you are going to put z equal to 0. So, you are going to get n plus 1 factorial bn plus 1x and then the middle term you are going to apply the Leibniz rule. So, all derivatives falling on here. So, n derivatives falling here and you are going to put z equal to 0. So, n factorial times bn I am going to put z equal to 0 remember. So, what am I going to get? I am going to get n factorial 2x bnx and then what else the third term n minus 1 derivatives here and 1 derivative here, but they are going to be a binomial coefficient n choose 1. One differentiation here means a minus 2 and n minus 1 differentiation here and z equal to 0. So, I am going to get the n minus 1st coefficient and I am going to get a n minus 1 factorial times bn minus 1. So, do a little rearrangement and you will get the simplified version n plus 1 bn plus 1 minus 2x bnx plus 2 bn minus 1 is 0. This is different from the three term recursion formula for the Hermite polynomials of course it is going to be different because the relationship is not bn equal to hn rather it is n factorial bn equal to hn. So, use induction we are checked that n factorial bn equal to hn for n equal to 0 and n equal to 1. Now, use induction to prove that this is true for all n's. You need the three term recursion formula for the b's and the three term recursion formula for the h capital hn's and comparing the two you establish this. Therewith we have proved this fundamental equation 7.2 the exponential generating series for the sequence of Hermite polynomials. In the next capsule we are going to critically use 7.2 to prove that the Hermite functions form a complete orthogonal system for L2 of r. I think it is a very good place to stop this capsule here. Thank you very much.