The Hermite Polynomials
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In the essay Solving Hermite’s Equation I obtained a formula that generates the solutions of that equation, the Hermite polynomials. Rodrigues’ formula says that
in which dx represents the derivative operator
Now I want to look for alternative ways to generate the Hermite polynomials, ways perhaps easier and more convenient than carrying out multiple differentiations.
Start by calculating the first derivative of Equation 1;
Rearranging the terms in that equation produces
That equation states that anyone can generate the (n+1)-st Hermite polynomial by taking the n-th Hermite polynomial, multiplying it by 2x, and subtraction its first derivative from that product. Starting with H0(x)=1, that procedure yields
Using the recursion process anyone can generate all of the Hermite polynomials that they want. But if someone wants an Hermite polynomial that bears a large index number and few of the Hermite polynomials with smaller index numbers, that person wants something more efficient, something that does not oblige them to generate all of the Hermite polynomials preceding the one they want; specifically, they want some procedure that would enable them to calculate the desired Hermite polynomial directly. An inspection of Equations 5 reveals that the desired procedure consists of calculating a power series and the observation that the n-th polynomial in that set begins with the n-th power of 2x implies a series in powers of 2x, so the n-th Hermite polynomial becomes
To determine the form of the coefficients ak, substitute that equation into Hermite’s Equation to get
In that equation we must haveλ=n, an integer equal to the index on the polynomial, to ensure that this derivation yields one pure Hermite polynomial (λ=n+½ only applies to the solutions of the Schrödinger version of Hermite’s Equation: multiplying the solutions of Schrödinger’s version by exp(x2) to convert them into the solutions of Hermite’s Equation removes the ½ term). Because the process of differentiation commutes with the process of addition, the derivative operators in Hermite’s Equation pass from left to right across the summation symbol without changing it.
The derivative operators reduce the first term in the sum by two powers of 2x. At the lower end of the sum that reduction causes no problem because the k=0 components of the sum represent constants, which differentiation zeros out; thus, the sum contains no terms corresponding to k<0. At the upper end of the sum two terms seem to have gone missing, the ones with coefficients an+1 and an+2. But because the n-th power of 2x stands as the highest term in the series, those two coefficients must, necessarily, each equal zero, so we can add to Equation 7 the components that include them without changing the value of the equation. Equation 7 then becomes
For each value of k the sum of terms must satisfy Equation 8 separately, which fact leads to the statement that
As inferred above, when k=n, an+2=0. That equation also displays the fact that each coefficient comes out as a multiple of its second predecessor. But Equation 9 yields only a ratio of neighboring coefficients: a proper representation of the power series requires a formula for calculating each coefficient directly. That formula comes from compounding the formula of Equation 9.
First, I want to confirm the statement that each coefficient in the power series stands as a multiple of its second predecessor. The recursion process for generating the Hermite polynomials begins with H0(x)=1 and generates each subsequent Hermite polynomial by multiplying the previous polynomial by 2x and then subtracting the first derivative of that previous polynomial from the product: Hn+1=2xHn-dHn/dx. That procedure raises each power of x by one and also lowers it by one, so the procedure consistently produces a series whose terms lie two powers of x from their immediate neighbors. Thus, if one term in a given Hermite polynomial contains an odd power of x, then all of the terms in that polynomial contain odd powers of x; if one term in the given polynomial contains an even power of x, then all of the terms in that polynomial contain even powers of x.
Starting with an=1, I can calculate the value of an-2, then from that coefficient I can calculate the value of an-4, and so on until I have calculated the complete set of the coefficients that multiply the powers of 2x in the n-th Hermite polynomial. To that end I define m=k+2 and rewrite Equation 9 as
Using that equation, I calculate
for all values of p from zero to [n/2] (the symbol [n/2] designates the largest integer equal to or less than n/2). Inverting Equation 10 yields
which I can use to rewrite Equation 11 as
in which Cy represents the coefficient on ay and Cn=1. Thus I get
With that formula I can now write out explicitly the power series for the n-th Hermite polynomial as
If I want to calculate Hn(0) (and I do for what comes next), then I only get a nonzero result for the case n-2p=0. I thus get two possibilities;
with q representing the maximum value of p in Equation 15. I thus obtain the set that consists of all of the terms, in all of the Hermite polynomials, that do not contain x as a factor.
That fact stands as an important component in the production of the Taylor series expansion of the function exp(-x2). Taylor’s theorem states that for some function f(x) the relationship
stands true to mathematics. Rearranging Equation 1 yields
Making the appropriate substitution from that equation into Equation 17 yields
Equation 16 encodes the fact that only even-numbered values of n yield a non-zero term in that sum, so, taking n=2q as above, I rewrite Equation 19 as
Because all of the terms in an odd-numbered value of n drop out of the sum, only the terms involving an integer value of q survive, so I changed the limits on the sum from n to q to reflect that fact. But that equation yields the same result that would have come from substituting y=x2 into the standard series representation of the negative exponential function,
The result encoded in Equation 20 thus seems counterintuitive, because the derivatives of exp(-y) don’t have the same algebraic form as do the derivatives of exp(-x2), so we would not expect the two Taylor series to have identical forms.
I feign to calculate the n-th Hermite polynomial at a point a distance t in the negative direction from the chosen point x and then let t go to zero. By Rodrigues’ formula that calculation comes out as
In setting up that calculation I did not modify the argument of the exponential on the left side of the n-fold differentiation operation, because that exponential does not form part of the solution of Hermite’s Equation, but merely serves as a conversion factor that transforms the solutions of Schrödinger’s version of Hermite’s Equation into the solutions of the proper Hermite’s Equation. For a function of the difference between two variables it stands true to mathematics that
so Equation 22 becomes
Multiplying that equation by tn/n! and then summing over all the values of n from zero toward infinity yields
The expression on the right side of that equation displays the Taylor series expansion of the exponential, so I can express the equation more simply as
Mathematicians refer to that exponential function as the generating function of the Hermite polynomials. That name encodes the fact that anyone can use Equation 26 to generate the n-th Hermite polynomial: they need only differentiate the equation n times and let t=0. Applying that procedure to the left side of the equation leaves only the designation, Hn(x), standing by itself and applying the procedure to the right side of the equation yields the Hermite polynomial. Verification of that statement follows from the fact that applying the process of differentiation to the exponential in that equation yields the exponential multiplied by some function of x and when t=0 that function becomes the Hermite polynomial. Each differentiation, after t goes to zero, multiplies that function by 2x and then adds the first derivative of the function itself to that product, thereby recapitulating the recursion process described at the top of this essay.
Rodrigues’ formula provides another way to create a description of the Hermite polynomials. So far I have used only real numbers in my calculations, but the line representing the set of the real numbers bisects the complex plane. That fact means that, if I can contrive the necessary singularities, I can use Cauchy’s Integral Formula, which represents derivatives as complex contour integrals. In the case of the Hermite polynomials Cauchy would rewrite Rodrigues’ formula as
That equation displays the instruction to calculate the seed exponential at a point z and then contrive a singularity at the point x by dividing the exponential by the appropriate power of z-x. To exploit Cauchy’s Integral Theorem integrate around a contour that makes a tight circle around the singularity. Making the replacement z=x-t with t=Rexp(iθ), which represents that small circle, yields
I can then use the calculus of residues to evaluate that integral; however, that calculation would take me too deep into complex analysis for this essay, so I will leave the evaluation to the interested reader.
However, I can still extract from that equation one more useful fact about Hermite polynomials. Because the process of differentiation commutes with the process of integration, calculating the first derivative of Equation 28 with respect to x yields
That result can replace the derivative in the recursion process for generating the Hermite polynomials and convert that process into only multiplications.
As implied above, there exist two different sets of Hermite polynomials. The standard Hermite polynomials, Hn(x), solve the proper Hermite’s Equation. And the Hermite functions solve Schrödinger’s version of Hermite’s Equation. The normalized Hermite functions come out as
which functions solve the Schrödinger-Hermite Equation in the form
The Hermite functions conform to the statement that
in which the Kronecker delta takes the value one when m=n and takes the value zero otherwise. To prove and verify that statement, start by substituting Equation 30 into Equation 32 and then replace one of the Hermite polynomials with its Rodrigues’ formula representation;
For convenience I have defined the constant
Also, conveniently enough, exp(-x2) and the exp(x2) from Hn(x) cancel each other out. Integrating Equation 33 by parts yields
The second term on the right side of that equation vanished because the function
goes to zero in the limit as the value of x goes toward plus or minus infinity. Proof and verification of that statement come from the twin facts that p(n,x) represents an n-th order polynomial in x and that exp(x2) represents, by way of its Taylor series expansion, an infinite-order polynomial in x. Thus, there will always exist some finite value of x for which the value of the exponential will exceed the value of the polynomial. That fact necessitates that the ratio
in the limit as x goes toward positive or negative infinity.
Each integration by parts shifts the derivative operator from acting on exp(-x2) to acting on Hm(x) and multiplies the integral by minus one. Carrying out the integration of Equation 33 n times thus yields
In devising that equation I used the fact, from Equation 29, that
and applied it n times (assuming, of course, that nm). If n=m, then
And if n≠m, then
And that proves and verifies the statement given in Equation 32.
That result makes the Hermite functions orthogonal to each other in a way analogous to the orthogonality of the unit vectors in a Cartesian space. The Hermite functions thus form the basis of an abstract infinite-dimensional vector space. That fact means that any linear superposition of the Hermite functions forms a proper solution of Equation 31, one that eliminates any need to consider the polynomial analogue of the distinction between covariant and contravariant vectors.
This presentation provides all of the information about Hermite polynomials that work in physics requires. Mathematicians, of course, can go much further in exploring the properties of these functions, but I will stop here.
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