The Hermite Polynomials

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

(Eq’n 1)

in which d_{x} represents the derivative operator

(Eq’n 2)

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;

(Eq’n 3)

Rearranging the terms in that equation produces

(Eq’n 4)

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 H_{0}(x)=1, that
procedure yields

(Eq’ns 5)

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

(Eq’n 6)

To determine the form of the coefficients a_{k},
substitute that equation into Hermite’s Equation to get

(Eq’n 7)

In that equation we must have λ=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(x^{2}) 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 a_{n+1 }and a_{n+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

(Eq’n 8)

For each value of k the sum of terms must satisfy Equation 8 separately, which fact leads to the statement that

(Eq’n 9)

As inferred above, when k=n, a_{n+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 H_{0}(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: H_{n+1}=2xH_{n}-dH_{n}/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 a_{n}=1, I can calculate the value
of a_{n-2}, then from that coefficient I can calculate the value of a_{n-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

(Eq’n 10)

Using that equation, I calculate

(Eq’n 11)

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

(Eq’n 12)

which I can use to rewrite Equation 11 as

(Eq’n 13)

in which C_{y} represents the coefficient on a_{y} and C_{n}=1.
Thus I get

(Eq’n 14)

With that formula I can now write out explicitly the power series for the n-th Hermite polynomial as

(Eq’n 15)

If I want to calculate H_{n}(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;

(Eq’n 16)

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(-x^{2}).
Taylor’s theorem states that for some function f(x) the relationship

(Eq’n 17)

stands true to mathematics. Rearranging Equation 1 yields

(Eq’n 18)

Making the appropriate substitution from that equation into Equation 17 yields

(Eq’n 19)

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

(Eq’n 20)

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=x^{2} into the standard
series representation of the negative exponential function,

(Eq’n 21)

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(-x^{2}), 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

(Eq’n 22)

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

(Eq’n 23)

so Equation 22 becomes

(Eq’n 24)

Multiplying that equation by t^{n}/n! and then summing over all the
values of n from zero toward infinity yields

(Eq’n 25)

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

(Eq’n 26)

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, H_{n}(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

(Eq’n 27)

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

(Eq’n 28)

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

(Eq’n 29)

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, H_{n}(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

(Eq’n 30)

which functions solve the Schrödinger-Hermite Equation in the form

(Eq’n 31)

The Hermite functions conform to the statement that

(Eq’n 32)

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;

(Eq’n 33)

For convenience I have defined the constant

(Eq’n 34)

Also, conveniently enough, exp(-x^{2}) and the exp(x^{2})
from H_{n}(x) cancel each other out. Integrating Equation 33 by parts
yields

(Eq’n 35)

The second term on the right side of that equation vanished because the function

(Eq’n 36)

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(x^{2})
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

(Eq’n 37)

in the limit as x goes toward positive or negative infinity.

Each integration by parts shifts the derivative operator
from acting on exp(-x^{2}) to acting on Hm(x) and multiplies the
integral by minus one. Carrying out the integration of Equation 33 n times thus
yields

(Eq’n 38)

In devising that equation I used the fact, from Equation 29, that

(Eq’n 39)

and applied it n times (assuming, of course, that n m). If n=m, then

(Eq’n 40)

And if n≠m, then

(Eq’n 41)

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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