When solving for the wave function under a harmonic potential $V(x)=\frac{1}{2}kx^2$, we attempt a solution in the form:
$$\psi(y)=H(y)e^{-\frac{1}{2}y^2},\quad y:=\left(\frac{m\omega_o}{\hbar}\right)^{\frac{1}{2}}x,\quad \alpha:=\frac{2E}{\hbar \omega_o},$$
where $x$ is position, $E$ is the energy. It follows by substitution into the time-independent Schrödinger equation that $H(y)$ is subject to the following:
$$H''-2yH'+(\alpha-1)H=0,$$
which is the Hermite equation. When solving this equation using Frobenius' method, we end up with the recursive relation:
$$a_{n+2}=\frac{2n+1-\alpha}{(n+1)(n+2)}a_n \quad\text{where}\quad H(y)=\sum_{n=0}^{\infty}a_ny^n.$$
When proving that this series must terminate at some finite order $n$ for the final wave function to be square-integrable, it is common to compare the limiting behavior of the ratio between successive terms in this series with that of the function $e^{2y^2}$:
$$H_\infty:\quad \frac{a_{n+2}}{a_n}\sim\frac{2}{n},$$
$$e^{2y^2}:\quad \frac{b_{n+2}}{b_n}\sim\frac{2}{n},$$
since the power series expansion for the latter is:
$$e^{2y^2}=\sum_{n=0}^\infty b_ny^n=\sum_{n=0}^\infty \frac{2^ny^{2n}}{n!}.$$
From here we assume that:
$$\frac{a_{n+2}}{a_n}\sim\frac{b_{n+2}}{b_n}\implies H_\infty(y)\sim e^{2y^2},$$
and therefore the wave function diverges as $y\to \infty$.
My question is: how do we prove that, if the ratios between successive terms in two different power series are asymptotically equivalent, then the two function defined by the power series are also asymptotically equivalent?
Best Answer
This material can be found in Bender and Orszag, Chapter 3. (They use the probabilist's convention, though, and derive $He(x)$ rather than $H(x)$.)
There is an alternative way to go about analyzing this problem in which we compute an asymptotic series about the irregular point at infinity. This first involves extracting the leading order behavior of the solutions to the original (scaled) differential equation, given by $$ -\frac{1}{2}y''+\frac{1}{2}x^2y-\epsilon y=0\,, $$ by first making the transformation $y(x)=e^{S(x)}$ and then using the method of dominant balance. This method is systematic if not entirely rigorous, and the result is $$ y(x)\sim e^{-x^2/2}x^{\nu}\,, $$ where $\nu=\epsilon-1/2$. At this point, we guess that the solution can be written as $$ y(x) = e^{-x^2/2}x^{\nu}\sum_{n=0}^{\infty}a_nx^{-n}\,. $$ (Note that this is a power series in negative powers of $x$!) This results in a recurrence relation for the $a_n$'s, given by $$ a_1=0,~~~~~ a_{n+2} = -\frac{(n-\nu)(n-\nu+1)}{2(n+2)}a_n\,, $$ so that the solutions looks like $$ y(x) = e^{-x^2/2}x^{\nu} \left( 1-\frac{\nu(\nu-1)}{4x^2} + \frac{\nu(\nu-1)(\nu-2)(\nu-3)}{32x^4} - \frac{\nu(\nu-1)(\nu-2)(\nu-3)(\nu-4)(\nu-5)}{384x^6} + \cdots \right) $$ Crucially, we can compute the radius of convergence of this series, and it is zero. It is still a good asymptotic solution for any $\nu$, but it is not good for the situation we are concerned with, which is for the function to go to zero at infinity. This computation of the radius of convergence of the asymptotic series replaces the hand-wavy "the power series acts like $e^{x^2}$" with a slightly less hand-wavy "the radius of convergence of the series expansion about $x\to\infty$ is zero".
In any case, this means that we only get a "good" solution when the power series truncates to a finite number of terms (because in that case the radius of convergence is of course infinite!), and this occurs when $\nu$ is a non-negative integer, which can be clearly seen from both the recursion relation and the first few terms of the expansion. After multiplying through by the factor of $x^{\nu}$, we get exactly the physicist Hermite polynomials $H_n(x)$, as we should.
The problem is that it's not clear to what extent the radius of convergence being zero is really related to the function blowing up at infinity, so this doesn't entirely answer the question, unfortunately.