Which conditions must fulfill real-valued power series in order represent solutions that vanishes at infinity

Question

Which conditions must fulfill real-valued power series in order represent solutions that vanishes at infinity

Intro______________

In this really interesting video 1 profesor Barton Zwiebach explains how having a power series as solution of the time independent Shrödinger equation leads to the quantization of energy, at least in the sense of modeling it (quantization comes from experimental results like the blackbody radiation, the photoelectric effect, spectral lines of hydrogen, among others). The discussion started previously on these other videos: video 2 and video 3.

The argument presented is that a power series will diverge unless it decays as vanishing at infinity (which is true since they cannot match a constant value on a non-zero measure interval or it will violate the Identity theorem), and this behavior is a requirement in order of having normalizable solutions (otherwise the energy of the solutions will be always infinite).

Then it is explained that for achieving the vanishing at infinity behavior, it is required than the coefficients of the power series must behave such as after some index $N$ "it must truncate": meaning here if the power series is $\sum_{n=0}^{\infty} a_n x^n$ then there is some $0<N<\infty$ such as $a_n=0,\,\forall n\geq N$ (calculating $a_N = 0$ leads to the quantization of energy in the example of the lecture).

I don't know if the argument is a general one, or instead it only applies for the specific problem of the lecture, but I took it as a truth, until I realize the following:

If I take the Taylor series of a non-negative continuous bounded function that vanishes at infinity like $f(x)=e^{-x^4}$ which Taylor series is given by $T_f(x) = \sum_{n=0}^{\infty} \frac{(-1)^{4n}}{n!}x^{4n}$, I could realize that the Taylor series must be vanishing at infinity as the function $f(x)$ does (I think it is analytic - correct me please if I am mistaken here), but clearly their coefficients aren't truncating as mentioned in the lecture, so I am wondering now the extent to which the argument of truncated coefficients holds.

Question___________

1. Will always converge as vanishing at infinity a power series with truncated coefficients?
2. Which conditions must hold the coefficients of a power series in order to converge as vanishing at infinity?

PS: the lectures are master piece if you want to introduce yourself on the maths of quantum mechanics.

Answer 1

Its a general fact (Staeckel, Robinson, Eisenhart), that the tree of the separated Schrödinger eigenvalue equation in euclidean n-space, charted in set of general elliptic or parabolic orthogonal coordinate charts and their special more symmetrical subsets, produces a set of linear homogenous ODE's for factors $X_i(x_i)$ coupled by constants of separations $c_{ik}$ with a matrix of functions of a single variable $\Phi_{ik}(x_i)$, the so called Stäckel matrix.

$$\left( - \left(\prod_k h_k(x_k)\right)^{-1}\sum_i \frac{1}{h_i(x_i) }\ \partial_i \left(\prod_k h_k(x_k) \quad \frac{1}{h_i(x_i) }\partial_i \right) + V(x) - \omega\right) \ \prod_l X_l(x_l) \ = \ 0$$ The $h_i$ are the scale factors of the different dimensions and their product is the volume factor in the Hilbert spcace norm.

The separated equations are $$a_i(x_i)\ \partial_i \left(b_i(x_i) \ \partial_i X_i(x_i)\right)+ \sum_k c_{ik} \ \Phi_{ik}(x_i) \ X_i(x_i) =0$$

As linear ODE of second order for a general spectral set of separation constants $c_{ik}$ the first general solution is a Taylor series with a three term recursion over three adjacent powers, if the coefficients are linear in all $x_i$ at most.

This solution is regular at $x_i=0$ and yields a set of hypergeometric series involving the spectral constants. The second linear independent term set involves a log term.

For the three toy systems of quantum physics: linear oscillator, angular momentum on the unit sphere and the hydrogen atom with Coulomb potential in spherical coordinates, one is seeking the spectral values of the separation constants by the condition, that the product solution is in the Hilbert space of over that manifold.

That follows from the mathematical condition to establish a maximal extesnion of the formally symmetric Schrödinger operator as an unbounded operator in Hilbert space, bounded below by its ground state solution, as a positive (wrt to its lowest eigenvalue), essentially selfadjoint operator.

It follows that the solutions of the separated particular solutions are elements of their tensor factor $$X_i(x_i)\in \mathcal L_2((a_i,b_i), h_i (x_i) \ dx_i)$$.

The further path determining the spectrum has to consider the intervals of the ODE operators in $(a_i,b_i)$ of the different 1-d separated, symmetric second order operators.

The domains for our toy systems are Oscillator in $\mathcal L_2(\mathbb R, dx)$, sphere in $\mathcal L_2 \left((0,\pi)\times(0,2\pi), d \cos \theta\wedge d\phi\right) $, hydrogen in $\mathcal L_2(\mathbb R_+\times \mathcal S^2)$

For any of these three cases its possible to find the lowest eigenvalue state in Hilbert spaces, a Gaussian for the oscillator, a power $e^{i l \phi} $ respecting smoothness on the sphere for a given total angular momentum number $l$, the radial decaying exponential function $e^{-\omega r }$ for $\mathbb R_+$

If one uses an ansatz for the eigenvalue equation as $$\text{ground state}\times \text{Taylor series},$$ one finally ends with the condition, that the Taylor series must be a polynomial, because both linear indepenendent solutions of the hypergeometric equations have a non-square-integrable singularity at one of the two ends of the interval for a general spectral set of the separation constants.

Caveat: The hypergeometric ODE is a very special case. In general, finding the ground state and its exited states over the spectral series is possible numerically only with sophisticated methods to reduce the dimensions of the parameter space of search. Symmetries and group theory methods and perturbation theory are helpful.

Answer 2

The point of departure is the quantized harmonic oscillator, after normalization of constants $$ (\hat p^2 + \hat x^2)\phi=ε\phi\\ -\phi''(u)+u^2\phi(u) = \varepsilon\phi(u). $$

The WKB approximation for the far-field behavior has the basis functions $$ \phi_{1,2}(u)\asymp |u|^{-1/2}e^{\pm u^2/2} $$

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Obviously, one does not want any part of the exploding solution (shooting methods will always end up with a part of the exploding solution due to floating point errors also accumulating in its coefficient).

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Thus one approach is to factorize the solution as $\phi(u)=h(u)e^{-u^2/2}$ and try to prevent $h(u)$ growing like $e^{u^2}$.

For the differential equation one gets $$ \phi'(u)=(h'(u)-uh(u))e^{-u^2/2},\\ \phi''(u)=(h''(u)-2uh'(u)+(u^2-1)h(u))e^{-u^2/2}. $$ Thus the equation transforms to $$ -h''(u)+2uh'(u)+(1-ε)h(u)=0 $$

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With a power series for $h(u)\sum a_ju^j$ (assuming that it is an entire function) we get $$ u^j: -(j+1)(j+2)a_{j+2}+(2j+1-ε)a_j = 0. $$ The equations for $j=-1,-2$ are trivial if one sets $a_{-k}=0$, $k=1,2,..$.

For the propagation we get $$ a_{j+2}=\frac{2j+1-ε}{(j+1)(j+2)}a_j. $$ If any of the numerators in the chain of even or odd $j$ becomes zero, the chain breaks there and is zero forthwith, giving a polynomial solution which obviously grows slower than any exponential. This would require $ε=2n+1$ for any non-negative integer $n$.

In the opposite case the asymptotics of the relation is $\frac{a_{j+2}}{a_j}\asymp\frac2{j}$ which gives indeed an infinite radius of convergence and an asymptotic behavior of $h(u)\asymp e^{2u^2}$ which is clearly growing too fast to have the solution $\phi$ fall to zero at infinity.

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For further research study the more general class of Sturm-Liouville eigenvalue problems. This is mostly about boundary value problems on finite intervals, but explains in great detail why the resulting spectrum is discrete and countable.