I'm working my way through a toy model just to improve my practical understanding of Sturm-Liouville ode theory and to get better at teaching quantum mechanics, but I ran into something strange that I want more info on. To set up, the toy model is a one-dimensional Hydrogen atom with the proton at the origin. So, we have an attractive Coulomb potential\begin{align} U(x)=-\dfrac{q^2}{4\pi\epsilon_0|x|}\end{align} and so the time-independent Schrödinger equation takes the usual form \begin{align} -\dfrac{\hbar^2}{2m}\dfrac{d^2\psi(x)}{dx^2}-\dfrac{q^2}{4\pi\epsilon_0|x|}\psi(x)=E\psi(x)\end{align} Anticipating that the Bohr radius $a_0=\dfrac{4\pi\epsilon_0\hbar^2}{mq^2}$ will come in as a natural size scale and the Rydberg energy $R=\dfrac{\hbar^2}{2ma_0^2}$ as a natural energy scale, the substitutions $x=a_0 y$ and $\lambda=\dfrac{E}{R}$ transform the Schrödinger equation into the form \begin{align}\dfrac{d^2\psi(y)}{dy^2}+\dfrac{2}{|y|}\psi(y)+\lambda \psi(y)=0 \end{align} Without even solving this equation yet, I'll go through the following simple argument: Assuming that the bound-state eigenfunction solutions $\psi_n$ are twice differentiable and bounded, the Schrödinger equation immediately implies that \begin{align}|\psi(y)|=|y|\left|\dfrac{1}{2}\dfrac{d^2\psi(y)}{dy^2}+\dfrac{\lambda}{2}\psi(y)\right|\leq \left|\sup\dfrac{d^2\psi(y)}{dy^2} +\lambda \sup\psi(y)\right||y|\end{align} which means $\psi(y)=\mathcal{O}(y)$ as $y\to 0$. By continuity, for every eigenstate $\psi_n$, we then know that $\boxed{\psi_n(0)=0}$
Anyone familiar with the node theorem from basic quantum mechanics should see the red flag here. The node theorem states that the ground state of the one-dimensional Schrödinger equation should have zero nodes.
Moreover, for bound states, we want $\lambda =-k^2$ for some $k$. Asymptotically, we can see that, as $|y|\to \infty$, we need \begin{align}\dfrac{d^2\psi(y)}{dy^2}-k^2\psi(y)\sim 0\Rightarrow \psi(y)\sim e^{-k|y|} \end{align} Substituting $\psi(y)=u(y)e^{-k|y|}$ into the Schrödinger equation transforms it to an equation for $u$ that looks like \begin{align} \dfrac{d^2 u}{dy^2} -2k\text{sgn}(y)\dfrac{du}{dy}+\dfrac{2}{|y|}u=0 \end{align} Expecting that the energy eigenvalues are the same as for the full Hydrogen atom, we could guess that $k_n=\dfrac{1}{n}$ which gives \begin{align} \dfrac{d^2 u}{dy^2} -\dfrac{2\text{sgn}(y)}{n}\dfrac{du}{dy}+\dfrac{2}{|y|}u=0 \end{align} It is then relatively straightforward to show than an odd-parity $n$th degree polynomial solves this equation whose coefficients satisfy a nice recurrence relation. The normalized ground state can be readily found to be $\boxed{\psi_1(y)=\sqrt{2}ye^{-|y|}}$
I found this to be all quite beautiful and intuitively pleasing, as the probability density must go to zero near the origin by the uncertainty principle. But based on the node theorem, the ground state shouldn't have any nodes at all, which confuses me.
After working this out, I later found this resource that confirms these solutions, but also claims that the one-dimension Hydrogen atom is apparently an unpleasant probelm. The eigenfunctions only being odd-parity is problematic for building arbitrary states from the eigenspace. Though I don't really see that as an issue since this is just a pedagogical toy model.
So, my two primary questions are:
Is the node theorem violated for certain potentials? Most proofs seem to assume that $V(x)\ge 0$ and is nice in a certain sense, such as being continuous. The Coulomb potential clearly violates these assumptions. Moreover, I don't think the ground state should be nodeless at the origin according to the uncertainty principle which alone predicts that the electron should tend to keep at least some nonzero distance from the nucleus.
What exactly is so ugly about this model? It has basically every nice feature I was hoping to see. A large volume of literature seems to exist on this problem, but most of it concludes that the basic model is unacceptable.
