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A diatomic-like setup:

There are 3 regions: $x < x', x \in [x', x''], x > x''$ with potential being $V\left(x\right) > E$, $V\left(x\right) < E$ and $V\left(x\right) > E$, respectively, and in all regions the eigenfunction $\psi\left(x\right) > 0$ where $x$ describes the distance between two particles in the diatomic setup.

The time-independent Schrodinger equation is given as

$\frac{d^{2}\psi}{dx^{2}} = \frac{2m}{\hbar^{2}}\left[V\left(x\right) - E \right] \psi$

In $x < x''$:

The potential $V\left(x\right)$ encodes the distance $x$, as a denominator term, between the two particles resulting in the potential $V\left(x\right)$ tending to $\infty$ as the distance $x$ tends to 0. This gives $\frac{d^{2}\psi}{dx^{2}} > 0$ so the second spatial derivative of $\psi$ is concave upwards. So $\psi$ is infinitely large in this region as tends to infinity for $x$ closer to 0.

In $x \in [x', x'']$:

In this region, the second spatial derivative of $\psi$ would be negative.

In $x > x''$:

But if this were a potential with a limiting value, $V_{l}$ then there exists some $x > x''$ such that $V$ converges to $V_{l}$. In this region, just after $x > x''$, it is true that $V\left(x\right) - E > 0$ and $V\left(x\right) - E$ increases the further $x$ is from $x''$ and only for $V\left(x\right) - E$ to be constant at some $x > x''$ once $V\left(x\right)$ converges to $V_{l}$. Depending on the specificity of the potential, in this case depending at where $V\left(x\right)$ converges to $V_{l}$, the second spatial derivative for $\psi$, $\frac{d^{2}\psi}{dx^{2}}$, can be infinitely large. However, it would necessarily be positive based on the governing potential outlined here.

Am I right in my understanding of the behaviour for $\psi$ in the third region? Most authors seem to conclude that $\psi$ would tend to infinity in the third region given this specific potential. My understanding is that it may tend to infinity.

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At a region where the second derivative of a function is positive. It doesn't lead to a ascending function. For example $f(x) = e^{-x}$ at region $x>0$

\begin{equation} \frac{d^2 f(x)}{dx^2} = + e^{-x} \gt 0 \end{equation}

But $e^{-x}$ is descending function for $x>0$.

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