# Linear Dependence of Energy-Degenerate Solutions to 1-D Schrödinger Equation

I'm trying to show that no energy-degeneracies can occur for the 1-dimensional Schrödinger equation (for a particle of mass $m$) for non-singular potential. My approach is as follows: Assume two stationary-states, $\psi_1\equiv\psi_1(x)$ and $\psi_2\equiv\psi_2(x)$, for a particular Hamiltonian (with non-singular potential) which correspond to the same eigenvalue. Then if we take a slightly "modified Wronskian" of $\psi_1$ and $\psi_2$, we can test their linear independence.

\begin{align*} \tilde{W}(\psi_1, \psi_2) &= \frac{-\hbar^2}{2m}\begin{vmatrix}\psi_1 & \psi_2 \\ \frac{d^2 \psi_1}{dx^2} & \frac{d^2 \psi_2}{dx^2} \end{vmatrix}\\ &=\psi_1\left(\frac{-\hbar^2}{2m}\frac{d^2 \psi_2}{dx^2}\right)-\psi_2\left(\frac{-\hbar^2}{2m}\frac{d^2 \psi_1}{dx^2}\right)\\ &=\psi_1\left(E-V(x)\right)\psi_2-\psi_2(E-V(x))\psi_1\\ &=0 \end{align*}

Thus, $\psi_1$ and $\psi_2$ are linearly dependent, which means they differ by a constant. But $\psi_1$ and $\psi_2$ are determined up to a phase factor, so that means they can only differ by a phase, which means they are indistinguishable, so there is no degeneracy.

What is wrong with this? What caveats are there in what I've said? In this paper, they give an example of degenerate wavefunctions which would render my argument invalid, though I can't see the problem with my argument. I don't see how this even requires anything else besides non-singularity of the potential.

A problem with your argument is that the vanishing of your `modified Wronskian' does not imply linear independence. Indeed, for a given $\psi_1$, there are two linearly independent solutions to $\psi_1 \psi_2''-\psi_2\psi_1''=0$ for $\psi_2$, being $\psi_2=\psi_1$ and $\psi_2=\psi_1\int\frac{1}{\psi_1^2}$.
Indeed, there are always two solutions to the Schr\"odinger equation locally, for any energy, and the eigenvalues allowed are determined only when boundary conditions are imposed. Your modified Wronskian $\tilde{W}$ is the derivative of the usual Wronskian $W$, so you've shown that $W$ is constant. More global data (e.g. vanishing of the wavefunctions at infinity) is needed to show that the constant is zero, so the physical (normalisable) solutions are in fact linearly dependent.
Your modified Wronskian can be zero and the functions could still be linearly independent. For example, $sin x$ and $cos x$.