Degeneracy in one dimension

(1). $\psi_1\psi_2^"-\psi_2\psi_1^"=\frac{d}{dx}(\psi_1\psi_2'-\psi_2\psi_1')=0$.

(2). At infinity, it is zero, so the constant must be 0

(3). Integrate $\psi_1\psi_2'=\psi_2\psi_1'$, you will get that

I agree with the answers to 1) and 3), but disagree with the proposed answer to 2) of buzhidao and levitopher (the same argument is used in well-known textbooks such as Shankar, p.176). It is incorrect to say that "at infinity, the function is zero". This is never true for piecewise continuous potentials: if the wave function would actually be zero from a finite point onwards, it would have to be zero everywhere because of the uniqueness property of second order ordinary differential equations (e.g. http://www.emis.de/journals/DM/v5/art5.pdf). What you actually want from a solution is for it to be square-integrable on the real line. This implies that it goes to zero at infinity but not that it attains zero.

The problem with the argument is that there exist square-integrable solutions to the Schrödinger equation where the first derivative is unbounded at infinity. An example is $\psi(x) = N \sin(x^3)/x$ for $x \geq x_0$ with $x_0$ some positive number and $N$ a suitable normalization. The derivative oscillates and is unbounded as $x \rightarrow \infty$. This is a solution to the Schrödinger equation with zero energy and potential $V(x) = 2/x^2 - 9x^4$ (setting $m=1/2, \hbar = 1$. I only mention the behavior for $x \geq x_0$ because we are interested in that regime, and my proposed $\psi$ would not be square-integrable near zero. This could be fixed by making the potential infinite for e.g. $x \leq x_0 = (2 \pi)^{1/3}$ so that $\psi(x \leq x_0) = 0$). So it is false that $\psi_1 \psi_2'$ tends to zero because $\psi_1$ and $\psi_2$ tend to zero. We have to be assured that the derivative does not mess things up by becoming infinite. To this end we should impose a constraint on the set of potentials $V$ we consider, such that this constraint ensures that $\psi'$ is bounded as $x \rightarrow \infty$ as a consequence of $\psi$ satisfying the Schrödinger equation.

In a recent article http://arxiv.org/pdf/0706.1135v2.pdf one refers to Messiah's quantum mechanics book (https:// archive.org/details/QuantumMechanicsVolumeI: the relevant pages are 98-106) for a proof that if $V(x) - E \geq M^2 > 0, \forall x > x_0$, for some numbers $M,x_0$, then the eigenlevel corresponding to $E$ is non-degenerate (assuming there is such an eigenlevel). He shows that in this case the eigenfunction satisfies $\psi(x) = \mathcal{O}(e^{-Mx})$ and $\psi'(x) = \mathcal{O}(e^{-Mx})$ as $x \rightarrow \infty$ which is more than enough to guarantee that buzhidao and levitopher's argument works. Note that the condition on $V$ is stronger than simply being bounded from below. Note also that this does not prove that the potential has to be bounded from below to ensure that the energy eigenlevels are non-degenerate.