The theorem on the number of zeros of a solution to a Sturm-Liouville equation is a well-know result in quantum mechanics. It doesn't seem to have a special name in the mathematics literature, but it is well-known as the Node Theorem in the physics literature. Without trying to be too precise - as I am not even sure about the conditions imposed for it to be true - the Node Theorem states that
any eigenfunction $\psi_n(x)$ corresponding to the $n$th eigenvalue of the one-dimensional Schrödinger equation, ordered in increasing magnitude, has exactly $n$ zeros.
For the reader's convenience, the one-dimensional Schrödinger equation is given by
$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2}+V(x)\psi(x)=E\psi(x)$,
where $V(x)$ is commonly known as the potential of the quantum system. The Schrödinger equation belongs in a class of eigenvalue problems known as the Sturm-Liouville equations:
$-(p(x)y')'+(q(x)-zr(x))y=0$.
I will not specify the subtleties in the study of this equation; the interested reader is referred to the very nice text of Gerald Teschl, Ordinary Differential Equations and Dynamical Systems.
I have seen an intuitive argument for the theorem in several physics sources (e.g. Nodes of wavefunctions, M. Moriconi). The argument goes roughly as follows:
Firstly, it is instructive to know that if $V(x)$, defined only on $(a, b)$, is a constant, then along with the boundary conditions $\psi(a)=0=\psi(b)$, the eigenvalues are "quantized" $E_0<E_1<E_2<\cdots$ and the eigenfunctions can be taken to be the cosine functions. Furthermore, any eigenfunction (a cosine function) corresponding to $E_0$ has no zero (i.e. no node in physics terminology), any eigenfunction (again a cosine function) corresponding to $E_1$ has one node, and so on. Such a potential is called an infinite potential well or a box in the physics literature and I will denote it by $U(x)$ below.
Now, the argument goes like this. Given any potential $V(x)$, place an infinite potential well $U(x)=V(x_0)$ with $x, x_0\in(a, b)\subseteq\text{Dom}(V(x))$ at a relatively flat portion (say the neighborhood of $x=x_0$) of $V(x)$. The boundary condition is for $\psi(a)=0=\psi(b)$, as is the case above. For narrow enough width of the box place in the relatively flat region of $V(x)$, the wavefunctions for $V(x)$ restricted to the domain $(a, b)$ also have similar behavior under the same boundary conditions. To argue that this behavior does not change as one widens the box, one notes that a "continuous" widening of the box would result in a "continuous" deformation of the wavefunction of the box to the wavefunction of $V(x)$ within the width of the box. Hence, suppose that for a certain width $(a, b)$, the wavefunction $\psi_n(x)$ has $n$ nodes because it is very well approximated by the narrow box. Now, it is a property of the $\psi_n(x)$ of the Schrödinger equation that $\psi(x^*)$ and $\psi'(x^*)$ cannot be both $0$ at any point $x^*\in(a, b)$. Thus, if while expanding the box, we introduce a node to the continuously deformed $\psi_n(x)$, then at some intermediate width the wavefunction $\psi_n(x)$ would have achieved $\psi_n(x^*)=0=\psi_n'(x^*)$, which could not have happened as we just noted. In conclusion, the number of nodes (i.e. zeros) don't change under such transformations.
Question: I have combed through the mathematics literature but I haven't come across a proof that uses the intuitive idea above. The standard proof seems to be the one that introduces the Prüfer variables and an analysis on them (cf. Teschl, Sec. 5.5) and it's a really nice piece of math. However, I would like to know if the argument from physics actually works, whether it has been done rigorously before or if there're flaws or counterexamples to the argument? Perhaps restricting the class of potentials $V(x)$ would do, but how does the proof go? Some references would be appreciated.
