When does the $n$th bound state of a 1D quantum potential have $n$ maxima/minima? In Moore's introductory physics textbook Six Ideas that Shaped Physics, he describes a set of qualitative rules that first-year physics students can use to sketch energy eigenfunctions in a 1D quantum-mechanical potential.  Many of these are basically the WKB formalism in disguise;  for example, he introduces a notion of "local wavelength", and justifies the change in amplitude in terms of the classical particle spending more time there.  He also notes that the wavefunction must be "wave-like" in the classically allowed region, and "exponential-like" in the classically forbidden region.
However, there is one rule that he uses which seems to work for many (but not all) quantum potentials:

The $n$th excited state $\psi_n(x)$ of a particle in a 1D potential has $n$ extrema.

This is true for the particle in a box (either infinite or finite), the simple harmonic oscillator, the bouncing neutron potential, and presumably a large number of other 1D quantum potentials.  It is not true, however, for a particle in a double well of finite depth;  the ground state, which has a symmetric wavefunction, has two maxima (one in each potential well) and one minimum (at the midpoint between the wells).
The following questions then arise:


*

*Are there conditions can we place on $V(x)$ that guarantee the above quoted statement is true?  For example, is the statement true if $V(x)$ has only one minimum?  Is the statement true if the classically allowed region for any energy is a connected portion of $\mathbb{R}$?  (The second statement is slightly weaker than the first.)

*Can we generalize this statement so that it holds for any potential $V(x)$?  Perhaps there is a condition on the number of maxima and minima of $V(x)$ and $\psi_n(x)$ combined?
I suspect that if a statement along these lines can be made, it will come out of the orthogonality of the wavefunctions with respect to some inner product determined by the properties of the potential $V(x)$.  But I'm not well-enough versed in operator theory to come up with an easy argument about this.  I would also be interested in any interesting counterexamples to this claim that people can come up with.
 A: I) We consider the 1D TISE
$$ -\psi^{\prime\prime}_n(x) +V(x)\psi_n(x) ~=~ E_n\psi_n(x) .\tag{1}$$
II) From a physics$^{\dagger}$ perspective, the most important conditions are: 


*

*That there exists a ground state $\psi_1(x)$.

*That we only consider eigenvalues 
$$ E_n ~<~\liminf_{x\to \pm\infty}~ V(x). \tag{2}$$
Eq. (2) implies the boundary conditions
$$ \lim _{x\to \pm\infty} \psi_n(x)~=~0 .\tag{3}$$
We can then consider $x=\pm\infty$ as 2 boundary nodes. (If the $x$-space is a compact interval $[a,b]$, the notation $\pm\infty$ should be replace with the endpoints $a$ & $b$, in an hopefully obvious manner.)
Remark: Using complex conjugation on TISE (1), we can without loss of generality assume that $\psi_n$ is real and normalized, cf. e.g. this Phys.SE post. We will assume that from now on.
Remark: It follows from a Wronskian argument applied to two eigenfunctions, that the eigenvalues $E_n$ are non-degenerate.
Remark: A double (or higher) node $x_0$ cannot occur, because it must obey $\psi_n(x_0)=0=\psi^{\prime}_n(x_0)$. The uniqueness of a 2nd order ODE then implies that $\psi_n\equiv 0$. Contradiction.
III) Define
$$ \nu(n)~:=~|\{\text{interior nodes of }\psi_n\}|,\tag{4}$$
$$ M_+(n)~:=~|\{\text{local max points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)>0\}|,\tag{5}$$
$$ M_-(n)~:=~|\{\text{local max points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)<0\}|,\tag{6}$$
$$ m_+(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)>0\}|,\tag{7}$$
$$ m_-(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)<0\}|,\tag{8}$$
$$ M(n)~:=~|\{\text{local max points for }|\psi_n|\}|~=~M_+(n)+M_-(n), \tag{9}$$ 
$$  m(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)\neq 0\}|~=~m_+(n)+m_-(n), \tag{10}$$ 
$$\Delta M_{\pm}(n)~:=~M_{\pm}(n)-m_{\pm}(n)~\geq~0.\tag{11} $$

Observation. Local max (min) points for $|\psi_n|\neq 0$ can only occur in classical allowed (forbidden) intervals, i.e. oscillatory (exponential) intervals, respectively.

Note that the roles of $\pm$ flip if we change the overall sign of the real wave function $\psi_n$.

Proposition.
$$ \begin{align}\Delta M_+(n)+\Delta M_-(n)~=~&\nu(n)+1, \cr
 |\Delta M_+(n)-\Delta M_-(n)|~=~&2~{\rm frac}\left(\frac{\nu(n)+1}{2}\right).\end{align}\tag{12} $$

Sketched Proof: Use Morse-like considerations. $\Box$
IV) Finally let us focus on the nodes.

Lemma. If $E_n<E_m$, then for every pair of 2 consecutive nodes for $\psi_n$, the eigenfunction $\psi_m$ has at least one node strictly in-between.  

Sketched Proof of Lemma: Use a Wronskian argument applied to $\psi_n$ & $\psi_m$, cf. Refs. 1-2. $\Box$

Theorem. With the above assumptions from Section II, the $n$'th eigenfunction $\psi_n$ has 
  $$\nu(n)~=~n\!-\!1.\tag{13}$$

Sketched proof of Theorem:


*

*$\nu(n) \geq n\!-\!1$: Use Lemma. $\Box$

*$\nu(n) \leq n\!-\!1$: Truncate eigenfunction $\psi_n$ such that it is only supported between 2 consecutive nodes. If there are too many nodes there will be too many independent eigenfunctions in a min-max variational argument, leading to a contradiction, cf. Ref. 1. $\Box$
Remark: Ref. 2 features an intuitive heuristic argument for the Theorem: Imagine that $V(x)=V_{t=1}(x)$ belongs to a continuous 1-parameter family of potential $V_{t}(x)$, $t\in[0,1]$, such that $V_{t=0}(x)$ satisfies property (4). Take e.g. $V_{t=0}(x)$ to be the harmonic oscillator potential or the infinite well potential. Now, if an extra node develops at some $(t_0,x_0)$, it must be a double/higher node. Contradiction.
References:


*

*R. Hilbert & D. Courant, Methods of Math. Phys, Vol. 1; Section VI. 

*M. Moriconi, Am. J. Phys. 75 (2007) 284, arXiv:quant-ph/0702260.
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$^{\dagger}$ For a more rigorous mathematical treatment, consider asking on MO.SE or Math.SE.
