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## 1 comments

## @Qmechanic 2016-12-01 15:19:00

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

$$ \nu(n)~:=~|\{\text{interior nodes of }\psi_n\}|,\tag{4}$$

$$ M_+(n)~:=~|\{\text{local max points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)>0\}|,\tag{5}$$

$$ M_-(n)~:=~|\{\text{local max points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)<0\}|,\tag{6}$$

$$ m_+(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)>0\}|,\tag{7}$$

$$ m_-(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)<0\}|,\tag{8}$$

$$ M(n)~:=~|\{\text{local max points for }|\psi_n|\}|~=~M_+(n)+M_-(n), \tag{9}$$

$$ m(n)~:=~|\{\text{local min points $x_0$ for $|\psi_n|$ with }\psi_n(x_0)\neq 0\}|~=~m_+(n)+m_-(n), \tag{10}$$

$$\Delta M_{\pm}(n)~:=~M_{\pm}(n)-m_{\pm}(n)~\geq~0.\tag{11} $$

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

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

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

--

$^{\dagger}$ For a more rigorous mathematical treatment, consider asking on MO.SE or Math.SE.

## @Michael Seifert 2016-12-03 15:17:23

Nice answer. Note that if the potential $V(x)$ only has one minimum (as is the case for all of the cases I cited above for which the proposition holds, but not for the double finite square well), then there will never be a classically forbidden region for any $E$ that doesn't include either $+\infty$ or $-\infty$. I think that this then implies that there will be no local minima of $|\psi_n(x)|$, and therefore that $\Delta M_+ + \Delta M_- = M = \nu + 1$.

## @Qmechanic 2016-12-03 15:34:38

$\uparrow$ Yes.