#### [SOLVED] 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:

1. 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.)

2. 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. #### @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:

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

2. 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 $$E_n, then for every pair of 2 consecutive nodes for $$\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:

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

2. $$\nu(n) \leq n\!-\!1$$: 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:

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

2. 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. #### @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.