4
$\begingroup$

A well-known result of quantum mechanics is that for a single particle in one dimension in a bounding potential $V(x)$ that goes to $+\infty$ as $x \to \pm \infty$, the energy eigenfunctions are discrete and the $n$th eigenfunction has exactly $n-1$ nodes at which $\psi(x) = 0$. (Moreover, we can say more - for example, between any two consecutive nodes in the $n$th eigenfunction, there exists a node in the $(n+1)$th eigenfunction.)

Do any similar results apply for single particles in higher than one dimension, or for multiparticle systems (for which the wave function is defined on configuration space rather than real space)? If not, is there an explicit example of a higher-dimensional or multiparticle system whose ground state wavefunction has a node?

$\endgroup$
2
  • 1
    $\begingroup$ Can we assume that the configuration space is $\mathbb{R}^n$ and the potential $V$ is spherically symmetric? $\endgroup$
    – Qmechanic
    Nov 4 '18 at 19:08
  • $\begingroup$ @Qmechanic No.. $\endgroup$
    – tparker
    Nov 4 '18 at 21:08
2
$\begingroup$

The result is actually applicable to 1d-equivalent motion, and as such is applicable to the radial part of the Schrodinger equation in any dimension if this radial part can be separated.

In general, the twist is that the equivalent 1d motion depends on the effective potential - in the case of a 3D central potential the effective 1d potential would include the centrifugal part proportional to $\ell(\ell+1)/r^2$ - so the result may be angular-momentum dependent or might depend on other parameters in the effective potential.

$\endgroup$
2
  • $\begingroup$ Are there counterexamples if the system isn't effectively 1D? $\endgroup$
    – tparker
    Nov 5 '18 at 0:38
  • $\begingroup$ @tparker I do not know. I found this older answer of mine: physics.stackexchange.com/a/319351/36194 which gives the proof so it's really a property of differential equations and - IIRC - of Sturm-Liouville systems. I do not know if there's a 2d version of this result but it doesn't seem so easy to reproduce the proof in 2d as there is an exact derivative that allows integration over $d\xi$: it seems problematic to reproduce this step in 2d. $\endgroup$ Nov 5 '18 at 1:15
0
$\begingroup$

I found the answer on Chemistry SE. Apparently, for systems of one particle in higher than one dimension, the only known general result is that the number of nodes in the $n$th eigenfuntion is $\leq n-1$, and only in 1D is the inequality always saturated. (See the answer for more info and precise term definitions.) Moreover, in higher dimensions the sequence of node numbers is not necessarily increasing; in fact, the question gives a simple explicit example of a one-particle system in 3D such that a higher excited state has fewer nodes than a lower excited state. For multiparticle systems, I don't know of any general results at all.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.