A A particle locked inside an arrangement of Dirac delta potentials

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The discussion explores the feasibility of trapping a particle, such as an electron, within a polyhedron using repulsing Dirac delta potentials placed at each vertex. It highlights that while a delta function can create a bound state, it requires a negative multiplier, and a positive delta potential will not trap a particle, leading to eventual tunneling out. The conversation also touches on the implications of using multiple delta potentials and compares this scenario to quantum dots, suggesting that a 3D lattice with point interactions may not guarantee a bound state. The complexity of tunneling times is acknowledged, depending on the distance and mass of the particle. Overall, the arrangement of delta potentials may not effectively confine a particle within the polyhedral surface.
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TL;DR
Would a dense enough set of point interaction potentials around a particle keep it confined in a limited space?
Suppose I have a 3D polyhedron with a large number of faces, and put a repulsing Dirac delta potential, ##c\delta (\mathbf{x} - \mathbf{x}_i )## with ##c>0## at each vertex point ##\mathbf{x}_i## of the polyhedron. Could this kind of an arrangement of delta potentials keep a particle such as an electron trapped inside that polyhedral surface as a bound state, or would probability density leak out even with a really densely spaced set of Dirac deltas?

Where I got this idea is this recent MIT study about a possibility to keep neutron locked inside a quantum dot, despite it interacting only with the (highly localized) nuclei and not the electrons.

https://pubs.acs.org/doi/10.1021/acsnano.3c12929

A quantum dot is often modelled as a "particle in 2D or 3D box", so I would guess some kind of arrangement of point interaction potentials is how the neutron quantum dot would be described with a theoretical model, but I can't access the full text of that publication yet.

The problem I posed could be investigated with a numerical calculation by approximating the delta functions with sharp gaussian spikes, but that would be less time consuming with an equivalent 2D version.
 
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A delta function already has one bound state.
 
Vanadium 50 said:
A delta function already has one bound state.
Yes, but only if it has a negative multiplier in front of it.
 
Are you saying that if you have a system with a potential of zero everywhere except a number of regions where it is greater than zero, does it have any bound states? It does not.

If you imagine a free particle, with a positive delta function at r = R, and a particle originally in the interior, it will eventually tunnel out. If you want to discuss how long this takes, that depends on the details of R and m. As you would expect, as R gets large (especially with respect to 1/m) the time gets long.

But this gets us into the question of "how almost is almost".
 
Vanadium 50 said:
Are you saying that if
If you imagine a free particle, with a positive delta function at r = R, and a particle originally in the interior, it will eventually tunnel out. If you want to discuss how long this takes, that depends on the details of R and m. As you would expect, as R gets large (especially with respect to 1/m) the time gets long.
Actually now that makes sense to me, even if a whole spherical surface acts as a delta potential, the particle will escape. Maybe it's a different situation if there's an infinite number of shells at ##r=R##, ##r=2R##, ##r=3R## and so on. Or a 3D lattice with point interactions at each lattice site and an empty vacant space somewhere for the particle to stay in.

Edit: In fact, if you look at Fig. 4 of this article, in a 1D imperfect Kronig-Penney lattice with a larger distance between one of the pairs of neighboring potential energy barriers, a particle seems to have a ground state where it's quite localized inside the largest interval available for it. An equivalent 3D version would probably not necessarily have a bound state, because a particle-in-sphere model with finite potential step (##V(r)=0## when ##r<R## and ##V(r)=V_0## when ##r\geq R##) doesn't have one either if ##V_0## isn't large enough compared to how small the volume of the sphere is.
 
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