I need to find the wavefunctions of the stationary states of a 3d square potential well with its boundaries defined by a triangular prism - like the one illustrated on the wikipedia page: https://en.wikipedia.org/wiki/Triangular_prism

Triangular prism

The potential well (viewed in 1-d cross section) is a simple square potential well, and can be either finite (0 outside, -V inside) or infinite (0 inside, ∞ outside), both would be reasonable approximations for my purposes.

I.e. the potential is something like this in cross-section, but its full 3-D shape is that of the triangular prism:

potential cross section

[Any solution for a close approximation of this geometry may also be helpful (for example, if the problem is easier to solve for a prism with a Reuleaux triangular cross-section instead of equilateral, or for a potential well described by a continuous function or something, it may be close enough).]

Because of the reduced symmetry compared to the textbook cylindrical or spherical cases, I am not sure how to approach this.

Is anyone able to point me in the direction of a solution? Many thanks!

  • $\begingroup$ Have you read this en.wikipedia.org/wiki/Airy_function. This could be of no value if it is 1D, but if it is a 3 D function, it might give you some clue to the correct coordinate system to use. $\endgroup$ – user154420 Jul 3 '17 at 20:14
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    $\begingroup$ I have, these are the solutions for a triangular potential though right? My potential is square, but the volume defining its boundaries is a triangular prism. I edited the original post to be clearer - thanks! $\endgroup$ – Ben Jul 3 '17 at 20:35
  • $\begingroup$ In the axial direction, at least, it should be a simple particle-in-a-box solution. For the other two components, maybe the fact that the Hamiltonian is invariant under 120 degree rotations means that the eigenstates also have this symmetry up to a constant? $\endgroup$ – probably_someone Jul 3 '17 at 20:39
  • $\begingroup$ I'm not sure. Because of the symmetry of the spherical and cylindrical problems, the radial and angular parts can be separated, but in this case I am not sure they can - the radial potential well changes in its width with angle, and also the centre of the triangle does not coincide with the centre of the well. $\endgroup$ – Ben Jul 3 '17 at 21:55
  • $\begingroup$ Take a look at this here: en.wikipedia.org/wiki/File:TriangleBarycentricCoordinates.svg I would imagine that the groundstate will surely have maximum amplitude at the point (1/3, 1/3, 1/3), but this is 2/3 along the 1-D cross section. This lack of symmetry is what I think makes the problem more complicated. $\endgroup$ – Ben Jul 3 '17 at 21:58

I'm unsure about the case with a finite well depth, but if the walls are infinitely hard this problem can be solved exactly. The solution is detailed in the papers

Other papers with relevant solutions are here, here and here.

The loss of continuous rotational symmetry means that you do need to fully solve a two-dimensional PDE, but the discrete symmetry does help, in that the solutions are required to carry representations of the $D_3$ symmetry group. This means that there are strict relations between the values of the eigenfunctions at the different edges, and those can be exploited to 'stitch' together multiple copies of the domain to make up a translationally-invariant region,

and you therefore expect the solutions to be plane-wave exponentials in that expanded region, which then project back down to sums of exponentials on the inside of the triangle.

I'm not sure to what extent those methods carry over to the finite-depth version of the well, this paper uses numerical diagonalization to solve the problem, and googling for "triangular quantum dot" (probably the most helpful starting point) doesn't immediately yield anything promising, and neither of those bode very well for the existence of closed solutions. (Ditto for their absence in this review.) Since you state that the infinite-walls problem is OK for your purposes, I'd encourage you to just stick to that.

  • $\begingroup$ He probably could follow the argument of the articles you mentioned by just using separation of variables for the infinite wall case. Very cool article, by the way. I figured somebody would have solved that. $\endgroup$ – Vendetta Jul 5 '17 at 20:45
  • $\begingroup$ @Vendetta The references cited appear to claim that the eigenstates are not separable, so it's likely a fair bit more complicated than that, but ultimately if people want the explicit solutions then there is the reference trail and it's up to them to chase it up. $\endgroup$ – Emilio Pisanty Jul 5 '17 at 20:53
  • $\begingroup$ The x and y directions are not separable, but I was referring to the z direction. $\endgroup$ – Vendetta Jul 10 '17 at 15:13
  • $\begingroup$ @EmilioPisanty The link to the JMP paper sends to your library rather than the journal. If I may: this is the kind of fantastic answer that keeps me around PhysicsSE. $\endgroup$ – ZeroTheHero Jul 11 '17 at 19:35
  • $\begingroup$ Hmm. Maybe you could actually generalize their method for any triangle, as this type of tiling can be done for any triangle (you can always use two triangles to make a parallelogram, as the pics show), you just deform the symmetry, recovering their relation in the equilateral case. This certainly will be significant for the spectrum and wavefunctions, but it seems a lot of their method might still hold. Also using this line of thought it might be possible to employ such reasoning to the isosceles right triangle easily, as two triangles make a square potential. $\endgroup$ – Vendetta Jul 12 '17 at 17:14

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