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I need to solve, as mentioned, Schrodinger's equation in a Bunimovich stadium-shaped infinite potential well with Dirichlet BC Numerically (this isn't possible analytically). In order to do so, I need to define the discrete Hamiltonian matrix of the system. Since the shape isn't defined well in any coordinates system, I have no idea how to do so.

Normally (as I did with a square well), I'd define the size of the well and then solve 1D TISE for each dimension using MATLAB eig function. Can it be solved similarly?

Any help will be greatly appreciated.

ScubaDiving
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  • Can you describe the equations and the form of the potential you want to solve for? This may allow people to understand the question who are not already familiar with the Bunimovich Stadium. – Wolfgang Bangerth Apr 28 '16 at 14:32
  • Bunimovich stadium is a rectangle capped by two half-circles on each side. This shape has no analytical expression, so I've used some tricks to create a discrete matrix of V(x,y) to have this shape. It looks like this: https://i0.wp.com/upload.wikimedia.org/wikipedia/commons/thumb/8/83/BunimovichStadium.png/800px-BunimovichStadium.png – ScubaDiving Apr 28 '16 at 14:38
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    I don't see how is different from this question. You can use Finite Differences or Finite Methods for that purpose. BTW, the drawing you are referring to is really confusing. – nicoguaro Apr 28 '16 at 19:50
  • @nicoguaro This is completely different as this shape has no mathematical expression. The only way to implement the potential into the solution is to create a matrix which is a discrete representation of the shape. Then I used an N^2 by N^2 matrix similar to the normal 1D matrix, only used to solve for eigenfunctions transformed into vectors.

    I know this explanation isn't very good; I can post my code below if it'll be of any help.

     – ScubaDiving Apr 28 '16 at 21:01
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    Not different at all. In the approaches I mentioned (and in the codes) the potentials are turn into a discrete representation. – nicoguaro Apr 28 '16 at 21:06
  • I fail to see how the matrix stays the same in the other solution in a 2D system. – ScubaDiving Apr 28 '16 at 21:14
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    You still discretize the 2D domain, and the row and column numbers correspond to your discretization process. It depends if you use Finite Differences or Finite Elements but in essence is the same. – nicoguaro Apr 28 '16 at 23:15
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    That's right. You replace your domain by a set of triangles or quadrilaterals, and then you just do the regular finite element approach on it (or finite differences, if you're so inclined). – Wolfgang Bangerth Apr 29 '16 at 02:45

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