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We have the Schrodinger eqn \begin{equation}(−\Delta+V(r))R(r)=E R(r)\end{equation}

where we can take $V(r)=-k/r$ for the beginning and we impose on the reduced radial function $u(r)=r R(r)$ the boundary conditions $u(0)=u(\infty)=0$.

The problem is to:

Solve numerically inwards and outwards the radial equation, and determine the eigenenergy by imposing the vanishing of the wronskian at some intermediate point.

How to proceed?

1) What is meant by solving "inwards and outwards" and what does "imposing the vanishing of the wronskian at some intermediate point" mean for us?

2) What numerical method shall I use? The Numerov method is probably not applicable, right?

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  • $\begingroup$ Is this homework? $\endgroup$ – Wolfgang Bangerth Jan 13 '15 at 2:00
  • $\begingroup$ @ Wolfgang Bangerth It's an exercise for me, then I want to use it for a more complicated potential. I've updated the question. $\endgroup$ – wondering Jan 13 '15 at 2:12
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To solve numerically you can use finite differences in MATLAB or GNU Octave on a finite interval. The Laplacian $\Delta$ in your equation is just the 2nd derivative $R''$. If you use 2nd order symmetric differences you get the stencil $$y''_i=\frac{-y_{i-1}+2y_{i}-y_{i+1}}{h^2}\,.$$ If you discretize the other term on the left-hand side (diagonal matrix) you will get a matrix Eigenvalue problem if the form $$Ay_h=\lambda y_h\,.$$ Don't forget to think about the boundary conditions as well.

Sorry but I don't how to do it on infinite intervals. You may use a transformation on a finite intervall e.g. $(0,1)$ such as $$z=\frac{x}{1-x}\,.$$

Hint: In theory of ODEs the Wronskian is the determinate of the fundamental matrix $\Phi$.

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  • $\begingroup$ +1 I have my answer below, but your answer helped me sth else, thanks. $\endgroup$ – wondering Mar 25 '15 at 15:13
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This is what brought me meanwhile to the solution:

1) "Solving inwards and outwards" simply means numerically integrating the mentioned differential equation:

  • from some small $r_1$ outwards to some $r_2>r_1$ (suitable BC in our case: $u_{out}(r_1)=r_1, u_{out}'(r_1)=1$ for example)

and

  • from some larger $r_3>r_2$ inwards back to $r_2$ (suitable BC for example: $u_{in}(r_3)= e^{-\sqrt{E} r_3}, u_{in}'(r_3) = -\sqrt{E} u(r_3)$, as we expect exponential decay)

2) If the wronskian $$W(E):=u_{out}'(r_2) u_{in}(r_2) - u_{out}(r_2) u_{in}'(r_2)$$

vanishes at point $r_2$, it means that the two solutions are linerly dependent and we get the energy eigenvalue(s) $E$ we were looking for (we have to find the roots of the function W(E)=0).we expect exponential decay,

3) In Mathematica, for instance, the "ImplicitRungeKutta" method for solving the equation for $u_{out}$ and $u_{in}$, i.e. $$-u''(r) - u(r)/r= -E u(r)$$ seems to be quite suitable.

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