What are the normalizations constants of psi(x) in a finite potential box?

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There's no closed form solution. Let's set V(x)=0 for -a<x<a and V(x)=\hbar^2 \gamma^2/2m, where \gamma is a constant, for |x|>a. Then, even parity solutions are of the form \psi(x)=A\cos(kx) for |x|<a and B\exp(-\kappa|x|) for |x|>a, where k^2+\kappa^2=\gamma^2, and the energy eigenvalue is E=\hbar^2 k^2/2m. Matching \psi(x) and \psi'(x) at x=a yields the eigenvalue condition \kappa=k\tan(ka) (which must be satisfied together with k^2+\kappa^2=\gamma^2; this pair of equations can be solved graphically or numerically, but not analytically) and B/A = \exp(\kappa a)\cos(ka). You can then get the value of A by normalizing \psi(x), which will give you some expression in terms of k, \kappa, and a, with k and \kappa being fixed by the energy eigenvalue (which can only be determined graphically or numerically).
 
what is the best way to solve it computationally?
 

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