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

1. Nov 19, 2008

### j-lee00

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

is there an algebraic or analytical solution.

I have the solution to the infinite box but it seems the two cases are not comparable. As in the finite box the particle does not need to be in the box?

2. Nov 19, 2008

### Avodyne

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).

3. Nov 19, 2008

### j-lee00

what is the best way to solve it computationally?