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

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

Similar threads

  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 12 ·
Replies
12
Views
4K
  • · Replies 21 ·
Replies
21
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K