Solving Schrodinger for eingenvalues

[Lee]

I'm looking to solve schrodingers equation for the bound energy state energy eigenfunctions, for a particle in a 1-D potential well (E<Vo), where;

V(x)= (inf, x<0)
(0, 0<x<a)
(Vo, x>a)so we know when x>a our function is going to be a negative exp, and between 0 and a the function is going to be equal to An.sin(kn.z)? from here can we ahead and starting finding solutions to the eqn?

 

[quasar987]

You found the solution in each particular regions; now you going to tie them up together using the boundary conditions and continuity conditions of psi (psi must be a continuous function)

for instance, you need that in the limit of psi as x approaches a- equals the limit as psi approaches a+.

 

[OlderDan]

You found the solution in each particular regions; now you going to tie them up together using the boundary conditions and continuity conditions of psi (psi must be a continuous function)

for instance, you need that in the limit of psi as x approaches a- equals the limit as psi approaches a+.

I just found this cute Java device for exploring solutions. I probably don't have it totally figured out, but you can model the situation pretty nicely by setting the piecewise constant potentials and choosing the functions in each region. You can't go to infinity, but you can go huge in one region.

From
http://web.phys.ksu.edu/vqm/AVQM Website/AVQMweb.htm

http://web.phys.ksu.edu/vqm/AVQM Website/WFEApplet.html

I had to close my browser window to restart it if I closed it. Expanded it to full screen it is quite useful.

 

[Lee]

so making sure An.sin(kn.a)=exp(-alpha.a) ?

 

[Lee]

I think my function for the mid section is wrong, as it will be discontinues if I pick it to be An.sin(kn.a) it shall equal 0 at 0 and a, which is good for the infinite wall to the left (as this is expected) but it that would mean exp(-alpha.a) would have to equal 0.

Can anyone make a better suggestion for the function for 0<x<a? Maybe a exp function?

 

[quasar987]

http://scienceworld.wolfram.com/physics/Half-InfiniteSquarePotentialWell.html

 

[Lee]

Perfect. Thanks.