Finite square well

1. Mar 3, 2005

broegger

Hi,

I have a problem with the finite square well. I have to analyze the odd bound states of the finite square well,

$$V(x)= \begin{cases} -V_0 & \text{for } -a<x<a\\ 0 & \text{otherwise} \end{cases}.$$​

Specifically, I have to examine the limiting cases (wide, deep well and narrow, shallow well) and find out, if there is always at least one odd bound state.

When I try to determine the energies of these odd states, I find that E=0 is always a solution. Is E=0 a bound state, a scattering state or what?

Also, what exactly are scattering states?

2. Mar 3, 2005

dextercioby

Okay,did you find the "normal" /general sollution...?Odd states refer to the behavior of wavefunctions under parity...In your case,depending on the E:V,it could be only $\sin$ or [itex] \sinh [/tex].

Bound states are normalizable states,physical states according to I-st postulate...In your case,which would be those...??
Scattering states would correspond to nonnormalizable states...

Daniel.

3. Mar 3, 2005

reilly

Scattering states are those that satisfy the Somerfeld Radiation Condition, which is obeyed if a state behaves like a plane wave at infinity. Drop a rock in a water wave, say from a speed boat. The wave pattern will settle down and stabilize-- the original wave will still be going, as will "scattered waves" generated by the rock -- and these waves will behave like free waves, once they've gone out a bit from the rock's splash.
Regards,
Reilly Atkinson

4. Mar 3, 2005

broegger

My question is really, does E=0 correspond to a bound state? I hope not, because this implies that there is always one odd, bound state (and there isn't according to the book).

5. Mar 3, 2005

dextercioby

No,the wave function is a "sine",which is not a bound state...It's not normalizable...

Daniel.

6. Mar 3, 2005

Ok, thanks!