# 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!