# Is the particle able to penetrate finite barrier?

1. May 17, 2010

### J.Asher

Let's have a kind of step potential,

V=infinity, where x$$\leq$$0,
V=constant, where 0$$\leq$$x$$\leq$$a,
V=zero , where x$$\geq$$a.

Then, my question is that when the energy of the particle is lesser than zero, which

means that the particle is in the bound state, is there a probability of getting the

particle where x$$\geq$$a ??

Last edited: May 17, 2010
2. May 17, 2010

### ZapperZ

Staff Emeritus
This is not an example of a particle "in a bound state". Your particle doesn't exist for x<0 since the wavefunction from where it is would have sufficiently decayed away in the barrier.

Zz.

3. May 17, 2010

### J.Asher

I acknowledge that I had mistaken about which the state particle is under

but I didn't ask for the particle in the region x<<0, but x>>a.

After I upload this, I got some approaching for this problem

that in the region 0<<x<<a, the wave funtion for the particle would have sin and cos and

in the region x>>a, the particle would have an exponentioal funtion(real).

and from the continuity equation, which says wave function always continuos and the

deravative of the wave funtion would countinuous if the potential is not infinitive,

I could get the transcedental equation that has tangent form.

Does it make sense?

4. May 17, 2010

### phyzguy

V=infinity x<0
V=0 0<x<a
V=V a<x<b
V=0 x>b

With the particle initially in the region 0<x<a, is there a possibility of the particle escaping to the region x>b? If this is your question, the answer is yes, as long as V is finite.

5. May 17, 2010

### J.Asher

Ah im sorry....
I had a mistake on saying the region in which the potential finite.

V=infinity, where x<0
V=-V (negative constant), where 0<x<a
V=0, where x>a

maybe the problem doesn't give any gurauntee that the particle is in 0<x<a initially.

anyway, what I get thru this problem is particle cannot move into region x>a,where the potential goes up, finitely, from the region like finite well, 0<x<a, but if the particle in on the region, x>a, initially, particle can move into the region, o<x<a.
(I assuemd that the energy is negative and more than -v, 2nd boundary is given by the fact that the energy must be more than minimum of the potential)

So wherever the particle is there, it always go to the well, 0<x<a, which means
bound state, consequently.