# Step potential

1. Nov 20, 2011

### Thyestean

1. The problem statement, all variables and given/known data
I'm working through a step potential and I am confused as to why one of the coefficients doesn't go to zero.

V(x) = 0 when x < 0;
V(x) = V_not when x > 0;

a. Calculate reflection coefficient when E < V_not
I can solve the reflection part, it is a step towards it that i am confused about.
2. Relevant equations

3. The attempt at a solution
Now i know the solutions solve to:

Aexp(ikx)+Bexp(-ikx) where k= sqrt(2mE)/hbar when x<0
Cexp(lx) where l=sqrt(-2m(E-V_not)/hbar when x>0

So my question is why doesn't B=0? Because when x->-infinity it goes to infinity so B has to be 0. The only reason i can think it wouldn't is because of tunneling. If this is the case how do I spot this. Is it only relevant in step potentials?

2. Nov 20, 2011

### vela

Staff Emeritus
e-ikx is oscillatory. It doesn't blow up as x goes to -∞.

3. Nov 20, 2011

### Thyestean

Ah thank you. Now that brings up another question for the same problem but now E>Vo.

The wave equations go to:
Aexp(ikx) + Bexp(-ikx) when x < 0. k=sqrt(2mE)/hbar
Cexp(ilx) + Dexp(-ilx) when x > 0. l=sqrt(2m(E-Vo))/hbar

Now in this case why does D=0.

4. Nov 20, 2011

### vela

Staff Emeritus
It's a boundary condition essentially. The idea here is you have an incident wave coming from the left. That corresponds to the A term. When it hits the potential step, you get a reflection, the B term, and a transmitted wave, the C term. The D term would correspond to a wave traveling to the right from x=+∞. You could certainly solve a problem with D not equal to 0, but it would be a different physical situation than the one you're interested in.