# Standing waves in a funny Potential distribution

1. Dec 18, 2008

### frankcastle

1. The problem statement, all variables and given/known data

The description of the potential distribution is given in the attached image.
The particle arrives from the left with E>V0.

write the solutions to the S.E in regions x<o and x between o and a

2. Relevant equations
I believe psi(x)= e^ikx+Re^-ikx in x<0
and psi(x)=Ae^iqx+Be^-iqx for x b/w o and a.

3. The attempt at a solution
My question is, since there is complete reflection occuring at x=a, can A=B in region x b/w 0 and a? If so, there will be destructive interference in the region, giving R=1, which is what we are asked to prove in the question. Is this approach of equating coefficients of wave traveling in +-x directions in this region applicable?

#### Attached Files:

• ###### qphy.jpg
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2. Dec 19, 2008

### turin

The potential:
V=0 for x<0
V=V0 for 0<x<a
V->infinity for x>a
You have one boundary condition at x=a that relates A and B. You have two boundary conditions at x=0 that relate A, B and R. Once you have determined k and q (which I'm assuming you know how to do), then I believe you simply apply these boundary conditions.

3. Dec 19, 2008

### frankcastle

thanks turin, I understand the problem well. My question is regarding the relation of the coefficients, A and B; with the respective intensities.

Since R=1 at x=a, I would immediately assume that B=A
instead of having to use Boundary conditions to find coefficients.
Would this be correct logic?

4. Dec 19, 2008

### turin

No. You're contradicting yourself. You specified R for x<0, and now you want to talk about R at x=a, which doesn't even make sense, unless this is somehow a different R than the coefficient of the exponential that you originally gave. It's been a while since I solved one of these problems, so I can't remember if it should turn out that A=B is, in fact true; however, your logic to arrive at this conclusion is flawed.