# Analytical solutions of S.E. for unusual potential

1. Mar 7, 2007

### DevoBoy

Analytical solutions of S.E. for "unusual" potential

Hi,

Given the potential

V(x)=0, when x<0 (region I)
V(x)=V_0, when 0<=x<=a, V_0=real constant (region II)
V(x)=infinite when x>a (region III)

What would be the general form on the solutions for each region?

I would think the solution for region I would be on the form A*exp(ikx) rather than A*exp(ikx)+B*exp(-ikx), since the exp(-ikx) term would "blow up" when x -> -infinite ?? But in the problem text I'm asked later "name the coefficient for the right-moving term A, and the left-moving term B", implying that I should use two terms for the solution ... ?

2. Mar 7, 2007

### dextercioby

No term would "blow up" in the I region simply because both the sine and the cosine are bounded functions.

Last edited: Mar 8, 2007
3. Mar 7, 2007

### Staff: Mentor

$e^{ikx}$ and $e^{-ikx}$ both oscillate. Their magnitude (absolute value) never gets larger than 1.

4. Mar 7, 2007

### DevoBoy

Ofcourse, my mistake.

So I end up with solutions:

psi_I(x) = A*exp(-ikx) + B*exp(ikx)
psi_II(x) = C*exp(-iqx) + D*exp(iqx)
psi_III(x) = 0

with boundary conditions

psi_I(0) = psi_II(0)
(d/dx)psi_I(0) = (d/dx)psi_II(0)
psi_II(a) = 0 (loss of continuinity here, right?)

Which gives the relations:

A+B=C+D
k(-A+B)=q(-C+D)
C*exp(-iqa)+D*exp(iqa)=0

Somehow I suspect I've missed something. A possibility of eliminating the exponentials in that last formula, or missed a boundary condition, or ... ???

5. Mar 7, 2007

### Mentz114

Devoboy,

I assume you know that

exp(ia) = cos(a) + isin(a)

.