Asymmetric potential well problem

1. Oct 30, 2008

youngoldman

The task is to find a wavefunction (doesn't need to be normalised) and the energy levels of a particle in an asymmetric potential well (a Schrodinger problem). i.e V = V1 for x<0, 0 for 0<x<d and V = V2 for x>d.

What I've got so far is

Let α^2= 2m (E-V1)/ℏ^2
β^2= 2mE/ℏ^2
T^2= 2m (E-V2)/ℏ^2

Using these substitutions in Shrodinger's Equations and keeping Ψ finite yields

Ψ = A exp (αx) x<0
C exp (iβx) + D exp (-iβx) 0<x<d
G exp (-Tx)

Boundary conditions give

A = C + D
Aα = (C - D) iβ
Gexp(-Td) = C exp(iβd) + D exp(-iβd)
-TGexp(-Td) = Ciβexp(iβd) - Diβexp(-iβd)

Eliminating A and G:
(C + D) α = (C - D) iβ
-T(Cexp(iβd) + Dexp(-iβd)) = iβ(Cexp(iβd) - Dexp(-iβd))

Shifting all these terms of these side onto one side ( so get 0 = .....) and putting into a matrix and setting determinant = 0 yields:

(αT - αiβ - Tiβ - β^2)(exp(-iβd)) = (αT + αiβ + Tiβ - β^2)(exp(iβd))

exp(-2iβd) = (αT - αiβ - Tiβ - β^2)/(αT + αiβ + Tiβ - β^2)

cos(2βd) -i sin(2βd) = (αT - αiβ - Tiβ - β^2)/(αT + αiβ + Tiβ - β^2)

at which point I am stuck. Hints??

2. Oct 30, 2008

Avodyne

This seems a like a very messy problem, but it would probably be easier to use a linear combination of sin and cos in the middle, since we know that the eigenfunctions are real.