- #1
hnicholls
- 49
- 1
I am able calculate the solution for the TISE relating to the interval to the right of a step potential; however, I am unclear as to what this solutions represents physically for this scattering problem.
So, starting with
(-ħ2/2m)d2/dx2Ψ2(x) + V(x)Ψ(x)2 = EΨ(x)2
where V(x)Ψ(x)2 = EΨ(x)2
So, d2/dx2Ψ2(x) = 0
In order for d2/dx2Ψ2(x) to equal 0
d/dx Ψ(x)2 = A2 and Ψ(x)2 = A2x + B2
Again, was does this tell us about what happens to the incident wave function as it hits the a step potential exactly equal to the energy of the incident wave function?
So, starting with
(-ħ2/2m)d2/dx2Ψ2(x) + V(x)Ψ(x)2 = EΨ(x)2
where V(x)Ψ(x)2 = EΨ(x)2
So, d2/dx2Ψ2(x) = 0
In order for d2/dx2Ψ2(x) to equal 0
d/dx Ψ(x)2 = A2 and Ψ(x)2 = A2x + B2
Again, was does this tell us about what happens to the incident wave function as it hits the a step potential exactly equal to the energy of the incident wave function?