Full reflection of wave function for E = V0

[hnicholls]

I have seen discussions which suggests that there is no solution for the interval after the step in a step potential where E = V0.

The set up is a potential step where E = V0, with an interval 1 defined as x < 0 before the step and an interval 2 as x > 0 after the step.

Is the following correct?

Wave equation for interval 1: Ψι = A₁e^ikx + B₁e^-ikx

Wave equation for interval 2:

-ħ²/2m d²/dx²Ψ₂(x) + v(x)Ψ(x)₂ = EΨ(x)

v(x)Ψ(x)₂ = EΨ(x)

So, d²/dx²Ψ(x)₂ = 0

In order for d²/dx²Ψ(x)₂ to equal 0

d/dx Ψ(x)₂ = A₂ and Ψ(x)₂ = A₂x + C

So,

Ψι = A₁e^ikx + B₁e^-ikx

Ψ₂ = A₂x

Continuity conditions:

Ψι = Ψ₂

d/dx Ψ₁ = d/dx Ψ₂

So,

A₁e^ikx + B₁e^-ikx = A₂x

at interval boundary x = 0

So.

A₁ + B₁ = 0

A₁ = B₁

i.e. full reflection of the incident wave function.

d/dx [A₁e^ikx + B₁e^-ikx] = d/dx [A₂x]

ikA₁e^ikx + ikB₁e^-ikx = A₂

at interval boundary x = 0

ikA₁ - ikB₁ = A₂

A₁ = B₁

So,

ik[A₁ - A₁] = A₂

0 = A₂

i.e. no transmission of the incident wave function.

 

[Simon Bridge]

$\psi_I=Ae^{ik_1x}+Be^{-ik_1x}\\ \psi_{II}=Ce^{ik_2x}+De^{-ik_2x}$

The wave-numbers are related to the energy of the wave:
So for a wave of energy E incoming:

$$k_1=\sqrt{2mE/\hbar}\\
k_2=\sqrt{3m(E-E_0)/\hbar}$$
So, for $E=E_0$, $k_2=0$ and $\psi_{II}$ becomes a constant.
The next step is to apply the boundary conditions and check the normalization.