# 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 = A1eikx + B1e-ikx

Wave equation for interval 2:

2/2m 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 = A1eikx + B1e-ikx

Ψ2 = A2x

Continuity conditions:

Ψι = Ψ2

d/dx Ψ1 = d/dx Ψ2

So,

A1eikx + 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 [A1eikx + B1e-ikx] = d/dx [A2x]

ikA1eikx + 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.

$$k_1=\sqrt{2mE/\hbar}\\ k_2=\sqrt{3m(E-E_0)/\hbar}$$