Step potential, continuous wave function proof

[ope211]

Homework Statement

I am being asked to show that the wave function ψ and dψ/dx are continuous at every point of discontinuity for a step potential. I am asked to make use of the Heaviside step function in my proof, and to prove this explicitly and in detail.

Homework Equations

d^2ψ/dx^2=2m(V(x)-E)/ħ^2Ψ
Θ(x)=1 if x>0, 0 if x<0

The Attempt at a Solution

I assumed this meant that the potential can be written as V(x)=V_0Θ(x). Therefore, if x>0,
d^2ψ/dx^2=2m(V_0-E)/ħ^2Ψ
and if x<0:
d^2ψ/dx^2=2m(-E)/ħ^2Ψ

Now, I figured that I need to show ψ_1(0)=ψ_2(0) and the two first derivatives equal each other. So:
lim ε->0 (∫ d^2ψ_1/dx^2 dx (from -ε to +ε)=∫dx ψ_1(x)2m(V_0-E)/ħ^2 (from -ε to +ε))
and
lim ε->0 (∫ d^2ψ_2/dx^2 dx (from -ε to +ε)=∫dx ψ_2(x)2m(-E)/ħ^2 (from -ε to +ε))

Taking the limit of these integrals should show that they all go to 0, but I have no idea if this is sufficient. I do not know of any other way to show this.

 

[Charles Link]

One suggestion I have is to use the fact that the second derivative of the wave function is always finite. (You must show and/or assume the wave function is finite on the right side). Even though the second derivative is discontinuous, if it remains finite, the first derivative, which is the integral of the second derivative, will be continuous. I don't know if my suggestion is a sufficient proof. If it is, the homework problem is almost too simple, and should simply be presented as part of the lecture. $\\$ According to the Physics Forums rules, homework helpers are not supposed to provide the complete answer, but this one could have you spinning your wheels trying to do an elaborate proof, when the solution of this potential problem makes the assumption that the first derivative of the wave function, along with the wave function, is continuous in order to solve the boundary problem. You can't really solve the boundary problem without making this assumption, and the assumption is justified by the reason I gave above.