# Wavefunction properties tunnel effect

$$\psi_E(x)=\begin{cases} e^{ikx}+Ae^{-ikx} \quad \quad x \le0 \\ Ce^{ikx} \quad \quad x\ge a \\ \end{cases}$$ where $$k^2=\frac{2mE}{\hbar^2}$$.

Now what I read in my notes is

"since the eigenfunctions of SE equation must not be equal to zero in a point with their first derivatives, then $$C \neq 0$$".

How can I prove this statement? Is due to the fact that the SE is $$\psi''(x)=-\frac{2mE}{\hbar^2}\psi(x)$$ so that the eigenfunction must have concavity facing upward when it's negative and concavity facing downward when it's positive?

The eigenfunction of the 1D Schrödinger equation satisfy a second order linear differential equation. If there is a point $$x$$ where $$\psi(x) = \psi'(x) = 0$$, then $$\psi$$ is zero everywhere, which is not allowed.
• In your particular case, the solutions are of the form $\psi(x) = A e^{ikx} + Be^{-ikx}$. It is not hard to see that the Cauchy problem has a unique solution. In general, you may need to assume some regularity for the potential $V(x)$. May 31, 2022 at 16:16