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$$ \psi_E(x)=\begin{cases} e^{ikx}+Ae^{-ikx} \quad \quad x \le0 \\ Ce^{ikx} \quad \quad x\ge a \\ \end{cases} $$ 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?

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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.

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  • $\begingroup$ I can't understand. Why would be zero everywhere? I just have a Cauchy's problem that "select" one of the solutions of the TISE. $\endgroup$
    – Salmon
    Commented May 31, 2022 at 11:27
  • $\begingroup$ The constant zero function is a solution of the TISE and has the same Cauchy data $\endgroup$
    – SolubleFish
    Commented May 31, 2022 at 11:54
  • $\begingroup$ Ok thank you. Are we assuming that there is a unique solution for Cauchy problem? Otherwise, why the fact that the constant zero is a solution of the TISE with same Cauchy data would be a problem? I mean: couldn't there be other functions that solve the TISE and Cauchy problem other than zero? $\endgroup$
    – Salmon
    Commented May 31, 2022 at 15:30
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    $\begingroup$ 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 $V(x)$. $\endgroup$
    – SolubleFish
    Commented May 31, 2022 at 16:16

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