quantum mechanics - Wavefunction and its derivative vanish at same point

It is stated that if $\psi(x_0)=\psi'(x_0)=0$ at some $x_0$, then $\psi(x)=0,\forall x$.

Hence, we cannot have the wavefunction and its derivative vanishing at the same point.

My question is, how to prove the above statement?

Answer

It is stated that if $\psi(x_0)=\psi'(x_0)=0$ at some $x_0$, then $\psi(x)=0, \ \forall x$. Hence, we cannot have the wavefunction and its derivative vanishing at the same point.

This is not true, but I believe I know what you mean.

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In 1D, the time-independent Schrodinger equation is a linear, 2nd order ODE of the form

$$-\psi'' + V(x) \psi = E\psi$$

or

$$\psi'' + (E-V)\psi = 0$$

The existence and uniqueness theorem for a linear, 2nd order ODE is usually stated as follows:

Let $p(x), q(x),$ and $g(x)$ be continuous functions on some open interval $I$. Given any $x_0\in I$, there is a unique function $y(x)$ defined on $I$ which satisfies the differential equation $$y''(x) + p(x) y'(x) + q(x) y(x) = g(x)$$ and the initial conditions $y(x_0)=y_0$, $y'(x_0)=y'_0$.

The proof of this is fairly involved. If you want to understand it, Coddington's book on ODEs is pretty inexpensive and discusses it pretty thoroughly (Chapter 6, Sections 7 and 8).

If we take that at face value, then the result you're looking for is immediate. Note that if $y_0=y'_0=0$, then the trivial function $y(x)=0$ satisfies both the Schrodinger equation and the initial conditions; therefore, it follows from the uniqueness theorem that it is the only function which does so, and must therefore be the solution we're looking for.

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That being said, the reason that the original claim is not true is simply that in general, a wave function need not satisfy the time-independent Schrodinger equation. The TISE is only satisfied by energy eigenfunctions.