What is the condition for $E$ in the time-independent Schrödinger equation?

[Ackbach]

Here is this week's POTW:

-----

The time-independent Schrödinger equation in one spatial dimension is
$$E \, \psi(x)=\left[-\frac{\hbar^2}{2m} \, \frac{d^2}{dx^2}+V(x)\right]\psi(x).$$
Show that $E$ must exceed the minimum value of $V(x)$ for every normalizable solution.

-----

Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Ackbach]

No one answered this week's POTW. My solution is below:

Spoiler

We can rewrite the Schrödinger equation as
\begin{align*}
E \, \psi(x)&=\left[-\frac{\hbar^2}{2m} \, \frac{d^2}{dx^2}+V(x)\right]\psi(x) \\
-\frac{\hbar^2}{2m} \, \frac{d^2 \psi}{dx^2}&=(E-V) \, \psi \\
\frac{d^2 \psi}{dx^2}&=-\frac{2m}{\hbar^2} \, (E-V) \, \psi.
\end{align*}
If $E$ is not greater than the minimum of $V$, then $\psi$ and its second derivative must have the same sign. Assuming sufficient differentiability (which we can get from physical arguments), this would mean a non-normalizable wave function, as the area under $|\psi|^2$ must be infinite. (N.B. that the norm we are talking about is the $L^2$ norm: $\displaystyle\|\psi\|_2=\int_{-\infty}^{\infty}|\psi|^2 \, dx.$)