Show that E must exceed the minimum value of V(x) , for every normalizable solution to the time- independent Schrödinger equation. What is the classical analog to this statement? Hint: Rewrite Equation 2.5 in the form
-\frac{h^{2}}{2 m} \frac{d^{2} \psi}{d x^{2}}+V \psi=E \psi (2.5).
\frac{d^2ψ}{dx^2} = \frac{2m}{\hbar ^2}[V(x) -E] ψ ;
if E< V_{min} , then ψ and its second derivative always have the same sign—argue that such a function cannot be normalized.
Given \frac{d^2ψ}{dx^2} = \frac{2m}{\hbar ^2}[V(x) -E] ψ , if E< V_{min} , then \psi ^{\prime \prime } and ψ always have the same sign: If ψ is positive(negative), then \psi ^{\prime \prime } is also positive(negative). This means that ψ always curves away from the axis (see Figure). However, it has got to go to zero as x →-∞ (else it would not be normalizable). At some point it’s got to depart from zero (if it doesn’t, it’s going to be identically zero everywhere), in (say) the positive direction. At this point its slope is positive, and increasing, so ψ gets bigger and bigger as x increases. It can’t ever \turn over” and head back toward the axis, because that would require a negative second derivative-it always has to bend away from the axis. By the same token, if it starts out heading negative, it just runs more and more negative. In neither case is there any way for it to come back to zero, as it must (at x →∞) in order to be normalizable. QED.
