Show that E must be exceed the minimum value of $\mathit{V}\mathbf{\left(}\mathit{x}\mathbf{\right)}$ ,for every normalizable solution to the time independent Schrodinger equation what is classical analog to this statement?

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{\Psi }}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{2}\mathbf{m}}{{\mathbf{h}}^{\mathbf{2}}}\mathbf{\left[}\mathbf{V}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{-}\mathbf{E}\mathbf{\right]}\mathbf{\Psi }$;

If$\mathbf{E}\mathbf{<}{\mathbf{V}}_{\mathbf{min}}$ then$\mathbf{\Psi }$ and its second derivative always have the same sign. Is it normalized?

A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field.

Its answer is related to a particle's probability density in space and time.

Write equation (2.5).

$\frac{{\mathrm{d}}^2\mathrm{\psi }}{{\mathrm{dx}}^2}=\frac{2\mathrm{m}}{{\mathrm{\hslash }}^2}\left[\mathrm{V}\right(\mathrm{x})-\mathrm{E}]\mathrm{\psi }$

If the energy $E$ was less than the minimum value of $V\left(x\right)$, then $[{\mathrm{V}}_{\min }-\mathrm{E}]>0$, which is mean that the time-independent wave-function $\mathrm{\psi }$ and its second derivative will have the same sign everywhere, and therefore $\mathrm{\psi }$will diverge from zero as $x\to$$\pm \mathrm{\infty }$, which is mean $\mathrm{\psi }$ is not normalized;

So that $E$ must be greater than $V_{\min }$ to have a normalized value of $\psi$.

The classical analog for this is the total energy of some particle (i.e., kinetic energy plus potential energy) where it must be greater than the minimum value of the potential energy just like the energy$E$and the potential energy$V\left(x\right)$in quantum mechanics

Thus, when $E<V_{\min }$, then the wave function will diverge from zero as $x$ goes to $\pm \mathrm{\infty }$, and therefore the wave function is not normalized, so to get a normalized wave function $\mathrm{E}>{\mathrm{V}}_{\min }$ must be true.