For the harmonic oscillator potential energy, $\mathit{U}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{x}}^{\mathbf{2}}$, the ground-state wave function is $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathit{A}{\mathit{e}}^{\mathbf{-}\mathbf{(}\sqrt{\mathbf{m}\mathbf{k}}\mathbf{/}\mathbf{2}\mathbf{\hslash }\mathbf{)}{\mathbf{x}}^{\mathbf{2}}}$, and its energy is $\frac{\mathbf{1}}{\mathbf{2}}\mathit{\hslash }\sqrt{\mathbf{k}\mathbf{/}\mathbf{m}}$.

(a) Find the classical turning points for a particle with this energy.

(b) The Schrödinger equation says that $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ and its second derivative should be of the opposite sign when E > Uand of the same sign when E < U . These two regions are divided by the classical turning points. Verify the relationship between $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and its second derivative for the ground-state oscillator wave function.

(Hint:Look for the inflection points.)

The potential is

$U=\frac{1}{2}k{x}^2$ .....(I)

The total energy is

$E=\frac{1}{2}\hslash \sqrt{k/m}$ .....(II)

The wave function is

$\psi \left(x\right)=A{e}^{-\left(\sqrt{mk}/2\hslash \right)x^2}$ .....(III)

The classical turning points are when the potential energy is equal to the total energy, that is

E = U .....(IV)

The turning points are obtained from equations (I), (II) and (IV) as follows

$\begin{array}{rcl}\frac{1}{2}\hslash \sqrt{k/m}& =& \frac{1}{2}k{x}^2\\ x& =& \pm {\left(\frac{{\hslash }^2}{mk}\right)}^{1/4}\\ & & \end{array}$

Thus the turning points are at $\pm {\left(\frac{{\hslash }^2}{mk}\right)}^{1/4}$.

The double derivative of equation (IV) gives

$\begin{array}{rcl}\frac{d^2}{d{x}^2}\psi \left(x\right)& =& \frac{d^2}{d{x}^2}A{e}^{-\left(\sqrt{mk}/2\hslash \right)x^2}\\ & =& \frac{d}{dx}\left(-A\frac{x\sqrt{mk}}{\hslash }{e}^{-\left(\sqrt{mk}/2\hslash \right)x^2}\right)\\ & =& -A\frac{\sqrt{mk}}{\hslash }{e}^{-\left(\sqrt{mk}/2\hslash \right)x^2}+A\left(\frac{x\sqrt{mk}}{\hslash }\right)\left(\frac{x\sqrt{mk}}{\hslash }\right)e^{-\left(\sqrt{mk}/2\hslash \right)x^2}\\ & =& -A\frac{\sqrt{mk}}{\hslash }{e}^{-\left(\sqrt{mk}/2\hslash \right)x^2}+\frac{A{x}^{2}mk}{{\hslash }^2}{e}^{-\left(\sqrt{mk}/2\hslash \right)x^2}\end{array}$

The double derivative is zero at

$\begin{array}{rcl}-A\frac{\sqrt{mk}}{\hslash }{e}^{-\left(\sqrt{mk}/2\hslash \right)x^2}+\frac{A{x}^{2}mk}{{\hslash }^2}{e}^{-\left(\sqrt{mk}/2\hslash \right)x^2}& =& 0\\ \frac{\sqrt{mk}}{\hslash }& =& \frac{x^{2}mk}{{\hslash }^2}\\ x& =& \pm {\left(\frac{{\hslash }^2}{mk}\right)}^{1/4}\end{array}$

that is exactly at the turning points.

At x = 0 the double derivative is

$-A\frac{\sqrt{mk}}{\hslash }{e}^{-\left(\sqrt{mk}/2\hslash \right)0}+\frac{A{0}^{2}mk}{{\hslash }^2}{e}^{-\left(\sqrt{mk}/2\hslash \right)0^2}=-A\frac{\sqrt{mk}}{\hslash }$

that is negative.

For large x the double derivative is

$-A\frac{\sqrt{mk}}{\hslash }{e}^{-\left(\sqrt{mk}/2\hslash \right)\left(\infty \right)}+\frac{A{\left(\infty \right)}^{2}mk}{{\hslash }^2}{e}^{-\left(\sqrt{mk}/2\hslash \right){\left(\infty \right)}^2}=\infty$

that is positive.