91影视

The potential is

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

The total energy is

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

The wave function is

$蠄\left(x\right)=A{e}^{-\left(\sqrt{mk}/2\mathrm{鈩�}\right)x^2}$ .....(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

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

Thus the turning points are at $卤{\left(\frac{{\mathrm{鈩�}}^2}{mk}\right)}^{1/4}$.

The double derivative of equation (IV) gives

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

The double derivative is zero at

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

that is exactly at the turning points.

At x = 0 the double derivative is

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

that is negative.

For large x the double derivative is

$-A\frac{\sqrt{mk}}{\mathrm{鈩�}}{e}^{-\left(\sqrt{mk}/2\mathrm{鈩�}\right)\left(鈭�\right)}+\frac{A{\left(鈭�\right)}^{2}mk}{{\mathrm{鈩�}}^2}{e}^{-\left(\sqrt{mk}/2\mathrm{鈩�}\right){\left(鈭�\right)}^2}=鈭�$

that is positive.