The particle has $\mathit{E}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$.

(a) Show that the potential energy for x>0is given by

$\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{-}\frac{{\mathbf{\hslash }}^{\mathbf{2}}\mathbf{a}}{\mathbf{m}}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{+}\frac{{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}$

(b) What is the potential energy for x<0?

The time independent Schrodinger equation relates the energy of the particle, Eto the potential energy, U, the mass of the particle, m, the position of the particle, X, and the wave function, $\psi$

$\mathbf{-}\frac{\mathbf{\hslash }}{\mathbf{2}\mathbf{m}}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{E\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

If the energy of the particles is 0 , then the potential energy can be is isolated by adding the first term to the other side and divided by the wave function.

The wave function of the particle is$\mathrm{\psi }\left(\mathrm{x}\right)$ .

$\mathbf{\psi }\left(x\right)\mathbf{=}\{\begin{array}{l}2\sqrt{a^3{\mathrm{xe}}^{-\mathrm{ax}}}x>0\\ 0x<0\end{array}$

The time-independent Schrodinger equation relates the energy of the particle, E to the potential energy, U, the mass of the particle, m, the position of the particle, X, and the wave function,$\psi$

$\mathbf{-}\frac{\mathbf{\hslash }}{\mathbf{2}\mathbf{m}}\frac{\mathbf{d}}{\mathbf{dx}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{E\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$

If the energy of the particles is 0 , then the potential energy can be isolated by adding the first term to the other side and divided by the wave function.

$\begin{array}{rcl}\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}& \mathbf{=}& \frac{\mathbf{1}}{\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}}\frac{{\mathbf{\hslash }}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{d}}^{\mathbf{2}}}{{\mathbf{dx}}^{\mathbf{2}}}\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\\ & \mathbf{=}& \frac{{\mathbf{e}}^{\mathbf{ax}}}{\mathbf{2}\sqrt{{\mathbf{a}}^{\mathbf{3}}}\mathbf{x}}\frac{{\mathbf{\hslash }}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\frac{{\mathbf{d}}^{\mathbf{2}}}{{\mathbf{dx}}^{\mathbf{2}}}\left(2\sqrt{a^3}{\mathrm{xe}}^{-\mathrm{ax}}\right)\\ & \mathbf{=}& \frac{{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{e}}^{\mathbf{ax}}}{\mathbf{4}\mathbf{mx}\sqrt{{\mathbf{a}}^{\mathbf{3}}}}\mathbf{2}\sqrt{{\mathbf{a}}^{\mathbf{3}}}\left[-2{\mathrm{ae}}^{-\mathrm{ax}}+{\mathrm{xa}}^2{e}^{-\mathrm{ax}}\right]\\ & \mathbf{=}& \mathbf{-}\frac{{\mathbf{\hslash }}^{\mathbf{2}}\mathbf{a}}{\mathbf{mx}}\mathbf{+}\frac{{\mathbf{\hslash }}^{\mathbf{2}}{\mathbf{a}}^{\mathbf{2}}}{\mathbf{2}\mathbf{m}}\end{array}$

Therefore, the potential energy for X > 0 is $\mathrm{U}\left(\mathrm{x}\right)=-\frac{{\mathrm{\hslash }}^2\mathrm{a}}{\mathrm{mx}}+\frac{{\mathrm{\hslash }}^2{\mathrm{a}}^2}{2\mathrm{m}}$.

The wave function of the particle is $\psi \left(x\right)$.

$\mathbf{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{l}\mathbf{2}\sqrt{{\mathbf{a}}^{\mathbf{3}}{\mathbf{xe}}^{\mathbf{-}\mathbf{ax}}}\mathbf{}\mathbf{}\mathbf{x}\mathbf{>}\mathbf{0}\\ \mathbf{0}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{x}\mathbf{<}\mathbf{0}\end{array}$

Calculation:

The time-independent Schrodinger equation relates the energy of the particle, E to the potential energy, U, the mass of the particle, m, the position of the particle, X, and the wave function, $\psi$

$-\frac{\mathrm{\hslash }}{2\mathrm{m}}\frac{\mathrm{d}}{\mathrm{dx}}\mathrm{\psi }\left(\mathrm{x}\right)+\mathrm{U}\left(\mathrm{x}\right)=\mathrm{E\psi }\left(\mathrm{x}\right)$

If the energy of the particles is 0 , then the potential energy can be is isolated by adding the first term to the other side and divided by the wave function.

$\mathrm{U}\left(\mathrm{x}\right)=\frac{1}{\mathrm{\psi }\left(\mathrm{x}\right)}\frac{{\mathrm{\hslash }}^2}{2\mathrm{m}}\frac{{\mathrm{d}}^2}{{\mathrm{dx}}^2}\mathrm{\psi }\left(\mathrm{x}\right)$

For

$$\begin{gathered} \mathrm{x}<0,\mathrm{\psi }\left(\mathrm{x}\right)=0 \\ \mathrm{U}\left(\mathrm{x}\right)=0 \end{gathered}$$

Thus, the potential energy is zero for X<0.