Given that the particle’s total energy is$\mathbf{0}$, show that the potential energy is role="math" localid="1657529957489" $\mathbf{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\frac{{\mathbf{h}}^2}{\mathbf{2}{\mathbf{mb}}^4}{\mathbf{x}}^2\mathbf{-}\frac{\mathbf{3}{\mathbf{h}}^2}{\mathbf{2}{\mathbf{mb}}^2}$.

The potential energy is described as the energy held by a particular object due to its position relative to the other objects. The potential energy is theenergy attained by the objects at rest.

The equation of the kinetic energy of the particle is expressed as:

$K\left(x\right)=\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}$ …(i)

Here,$K\left(x\right)$is the kinetic energy as a function of$x$ , $\hslash$is the Planck’s constant, $m$is the mass of the particle and $\frac{d^2}{d{x}^2}$is the double derivative.

The equation of the wave function is expressed as:

$\psi \left(x\right)=Ax{e}^{\frac{x^2}{2b^2}}$

Here,$\psi \left(x\right)$ is the wave function as a function of$x$ , $A$and$b$ are the constants and $x$is the distance between the particle and the wave.

The equation of the total energy is expressed as:

$E\psi \left(x\right)=K\left(x\right)\psi \left(x\right)+U\left(x\right)\psi \left(x\right)$

…(ii)

Here,$E\psi \left(x\right)$ is the total energy, $K\left(x\right)\psi \left(x\right)$is the kinetic energy and $U\left(x\right)\psi \left(x\right)$is the potential energy.

Substitute $0$for $E\psi \left(x\right)$and$\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}$ for $K\left(x\right)$as obtained from the equation (i).

$\begin{array}{c}0=\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}\left(\psi \left(x\right)\right)+U\left(x\right)\psi \left(x\right)\\ U\left(x\right)\psi \left(x\right)=-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}\left(\psi \left(x\right)\right)\end{array}$

Substitute$Ax{e}^{\frac{x^2}{2b^2}}$ for$\psi \left(x\right)$ in the above equation.

$\begin{array}{c}U\left(x\right)\psi \left(x\right)=-\frac{{\hslash }^2}{2m}\frac{d^2}{d{x}^2}\left(Ax{e}^{\frac{x^2}{2b^2}}\right)\\ =-\frac{{\hslash }^2}{2m}(-\frac{3}{b^2}+\frac{1}{b^4}{x}^2)\left(Ax{e}^{\frac{x^2}{2b^2}}\right)\\ =-\frac{{\hslash }^2}{2m}(-\frac{3}{b^2}+\frac{1}{b^4}{x}^2)\psi \left(x\right)\\ U\left(x\right)=\frac{{\hslash }^2}{2m{b}^4}{x}^2-\frac{3{\hslash }^2}{2m{b}^2}\end{array}$

Thus, the potential energy is $\frac{{\hslash }^2}{2m{b}^4}{x}^2-\frac{3{\hslash }^2}{2m{b}^2}$proved.