A particle of mass m has the wave function$\psi \left(x\right)=Axexp\left(\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$a^2$}\right.\right)$ when it is in an allowed energy level with $E=0$.

a. Draw a graph of $\psi \left(x\right)$versus$x$.

b. At what value or values of $x$is the particle most likely to be found?

c. Find and graph the potential-energy function $U\left(x\right)$.

We need to find a graph of $\psi \left(x\right)$versus$x$.

Now, divide both sides of the given wave function as $aA$, and take the term $\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$a$}\right.$to be a new variable $u$:

$\frac{\psi \left(x\right)}{aA}=\frac{x}{a}exp\left(\raisebox{1ex}{$-x^2$}\!\left/ \!\raisebox{-1ex}{$a^2$}\right.\right)$

Here, replace $\left(\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$a$}\right.\right)$with $u$, the wave function is still a function of $x$because $a$is just a constant and we used the $u$for simplicity

$\frac{\psi \left(x\right)}{aA}=uexp\left(-u^2\right)$

Therefore, we have $\left(\frac{\psi \left(x\right)}{aA}\right)$on the $y-axis$and $u$on $x-axis$.

We need to find the value of$x$.

The particle is most likely to be found at the maxima of the probability density function $|\psi \left(x\right){|}^2$there, this can be described mathematically by setting the first derivative of $|\psi \left(x\right){|}^2$equation for $x$:

$\begin{array}{c}|\psi \left(x\right){|}^2={\left(Ax{e}^{-x^2/a^2}\right)}^2=A^2{x}^2{e}^{-2x^2/a^2}\\ \frac{d}{dx}|\psi \left(x\right){|}^2=A^2\left[2x{e}^{-2x^2/a^2}-\frac{4x^3}{a^2}{e}^{-2x^2/a^2}\right]=0\end{array}$

divide both sides by $2A{e}^{-2x^2/a^2}$and take $x$as a common factor:

$x\left[1-\frac{2x^2}{a^2}\right]=0$

Hence, one of the solutions to this equation is $\left(x=0\right)$and we can see from the graph in part (a) that the value of the wave function at $x=0$in minimum and so the probability density at is minimum. The two other solutions are ,$x=\pm \raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$\sqrt{2}$}\right.$and these are the points at which the particle is most likely to be found.

We need to find value or values of $x$is the particle most likely to be found.

The potential energy function can be found using the Schrodinger equation:

$-\frac{{ħ}^2}{2m}\frac{d^2}{d{x}^2}\psi \left(x\right)+U\left(x\right)\psi \left(x\right)=E\psi \left(x\right)$

There the energy is $E=0$. Substitute the wave function's expression and find its second derivative, then by rearranging the equation, we can get the potential energy as a function of $x$. Lets first find the second derivative of the wave function:

$$\begin{gathered} \frac{d^2}{d{x}^2}\psi \left(x\right)=\frac{d}{dx}\left(\frac{d}{dx}\psi \left(x\right)\right)=\frac{d}{dx}\left[\frac{d}{dx}\left(Ax{e}^{-x^2/a^2}\right)\right] \\ \frac{d^2}{d{x}^2}\psi \left(x\right)=A\frac{d}{dx}\left[e^{-x^2/a^2}-\frac{2x^2}{a^2}{e}^{-x^2/a^2}\right] \\ \frac{d^2}{d{x}^2}\psi \left(x\right)=\left[-\frac{2x}{a^2}{e}^{x^2/a^2}-\frac{4x}{a^2}{e}^{-x^2/a^2}+\frac{4x^3}{a^2}{e}^{x^2/a^4}\right] \\ \frac{d^2}{d{x}^2}\psi \left(x\right)=A\left[-\frac{2x}{a^2}-\frac{4x}{a^2}+\frac{4x^3}{a^4}\right]\psi \left(x\right) \end{gathered}$$

Since, the Schrodinger equation becomes

$\begin{array}{c}-\frac{{ħ}^2}{2m}A\left[-\frac{2x}{a^2}-\frac{4x}{a^2}+\frac{4x^3}{a^4}\right]\psi \left(x\right)+U\left(x\right)\psi \left(x\right)=0\\ U\left(x\right)\psi \left(x\right)=\frac{{ħ}^2}{2m}A\left[-\frac{2x}{a^2}-\frac{4x}{a^2}+\frac{4x^3}{a^4}\right]\psi \left(x\right)\\ U\left(x\right)=\frac{{ħ}^2}{2m}A\left[\frac{4x^3}{a^4}-\frac{6x}{a^2}\right]\end{array}$

The graph is given below.