Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are given that
$$ \psi(x) = A x e^{-x^2 / a^2} $$
where $ \psi(x) $ has the following properties:
$\Rightarrow$ The function has a node at $ x = 0 $, meaning that $ \psi(0) = 0 $.
$\Rightarrow$ As $ x $ moves away from 0, $ \psi(x) $ increases to a peak and then decays exponentially as $ x \to \infty $.
See the figure below.
$$\color{blue}{\bf [b]}$$
The particle is most likely to be found where the probability density $ |\psi(x)|^2 $ is maximized. To find this point, we need to take the derivative of $ |\psi(x)|^2 $ with respect to $ x $, set it equal to zero, and solve for $ x $.
The probability density is given by:
$$
|\psi(x)|^2 = A^2 x^2 e^{-2x^2 / a^2}
$$
To maximize $ |\psi(x)|^2 $, we need to solve:
$$
\frac{d}{dx} |\psi(x)|^2 =\frac{d}{dx} \left(A^2 x^2 e^{-2x^2 / a^2} \right) = 0
$$
$$
A^2 \left( 2x e^{-2x^2 / a^2} - \frac{4x^3}{a^2} e^{-2x^2 / a^2} \right)=0
$$
$$
2x e^{-2x^2 / a^2} \left(1 - \frac{2x^2}{a^2} \right) = 0
$$
$$1 - \frac{2x^2}{a^2} =0$$
Thus,
$$\boxed{ x = \pm \frac{a}{\sqrt{2}} }$$
The particle is most likely to be found at $ x = \pm \frac{a}{\sqrt{2}} $.
$$\color{blue}{\bf [c]}$$
The Schrödinger equation in one dimension is given by:
$$
\frac{d^2 \psi}{dx^2} = \frac{2m}{\hbar^2} [U(x) - E] \psi(x)
$$
Since the energy $ E = 0 $,
$$
\frac{d^2 \psi}{dx^2} = \dfrac{2m}{\hbar^2} U(x) \psi(x)
$$
Thus,
$$
U(x) = \dfrac{\hbar^2}{2m} \dfrac{\dfrac{d^2 \psi}{dx^2}}{\psi(x)}\tag 1
$$
Now we need to find the first derivative of $ \psi(x) $;
$$
\psi(x) = A x e^{-x^2 / a^2}
$$
$$
\frac{d\psi}{dx} = A \left( e^{-x^2 / a^2} - \frac{2x^2}{a^2} e^{-x^2 / a^2} \right)
$$
Now we need to find the second derivative of $ \psi(x) $;
$$
\frac{d\psi}{dx} = A \left( e^{-x^2 / a^2} - \frac{2x^2}{a^2} e^{-x^2 / a^2} \right)
$$
Differentiate the first term;
$$
\frac{d}{dx} \left( e^{-x^2 / a^2} \right) = e^{-x^2 / a^2} \cdot \left( -\frac{2x}{a^2} \right) $$
$$\frac{d}{dx} \left( e^{-x^2 / a^2} \right)= -\frac{2Ax}{a^2} e^{-x^2 / a^2}\tag 2
$$
Differentiate the second term;
which is a product of two functions: $ -\frac{2x^2}{a^2} $ and $ e^{-x^2 / a^2} $. We will use the product rule to differentiate it $\left[
\frac{d}{dx} \left( u v \right) = u' v + u v'\right]$
In this case:
$ u = - \frac{2x^2}{a^2} $ and $ v = e^{-x^2 / a^2} $
Let's differentiate each term:
$ u' = \frac{d}{dx} \left( -\frac{2x^2}{a^2} \right) = -\frac{4x}{a^2} $
$ v' = \frac{d}{dx} \left( e^{-x^2 / a^2} \right) = -\frac{2x}{a^2} e^{-x^2 / a^2} $
Now apply the product rule:
$$
\frac{d}{dx} \left( - \frac{2x^2}{a^2} e^{-x^2 / a^2} \right) = \left( -\frac{4x}{a^2} \right) e^{-x^2 / a^2} + \left( -\frac{2x^2}{a^2} \right) \left( -\frac{2x}{a^2} e^{-x^2 / a^2} \right)
$$
$$
\frac{d}{dx} \left( - \frac{2x^2}{a^2} e^{-x^2 / a^2} \right) =\left( -\frac{4x}{a^2} \right) e^{-x^2 / a^2} + \left( \frac{4x^3}{a^4} \right) e^{-x^2 / a^2}
$$
Now, factor out $ e^{-x^2 / a^2} $:
$$
\frac{d}{dx} \left( - \frac{2x^2}{a^2} e^{-x^2 / a^2} \right) = e^{-x^2 / a^2} \left( -\frac{4x}{a^2} + \frac{4x^3}{a^4} \right)\tag 3
$$
Thus, from (2) and (3);
$$
\frac{d^2 \psi}{dx^2} = -\frac{2Ax}{a^2} e^{-x^2 / a^2} + A e^{-x^2 / a^2} \left( -\frac{4x}{a^2} + \frac{4x^3}{a^4} \right)
$$
Factor out $ A e^{-x^2 / a^2} $ from both terms:
$$
\frac{d^2 \psi}{dx^2} = A e^{-x^2 / a^2} \left( -\frac{2x}{a^2} - \frac{4x}{a^2} + \frac{4x^3}{a^4} \right)
$$
$$
\frac{d^2 \psi}{dx^2} = A e^{-x^2 / a^2} \left( -\frac{6x}{a^2} + \frac{4x^3}{a^4} \right)
$$
Hence, the second derivative of $ \psi(x) $ is:
$$
\frac{d^2 \psi}{dx^2} = A e^{-x^2 / a^2} \left( \frac{4x^3}{a^4} - \frac{6x}{a^2} \right)\tag 4
$$
Substituting (4) into (1);
$$
U(x) = \frac{\hbar^2}{2m}\dfrac{A e^{-x^2 / a^2} \left( \frac{4x^3}{a^4} - \frac{6x}{a^2} \right) }{ A x e^{-x^2 / a^2} }= \frac{\hbar^2}{2m} \left( \frac{4x^2}{a^4} - \frac{6}{a^2} \right)
$$
Thus,
$$
\boxed{U(x) = \frac{2 \hbar^2}{m a^2} \left( \frac{x^2}{a^2} - \frac{3}{2} \right)}
$$
We can see that this harmonic oscillator's potential energy is been shifted downward by a distance of $ \dfrac{-3 \hbar^2}{m a^2}$, see the graph below.
