Question

A particle experiences a potential energy given by$\mathit{U}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{(}{\mathit{x}}^{\mathbf{2}}\mathbf{-}\mathbf{3}\mathbf{)}{\mathit{e}}^{\mathbf{-}{\mathbf{x}}^{\mathbf{2}}}$

1. Make a sketch of of$\mathit{U}\mathbf{\left(}\mathit{x}\mathbf{\right)}$, with numerical values of minima and maxima.
2. What is the maximum energy that a particle can have and yet be bound?
3. What is the maximum energy that a particle can have and yet be bound for a considerable amount of time?
4. Is it possible for a particle to have more energy than in part (c) and still be bound for some period of time?

Short answer

a.

(b)must be no greater than zero.

(c)Energy is between 0 and${\mathit{e}}^{\mathbf{-}\mathbf{4}}$.

(d)Yes,it might “bounce back and forth” for some time.

Step-by-step solution

Definition of bound state

Bound state is a state in quantum physics in which when a particle is provided with a potential energy, it tends to remain in the one or more regions of the state.

The maxima and minima of a given function$f\left(x\right)$, can be easily calculated by finding solution of the equation$f\text{'}\left(x\right)=0$,where, $f\text{'}\left(x\right)$represents the differential of function$f\left(x\right)$.

Given equations

In the given scenario, particle is subject to the following potential energy:

$U\left(x\right)=(x^2-3)e^{-x^2}$

Calculations

In order to find the maximum and minimum value of the energy, we need to equate the differential of the given energy to zero.

$\begin{array}{rcl}\frac{d}{dx}\left(x^2-3\right)e^{-x^2}& =& \left[2x+\left(x^2-3\right)\left(-2x\right)\right]e^{-x^2}\\ \frac{d}{dx}\left(x^2-3\right)e^{-x^2}& =& \left(-2x^2+8\right)x{e}^{-x^2}\\ & =& 0\\ x& =& 0,\pm 2,\pm \infty \\ & & \end{array}$

Hence, the energy will have its extreme positions at$x=\pm \infty$, $x=0$and$x=\pm 2$

At$x=\pm \infty$,$\psi \left(x\right)=0$.

At$x=0$,$\psi \left(x\right)=-3$.

At $x=\pm 2$, $\psi \left(x\right)=e^{-4}$.

Step 3: (a) Explanation

By taking into account the values of energies on the extremities of the energy given, the following graph can be plotted.

It can be clearly seen in the above-mentioned graph that the extremities lie on the values -2, 0 and +2 and those are$e^{-4}$, -3 and $e^{-4}$respectively.

(b) Explanation

It would only be confined eternally if it was "below the walls" on both sides and unable to tunnel, meaning that the walls did not subsequently dip below the total energy. E has to be less than zero for this to hold true.

Hence, Energy must not be greater than zero.

(c) Explanation

The particle would be bound classically but could quantum mechanically tunnel if it were below the walls but above the point at which the walls extend further out. This is the case if the energy is between 0 and $e^{-4}$.

Hence, Energy should be between 0 and$e^{-4}$

(d) Explanation

Yes. Even if it is conventionally unbound over the tops of the walls, there is a quantum mechanical possibility of reflection at potential energy changes, thus it may "bounce back and forth" for a while.

Hence, yes, it might “bounce back and forth” for some time.