Particles of energy Eare incident from the left, where U(x)=0, and at the origin encounter an abrupt drop in potential energy, whose depth is -3E.

1. Classically, what would the particles do, and what would happen to their kinetic energy?
2. Apply quantum mechanics, assuming an incident wave of the form${\mathbf{\psi }}_{\mathbf{inc}}\mathbf{=}{\mathbf{e}}^{\mathbf{ikx}}$, where the normalization constant has been given a simple value of 1, determine completely the wave function everywhere, including numeric values for multiplicative constants.
3. What is the probability that incident particles will be reflected?

Energy contained by the object by the virtue of its motion is called kinetic energy.

Potential energyis the energy contained by an object by the virtue of its position.

Classical particles would continue to the right, their kinetic energy abruptly increasing from E to 4E $\left(\left[\mathrm{E}-\left(-3\mathrm{E}\right)\right]\right)$.

Consider the function as:

${\mathrm{\psi }}_{\mathrm{inc}}={\mathrm{e}}^{\mathrm{ikx}}$

To the left of the drop,$\mathrm{x}<0$solve as:

$\begin{array}{rcl}\mathrm{\psi }& =& {\mathrm{\psi }}_{\mathrm{inc}}+{\mathrm{\psi }}_{\mathrm{refl}}\\ & =& {\mathrm{e}}^{\mathrm{ikx}}+{\mathrm{Be}}^{-\mathrm{ikx}}\end{array}$

To the right of the drop, (x>0):$\mathrm{\psi }={\mathrm{Be}}^{{\mathrm{ik}}^{\text{'}}\mathrm{x}}$

Here,

$\begin{array}{rcl}{\mathrm{k}}^{\text{'}}& =& \frac{\sqrt{2\mathrm{m}\left[\mathrm{E}-\left(-3\mathrm{E}\right)\right]}}{\mathrm{h}}\\ & =& 2\frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}\end{array}$

Here,data-custom-editor="chemistry" $\mathrm{\psi }$ must be continuous at data-custom-editor="chemistry" $\mathrm{x}=0$.

$$\begin{gathered} {\mathrm{e}}^{\mathrm{o}}+{\mathrm{Be}}^{\mathrm{o}}={\mathrm{Ce}}^{\mathrm{o}} \\ 1+\mathrm{B}=\mathrm{C} \end{gathered}$$

Here,$\frac{d\mathrm{\psi }}{d\mathrm{x}}$ must be continuous at$\mathrm{x}=0$.

$$\begin{gathered} {\mathrm{ike}}^{\mathrm{o}}-{\mathrm{ikBe}}^{\mathrm{o}}=-{\mathrm{\alpha Ce}}^{\mathrm{o}} \\ \mathrm{k}\left(1-\mathrm{B}\right)={\mathrm{k}}^{\text{'}}\mathrm{C} \end{gathered}$$

From the first and second conditions solve as:

$$\begin{gathered} \mathrm{k}\left(1-\mathrm{B}\right)={\mathrm{k}}^{\text{'}}\left(1+\mathrm{B}\right) \\ \mathrm{B}=\frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}} \\ \mathrm{B}=\frac{{\displaystyle \frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}}-{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-\left(-3\mathrm{E}\right)\right)}}{\mathrm{h}}}}{{\displaystyle \frac{\sqrt{2\mathrm{mE}}}{\mathrm{h}}}+{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-\left(-3\mathrm{E}\right)\right)}}{\mathrm{h}}}} \\ \mathrm{B}=-\frac{1}{3} \end{gathered}$$

So, data-custom-editor="chemistry" $\mathrm{C}=\frac{2}{3}$.

Hence the required wave functions are given by,

$$\begin{gathered} {\mathrm{\psi }}_{\mathrm{refl}}=-\frac{1}{3}{\mathrm{e}}^{-\mathrm{ikx}} \\ {\mathrm{\psi }}_{\mathrm{x}>0}=\frac{2}{3}{\mathrm{e}}^{{\mathrm{ik}}^{,}\mathrm{x}} \end{gathered}$$

The required probability can be calculated by

$\frac{{\mathrm{B}}^{*}\mathrm{B}}{{\mathrm{A}}^{*}\mathrm{A}}=\frac{1}{9}$