A beam of particles of energy incident upon a potential step of${\mathit{U}}_{\mathbf{0}}\mathbf{=}\left(3/4\right)$,is described by wave function:${\mathit{\psi }}_{\mathbf{i}\mathbf{n}\mathbf{c}}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}{\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{x}}$

The amplitude of the wave (related to the number of the incident per unit distance) is arbitrarily chosen as 1.

1. Determine the reflected wave and wave inside the step by enforcing the required continuity conditions to obtain their (possibly complex) amplitudes.
2. Verify the explicit calculation of the ratio of reflected probability density to the incident probability density agrees with equation (6-7).

A particle is defined by the wave function:${\mathbf{Be}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$for$\mathit{x}\mathbf{<}\mathbf{0}$and$\mathit{C}{\mathit{e}}^{\mathbf{4}\mathbf{x}}$for$\mathit{x}\mathbf{>}\mathbf{0}$. For the given wave function to becontinuous, at$\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{B}\mathbf{=}\mathit{C}$

To the left of the step $x<0$solve as:

$$\begin{gathered} \psi ={\psi }_{inc}+{\psi }_{ref} \\ \psi =e^{ikx}+B{e}^{-ikx} \end{gathered}$$

To the right of the step $x>0$solve as:

$\psi =C{e}^{ik\text{'}x}$

Considermust be continuous at $x=0$solve as:

$$\begin{gathered} e^0+B{e}^0=C{e}^0 \\ 1+B=C \end{gathered}$$

Consider $\frac{d\psi }{dx}$must be continuous at $x=0$.

$$\begin{gathered} ik{e}^0-ikB{e}^0=-\alpha C{e}^0 \\ k\left(1-B\right)=k\text{'}C \end{gathered}$$

From the first and second conditions solve as:

$$\begin{gathered} k\left(1-B\right)=k\text{'}\left(1+B\right) \\ \frac{k-k\text{'}}{k+k\text{'}}=\frac{\frac{\sqrt{2mE}}{h}-\frac{\sqrt{2m\left(E-\frac{3}{4}E\right)}}{h}}{\frac{\sqrt{2mE}}{h}+\frac{\sqrt{2m\left(E-\frac{3}{4}E\right)}}{h}} \end{gathered}$$

Divide by $\frac{\sqrt{2mE}}{h}$everywhere:

$$\begin{gathered} B=\frac{1-\sqrt{\left(1-\frac{3}{4}\right)}}{1+\sqrt{\left(1-\frac{3}{4}\right)}}=\frac{1-\frac{1}{2}}{1+\frac{1}{2}} \\ B=\frac{1}{3} \end{gathered}$$

Putting this in C, we get $C=\frac{4}{3}$

(a)

Write the components of the function as:

$$\begin{gathered} {\psi }_{refl}=B{e}^{-ikx} \\ =\frac{1}{3}{e}^{-ikx} \end{gathered}$$

Also:

$$\begin{gathered} {\psi }_{x>0}=C{e}^{ik\text{'}x} \\ =\frac{4}{3}{e}^{ik\text{'}x} \end{gathered}$$

$Here,k=\frac{\sqrt{2mE}}{h}andk\text{'}=\frac{\sqrt{2m\left(\frac{1}{4}E\right)}}{h}.$

(b)

If it is given that, ${\psi }_{inc}=e^{ikx}$and ${\psi }_{ref}=B{e}^{-ikx}$solve as:

$$\begin{gathered} \frac{{\left|{\psi }_{refl}\right|}^2}{{\left|{\psi }_{inc}\right|}^2}=\frac{{\left(\frac{1}{3}\right)}^2}{1^2} \\ =\frac{1}{9} \end{gathered}$$

From equation (6-7), $R=\frac{1}{9}$