A beam of particles of energy E incident upon a potential step of${\mathit{U}}_{\mathbf{0}}\mathbf{=}\left(5/4\right)$E is described by wave function:${\mathit{\psi }}_{\mathbf{i}\mathbf{n}\mathbf{c}}\left(x\right)\mathbf{=}{\mathit{e}}^{\mathbf{i}\mathbf{k}\mathbf{x}}$

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 the ratio of reflected probability density to the incident probability density is 1.

A particle is defined by the wave function: $\mathit{B}{\mathit{e}}^{\mathbf{-}\mathbf{2}\mathbf{x}}$ 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 left of the step $x<0$as:

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

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

$\psi =C{e}^{-\alpha x}$

Her, $\psi$must be continuous at $x=0$.

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

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

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

From first and second condition:

$$\begin{gathered} ik\left(1-B\right)=\alpha \left(1+B\right) \\ \frac{ik+\alpha }{ik-\alpha }=\frac{i\frac{\sqrt{2mE}}{h}+\frac{\sqrt{2m\left(\frac{5}{4}E-E\right)}}{h}}{i\frac{\sqrt{2mE}}{h}-\frac{\sqrt{2m\left(\frac{5}{4}E-E\right)}}{h}} \end{gathered}$$

Divide by √2mE/$\hslash$everywhere:

$$\begin{gathered} B=\frac{i+\sqrt{\left(\frac{5}{4}-1\right)}}{i-\sqrt{\left(\frac{5}{4}-1\right)}}=\frac{i+\frac{1}{2}}{i-\frac{1}{2}} \\ B=\frac{3}{5}-i\frac{4}{5} \end{gathered}$$

Substitute this in C, the equation obtained is:

$C=\frac{8}{5}-i\frac{4}{5}$

If,${\psi }_{refl}=B{e}^{-\alpha x}$

By putting value of B from Step 3 in the above-mentioned equation and solve:

${\psi }_{refl}=\left(\frac{3}{5}-i\frac{4}{5}\right)e^{-\alpha x}$

If, ${\psi }_{x>0}=C{e}^{-\alpha x}$

By putting value of C from Step 3 in the above-mentioned equation solve as:

${\psi }_{x>0}=\left(\frac{8}{5}-i\frac{4}{5}\right)e^{-\alpha x}$

Hence, the reflected wave is given by, ${\psi }_{refl}=\left(\frac{3}{5}-i\frac{4}{5}\right)e^{-\alpha x}$ and the wave inside the step can be written as, ${\psi }_{x>0}=\left(\frac{8}{5}-i\frac{4}{5}\right)e^{-\alpha x}$

The ratio of reflected and incident probability can be calculated by the following equation,

$$\begin{gathered} B^{*}B=\left(\frac{3}{5}+i\frac{4}{5}\right)\left(\frac{3}{5}-i\frac{4}{5}\right) \\ =1 \end{gathered}$$