Question

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).

Short answer

1. Therefore, the reflected wave is${\psi }_{refl}=\frac{1}{3}{e}^{-ikx}$ and inside the step is ${\psi }_{x>0}=\frac{4}{3}{e}^{ik\text{'}x}$.
2. The verified value is$R=\frac{1}{9}$

Step-by-step solution

Concept involved

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$\mathit{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{B}\mathbf{=}\mathit{C}$

Determining value of B and 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 $C=\frac{4}{3}$

Determining reflected wave and the wave inside step

(a)

Write the components of the function as:

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

Also:

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

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

Determining ratio of incident to reflected probability density

(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), $R=\frac{1}{9}$