Chapter 6, Problem 17E

(a)

To determine

The wave function of the reflected wave.

(b)

To determine

To verify: The ratio of reflected to incident probability density is $\boxed{1}$

Formula Used:

The coefficient of reflected wave is

  $2ik=\left(ik-\alpha \right)C$

Calculation:

Now go back to the two conditions of a valid solution and eliminate the coefficient of the reflecred wave, B.

  $2ik=\left(ik-\alpha \right)C$

Solve the combined conditions for the coefficient of the decaying wave, C.

  $\begin{array}{l}C=\frac{2ik}{\left(ik-\alpha \right)}\\ =\frac{2ik}{\left(ik-\alpha \right)}\frac{\left(ik-\alpha \right)}{\left(-ik-\alpha \right)}\\ =\frac{2k^2-2ik\alpha }{k^2+{\alpha }^2}\\ \end{array}$

Use the coefficient of the decaying wave, C, the wave number of the incident wave, K, and the decay constant of the decaying wave, $\alpha$ , to complete the wave function, ${\psi }_{dee}$ .

  ${\psi }_{dec}=\frac{2k^2-2ik\alpha }{k^2+{\alpha }^2}{e}^{-ax}$

Substitute $\frac{2mE}{{\hslash }^2}for{k}^{2}and\frac{mE}{2{\hslash }^2}for{\alpha }^2$

  ${\psi }_{dec}=\frac{\left(\frac{2mE}{{\hslash }^2}\right)-2\left(\frac{mE}{{\hslash }^2}\right)i}{\frac{2mE}{{\hslash }^2}+\frac{mE}{2{\hslash }^2}}{e}^{-ax}$

Now go back to the two conditions of a valid solution and eliminate the coefficient of the reflected wave, B.

  $2ik=\left(ik-\alpha \right)C$

Solve the combined conditions for the coefficient of the decaying wave, C.

  $\begin{array}{l}C=\frac{2ik}{\left(ik-\alpha \right)}\\ =\frac{2ik}{\left(ik-\alpha \right)}\frac{\left(-ik-\alpha \right)}{\left(-ik-\alpha \right)}\\ =\frac{2k^2-2ik\alpha }{k^2+{\alpha }^2}\end{array}$

Use the coefficient of the decaying wave, C, the wave number of the incident wave, K, and the decay constant of the decaying wave, $\alpha$ , to complete the wave function, ${\psi }_{dec}$ .

  ${\psi }_{dec}=\frac{2k^2-2ik\alpha }{k^2+{\alpha }^2}$

Substitute $\frac{2mE}{{\hslash }^2}for{k}^{2}and\frac{mE}{2{\hslash }^2}for{\alpha }^2$

  ${\psi }_{dec}=\frac{\left(\frac{2mE}{{\hslash }^2}\right)-2\left(\frac{mE}{{\hslash }^2}\right)i}{\frac{2mE}{{\hslash }^2}+\frac{mE}{2{\hslash }^2}}{e}^{-ax}$

  $\begin{array}{l}=\frac{\left(\frac{4mE}{{\hslash }^2}\right)-\left(\frac{2mE}{{\hslash }^2}\right)i}{\frac{5mE}{2{\hslash }^2}}{e}^{-ax}\\ =\left(\frac{8}{5}-\frac{4}{5}i\right)e^{-ax}\end{array}$

The ratio of the reflected probability density to the incident probability density can be found by diving the magnitude squared of the reflected wave, ${\left|{\psi }_{ref}\right|}^2$ by the magnitude squared of the incident wave, ${\left|{\psi }_{inc}\right|}^2$ .

  $\begin{array}{l}\frac{{\left|{\psi }_{ref}\right|}^2}{{\left|{\psi }_{inc}\right|}^2}=\frac{\left[\left(\frac{3}{5}+\frac{4}{5}i\right)e^{i\sqrt{2me/{}^2}x}\right]\left[\left(\frac{3}{5}-\frac{4}{5}i\right)e^{-i\sqrt{2me/{}^2}x}\right]}{\left[e^{-i\sqrt{2me/{}^2}x}\right]\left[e^{i\sqrt{2me/{}^2}x}\right]}\\ =\left(\frac{3}{5}+\frac{4}{5}i\right)\left(\frac{3}{5}-\frac{4}{5}i\right)\\ =\frac{9}{25}+\frac{16}{25}\\ =1\end{array}$

Conclusion:

The ratio of reflected probability density to the icidemt prpability density is $\boxed{1}$