Question

Herewetake direct approach to calculate reflection probability for tunneling mean while obtaining relationship applying in further exercise.

1. Write out thesmoothness condition oftheboundaries between regions for the $\mathit{E}\mathbf{<}{\mathit{U}}_{\mathbf{0}}$barrier from them. Show that the coefficient H of reflected wave is given by,
   $$\begin{gathered} \mathit{B}\mathbf{=}\mathit{A}\frac{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{h}\left(\alpha L\right)}{\sqrt{\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{}{\mathbf{h}}^{\mathbf{2}}\left(\alpha L\right)\mathbf{+}\mathbf{4}{\mathbf{\alpha }}^{\mathbf{2}}{\mathbf{k}}^{\mathbf{2}}\mathbf{/}{\left(k^2+{\alpha }^2\right)}^{\mathbf{2}}}}{\mathit{e}}^{\mathbf{-}\mathbf{t}\mathbf{\beta }} \\ \mathit{W}\mathit{h}\mathit{e}\mathit{r}\mathit{e}\mathbf{,} \\ \mathit{\beta }\mathbf{=}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{-}}\left(\frac{2\alpha k}{k^2-{\alpha }^2}coth\left(\alpha L\right)\right) \end{gathered}$$
   
2. Verify that the reflection probability R given in equation (6.16) follows from this result.

Short answer

1. ${\psi }_{x<0}\left(x\right)={\psi }_{x>0}\left(x\right)and{\overline{)\frac{d{\psi }_{x<0}}{dx}}}_{x-0}={\overline{)\frac{d{\psi }_{x>0}}{dx}}}_{x-0}$
2. Proved

Step-by-step solution

Step 1(a): Determine the smoothness condition of the boundaries between regions for the E < U0 barrier from them

Boundary conditions for E < U₀ are as follows.

1. Due to the requirement that the wave function value for the left side of the barrier must equal the wave function value for the right side of the barrier, ${\psi }_{X<0}\left(x\right)={\psi }_{X>0}\left(x\right)$
2. The first derivative on the left side must match the derivative on the right side since the wave function should be smooth.
   role="math" localid="1660028357372" ${\overline{)\frac{d{\psi }_{x<0}}{dx}}}_{x-0}={\overline{)\frac{d{\psi }_{x>0}}{dx}}}_{x-0}$
   

Step 2(b): Verify that the reflection probability R given in equation (6.16)

Consider the wave function which is given in equation 6.15

$\begin{array}{rcl}{\psi }_{x<0}\left(x\right)& =& A{e}^{+ikx}+B{e}^{-ikx}\\ {\psi }_{0<x-<L}\left(x\right)& =& C{e}^{+\alpha x}+D{e}^{-x}\\ {\psi }_{x>L}& =& F{e}^{+ikx}\end{array}$

Apply condition (i),

$\begin{array}{rcl}{\psi }_{x<0}\left(x\right)& =& {\psi }_{x>0}\left(x\right)\\ A{e}^{+ikx}+{\overline{)B{e}^{-ikx}}}_{x=0}& =& C{e}^{+\alpha x}+{\overline{)D{e}^{-\alpha x}}}_{x=0}\\ A+B& =& C+D\end{array}$

Apply condition (ii),

$\begin{array}{rcl}{\overline{)\frac{d{\psi }_{x<0}}{dx}}}_{x-0}& =& {\overline{)\frac{d{\psi }_{x>0}}{dx}}}_{x-0}\\ ikA{e}^{+ikx}-{\overline{)ikB{e}^{-ikx}}}_{x-0}& =& {\overline{)\alpha C{e}^{+\alpha x}-\alpha D{e}^{-\alpha x}}}_{x-0}\\ ik\left(A-B\right)& =& \alpha \left(C-D\right)\end{array}$

Now, apply condition for x = L,

$\begin{array}{rcl}{\psi }_{L<x}\left(L\right)& =& {\psi }_{x>L}\left(L\right)\\ C{e}^{+\alpha L}+D{e}^{-\alpha L}& =& F{e}^{+ikL}\\ {\overline{)\frac{d{\psi }_{L<x}\left(x\right)}{dx}}}_{x=L}& =& {\overline{)\frac{d{\psi }_{L>x}\left(x\right)}{dx}}}_{x=L}\\ \alpha C{e}^{+\alpha L}-\alpha D{e}^{-L}& =& ikF{e}^{+ikL}\end{array}$

Write the expression for reflection coefficient of wave.

$R=\frac{B^{*}B}{A^{*}A}......\left(1\right)$

Write the relation between B and A.

role="math" localid="1660029338206" $B=A\frac{\sin h\left(\alpha L\right)}{\sqrt{\sin h^2\left(\alpha L\right)+4{\alpha }^2{k}^2/{\left(k^2+{\alpha }^2\right)}^2}}{e}^{-i\beta }$

Determine $\frac{B}{A}$and $\frac{B^{*}}{A^{*}}$from above equation.

$\begin{array}{rcl}B& =& A\frac{\sin h\left(\alpha L\right)}{\sqrt{\sin h^2\left(\alpha L\right)+4{\alpha }^2{k}^2/{\left(k^2+{\alpha }^2\right)}^2}}{e}^{-i\beta }\\ \left(\frac{B^{*}}{A}\right)& =& \frac{\sin h\left(\alpha L\right)}{\sqrt{\sin h^2\left(\alpha L\right)+4{\alpha }^2{k}^2/{\left(k^2+{\alpha }^2\right)}^2}}{e}^{-i\beta }\end{array}$

Now, substitute the all values in equation (1)

$\begin{array}{rcl}R& =& \left(\frac{\sin h\left(\alpha L\right)}{\sqrt{\sin h^2\left(\alpha L\right)+4{\alpha }^2{k}^2/{(k^2+{\alpha }^2)}^2}}{e}^{-i\beta }\right)\left(\frac{\sin h\left(\alpha L\right)}{\sqrt{\sin h^2\left(\alpha L\right)+4{\alpha }^2{k}^2/{(k^2+{\alpha }^2)}^2}}{e}^{-i\beta }\right)\\ & =& \frac{\sin h^2\left({\displaystyle \frac{\sqrt{2m\left(U_0-E\right)}}{h}}L\right)}{\sin h^2\left(\frac{\sqrt{2m\left(U_0-E\right)}}{h}L\right)+4\left(\frac{\sqrt{2m\left(U_0-E\right)}}{h}L\right)k^2/{\left(k^2+{\left(\frac{\sqrt{2m\left(U_0-E\right)}}{h}L\right)}^2\right)}^2}\\ & =& \frac{\sin h^2\left({\displaystyle \frac{\sqrt{2m\left(U_0-E\right)}}{h}L}\right)}{\sin h^2\left(\frac{\sqrt{2m\left(U_0-E\right)}}{h}L\right)+4\left(\frac{2m\left(U_0-E\right)}{h}L\right)k^2/{\left(k^2+{\left(\frac{2m\left(U_0-E\right)}{h}L\right)}^2\right)}^2}\end{array}$

$k=\frac{\sqrt{2mE}}{h}and{\alpha }^2+k^2=\frac{2m{U}_0}{h^2}givesRinequation(6-16)$

Hence, it is verified that he reflection probability R given in equation (6.16) follows from this result.