For the E>U₀ potential barrier, the reflection, and transmission probabilities are the ratios:

$$\begin{gathered} \mathbf{R}\mathbf{=}\frac{{\mathbf{B}}^{\mathbf{*}}\mathbf{B}}{{\mathbf{A}}^{\mathbf{*}}\mathbf{A}} \\ \mathbf{T}\mathbf{=}\frac{{\mathbf{F}}^{\mathbf{*}}\mathbf{F}}{{\mathbf{A}}^{\mathbf{*}}\mathbf{A}} \end{gathered}$$

Where A, B, and F are multiplicative coefficients of the incident, reflected, and transmitted waves. From the four smoothness conditions, solve for B and F in terms of A, insert them in R and T ratios, and thus derive equations (6-12).

Tunneling is a phenomenon in which a wavefunction tunnels or propagates through a potential barrier. Reflection happens if the wavefunction is not enough for tunneling.

Dividing the 4^th condition by the 3^rd and eliminating F, we get:

$\mathrm{k}=\frac{{\mathrm{k}}^{\text{'}}\left({\mathrm{Ce}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}-{\mathrm{De}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)}{\left({\mathrm{Ce}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}+{\mathrm{De}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)}$

This further gives $\mathrm{D}=-\frac{{\mathrm{k}}^{\text{'}}-\mathrm{k}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\mathrm{C}$

Put this in the condition in 1^st and 2^nd conditions, we get:

$\begin{array}{rcl}\mathrm{A}+\mathrm{B}& =& \left(1-\frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)\mathrm{C}\\ \frac{\mathrm{k}}{{\mathrm{k}}^{\text{'}}}\left(\mathrm{A}-\mathrm{B}\right)& =& \left(1+\frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)\mathrm{C}\end{array}$

Dividing these two $\frac{\mathrm{k}\left(\mathrm{A}-\mathrm{B}\right)}{{\mathrm{k}}^{\text{'}}\left(\mathrm{A}+\mathrm{B}\right)}=\frac{\left(1-{\displaystyle \frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)}{\left(1+{\displaystyle \frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)}$

Now, $\mathrm{B}=-\frac{\left(1+{\displaystyle \frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)-{\displaystyle \frac{\mathrm{k}}{{\mathrm{k}}^{\text{'}}}}\left(1-{\displaystyle \frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)}{\left(1+{\displaystyle \frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)+{\displaystyle \frac{\mathrm{k}}{{\mathrm{k}}^{\text{'}}}}\left(1-{\displaystyle \frac{\mathrm{k}-{\mathrm{k}}^{\text{'}}}{\mathrm{k}+{\mathrm{k}}^{\text{'}}}}{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)}\mathrm{A}$

Multiplying by ${\mathrm{k}}^{\text{'}}\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)$ to numerator and denominator and solving:

$\mathrm{B}=-\frac{\left({\mathrm{k}}^{\text{'}2}-{\mathrm{k}}^2\right)\left(1-{\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)}{{\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)}^2{\mathrm{e}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}-\left(\mathrm{k}-{\mathrm{k}}^{\text{'}}\right){\mathrm{e}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}}\mathrm{A}$

Multiply by ${\mathrm{e}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}$to numerator and denominator:

$$\begin{gathered} \mathrm{B}=-\frac{\left({\mathrm{k}}^{\text{'}2}-{\mathrm{k}}^2\right)\left(-2\mathrm{isin}\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)\right)}{{\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)}^2-\left(\mathrm{k}-{\mathrm{k}}^{\text{'}}\right){\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}}\mathrm{A} \\ \frac{{\mathrm{B}}^{*}\mathrm{B}}{{\mathrm{A}}^{*}\mathrm{A}}=\frac{2\mathrm{i}\left({\mathrm{k}}^{\text{'}2}-\mathrm{k}\right)\sin \left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)}{{\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)}^2{\mathrm{e}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}-{\left(\mathrm{k}-{\mathrm{k}}^{\text{'}}\right)}^2{\mathrm{e}}^{-{\mathrm{ik}}^{\text{'}}\mathrm{L}}}\frac{-2\mathrm{i}\left({\mathrm{k}}^{\text{'}2}-\mathrm{k}\right)\sin \left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)}{{\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)}^2{\mathrm{e}}^{-{\mathrm{ik}}^{\text{'}}\mathrm{L}}-{\left(\mathrm{k}-{\mathrm{k}}^{\text{'}}\right)}^2{\mathrm{e}}^{\mathrm{ikL}}} \\ \frac{{\mathrm{B}}^{*}\mathrm{B}}{{\mathrm{A}}^{*}\mathrm{A}}=\frac{4{\left({\mathrm{k}}^{\text{'}2}-{\mathrm{k}}^2\right)}^2{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)}{{\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)}^4+{\left(\mathrm{k}-{\mathrm{k}}^{\text{'}}\right)}^4-{\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)}^2{\left(\mathrm{k}-{\mathrm{k}}^{\text{'}}\right)}^2\left({\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}-{\mathrm{e}}^{-2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)} \end{gathered}$$

Put$\left({\mathrm{e}}^{2{\mathrm{ik}}^{\text{'}}\mathrm{L}}-{\mathrm{e}}^{-2{\mathrm{ik}}^{\text{'}}\mathrm{L}}\right)=\cos 2\mathrm{\theta }=1-2{\sin }^2{\mathrm{k}}^{\text{'}}\mathrm{L}$ and solve as:

$\frac{{\mathrm{B}}^{*}\mathrm{B}}{{\mathrm{A}}^{*}\mathrm{A}}=\frac{4{\left({\mathrm{k}}^{\text{'}2}-{\mathrm{k}}^2\right)}^2{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)}{{\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)}^4+{\left(\mathrm{k}-{\mathrm{k}}^{\text{'}}\right)}^4-{\left(\mathrm{k}+{\mathrm{k}}^{\text{'}}\right)}^2{\left(\mathrm{k}-{\mathrm{k}}^{\text{'}}\right)}^2\left(1-2{\sin }^2{\mathrm{k}}^{\text{'}}\mathrm{L}\right)}$

Arranging and solving you get:

$\mathrm{R}=\frac{{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)}{{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)+4{\displaystyle \frac{{\mathrm{k}}^{\text{'}2}{\mathrm{k}}^2}{{\left({\mathrm{k}}^{\text{'}2}-{\mathrm{k}}^{\text{'}2}\right)}^2}}}$

Since, data-custom-editor="chemistry" $\mathrm{R}+\mathrm{T}=1$. So, $\mathrm{T}=\frac{4{\displaystyle \frac{{\mathrm{k}}^{\text{'}2}{\mathrm{k}}^2}{{\left({\mathrm{k}}^{\text{'}2}-{\mathrm{k}}^{\text{'}2}\right)}^2}}}{{\sin }^2\left({\mathrm{k}}^{\text{'}}\mathrm{L}\right)+4{\displaystyle \frac{{\mathrm{k}}^{\text{'}2}{\mathrm{k}}^2}{{\left({\mathrm{k}}^{\text{'}2}-{\mathrm{k}}^{\text{'}2}\right)}^2}}}$.