In the $\mathbf{E}\mathbf{>}{\mathbf{U}}_{\mathbf{o}}$potential barrier, there should be no reflection when the incident wave is at one of the transmission resonances. Prove this by assuming that a beam of particles is incident at the first transmission resonance, $\mathbf{E}\mathbf{=}{\mathbf{U}}_{\mathbf{o}}\mathbf{+}\left({\pi }^2{h}^2/2{\mathrm{mL}}^2\right)$, and combining continuity equations to show that$\mathbf{B}\mathbf{=}\mathbf{0}$. (Note: k’ is particularly simple in this special case, which should streamline your work.)

Classically, the particle shouldn’t reflect. It should slow down between the barriers. But quantum mechanically, we apply the Schrödinger equation and get the solution to be of the form:${\mathbf{e}}^{\mathbf{\pm }\mathbf{ikx}}$. So,

$\begin{array}{rcl}{\mathbf{\psi }}_{\mathbf{x}\mathbf{<}\mathbf{0}}\left(x\right)& \mathbf{=}& {\mathbf{Ae}}^{\mathbf{ikx}}\mathbf{+}{\mathbf{Be}}^{\mathbf{-}\mathbf{ikx}}\\ {\mathbf{\psi }}_{\mathbf{0}\mathbf{<}\mathbf{x}\mathbf{<}\mathbf{L}}\left(x\right)& \mathbf{=}& {\mathbf{Ce}}^{{\mathbf{ik}}^{\mathbf{\text{'}}}\mathbf{x}}\mathbf{+}{\mathbf{De}}^{\mathbf{-}{\mathbf{ik}}^{\mathbf{\text{'}}}\mathbf{x}}\\ {\mathbf{\psi }}_{\mathbf{x}\mathbf{>}\mathbf{0}}\left(x\right)& \mathbf{=}& {\mathbf{Fe}}^{\mathbf{ikx}}\end{array}$

Here, α in second equation signifies the real nature.data-custom-editor="chemistry" $\mathbf{\psi }$= Wave Functions, A,B,C,D,F are the arbitrary multiplicative constants.

Applying boundary conditions for 0<x<L and x<0:

If, L= Width of potential barrier, $\mathrm{k}=\frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{V}}_{\mathrm{o}}\right)}}{{\mathrm{h}}^2}$and ${\mathrm{k}}^{\text{'}}=\frac{\sqrt{2\mathrm{mE}}}{{\mathrm{h}}^2}$

$$\begin{gathered} {\mathrm{\psi }}_{\mathrm{x}<0}\left(0\right)={\mathrm{\psi }}_{0<\mathrm{x}<\mathrm{L}}\left(0\right): \\ {\mathrm{Ae}}^{\mathrm{ik}0}+{\mathrm{Be}}^{-\mathrm{ik}0}={\mathrm{Ce}}^{{\mathrm{ik}}^{\text{'}}0}+{\mathrm{De}}^{-{\mathrm{ik}}^{,}0} \\ \mathrm{A}+\mathrm{B}=\mathrm{C}+\mathrm{D} \end{gathered}$$

Solve as:

$$\begin{gathered} \frac{d{\mathrm{\psi }}_{\mathrm{x}<0}}{d\mathrm{x}}{|}_{\mathrm{x}=0}=\frac{d{\mathrm{\psi }}_{0<\mathrm{x}<\mathrm{L}}}{d\mathrm{x}}{|}_{\mathrm{x}=0}: \\ {\mathrm{ikAe}}^{\mathrm{ik}0}-{\mathrm{ikBe}}^{-\mathrm{ik}0}={\mathrm{ik}}^{\text{'}}{\mathrm{Ce}}^{{\mathrm{ik}}^{\text{'}}0}-{\mathrm{ik}}^{\text{'}}{\mathrm{De}}^{{\mathrm{ik}}^{\text{'}}0} \\ \mathrm{k}\left(\mathrm{A}-\mathrm{B}\right)={\mathrm{k}}^{\text{'}}\left(\mathrm{C}-\mathrm{D}\right) \end{gathered}$$

Now applying boundary conditions for 0<x<L and x>L:

$$\begin{gathered} {\mathrm{\psi }}_{0<\mathrm{x}<\mathrm{L}}\left(\mathrm{L}\right)={\mathrm{\psi }}_{\mathrm{x}>0}\left(\mathrm{L}\right): \\ {\mathrm{Ce}}^{-{\mathrm{ik}}^{\text{'}}\mathrm{L}}+{\mathrm{De}}^{-{\mathrm{ik}}^{\text{'}}\mathrm{L}}={\mathrm{Fe}}^{\mathrm{ikL}} \end{gathered}$$

Also.

$$\begin{gathered} \frac{d{\mathrm{\psi }}_{0<\mathrm{x}<\mathrm{L}}}{d\mathrm{x}}{|}_{\mathrm{x}=\mathrm{L}}=\frac{d{\mathrm{\psi }}_{\mathrm{x}>0}}{d\mathrm{x}}{|}_{\mathrm{x}=\mathrm{L}}: \\ {\mathrm{ik}}^{\text{'}}\left({\mathrm{Ce}}^{{\mathrm{ik}}^{\text{'}}\mathrm{L}}-{\mathrm{De}}^{{\mathrm{ik}}^{,}\mathrm{L}}\right)={\mathrm{ikFe}}^{\mathrm{ikL}} \end{gathered}$$

From the second condition:$\mathrm{C}=\frac{\mathrm{k}}{{\mathrm{k}}^{\text{'}}}\left(\mathrm{A}-\mathrm{B}\right)+\mathrm{D}$put in first to obtain:

data-custom-editor="chemistry" $$\begin{gathered} \mathrm{C}=\frac{{\mathrm{k}}^{\text{'}}\left(\mathrm{A}+\mathrm{B}\right)+\mathrm{k}\left(\mathrm{A}-\mathrm{B}\right)}{2{\mathrm{k}}^{\text{'}}} \\ \mathrm{D}=\frac{{\mathrm{k}}^{\text{'}}\left(\mathrm{A}+\mathrm{B}\right)-\mathrm{k}\left(\mathrm{A}-\mathrm{B}\right)}{2{\mathrm{k}}^{\text{'}}} \end{gathered}$$

Substitute these values in the definition of reflection, the coefficient of reflection is given by:

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

Now, substituting the values in the above equation,$\mathrm{k}=\sqrt{\frac{2\mathrm{m}\left(\mathrm{E}-{\mathrm{V}}_{\mathrm{o}}\right)}{{\mathrm{h}}^2}}$and${\mathrm{k}}^{\text{'}}=\sqrt{\frac{2\mathrm{mE}}{{\mathrm{h}}^2}}$

Hence,$\mathrm{R}=\frac{{\sin }^2\left[{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}+{\mathrm{U}}_{\mathrm{o}}\right)}\mathrm{L}}{\mathrm{h}}}\right]}{{\sin }^2\left[{\displaystyle \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}+{\mathrm{U}}_{\mathrm{o}}\right)}\mathrm{L}}{\mathrm{h}}}\right]+4\left({\displaystyle \frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}}\right)\left[\left({\displaystyle \frac{\mathrm{E}}{{\mathrm{U}}_{\mathrm{o}}}}\right)+1\right]}$

Now, R is non-zero in general because the numerator contains sine term.

In order to consider no reflection, the numerator becomes zero and is equated to nπ, the condition is:

$$\begin{gathered} \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{U}}_{\mathrm{o}}\right)}}{\mathrm{h}}\mathrm{L}=\mathrm{n\pi } \\ \frac{\sqrt{2\mathrm{m}\left(\mathrm{E}-{\mathrm{U}}_{\mathrm{o}}\right)}}{\mathrm{h}}\mathrm{L}=\mathrm{\pi } \end{gathered}$$OR

(n=1 for first transmission resonance)

$\mathrm{E}={\mathrm{U}}_{\mathrm{o}}+\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}$

Here, E= Kinetic energy, U_o= Potential energy, h= Modified Plank’s constant

$\mathrm{E}={\mathrm{U}}_{\mathrm{o}}+\frac{{\mathrm{\pi }}^2{\mathrm{h}}^2}{2{\mathrm{mL}}^2}$

This expression is not a quantization condition but it defines the transmission resonance condition such that it occurs only at certain energies no matter what incident energy is.

Hence, there should be no reflection when the incident wave is at one of the transmission resonances