Exercise 39 gives a condition for resonant tunneling through two barriers separated by a space width of$\mathbf{2}\mathbf{s}$, expressed I terms of factor$\mathbf{\beta }$given in exercise 30. Show that in the limit in which barrier width$\mathbf{L}\mathbf{\to }\mathbf{\infty }$, this condition becomes exactly energy quantization condition (5.22) for finite well. Thus, resonant tunneling occurs at the quantized energies of intervening well.

Resonant tunneling is a phenomenon in which an electron enters from one side of a double barrier structure, and travels across it at or near the metastable levels in the quantum well.

The condition for resonant tunneling from the Exercise 39 is given by,

$2\mathrm{s}=\mathrm{\beta }/\mathrm{k}$

Where, $\mathrm{\beta }={\tan }^{-1}\left(\frac{2\mathrm{\alpha k}}{{\mathrm{k}}^2-{\mathrm{\alpha }}^2}\coth \left(\mathrm{\alpha L}\right)\right)$.

$k$= wave number

$L$ = barrier width

$\alpha$ =constant

In the limit $\mathrm{L}\to \mathrm{\infty }$$\coth \to 1$ and so:

$\begin{array}{c}\mathrm{\beta }={\tan }^{-1}\left(\frac{2\mathrm{\alpha k}}{{\mathrm{k}}^2-{\mathrm{\alpha }}^2}\right)\\ 2\mathrm{sk}={\tan }^{-1}\left(\frac{2\mathrm{\alpha k}}{{\mathrm{k}}^2-{\mathrm{\alpha }}^2}\right)\\ \tan \left(2\mathrm{sk}\right)=\left(\frac{2\mathrm{\alpha k}}{{\mathrm{k}}^2-{\mathrm{\alpha }}^2}\right)\end{array}$

Rearranging, $2\cot \left(\mathrm{k}.2\mathrm{s}\right)=\frac{\mathrm{k}}{\mathrm{\alpha }}-\frac{\mathrm{\alpha }}{\mathrm{k}}$

2s is the distance between the barriers; this is same as the expression (5.22) which is the energy quantization condition.

Here, the condition of resonant tunneling is proved to be equal to the energy quantization condition, when$\mathrm{L}\to \mathrm{\infty }$ . Thus, resonant tunneling occurs at the quantized energies of intervening well.