Question

Use equation 8.22 calculate the approximate transmission probability for a particle of energy E that encounters a finite square barrier of height V0 > E and width 2a. Compare your answer with the exact result to which it should reduce in the WKB regime T << 1.

Short answer

The solution of Transmission probability is $T~e^{-2y}$.

Step-by-step solution

To Calculate Transmission Probability.

Consider a finite square barrier of height $V_0>E$and width 2a, we find

$$\begin{gathered} \gamma =\frac{1}{ħ}{\int }_0^{2a}\sqrt{2m\left(V_0-E\right)}dx=\frac{2m}{ħ}\sqrt{2m\left(V_0-E\right)} \\ \mathrm{And},\mathrm{the}\mathrm{probability}\mathrm{of}\mathrm{transmission}\mathrm{is} \\ {\mathrm{T}}_{\mathrm{WKB}}\approx \exp \left[-\frac{4\mathrm{a}}{\mathrm{ħ}}\sqrt{2\mathrm{m}\left({\mathrm{V}}_0-\mathrm{E}\right)}\right] \\ \mathrm{Therefore},\mathrm{the}\mathrm{exact}\mathrm{result}\mathrm{is} \\ \mathrm{T}=\frac{1}{1+{\mathrm{Ksinh}}^2\mathrm{\gamma }} \\ \mathrm{Where}\mathrm{K}={\mathrm{V}}_0^2/4\mathrm{E}\left({\mathrm{V}}_0-\mathrm{E}\right). \\ \mathrm{Now}\mathrm{remember}\mathrm{that}\sinh \mathrm{x}=\left({\mathrm{e}}^{\mathrm{x}}-{\mathrm{e}}^{\mathrm{x}}\right)/2,\mathrm{then}\mathrm{we}\mathrm{find}\mathrm{in}\mathrm{the}\mathrm{WKB}\mathrm{regime}\mathrm{y}>>1. \\ {\sinh }^2\mathrm{y}=\left(\frac{{\mathrm{e}}^{\mathrm{y}}-{\mathrm{e}}^{-\mathrm{y}}}{2}\right)\approx \frac{{\mathrm{e}}^{2\mathrm{y}}}{4} \\ \mathrm{which}\mathrm{leads}\mathrm{to} \\ \mathrm{T}\approx \frac{1}{1+\left(\mathrm{K}/4\right){\mathrm{e}}^{2\mathrm{y}}} \\ \mathrm{In}\mathrm{the}\mathrm{WKB}\mathrm{regime}\mathrm{T}<<1\left(\mathrm{y}>>1\right),\mathrm{we}\mathrm{can}\mathrm{neglect}\mathrm{the}\mathrm{one}\mathrm{in}\mathrm{the}\mathrm{denominator}\mathrm{and}\mathrm{so} \\ \mathrm{we}\mathrm{show}\mathrm{this}\mathrm{result}\mathrm{reduces}\mathrm{to}\mathrm{the}\mathrm{WKB}\mathrm{transmission}\mathrm{probability}, \\ \mathrm{T}~{\mathrm{e}}^{-2\mathrm{y}} \\ \mathrm{Where}, \\ \mathrm{Y}=\left(2\mathrm{a}/\mathrm{ħ}\right)\sqrt{2\mathrm{m}\left({\mathrm{V}}_{-}\mathrm{E}\right)} \end{gathered}$$

Draw the graph.

To draw the graph of the finite square barrier of height $V_0>E$and width 2a .

For graph, finite square barrier of height$V_0>E$and width ,