Chapter 8: Q3P (page 324)

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

Expert verified

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

Step by step solution

To Calculate Transmission Probability.

Consider a finite square barrier of height $V_{0} > E$ and width 2a, we find

$γ = \frac{1}{ħ} \int_{0}^{2 a} \sqrt{2 m (V_{0} - E)} d x = \frac{2 m}{ħ} \sqrt{2 m (V_{0} - E)} And, the probability of transmission is T_{WKB} \approx \exp [- \frac{4 a}{ħ} \sqrt{2 m (V_{0} - E)}] Therefore, the exact result is T = \frac{1}{1 + {Ksinh}^{2} γ} Where K = V_{0}^{2} / 4 E (V_{0} - E) . Now remember that \sinh x = (e^{x} - e^{x}) / 2, then we find in the WKB regime y > > 1 . \sinh^{2} y = (\frac{e^{y} - e^{- y}}{2}) \approx \frac{e^{2 y}}{4} which leads to T \approx \frac{1}{1 + (K / 4) e^{2 y}} In the WKB regime T < < 1 (y > > 1), we can neglect the one in the denominator and so we show this result reduces to the WKB transmission probability, T ~ e^{- 2 y} Where, Y = (2 a / ħ) \sqrt{2 m (V_{-} E)}$