Discussion
Using WKB approximation, the transmission probability through a parabolic barrier :
\[ V(x) = \begin{cases} V_{0}\left(1 - \frac{x^{2}}{a^{2}}\right) & \text{for } |x| \lt a \\ 0 & \text{for } |x| \lt a \end{cases} \]
is :
Answer: B
The transmission probability is the second option, \(T = e^{-\pi\sqrt{2mV_{0}a^{2}/\hbar^{2}}\,(1 - E/V_{0})}\).
The barrier is \(V_{0}(1 - x^{2}/a^{2})\) inside \(|x| \lt a\) and zero outside (the paper prints \(|x| \lt a\) for both regions; the second is meant to be the outside). WKB gives \(T = \exp(-2\int\kappa\,dx)\) between the turning points \(\pm x_{0}\), where \(x_{0}^{2} = a^{2}(1 - E/V_{0})\):
\[ \begin{aligned} \kappa &= \frac{\sqrt{2mV_{0}}}{\hbar a}\sqrt{x_{0}^{2} - x^{2}} \\ \int_{-x_{0}}^{x_{0}}\kappa\,dx &= \frac{\sqrt{2mV_{0}}}{\hbar a}\cdot\frac{\pi x_{0}^{2}}{2} \\ 2\int\kappa\,dx &= \pi\sqrt{\frac{2mV_{0}a^{2}}{\hbar^{2}}}\left(1 - \frac{E}{V_{0}}\right) \end{aligned} \]
The first option has a positive exponent, which gives \(T \gt 1\). The last two use \(\frac{E}{V_{0}} - 1\), which is negative below the barrier top and again makes \(T\) exceed 1.
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