Discussion
A particle in one-dimension is moving under the potential described by :
\[ V(x) = \begin{cases} \infty & \text{for } x \lt 0 \\ -V_{0} & \text{for } 0 \le x \le a,\ V_{0} \gt 0 \\ 0 & \text{for } x \gt a \end{cases} \]
The minimum depth of the potential for at least one bound state is :
Answer: B
The minimum depth is \(\frac{\hbar^{2}\pi^{2}}{8ma^{2}}\).
The infinite wall at \(x = 0\) forces \(\psi(0) = 0\), so inside the well \(\psi = A\sin kx\) with \(k = \sqrt{2m(V_{0} + E)}/\hbar\), and outside it decays as \(e^{-\kappa x}\). A bound state just appears when E reaches zero from below and the inside wave has turned through a quarter wavelength by \(x = a\), so that it can join a flat exterior solution:
\[ \begin{aligned} ka &= \frac{\pi}{2},\quad k^{2} = \frac{2mV_{0}}{\hbar^{2}} \\ V_{0} &= \frac{\pi^{2}\hbar^{2}}{8ma^{2}} \end{aligned} \]
This is the odd-parity condition of a finite well of width 2a, so unlike a symmetric well it needs a minimum depth. \(\frac{\hbar^{2}\pi^{2}}{2ma^{2}}\) is the ground-state energy of an infinite well of width a, a different quantity.
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