Discussion
A particle is constrained to move in a truncated harmonic potential well (as shown below). Which one of the following statements is correct ?
The figure shows V(x) = \(\infty\) for x \(\lt\) 0 (a wall at the origin) and \(V = \frac{1}{2}m\omega^{2}x^{2}\) for x \(\gt\) 0.
Answer: D
The correct statement is that the energy of the first excited state is \(\frac{7}{2}\hbar\omega\).
The infinite wall forces \(\psi(0) = 0\). For x \(\gt\) 0 the potential is that of the full oscillator, so the allowed states are exactly the oscillator states that vanish at the origin: the odd ones, n = 1, 3, 5, and so on.
\[ E_{n} = (n + \tfrac{1}{2})\hbar\omega,\ n = 1, 3, \dots \]
The ground state is n = 1 with \(\frac{3}{2}\hbar\omega\), and the first excited state is n = 3 with \(\frac{7}{2}\hbar\omega\). So \(\frac{1}{2}\hbar\omega\) is wrong, and both parity statements fail because the surviving states are the odd-parity ones.
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