Question

A particle of mass \( m \) is in a potential \( V(x) = \frac{1}{2} m \omega^2 x^2 \) for \( x > 0 \), and \( V(x) = \infty \) for \( x \leq 0 \), where \( \omega \) is the angular frequency. The ratio of the energies corresponding to the lowest energy level to the next higher level is:

• A. \( \frac{3}{7} \)
• B. \( \frac{1}{2} \)
• C. \( \frac{1}{3} \)
• D. \( \frac{3}{5} \)

Hint:

For a half-harmonic oscillator, the energy levels differ slightly from the full harmonic oscillator due to the boundary condition at \(x = 0\).

Answer 1

The Correct Option is A

In the case of a particle in a potential of the form \( V(x) = \frac{1}{2} m \omega^2 x^2 \), the energy levels for the quantum harmonic oscillator are given by:
\[ E_n = \left(n + \frac{1}{2}\right) \hbar \omega \quad \text{where} \quad n = 0, 1, 2, \dots \]
For the potential well defined on \( x > 0 \) (i.e., a half-harmonic oscillator), the energy levels are slightly altered due to the boundary condition at \( x = 0 \) (the infinite potential barrier). In this case, only the odd wavefunctions of the full harmonic oscillator are allowed, so the energy levels become:
\[ E_n = \left(n + \frac{1}{2}\right) \hbar \omega \quad \text{for odd } n \]
The lowest energy level corresponds to \( n = 1 \):
\[ E_1 = \frac{3}{2} \hbar \omega \]
The next higher level is at \( n = 3 \):
\[ E_3 = \frac{7}{2} \hbar \omega \]
The ratio of the energies is:
\[ \frac{E_1}{E_3} = \frac{\frac{3}{2} \hbar \omega}{\frac{7}{2} \hbar \omega} = \frac{3}{7} \]