Question

Consider a state described by \[ \psi(x, t) = \psi_2(x, t) + \psi_4(x, t), \] where} \( \psi_2(x, t) \) and \( \psi_4(x, t) \) are respectively the second and fourth normalized harmonic oscillator wave functions and \( \omega \) is the angular frequency of the harmonic oscillator. The wave function \( \psi(x, t = 0) \) will be orthogonal to \( \psi(x, t) \) at time \( t \) equal to
 

• A. \( \frac{\pi}{2\omega} \)
• B. \( \frac{\pi}{\omega} \)
• C. \( \frac{\pi}{4\omega} \)
• D. \( \frac{\pi}{6\omega} \)

Hint:

Orthogonality in quantum mechanics for energy eigenstates occurs when the phase difference between them is an odd multiple of \( \pi \).

Answer 1

The Correct Option is A

The given state \( \psi(x, t) \) is a linear superposition of the second and fourth normalized harmonic oscillator wave functions, i.e., \( \psi_2(x, t) \) and \( \psi_4(x, t) \), which are energy eigenstates of the quantum harmonic oscillator. The energy eigenvalues corresponding to these states are:
- Energy of state \( \psi_2(x, t) \) is \( E_2 = \left( 2 + \frac{1}{2} \right) \hbar \omega = \frac{5}{2} \hbar \omega \),
- Energy of state \( \psi_4(x, t) \) is \( E_4 = \left( 4 + \frac{1}{2} \right) \hbar \omega = \frac{9}{2} \hbar \omega \).
For the state \( \psi(x, t) = \psi_2(x, t) + \psi_4(x, t) \), the time-dependent wave functions are: \[ \psi_2(x, t) = \psi_2(x) e^{-i E_2 t / \hbar} = \psi_2(x) e^{-i \frac{5}{2} \omega t}, \] \[ \psi_4(x, t) = \psi_4(x) e^{-i E_4 t / \hbar} = \psi_4(x) e^{-i \frac{9}{2} \omega t}. \] At \( t = 0 \), we have: \[ \psi(x, t = 0) = \psi_2(x) + \psi_4(x). \] Now, for the wave functions to be orthogonal at some time \( t \), the phase difference between the two states must be an odd multiple of \( \pi \). The phase difference between the two energy eigenstates is: \[ \Delta \phi = \left( \frac{9}{2} \omega - \frac{5}{2} \omega \right) t = 2 \omega t. \] For orthogonality, we require: \[ 2 \omega t = \pi $\Rightarrow$ t = \frac{\pi}{2 \omega}. \] Thus, the wave function \( \psi(x, t = 0) \) will be orthogonal to \( \psi(x, t) \) at time \( t = \frac{\pi}{2\omega} \), which corresponds to option (A).