Question

Consider a one-dimensional harmonic oscillator of angular frequency \( \omega \). If \( N \) identical particles occupy the energy levels of this oscillator at zero temperature, which of the following statement(s) about their ground state energy \( E_0 \) is (are) correct?

• A. If the particles are electrons, \( E_0 = \frac{13}{2} \hbar \omega \)
• B. If the particles are protons, \( E_0 = \frac{25}{2} \hbar \omega \)
• C. If the particles are spin-less fermions, \( E_0 = \frac{25}{2} \hbar \omega \)
• D. If the particles are bosons, \( E_0 = \frac{5}{2} \hbar \omega \)

Hint:

For fermions, the ground state energy is affected by the Pauli exclusion principle, while for bosons, the energy depends on Bose-Einstein condensation.

Answer 1

The Correct Option is A, C, D

Step 1: Understanding the ground state energy.
For a quantum harmonic oscillator, the energy levels are quantized, and the ground state energy depends on the particle type and statistics. The ground state energy for spin-less fermions is determined by the Pauli exclusion principle, leading to \( E_0 = \frac{25}{2} \hbar \omega \). For electrons, the energy is different due to the spin and statistical properties of fermions, yielding \( E_0 = \frac{13}{2} \hbar \omega \). For bosons, the energy levels are not affected by the Pauli exclusion principle, and their ground state energy is lower, resulting in \( E_0 = \frac{5}{2} \hbar \omega \).
Step 2: Conclusion.
Thus, the correct answers are options (A), (C), and (D).