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**[SOLVED] Identical Particles in a 1-D Harmonic Oscillator**

## Homework Statement

Three particles are confined in a 1-D harmonic oscillator potential. Determine the energy and the degeneracy of the

*ground state*for the following three cases.

**(a)**The particles are identical bosons (say, spin 0).

**(b)**The particles are identical fermions (say, spin 1/2).

**(c)**The particles are distinguishable spin 1/2 particles but have the same mass (e.g. a proton and a neutron).

## The Attempt at a Solution

*How does this sound...*

**(a)**The energy levels for a single particle in a 1-D harmonic oscillator are given by

[tex]E_{n}=\hbar \omega (n+\frac{1}{2})[/tex]

Since bosons can all occupy the same state, I presume they will. Therefore they will all be in the ground state with degeneracy 3. The energy will just be three times the ground state energy for one particle,

[tex]E_{gnd,tot}=3 \hbar \omega (0+\frac{1}{2}) = \frac{3}{2} \hbar \omega[/tex]

**(b)**Because of the Pauli Exclusion Principle, more than one fermion cannot occupy the same state. Therefore, there will be two particles in the ground state (one spin-up, one spin-down), and one particle in the first excited state.

Hence, [itex]E_{0}[/itex] has degeneracy 2, and [itex]E_{1}[/itex] has degeneracy 1. The total energy will be,

[tex]E_{gnd,tot}=2 \hbar \omega (0+\frac{1}{2}) + \hbar \omega (1+\frac{1}{2}) = \frac{5}{2} \hbar \omega[/tex]

**(c)**I have no idea. They behave like two bosons since they're distinguishable?