# Identical Particles in a 1-D Harmonic Oscillator

• logic smogic

#### logic smogic

[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

$$E_{n}=\hbar \omega (n+\frac{1}{2})$$

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,

$$E_{gnd,tot}=3 \hbar \omega (0+\frac{1}{2}) = \frac{3}{2} \hbar \omega$$

(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, $E_{0}$ has degeneracy 2, and $E_{1}$ has degeneracy 1. The total energy will be,

$$E_{gnd,tot}=2 \hbar \omega (0+\frac{1}{2}) + \hbar \omega (1+\frac{1}{2}) = \frac{5}{2} \hbar \omega$$

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

## Answers and Replies

Would you really say having three identical bosons in the ground state was threefold degenerate? I wouldn't. They are indistinguishable. Rethink your degeneracy numbers. For c) fermi exclusion only applies between identical particles. You have two of one type and one of the other.

Would you really say having three identical bosons in the ground state was threefold degenerate? I wouldn't. They are indistinguishable. Rethink your degeneracy numbers. For c) fermi exclusion only applies between identical particles. You have two of one type and one of the other.

a.) Oh, I think I see. I was thinking of degeneracy wrong? It refers to the number of possible states with a particular energy, not the number of particles occupying a specific energy level, right? How would I go about determining the degeneracy of a system like this? (of course, I'll do the work - I'm just not sure what I'm supposed to do..)

c.) Well, presumably all three are distinguishable (say an electron, proton, and neutron). So I treat them as three noninteracting particles in a harmonic oscillator?

For a) I would say the degeneracy is 1. There's only one ground state possibility. For c), yes, I think all three distinguishable. How many different ground states (think spin)? Also rethink b).

So (a) is one since they will all be in the same state. But (c) will be $E_{0}$ has degeneracy 2, since they will each be in the ground state, but each can either be up or down (or since the Hamiltonian is just a superposition of the different individual particles Hamiltonians, is it (3!=6)?).

As for (b), I'm changing my answer to $E_{0}$ has degeneracy 2 and $E_{1}$ has degeneracy 2 (both can be occupied, and both have spin-up, spin-down options).

PS And thanks Dick! This is really instructive.

Think about b) again. You don't count degeneracies for each level and add them! You have to figure out how many different versions of the ground state there are. The E0 state will have two opposite spin fermions. There's only one way to do that! For c), I think you have the right picture but 3 particles with a choice of up/down does NOT give you 3! possibilities.

Wow, Logic Smogic, I had this same exact question on my final yesterday, the 13th. I wish I had seen this post beforehand. You don't happen to attend UPenn, do you? I think you'd make the ultimate study partner.