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## Homework Statement

Five non-interacting particles are placed in a three dimensional harmonic oscillator potential for which the single-particle energy is:

[tex]E_n = (n_x + n_y + n_z +3/2)\hbar\omega[/tex]

What is the lowest energy of the five-particle system when the particles are:

a) Distinguishable, spinless bosons

b) Identical, spinless bosons

c) Identical fermions each with spin s=1/2

d) Identical fermions each with spin s=3/2

## Homework Equations

I think effectively just the above equation thats given in the question.

## The Attempt at a Solution

From what I can tell so far:

The total energy will be a summation of the (lowest) energies of all the different particles.

For part a) because spin = 0, then the energy of the bosons will be:

[tex]E = E_n + E_{n'\prime} = (n + n\prime +1)\hbar\omega[/tex] (this is for 2 particles of different energies, would just add the other 3 terms into the brackets.)

Here the various values of n would be 1->5 as they are the lowest non-identical energy levels (correct?)

b) For this I would use the same equation, just with n=1 for all of them.

c) and d) Again I think I use the same equation, except that this time n will be dependent on the spin of the particle.

Any hints would be appreciated.

Thanks,

Brewer