- #1
Brewer
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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 that's 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