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In a box which has length $L$, there are 3 distinguishable particles which have same mass $m$ and 1/2 spin.

When one is in search for lowest energy level and degeneracies, I personally think one must consider the Pauli exclusion principle.

So the lowest level has two particles (up and down) have ground state energy level and 1st excited state particle (regardless of up and down).

But many solutions say all the particles are in ground state.

2x2x2 = 8 degeneracies from 3up-down degrees of freedom exist.

In this case, the Pauli exclusion principle should not be considered? why isn't the principle needed?

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  • $\begingroup$ Is it a cubic box: L x L x L? Follow on question: box L1 x L2 x L3. Does Pauli Exclusion apply there? $\endgroup$
    – isometry
    Commented Sep 29, 2018 at 9:37
  • $\begingroup$ L x L x L cubic box and so each particle has a n-vector(n1,n2,n3) for energy eigenvalue. $\endgroup$
    – Summal
    Commented Sep 29, 2018 at 10:16
  • $\begingroup$ So you've answered your own question. Count how many different values of vector (n1,n2,n3) all give the same lowest energy. Then apply Pauli exclusion separately to each of these states. $\endgroup$
    – isometry
    Commented Sep 29, 2018 at 10:45

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The particles are distinguishable. The exclusion principle applies only to indistinguishable particles. That's all there is to it.

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  • $\begingroup$ Roger, do you happen to know why does it only apply to indistinguishable particles? I have seen the statement several times, but I have never seen a deeper reason for it. Is it something that was added simply to match experimental data? $\endgroup$
    – user137661
    Commented Apr 23, 2019 at 3:14

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