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We are given five idential spin-half particles in Linear Isotropic Harmonic Oscillator potential. It is required to find the degeneracy of the ground state of the system.

I understood the approach given in the Energy degeneracy of two particles in isotropic harmonic potential

But I am not sure how to get spin wave function for 5 identical particles. I mean currently, I do not have working knowledge of how to get Clebsch Gordan coefficients and things like that.

But generally, I have seen some of my collegues use combinatorial approach to deal with such problems. I am not finding this combinatorial approach in any textbook, this is why I suspect whether it is valid or not.

What they do is, Energy degeneracy of two particles in isotropic harmonic potential , for this example, for spin-1/2 particle to find degeneracy for first excited state is, after first energy level having three splits (100),(010),(001) , each split can have two possible states. So, it is $3 \times 2=6$ and also first electron in ground state can have 2 possible states $2 \times 1 = 2$. So, total degeneracy $6 \times 2 =12$.

Similarly, for second excited state, they divided it into two sub-possibilities. One particle in ground state, other in second excited state. Whereas, second possibility is, both particles is in first excited state. And we need to add both the possibilities.

For first case, we have 2 states spin up and down for particle in ground state, and $6 \times 2=12$ possible states for particle in second excited state. Because there will be 6 split for second excited energy level in 3d harmonic oscillator. Therefore, for both particles, there will be $12 \times 2=24$ degeneracy.

If both particles kept in first excited state, then there will be $^6C_2 = 15$ possibilities.

So, total $15+24=39$ degeneracy. Answer matches with what obtained in the given link.

Now, my objection with this method is, they do not even consider wave function symmetrization or anti-symmetrization to calculate degeneracies.

By the way, coming back to the original question posed above, the answer of above question posed is 20. They got this by taking two particles at ground state and now three particles remaining to put in 6 available states in first excites state. So, $^6C_3 = 20$. Here also how can we take wave function anti symmetrization into account ? Also, answer $20$ is given correct one. But I am not sure whether it is actually correct or not following standard symmetrization constraint problem as described in the link I put above.

Is combinatorial approach acceptable one in general?

Any suggestion would be helpful.

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    $\begingroup$ You need Young diagram techniques for this. In particular, there will be repeated values of total $S$. Probably the best source is Ping J, Wang F, Chen JQ. Group representation theory for physicists. World Scientific Publishing but if you do not understand CG technology this might be a struggle. See this post for some details with $4$ particles. $\endgroup$
    – ZeroTheHero
    Commented Feb 25, 2023 at 13:13
  • $\begingroup$ So combinatorial approach as described above will not work ? Thanks for resource suggestion $\endgroup$
    – curious_mind
    Commented Feb 25, 2023 at 13:44
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    $\begingroup$ It might work but is likely intractable for larger number of particles. Plus there are subtleties with repeated irreps which, unless you know how many irreps there are, will certainly defeat you. For two particles it’s easy enough but even with 3 you’d have to do quite a bit of work. $\endgroup$
    – ZeroTheHero
    Commented Feb 25, 2023 at 14:11
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    $\begingroup$ In your original question answer you already did take antisymmetrization into account! You realized there is only room for 2 fermions in the ground state, and you were lucky there was space for the other 3 in the first excited state, which is a constraint imposed by antisymmetry. I believe that you were just lucky the ground state does not block the first excited state fermions for the lowest total energy state, but you may not always be this lucky.... $\endgroup$
    – Cosmas Zachos
    Commented Feb 25, 2023 at 17:03
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    $\begingroup$ Yes. Are you familiar with Nakahara's book referenced here? As for your first question, I don't know how your source can bypass symmetry, as you claim: maybe you have to chase down its reference and review it here... $\endgroup$
    – Cosmas Zachos
    Commented Feb 26, 2023 at 14:59

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