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I'm uncomfortable with the concepts of $N$ number of identical fermions and identical bosons and their degeneracies and energies.

For example, lets say we a have a 3-dimensional isotropic harmonic oscillator that has the eigenstates and eigenenergies as:

$$\begin{align} |n_{x},n_{y},n_{z} \rangle &= |n_{x}\rangle |n_{y}\rangle |n_{z} \rangle \\ E_{n_{x},n_{y},n_{z}} &= \hbar \omega(n_{x}+n_{y}+n_{z} + \frac{3}{2}) \end{align}$$

Lets say we have $N=5$ particles and we are looking for the ground state energy:

For a spin-1 boson: each $N$ particle will occupy its ground state. So if the ground state is just only $E_{tot,GR} = 5 \times E_{GR} = \frac{15}{2}\hbar \omega$. And would have a degeneracy of 1. Is the correct notion for $N$-identical bosons?

For a spin-1/2 fermion: This is where I am most uncomfortable thinking about. Identical fermions cannot occupy the same state. So one spin and and down fill the ground state, the next two fill up the next state, and the last one is in a unfilled state. How would you compute the ground state energy and degeneracies from this (assuming what I worked is right)?

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First you need to establish how many states have the same energy.

  1. For the ground state there is only one: $\vert 000\rangle$.
  2. For the first excited state there are three: $\vert 100\rangle$, $\vert 010\rangle$ and $\vert 001\rangle$.

Next, for bosons you can condensate all of them in the ground state so the ground state energy is $5\times \textstyle\frac{3}{2}\hbar\omega$.

For the fermions, you can have two in the ground state, and the remaining three in - say - one distinct h.o. state given above with $E_1=\textstyle\frac{5}{2}\hbar \omega$. Thus you get \begin{align} E&=2\times \textstyle\frac{3}{2}\hbar \omega + 3\times \textstyle\frac{5}{2}\hbar \omega \, ,\\ &=\left(2\times \textstyle\frac{3}{2}+ \textstyle\frac{15}{2}\right)\hbar \omega\, ,\\ &=\textstyle\frac{21}{2}\hbar \omega\, . \end{align}

Obviously you could put three more in the $N=1$ shell, and then you'd have to go to the $N=2$ shell, which contains 6 states: $$ \vert 200\rangle\, , \quad \vert 110\rangle\, , \quad \vert 101\rangle \, , \quad \vert 020\rangle\, , \quad \vert 011\rangle\, , \quad \vert 002\rangle \, , $$ where you could place a maximum of $12$ fermions etc.

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  • $\begingroup$ So the main take away is this: For any $N$ number of identical particles, bosons will always fill up the ground state and half spin fermions will fill up for ever increasing energy level configurations via pauli exclusion principle. So If I then think of $N=3$ identical particles, then all three bosons fill up the ground state, thus E = 9\hbar\omega/2 and is non-degenerate. For 1/2- spin fermions, Two will fill the ground state and one will fill the 1st excited state. Giving $3\hbar \omega + 5\hbar\omega/2 = 11\hbar\omega/2$ with a 3-fold degeneracy since their are three possible configuration $\endgroup$ – iron2man Apr 3 '17 at 1:53
  • $\begingroup$ for the first excited state $\endgroup$ – iron2man Apr 3 '17 at 1:54
  • $\begingroup$ something like this but degeneracy applies to energy levels, and not necessarily filled levels. Also if you count spin the $N=1$ are 6-fold degenerate you can can fix $6$ fermions on there, $\endgroup$ – ZeroTheHero Apr 3 '17 at 1:59
  • $\begingroup$ It is because that it can either be spin up or spin down, correct? $\endgroup$ – iron2man Apr 3 '17 at 2:28
  • $\begingroup$ @DarthLazar precisely. $\endgroup$ – ZeroTheHero Apr 3 '17 at 3:25

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