Given a system with N identical Fermions, with spin $\frac{1}{2}$ and mass $m$, subject to the potential:
$V(\vec{r}) = \frac{1}{2}m\omega^{2}(x^{2}+y^{2}+z^{2})$
and the single particle energy levels:
$E_{\vec{n}} = \hbar\omega(n_{x}+n_{y}+n_{z}+\frac{3}{2})$,
Calculate the Density of states $g(\epsilon)$.
I know that the density of states is the number of single particle states with energy between $(\epsilon,\epsilon +d\epsilon)$, per unit energy.
If we define $F(\epsilon)$ = #States with energy $<\epsilon$,
then $ g(\epsilon) = \frac{dF(\epsilon)}{d\epsilon}$
I am unsure how to get this #States in either case.
Best Answer
In this case, single particle states are denoted with $|n_x,n_y,n_z,\pm\rangle$. We first want to find the number of single particle states that have a specific energy $\epsilon(n) =\hbar \omega \left( \frac{3}{2} + n \right)$. These states are characterized by the constraint $n = n_x + n_y + n_z$. The fundamental realization is that we can list all the states with energy $\epsilon(n)$ as
\begin{equation} \left\{ \, |n_x,n_y,n-n_x-n_y,s\rangle \quad \mathrm{such \, that} \quad n_y =0,...,n-n_x \, ; \, n_x = 0,...,n \, ; \, s=\pm \, \right\} \,. \end{equation}
This means we can compute the degeneracy for single-particle states with energy $\epsilon(n)$ with the sum
\begin{equation} d(n)=2\times \sum^{n}_{n_x=0} \sum^{n-n_x}_{n_y=0} 1 \, . \end{equation}
Now it just remains to compute the number of states with energy less or equal than $\epsilon$, $F(\epsilon)$. More explicitly, we can use the notation we have been using to write
\begin{equation} F(\epsilon) = \sum^{n_f}_{n=0} d(n) \, , \end{equation}
with $n_f = -\frac{3}{2}+\frac{\epsilon}{\hbar \omega}$ (notice that $n_f$ is the radius of the Fermi sphere). Compute the sums and you'll get the result you want.