# Calculating density of states for $N$ identical Fermions subject to a potential

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.

• I think you're finding an energy spectrum (including degeneracies)--so the question is what values do $E_n$ take for various $n_i$ combos. In my opinion, the density of states refers to a continuum of states, which you don't have. The isotropic harmonic oscillator (that's a search hint) has bound states in a discrete spectrum. – JEB Feb 10 '18 at 15:12
• @JEB No, he is looking for a density of states, i'm according to you on the fact that here there is not a "continuum of states" but usually in this cases you proceed with the semi-classical approximation which means that you consider the real discrete state of the system like a continuum -> the sum over quantum states becomes an integral over the classical phase space. – MRT Aug 5 '18 at 12:11

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

$$\left\{ \, |n_x,n_y,n-n_x-n_y,s\rangle \quad \mathrm{such \, that} \quad n_y =0,...,n-n_x \, ; \, n_x = 0,...,n \, ; \, s=\pm \, \right\} \,.$$

This means we can compute the degeneracy for single-particle states with energy $\epsilon(n)$ with the sum

$$d(n)=2\times \sum^{n}_{n_x=0} \sum^{n-n_x}_{n_y=0} 1 \, .$$

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

$$F(\epsilon) = \sum^{n_f}_{n=0} d(n) \, ,$$

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.

First of all the single particle hamiltonian in this framework is: $$h_{s}= \frac{p^2}{2m} + \frac{1}{2} m \omega^{2} r^{2}$$ i make the semi-classical approximation, in this way the density of state can be written as: $$g(\epsilon) = \frac{(2 S +1)}{(2\pi \hbar)^{3}}\int d^{3}\vec{r}d^{3}\vec{p} \delta(\epsilon - h_{s}) = \frac{2(4\pi)^{2}}{(2\pi\hbar)^{3}}\int_{0}^{r_{max}} r^{2} dr\int_{0}^{p_{max}} p^{2} dp \delta( \epsilon -h_{s})$$ where in the last equality i have intergrated over the angles in sperichal coordinates and setted $S=\frac{1}{2}$ interpretating fermions as electrons of spin $\frac{1}{2}$. In order to compute the integral i give to you the following input: consider the argument of the delta function as a function of p: $$f(p) = \epsilon - \frac{p^{2}}{2 m} - \frac{m \omega^{2} r^{2}}{2}$$ The roots with respect to $p$ are: $$p_{+/-} = \sqrt{ 2m\epsilon - m^{2} \omega^{2}r^{2}}$$ then: $$|f'(p_{+/-})| =\left|\frac{-p_{+/-}}{m}\right|$$ is equal for both the roots. You can now use the following delta function property: $$\delta\left(\epsilon - \frac{p^{2}}{2 m} - \frac{m \omega^{2} r^{2}}{2}\right)= \frac{\delta(p-p_{+}) + \delta(p-p_{-})}{|f'(p_{+/-})|}$$ the action of the delta function solve the integral in $p$ and at this point you have: $$\frac{(4 m) (4 \pi)^{2}}{(2 \pi \hbar)^{3}} \int_{0}^{r_{max}} dr r^{2} \sqrt{ 2 m \epsilon - m^{2} \omega^{2} r^{2} }$$ Where $r_{max}$ is given by the domain of the square root (i.e. $r_{max}= \sqrt{\frac{2\epsilon} {m \omega^{2}}}$)

The computation of this last integral will take you to the following result: $$g(\epsilon) = \frac{2 \epsilon ^{2} } {(\hbar \omega)^{3}}$$ You can finally observe that it is dimensionally correct: $$[g] = \frac{ E^{2} }{\frac{E^{3}t^{3}}{t^{3}}} = \frac{1}{E}$$ which is the correct dimension for a density of states.