Angular-momentum resolved density of states It is well known that in 3d, for a non-relativistic, free particle the density of states scales as $D(E) \propto E^{1/2}$. The problem is, we can classify the eigenstates according to the total angular momentum $l$. What is the density of state for the $l$-states?
 A: Your question could be interpreted in two different ways: derive $D(E)$ from the spherical basis or what is the density of states for the $L^2$ operator.
I’ll use the Hamiltonian of a free particle in infinite space (setting $m=\hbar=1$):
$$
H=\frac{\vec p^2}{2}
$$

*

*You can diagonalize $H$ by using angular momentum quantum numbers, $L_z,L^2$. However, you’ll still need an extra one $E$ or $p$ ($E=\frac{p^2}{2}$). You’ll get the eigenstates:
\begin{align}
\langle r,\theta, \phi |p,l,m\rangle &\propto e^{im\phi}j_l(p r) \\
H|p,l,m\rangle&= \frac{p^2}{2}|p,l,m\rangle \\
\langle p’,l’,m’|p,l,m\rangle &\propto \frac{1}{p^2}\delta(p-p’)\delta_{ll’}\delta_{mm’}
\end{align}
with $j_l$ the spherical Bessel function of the first kind (spherical Bessel transform).
The key is that the eigenvalues and the dot propucts do not depend on $l,m$, only on $p$. This means that for the density of states, the $l,m$ sums will only contribute to an overall multiplicative (formally infinite) factor and you still get
$$D(E)dE\propto p^2dp$$


*The spectrum of $L^2$ is discrete, so the density of states is a sum of Dirac peaks. Using the spherical Bessel transform, you can see that a $L^2$ eigenspace has degeneracy $2l+1$ by varying $m=-l,-l+1,…,l$ as well as a “constant infinite” degeneracy by varying $p$, so you get:
$$
D(L^2)\propto\sum_{l\in\mathbb N}(2l+1)\delta(L^2-l(l+1))
$$
Hope this helps.
