Assume a Hamiltonian $H$ with $N$ orthonormal eigenstates $\{\vert n\rangle\}$ of energies $\epsilon_n$. One can define a density of states, \begin{align} \rho(E)&=\mathrm{tr}\,\hat{\delta}(E-\hat{H})\\ &=\sum_{n=1}^{N}\langle n\vert\hat{\delta}(E-\hat{H}\vert n\rangle\\ &=\sum_{n=1}^N \delta(E-\epsilon_n). \end{align} Here, the second line can be taken to be the definition of the "operator delta function" $\hat{\delta}$ in the first line.
However, one also typically sees delta functions with "matrix arguments" in path integrals of matrix models, such as
\begin{align} \tilde\delta(E-H)=\int DT \,e^{i\mathrm{tr}[T(E-H)]}, \end{align}
where $T$ is a Hermitian matrix. This is discussed, for instance, in arXiv:1607.02871. However, this path integral computes (p.16, proposition 3.7 of the above reference), setting $X\!=(\!E\!-\!H)$,
\begin{align} \tilde\delta(X)\propto \prod_j \delta(X_{jj}) \prod_{k<j}\delta(\mathrm{Re}X_{jk})\delta(\mathrm{Im}X_{jk}). \end{align}
So are the operator delta function $\hat{\delta}(X)$ used to define the density of states, and the delta function of matrix argument $\tilde{\delta}(X)$ defined via a matrix model path integral, different? I don't see how to "trace" over the latter $\tilde{\delta}(X)$ and regain the DOS.
