The spectral theorem states that for any self adjoint operator $H$ on some Hilbert space $\mathcal{H}$, there exists a projection-valued measure $E_H$ such that $$H= \int_{\mathbb{R}} \lambda \mathrm{d}E_H(\lambda).$$ We also have that, restricting our attention to $L^2(\mathbb{R})$ and Schrodinger operators, (i.e. $H=-\nabla+V$), that we can define the density of states by taking a sufficiently regular nesting set of intervals $I_1\subset I_2\subset \cdots,$ $$g([a,b]):= \lim_{j\to \infty}\frac{1}{\mu(I_j)}\mathrm{Tr}[E_{H\vert_{I_j}}([a,b])],$$ where $H_{I_j}$ is the natural restriction of $H$ to $L^2(I_j)$ with appropriate boundary conditions.
However, this answer states that "For the absolutely continuous part of the spectrum of a self-adjoint operator $$, the "density of states" is provided by the Radon-Nikódym derivative of the spectral measure of $_{}$ with respect to Lebesgue measure."
Now, this doesn't correspond immediately to the above definition, firstly because the Radon-Nikódym derivative will be an operator-valued function. However, is there a way to make it correspond? For instance, I was thinking maybe we could let the absolutely continuous density of states be defined as: $$\tilde{g}([a,b]) := \int_{[a,b]} \mathrm{Tr}\left[\frac{\mathrm{d}E_{HP_{ac}}}{\mathrm{d}\mu}\right]\mathrm{d}\mu(x).$$ Thus, we reach my question: does this, or some similar definition that sidesteps taking the thermodynamic limit and just works in it, correspond to the above definition? Could we work it out in the case of the free particle for instance?
