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For operators with pure point spectra it is clear how to count number of states corresponding to a given eigenvalue - one can just calculate the dimension of eigenspaces. I am wondering how to do it for continuous spectra. What I usually saw in my undergrad classes is the classical trick of putting the system into an artificial box to quantize the momenta. In a way it is an ingenious idea, but it is not very elegant and looks dirty. But what's the worst it gives wrong answer in some cases. I am wondering if there is more rigorus way to do it.

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For the absolutely continuous part of the spectrum of a self-adjoint operator $H$, the "density of states" is provided by the Radon-Nikódym derivative of the spectral measure of $HP_{ac}$ with respect to Lebesgue measure, where $P_{ac}$ is the orthogonal projection onto the absolutely continuous subspace of the domain of $H$. This formula is well defined precisely because of the definition of absolutely continuous spectrum.

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I am not sure I understand perfectly you question but formally in the canonical ensemble we can write the partition function $Q(\beta)$ as being:

\begin{equation} Q(\beta) = \int\cdot \cdot \int d\mu(x)\: e^{-\beta H(x)} = \int_0^{+\infty} dE \: \rho(E)e^{-\beta E} \end{equation} where $d\mu(x)$ is the volume measure for the micro states in the system, $H(x)$ the hamiltonian, $E$ energy values of the hamiltonian and $\rho(E)$ is the density of states. It is not difficult to see that it is in fact the Laplace transform of the the density of state so that $Q(\beta) = \mathcal{L}[\rho(E)]$.

In this context, it follows that the density of state is nothing but the inverse Laplace transform of the partition function:

\begin{equation} \rho(E) = \mathcal{L}^{-1}[Q(\beta)] \end{equation}

This general relation enables one to derive exact results on density of states for particular model cases. The drawback of course is that inverse Laplace transforms are not as easy to compute as inverse Fourier transforms.

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