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The density of states (DOS) is defined as $$\mathcal{N}\left(\lambda\right)=\sum_{n=1}^{M}\delta\left(\lambda-\lambda_{n}\right).$$ We can then get $$\int d\lambda\mathcal{N}\left(\lambda\right)=M,$$ i.e. as total we have $M$ eigenstates.

This is true when there is no degenerated states for each eigenvalue $\lambda_{n}$; However, it is obviously wrong when any one eigenvalue has degeneracy. How should deal with this situations?

Should we rewrite it as $$\mathcal{N}\left(\lambda\right)=\sum_{n=1}^{M}g_{n}\delta\left(\lambda-\lambda_{n}\right)~?$$ With $g_{n}$ the corresponding degeneracy.

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No, your formula is not wrong.

You have to be aware that a typical mistake when starting with this kind of sums is to make confusion about the meaning of the index $n$.

One has to decide if $n$ is labeling an eigenstate o an eigenvalue. Your formula $\mathcal{N}\left(\lambda\right)=\sum_{n=1}^{M}\delta\left(\lambda-\lambda_{n}\right)$ is not obviously wrong if $n$ is the index of the states (which implicitly your case, once you say that $M$ is the number of eigenstates). Actually the formula works as it is, even in the case of degeneracy, the only effect of a degeneracy (of eigenvalues) being the presence of two or more delta's with the same argument. But this is harmless and gives the required result for the sum rule.

A formula like $\mathcal{N}\left(\lambda\right)=\sum_{n=1}^{M_l}g_{n}\delta\left(\lambda-\lambda_{n}\right)$, with $g_{n}$ indicating the degeneracy of the $n-th$ eigenvalue, works as well, but notice that it is a sum over the $M_l \neq M$ eigenvalues. A more transparent way of writing this sum would be $$ \mathcal{N}\left(\lambda\right)=\sum_{\lambda_n}g(\lambda_n)\delta\left(\lambda-\lambda_{n}\right) $$ where the fact that one is summing over the eigenvalues is more evident.

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The answer to your question is simply ‘yes’. You figured out the right way to do it.

The case you mention, however, covers only the case of a discrete spectrum. In solid state physics we are often concerned with hamiltonians having continuous spectra, in which case the density of states will be normalized to the system volume.

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  • $\begingroup$ Can the person downvoting please leave a comment with the reason? $\endgroup$ – flaudemus Mar 3 at 11:10

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