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.


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.


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.

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

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