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I came across a reference here where in Eq. 2.12 one seems to be concerned with the derivative of the density of state which is given by

$$\begin{aligned} -\operatorname{Tr} \delta^{\prime}(H-\mu) & \equiv \operatorname{Im} \frac{1}{\pi} \int \mathrm{~d} \lambda \frac{\partial}{\partial \mu} R(\mu-i \epsilon ; \lambda, \lambda) \\ & =\frac{1}{2 \pi} \operatorname{Im} \int_0^{\infty} \mathrm{d} s \mathrm{e}^{-i \mu s} \frac{s}{\operatorname{sh} s / 2} \\ & =\frac{\partial}{\partial \mu}\left[\frac{1}{2 \pi} \operatorname{Re} \Psi\left(\frac{1}{2}+i \mu\right)\right]. \end{aligned}\tag{2.12}$$

Can someone explain to me what kind of observables are dependent on the derivative of the density of states in the context of this reference or otherwise in general? What is the physical meaning of the derivative of density of states?

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Anything that depends on the difference of density-of-states (DOS) at different energies.

An obvious example is the absorption in a material under influence of a periodic perturbation. The Fermi golden rule gives us the absorption/emission probability: $$ W(E_f, E_i, \pm\hbar\omega)=\frac{2\pi}{\hbar}|V|^2\delta(E_f-E_i\pm\hbar\omega), $$ (for simplicity I assume matrix element independent on energy.)
However the absorption/emission rate depends on the number of electrons available for transitions, which is dependent on the DOS and the Fermi distribution: $$ R_{absorption} = \frac{2\pi}{\hbar}|V|^2\int dE_f\int dE_i\delta(E_f-E_i-\hbar\omega)\rho(E_i)\rho(E_f)f(E_i)\left[1-f(E_f)\right]=\\ \frac{2\pi}{\hbar}|V|^2\int dE_i\rho(E_i)\rho(E_i+\hbar\omega)f(E_i)\left[1-f(E_i+\hbar\omega)\right],\\ R_{emission} = \frac{2\pi}{\hbar}|V|^2\int dE_f\int dE_i\delta(E_f-E_i+\hbar\omega)\rho(E_i)\rho(E_f)f(E_i)\left[1-f(E_f)\right]=\\ \frac{2\pi}{\hbar}|V|^2\int dE_i\rho(E_i)\rho(E_i-\hbar\omega)f(E_i)\left[1-f(E_i-\hbar\omega)\right] $$ The net result is then $$ R_{absorption} - R_{emission} = \frac{2\pi}{\hbar}|V|^2\int dE_i\rho(E_i)f(E_i)\left\{ \rho(E_i+\hbar\omega)\left[1-f(E_i+\hbar\omega)\right]- \rho(E_i-\hbar\omega)\left[1-f(E_i-\hbar\omega)\right]\right\}\approx \frac{2\pi}{\hbar}|V|^2\int dE_i\rho(E_i)f(E_i)\frac{d}{dE_i}\left\{\rho(E_i)\left[1-f(E_i)\right]\right\}2\hbar\omega $$ In fact, $\omega\rightarrow0$ limit gives us polarizability of the material. Such calculations often (although not necessarily) result from applying Kubo formula, which is easier evaluated for finite frequency, but is often used for deriving static properties (polarizability, conductivity, etc.) I don't cite here WIkipedia, since its article on Kubo formula is very deficient, but the derivations can be found in any textbook on QFT in solid state physics (Fetter&Walecka, Mahan, AGD.)

Very similar calculations also emerge when calculating linear conductance, e.g., as discussed in this answer (although this particular calculation assumes for simplicity flat DOS - so-called "broad band limit".) Note that although the example considers a tunneling problem, similar calculations arise when calculating conductance for bulk materials. The difference with the earlier example is that instead of frequency, we take to zero the potential difference between two Fermi seas - in a way this is even more relevant, since it reduces the problem to an equilibrium one. In fact, one often works in a different order - first obtaining a general formula in the zero-bias limit, and only the actually evaluating the DOS using equilibrium perturbation theory.

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  • $\begingroup$ These are all very good and relevant examples. But I somehow feel they are restricted to semi-conductor physics. Do you think you can tell me other examples that are more broader and general? $\endgroup$ Commented Oct 28 at 15:18
  • $\begingroup$ @Dr.user44690 my field is condensed matter - venturing beyond that would involved too much guesswork on my side. Nevertheless, I think you need to look for applications rather than theoretical significance. Any kind of susceptibility/response calculations is likely to run into similar situations. $\endgroup$ Commented Oct 28 at 15:34
  • $\begingroup$ Do you think you can suggest some papers that has these computation as a starting point? I am unable to find it by myself. $\endgroup$ Commented Oct 29 at 2:02
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I think a much simpler answer can be by utilizing the following definition of entropy $$S = k\ln \rho(E)$$ Then by making use of $$\beta = \frac{\partial S}{\partial E}$$ we get $$\beta = k \frac{\rho'(E)}{\rho(E)}$$

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  • $\begingroup$ I think it is worth spelling, in which case this formula for entropy might apply. It resembles traditional Boltzmann expression, but the latter contains the total number of many-particle states rather than the single-particle DOS. Still, perhaps there are situations/limits where your expression applies... but I don't know them. $\endgroup$ Commented Oct 31 at 13:18

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