Timeline for A naive question about non-Hermitian random matrix
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 26 at 10:02 | vote | accept | stopro | ||
| Nov 26 at 10:01 | comment | added | stopro | I see. Still not sure why we can express the resolvent as a formal series (ref Brézin and Zee) despite being singular but I get the idea | |
| Nov 24 at 11:34 | comment | added | Carlo Beenakker | the relation between the spectral density $\rho(z)=\sum_{\lambda\in\sigma(A)}\delta(z-\lambda)$ and the resolvent $G(z)=(A-z)^{-1}$ is $$\rho(z)=-\frac{1}{\pi}\frac{\partial}{\partial z^\ast}\operatorname{Tr}G(z).$$ This equation is formally correct, but not a useful relation because the right-hand-side is singular [it vanishes when $z\notin\sigma(A)$ and diverges when $z\in\sigma(A)$]. If you average the right-hand-side, the divergence is regularized, this is how Feinberg and Zee proceed. | |
| Nov 24 at 5:24 | comment | added | stopro | Thanks for the reference, and I never realized it was Girko that introduced this Hermitization method! I always thought it was Feinberg and Zee... However this doesn't entirely answer my question: While I cannot use the inverse $(A-z)^{-1}$, it seems like the relation between $\rho(z)$ and $(A-z)^{-1}$ still holds. Even when $(A-z)^{-1}$ is not defined. Am I mistaken? Second, in the paper by Brézin and Zee I referred, they express $(A-z)^{-1}$ as a formal series, which should not be possible since $(A-z)^{-1}$ is not analytic? | |
| Nov 23 at 13:38 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Nov 23 at 13:32 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 175 characters in body
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| Nov 23 at 13:27 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |