Timeline for Alternative formula of a Green's function for average density of eigenvalues of random matrix
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| S Dec 29, 2015 at 19:18 | history | suggested | Marcel | CC BY-SA 3.0 |
Deleted information unrelated to the question and improved style
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| Dec 29, 2015 at 18:31 | vote | accept | Xingdong Zuo | ||
| Dec 29, 2015 at 18:14 | review | Suggested edits | |||
| S Dec 29, 2015 at 19:18 | |||||
| Dec 29, 2015 at 18:11 | answer | added | Marcel | timeline score: 1 | |
| Dec 29, 2015 at 17:22 | comment | added | Marcel | @XingdongZuo Almost. You should exchange $E\left[\frac{1}{N}\sum_\lambda f(\lambda)\right]$ for $\int d^2\lambda \rho(\lambda)f(\lambda)$. This is the meaning of a probability density. | |
| Dec 29, 2015 at 17:11 | comment | added | Xingdong Zuo | @Marcel If I understand you correctly, should it be $\frac{1}{N}\mathrm{E}\big[ \sum_\lambda\frac{1}{\omega - \lambda} \big] = \frac{1}{N}\sum_\lambda \int d^2\lambda \frac{\rho(\lambda)}{\omega - \lambda} = \frac{1}{N}N\int d^2\lambda \frac{\rho(\lambda)}{\omega - \lambda} = \int d^2\lambda \frac{\rho(\lambda)}{\omega - \lambda}$ | |
| Dec 29, 2015 at 17:04 | comment | added | Marcel | Which equality are you asking about? The last one just follows from the definition of spectral density $\rho(\lambda)$. From your comment, it seems you are confused with the notation $d^2\lambda$. This is not a double sum over $\lambda$; it only means the integral is over the complex plane. | |
| Dec 29, 2015 at 16:14 | comment | added | Xingdong Zuo | @CarloBeenakker Is it possible to reason as $\frac{1}{N}\mathrm{E}\big[ \sum_\lambda\frac{1}{\omega - \lambda} \big] = \frac{1}{N}\sum_\lambda(\sum_\lambda\frac{1}{\omega - \lambda})\rho(\lambda) = \frac{1}{N} \sum_{\lambda^2}\frac{1}{\omega - \lambda}\rho(\lambda) = \frac{1}{N}\int d^2\lambda \frac{\rho(\lambda)}{\omega - \lambda}$, now only $\frac{1}{N}$ is still left to be canceled | |
| Dec 29, 2015 at 15:58 | comment | added | Carlo Beenakker | you need to give $\omega$ an infinitesimal imaginary part and then the imaginary part of the expectation value will pick up the eigenvalues of $J$ | |
| Dec 29, 2015 at 15:03 | history | asked | Xingdong Zuo | CC BY-SA 3.0 |