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A Green's function is defined as follows: $$G(\omega) = \frac{1}{N}\mathrm{E}\big[ \mathrm{Tr}\frac{1}{I\omega - J} \big]$$, where $I$ is the $N$-dimensional identity and $E$ means expectation value with respect to the random matrix $J$.

It follows that $$ G(\omega) = \frac{1}{N}\mathrm{E}\big[ \sum_\lambda\frac{1}{\omega - \lambda} \big] = \int d^2\lambda\frac{\rho(\lambda)}{\omega - \lambda},$$ where $\rho(\lambda)$ is the average density of eigenvalues $\lambda$ of $J$ in the complex plane.

I am confused as to how this equality is obtained. Can someone help to explain or present the idea behind it ?

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  • $\begingroup$ you need to give $\omega$ an infinitesimal imaginary part and then the imaginary part of the expectation value will pick up the eigenvalues of $J$ $\endgroup$
    – Carlo Beenakker
    Commented Dec 29, 2015 at 15:58
  • $\begingroup$ @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 $\endgroup$
    – Xingdong Zuo
    Commented Dec 29, 2015 at 16:14
  • $\begingroup$ 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. $\endgroup$
    – Marcel
    Commented Dec 29, 2015 at 17:04
  • $\begingroup$ @Marcel If I understand you correctly, should it be $\frac{1}{N}\mathrm{E}\big[ \sum_\lambda\frac{1}{\omega - \lambda} \big] = \frac{1}{N}\sum_\lambda \int d^2\lambda \frac{\rho(\lambda)}{\omega - \lambda} = \frac{1}{N}N\int d^2\lambda \frac{\rho(\lambda)}{\omega - \lambda} = \int d^2\lambda \frac{\rho(\lambda)}{\omega - \lambda}$ $\endgroup$
    – Xingdong Zuo
    Commented Dec 29, 2015 at 17:11
  • $\begingroup$ @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. $\endgroup$
    – Marcel
    Commented Dec 29, 2015 at 17:22

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I'll post an answer to spell out all the details.

You have $$G(ω)=\frac{1}{N}E\left[{\rm Tr}\frac{1}{Iω−J}\right]=\frac{1}{N}E\left[\sum_\lambda\frac{1}{ω−\lambda}\right]$$

This can be written as $$G(ω)=\frac{1}{N}E\left[\int d^2z \sum_\lambda\frac{\delta(z-\lambda)}{ω−z}\right],$$ where $\delta$ is the delta-function/distribution in the complex plane.

Now define $$\rho(z)=\frac{1}{N}E\left[\sum_\lambda\delta(z-\lambda)\right]$$, which is the spectral density. Then it follows that $$G(\omega)=\int d^2z \frac{\rho(z)}{\omega-z}=$$

Or maybe you mean it the other way around. If you start with $$\frac{1}{\omega+i\epsilon-\lambda}=\frac{\omega-i\epsilon-\lambda}{(\omega-\lambda)^2+\epsilon^2}$$, the imaginary part gives, in the limit of vanishing $\epsilon$ (as Beenakker mentioned), $$\lim_{\epsilon\to 0}{\rm Im}E\left[\sum_\lambda\frac{1}{\omega+i\epsilon-\lambda}\right]=\frac{1}{\pi}E\left[\sum_\lambda\delta(\omega-\lambda)\right]$$, which means that $\lim_{\epsilon\to 0}{\rm Im}G(\omega+i\epsilon)=\rho(\omega)N/\pi$.

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