Different definitions of resolvent in matrix model

Question

When I study the matrix models, I get confused of different definations of resolvent. After we define the partition function as $$Z=\int[dM]e^{-NTrV(M)},$$ where $V(M)$ is a matrix valued function of $M$ and $[dM]$ is a measure factor, the saddle point equation reads $$V'(\lambda_I)=\frac{2}{N}\sum_{J\ne I}(\lambda_I-\lambda_J)^{-1}.$$

To solve this problem, in some papers (eg. (3.17) in https://arxiv.org/abs/2004.01171, (2.5) in https://arxiv.org/abs/hep-th/9306153 ) we introduce the resolvent as

$$R(z)\equiv \frac{1}{N} \text{Tr} (z\mathcal{I}-M)^{-1}=\frac{1}{N}\sum_{I=1}^N\frac{1}{z-\lambda_I}=\frac{1}{N}\sum_{k=0}^{\infty}\frac{\text{Tr}M^k}{z^{k+1}}\tag1,$$

which becomes a continuous function

$$R(z)=\int^a_{-a}d\mu\frac{\rho(\mu)}{z-\mu}$$ in the large $N$ limit. Here $\mathcal{I}$ is the identity matrix and $\rho(\mu)$ is the normalized eigenvalue distribution $$\rho(\mu)=\frac{1}{N}\sum^N_{I=1}\delta(\lambda-\lambda_I).\tag2$$

However I have seen another definition of resolvent which play a role as a generation function of moments as

$$R(z)\equiv \frac{1}{N}\left < \text{Tr}(z\mathcal{I}-M)^{-1}\right>=\frac{1}{N}\sum_{I=1}^N\frac{1}{z-\lambda_I}=\frac{1}{N}\sum_{k=0}^{\infty}\frac{\left<\text{Tr}M^k\right>}{z^{k+1}}\tag3$$

in other papers (eg. (1.24) in https://arxiv.org/abs/hep-th/9605140, (8.2.45) in < https://doi.org/10.1017/CBO9781107705968>), where the eigenvalue distribution is also sometimes defined as above and some times as

$$\rho(\mu)=\frac{1}{N}\sum^N_{I=1}\left<\delta(\lambda-\lambda_I)\right>.\tag4$$

I cannot understand why (1) and (3) are equal. I try to derive that and find I may need $$\frac{1}{N}\left < \text{Tr}(z\mathcal{I}-M)^{-1}\right> =\frac{1}{N} \frac{1}{Z}\int [dM] \text{Tr}(z\mathcal{I}-M)^{-1}e^{-NTrV(M)}\\ =\frac{1}{N} \frac{1}{Z}\int [dM] \sum_{I=1}^N\frac{1}{z-\lambda_I}e^{-NTrV(M)}\\ =\frac{1}{N} \frac{Z}{Z} \sum_{I=1}^N\frac{1}{z-\lambda_I}\\ =\frac{1}{N} \text{Tr}(z\mathcal{I}-M)^{-1}.\tag5\\ $$

But I don't think that's true, because in the second line the eigenvalues are related to the matrix element $M$, so we cannot simply put them outside the integral.

Another possible solution is to use the formula $$\left<Tr M^{k} \right>=\int^a_{-a} d\mu \rho(\mu) \mu^k$$ in the large $N$ limit, which is proposed in (24) of https://doi.org/10.1007/BF01614153. This is equal to $$\left<Tr M^{k} \right>=N\int^a_{-a} d\mu \rho(\mu) \mu^k=\sum_I \lambda_I^k=TrM^k\tag6$$ So the (1) (3) is equal.

My question is :

1. Are (1) (3) and (2) (4) equal definitions, or they are different definitions and I have missed the conditions they satisfied?

2. If they are or not different definitions, are my proposals (5) and (6) wrong?

3. If (6) is right, does it mean we could remove all the matrix ensemble average $\left< \cdot \right >$ in large $N$ limit?

Maybe my questions are evident, but they have confused me for a while because I haven't find any note to clarify this.

Answer 1

This is due to self-averaging properties in the thermodynamic limit. It's a more involved case of the law of large numbers. Recall that for the latter, say you have integrable iid random variables $(X_n)$ with the associated cumulative empirical average $\bar X_n$, then $\bar X_n$ converges to its expected value $\langle \bar X_n\rangle$ almost surely. The thermodynamic limit here is when $n\to\infty$. This is why a single realisation (which is what you are interested in) is the same in the thermodynamic limit as the expected value (which is what you calculate).

The same occurs for $R$ which "converges" to $\langle R\rangle$. In this case, it is more difficult to qualify the convergence, since we are not talking about a single number, but rather a whole function. For (2), (4), for example, the convergence is weak since for finite $n$, $R$ is discrete. Similarly for $(1),(3)$, the convergence cannot be uniform since you get a branch cut when $N\to\infty$.

Your attempts (5) and (6) are false because you cannot prove the quality for finite $N$ using simple algebra. It is only in the limit $N\to\infty$. It is the analogous to trying to prove for finite $n$ that $\bar X_n =\langle \bar X_n\rangle$ in the case of the law of large numbers.

Hope this helps.

Answer 2

For what it's worth, averaging/integrating over $M$ $$\langle\ldots \rangle$$ produces different quantities: Before the quantity depends on a matrix $M$ (or at least its eigenvalues, due to the presence of the trace); afterwards it doesn't. Even for large $N$.