The "replica trick" initial formula In Spin-glass theory for pedestrians by Castellani and Cavagna, the initial formula used to introduce the replica trick is written as:
$$\overline{\log Z}=\lim_{n\rightarrow0}\frac{1}{n}\log\overline{Z^{n}}\qquad(1)$$
where the overbar denotes average over quenched disorder. I don't know how to prove this formula.
In other treatments I have seen of the replica method (wiki, for example), one starts from:
$$\log Z = \lim_{n\rightarrow 0} \frac{Z^n-1}{n}\qquad(2)$$
which I understand. How are (2) and (1) connected? What's the proof of (1)?
 A: Try to look at Introduction to the Replica Theory of Disordered Statistical Systems by V. Dotsenko. In the following, I've written a possible answer to your question:
\begin{equation}
f=-\lim_{N\rightarrow\infty}\frac{1}{\beta N}\mathbb{E}\left[\ln Z_{J}\right]
\end{equation}
where:


*

*$\mathbb{E}\left[\mathcal{O}\right]=\left(\prod_{\left\{ i,j\right\} }\int dJ_{ij}\right)P\left[J\right]\mathcal{O}
 $

*$Z_{J}=\sum_{\sigma}e^{-\beta H\left[J,\sigma\right]}
 $


Then labelling with $a$ the replicas:
\begin{equation}
Z_{J}^{n}=\left(\prod_{a=1}^{n}\sum_{\sigma^{a}}\right)e^{-\beta\sum_{a=1}^{n}H\left[J,\sigma_{a}\right]}
\end{equation}
Thus, remember that $\ln x=\lim_{n\rightarrow0}\frac{1}{n}\left(x^{n}-1\right)$:
\begin{equation}
f=-\lim_{N\rightarrow\infty}\frac{1}{\beta N}\mathbb{E}\left[\ln\left(Z_{J}\right)\right]=-\lim_{N\rightarrow\infty}\lim_{n\rightarrow0}\frac{1}{\beta N}\mathbb{E}\left[\frac{\left(Z_{J}^{n}-1\right)}{n}\right]=-\lim_{N\rightarrow\infty}\lim_{n\rightarrow0}\frac{1}{\beta nN}\mathbb{E}\left[Z_{J}^{n}\right]
\end{equation}
but in general there are many issues concerning the commutation of the two limits.
A: For an even more direct answer to how the two replica tricks you listed are related, simply note that we expect $Z^n \rightarrow 1$ as $n \rightarrow 0$. Then,
$$ \lim_{n \rightarrow 0} \frac{1}{n} \log  \overline{Z^n} = \lim_{n \rightarrow 0} \frac{1}{n} \log \Big( 1 + (\overline{Z^n} - 1) \Big) = \lim_{n \rightarrow 0} \frac{\overline{Z^n} - 1}{n} = \overline{\log Z}$$
A: I realize this thread is old, but I was also reading the "for Pedestrians" introduction and I had the same question. Although the previous response seems reasonable, it does not address the original question. However, using a very similar approach, it is possible to derive (1) from (2).
Let's start from the right hand side of (1):
$$F = \lim_{n \rightarrow 0} \frac{1}{n} \log \left( \overline{Z^n} \right),$$
where the overbar denotes average over the quenched disorder. Rewriting (2), we have:
$$Z^n \simeq 1 + n \log Z$$
Inserting into the above expression for $F$ and using the fact that $\overline{1} = 1$, we have:
$$F = \lim_{n \rightarrow 0} \frac{1}{n} \log \left( 1 + n \overline { \log Z } \right)$$
Since $n$ is small, we can expand the outer logarithm around $1$ and find:
$$F = \overline{ \log Z }$$
which is the left hand side of (1). This answer was inspired by "Spin Glass Theory and Beyond" by Mezard, Parisi and Virasoro, which I've found to be a very useful resource for this topic.
