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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)?

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  • $\begingroup$ I think you are missing an $1/n$ in the first equation. $\endgroup$ – Antonio Ragagnin May 9 '14 at 15:49
  • $\begingroup$ @AntonioRagagnin fixed. $\endgroup$ – becko May 9 '14 at 17:01
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I realize this thread is old, but I was also reading the "for Pedestrians" introduction and I had the same question. Although the above 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.

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  • $\begingroup$ Thanks. Do they make this calculation in "Spin glass theory and beyond"? Where? $\endgroup$ – becko Sep 3 '18 at 14:14
  • $\begingroup$ If I remember correctly, one of the identities used in the calculation shows up in a footnote early on (the first chapter?) $\endgroup$ – farzan Sep 3 '18 at 21:43
  • $\begingroup$ They just spit it out in Eqs. I.5 & I.6, but they don't give the explicit derivation... too trival :P $\endgroup$ – becko Sep 3 '18 at 21:58
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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.

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