# Replica Trick for Bimodal Disorder Distribution

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Given a partition function of the form $Z\sim \Sigma_{\sigma} \exp(J_{ij}\sigma_{i}\sigma_{j})$

and a bimodal distribution of the couplings,

\begin{equation}P(J_{ij})=p\,\delta(J_{ij}+J)+(1-p)\,\delta(J_{ij}-J)\end{equation}

I'm attempting to calculate $\langle Z^{n}\rangle$ using the replica method, however it seems that the replicas under this distribution (as opposed to a Gaussian distribution, for example) don't become coupled together, leaving me with the result that $\langle Z^{n}\rangle =\langle Z\rangle ^{n}$ which doesn't seem right. (Maybe it is, but it strikes me as too simple of a solution)

My attempt:

$H^{a}=\sum_{ij} J^{a}_{ij}\sigma^{a}_{i}\sigma^{a}_{j}$ ($a$ is the replica index).

Then after averaging over the coupling disorder, I have the thermal average of something like

$\prod_{a=1}^{n}\prod_{\langle ij\rangle}\left[p\exp(-J^{a}\sigma^{a}_{i}\sigma^{a}_{j})+(1-p)\exp(J^{a}\sigma^{a}_{i}\sigma^{a}_{j})\right]$

Do I need to perform an additional average for each replica? As it stands, the replicas are independent so $\langle Z^{n}\rangle$ seems wrong (it indicates the system is self-averaging which it shouldn't be). For other distributions, you typically get a term of the form $H\sim J\sigma^{a}_{i}\sigma^{b}_{i}$ or something similar.

asked Aug 16, 2016
reopened Aug 22, 2016

## 1 Answer

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I've returned to this with an answer, so I thought I'd write it up for posterity's sake.

The average as it's written in the question is wrong, it should be

$\prod_{\langle ij\rangle}\left[p\exp{(-\sum_{a=1}^{n}J^{a}\sigma^{a}\sigma^{a})}+(1-p)\exp{(\sum_{a=1}^{n}J^{a}\sigma^{a}\sigma^{a})}\right]$

which actually does couple the replicas upon expanding the product.

answered Jun 27, 2017 by (55 points)

Nice. Yeah it's always the Boltzmann weights that get averaged. It's like having another field in the path integral, in this case a percolating field.

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