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  Replica Trick for Bimodal Disorder Distribution

+ 4 like - 0 dislike

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,


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 in Open problems by TDowning (55 points) [ revision history ]
reopened Aug 22, 2016 by dimension10

1 Answer

+ 2 like - 0 dislike

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 TDowning (55 points) [ no revision ]

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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