1
$\begingroup$

I'm going through some lecture notes dealing with the replica method and I feel like that I did not fully understand the concept of the replica symmetry (RS) Ansatz. At some point in the notes we come to an integral looking like this (I simplified it a little bit for this question): $$\prod_{\alpha=1}^{n}\left(\int dm^{(\alpha)}\right)\exp\left[-\sum_{\alpha=1}^{n}m^{(\alpha)}\right]$$ where $\alpha$ is my index for all the $n$-replicas I have.

To simplify this expression we now use RS, saying that $m^{(\alpha)} = m$. The notes say that using this RS we essentially get: $$\int dm\exp\left[-nm\right]$$

The thing that is confusing me now is why we are left with only one integral. Naively I would have thought we would get something like this: $$\left(\int dm\right)^n\exp\left[-nm\right]$$ but I guess this is wrong and the reason that I'm wrong is probably some misunderstanding on what the RS Ansatz really is. That's the reason I wanted to try and ask here if anyone could help me and give me a better understanding on this topic. Thank you!

$\endgroup$

1 Answer 1

1
$\begingroup$

The key point is precisely that, once you assume that the integral is dominated by replica-symmetric configurations, the variables $m^{(\alpha)}$ on each replica are no longer independent; in particular, when performing the integral, you just need one single representative $m$. If you want to wrap a few equations around this just to make it slightly more explicit, you can imagine that the replica symmetry ansatz amounts to replacing the integral

$$\prod_{\alpha=1}^n (\int dm^{(\alpha)}) \,\exp\left[-\sum_{\alpha=1}^nm^{(\alpha)}\right] $$

with the integral

$$\int dm \exp(-m)\times \prod_{\alpha=2}^n (\int dm^{(\alpha)})\delta(m^{(\alpha)}-m) \,\exp\left[-\sum_{\alpha=2}^nm^{(\alpha)}\right]$$ where I chose the first replica as the representative. Then, the result follows immediately.

$\endgroup$
2
  • 1
    $\begingroup$ Oh wow, thank you! That's a really nice explanation in my opinion. Maybe a short follow up question to this: I wrote down your expression for the simple case of $n=2$ replicas and this thought came to my mind: Before doing the RS Ansatz, $m^1$ and $m^2$ could be any value. But after doing the Ansatz I constrain their value to be on this $45$ degree diagonal between them (if draw a graph with $m^1$ on the $x$ and $m^2$ on the $y$ axis). And this constraining on the diagonal is what happens with this delta distribution. Would this be a valid way to talk about this? $\endgroup$ Jul 11, 2023 at 15:27
  • 1
    $\begingroup$ Yes, that is pretty much what is happening. $\endgroup$ Jul 11, 2023 at 16:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.