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In a system with quenched disorder one is usually looking for self-averaging quantities, i.e., quantities such that the average over the couplings produces a ``typical" configuration in the thermodynamic limit. Free energy is such a self-averaging quantity. Therefore, one should average over the log of the partition function \begin{equation} \bar{F}=\int dJ\, P(J) F_J=\int d J\,P(J)\log Z_J, \end{equation} where $F_J$ and $Z_J$ are free energy and partition function at a fixed value of $J$. This should be contrasted with systems with annealed disorder where one would calculate

\begin{equation} \bar{Z}=\int d J\, P(J) Z_J \Rightarrow \bar{F}=\log \bar{Z}. \end{equation}

Are there conditions under which one is allowed to calculate the free energy of a quenched system using the latter equation? Does it depend on the absence of a spin-glass phase transition? Using the replica trick to calculate $\bar{F}$ in the first way and allowing only for replica diagonal solutions, does it boil down to do the averaging as in the second case?

In particular in discussions of the SYK model,i.e., a model with Hamiltonian \begin{equation} H\propto \sum_{ijkl}J_{ijkl}\psi_i\psi_j\psi_k\psi_l, \end{equation} where $J_{ijkl}$ are random couplings with distribution $P(J_{ijkl})$ and $\psi_i, i=1,...,N$ are Majorana fermions, it often appears that people calculate the average of the partition function, although the system is said to be a quenched system. (e.g. here around minute 17.)

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If we consider the replica trick: $$ \overline{\log Z}= \lim_{n \rightarrow 0} \frac{\overline{Z^n}-1}{n} $$ Then self-averaging is equivalent to equation $\overline{Z^n}=\overline{Z}^n$.

In general, this is not true for SYK, but in a certain limit, e.g. $N\gg \beta J \gg 1$, this is true in the leading order of $1/N$ expansion, e.g. see the diagrams in figure 3 and the discussions in Section 3.2 of paper (https://link.springer.com/article/10.1007/JHEP05(2017)125).

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  • $\begingroup$ Are you saying that even if one consider $N\to\infty$, that there is still only a limited range of $\beta J > c$, for which the model is self averaging? $\endgroup$ Commented Apr 3, 2022 at 15:41
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    $\begingroup$ If you consider strict infinite N, then it should be self averaging as far as I know, based on diagrammatic expansion. BTW, here is an interesting paper discussing related topic: arxiv.org/pdf/1806.10145.pdf $\endgroup$
    – Yingfei Gu
    Commented May 5, 2022 at 23:32

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