Replica trick

Remarks

The identity holds because $(Z^n - 1)/n$ decreases to $\log Z$ as $n \downarrow 0$, by convexity of $n \mapsto Z^n$, and is bounded by $(Z^\varepsilon - 1)/\varepsilon$ for $n \le \varepsilon$; monotone convergence then applies. Step 3 is not justified in general: it exchanges the limits $N \to \infty$ and $n \to 0$, and the values of $\phi$ at the integers do not determine its continuation.

Example: let $Z = e^{\beta X}$ with $X$ standard Gaussian. Then $\mathbb{E} Z^n = e^{n^2 \beta^2 / 2}$, and $(e^{n^2 \beta^2/2} - 1)/n \to 0 = \mathbb{E} \log Z$, as the identity claims.

Non-example: if $\log Z$ has a Cauchy law, then $\mathbb{E} Z^n = \infty$ for every $n > 0$ and $\mathbb{E} \abs{\log Z} = \infty$, so the hypotheses fail and the formula has no meaning.

In spin glass theory, the order parameter of the replicated system is the $n \times n$ overlap matrix $Q_{ab} = \frac{1}{N} \sum_i \sigma^a_i \sigma^b_i$. The choice of its form in step 3 is called a replica symmetric or a replica symmetry breaking ansatz.