## 1 Correction to: J Stat Phys (2013) 153:363–375 https://doi.org/10.1007/s10955-013-0825-6

In [1], on p. 366, line 5 and on p. 369, line 12, replace

\begin{aligned} \text {}\alpha =2\Vert \sigma \Vert _{{L^{1}}(\mathbb {T})}.\text {''} \end{aligned}

with

\begin{aligned} \sigma =2\hbox { min}\{\Vert \sigma \Vert _{{L^{1}}(\mathbb {T})},-D(0)\}, \end{aligned}

where D(0) is defined in [10], p. 4.”.

We underline that the strategy of proof described in [1] is independent of the value of $$\alpha$$.

The precise formulation of D(0) is recalled below. Let

\begin{aligned} A_\sigma = \begin{pmatrix} 0 &{}\quad 1 \\ \partial _{xx} &{}\quad -2\sigma \\ \end{pmatrix} \end{aligned}

with

$$\mathcal {D}(A_\sigma )= \left( H^1_0 \cap H^2\right) \oplus H^1_0,$$

and denote with $$\text {sp}(A_\sigma )$$ the spectrum of $$A_\sigma$$. For $$R>0$$, consider

$$D(R)=\sup \{ \mathcal {R}e(\lambda _j),\ \lambda _j \in \text {sp}(A_\sigma ), \vert \lambda _j \vert \ge R \}.$$

The value of D(0) is then obtained in the limit $$D(0)=\lim _{R\rightarrow 0^+} D(R)$$.