Motion of Lee–Yang Zeros

Thermodynamic limit of the first Lee‐Yang zero

“…where 0 < 𝛼 1 (Λ, 𝛽) ≤ 𝛼 2 (Λ, 𝛽) ≤ … are all the positive zeros of 𝑍 Λ,𝛽,𝑖ℎ as a function of ℎ (listed according to their multiplicities), and ∑ ∞ 𝑗=1 𝛼 −2 𝑗 (Λ, 𝛽) < ∞; see Lemma 6 of [11]. Combining this with (1.4) and (2.4), we have…”

“…Proof of Proposition The Hadamard factorization theorem and the fact that $$Z_{\Lambda ,\beta ,h}$$ is a function of h of exponential order 1 imply that $$$$\begin{equation} Z_{\Lambda ,\beta ,h}=Z_{\Lambda ,\beta ,0}\prod _{j=1}^{\infty }{\left(1+\frac{h^2}{\alpha _j^2(\Lambda ,\beta )}\right)}, \forall h\in \mathbb {C}, \end{equation}$$$$where $$0&lt;\alpha _1(\Lambda ,\beta )\le \alpha _2(\Lambda ,\beta )\le \dots$$ are all the positive zeros of $$Z_{\Lambda ,\beta ,ih}$$ as a function of h (listed according to their multiplicities), and $$\sum _{j=1}^{\infty }\alpha _j^{-2}(\Lambda ,\beta )&lt;\infty$$; see Lemma 6 of [11]. Combining this with () and (), we have …”

“…\end{equation}$$Using the Taylor series for $$\ln (1+x)$$, we get $$$$\begin{equation} u_{2k-1}(M_{\Lambda ,\beta ,0})=0, u_{2k}(M_{\Lambda ,\beta ,0})=(-1)^{k-1}\frac{(2k)! }{k}\sum _{j=1}^{\infty }\frac{1}{\alpha _j^{2k}(\Lambda ,\beta )}, \forall k\in \mathbb {N}.\end{equation}$$$$Note that, as a function of h , $$Z_{\Lambda ,\beta ,ih}$$ has a period 2π and it has exactly $$2|\Lambda |$$ zeros in [0, 2π); see the discussion before Theorem 1 of [11] for more details. So we have (to simplify notation, we write $$\alpha _j$$ for $$\alpha _j(\Lambda ,\beta )$$ ) …”

“…Note that, as a function of ℎ, 𝑍 Λ,𝛽,𝑖ℎ has a period 2𝜋 and it has exactly 2|Λ| zeros in [0, 2𝜋); see the discussion before Theorem 1 of [11] for more details. So we have (to simplify notation, we write 𝛼 𝑗 for 𝛼 𝑗 (Λ, 𝛽)…”