Phase Transition in the Peierls Model for Polyacetylene

Abstract

We consider the Peierls model for closed polyactetylene chains with an even number of carbon atoms as well as infinite chains, in the presence of temperature. We prove the existence of a critical temperature below which the chain is dimerized and above which it is 1-periodic. The chain behaves like an insulator below the critical temperature and like a metal above it. We characterize the critical temperature in the thermodynamic limit model and prove that it is exponentially small in the rigidity of the chain. We study the phase transition around this critical temperature.

Appendix A: Gain of Energy in the Thermodynamic Limit

In this section, we prove that the gain of energy due to Peierls dimerization is exponentially small in $\mu$. We focus on the thermodynamic limit case (although the proof is similar in the $L \in 2 \mathbb {N}$ case). We also focus only on the null temperature case $\theta = 0$. In this case, the thermodynamic energy reads

$$\begin{aligned} g_0(W,\delta ) = \frac{\mu }{2} ((W - 1)^2 + \delta ^2) - \frac{4}{\pi }\int _{0}^{\pi /2}\sqrt{W^2\sin ^2{(s)} + \delta ^2\cos ^2{(s)}} {\textrm{d}}s.\qquad \end{aligned}$$

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We introduce

$$\begin{aligned} f_0:= \min \left\{ g_0(W, \delta ), \ W \ge 0, \ \delta \ge 0\right\} , \quad \text {and} \quad f_{0, \textrm{per}}:= \min \left\{ g_0(W, 0), \ W \ge 0 \right\} . \end{aligned}$$

In other words, $f_0$ is the minimum of $g_0$ over 2–periodic (and all) configurations, and $f_{0, \textrm{per}}$ is the minimum over 1-periodic configurations. We prove the following

Theorem A.1

There is $C > 0$ such that, for all $\mu$ large enough,

$$\begin{aligned} 0 < f_{0, \textrm{per}} - f_0 \le C {\textrm{e}}^{- \frac{\pi }{2} \mu }. \end{aligned}$$

In other words, the energy gained by the Peierls distorsion is exponentially small in the $\mu$ parameter. The first inequality states that in the thermodynamic limit at null temperature, the minimizers are always dimerized, as first proved by Kennedy and Lieb [5].

Proof

Let us first compute $W_1$, the optimizer of $g_0(W, 0)$. This is simply the minimum of

$$\begin{aligned} g_0(W,0) = \frac{\mu }{2} (W - 1)^2 - \frac{4}{\pi }\int _0^{\pi /2}\sqrt{W^2\sin ^2(s)} {\textrm{d}}s = \frac{\mu }{2}\mu (W - 1)^2 - \frac{4}{\pi }W. \end{aligned}$$

The minimizer satisfies $\mu (W_1 - 1) = \frac{4}{\pi }$, hence $W_1 = 1 + \frac{4}{\pi \mu }$. In particular,

$$\begin{aligned} f_{0, \textrm{per}} = -\frac{4}{\pi } - \frac{8}{\pi ^2\mu }. \end{aligned}$$

We now compute the energy gain from the breaking of periodicity. For $(W, \delta )$ a trial pair, we write $W = W_1 + \varepsilon .$ We assume that $g_0(W,\delta )< g_0(W_1,0). { \mathrm Then}$

$$\begin{aligned}&g_0 (W, \delta ) - g_0(W_1, 0)\\&\quad = \frac{\mu }{2}(\varepsilon ^2 + \delta ^2) - \frac{4W_1}{\pi }\int _0^{\pi /2}\left[ \sqrt{ \frac{(W_1 + \varepsilon )^2}{W_1^2} + \frac{\delta ^2}{W_1^2}\cot ^2(s)} - 1 - \frac{\varepsilon }{W_1} \right] \sin (s) {\text {d}}s\\&\quad \ge \frac{\mu }{2}(\varepsilon ^2 + \delta ^2) - \frac{4\delta }{\pi },\;\hbox {so}\;\delta<\frac{8}{\pi \mu }\;\hbox {and}\;|\varepsilon |<\frac{4}{\pi \mu }. \end{aligned}$$

To compute the integral, we make the change of variable $u=\cos (s),$ and get that the integral equals

$$\begin{aligned} \frac{W_1 + \varepsilon }{W_1}\left( \int _0^1\sqrt{1+ \frac{au^2}{1-u^2}} {\textrm{d}}u -1 \right) , \text{ with } a:= \left( \frac{\delta }{W_1 + \varepsilon }\right) ^2. \end{aligned}$$

Using that

$$\begin{aligned} \int _0^1\sqrt{1+ \frac{au^2}{1-u^2}} {\textrm{d}}u = E(1 - a) = 1 + \left( \frac{-\ln (a)}{4} -\frac{1}{4} + \ln (2) \right) a + O(a^2), \end{aligned}$$

where E is a complete elliptic integral of the second kind, we get

$$\begin{aligned} g_0(W, \delta ) - g_0(W_1, 0)&= \frac{\mu }{2}(\varepsilon ^2 + \delta ^2) - \frac{4\delta ^2}{\pi (W_1 + \varepsilon )} \left[ -\frac{1}{2}\ln \left( \frac{\delta }{W_1 + \varepsilon }\right) -\frac{1}{4} + \ln (2) + O(a)\right] \\&= \frac{1}{2}\mu (\varepsilon ^2 + \delta ^2) - \frac{2}{\pi W_1 }\delta ^2\ln (\delta ^{-1})(1+ o(1)). \end{aligned}$$

We now minimize the right-hand side. For large $\mu$, we have $W_1 \approx 1$ and the minimization in $\varepsilon$ gives $\varepsilon = 0$. So

$$\begin{aligned} g_0(W, \delta ) - g_0(W_1, 0) \ge \delta ^2\left( \frac{\mu }{2} - \frac{2\ln (\delta ^{-1})}{\pi } (1+o(1))\right) . \end{aligned}$$

We optimize the right-hand side by taking $\delta = {\textrm{e}}^{-(\frac{\pi }{4}\mu + \frac{1}{2})}$, and this completes the proof. $\square$