Aubry transition with small distortions

In the absence of periodic potential, $K=0$, the solution is simply given by

where the phase $\phi$ is arbitrary. This is the sliding phase of a trivial integrable model with continuous translation symmetry. This state satisfies the balance of forces and becomes the ground state when $\mu$ equals the lattice constant $\ell$.

When $K\neq 0$, the problem is no longer simple but some exact properties of the ground states are known aubry_ledaeron (see also bangert and appendix B). For a general class of models that includes the model we consider here (see appendix B for the general conditions), Aubry and Le Daeron have shown that the ground state can be written

with a well-defined lattice constant,

that can take any real value provided that $\mu$ is appropriately tuned. It is thus possible to work at fixed $\ell$. Importantly, the distortions $\{u_{n}\}$ are bounded for the ground state and satisfy (see appendix B):

In the ground state, the bond lengths are constrained not to be far from the average $\ell$. $\ell$, which is in units of the period of the potential, can be a rational or irrational number. In the first case,

where $r$ and $s$ are two coprime integers, one has

thus $u_{n}$ is periodic with period $s$, $u_{n+s}=u_{n}$. In that case, the ground state is said to be commensurate and has a unit-cell of size $r$ admitting $s$ atoms at positions $x_{n}$ ($n=1,\dots,s$).

In the second case, when $\ell$ is an irrational number, the ground state is said to be incommensurate and can be viewed, physically, as a commensurate solution with a very large period $s$. The distortions can be written

where $g$ is a periodic function with period 1 which is defined everywhere since $n\ell+\phi$ takes, modulo 1, all values in $[0,1]$. The ground state then takes the special form

where $f(x)=x+g(x)$ is a strictly increasing function, called the envelope function. $f$ depends on $\ell$ and on the various model parameters. The form (15) is exact for incommensurate ground states provided that the model fulfills the properties given in appendix B. Importantly, $f$ can be continuous or discontinuous: the change of regularity with model parameters is Aubry’s breaking of analyticity (Aubry transition) Aubry ; aubry_ledaeron .

For the model we consider, we define $K_{c}(\alpha)$ as the threshold of the Aubry transition: for $K<K_{c}(\alpha)$, the ground state is characterized by a continuous function $f$. This is the sliding phase. For $K>K_{c}(\alpha)$, $f$ is discontinuous: this is the pinned phase. The threshold $K_{c}(\alpha)$ depends on the irrational $\ell$. For the standard Frenkel-Kontorova model ($\alpha\rightarrow\infty$), it is empirically known that the maximal threshold occurs for $\ell=\frac{3-\sqrt{5}}{2}=2-\varphi\approx 0.3819660\dots$ (where $\varphi$ is the golden number) or, equivalently, at $\ell=\varphi-1$, and $K_{c}(\infty)=0.9716$ Greene . In the following we examine the Aubry transition for $\ell=2-\varphi$.

At small $K$, the form of the solution (15) is coherent with KAM theorem which applies to the dynamical system, Eq. 7. KAM theorem ensures that, if $\ell$ is “sufficiently” irrational (there is a Diophantian condition Remark ), the solution (8) of the integrable model ($K=0$) remains a stable trajectory when the nonlinear potential is small enough, up to a change of variable $f$ of the form (15). In this case, $f$ is a continuous bijection. As a consequence, $x_{n}~{}$mod 1 takes all values in $[0,1]$ just as in the $K=0$ case: the KAM torus (here the circle $[0,1]$) with that irrational $\ell$ is preserved. It is known that KAM theorem does not hold for arbitrary large $K$ although (15) remains true with a discontinuous $f$. The threshold at which all KAM tori cease to exist signals the transition to “stochasticity” Greene which is also the Aubry transition for $\ell=2-\varphi$.

We first mention that the continuous degeneracy of the ground states (8) at $K=0$ is lifted at first order in perturbation theory for commensurate states but not for incommensurate states. We approximate the irrational number $\ell$ by a rational number $\ell=r/s$ (the larger the $s$, the better the approximation). At first order, the energy per atom of a unit-cell noted $w^{(1)}$, assuming $\mu=\ell$, is given by

which goes to zero for all $\phi$ when $s\rightarrow+\infty$ ($C$ is a constant independent of $\phi$). It means that the energy barrier for translating the undistorted state by $\phi$ vanishes for an ideal incommensurate state at the lowest order (it is remarkable that it remains true at higher orders as we will see below). Note that for a finite $s$, the energy depends on $\phi$ and the extrema are at $\phi=\frac{m}{2s}$ where $m$ is an integer.

We now solve (6) by a perturbation theory at small $K$. In order to determine the distortions $u_{n}=g(n\ell+\phi)$ one can rewrite the extremal condition Eq. (6) as

where $x=n\ell+\phi$, and use perturbation theory in $K$ to determine the periodic function $g$. By using Fourier series, one formally gets,

where $A_{p}$ are some coefficients given, at first and second order in $K$, by

where the last sum excludes all $n$ such that $n-p=ks$ where $k$ is an integer, if $\ell=r/s$.

Does the series Eq. (18) converge and what is the amplitude of the distortion?

For rational $\ell=r/s$, $g$ has denominators $1-\cos(2\pi p\ell)=0$ when $p$ is a multiple of $s$ ($p=ks$), so that (18) is infinite, which means that Eq. (17) cannot be satisfied for all $x$ (or all $\phi$). The only possibility is to choose the special values $\phi=\frac{m}{2s}$, where $m$ is any integer, which correspond to the extrema of $w^{(1)}$. In this case, when the denominator vanishes (for $p=ks$), the numerator vanishes too: $\sin(2\pi pn\ell+2\pi p\phi)=\sin(2\pi kn+\pi km)=0$. Consequently (17) can be satisfied at these special points but not for all $x$ or all $\phi$, i.e. $g$ is not defined everywhere, as expected for a commensurate solution.

For irrational $\ell$, however, there are “small denominators” but they never vanish. For $\ell$ “sufficiently” irrational (this is where the Diophantian condition is important), it can be shown that the first-order series (18) converges, thanks to the exponential decrease of the harmonics Wayne . The KAM theorem further ensures that higher-order series in $K$ are convergent as well, provided that $K$ remains small enough. In this case, $g$ is a continuous function.

The amplitude of the distortions $\delta$ defined as

depends on $K$, $\alpha$ and $\ell$. It is not simply proportional to $K$ but depends on both $\alpha$ and $\ell$ in a complicated way because of the small denominators in (18). We now compute the distortions numerically, without relying on perturbation theory.

In Fig. 2, we plot two examples, for two different values of $\alpha$ at small $K$, of envelope functions $g$ obtained numerically by a gradient descent algorithm which searches a zero of Eq. 6. When the algorithm converges (i.e. when the gradient is small), it gives a local mimimum. In the regime of small $K$, there is no metastable states and starting from random configurations always produces the same state characterized by the envelope functions $f$ and $g$ as expected for the ground state. We thus obtain in this way, for a rational approximant of $2-\varphi$ (typically $\ell=r/s=377/987$), a periodic configuration $\{x_{n}\}$ ($n=1,\dots,s$) and distortions $\{u_{n}\}$ which we plot as a function of $n\ell$ mod 1 to define $g$. We then compare them with the perturbative results given by Eq. (18) up to second order. Perturbation theory is in principle limited to $K\ll K_{c}(\alpha)$. By taking $K=40\%K_{c}(\alpha)$, we observe in both cases of Fig. 2 ($\alpha=0.4$ on the left and $\alpha=3$ on the right) that perturbation theory is accurate, especially at second-order. Small deviations are visible and disappear for smaller values of $K\ll K_{c}(\alpha)$, or, conversely, are amplified when $K\rightarrow K_{c}(\alpha)$. Note that the $y$ scale is not the same, so that the amplitude of the distortions $\delta$ is much smaller for small $\alpha$ (we have to take smaller values of $K$ as well as to remain at a fixed distance from the transition). The shape of the distortions also depends on $\alpha$. When $\alpha\rightarrow+\infty$, we get the usual Frenkel-Kontorova model and only the first harmonic $p=1$ is retained in (18). Thus, $g(x)=\gamma\sin 2\pi x$, $\gamma\equiv\frac{K}{4\pi}\frac{1}{1-\cos 2\pi\ell}$. This is already close to the result for $\alpha=3$. On the other hand, for $\alpha\rightarrow 0$, more and more harmonics must be included, some of them with a large denominator but the numerical result remains small (thanks to smaller $K$). This is what is seen in Fig. 2 (left). The numerical result is thus in agreement with KAM theorem and well approximated by lowest order perturbation theory.

The distortions are small in this regime and particularly so when $\alpha$ is small and $K$ is appropriately reduced below the transition.

When $K$ increases further, however, the previous perturbation theory fails. One can consider instead the “anti-integrable” limit $K\rightarrow\infty$ AA and do perturbation theory in $1/K$. When $1/K=0$, the solutions of Eq. (6) are given by

so that $x_{n}$ must be integers or half-integers. Since there is no determination of $x_{n+1}$ from $x_{n}$, any series of integers (or half-integers or a mixing) $\{x_{n}\}$ is acceptable, i.e. can be random and very “chaotic”. Among the solutions, the following one,

where $[\dots]$ is the integer part, is special. It is a ground state since all the atoms are at the bottoms of the potential ($x_{n}~{}$mod 1=1/2) and its average bond length is $\ell$. It has the expected form given by Eq. (15) with a discontinuous envelope function given by $f(x)=[x]+1/2$. Moreover, when $1/K$ is small but nonzero, a perturbative calculation in $1/K$ starting from Eq. (23) gives the ground state, whereas the other configurations give higher energy metastable states AA . The ground state in perturbation in $1/K$ is thus obtained by

with coefficients $B_{n}$ vanishing for large $K$. At first and second order in $1/K$, we find

Thus $B_{n}=B_{n}^{(1)}+B_{n}^{(2)}$ (at second order) is a decreasing function of $n$. The perturbation theory is correct if the prefactor is small $\frac{1}{K}(\frac{1+\cosh\alpha}{\sinh\alpha})^{2}\ll 1$, which needs, in the limit of small $\alpha$, that $K\alpha^{2}/4\gg 1$.

In Eq. (24), the periodic function $[x+a]+[x-a]-2[x]$ is discontinuous at points $na$, for all $n$. Therefore $f$ is discontinuous at each point $x=\pm n\ell~{}$mod 1, i.e. everywhere (since $n\ell~{}$mod 1 is dense in $[0,1]$ for $\ell$ irrational), with discontinuities that are functions of $B_{n}$.

In Fig. 3, the points are the atomic positions $x_{n}$ mod 1 computed numerically for large $K$ and plotted as a function of $n\ell$ mod 1. The gradient descent, started with a configuration sufficiently close to (24), converges to the ground state. We thus obtain the function $f$ which is increasing and discontinuous, as expected for a ground state in this regime. For $\alpha=3$ (Fig. 3, right), the second order result from Eq. 24 is shown as well (dashed line) and is a good approximation of the numerical result with main discontinuities at zero (obtained at zeroth order), $\pm\ell$ (first-order) and $\pm 2\ell~{}$mod $1$ (second order). The other discontinuities obtained numerically are not reproduced at this order. The largest discontinuity at zero means that atoms avoid the maxima of the potential. For $\alpha=0.4$ (Fig. 3, left), the lowest order perturbation theory already fails even for $K=15$ because the prefactor is of order 1. The result shown by a dashed line is a fit that uses (24) and fitting parameters $B_{n}$ up to $n=3$, reproducing the main three discontinuities. The form of the solution (24) seems to remain accurate even though the parameters $B_{n}$ are no longer given by perturbation theory. In both cases, we see that the distortions (departure from the $y=x$ line) are strong,

The interpolation between the two previous regimes $K\rightarrow 0$ and $K\rightarrow\infty$ is through the Aubry transition.

The simplest way to show numerically the existence of an Aubry transition is to follow the discontinuities of the envelope functions AubryQuemerais . In Fig. 4, we show, as above, the numerically computed envelope function $f$. Since $K$ is reduced compared with the results of Fig. 3, the distortions are reduced and $f(x)$ gets closer to $f(x)=x$. In each figure (top), two examples of values of $K$ close to the threshold of the Aubry transition, $K_{c}(\alpha)$, are given. The orange (light gray) points ($K<K_{c}(\alpha)$) are the points of a continuous function (see insets for more clarity), just as in Fig. 2, and the black points ($K>K_{c}(\alpha)$) are that of a discontinuous function as in Fig. 3. We observe that the discontinuities close continuously so that the transition is a second-order transition. Furthermore, we have checked the convergence of the envelope function for successive rational approximants of $2-\varphi$. Two examples, one for $K<K_{c}(\alpha)$ and one for $K>K_{c}(\alpha)$, are given in Fig. 4 (bottom). Although, strictly speaking, there is no Aubry transition in commensurate systems, using such large values of $s$ ensures a sharp change of continuity as a function of $K$. By locating the value of $K$ for which the discontinuities close, we extract $K_{c}(\alpha)$ and the phase diagram showing the sliding phase $K<K_{c}(\alpha)$ and the pinned phase $K>K_{c}(\alpha)$ (Fig. 5). Note that $K_{c}(\alpha)$ vanishes when $\alpha\rightarrow 0$.

This can be simply understood by adapting an earlier argument aubry_minor on the equilibrium of forces, Eq. (6). From the existence of a bound on the distortions expressed by Eq. (11), one obtains for the first part of Eq. (6) that $|2x_{n}-x_{n+1}-x_{n-1}|\leq 2$. In order to satisfy Eq. (6), its second part given by $\frac{K}{(2\pi)^{2}}V_{\alpha}^{\prime}(x_{n})$ cannot be arbitrary large. For irrational $\ell$ and sliding solution, $x_{n}$ mod 1 takes all values in $[0,1]$ so that the maximum of $V^{\prime}_{\alpha}(x)$, noted $V^{\prime}_{m}$, is necessarily reached. Therefore, if $\frac{K}{(2\pi)^{2}}V^{\prime}_{m}>2$, some atoms in the sliding phase cannot be at equilibrium. In the limit of small $\alpha$, given the expression of the maximum of the derivative (4), we get that, for

it is impossible to maintain the balance of forces in the sliding phase. In particular, when $\alpha=0$ the right-hand side vanishes so that

At the limit $\alpha=0$ the potential is discontinuous at $x=0$ and the KAM torus is destroyed at that point. This qualitative argument confirms that in the limit of small $\alpha$, the pinned phase should be favored at small $K$. Importantly, one sees that if the derivative of the potential $V^{\prime}_{m}$ is somewhere strong enough, the sliding phase no longer exists because the atoms that experience that strong force (some necessarily do in the sliding phase) cannot be at equilibrium: the Aubry transition threshold of the pinned state is reduced. The bound in (28) is in fact a very crude estimate of $K_{c}(\alpha)$ (see the steep dashed line in Fig. 5). A better bound represented by the dashed curve in Fig. 5 will be given in section IV.

In order to get some quantitative insights into how large the distortions can be near and at the Aubry transition when the phase is pinned, we also compute numerically the bond lengths,

Recall that the average bond length is $\frac{1}{s}\sum_{n=1}^{s}\ell_{n}=\ell$ so that $(u_{n+1}-u_{n})/\ell$ measures the amplitude of the distortion with respect to the average. We define another envelope function $h$ by $(u_{n+1}-u_{n})/\ell=h(n\ell+\phi)$. Its amplitude

gives an idea of how much distorted the structure is. In Fig. 6, we show $(u_{n+1}-u_{n})/\ell$ as a function of $n\ell~{}$mod 1 (i.e. $h$) for two values of $K$, one for $K<K_{c}(\alpha)$ (continuous curves), the other for $K>K_{c}(\alpha)$ (discontinuous curve). For large values of $\alpha$ ($\alpha=3$ in Fig. 6, right), the amplitude is large. On the contrary, for small $\alpha$ ($\alpha=0.4$ in Fig. 6, left), we observe that the distortions are well within $1~{}\%$ of the bond length (the two horizontal lines correspond to $\pm 1~{}\%$). The amplitude $\xi$ is reported in Fig. 7 as a function of $\alpha$ at $K=K_{c}(\alpha)$. For large $\alpha$, the maximal distortion at the transition is 23% (Frenkel-Kontorova limit). For smaller values of $\alpha$, $\xi$ can be arbitrary small. It can also be seen qualitatively in Fig. 4 that for small $\alpha$ (left), the distortions are very small and $f$ is very close to the undistorted result $f(x)=x$, whatever the values of $K$ across and near the Aubry transition.

The important conclusion is that it is not necessary to have large distortions to have an Aubry transition or be in the pinned phase. In particular, if $\alpha=0$, so that the potential is discontinuous, $K_{c}(0)=0$ as we have shown above: the system is immediately in the pinned phase. Yet the potential is flat almost everywhere so that there is no distortion. This extreme situation remains somehow valid at small $\alpha$, as shown here: it is replaced by a pinned phase with small distortions.