path: root/hidden-order.tex

\documentclass[fleqn]{article}

\usepackage{amsmath,amssymb}
\usepackage{fullpage,graphicx}

\title{Elastic Lifshitz point}
\author{Jaron Kent-Dobias}

\begin{document}

\def\Aog{\mathrm A_{1\mathrm g}}
\def\Bog{\mathrm B_{1\mathrm g}}
\def\Btg{\mathrm B_{2\mathrm g}}
\def\Eg{\mathrm E_{\mathrm g}}
\def\X{\mathrm X}

\maketitle

The free energy density of our model has three terms, respectively corresponding to the strain, the order parameter, and the strain--order parameter coupling. The free energy density for the strain, taken to quadratic order under the assumption of small strains, is generically of the form $f_{\mathrm e}=\frac12\lambda_{ijk\ell}\epsilon_{ij}\epsilon_{k\ell}$. Symmetries of the strain tensor, which is symmetric, and those of the lattice restrict the form of $\lambda$. The most general form for a system with a tetragonal point group is
\[
  f_{\mathrm e}=\frac12\Big[\lambda_{1111}(\epsilon_{11}^2+\epsilon_{22}^2)+
    \lambda_{3333}\epsilon_{33}^2+
      2\lambda_{1133}(\epsilon_{11}+\epsilon_{22})\epsilon_{33}+
      2\lambda_{1122}\epsilon_{11}\epsilon_{22}+
      4\lambda_{1212}\epsilon_{12}^2+
      4\lambda_{1313}(\epsilon_{13}^2+\epsilon_{23}^2)\Big]
\]
A very convenient way to come by this form is to use knowledge of the point group of the lattice and the way that strain transforms under it. The following linear combinations of strains transform like a particular irreducible representation of the point group:
\begin{align*}
  \epsilon_{\Aog}^{(1)}=\epsilon_{11}+\epsilon_{22}
  &&
  \epsilon_{\Aog}^{(2)}=\epsilon_{33}
  \\
  \epsilon_{\Bog}^{(1)}=\epsilon_{11}-\epsilon_{22}
  &&
  \epsilon_{\Btg}^{(1)}=\epsilon_{12}
  &&
  \epsilon_{\mathrm E_{\mathrm G}}^{(1)}=\{\epsilon_{11},\epsilon_{22}\}
\end{align*}
The most general quadratic free energy density is built from combinations of strains that transform like scalars, or $\Aog$, yielding
\[
  \begin{aligned}
    f_{\mathrm e}&=\frac12\sum_{\X}\lambda_{\X}^{(ij)}\epsilon_{\X}^{(i)}\epsilon_{\X}^{(j)}\\
    &=\frac12\Big[\lambda^{(11)}_{\Aog}(\epsilon_{\Aog}^{(1)})^2+
      \lambda_{\Aog}^{(22)}(\epsilon_{\Aog}^{(2)})^2+
      2\lambda_{\Aog}^{(12)}\epsilon_{\Aog}^{(1)}\epsilon_{\Aog}^{(2)}+
      \lambda_{\Bog}^{(11)}(\epsilon_{\Bog}^{(1)})^2+
      \lambda_{\Btg}^{(11)}(\epsilon_{\Btg}^{(1)})^2+
    \lambda_{\Eg}^{(11)}\epsilon_{\Eg}^{(1)}\cdot\epsilon_{\Eg}^{(1)}\Big]
  \end{aligned}
\]
where the sum is over the irreducible representations of the point group.
We can now identify
\begin{align*}
  \lambda_{\Aog}^{(11)}=\tfrac12(\lambda_{1111}+\lambda_{1122})
  &&
  \lambda_{\Aog}^{(22)}=\lambda_{3333}
  &&
  \lambda_{\Aog}^{(12)}=\lambda_{1133}
  \\
  \lambda_{\Bog}^{(11)}=\tfrac12(\lambda_{1111}-\lambda_{1122})
  &&
  \lambda_{\Btg}^{(11)}=4\lambda_{1212}
  &&
  \lambda_{\Eg}^{(11)}=4\lambda_{1313}
\end{align*}
Consider a generic order parameter $\eta$. To write down its free energy, we in principle must know both how it transforms under symmetry operations of the lattice and how derivative operators transform. For derivative operators, the two independent combinations are $\nabla_\parallel=\{\partial_x,\partial_y\}$ which transforms like $\Eg$, and $\nabla_\perp=\partial_3$ which transform like $\Aog$. The irreducible symmetry that $\eta$ transforms like determines its coupling to strain. To quadratic order we have
\[
  f_{\X}=\tfrac12b^{(i)}\epsilon_{\X}^{(i)}\cdot\eta
  +\tfrac12e^{(i)}\epsilon_{\Aog}^{(i)}\eta^2
V\]
The total  free energy is 
\[
  F=\int d^3x\,(f_{\mathrm e}+f_{\mathrm o}+f_{\X})
\]
There are three distinct possibilities here: $\eta$ transforms like a one-component irreducible representation of the point group that is not $\Aog$ ($\Bog$ or $\Btg$), $\eta$ transforms like $\Aog$, and $\eta$ is a two-component vector that transforms like $\Eg$.

\section{$\Bog$ or $\Btg$ order parameter}

We will first tackle the case of a non-$\Aog$, one-component order parameter.
The most general quartic free energy density (discounting total derivatives) is
\[
  f_{\mathrm o}=\frac12\Big[r\eta^2+
                c_\parallel(\nabla_\parallel\eta)^2+
                c_\perp(\nabla_\perp\eta)^2+
                D(\nabla^2\eta)^2\Big]
                +u\eta^4
\]
independent of the symmetry of $\eta$. In principle we could have $D_\parallel\neq D_\perp$, but this does not affect the physics at hand. This is the free energy for a Lifshitz point, and so we expect to see that phenomenology in $\eta$. 
Before doing anything, we can minimize the free energy with respect to strain alone to find the strain in terms of $\eta$ exactly. We have
\[
  0=\frac{\delta F}{\delta\epsilon_{\mathrm X}^{(1)}(x)}=\lambda_{\mathrm X}^{(11)}\epsilon_{\mathrm X}^{(1)}(x)+\frac12b^{(1)}\eta(x)
\]
whence we immediately have $\epsilon_{\mathrm X}^{(1)}=-\frac{b^{(1)}}{2\lambda_{\mathrm X}^{(11)}}\eta(x)$. We also have
\[
  0=\frac{\delta F}{\delta\epsilon_{\Aog}^{(i)}(x)}
  =\lambda_{\Aog}^{(ij)}\epsilon_{\Aog}^{(j)}(x)+\frac12 e^{(i)}\eta^2(x)
\]
which is a linear system whose solutions are
\begin{align*}
  \epsilon_{\Aog}^{(1)}(x)=\frac12\frac{e^{(1)}\lambda_{\Aog}^{(22)}-e^{(2)}\lambda_{\Aog}^{(12)}}{((\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}\eta^2(x)
  &&
  \epsilon_{\Aog}^{(2)}(x)=\frac12\frac{e^{(2)}\lambda_{\Aog}^{(11)}-e^{(1)}\lambda_{\Aog}^{(12)}}{((\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}\eta^2(x)
\end{align*}
All other strain fields are extremized for $\epsilon_{\mathrm Y}^{(i)}(x)=0$. These solutions may be substituted into the free energy to arrive at one that is only a functional of $\eta(x)$ whose free energy density is identical to $f_{\mathrm o}$ but with
\begin{align*}
  r\to \tilde r=r-\frac{(b^{(1)})^2}{4\lambda_{\X}^{(11)}}
  &&
  u\to \tilde u=u+\frac18\frac{(e^{(1)})^2\lambda_{\Aog}^{(22)}+(e^{(2)})^2\lambda_{\Aog}^{(11)}-2e^{(1)}e^{(2)}\lambda_{\Aog}^{(12)}}{(\lambda_{(12)}^{\Aog})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}
\end{align*}
so that the total free energy is
\[
  F=\int dx\bigg(\frac12\Big[\tilde r\eta^2+c_\parallel(\nabla_\parallel\eta)^2+c_\perp(\nabla_\perp\eta)^2+D(\nabla^2\eta)^2\Big]+\tilde u\eta^4\bigg)
\]
Replacing $\eta$ with its inverse Fourier transform, the free energy is
\[
  F=\frac V2\sum_q\Big[\tilde r|\tilde\eta_q|^2+c_\parallel q_{\parallel}^2|\tilde\eta_q|^2
  +c_\perp q_{\perp}^2|\tilde\eta_q|^2+Dq^4|\tilde\eta_q|^2\Big]
  +V\tilde u\sum_q\sum_{q'}\sum_{q''}\tilde\eta_q\tilde\eta_{q'}\tilde\eta_{q''}\tilde\eta_{-(q+q'+q'')}
\]
The susceptibility of this system is
\[
  \begin{aligned}
    \chi^{-1}(x,x')
    &=\frac{\delta^2F}{\delta\eta(x)\delta\eta(x')}
    =\frac{\delta}{\delta\eta(x')}\Big[\tilde r\eta(x)-c_\parallel\nabla_\parallel^2\eta(x)-c_\perp\nabla_\perp^2\eta(x)+D\nabla^4\eta(x)+4\tilde u\eta^3(x)\Big] \\
    &=\Big[\tilde r-c_\parallel\nabla_\parallel^2-c_\perp\nabla_\perp^2+D\nabla^4+12\tilde u\eta^2(x)\Big]\delta(x-x')
  \end{aligned}
\]
or in Fourier space,
\[
  \chi^{-1}(q)
  =\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4+12\tilde u\sum_{q'}\tilde\eta(q')\tilde\eta(-q')
\]
To get the elastic susceptibility we must now extremize over $\eta$ and $\epsilon_{\Aog}^{(i)}$. The latter is done exactly as before, but the former givs
\[
  0=\frac{\delta F}{\delta\eta}=\frac{\delta F_\eta}{\delta\eta}+\frac12b^{(1)}\epsilon_\X^{(1)}
\]
whose solutions give $\eta[\epsilon_\X^{(1)}]$ implicitly as
\[
  \eta^{-1}[\eta(x)]=-\frac2{b^{(1)}}\frac{\delta F_\eta}{\delta\eta(x)}
\]
The inverse function theorem gives
\[
  \begin{aligned}
    \bigg(\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}\bigg)^{-1}
    =\frac{\eta^{-1}[\eta(x)]}{\eta(x')}
    =-\frac2{b^{(1)}}\frac{\delta^2F_\eta}{\delta\eta(x)\delta\eta(x')}
    =-\frac2{b^{(1)}}\bigg[\chi^{-1}(x,x')+\frac{(b^{(1)})^2}{4\lambda_\X^{(11)}}\delta(x-x')\bigg]
  \end{aligned}
\]
The elastic susceptibility then follows as
\begin{align*}
  (\chi_{\X}^{(11)}(x,x'))^{-1}
  &=\frac{\delta^2F}{\delta\epsilon_{\X}^{(1)}(x)\delta\epsilon_{\X}^{(1)}(x')} \\
  &=\lambda_\X^{(11)}\delta(x-x')+
  b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}+\frac12b^{(1)}\int dx''\,\epsilon(x'')\frac{\delta^2\eta(x)}{\delta\epsilon_\X^{(1)}(x')\delta\epsilon_\X^{(1)}(x'')} \\
  &\qquad+\int dx''\,dx'''\,\frac{\delta^2F_\eta}{\delta\eta(x'')\delta\eta(x''')}\frac{\delta\eta(x'')}{\delta\epsilon_\X^{(1)}(x)}\frac{\delta\eta(x''')}{\delta\epsilon_\X^{(1)}(x')} \\ 
  &=\lambda_\X^{(11)}\delta(x-x')+
  b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}
  -\frac12b^{(1)}\int dx''\,dx'''\,\bigg(\frac{\partial\eta(x'')}{\partial\epsilon_\X^{(1)}(x''')}\bigg)^{-1}\frac{\delta\eta(x'')}{\delta\epsilon_\X^{(1)}(x)}\frac{\delta\eta(x''')}{\delta\epsilon_\X^{(1)}(x')} \\ 
  &=\lambda_\X^{(11)}\delta(x-x')+
  b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}
  -\frac12b^{(1)}\int dx''\,\delta(x-x'')\frac{\delta\eta(x'')}{\delta\epsilon_\X^{(1)}(x')} \\
  &=\lambda_\X^{(11)}\delta(x-x')+
  \frac12b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}
\end{align*}
whence
\[
  \chi_\X^{(11)}(x,x')
  =\bigg(\lambda\delta(x-x')+\frac12b^{(1)}\frac{\delta\eta(x)}{\delta\epsilon_\X^{(1)}(x')}\bigg)^{-1}
  =\frac1{\lambda_\X^{(11)}}\bigg(\delta(x-x')+\frac{(b^{(1)})^2}{4\lambda_\X^{(11)}}\chi(x,x')\bigg)
\]
or
\[
  \chi_\X^{(11)}(q)=\frac1{\lambda_\X^{(11)}}\bigg(1+\frac{(b^{(1)})^2}{4\lambda_\X^{(11)}}\chi(q)\bigg)
\]
The same process can be used to get the susceptibility for the $\Aog$ components. Now extremizing over $\epsilon_{\X}^{(1)}$ and $\eta$, we have
\[
  0=\frac{\delta F}{\delta\eta}=\frac{\delta F_\eta}{\delta\eta}+e^{(i)}\epsilon_{\Aog}^{(i)}\eta
\]
Unfortunately we cannot use the inverse function theorem as before. I'm still figuring out how to do this cleanly.

We will assume that our system orders at some specific $q_\perp=q^*$. This ansatz is equivalent to
\begin{align*}
  \tilde\eta(q)=\tfrac12\eta_*\big[\delta_{q_\perp,q_*}+\delta_{q_\perp,-q_*}\big]\delta_{q_\parallel,0}
\end{align*}
For $q_*\neq0$, we have
\[
  F=\frac V2\bigg[\frac12(\tilde r+c_\perp q_*^2+Dq_*^4)\eta_*^2+\frac34\tilde u\eta_*^4\bigg]
\]
while for $q^*=0$ we have
\[
  F=\frac V2\big(\tilde r\eta_*^2+2\tilde u\eta_*^4\big)
\]
For $\tilde r>0$ and $c_\perp>0$ or $r>c_\perp^2/4D$ and $c_\perp<0$, there is a free energy minimizer with $\eta_*=0$. For $r<0$ there is a local minimum with $q_*=0$ and $\eta_*^2=-\tilde r/4\tilde u$. For $c_\perp<0$ and $r<c_\perp^2/4D$ there is a local minimum with $q_*^2=-c_\perp/2D$ and
\[
  \eta_*^2=\frac{c_\perp^2-4D\tilde r}{12D\tilde u}=-\frac{\tilde r-\tilde r_c}{3\tilde u}
\]
with $\tilde r_c=c_\perp^2/4D$. Between the disordered and either of these phases is a continuous phase transition, and between the nontrivial phases is an abrupt transition.

In the disordered phase, $\sum_{q}\tilde\tilde\eta(q)\tilde\eta(-q)=0$ and the susceptibility is
\[
  \chi^{-1}(q)
  =\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4
\]
In the ordered phase, $\sum_q\tilde\eta(q)\tilde\eta(-q)=\eta_*^2=-\tilde r/4\tilde u$, and
\[
  \chi^{-1}(q)
  =\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4-3\tilde r
  =-2\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2
\]
In the modulated phase, $\sum_q\tilde\eta(q)\tilde\eta(-q)=\frac12\eta_*^2=(c_\perp^2-4D\tilde r)/24D\tilde u$ and
\[
  \chi^{-1}(q)
  =\tilde r+c_\parallel q_\parallel^2+c_\perp q_\perp^2+Dq^4+c_\perp^2/2D-2\tilde r
  =-(\tilde r-\tilde r_c)+c_\parallel q_\parallel^2+D(q_\parallel^4+2q_\parallel^2q_\perp^2)+D(q_*^2-q_\perp^2)^2
\]
At the disordered--ordered transition with $\tilde r_c=0$, this yields
\[
  \chi(q)=\begin{cases}\big[Dq^4+c_\parallel q_\parallel^2+c_\perp q_\perp^2+|\Delta\tilde r|\big]^{-1}&\Delta\tilde r>0\\\big[Dq^4+c_\parallel q_\parallel^2+c_\perp q_\perp^2+2|\Delta\tilde r|\big]^{-1}&\Delta\tilde r<0\end{cases}=\frac1{c_\perp}\frac{\xi_\perp^2}{1+\xi_\parallel^2q_\parallel^2+\xi_\perp^2q_\perp^2+\xi_\perp^2(D/c_\perp)q^4}
\]
where the correlation lengths are
\begin{align*}
  \xi_\perp=\begin{cases}(|\Delta\tilde r|/c_\perp)^{-1/2}&\tilde r<\tilde r_c\\(2|\Delta \tilde r|/c_\perp)^{-1/2}&\tilde r>\tilde r_c\end{cases}
  &&
  \xi_\parallel=\begin{cases}(|\Delta\tilde r|/c_\parallel)^{-1/2}&\tilde r<\tilde r_c\\(2|\Delta \tilde r|/c_\parallel)^{-1/2}&\tilde r>\tilde r_c\end{cases}
\end{align*}
At the disordered--modulated transition with $\tilde r_c=c_\perp^2/4D$, this yields
\[
  \chi(q)=\big[c_\parallel q_\parallel^2+D(q_\parallel^4+2q_\parallel^2q_\perp^2)+D(q_*^2-q_\perp^2)^2+|\Delta\tilde r|\big]^{-1}=\frac1{D}\frac{\xi_\perp^4}{1+\xi_\parallel^2q_\parallel^2+\xi_\perp^4(q_\parallel^4+2q_\parallel^2q_\perp^2)+\xi_\perp^{4}(q_*^2-q_\perp^2)^2}
\]
where $\xi_\perp=(|\Delta\tilde r|/D)^{-1/4}$ and $\xi_\parallel=(|\Delta \tilde r|/c_\parallel)^{-1/2}$. 

The Ginzburg criterion gives the proximity $t_G$ of the critical point at which mean field theory is no longer self consistent, with
\begin{align*}
  t_G=\frac{k_B^2}{32\pi^2(\Delta c_V)^2\xi_0^6}
\end{align*}
Experiments give $\Delta c_V\sim1\times10^5\,\mathrm J\,\mathrm m^{-3}\,\mathrm K^{-1}$ \cite{fisher_specific_1990} and $T_c\sim17.5\,\mathrm K$. A fit of $\tilde\lambda$ to experimental data (Fig.~\ref{fig:B1g.fit}) yields
\begin{align*}
  \lambda=71\,\mathrm{GPa}-(0.10\,\mathrm{GPa}\,\mathrm{K}^{-1})T &&
  \frac{b^2}{4\lambda Dq_*^4}=0.084&&\frac a{Dq_*^4}=0.0038\,\mathrm K^{-1}
\end{align*}
with $|r-r_c|=a|T-T_c|$. We suspect that the modulation at the transition at very low pressure is on the order of the lattice spacing, which would give $q_*^{-1}\sim9.568\,\text{\r A}$. Combined with our fit, this gives
\[
  \xi_0=(aT_c/D)^{-1/4}=\big[T_c(a/Dq_*^4)q_*^4\big]^{-1/4}\sim19\,\text{\r A}
\]
and therefore $t_G\sim1.2\times10^{-6}$, which means one would need to get within about $\Delta T=T_ct_G\sim20\,\mu\mathrm K$ of the critical point to see mean field theory break down.

\begin{figure}
  \centering
  \includegraphics[width=0.5\textwidth]{cusp}
  \caption{Experimental data for the $\Bog$ effective elastic constant $\tilde\lambda_{\Bog}$ (blue dots), along with a fit of the above theory (orange line).}
  \label{fig:B1g.fit}
\end{figure}

\section{$\Aog$ order parameter}

For the case where $\eta$ transforms like $\Aog$, the most general free energy is
\[
  f_{\mathrm o}=\frac12\Big[r\eta^2+
                c_\parallel(\nabla_\parallel\eta)^2+
                c_\perp(\partial_3\eta)^2+
                g(\partial_3\eta)^3+
                D\big[(\nabla_\parallel^2+\partial_3^2)\eta\big]^2\Big]
                +v\eta^3
                +u\eta^4
\]
Extremizing over $\epsilon$ requires
\[
  0=\frac{\delta F}{\delta\epsilon_{\Aog}^{(i)}(x)}
  =\lambda_{\Aog}^{(ij)}\epsilon_{\Aog}^{(j)}(x)+\frac12 b^{(i)}\eta(x)+\frac12 e^{(i)}\eta^2(x)
\]
which is a system of equations with solution
\begin{align*}
  \epsilon_{\Aog}^{(1)}(x)&=-\frac1{2\big[(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}\big]}\bigg[
      \Big(b^{(2)}\lambda_{\Aog}^{(12)}-b^{(1)}\lambda_{\Aog}^{(22)}\Big)\eta(x)+\Big(e^{(2)}\lambda_{\Aog}^{(12)}-e^{(1)}\lambda_{\Aog}^{(22)}\Big)\eta^2(x)
    \bigg]\\
    \epsilon_{\Aog}^{(2)}(x)&=-\frac1{2\big[(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}\big]}\bigg[
      \Big(b^{(1)}\lambda_{\Aog}^{(12)}-b^{(2)}\lambda_{\Aog}^{(22)}\Big)\eta(x)+\Big(e^{(1)}\lambda_{\Aog}^{(12)}-e^{(2)}\lambda_{\Aog}^{(11)}\Big)\eta^2(x)
    \bigg]
\end{align*}
All other strains are zero. Like the previous case, substitution of this solution into the free energy results in a functional of $\eta$ alone. As before, the result is a free energy density with the form of $f_{\mathrm o}$ but with
\begin{align*}
  r&\to \tilde r=r+\frac18\frac{(b^{(1)})^2\lambda_{\Aog}^{(22)}+(b^{(2)})^2\lambda_{\Aog}^{(11)}-2b^{(1)}b^{(2)}\lambda^{(12)}_{\Aog}}{(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}} \\
  u&\to \tilde u=u+\frac18\frac{(e^{(1)})^2\lambda_{\Aog}^{(22)}+(e^{(2)})^2\lambda_{\Aog}^{(11)}-2e^{(1)}e^{(2)}\lambda^{(12)}_{\Aog}}{(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}
\end{align*}
Unlike before, there is now an additional term in $f$ proportional to $\eta^3(x)$ with coefficient
\[
  v\to\tilde v=v+\frac14\frac{b^{(1)}e^{(1)}\lambda_{\Aog}^{(22)}+b^{(2)}e^{(2)}\lambda_{\Aog}^{(11)}-\lambda_{\Aog}^{(12)}(b^{(1)}e^{(2)}+b^{(2)}e^{(1)})}{(\lambda_{\Aog}^{(12)})^2-\lambda_{\Aog}^{(11)}\lambda_{\Aog}^{(22)}}
\]
This implies that, unless the model is tuned to a place where this term vanishes, all transitions will be abrupt. We won't study this phenomenology any further.

\section{$\Eg$ order parameter}

For a $\Eg$ order parameter, the most general quartic free energy density is
\[
  f_{\mathrm o}=\frac12\Big[r\eta^2+c_\parallel(\nabla_\parallel\cdot\eta)^2+c_\perp(\nabla_\perp\eta)^2+D(\nabla^2\eta)^2\Big]+u\eta^4
\]
Minimizing the strain in $\Aog$ proceeds exactly as before, while $\Eg$ gives
\[
  0=\frac{\delta F}{\delta\epsilon_{\Eg i}^{(1)}(x)}=\lambda_{\Eg}^{(11)}\epsilon_{\Eg i}^{(1)}+\frac12b^{(1)}\eta_i(x)
\]
whence $\epsilon_{\Eg i}^{(1)}(x)=-b^{(1)}/2\lambda_{\Eg}^{(11)}\eta_i(x)$. This leads to the same shifts in $r$ and $u$ as before, and we will use $\tilde r$ and $\tilde u$ to take the same meanings as they did for the $\Bog$ case. We now have
\[
  \begin{aligned}
    \chi^{-1}_{ij}(x,x')
    &=\frac{\delta^2F}{\delta\eta_i(x)\delta\eta_j(x')}
    =\frac\delta{\delta\eta_j(x')}\Big[\tilde r\eta_i(x)-c_\parallel\partial_i^2\eta_i(x)-c_\perp\nabla_\perp^2\eta_i(x)+D\nabla^4\eta_i(x)+4\tilde u\eta_k(x)\eta_k(x)\eta_i(x)\Big] \\
    &=\Big\{\Big[\tilde r-c_\parallel\partial_i^2-c_\perp\nabla_\perp^2+D\nabla^4+4\tilde u\eta^2(x)\Big]\delta_{ij}+8\tilde u\eta_i(x)\eta_j(x)\Big\}\delta(x-x')
  \end{aligned}
\]
The Fourier transformed version is
\[
  \chi^{-1}_{ij}(q)=\Big[\tilde r+c_\parallel q_i^2+c_\perp q_\perp^2+Dq^4+4\tilde u\sum_{q'}\tilde\eta_i(q')\tilde\eta_i(-q')\Big]\delta_{ij}+8\tilde u\sum_{q'}\tilde\eta_i(q')\tilde\eta_j(-q')
\]

We will assume the system orders at some $q_\perp=q_*$ as before, with
\begin{align*}
  \tilde\eta_1(q)=\tfrac12\eta_*(\delta_{q_\perp,q_*}+\delta_{q_\perp,-q_*})\delta_{q_\parallel,0}
  &&
  \tilde\eta_2(q)=\tfrac12\eta_*(e^{-i\phi_*}\delta_{q_\perp,q_*}+e^{i\phi_*}\delta_{q_\perp,-q_*})\delta_{q_\parallel,0}
\end{align*}
where we have introduced a relative phase $\phi_*$ between the two components. For $q_*\neq0$ we have
\[
  F=\frac V2\big[\tilde r+c_\perp q_*^2+Dq_*^4\big]\eta_*^2+\frac V4\tilde u\big[5+\cos(2\phi)\big]\eta_*^4
\]
while for $q_*=0$ we have
\[
  F=\frac14\tilde r\big[3+\cos(2\phi)\big]\eta_*^2+\frac34\tilde u\big[3+\cos(2\phi)\big]^2\eta_*^4
\]
There is as before an $\eta^*=0$ solution corresponding to a trivial state, with the same boundaries as the $\Bog$ case. There is an ordered state with $\eta_*^2=-\tilde r/4\tilde u$ and $\phi_*=\pm\pi/2$. Finally, there is a modulated state with $q_*^2=-c_\perp/2D$, $\phi^*=\pm\pi/2$, and
\[
  \eta_*^2=\frac{c_\perp^2-4D\tilde r}{16D\tilde u}
\]
Now, however, the ordered state becomes unstable when $c_\perp<0$, and there is a continuous transition between the ordered and metastable states along the $c_\perp=0$, $r<0$ line.

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\bibliography{hidden_order.bib}

\end{document}