Spatially Modulated Phases and Competing Orders

The previous page described the simplest ordered phase: a charged scalar condenses at zero momentum, producing a homogeneous holographic superfluid. Real strongly correlated matter is rarely that polite. Charge density waves, spin density waves, pair density waves, stripes, nematicity, loop-current order, smectics, crystals, and superconductivity often appear close together and compete for the same low-energy degrees of freedom.

Holography has a crisp way to organize this zoo. A homogeneous charged black brane can become unstable not only to $k=0$ scalar hair, but also to a finite-momentum normalizable mode:

$$\boxed{ \text{finite-}k\text{ bulk instability} \quad\longleftrightarrow\quad \text{spontaneous spatial modulation in the boundary theory}. }$$

The important word is spontaneous. We are not imposing a lattice or a striped chemical potential from the boundary. The sources remain homogeneous. The vevs choose a wavevector, a phase, and a pattern.

Typical boundary profiles look like

$$\langle J^t(x)\rangle = \rho_0+\rho_1\cos(k_*x+\varphi), \qquad \langle O(x)\rangle = O_1\cos(k_*x+\varphi),$$

possibly accompanied by modulated currents such as

$$\langle J_y(x)\rangle = J_1\sin(k_*x+\varphi).$$

The phase $\varphi$ is not a decorative constant. It is the Goldstone coordinate of the broken translation symmetry. Shifting $\varphi$ slides the entire pattern.

A finite-momentum instability of a homogeneous charged brane. The boundary sources remain homogeneous, but the normalizable data develop spatial dependence. The preferred wavevector $k_*$ is found from the linearized instability near $T_c$ and, below $T_c$, from the nonlinear free energy $\Omega(T,\mu,k)$.

Throughout this page $d_s$ denotes the number of boundary spatial dimensions, so the boundary spacetime dimension is $d=d_s+1$. The radial coordinate is $z$, with the boundary at $z=0$. A wavevector along $x$ is denoted by $k$ or $Q$.

What it means to break translations

Consider a scalar operator $O$ and its Fourier components $O_Q$. Under a translation $x\mapsto x+a$,

$$O_Q\mapsto e^{iQa}O_Q.$$

Therefore a nonzero expectation value

$$\Phi_Q\equiv \langle O_Q\rangle\neq0$$

chooses a phase. In position space,

$$\langle O(x)\rangle = \Phi_Q e^{iQx}+\Phi_Q^*e^{-iQx} = 2|\Phi_Q|\cos(Qx+\varphi).$$

A continuous translation by arbitrary $a$ no longer leaves the state invariant. Only translations by integer multiples of the period survive:

$$a=\frac{2\pi n}{Q}.$$

This is the continuum version of a density wave. On a microscopic lattice, the story is slightly richer because only discrete translations were present to begin with. Then one must distinguish commensurate and incommensurate order. Most bottom-up holographic models begin in a continuum, so the finite-$Q$ state spontaneously breaks a continuous translation symmetry.

Different patterns break different symmetries:

Pattern	Boundary profile	Symmetry breaking
Stripe or smectic	$\rho(x)=\rho_0+\rho_1\cos Qx$	one translation, usually rotations
Crystal	$\rho(\vec x)$ periodic in two or more directions	translations to a lattice subgroup
Helix	$\langle J_y\rangle+i\langle J_z\rangle\propto e^{iQx}$	translation and rotation separately, but not a combined operation
Nematic	no density wave, anisotropic tensor vev	rotations, but not translations
Pair-density wave	$\langle O_{\rm pair}(x)\rangle\propto \cos Qx$	$U(1)$ and translations

The distinction between explicit and spontaneous translation breaking is the first thing to check in any holographic model. For a bulk field $X$ dual to an operator $O$ of dimension $\Delta$,

$$X(z,x) = z^{d-\Delta}J(x) + z^\Delta \langle O(x)\rangle +\cdots.$$

The modulation is explicit if the source $J(x)$ contains the wavevector. It is spontaneous if the source is homogeneous or zero while the normalizable coefficient is modulated:

$$J_{Q}=0, \qquad \langle O_Q\rangle\neq0.$$

That source-vev distinction is not bookkeeping. It determines the hydrodynamics. A spontaneously broken density wave has a sliding Goldstone mode; an explicitly imposed lattice generally does not.

The finite-momentum instability

The normal state is usually a charged black brane. At low temperature its deep interior is often controlled by an $AdS_2\times\mathbb R^{d_s}$ region, or by a related semi-local scaling geometry. Linear perturbations about this background can be decomposed by boundary momentum:

$$\delta X(z,x,t)=e^{-i\omega t+ikx}\,\delta X_k(z).$$

For a single uncoupled fluctuation, the near-horizon equation often reduces to an $AdS_2$ scalar problem with an effective mass $m_{\rm eff}^2(k)$. More generally several fields mix, and the IR problem is a matrix eigenvalue problem:

$$\left[ \mathcal D_{AdS_2} - \mathsf M^2(k) \right] \vec V_k=0.$$

Let $\lambda_i(k)$ be the eigenvalues of $\mathsf M^2(k)$. The corresponding IR scaling exponents are

$$\nu_i(k) = \sqrt{\frac14+L_2^2\lambda_i(k)}.$$

The $AdS_2$ Breitenlohner—Freedman stability condition is

$$L_2^2\lambda_i(k)\ge -\frac14.$$

A finite-momentum instability occurs when the most dangerous eigenvalue violates the bound first at nonzero momentum:

$$L_2^2\lambda_{\rm min}(k_*)<-\frac14, \qquad k_*\neq0.$$

Equivalently, $\nu(k_*)$ becomes imaginary. The normal state is then unstable to a condensate with wavevector $k_*$. At finite temperature the sharp $AdS_2$ criterion is replaced by a static normalizable zero mode of the full black-brane fluctuation problem:

$$\omega=0, \qquad k=k_c, \qquad \text{infalling/regular at the horizon, source-free at the boundary}.$$

The highest temperature at which such a mode exists is $T_c$. Near $T_c$ the order parameter is small, so the linear problem determines the onset. Below $T_c$ one must solve the nonlinear equations and minimize the free energy over possible wavevectors.

This mechanism is the strong-coupling cousin of familiar finite-wavevector instabilities in metals. In a weakly coupled Fermi liquid, density waves are often tied to nesting, hot spots, or soft particle-hole excitations. In holography, the immediate cause is instead the finite-$k$ spectrum of a charged horizon. The analogy is useful; the microscopic interpretation is different.

Why $AdS_2$ throats are fertile ground

Semi-local criticality has a peculiar property: the low-energy scaling is in time, while momentum remains a label. Operators can have $k$-dependent IR dimensions,

$$\Delta_{\rm IR}(k)=\frac12+\nu(k),$$

rather than dimensions that depend only on representation data at $k=0$. This makes it possible for a mode at nonzero $k$ to become unstable while the homogeneous mode remains stable.

That is why charged horizons are so good at producing modulated order. The IR geometry has a dense set of low-energy channels indexed by momentum. Couplings among scalars, gauge fields, metric perturbations, and topological terms can bend the function $\nu(k)$ so that its minimum occurs at $k\neq0$.

A schematic example is

$$L_2^2\lambda_{\rm min}(k) = a+b(k-k_*)^2+\cdots, \qquad b>0.$$

If $a<-1/4$, then a band of momenta around $k_*$ is unstable. The nonlinear solution chooses a specific periodic pattern from this band.

Topological terms and current stripes

One common way to generate finite-$k$ instabilities is to include parity-odd couplings. In four bulk dimensions, a neutral pseudoscalar $\phi$ can couple to the gauge field through

$$S = \int d^{4}x\sqrt{-g} \left[ R+\frac{6}{L^2} -\frac14\tau(\phi)F_{MN}F^{MN} -\frac12(\partial\phi)^2 -V(\phi) \right] - \frac12\int \vartheta(\phi)F\wedge F.$$

The Reissner—Nordström AdS solution with $\phi=0$ can be stable at $k=0$ but unstable at finite $k$. The reason is that the $\vartheta(\phi)F\wedge F$ term mixes scalar and gauge perturbations in a way that is odd in momentum. A typical fluctuation sector contains fields of the form

$$\delta\phi(z,x)=w(z)\cos kx, \qquad \delta A_y(z,x)=a_y(z)\sin kx, \qquad \delta g_{ty}(z,x)=h_{ty}(z)\sin kx.$$

The boundary interpretation is a striped state with a density modulation and a transverse current modulation:

$$\langle O_\phi(x)\rangle\propto \cos kx, \qquad \langle J^t(x)\rangle=\rho_0+\rho_1\cos kx, \qquad \langle J_y(x)\rangle\propto \sin kx.$$

Such states often break parity $P$ and time reversal $T$, because they contain spontaneous current patterns. They are therefore closer to loop-current or flux-like orders than to a simple charge-density wave.

In five bulk dimensions another canonical mechanism uses a Chern—Simons term,

$$S = \int d^5x\sqrt{-g} \left[ \frac{1}{2\kappa^2}\left(R+\frac{12}{L^2}\right) -\frac{1}{4e^2}F_{MN}F^{MN} \right] - \frac{\alpha}{6}\int A\wedge F\wedge F.$$

The Chern—Simons coupling can drive a helical instability. The resulting state has a spatially rotating current rather than a simple sinusoidal density wave.

Helical order: spatial dependence without PDEs

A helical phase in three boundary spatial dimensions can be described using Bianchi VII$_0$ one-forms,

$$\omega_1=dx,$$ $$\omega_2=\cos(kx)\,dy-\sin(kx)\,dz,$$ $$\omega_3=\sin(kx)\,dy+\cos(kx)\,dz.$$

A helical current profile is

$$\langle J_y(x)\rangle=J_0\cos kx, \qquad \langle J_z(x)\rangle=J_0\sin kx.$$

This profile breaks translations in $x$ and rotations in the $y$-$z$ plane separately, but preserves a combined operation: translate in $x$ and rotate in the transverse plane. That residual symmetry is a technical gift. A bulk ansatz such as

$$A=A_t(r)dt+B(r)\omega_2$$

and

$$ds^2 = -g(r)dt^2+ \frac{dr^2}{g(r)}+ e^{2v_1(r)}\omega_1^2+ e^{2v_2(r)}\omega_2^2+ e^{2v_3(r)}\omega_3^2$$

is spatially modulated from the boundary viewpoint but depends only on the radial coordinate $r$. The equations are ordinary differential equations rather than partial differential equations.

That makes helical phases excellent laboratories. They are not generic crystals, but they let us study finite-momentum order, anisotropic transport, and symmetry breaking in a controlled computational setting.

Stripes and fully inhomogeneous black branes

A unidirectional stripe is less symmetric than a helix. The bulk fields genuinely depend on both the radial coordinate and one boundary coordinate:

$$X=X(z,x), \qquad X(z,x+2\pi/k)=X(z,x).$$

A typical ansatz contains

$$A=A_t(z,x)dt+A_y(z,x)dy,$$

as well as metric components such as

$$g_{tt}(z,x),\quad g_{zz}(z,x),\quad g_{xx}(z,x),\quad g_{yy}(z,x),\quad g_{ty}(z,x).$$

The boundary conditions are the whole story:

$$\mu(x)=\mu, \qquad g_{\mu\nu}^{(0)}=\eta_{\mu\nu}, \qquad J_\phi(x)=0,$$

but

$$\rho(x),\quad \langle O_\phi(x)\rangle,\quad \langle J_y(x)\rangle,\quad \langle T_{\mu\nu}(x)\rangle$$

are allowed to become periodic functions. The equations are nonlinear elliptic PDEs, usually solved with spectral methods and a DeTurck gauge-fixing procedure.

At the onset of the transition, the critical wavevector $k_c$ is found by the linear zero-mode problem. Away from $T_c$, the preferred wavevector can drift. The physical value is obtained by minimizing the grand potential:

$$\Omega_{\rm striped}(T,\mu,k) = T I_E^{\rm ren}[X_{\rm striped}(z,x;k)].$$

The preferred branch satisfies

$$\frac{\partial \Omega_{\rm striped}}{\partial k}=0.$$

This step is essential. A linear instability tells us that the homogeneous phase is not the endpoint; it does not by itself tell us which nonlinear pattern wins.

Crystals and triangular lattices

A crystal requires modulation in at least two independent spatial directions:

$$\rho(\vec x) = \rho_0+ \sum_{a}\rho_a\cos(\vec Q_a\cdot\vec x+\varphi_a).$$

For a triangular lattice in two spatial dimensions, one may take three wavevectors separated by $120^\circ$ with

$$\vec Q_1+\vec Q_2+\vec Q_3=0.$$

The corresponding bulk problem depends on $(z,x,y)$ and is much harder than stripes. The reward is conceptual: a holographic crystal is the closest gravitational analogue of an ordinary solid. It has phonons, elastic moduli, and a spatially periodic stress tensor.

There is no universal rule that the most symmetric lattice wins. In many bottom-up models simple stripes are competitive or preferred; with different boundary conditions or interactions, rectangular or triangular patterns can appear. The free energy comparison is model-dependent. This is one of the places where holography is still more a toolkit than a classification theorem.

Pair-density waves and modulated superconductivity

A pair-density wave is a superconducting or superfluid order parameter at nonzero momentum:

$$\langle O_{\rm pair}(x)\rangle = \Psi_Q e^{iQx}+\Psi_{-Q}e^{-iQx}.$$

If $\Psi_Q=\Psi_{-Q}^*$, then

$$\langle O_{\rm pair}(x)\rangle =2|\Psi_Q|\cos(Qx+\varphi).$$

This breaks both the boundary $U(1)$ symmetry and translations. It can coexist with a uniform condensate,

$$\langle O_{\rm pair}(x)\rangle =\Psi_0+2|\Psi_Q|\cos(Qx+\varphi),$$

or exist without one. In ordinary condensed matter, pair-density waves naturally induce charge order at twice the wavevector:

$$|\langle O_{\rm pair}(x)\rangle|^2 \sim \text{constant}+\cos(2Qx+2\varphi).$$

In holography, modulated superconductivity can arise when a charged scalar is itself driven unstable at finite momentum, or when a homogeneous superconducting condensate coexists with a finite-$k$ neutral or current order. The details depend on the bulk couplings. The qualitative lesson is robust: once the IR geometry contains multiple nearly unstable channels, homogeneous superfluidity is only one possible way to reorganize charge outside the horizon.

Competing orders as coupled bulk hair

The phrase competing orders means that more than one order parameter wants to condense, and the condensates affect each other’s stability. Holographically, this is literal: several bulk fields can want to grow hair on the same charged black brane.

Let $\Psi$ denote a homogeneous charged scalar and $\Phi_Q$ a finite-momentum density-wave order. A simple Landau free energy is

$$\mathcal F = r_s|\Psi|^2+ \frac{u_s}{2}|\Psi|^4 + r_c|\Phi_Q|^2+ \frac{u_c}{2}|\Phi_Q|^4 + \lambda |\Psi|^2|\Phi_Q|^2+ \cdots.$$

The signs of $r_s$ and $r_c$ determine which instabilities are present. The mixed quartic coupling $\lambda$ determines whether the orders help or suppress one another:

$$\lambda>0 \quad\Rightarrow\quad \text{competition},$$ $$\lambda<0 \quad\Rightarrow\quad \text{cooperation or coexistence}.$$

This effective description is only the boundary shadow. In the bulk, the mechanisms are more concrete:

1. Both condensates may draw charge away from the horizon.
2. One condensate may remove the $AdS_2$ throat that made the other instability possible.
3. A modulated order can anisotropically gap spectral weight and change the transport channel seen by the superfluid.
4. A superfluid condensate can screen electric flux and thereby shift the finite-$k$ BF bound.

Thus the nonlinear geometry decides the phase diagram. Near a multicritical point, several branches must be compared:

$$\Omega_{\rm normal}, \qquad \Omega_{\rm SC}, \qquad \Omega_{\rm stripe}, \qquad \Omega_{\rm coexist}.$$

A very common mistake is to identify the first instability on cooling with the final low-temperature ground state. That need not be correct. The first instability tells us which phase appears continuously from the normal state. A lower-temperature first-order transition to another ordered phase can still occur.

Goldstone modes: phonons and phasons

For a unidirectional density wave,

$$\rho(x)=\rho_0+\rho_1\cos[k(x-x_0)],$$

all values of $x_0$ describe the same free energy. The coordinate $x_0$ is the phase of the density wave. A slowly varying phase field

$$\varphi(t,\vec x)=-kx_0(t,\vec x)$$

is the phason. In a true crystal it is the longitudinal phonon. For a stripe or smectic there are also subtleties associated with rotations and transverse fluctuations, but the central idea is the same: broken translations produce gapless sliding modes.

This has a sharp transport consequence. Spontaneously breaking translations does not automatically make the DC conductivity finite. In a perfectly clean system the density wave can slide. A uniform electric field can accelerate the sliding mode, producing an infinite conductivity.

One quick field-theory way to see this is to let the density wave slide with velocity $v=\dot x_0$:

$$\rho(x,t)=\rho_0+\rho_1\cos[k(x-x_0(t))].$$

The current

$$J_x(x,t)=v\rho(x,t)$$

satisfies the continuity equation

$$\partial_t\rho+\partial_xJ_x=0.$$

The spatial average is

$$\overline{J_x}=v\rho_0.$$

If there is no pinning or damping mechanism for $x_0$, the sliding mode carries current forever. In frequency space, this appears as a pole

$$\sigma(\omega)\sim \frac{D_{\rm slide}}{-i\omega}$$

and therefore a delta function in $\operatorname{Re}\sigma(\omega)$.

This statement often surprises people because the state visibly lacks continuous spatial homogeneity. But it is not explicit disorder. The conserved momentum of the full system has not disappeared; it has been reorganized into collective motion of the pattern.

Pinning, phase relaxation, and finite conductivity

Real density waves are rarely perfectly free to slide. A small explicit lattice, disorder, commensurability, or impurities can pin the phase. Then the phason acquires a small frequency $\omega_0$:

$$\omega_{\rm phason}(q=0)=\omega_0.$$

If dislocations or other defects relax the phase, one introduces a phase relaxation rate $\Omega$. A useful schematic form of the pinned density-wave conductivity is

$$\sigma(\omega) = \sigma_{\rm inc} + D\, \frac{-i\omega+\Omega}{ \omega_0^2- \omega^2-i\omega\Gamma+\Omega(\Gamma-i\omega) }.$$

Here $\Gamma$ is momentum relaxation, $\sigma_{\rm inc}$ is the incoherent conductivity, and $D$ is a spectral weight. The precise coefficients depend on the hydrodynamic frame and on the model, but the pole structure is robust: pinning moves the sliding pole away from $\omega=0$.

In holography, pinned density waves are studied by adding a small explicit lattice or disorder to a spontaneously modulated background. The quasinormal modes then reveal the pseudo-Goldstone pole. This is one of the cleanest ways to connect holographic order to measured optical conductivity.

Relation to explicit lattices

The previous page in the transport section already discussed explicit translation breaking: linear axions, Q-lattices, helical lattices, massive gravity, and disorder. Those models answer the question:

$$\text{What happens when translations are broken by sources?}$$

The present page asks a different question:

$$\text{Can translations break by themselves?}$$

The distinction is visible in the near-boundary data:

$$\begin{array}{ccl} \text{explicit lattice} &:& J_Q\neq0,\quad \langle O_Q\rangle\neq0,\\ \text{spontaneous stripe} &:& J_Q=0,\quad \langle O_Q\rangle\neq0. \end{array}$$

The distinction is also visible in low-energy dynamics:

$$\begin{array}{ccl} \text{explicit lattice} &:& \text{momentum can relax},\\ \text{spontaneous stripe} &:& \text{momentum conserved; phason slides},\\ \text{pinned stripe} &:& \text{pseudo-Goldstone mode and finite DC response}. \end{array}$$

Holographic models often combine both. One first finds a spontaneous density wave, then adds a weak explicit lattice to pin it. That is the setting closest to many experimental charge-density-wave materials.

Numerical gravity and the endpoint problem

Finite-momentum order is technically hard because it usually destroys the cohomogeneity-one structure of homogeneous black branes. The steps are:

1. Find a static normalizable zero mode at $T=T_c$ and $k=k_c$.
2. Construct nonlinear periodic black branes below $T_c$.
3. Holographically renormalize the on-shell action.
4. Compute $\Omega(T,\mu,k)$.
5. Minimize over $k$ and compare with competing phases.
6. Compute fluctuations around the ordered geometry to obtain phonons, conductivities, and spectral functions.

For stripes the PDEs depend on $(z,x)$. For crystals they depend on $(z,x,y)$. At zero temperature the problem is harder still, because the IR endpoint may be a new inhomogeneous scaling geometry rather than a smooth finite-temperature horizon.

This is why helical phases and other homogeneous tricks are valuable. They let us learn lessons about finite-$k$ order using ODEs. But one should not confuse technical convenience with genericity.

Control and caveats

Finite-momentum holographic order is powerful, but it is easy to over-interpret. A source-free striped solution is a controlled large-$N$ saddle of a particular quantum field theory or effective bottom-up model. It is not automatically a microscopic theory of cuprate stripes, heavy-fermion order, or charge order in any named material.

The most trustworthy claims are structural:

$$\text{finite-}k\text{ instability},\qquad \text{source-free modulated vev},\qquad \text{phonon/phason mode},\qquad \text{horizon computation of response}.$$

The less universal claims are quantitative details such as the preferred lattice geometry, the value of $k_*/\mu$, and the relative free energies of stripe, helix, and crystal branches. Those depend on the bulk action and on boundary conditions.

Large $N$ also suppresses fluctuations. A finite-temperature transition into a modulated order can look mean-field-like even in spatial dimensions where ordinary condensed matter would have strong thermal fluctuations. In low dimensions, true long-range order may be destroyed or replaced by quasi-long-range order once $1/N$ effects are included. The classical bulk calculation is best read as the saddle-point answer.

Finally, a continuum holographic density wave is not automatically commensurate with any microscopic lattice. Commensurability, lock-in transitions, and impurity pinning require extra ingredients: an explicit periodic source, disorder, or a top-down construction whose microscopic theory contains the relevant lattice-scale information. This is not a defect of the formalism; it is a reminder of what the model has and has not been asked to encode.

Common pitfalls

Pitfall 1: “A density wave automatically gives finite DC conductivity.”

Not if the density wave is spontaneous and clean. The sliding mode gives an infinite conductivity unless it is pinned or phase-relaxed.

Pitfall 2: “The critical wavevector from the linear instability is the final wavevector.”

It is the onset wavevector near $T_c$. Deeper in the ordered phase, the preferred $k$ is determined by minimizing the nonlinear free energy.

Pitfall 3: “Spatial modulation means the boundary source is modulated.”

Not for spontaneous order. The source must remain homogeneous or vanish in the modulated channel.

Pitfall 4: “Helical models are just ordinary lattices.”

A helical state can be spontaneous or explicit depending on boundary data. Its special feature is a combined translation-rotation symmetry that keeps the bulk ODE-like.

Pitfall 5: “Finite-$k$ order proves Fermi-surface physics.”

No. Finite-$k$ instabilities in holography are often enabled by semi-local critical spectral weight. They are analogous to Fermi-surface instabilities, not identical to them.

What to remember

The conceptual spine is short:

$$\text{homogeneous charged horizon} \quad\xrightarrow{\text{finite-}k\text{ instability}}\quad \text{striped/helical/crystalline black brane}.$$

The dictionary is equally short:

$$\text{normalizable finite-}k\text{ bulk mode} \quad\longleftrightarrow\quad \text{source-free spatially modulated vev}.$$

The physics is rich because the broken symmetry has a Goldstone mode. A spontaneous density wave is not merely a mechanism for momentum relaxation. It is an ordered phase with sliding collective motion, possible pinning, elastic response, and competition with superconductivity.

Exercises

Exercise 1: source or vev?

A scalar field dual to an operator of dimension $\Delta$ has the near-boundary expansion

$$X(z,x)=z^{d-\Delta}J(x)+z^\Delta V(x)+\cdots.$$

Suppose

$$J(x)=J_0, \qquad V(x)=V_0+V_1\cos Qx.$$

Is the $Q$-modulation explicit or spontaneous?

Solution

It is spontaneous. The source has no Fourier component at $Q$:

$$J_Q=0.$$

The modulated coefficient is the response:

$$\langle O_Q\rangle\propto V_1\neq0.$$

Therefore the boundary Hamiltonian is translation invariant, while the state is not.

Exercise 2: finite-momentum BF window

Assume the lowest IR eigenvalue near its minimum is

$$L_2^2\lambda_{\rm min}(k)=-\frac14-\epsilon+b(k-k_*)^2, \qquad \epsilon>0, \qquad b>0.$$

For which momenta is the $AdS_2$ BF bound violated?

Solution

The BF bound is violated when

$$L_2^2\lambda_{\rm min}(k)<-\frac14.$$

Using the given form,

$$-\frac14-\epsilon+b(k-k_*)^2<-\frac14,$$

so

$$b(k-k_*)^2<\epsilon.$$

Thus the unstable band is

$$|k-k_*|<\sqrt{\frac{\epsilon}{b}}.$$

Exercise 3: the sliding current

Let

$$\rho(x,t)=\rho_0+\rho_1\cos[k(x-x_0(t))]$$

and define $v=\dot x_0$. Show that, for constant $v$,

$$J_x(x,t)=v\rho(x,t)$$

satisfies charge conservation. What is the spatially averaged current?

Solution

First,

$$\partial_t\rho = \rho_1 k\dot x_0\sin[k(x-x_0)] = \rho_1 k v\sin[k(x-x_0)].$$

For $J_x=v\rho$ with constant $v$,

$$\partial_xJ_x = v\partial_x\rho = -v\rho_1 k\sin[k(x-x_0)].$$

Therefore

$$\partial_t\rho+\partial_xJ_x=0.$$

The spatial average of the oscillatory part vanishes, so

$$\overline{J_x}=v\rho_0.$$

A sliding density wave can therefore carry a uniform current.

Exercise 4: helical residual symmetry

Consider

$$\langle J_y(x)\rangle=J_0\cos kx, \qquad \langle J_z(x)\rangle=J_0\sin kx.$$

Show that a translation $x\mapsto x+a$ can be undone by a rotation in the $y$-$z$ plane.

Solution

After translation,

$$\begin{pmatrix} J_y\\ J_z \end{pmatrix} \mapsto J_0 \begin{pmatrix} \cos(kx+ka)\\ \sin(kx+ka) \end{pmatrix}.$$

This is obtained from the original vector by a rotation through angle $ka$:

$$\begin{pmatrix} \cos(kx+ka)\\ \sin(kx+ka) \end{pmatrix} = \begin{pmatrix} \cos ka&-\sin ka\\ \sin ka&\cos ka \end{pmatrix} \begin{pmatrix} \cos kx\\ \sin kx \end{pmatrix}.$$

Thus translation and transverse rotation are broken separately, but a combined operation remains a symmetry.

Exercise 5: coexistence from a two-order Landau theory

Take

$$\mathcal F = r_s\psi^2+\frac{u_s}{2}\psi^4 + r_c\phi^2+\frac{u_c}{2}\phi^4 + \lambda\psi^2\phi^2,$$

with $u_s>0$ and $u_c>0$. Find the coexistence solution with $\psi\neq0$ and $\phi\neq0$.

Solution

The stationarity equations are

$$\frac{1}{2\psi}\frac{\partial\mathcal F}{\partial\psi} = r_s+u_s\psi^2+\lambda\phi^2=0,$$

and

$$\frac{1}{2\phi}\frac{\partial\mathcal F}{\partial\phi} = r_c+u_c\phi^2+\lambda\psi^2=0.$$

Solving the two linear equations for $\psi^2$ and $\phi^2$ gives

$$\psi^2 = \frac{-r_s u_c+\lambda r_c}{u_s u_c-\lambda^2},$$ $$\phi^2 = \frac{-r_c u_s+\lambda r_s}{u_s u_c-\lambda^2}.$$

A stable coexistence phase requires

$$u_s u_c-\lambda^2>0$$

and both numerators positive. If $\lambda$ is too large and positive, coexistence is disfavored and the system tends to choose one order or the other.

Further reading

For the broad review treatment, see Hartnoll, Lucas, and Sachdev, Holographic Quantum Matter, especially the sections on symmetry-broken phases and spontaneous breaking of translation symmetry. For the condensed-matter-facing narrative and the relation to explicit translation breaking, helical models, massive gravity, and holographic crystallisation, see Zaanen, Liu, Sun, and Schalm, Holographic Duality in Condensed Matter Physics, Chapter 12. Foundational research examples include Nakamura—Ooguri—Park finite-momentum instabilities, Donos—Gauntlett helical and striped phases, and Withers’ fully backreacted striped black branes.