Quantum-Critical Transport

The previous page explained how a retarded Green function is computed from a Lorentzian black-brane boundary-value problem. This page uses that machinery for the observables that made holography famous in many-body physics: transport coefficients.

The setting is deliberately clean. We study a translation-invariant thermal quantum critical state at zero charge density. Equivalently, the bulk background is a neutral planar AdS black brane, and the gauge field dual to a conserved boundary current is treated as a fluctuation, not as part of the background. This is the simplest arena in which the horizon acts as a dissipative medium.

The central lesson is:

$$\text{transport coefficient} \quad\Longleftrightarrow\quad \text{low-frequency retarded correlator} \quad\Longleftrightarrow\quad \text{near-horizon absorption}.$$

This statement should be read with some restraint. Holography does not say that every quantum critical point in nature has the same conductivity or viscosity. It says that in large-$N$, strongly coupled theories with a semiclassical gravitational dual, transport can be computed by solving classical wave equations in a black-brane geometry. That is already a remarkable amount of control over a problem that is usually brutal.

Throughout this page, $d_s$ is the number of boundary spatial dimensions and

$$d=d_s+1$$

is the boundary spacetime dimension. The bulk has dimension $d+1=d_s+2$. We use units

$$\hbar=k_B=c=1.$$

The physical problem

A quantum critical state has no intrinsic energy scale at $T=0$. At nonzero temperature, the temperature itself is the only scale, so relaxation and transport often take the scaling form

$$\tau^{-1}\sim T, \qquad D\sim \frac{1}{T},$$

up to dimensionless constants and velocities. This is the basic reason quantum-critical transport is interesting: it is transport not controlled by long-lived quasiparticles, impurity scattering, or a weak-coupling mean free path.

The observables are retarded correlators of conserved currents and stress tensors. If a global $U(1)$ symmetry exists, the conserved current is

$$\partial_\mu J^\mu=0.$$

If translations are unbroken, the stress tensor obeys

$$\partial_\mu T^{\mu\nu}=0.$$

Small external sources couple as

$$\delta S_{\rm QFT} = \int d^d x\left(a_\mu J^\mu+\frac12 h_{\mu\nu}T^{\mu\nu}\right).$$

The source $a_\mu$ is a nondynamical background gauge field. The source $h_{\mu\nu}$ is a nondynamical background metric perturbation. In holography these sources are literally the boundary values of bulk fields:

$$a_\mu \longleftrightarrow A_\mu\big|_{\partial{\rm AdS}}, \qquad h_{\mu\nu} \longleftrightarrow \delta g_{\mu\nu}\big|_{\partial{\rm AdS}}.$$

The retarded functions

$$G^R_{J_\mu J_\nu}, \qquad G^R_{T^{\mu\nu}T^{\rho\sigma}}$$

then determine conductivity, diffusion, shear viscosity, sound attenuation, and the hydrodynamic pole structure.

Why zero density is the clean starting point

At zero charge density, the electric current has no overlap with the conserved momentum. A finite DC conductivity is therefore possible even with exact translation symmetry. At nonzero density this ceases to be true: the current can drag the conserved momentum, producing an infinite clean DC conductivity. That momentum bottleneck is the beginning of the metallic-transport story later in the course.

The transport dictionary for neutral holographic quantum critical matter. Boundary sources such as $A_x^{(0)}$ and $h_{xy}^{(0)}$ excite bulk Maxwell and metric perturbations. Infalling regularity at the future horizon selects the retarded correlator. Kubo limits and hydrodynamic poles then give $\sigma$, $D$, $\eta$, shear diffusion, and sound attenuation.

Linear response and Kubo formulas

Let $A$ and $B$ be bosonic operators. The retarded correlator is

$$G^R_{AB}(t,\vec x) = -i\theta(t)\langle[A(t,\vec x),B(0,\vec0)]\rangle.$$

With Fourier convention

$$G^R(\omega,\vec k) = \int dt\,d^{d_s}x\, e^{i\omega t-i\vec k\cdot\vec x}G^R(t,\vec x),$$

transport coefficients are low-frequency, long-wavelength limits of these correlators.

For an electric field in the $x$ direction, use the gauge $a_t=0$. Then

$$E_x=-\partial_t a_x \quad\Longrightarrow\quad E_x(\omega)=i\omega a_x(\omega)$$

for perturbations proportional to $e^{-i\omega t}$. Linear response gives

$$\delta\langle J_x\rangle = G^R_{J_xJ_x}(\omega,0)a_x.$$

Ohm’s law is

$$\delta\langle J_x\rangle=\sigma(\omega)E_x.$$

Therefore

$$\boxed{ \sigma(\omega) = \frac{1}{i\omega}G^R_{J_xJ_x}(\omega,\vec k=0) }$$

up to contact-term conventions. Equivalently,

$$\operatorname{Re}\sigma(\omega) = -\frac{1}{\omega}\operatorname{Im}G^R_{J_xJ_x}(\omega,0)$$

with the sign depending on the Fourier convention for $G^R$.

For shear viscosity, perturb the boundary metric by $h_{xy}(t)$. The Kubo formula is

$$\boxed{ \eta = -\lim_{\omega\to0} \frac{1}{\omega} \operatorname{Im}G^R_{T^{xy}T^{xy}}(\omega,\vec k=0) }$$

again in the same convention as above.

These two formulas are structurally identical. Conductivity is the absorptive response of a current to a vector source. Shear viscosity is the absorptive response of momentum flux to a metric source.

Order of limits matters

Transport coefficients are hydrodynamic limits. For DC conductivity and viscosity one usually takes $\vec k=0$ first and then $\omega\to0$. Diffusion constants are instead extracted from poles at small but nonzero $k$, such as $\omega=-iDk^2+\cdots$. Mixing up these limits is a classic way to create fake paradoxes.

The neutral black brane background

The gravitational dual of a neutral thermal CFT state is the planar AdS-Schwarzschild black brane

$$ds^2 = \frac{L^2}{z^2} \left[ -f(z)dt^2+d\vec x_{d_s}^{\,2}+\frac{dz^2}{f(z)} \right],$$

with

$$f(z)=1-\left(\frac{z}{z_h}\right)^d, \qquad T=\frac{d}{4\pi z_h}.$$

The entropy density is the horizon area density divided by $4G_N$:

$$s= \frac{1}{4G_N} \left(\frac{L}{z_h}\right)^{d_s}.$$

For a conformal fluid at zero chemical potential,

$$\epsilon=d_s P, \qquad \epsilon+P=sT, \qquad \zeta=0.$$

Here $\epsilon$ is energy density, $P$ is pressure, and $\zeta$ is bulk viscosity. The vanishing of $\zeta$ is not a holographic miracle; it follows from conformal invariance. A scale-invariant fluid cannot dissipate through uniform compression in the same way a generic nonconformal fluid can.

Current transport from a bulk Maxwell field

A conserved boundary current $J^\mu$ is dual to a bulk gauge field $A_M$. At zero density the background gauge field vanishes,

$$A_M^{\rm bg}=0,$$

so the leading current-current correlator is obtained from a quadratic Maxwell action on the fixed black-brane geometry:

$$S_A = -\frac{1}{4g_F^2} \int d^{d+1}x\sqrt{-g}\,F_{MN}F^{MN}.$$

More general bottom-up models replace $1/g_F^2$ by a scalar-dependent coupling $Z(\phi)/g_F^2$; for now set $Z=1$.

Choose radial gauge

$$A_z=0.$$

For a homogeneous electric perturbation,

$$A_x(z,t)=e^{-i\omega t}a_x(z),$$

the Maxwell equation is

$$\partial_z\left(\sqrt{-g}\,g^{zz}g^{xx}\partial_z a_x\right) - \omega^2\sqrt{-g}\,g^{tt}g^{xx}a_x=0.$$

Near the horizon one imposes the infalling condition. Near the boundary,

$$a_x(z)=a_x^{(0)}+a_x^{(1)}z^{d-2}+\cdots,$$

with possible logarithms in special dimensions. The leading coefficient $a_x^{(0)}$ is the source. The response is the renormalized radial electric flux

$$\langle J^x\rangle = \Pi_A^{x,{\rm ren}},$$

where

$$\Pi_A^x = -\frac{1}{g_F^2}\sqrt{-g}F^{zx} = -\frac{1}{g_F^2}\sqrt{-g}\,g^{zz}g^{xx}\partial_z a_x.$$

Thus

$$G^R_{J_xJ_x}(\omega,0) = \frac{\Pi_A^{x,{\rm ren}}}{a_x^{(0)}}$$

for the infalling solution normalized by $a_x^{(0)}$.

The boundary conductivity is therefore

$$\sigma(\omega) = \frac{1}{i\omega} \frac{\Pi_A^{x,{\rm ren}}}{a_x^{(0)}}.$$

This equation is mundane-looking, but it is doing real work. A strongly coupled many-body conductivity has been reduced to a classical wave-absorption problem.

DC conductivity from the horizon

The DC limit is especially transparent. Define the radial current

$$\mathcal J^x(z) = -\frac{1}{g_F^2}\sqrt{-g}F^{zx}.$$

At $\omega=0$ and $k=0$, the Maxwell equation says

$$\partial_z\mathcal J^x=0.$$

So the current can be evaluated at any radial position. The boundary current equals the horizon current.

For an isotropic metric written schematically as

$$ds^2=g_{tt}(z)dt^2+g_{zz}(z)dz^2+g_{xx}(z)d\vec x_{d_s}^{\,2},$$

with $g_{tt}<0$, horizon regularity in ingoing coordinates gives a local Ohm’s law at the horizon. The result is the membrane-paradigm formula

$$\boxed{ \sigma_{\rm dc} = \left. \frac{1}{g_F^2} \frac{\sqrt{-g}\,g^{xx}}{\sqrt{-g_{tt}g_{zz}}} \right|_{z=z_h} }$$

for the simple Maxwell action.

For the AdS black brane above,

$$\sigma_{\rm dc} = \frac{1}{g_F^2} \left(\frac{L}{z_h}\right)^{d-3} = \frac{1}{g_F^2} \left(\frac{4\pi L T}{d}\right)^{d_s-2}.$$

This matches the scaling expectation

$$[\sigma]=d_s-2.$$

In $d_s=2$ spatial dimensions, the conductivity is dimensionless:

$$\sigma_{\rm dc}=\frac{1}{g_F^2}.$$

This is the famous case most often associated with quantum critical transport in $2+1$ dimensions. In $d_s=3$, conductivity has dimension of energy and scales as $T$.

The meaning of $1/g_F^2$

The constant $1/g_F^2$ is not a universal number by itself. It fixes the normalization of the current two-point function, often denoted $C_J$. Different CFTs have different $C_J$. Holography makes the computation controlled, not model-independent.

Optical conductivity and electromagnetic self-duality

The full optical conductivity $\sigma(\omega)$ generally requires solving the radial Maxwell equation at finite $\omega$. In $2+1$ boundary dimensions with the simplest Maxwell action in $AdS_4$, something special happens: bulk electromagnetic duality implies a frequency-independent conductivity,

$$\sigma(\omega)=\frac{1}{g_F^2}.$$

This is elegant, but fragile. Higher-derivative terms such as

$$\Delta S \sim \int d^{4}x\sqrt{-g}\,C_{MNPQ}F^{MN}F^{PQ}$$

or couplings to additional bulk fields break the simple self-duality and produce nontrivial frequency dependence.

That fragility is good physics. A real quantum critical conductivity is constrained by scaling and symmetries, but it is not normally a single universal constant at all frequencies. The constant result is best understood as an exactly solvable benchmark, not as a generic prediction for all quantum critical matter.

Diffusion and the Einstein relation

Conductivity describes the response to an electric field. Diffusion describes the relaxation of an inhomogeneous density.

At zero background density, let

$$\rho=J^t$$

be a small charge-density fluctuation. Hydrodynamics gives the continuity equation

$$\partial_t\rho+\nabla\cdot\vec J=0.$$

The constitutive relation is Fick’s law,

$$\vec J=-D\nabla\rho.$$

Combining these equations gives the diffusion equation

$$\partial_t\rho-D\nabla^2\rho=0.$$

For a Fourier mode $e^{-i\omega t+i\vec k\cdot\vec x}$, the hydrodynamic pole is

$$\boxed{ \omega=-iDk^2+\cdots }$$

The same $D$ is related to the DC conductivity by the Einstein relation. Turn on a slowly varying chemical potential $\mu(\vec x)$. In local equilibrium,

$$\rho=\chi\mu,$$

where

$$\chi=\left(\frac{\partial\rho}{\partial\mu}\right)_T$$

is the static charge susceptibility. The electric field is

$$\vec E=-\nabla\mu.$$

Then Fick’s law becomes

$$\vec J =-D\nabla\rho =-D\chi\nabla\mu =D\chi\vec E.$$

Comparing with Ohm’s law $\vec J=\sigma\vec E$ gives

$$\boxed{ \sigma_{\rm dc}=D\chi }$$

or

$$\boxed{ D=\frac{\sigma_{\rm dc}}{\chi}. }$$

For the simple Maxwell field in an AdS$_{d+1}$ black brane, one finds

$$\chi = \frac{d-2}{g_F^2} \frac{L^{d-3}}{z_h^{d-2}},$$

and therefore

$$\boxed{ D= \frac{z_h}{d-2} = \frac{d}{4\pi T(d-2)} }$$

for $d>2$.

In $d=4$ boundary spacetime dimensions, this reduces to

$$D=\frac{1}{2\pi T}.$$

The important point is not the exact coefficient. The important point is the structure:

$$\text{diffusion constant} \sim \frac{1}{T}$$

in a thermal quantum critical state.

The density-density correlator

The diffusion pole appears directly in the retarded density correlator. Hydrodynamics predicts

$$G^R_{\rho\rho}(\omega,k) = \chi\frac{Dk^2}{Dk^2-i\omega} + \text{contact terms}.$$

This form satisfies two checks.

First, the pole is at

$$Dk^2-i\omega=0 \quad\Longrightarrow\quad \omega=-iDk^2.$$

Second, the static limit gives the susceptibility:

$$\lim_{k\to0}\lim_{\omega\to0}G^R_{\rho\rho}(\omega,k)=\chi,$$

up to the same contact-term convention.

The noncommutativity of limits is physical. At exactly $k=0$, total charge is conserved and cannot relax. At small nonzero $k$, charge can diffuse from one region to another.

Stress-tensor transport

The stress tensor of a relativistic fluid is organized in a derivative expansion. To first order,

$$T^{\mu\nu} = (\epsilon+P)u^\mu u^\nu +P\eta^{\mu\nu} +\tau^{\mu\nu},$$

where

$$\tau^{\mu\nu} = -\eta\,\sigma^{\mu\nu} -\zeta P^{\mu\nu}\partial_\lambda u^\lambda.$$

Here

$$P^{\mu\nu}=\eta^{\mu\nu}+u^\mu u^\nu$$

projects orthogonally to the velocity, and

$$\sigma^{\mu\nu} = P^{\mu\alpha}P^{\nu\beta} \left( \partial_\alpha u_\beta+ \partial_\beta u_\alpha - \frac{2}{d_s}\eta_{\alpha\beta}\partial_\lambda u^\lambda \right)$$

is the shear tensor.

The shear viscosity $\eta$ measures the dissipation of transverse momentum gradients. The bulk viscosity $\zeta$ measures dissipation under compression. For a conformal fluid,

$$\zeta=0.$$

The hydrodynamic modes of $T^{\mu\nu}$ are fixed by conservation,

$$\partial_\mu T^{\mu\nu}=0.$$

There are two types that matter here.

Shear diffusion

Take momentum along $x$ and a velocity perturbation transverse to it, say $u_y(t,x)$. Linearized hydrodynamics gives

$$(\epsilon+P)\partial_t u_y-\eta\partial_x^2u_y=0.$$

Therefore the shear mode has dispersion

$$\boxed{ \omega=-iD_\eta k^2+\cdots, \qquad D_\eta=\frac{\eta}{\epsilon+P}. }$$

At zero chemical potential,

$$\epsilon+P=sT,$$

so

$$D_\eta=\frac{\eta}{sT}.$$

Sound

Longitudinal energy and momentum fluctuations produce sound:

$$\boxed{ \omega = \pm v_s k -i\Gamma k^2+\cdots }$$

where

$$v_s^2=\left(\frac{\partial P}{\partial\epsilon}\right)_s.$$

For a conformal fluid,

$$v_s^2=\frac{1}{d_s}.$$

The damping coefficient is

$$\Gamma = \frac{1}{2(\epsilon+P)} \left[ \frac{2(d_s-1)}{d_s}\eta+\zeta \right].$$

For a conformal holographic plasma with $\zeta=0$ and $\eta/s=1/(4\pi)$,

$$D_\eta=\frac{1}{4\pi T},$$

and

$$\Gamma = \frac{d_s-1}{d_s}\frac{1}{4\pi T}.$$

Again the scaling is quantum-critical:

$$D_\eta\sim \Gamma\sim \frac{1}{T}.$$

The shear viscosity calculation in one page

Now we compute the most famous number in holographic transport. Consider a transverse metric perturbation

$$h_{xy}(z,t)=e^{-i\omega t}h_{xy}(z).$$

It is useful to write

$$h_{xy}=g_{xx}\varphi.$$

For an isotropic two-derivative Einstein gravity background, the fluctuation $\varphi$ obeys the same equation as a minimally coupled massless scalar at zero spatial momentum. Its quadratic action takes the schematic form

$$S^{(2)}_{\varphi} = -\frac{1}{32\pi G_N} \int d^{d+1}x\sqrt{-g}\,g^{MN}\partial_M\varphi\partial_N\varphi+ \cdots,$$

where the omitted terms either vanish at $\omega\to0$ or contribute contact terms for the Kubo limit.

The canonical radial momentum is

$$\Pi_\varphi = -\frac{1}{16\pi G_N} \sqrt{-g}\,g^{zz}\partial_z\varphi,$$

up to the conventional factor associated with the normalization of $h_{xy}$. The shear Green function is obtained from

$$G^R_{T^{xy}T^{xy}} = \frac{\Pi_\varphi^{\rm ren}}{\varphi^{(0)}}.$$

In the low-frequency limit, the imaginary part is controlled by the horizon flux. Infalling regularity gives the horizon relation between radial derivative and time derivative. The result is

$$\eta = \frac{1}{16\pi G_N} \left(\frac{L}{z_h}\right)^{d_s}.$$

Compare this with the entropy density

$$s= \frac{1}{4G_N} \left(\frac{L}{z_h}\right)^{d_s}.$$

Therefore

$$\boxed{ \frac{\eta}{s}=\frac{1}{4\pi}. }$$

This is the Kovtun-Son-Starinets result in its simplest two-derivative form.

Not a theorem of nature

The value $\eta/s=1/(4\pi)$ is universal for a broad class of isotropic, translationally invariant, two-derivative Einstein gravity duals in the classical limit. It is not a theorem for all quantum field theories. Higher-derivative bulk interactions, anisotropy, finite-coupling corrections, and other departures from the simplest setup can change the ratio.

Why the ratio is small

At weak coupling, kinetic theory gives a large shear viscosity. Roughly,

$$\eta\sim \epsilon\,\ell_{\rm mfp},$$

where $\ell_{\rm mfp}$ is a mean free path. If interactions are weak, particles travel far before colliding, so $\ell_{\rm mfp}$ is large and $\eta/s$ is large.

The holographic fluid has no such long mean free path. It equilibrates on a time of order

$$\tau\sim \frac{1}{T}.$$

That is why the dimensionless ratio $\eta/s$ is order one in natural units. More precisely, for classical Einstein gravity it is $1/(4\pi)$.

The ratio is the right object to compare across systems. The viscosity $\eta$ itself can be large if the entropy density is large. A hot large-$N$ plasma has many degrees of freedom, so both $\eta$ and $s$ scale like $N^2$. The small quantity is not $\eta$ by itself, but

$$\frac{\eta}{s}.$$

Hydrodynamic poles as quasinormal modes

On the previous page, quasinormal modes appeared as source-free infalling solutions. Hydrodynamic modes are special quasinormal modes forced toward the origin by conservation laws.

For charge diffusion,

$$\omega(k)=-iDk^2+\cdots.$$

For shear diffusion,

$$\omega(k)=-iD_\eta k^2+\cdots.$$

For sound,

$$\omega(k)=\pm v_s k-i\Gamma k^2+\cdots.$$

These are not optional features of the holographic model. They must be present because the boundary theory has conserved charge, energy, and momentum. The bulk gravitational constraints know about the boundary Ward identities.

Generic nonhydrodynamic QNMs have frequencies of order $T$:

$$\omega_n\sim T.$$

They describe fast relaxation. Hydrodynamic QNMs are parametrically slower at small $k$ because conservation laws prevent local densities from disappearing; they can only spread.

Ward identities and gauge-invariant variables

For current correlators, gauge invariance and charge conservation imply Ward identities. In Fourier space,

$$k_\mu G^R_{J^\mu J^\nu}(k)=0$$

up to contact terms. In the bulk, this is reflected in the Maxwell constraint equation.

For momentum along $x$, it is useful to decompose the gauge field into transverse and longitudinal channels. In radial gauge $A_z=0$, define

$$a_\perp=a_y,$$

and

$$a_\parallel=\omega a_x+k a_t.$$

This longitudinal combination is invariant under the residual gauge transformation

$$A_\mu\to A_\mu+\partial_\mu\lambda.$$

The diffusion pole lives in the longitudinal channel. The optical conductivity at $k=0$ can be computed from the transverse channel.

Metric perturbations similarly decompose into tensor, vector, and scalar channels under rotations transverse to $\vec k$. Shear diffusion lives in a vector channel. Sound lives in a scalar channel. The $h_{xy}$ perturbation used for the Kubo formula is a tensor channel at $k=0$, which is why the calculation is so clean.

What is universal, and what is not?

It is tempting to summarize holographic transport as a list of universal constants. That is the wrong moral. A better taxonomy is:

Quantity	Holographic lesson	Universality status
$\eta/s$	Horizon absorption of a transverse graviton	Universal for classical two-derivative Einstein gravity, not universal in general
$\sigma_{\rm dc}$ at $\rho=0$	Horizon electric flux	Depends on current normalization and bulk gauge couplings
$D=\sigma/\chi$	Diffusion from hydrodynamics and susceptibility	Coefficient model-dependent; scaling $D\sim1/T$ is natural at quantum criticality
Hydrodynamic poles	Boundary Ward identities become bulk constraints	Universal structure; coefficients depend on the theory
Constant $\sigma(\omega)$ in $2+1$ dimensions	Bulk electromagnetic self-duality	Special to simple Maxwell theory in $AdS_4$

The sharpest universal statement is structural: black branes turn strongly coupled real-time transport into horizon boundary conditions and radial fluxes.

Relation to condensed matter language

In condensed matter theory, a clean quantum critical point in $d_s=2$ often has a conductivity of the form

$$\sigma_Q =\frac{Q^2}{h}\,\Phi_\sigma,$$

where $Q$ is the charge of the carriers and $\Phi_\sigma$ is a dimensionless universal number of the critical theory. Holographic models compute the strong-coupling, large-$N$ analogue of $\Phi_\sigma$.

This is not Drude physics. The Drude formula

$$\sigma_{\rm Drude} =\frac{n q^2\tau}{m}$$

assumes identifiable carriers with density $n$, charge $q$, mass $m$, and lifetime $\tau$. At a relativistic charge-neutral quantum critical point, the current is carried by thermally excited particles and antiparticles, or more abstractly by charged critical degrees of freedom. There need not be a quasiparticle lifetime at all.

Holography replaces the Drude story with a geometric one:

$$\text{electric field at the boundary} \quad\to\quad \text{Maxwell wave in the bulk} \quad\to\quad \text{absorption by the horizon}.$$

That is the conceptual leap.

A preview of finite density

Everything above assumed zero background density. At finite density,

$$\rho\neq0,$$

the current contains a convective part

$$J^i\supset \rho u^i.$$

If translations are exact, the momentum density cannot decay. Since electric current overlaps with momentum, a constant electric field accelerates the whole fluid. The clean DC conductivity therefore contains a delta function:

$$\operatorname{Re}\sigma(\omega) \supset \pi \frac{\rho^2}{\chi_{PP}}\delta(\omega),$$

where $\chi_{PP}=\epsilon+P$ for a relativistic fluid.

The finite part left after subtracting momentum drag is often called the incoherent conductivity. At zero density, the incoherent conductivity is simply the ordinary conductivity:

$$\sigma_Q=\sigma_{\rho=0}.$$

This is why quantum-critical transport is the right warm-up for holographic metals. The finite-density story is not a replacement of this page; it is this page plus momentum.

Common pitfalls

Pitfall 1: calling $\eta/s=1/(4\pi)$ a universal lower bound. It is a robust result for classical two-derivative Einstein gravity, not an absolute theorem for all quantum matter.

Pitfall 2: expecting $\sigma$ to be as universal as $\eta/s$. The conductivity depends on the normalization and dynamics of the bulk gauge field. Even in a CFT, different conserved currents can have different $C_J$.

Pitfall 3: forgetting charge density. At $\rho=0$, clean DC conductivity can be finite. At $\rho\neq0$, translation invariance usually gives infinite DC conductivity.

Pitfall 4: identifying every pole with a quasiparticle. Hydrodynamic poles are long-lived because of conservation laws, not because they are Landau quasiparticles.

Pitfall 5: comparing $\eta$ instead of $\eta/s$. The dimensionless ratio is what measures fluidity in natural units. Large-$N$ systems can have large $\eta$ and still very small $\eta/s$.

Pitfall 6: ignoring contact terms. Kubo formulas depend on the absorptive part or the pole structure. Local counterterms can shift real analytic pieces of correlators.

Exercises

Exercise 1: Scaling of quantum-critical conductivity

Use dimensional analysis to show that the conductivity of a relativistic quantum critical theory in $d_s$ spatial dimensions scales as

$$\sigma\sim T^{d_s-2}.$$

What is special about $d_s=2$?

Solution

The current has scaling dimension

$$[J^i]=d_s,$$

because the conserved charge

$$Q=\int d^{d_s}x\,J^t$$

is dimensionless, so $[J^t]=d_s$, and current conservation gives the same scaling for $J^i$ in a relativistic theory.

The electric field has one derivative acting on a gauge potential. Since $a_\mu J^\mu$ appears in the action density and the action is dimensionless,

$$[a_\mu]+[J^\mu]=d_s+1,$$

so

$$[a_\mu]=1.$$

Therefore

$$[E_i]=[\partial_t a_i]=2.$$

Ohm’s law is

$$J^i=\sigma E_i.$$

Hence

$$[\sigma]=[J^i]-[E_i]=d_s-2.$$

At a quantum critical point, the only scale at nonzero temperature is $T$, so

$$\sigma\sim T^{d_s-2}.$$

For $d_s=2$, the conductivity is dimensionless. This is why quantum critical conductivity in $2+1$ dimensions is often expressed as a universal number times $Q^2/h$.

Exercise 2: Derive the Einstein relation

Starting from Fick’s law

$$\vec J=-D\nabla\rho$$

and the static susceptibility

$$\rho=\chi\mu,$$

show that

$$\sigma=D\chi.$$

Solution

A spatially varying chemical potential produces an electric field

$$\vec E=-\nabla\mu.$$

Using $\rho=\chi\mu$, assuming $\chi$ is constant in linear response,

$$\nabla\rho=\chi\nabla\mu.$$

Fick’s law becomes

$$\vec J=-D\chi\nabla\mu.$$

Since $\vec E=-\nabla\mu$,

$$\vec J=D\chi\vec E.$$

Comparing with Ohm’s law

$$\vec J=\sigma\vec E,$$

we get

$$\sigma=D\chi.$$

Exercise 3: Diffusion pole from hydrodynamics

Use

$$\partial_t\rho+\nabla\cdot\vec J=0, \qquad \vec J=-D\nabla\rho,$$

to derive the pole

$$\omega=-iDk^2.$$

Solution

Substitute Fick’s law into charge conservation:

$$\partial_t\rho-D\nabla^2\rho=0.$$

Take a Fourier mode

$$\rho(t,\vec x)=\rho_0 e^{-i\omega t+i\vec k\cdot\vec x}.$$

Then

$$\partial_t\rho=-i\omega\rho, \qquad \nabla^2\rho=-k^2\rho.$$

The diffusion equation becomes

$$-i\omega\rho+Dk^2\rho=0.$$

For a nonzero fluctuation,

$$-i\omega+Dk^2=0,$$

so

$$\omega=-iDk^2.$$

Exercise 4: Horizon formula for DC conductivity

Consider the Maxwell action

$$S_A=-\frac{1}{4g_F^2}\int d^{d+1}x\sqrt{-g}F^2$$

in an isotropic black-brane metric

$$ds^2=g_{tt}(z)dt^2+g_{zz}(z)dz^2+g_{xx}(z)d\vec x_{d_s}^{\,2}.$$

Show that the DC conductivity is

$$\sigma_{\rm dc} = \left. \frac{1}{g_F^2} \frac{\sqrt{-g}\,g^{xx}}{\sqrt{-g_{tt}g_{zz}}} \right|_{z=z_h}.$$

Solution

For a homogeneous perturbation $A_x(z,t)$, define the radial electric flux

$$\mathcal J^x =-\frac{1}{g_F^2}\sqrt{-g}F^{zx} =-\frac{1}{g_F^2}\sqrt{-g}\,g^{zz}g^{xx}\partial_z A_x.$$

In the DC limit, the Maxwell equation gives

$$\partial_z\mathcal J^x=0,$$

so $\mathcal J^x$ is independent of $z$. It may therefore be evaluated at the horizon.

Near the future horizon, use ingoing Eddington-Finkelstein regularity. A regular field depends on

$$v=t+\int^z dz'\sqrt{\frac{g_{zz}}{-g_{tt}}}$$

near the horizon. Thus the radial derivative and time derivative are related by

$$\partial_z A_x \simeq \sqrt{\frac{g_{zz}}{-g_{tt}}}\,\partial_v A_x.$$

The physical electric field is

$$E_x=-\partial_t A_x=-\partial_v A_x.$$

Up to the sign fixed by the convention for current flow,

$$\partial_z A_x \simeq -\sqrt{\frac{g_{zz}}{-g_{tt}}}E_x.$$

Plugging into $\mathcal J^x$ gives

$$\frac{\mathcal J^x}{E_x} = \left. \frac{1}{g_F^2} \sqrt{-g}\,g^{zz}g^{xx} \sqrt{\frac{g_{zz}}{-g_{tt}}} \right|_{z_h}.$$

Since $g^{zz}=1/g_{zz}$,

$$\sigma_{\rm dc} = \frac{\mathcal J^x}{E_x} = \left. \frac{1}{g_F^2} \frac{\sqrt{-g}\,g^{xx}}{\sqrt{-g_{tt}g_{zz}}} \right|_{z_h}.$$

Exercise 5: Shear diffusion constant

Starting from the linearized transverse momentum equation

$$(\epsilon+P)\partial_t u_y-\eta\partial_x^2u_y=0,$$

show that

$$D_\eta=\frac{\eta}{\epsilon+P}.$$

Then use $\epsilon+P=sT$ and $\eta/s=1/(4\pi)$ to obtain

$$D_\eta=\frac{1}{4\pi T}.$$

Solution

Take

$$u_y(t,x)=u_0 e^{-i\omega t+ikx}.$$

Then

$$\partial_tu_y=-i\omega u_y, \qquad \partial_x^2u_y=-k^2u_y.$$

The linearized equation becomes

$$-i\omega(\epsilon+P)u_y+\eta k^2u_y=0.$$

Dividing by the nonzero amplitude $u_y$ gives

$$-i\omega(\epsilon+P)+\eta k^2=0.$$

Thus

$$\omega=-i\frac{\eta}{\epsilon+P}k^2.$$

Comparing with

$$\omega=-iD_\eta k^2,$$

we find

$$D_\eta=\frac{\eta}{\epsilon+P}.$$

At zero chemical potential,

$$\epsilon+P=sT.$$

Therefore

$$D_\eta=\frac{\eta}{sT}.$$

Using

$$\frac{\eta}{s}=\frac{1}{4\pi},$$

we obtain

$$D_\eta=\frac{1}{4\pi T}.$$

Exercise 6: Why finite density changes the DC conductivity

At zero density, the current has no convective piece proportional to fluid velocity. At finite density,

$$J^i=\rho u^i+\cdots.$$

Explain why exact momentum conservation implies an infinite clean DC conductivity when $\rho\neq0$.

Solution

In a translation-invariant system, total momentum is conserved. A uniform electric field exerts a force on the charge density:

$$\partial_t P^i=\rho E^i.$$

If $\rho\neq0$, the electric field continuously pumps momentum into the fluid. Since there is no momentum relaxation, the fluid velocity grows rather than settling to a steady finite value.

The electric current contains the convective contribution

$$J^i=\rho u^i+\cdots.$$

Thus the growing momentum produces a growing current. In linear response, this appears as a delta function in the real part of the conductivity and a pole in the imaginary part:

$$\operatorname{Re}\sigma(\omega) \supset \pi\frac{\rho^2}{\chi_{PP}}\delta(\omega),$$ $$\operatorname{Im}\sigma(\omega) \supset \frac{\rho^2}{\chi_{PP}}\frac{1}{\omega}.$$

At $\rho=0$, this convective overlap is absent, so the clean DC conductivity can be finite.

Further reading

For quantum-critical transport and the holographic Maxwell calculation, see Hartnoll, Lucas, and Sachdev, Holographic Quantum Matter, especially the section on quantum critical charge dynamics. For hydrodynamics, Kubo formulas, diffusion, and shear viscosity, see Natsuume, AdS/CFT Duality User Guide, chapters on nonequilibrium physics and applications to plasmas. For a textbook treatment of linear response and holographic hydrodynamics, see Ammon and Erdmenger, Gauge/Gravity Duality. For the original viscosity calculation and its interpretation, see Policastro, Son, and Starinets, and the later Kovtun-Son-Starinets universality papers.