Holographic Transport and the Hall Angle Mike Blake - DAMTP - - PowerPoint PPT Presentation

Holographic Transport and the Hall Angle

Mike Blake - DAMTP

arXiv:1406.1659 with Aristomenis Donos arXiv:1310.3832 with David Tong and David Vegh arXiv:1308.4970 with David Tong

Part 1: Holographic Transport

AdS/CMT 101

RN Solution

AdS/CMT 102

δAx(ω) = Ex(ω) iω + hJx(ω)ir + ...

Jx(ω) = σ(ω)Ex(ω)

5 10 15 20 25 0.0 0.2 0.4 0.6 0.8 1.0 1.2 ΩêT Re@ΣD 5 10 15 20 25

• 0.5

0.0 0.5 1.0 1.5 2.0 wêT Im@sD

DC conductivity diverges!

Hartnoll

• To get finite DC conductivity need to have

momentum dissipation.

• Break translational invariance of boundary

theory weakly using irrelevant operator.

At → µ + λcos(kLx)

Horowitz, Santos & Tong

Solve very complicated PDEs!

Numerical Conductivity

σ(ω) = σDC 1 − iωτ

ρ ∼ T 2∆Jt(kL)

Result for resistivity agrees with memory matrix prediction

∆Jt(kL) = 1 2 q 5 + 2(kL/µ)2 − 4 p 1 + (kL/µ)2 − 1 2 Horowitz, Santos & Tong Hartnoll and Hofman

• Make progress by working perturbatively

in lattice strength

• Simplest thing is to add a background

scalar lattice on top of the RN solution

Analytic Conductivity

• To leading order can ignore the

backreaction on the metric and gauge field backgrounds.

MB, Tong and Vegh

• At leading order the conductivity calculation

simplifies enormously.

• In radial gauge we need only

consider

• The novel ingredient is scalar perturbation
• This is simply the phonon mode of the

lattice

(δAx, δgtx, δφ) (r, x, t) = ✏0(r)cos(kL[x − ⇡(r, t)]) δgrx = 0 = ✏kL0(r)⇡(r, t) δφ(r, x, t) = δφ(r, t)sin(kLx)

• After eliminating equations take the

form

δgtx M 2(r) = 1 2✏2k2

L2 0(r)

• These are the nothing but the perturbations

equations of massive gravity with a radially dependent graviton mass

Connection with Massive Gravity

MB and Tong

• Surprise is the existence of a massless mode

even at finite density.

• Whenever you have massless mode you can

use the Iqbal/Liu trick to show the membrane conductivity is constant.

• Evaluating at the horizon gives

δλ1 = ✓ 1 + µ2r2 M 2r2

h

◆1 δAx − µf iωrh π0

• σDC(r)

σDC = Q2r2

h

M 2(rh)

Universal Conductivity

MB and Tong

⇢ ∼ M 2(rh) ∼ ✏2k2

L0(rh)2

AdS2 × R2 φ0 ∼ ξ−∆O(kL) ξh ∼ T −1 ρ ∼ T 2∆O(kL)

• Zero temperature near horizon geometry

is

• At finite temperature the geometry

terminates at giving a resistivity

• Key result is that resistivity is determined

by value of the graviton mass at the horizon

Locally Critical Scaling

MB, Tong and Vegh

Related Work

• Can break momentum conservation in
• ther (simpler) ways:

Massive Gravity Linear Axions Q-lattices

χ ∼ kx φ1 ∼ sin(kLx) φ2 ∼ cos(kLx)

• In these models our method gives the

exact DC conductivity in terms of horizon data.

• Can use a similar approach to calculate

thermal and electrothermal conductivity.

Vegh, Davison Andrade and Withers, Gouteraux Donos and Gauntlett Donos and Gauntlett

Part II: The Hall Angle

Drude model

~ j(!) = (!) ~ E(!) ~ j = nq~ v

md~ v dt + m ⌧ ~ v = q( ~ E + ~ v × ~ B)

A puzzle...

B = 0 σDC = nq2τ m B 6= 0 θH = σxy σxx = qBτ m

Drude Strange Metal

θH ∼ 1 T 2 σDC ∼ 1 T

The strange metal experiments seem to imply different scattering times for electric and Hall currents.

Anderson Coleman, Schofield & Tsvelik

Q-Lattices

• Can use ‘Q-lattices’ to obtain analytic expression

for transport even when momentum dissipation is strong.

• Build lattices out of two complex scalar fields

χ1 → kx χ2 → ky

Donos and Gauntlett

ψ2 ∼ φeiχ2 ψ1 ∼ φeiχ1

• Stress tensor is homogeneous: can study exactly

using ODEs.

DC Transport

σDC =  Z(φ) + 4πQ2 k2Φ(φ)s

• rh

¯ α =  4πQ k2Φ(φ)

• rh

Z(φ)|rh

• is a new term that did not appear in
• ur perturbative lattice calculation.
• Compare with the electrothermal

conductivity

• Hence corresponds to excitations

that carry a current but no momentum.

Z(φ)|rh

Donos and Gauntlett

Weak Coupling Intuition

Jx Px

Holes Particles

• This is analogous to what happens at

`charge-conjugation symmetric’ critical points.

• Hence we define

σccs = Z(φ)|rh

Sachdev and Damle

DC Conductivity

• At finite density there are two additive

contributions to the conductivity - `Inverse Matthiessen Law’.

• In holography, is present even at low
• energies. This is not true for weakly

coupled particles at finite density.

σccs

σDC = σccs + Q2 E + P τL σccs = Z(φ)|rh τ −1

L

= s 4π k2Φ(φ) E + P

• rh

Hall angle

θH = 4πBQ k2Φs B2Z2 + Q2 + 8πZk2Φ/s B2Z2 + Q2 + 4πZk2Φ/s

• rh

Hall angle

θH ∼ BQ E + P τL

Hall angle

Holes Particles

B

No analogous term to

MB and Donos

σccs θH ∼ BQ E + P τL

• Weak momentum dissipation -
• Strong momentum dissipation -

τL → ∞ τL → 0 σDC = Q2 E + P τL θH = BQ E + P τL θH = 2BQ E + P τL

can now get different scalings! reproduces Drude-like results.

c.f. Hartnoll & Hofman etc

σDC = σccs

Comments

• Story can be applied more generally than to

the specific lattice models studied here e.g. to hydro, probe branes.

• Would be exciting to understand whether

mechanism can be applied to the cuprates or

• ther experimental systems.
• Supports recent suggestion that strange

metals are governed by incoherent transport.

Karch

σccs ∼ 1/T σdiss ∼ 1/T 2 σDC ∼ 1/T + 1/T 2 θH ∼ 1/T 2

= ⇒

Hartnoll

`` Over broad regions of doping, the two kinds of relaxation rates, the one for the conductivity and the

• ne for the Hall rotation, seem to add as inverses:

Conductivity is proportional to 1/ T + 1/ T 2—that is, it

• beys an anti-Matthiessen law.’’

P.

• W. Anderson - Physics

Today

Thank you!