A Quantum Quench of the Sachdev-Ye-Kitaev Model Julia Steinberg - PowerPoint PPT Presentation

A Quantum Quench of the Sachdev-Ye-Kitaev Model Julia Steinberg Harvard University arXiv:1703.07793 [cond-mat.str-el] Chaos, Topology, and Dualities in Condensed Matter Theory UIUC November 4, 2017

Collaborators Valentin Kasper Subir Sachdev Andreas Eberlein Harvard University Harvard University Harvard University Perimeter Institute of Theoretical Physics

Quantum matter without quasiparticles • Want to study properties of systems without quasiparticles • First: what is a quasiparticle? • Long lived additive excitation with same quantum numbers as free particle • How do we identify systems without quasiparticles? ~ • Fastest relaxation τ eq ≥ C k B T , T → 0 • No long lived excitations in any basis • “Too fast”: cannot study long time behavior with conventional techniques

The SYK model: a solvable system without quasiparticles • Model of N flavors of Majorana fermions with infinite range q-body interactions i 1 ...i q i = J 2 ( q � 1)! q X H = ( i ) j i 1 i 2 ...i q ψ i 1 ψ i 2 ... ψ i q h j 2 2 . N q − 1 1 ≤ i 1 <i 2 <...<i q ≤ N • Solvable in large N limit • Maximally chaotic • Disorder average → melon diagrams, only keep one • Full propagator + =

Strong coupling behavior • For → “Emergent reparameterization symmetry β J � 1 ! 2 ∆ 1 ∆ = 1 iG R ( t ) = C ( J, ∆ ) θ ( t ) β sinh π t q β • Spontaneously and explicitly broken → Schwarzian action • SYK is “dual” to nearly AdS 2 ↔ Escher “Heaven and hell”

Quantum quenches and thermalization • Quench of SYK → probe nonequilibrium dynamics without quasiparticles • Possibly reach thermal state B • What is a thermal state? A • System is its own heat bath for subsystems • Steady state, observables reach thermal values • For SYK, our working definition of thermalization: 2-point function obeys KMS, T from energy conservation

Quench procedure • Start with SYK model with q and pq interactions • Turn off pq term instantaneously • Track evolution of Green’s function • Does SYK Greens function thermalize? • How long does it take (scaling dependence on T)? • What is the best way to do this?

Green’s Functions on the Closed-Time- Contour • Out of equilibrium, must study full evolution along contour • Two Greens functions: and 1 , t + G < ( t 1 , t 2 ) ≡ G ( t + G > ( t 1 , t 2 ) ≡ G ( t − 2 ) 2 ) 1 , t − • Use to form 2 by 2 matrix • Dyson (matrix) equation from disorder average: G − 1 0 ( t 1 , t 2 ) − G − 1 ( t 1 , t 2 ) = Σ( t 1 , t 2 ) X i q i J 2 q i G ( t 1 , t 2 ) q i − 1 Σ ( t 1 , t 2 ) = i

The Kadanoff-Baym equations • How do we study 2-pt function with no time translation? • Kadanoff-Baym equation directly from Dyson equation • Go to real time plane get two integro-differential equations: G R ⊗ Σ > + G > ⊗ Σ A � G > ⊗ G − 1 � = 0 ⊗ G > = Σ R ⊗ G > + Σ > ⊗ G A � G − 1 � 0 • For Majorana fermions have condition: G > ( t 1 , t 2 ) = − G < ( t 2 , t 1 ) • Always true → everything from ! G > ( t 1 , t 2 ) =

Causal Structure • Evolution from integral structure of Kadanoff Baym equations t 2 • Rewrite everything in terms of G > ( t 1 , t 2 ) = • Step functions → limits of integration > ( t 1 , t 2 ) = • For point → integrate “rectangle region” > ( t 1 , t 2 ) = t 1 • Pre-quench t 1 ≤ 0 , t 2 ≤ 0 • Post quench t 1 ≥ 0 , t 2 ≥ 0 • Pass through other quadrants → causal effect

Numerics model • Consider the SYK+random matrix model (p=1/2 q=4) X X H ( t ) = i j 2 ,ij f ( t ) ψ i ψ j − j 4 ,ijkl g ( t ) ψ i ψ j ψ k ψ l i<j i<j<k<l • Only pq: integrable • Both terms: “Fermi liquid” τ − 1 eq ∼ T 2 τ − 1 • Only q: “strange metal” eq ∼ T • Use and to specify quench ,ij f ( t ) ψ kl g ( t ) ψ • Solve full Kadanoff-Baym equations numerically • Use Majorana condition → solve for G > ( t 1 , t 2 ) =

Procedure • Solve Dyson equation self-consistently for thermal initial state • Use as BC for quench • Immediately post quench, no time translation invariance, define T = t 1 + t 2 • Absolute time: 2 • Relative time: t = t 1 − t 2 • Near equilibrium, varies slowly with look at low frequency behavior T • Wigner transform f ( t 1 , t 2 ) → f ( T , ω ) • Also define G K ( t 1 , t 2 ) = G > ( t 1 , t 2 ) + G < ( t 1 , t 2 )

Thermal state from the KMS condition J 2 ,i = 0 . 5, J 2 ,f = 0, J 4 ,i = J 4 ,f = 1, T i = 0 . 04 J 4 • “Thermal” 2-pt function obeys KMS iG K ( T , ω ) /A ( T , ω ) 1 • KMS → FDT iG K ( T , ω ) ✓ β ( T ) ω ◆ 0 = tanh A ( T , ω ) 2 T J 4 = − 50 T J 4 = 0 • “Effective inverse T”: β ( T ) − 1 T J 4 = 50 • Start with thermal state − 2 . 5 0 2 . 5 ω /J 4 ,f • Right after quench out of equilibrium , varies slowly → “thermal” • T → ∞ β ( T )

Effective temperature J 2 ,i = 0 . 0625, J 2 ,f = 0, J 4 ,i = J 4 ,f = 1 • from fit of FDT relation T eff 0 . 1 • Relaxes exponentially 0 . 075 • Check throughout quench T e ff h H i = E f 0 . 05 • Determines T i = 0 . 04 J 4 T eff T i = 0 . 08 J 4 • Depends on only through J 2 E f − 50 0 50 T

Relaxation rate • Know final temperature from energy conservation • How long does it take to reach final temperature for ? β J � 1 • Exponential rate Γ ∝ T • Higher temperature, controlled by Γ ∝

The large q limit • Consider large q interaction (after Large N) • Expand G > ( t 1 , t 2 ) =  1 + 1 � G > ( t 1 , t 2 ) = � i h ψ ( t 1 ) ψ ( t 2 ) i = � i q g ( t 1 , t 2 ) + . . . 2 exponential form of self energy • q → ∞ • Derivatives of KB eqns → Lorentzian-Liouville eqn ∂ 2 g ( t 1 , t 2 ) = 2 J ( t 1 ) J ( t 2 ) e g ( t 1 ,t 2 ) + 2 J p ( t 1 ) J p ( t 2 ) e pg ( t 1 ,t 2 ) ∂ t 1 ∂ t 2 J 2 ( t ) = qJ 2 ( t )2 1 − q J 2 p ( t ) = qJ 2 p ( t )2 1 − pq , • Exact solution for p=1/2, or 2

Quench Regions • General solution in all regions " # 0 0 1 ( t 1 ) h 2 ( t 2 ) − h g ( t 1 , t 2 ) = ln J 2 ( h 1 ( t 1 ) − h 2 ( t 2 )) 2 • Majorana condition → g ( t, t ) = 0 • For equilibrium solution g ( t 2 , t 1 ) = [ g ( t 1 , t 2 )] ∗ t 1 ≤ 0 , t 2 ≤ 0 • Structure of integrals in KB equations show this is always true • Need to solve in 5 regions

Quench time plane Z t 2 Z t 1 ∂ h e g ( t 1 ,t 3 ) + e g ( t 3 ,t 1 ) i d t 3 J ( t 1 ) J ( t 3 ) e g ( t 1 ,t 3 ) − g ( t 1 , t 2 ) = 2 d t 3 J ( t 1 ) J ( t 3 ) ∂ t 1 −∞ −∞ Z t 2 Z t 1 d t 3 J p ( t 1 ) J p ( t 3 ) e pg ( t 1 ,t 3 ) − h e pg ( t 1 ,t 3 ) + e pg ( t 3 ,t 1 ) i +2 d t 3 J p ( t 1 ) J p ( t 3 ) −∞ −∞ Z t 1 Z t 2 ∂ h e g ( t 3 ,t 2 ) + e g ( t 2 ,t 3 ) i d t 3 J ( t 3 ) J ( t 2 ) e g ( t 3 ,t 2 ) − g ( t 1 , t 2 ) = 2 d t 3 J ( t 3 ) J ( t 2 ) ∂t 2 −∞ −∞ Z t 1 Z t 2 d t 3 J p ( t 3 ) J p ( t 2 ) e pg ( t 3 ,t 2 ) − h e pg ( t 3 ,t 2 ) + e pg ( t 2 ,t 3 ) i +2 d t 3 J p ( t 3 ) J p ( t 2 ) −∞ −∞

Boundary Conditions • Need 3 BCs • Get , and from , , , h 0 h 0 B 2 (0) g C ( t ) h A 1 (0) h A 2 (0) A 1 (0) h B 1 ( −∞ ) h B 2 (0) • Too many BCs h ( t ) → a h ( t )+ b • SL(2,C) invariance: c h ( t )+ d • Constraint: ad − bc = 1 • Result independent of choices for , , h 0 h B 1 ( −∞ ) h B 2 (0) B 2 (0) • SL(2,R) invariance → “gauge” choice!

Post-Quench Solution h A 1 ( t ) = ae σ t + c h A 2 ( t ) = ae − 2 i θ e σ t + b • Choose ansatz , ce σ t + d ce − 2 i θ e σ t + d e − 4 i θ = ( b − dh B 1 ( −∞ ))( a ∗ − c ∗ h ∗ B 1 ( −∞ )) σ = 2 J sin θ ( b ∗ − d ∗ h ∗ B 1 ( −∞ ))( a − ch B 1 ( −∞ )) • Find solution for p=1/2 − σ 2  � g A ( t 1 , t 2 ) = ln 4 J 2 sinh 2 ( σ ( t 1 − t 2 ) / 2 + i θ ) β f = 2( π − 2 θ ) • From KMS: σ • Only depends on relative time: instant thermalization!

Relation to the Schwarzian Action • SYK described by Schwarzian for β J � 1 ◆ 2 L [ h ( t )] = h 000 ( t ) h 0 ( t ) − 3 ✓ h 00 ( t ) 2 h 0 ( t ) • Take from KB equations h ( t ) is solution to Schwarzian EOM • h ( t ) [ h 0 ( t )] 2 h 0000 ( t ) + 3 [ h 00 ( t )] 3 − 4 h 0 ( t ) h 00 ( t ) h 000 ( t ) = 0 • Thermalization connected to reparameterization modes • Schwarzian should also exhibit instantaneous thermalization

Final Remarks • Low energy limit, rate linear in T as expected • Also depends on q, what do 1/q 2 corrections look like? • 2-point function instantly thermalizes, but other quantities do not • Which quantities thermalize and on what timescale (some do not)? • When does the large N limit break down? • What other consequences does this have in gravity?