Holographic Entanglement and Interaction Shigenori Seki RINS, Hanyang University and Institut des Hautes Études Scientifiques Intrication holographique et interaction à l IHES le 30 janvier 2014 1
Contents 1. Quantum entanglement and holography [S. Ryu and T. Takayanagi, Phys.Rev.Lett. 96 (2006) 181602] 2. EPR = ER conjecture [J. Maldacena and L. Susskind, arxiv:1306.0533] 3. Accelerating quark and anti-quark [K. Jensen and A. Karch, Phys.Rev.Lett. 111 (2013) 211602] 4. Gluon scattering [SS and S.-J. Sin, to appear] 2
Quantum entanglement and holography 3
Quantum entanglement Let s consider two systems: A and B Hilbert space: H A H B { i A } { i B } basis A general state in total system is denoted by ψ tot = i,j c ij i A j B If If c ij = c A i c B j c ij = c A i c B j, it is a non-entangled state., it is an entangled state. What is an order parameter of entanglement? Entanglement Entropy 4
Density matrix: for Ψ H A H B Reduced density matrix ρ tot = ΨΨ ρ A := tr B ρ tot = j Bj ( ΨΨ ) j B The entanglement entropy is defined by Von Neumann entropy as S A := tr(ρ A log ρ A ) 5
EPR pair Einstein-Podolsky-Rosen pair [Einstein-Podolsky-Rosen, Phys.Rev. 47 (1935) 777] entangled two particles A B e.g. a spin-0 particle decays to two spin-1/2 particles. Separate them from each other at long distance A B Observe A The state of B is determined. A and B are still entangled. 6
e.g. Consider the state Ψ = 1 2 ( A B A B ) Since the reduced density matrix is ρ A = 1 2 ( A A + A A ) we obtain the entanglement entropy S A = tr(ρ A log ρ A ) = log 2 Note that the pure state ρ P = 1 2 ( A + A )( A + A ) has zero entanglement entropy S P = tr(ρ P log ρ P )=0 So the entanglement entropy works as an order parameter. 7
Holographic entanglement entropy AdS/CFT correspondence (Holography) Field theory on boundary Gravity theory in bulk Partition function Wilson loop Z CFT = e S grav W = e Area(γ) Entanglement entropy S A = 1 4G N Area(γ) A z AdS A z A γ A γ boundary minimal surface A [Ryu-Takayanagi, PRL 96 (2006) 181602] 8
EPR = ER conjecture 9
ER bridge [Maldacena-Susskind, arxiv:1306.0533] Consider the eternal AdS-Schwarzschild black hole. There are two boundaries and two CFTs. H L,R n L,R = E n n L,R n L Future interior This eternal BH is described by the entangle state, Ψ = n e βe n/2 n L n R We can interpret this state in two ways: 1. a single black hole in thermal equilibrium [Israel, Phys.Lett.A57 (1976) 107] Left boundary Left exterior Right exterior Right boundary n Past interior R Penrose diagram 2. two black holes in disconnected spaces with a common time [Maldacena, hep-th/0106112] 10
1. a single black hole in thermal equilibrium Define a fictitious thermofield Hamiltonian, H tf = H R H L, u(i which generates boosts. u() Ψ = n e βe n/2 n L n R is an eigenvector of H tf with eigenvalue zero. U t = const 2. two black holes in disconnected spaces with a common time The time evolution is generated by H = H R + H L Ψ(t) = n e βe n/2 e 2iE nt n L n R The state Ψ describes the two black holes at a specific instant, t =0. Even though the two black holes exist in separate non-interacting worlds, their geometry is connected by an Einstein-Rosen bridge (a wormhole). [Einstein-Rosen, Phys.Rev. 48 (1935) 73] 11
EPR = ER conjecture From the example of eternal black hole, we studied Quantum mechanics entanglement geometric interpretation Gravity two systems connected by ER bridge By extending this concept, Maldacena and Susskind conjectured pairs of entangled particles separate at a long distance ER bridge BH BH 12
Further extension Even for an entangled pair of particles, in a quantum theory of gravity, there must be a Planckian bridge between them. ER bridge Black Black hole hole Hawking radiation. 13
Accelerating quark and anti-quark 14
The holographic surface of accelerating quark and anti-quark 0.1,23(0$4! 3(0$4! x 2 = t 2 + b 2 z 2 [Xiao, PLB 665 (2008) 173]! "#$%&!'#%()*!+#$,-#./!! The trajectories of quark and anti-quark are causally disconnected on the world-sheet. t x z The quark and anti-quark are entangled by the wormhole that the open string goes through. [Jensen-Karch, PRL 111 (2013) 211602] 15
The entanglement of quark and anti-quark (EPR pair) The interaction between quark and anti-quark ER bridge (wormhole) on world-sheet Do other interacting particles also have ER bridge on world-sheet? Fortunately, we know the minimal surface in AdS that describes a gluon-gluon scattering.
Gluon scattering 17
Minimal surface solution for gluon scattering AdS 5 (momentum space) [Alday-Maldacena, JHEP 0706 (2007) 064] ds 2 = R2 r 2 (η µνdy µ dy ν + dr 2 ) y µ =2πk µ The solution of Nambu-Goto action r =0 IR boundary condition y 0 = y 1 = y 2 = α 1+β 2 sinh u 1 sinh u 2 cosh u 1 cosh u 2 + β sinh u 1 sinh u 2, α sinh u 1 cosh u 2 cosh u 1 cosh u 2 + β sinh u 1 sinh u 2, α cosh u 1 sinh u 2 cosh u 1 cosh u 2 + β sinh u 1 sinh u 2, y 3 =0, α r =, cosh u 1 cosh u 2 + β sinh u 1 sinh u 2 Mandelstam variables: s(2π) 2 = t(2π) 2 = 8α2 (1 β) 2, 8α2 (1 + β) 2. 18
AdS 5 (momentum space) AdS 5 T-dual transformation: m y µ = R2 z 2 mn n x µ, (position space) z = R2 r ds 2 = R2 z 2 (η µνdx µ dx ν + dz 2 ) The Alday-Maldacena solution is mapped to x 0 = R2 1+β2 sinh u + sinh u, 2α x + := x 1 + x 2 2 = R2 2 2α [(1 + β)u +(1 β) cosh u + sinh u ], x := x 1 x 2 2 = R2 2 2α [(1 β)u + +(1+β)sinhu + cosh u ], x 3 =0, where u ± := u 1 ± u 2 z = R2 2α [(1 + β) cosh u + +(1 β) cosh u ]. For later convenience, we introduce X µ := α R 2 x µ (µ =0, +,, 3), Z := α z ( 1) R2 19
Causal structure on world-sheet The induced metric on world-sheet [SS-Sin, to appear] ds 2 ws = R 2 g ++ du 2 + +2g + du + du + g du 2 g ++ = 4(1 + β)2 sinh 2 u + + 4(1 + β 2 ) [(1 + β) cosh u + (1 β) cosh u ] 2 2[(1+β) cosh u + +(1 β) cosh u ] 2, g + = 2(1 β 2 )sinhu + sinh u [(1 + β) cosh u + +(1 β) cosh u ] 2, g = 4(1 β)2 sinh 2 u + 4(1 + β 2 ) [(1 + β) cosh u + (1 β) cosh u ] 2 2[(1+β) cosh u + +(1 β) cosh u ] 2. Horizons g ++ =0: (1 β) cosh u =(1+β) cosh u + +2 (1 + β) 2 sinh 2 u + +1+β 2 g =0: (1+β) cosh u + =(1 β) cosh u +2 (1 β) 2 sinh 2 u +1+β 2 20
0 β < 1 β =0 X ± (, + ) ˆX ± := 2 π arctan X ± [ 1, 1] 10 1.0 Z =1 5 0.5 Z = 10 5 5 10 1.0 0.5 0.5 1.0 5 0.5 β =1/2 10 10 1.0 1.0 red: g =0, dashed red: g ++ =0, dotted blue: g ++ = g 5 0.5 10 5 5 10 1.0 0.5 0.5 1.0 5 0.5 10 1.0 21
β =1 Regge limit: s with t fixed. X 0 = 1 2 sinh u + sinh u, X + = 1 2 u, X = 1 2 sinh u + cosh u, X 3 =0, Z = cosh u +. 10 1.0 5 0.5 10 5 5 10 1.0 0.5 0.5 1.0 5 0.5 10 1.0 While g ++ is positive definite, g is negative in cosh u + > 2. 22
EPR = ER = Interaction? The entanglement is given by a wormhole. 1.0 0.5 Incoming gluons: g 1 (t 1 ) = A L (t 1 ) A R (t 1 ) g 2 (t 1 ) = B L (t 1 ) B R (t 1 ) 1.0 0.5 0.5 1.0 0.5 Outgoing gluons: g 3 (t 2 ) = A L (t 2 ) B R (t 2 ) 1.0 g 4 (t 2 ) = B L (t 2 ) A R (t 2 ) The entanglement is induced by gluonic interaction. 23
There are two ways to see entanglement. 1. Internal entanglement g 1 (t 1 ) = A L (t 1 ) A R (t 1 ) 1 0 The open string endpoints in each gluon are entangled by the open string going through the wormhole. This is in the same way as the entanglement of quark and anti-quark. 24
2. Entanglement of gluons There are two channels. 1.0 0.5 1.0 0.5 0.5 1.0 0.5 1.0 1.0 0.5 1.0 0.5 0.5 1.0 0.5 1.0 25
How can we measure the entanglement of gluons? i ) (naively) log of scattering amplitude The scattering amplitude corresponds to the Wilson loop which is given by the area of minimal surface. λ A e Area, S log A = log 1+β 2 + (divergent part) 2π 1 β ii) the length between boundaries at the contacting points We consider + (β) =R +u+ u + du + g++ u =0, (β) =R where we introduced the cutoff, z ( ). +u u du g u+ =0 2 αz R 2 =(1+β) cosh u + +1 β =(1 β) cosh u +1+β 6 2αz ± (β) =R log R 2 + 1 1 6 log 1 ± β + O z 6log 1 S + (β)+ (β) =R + (divergent part) 1 β2 Anyway, diverges at the Regge limit, and vanishes at. β =1 S β =0 26
EPR = ER = Interaction 27