Complete Einstein Equa?on from Generalized 1 st law of Entanglement - - PowerPoint PPT Presentation

Sang-Jin Sin (Hanyang)

2017.11@KEK

Complete Einstein Equa?on from Generalized 1st law of Entanglement

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arXiv:1709.05752 Es.Oh, Iy. Park + SJS

disClaim

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We will not derive holography. We assume holography: CFT dual exist and the dual of vacuum is AdS. Question : when a physical configuration satisfy certain entanglement entropy relation, what is the equation that is satisfied by its gravity dual?

Quantum Entanglement = connectivity by quantum superposition

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If No superposi?on, àNo wave func?on collapse à No affec?on by measurement à No entanglement

Non-local connec?on

Classical connec?vity from quantum superposi?on

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Raamsdonk : Spacetime is also consequence of

quantum entanglement. [arXiv:1005.3035] |Ψ⟩ = |Ψ1⟩ |Ψ2⟩ ßà disconnected pair of space?mes. |ψ(β)⟩ = Σi exp{−βEi/2 }|Ei ⟩ |Ei ⟩ ßàconnected Eternal BH

i

E

i

E

i

E

=

Σe

−β

Tr2(|ψ⟩⟨ψ|) =

• i

e−βEi|Ei⟩⟨Ei| = ρT .

Space-Time connec?vity by entangling the d.o.f living in the two components.

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Quantification of entanglement by Ent. Entropy

ρA = TrB(|Ψ⟩⟨Ψ|), S(A) = −Tr(ρA logρA) S(A)= Minimal surface/4GN

Ryu-Takayanagi proposal

Holography on quantum Entanglement (Raamdonk)

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A B A ~

A B A B

Decreasing Ent.à Decreasing Area.

è Pinch off if Ent à0

First Law of Entanglement

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H = HA HB , HB is the Hilbert space of local fields over ball B. S(B) = −TrB (ρB logρB) , delete B if no confusion. Take a reference state ρ0 =vacuum Modular Hamiltonian : H0 = − Log ρ0 ⟨H0 ⟩ = −Tr ρ log ρ0 call it `energy’

∆E ∆S = S(ρ|ρ0), (1) where ∆E = Tr(ρ ρ0) ln ρ0, (2) ∆S = Trρ ln ρ + Trρ0 ln ρ0, (3) S(ρ|ρ0) = Trρ ln(ρ/ρ0). (4)

First Law of Entanglement

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Positivity of relative entropyà S (ρ|ρ0 ) is minimal at ρ.

∆E − ∆S = S(ρ|ρ0), ρ0 Extremality condi?on : δE − δS = 0 à 1st law

Entanglement First Law in gravity version

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δE − δS = 0

Express δE , δS in terms of gravity (geometry).

δE : Use Casini et.al, δS : Ryu-Takayanagi.

δS = R 8GN

• ddx(δij − 1

R2xixj)Hij

δEhyp

A

= d 16GN

• A

ddxR2 − r2 R δH00(0, x)

ds2 = 1 z2(dz2 + dxµdxµ + zdHµν(x, z)dxµdxν)

tµν(x) = d 16πGN Hµν(z = 0, x)

• Ent. First law & Einstein Eq.

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Lashkari+Macdermott+Raamsdonk [arXiv:1308.3716]

δE − δS = 0 is ßà Linearized Einstein equa?on.

which is equiv. to linearization of

∂x0∂y0δE = 3 2

• ∂DR

ˆ A = −3

• ∂DR

(xHxxdx + yHyydy) = ∂x0∂y0δS

ˆ A = −2xHxx + (R2 − x2 − y2)∂xHyy dx + −2yHyy + (R2 − x2 − y2)∂yHxx dy .

Hα

α = 0

∂µHµν = 0 1 z4∂z

• z4∂zHµν
• + ∂2Hµν = 0

Wµν = Rµν − 1 2gµνR − 3gµν = 0 ,

Σt ˜ B B ξA vA

t z X

Reformulation

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• T. Faulkner, M. Guica, T. Hartman, R. C. Myers and M. Van Raamsdonk, “Gravita?on

from Entangle- ment in Holographic CFTs,” JHEP 1403, 051 (2014) [arXiv:1312.7856].

There exists a d − 1 form χ Entanglement first law ßà δEab = 0.

Egrav

B

Sgrav

B

= Z

B− ˜ B

= Z

Σ

d,

d = 2⇠a

BEab✏b,

Z

B

= Sgrav

B

, and Z

˜ B

= Egrav

B

Brief review

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• This is off-shell Noether theorem
• Eq. of mo?onßàconserva?on law

S = Z dtL[q, ˙ q] Symmetry ⇒ δS = 0 pf − pi = Z dtδq · ˆ E where, p = ∂L ∂ ˙ q , ˆ E = d dt ∂L ∂ ˙ q − ∂L ∂q

B B~ t1 t2

Full version of the story?

• δE − δS = 0 ßà Linearized Einstein eq. δEab = 0.
• Full Einstein eq?

it has been Ques?oned during 2013-2017.

• Most recencent work: proof for second order in δg.

By Faulkner et.al 1705.03026

• T. Ugajin’s talk Yesterday 1705.01486
• Claim : Full eq. can be inferred from
• And above eq. is the finite difference version of first law

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∆E ∆S = S(ρ|ρ0),

Noether identity of Holland-Wald and Full Einstein Eq.

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On-shellà E^ =0. Then Conversely, for metric satisfying above eqà E^=0 or Since

C[g0 + g] C[g0] = 0.

Ca = 2Eg

ab✏b.

E[g(")] = E[g0] = 0.

ε g0

Egrav

B

Sgrav

B

• Z

Σ

!(g0; g, ξBg) = Z

Σ

ˆ E where ˆ E = ⇠a

B

• ✏aEg

bc[g0]gbc + Ca

• .

Egrav

B

Sgrav

B

= Z

Σ

!(g0; g, ξBg) ✏aEg

bc[g0]gbc + Ca = 0

for all order in ". g = AdS metric. for all order in ". g = AdS metric.

Deriva?on of Holland-Wald

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L ⌘ L[]✏

L[] = Eφ + dΘ[],

ariation is a diffeomorphism generat ⇠, ξL = d(⇠ · L) since ξ = iξd + diξ

ector field ⇠, he top form. ion 1 form

d by a d L is

J[⇠] = Θ[ξ] ⇠ · L, dJ[⇠] = E[]ξ,

• at J is the closed form for the fields at on-shell.

J[⇠] = dQ[⇠] + ⇠aCa,

J[⇠] = !(, ξ) + d(⇠ · Θ) ⇠ · Eφ !(; 1, 2) = 1Θ(2) 2Θ(1)

d = !(, ξ) ⇠ · (C + Eφ),

h = Q[⇠] ⇠ · Θ. he explicit form of No

Q = 1 16⇡GN ra⇠b✏ab, Ca = 2Eg

ab✏b.

Eg

ab =

1 8⇡GN (Rab 1 2gabR) T m

ab.

Generalized first law i) gravity version

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When and Why Claim: This is the gravity version of the identity interpreted as the finite difference version of the first law.

We need and

• N. Lashkari, J. Lin, H. Ooguri, B. Stoica and M. Van Raamsdonk [arXiv:1605.01075] .

Smarr Rela?on:

Z

Σ

!(g0; g, ξBg) = S(⇢|⇢0),

Egrav

B

Sgrav

B

= Z

Σ

!(g0; g, ξBg)

15–17]. If Wξ is defined by

Wξ = Z

Σ

!(g0; g, ξBg),

R S(⇢|⇢0) = Wξ[M] Wξ[AdS] := W,

∆E ∆S = S(ρ|ρ0),

Egrav − Sgrav = Wξ

Generalized 1st Law of Entanglement: ii) CFT version

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Suppose the density opereator depends on parameters R1, R2, · · · , RM, which we simbolically denote by a vec- tor R. Then ρ0 = ρ(R0) and ρ = ρ(R1) for some R0, R1.

−1 ∂ 1

g Fα = rα ln ρ = ρ−1

∂ ∂Rα ρ,

space, we can express the rel

S(ρ|ρ0) = h Z

C

dR · Fi,

Work done on the system ∆E ∆S = W. all ‘generalized entanglement first law’. gravity version, as we will see later. O

Remarks

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• 1. I assumed

following [arXiv:1605.01075; 1508.00897] However, it is worth to prove it by using bulk-boundary correspondence and actually this is done to second order in metric perturba?on by Sarosi+Ugajin:1611.02959, 1705.01486 and also by Faulkner Haehl, Hijano,Parrikar,Rabideau, Raamsdonk 1705.03026

• 2. I used Wald-Holland gauge:

form invariance of the B~ and killing field.

Z

Σ

!(g0; g, ξBg) = S(⇢|⇢0),

Goal: construct a vector field VE such that

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Entanglement Vector Field

Z

B0 V a EdΣa =

Z

˜ B

V a

EdΣa = SB.

16⇡GNQ = ra⇠b✏ab = 2ra⇠tpgtt✏a := Va✏a.

V = 4⇡ Rz h (R2 z2 ~ x2 2z + z)dz + xidxii . R

h i It is easy to check that R

˜ B Va✏a = 4⇡Area[ ˜

B]. it is tempting to call V as entanglement ve

raV a = 2⇡d Rz (z2 + ~ x2 R2) = (2d)n · ⇠,

Goal: construct a vector field V0 such that Answer : Finally

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Entanglement Vector Field

V0 = 2⇡d R (r R)2 r2 cos3 ✓ dr,

Therefore we look for a at ra(V a V a

0 ) = 0

VE = 2⇡ R hr2 + R2 r2 cos ✓ dr (R2 r2) r tan ✓ cos ✓ d✓ i V0

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• 2
• 1

1 2 0.0 0.5 1.0 1.5 2.0 x z

• 2
• 1

1 2 0.0 0.5 1.0 1.5 2.0 x z

B B

Σt ˜ B B ξA vA

t z X

Flux lines of V and VE

Sewing the space with flux line of VE

flux lines of VE look like sewing the two regions B and B ̄ along their interface.

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B B

Wanted feature in ref.

• M. Freedman and M. Headrick,

“Bit threads and holo- graphic entanglement,” arXiv:1604.00354 [hep-th].

Conclusion

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• 1. Generalized Entanglement fist law,

çè Full Einstein equation.

• 2. entanglement mostly through the edge

B B

Egrav

B

Sgrav

B

• Z

Σ

!(g0; g, ξBg) = Z

Σ

ˆ E where ˆ E = ⇠a

B

• ✏aEg

bc[g0]gbc + Ca

• .

감사합니다.

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