Remnants, Fuzzballs or Wormholes ?

Samir D. Mathur

The Ohio State University

General Relativity: Black holes form

Quantum mechanics:entangled pairs are created

There is a problem near theendpoint of evaporation

The black hole information paradox

1p2(00 + 11)

Remnant

A remnant would have an unbounded number of states for given mass and volume ….baby Universes?

In string theory we believe we understand all the states at the planck scale …

The dual CFT does not allow an unbounded number of states for a given energy …

Thus the remnant solution to the information paradox does not look correct for string theory …

. . .

Two options:

(1) Violate the no hair “theorem”, so we do not have the vacuum at the horizon Fuzzballs

(2) Alter some quantum mechanical principle; e.g. (a) Nonlocal identification of Hilbert spaces: Wormholes (Maldacena+Susskind)

( Conflicts with AdS/CFT ? (SDM 2014) )

(b) Final state boundary condition (Horowitz+Maldacena)

(c) Nonlinear quantum mechanics (Papadodimas+Raju …)

(d) Computability limits in quantum theory (Hayden-Harlow)

(e) Others …

This talk:

(1) Summarize the fuzzball program

(2) How is the semiclassical approximation violated ?

(Use the Kraus-Wilczek-Parikh computation to get an exact cancellation)

(3) The idea of Fuzzball complementarity

The hole in the AMPS argument

(4) A problem with wormholes: Conflict with AdS/CFT?

� = Ylm(✓,�) �(r)

⇤� = 0

0 l . R

lp

If we found good functions , then we could populate these mode functions with quanta

to make quantum states of the form

�(r)

a†l1m1a†l2m2

. . . a†lkmk|0i

Such states would give the correct order of entropy Sbek ⇠ (R/lp)2

For we cannot do any easy calculation, but if we had found

normalizable for … then there would be no information paradox ….

l ⇠ R/lp

l = 0, 1, 2, 3,�(r)

At least the Hawking argument fora paradox would have to be rethought …

The no hair “theorem” :

1p2(00 + 11) ??

Rough analogy: Witten’s bubble of nothing (1982)

Minkowski space with an extra compact circle is unstable to tunneling into a ‘bubble of nothing’

not part of spacetime

compact circle pinches off

|n⇧total = (J�,total�(2n�2))

n1n5(J�,total�(2n�4))

n1n5 . . . (J�,total�2 )n1n5 |1⇧total (5)

A

4G= S = 2⇥

⌥n1n2n3

�E =1

nR+

1

nR=

2

nR

�E =2

nR

S = ln(1) = 0 (6)

S = 2⌥

2⇥⌥

n1n2 (7)

S = 2⇥⌥

n1n2n3 (8)

S = 2⇥⌥

n1n2n3n4 (9)

n1 ⇤ n5 ⇤ n

⇤ n14 lp

⇤ n12 lp

⇤ n lp

M9,1 ⌅ M4,1 ⇥K3⇥ S1

A

4G⇤

�n1n5 � J ⇤ S

A

4G⇤⌥

n1n5 ⇤ S

e2�⇥

2⇥

n1np

1 +Q1

r2

1 +Qp

r2

e2�⇥

2⇥

n1n5

w = e�i(t+y)�ikz w(r, �, ⇤) (10)

B(2)MN = e�i(t+y)�ikz B(2)

MN(r, �, ⇤) , (11)

2

horizon

singularity

weak coupling strong coupling

String theory :

??

NO !!

weak coupling: “supertube”

low redshift

high redshift

strings

fluxes

D1D5 system :

S = 2π√

n5(√

n1 +√

n1)(√

np +√

np) (57)

= 2π√

n5(E

√m1mp

) (58)

S = 2π√

n1n5npnkk (59)

S = 2π√

n1n5nkk(√

np +√

np) (60)

= 2π√

n1n5(E

√mpmkk

) (61)

S = 2π√

n1n5(√

np +√

np)(√

nkk +√

nkk) (62)

∼ lp (63)

∼ n1

6 lp (64)

M9,1 → M4,1 × T 4 × S1 (65)

E/(2mkk) = 0.5 (66)

E/(2mkk) = 1.2 (67)

Lz ∼ [g2α′4√n1n5np

V R]1

3 ∼ Rs (68)

∆S (69)

eS (70)

eS+∆S (71)

S = 2π√

n1n5np(1 − f) + 2π√

n1n5npf(√

nk +√

nk) (72)

nk = nk =1

2

∆E

mk=

1

2Dmk(73)

D ∼ [

√n1n5npg2α′4

V Ry]1

3 ∼ RS (74)

∆S = S − 2π√

n1n5np = 1 (75)

S =A

4G(76)

mk ∼ G5

G24

∼ D2

G5(77)

D ∼ G1

3

5 (n1n5np)1

6 ∼ RS (78)

∼ Nα lp (79)

eS (80)

5

‘Effective string’ withtotal winding number

+

L =⇥

dx[�14F a

µ⇥Fµ⇥a +i

2⇥⌅⇥ + . . .]

P =2�np

L=

2�(n1np)LT

p =2�k

LT

�

k

knk = n1np

e2⇤�

2�

n1np

S = 2�⇧

2⇧n1np

LT = n1L

L

M9,1 ⇤ M4,1 ⇥ T 4 ⇥ S1

D1 D5 P

n1 n5 n1n5 T 4 S1

1

L =⇥

dx[�14F a

µ⇥Fµ⇥a +i

2⇥⌅⇥ + . . .]

P =2�np

L=

2�(n1np)LT

p =2�k

LT

�

k

knk = n1np

e2⇤�

2�

n1np

S = 2�⇧

2⇧n1np

LT = n1L

L

M9,1 ⇤ M4,1 ⇥ T 4 ⇥ S1

D1 D5 P

n1 n5 n1n5 T 4 S1

1

D1 branes D5 branes

D ∼ G1

3

5 (n1n5np)1

6 ∼ RS (81)

∼ Nα lp (82)

eS (83)

S ∼ E ∼√

E√

E (84)

n1 n1 np np (85)

S = 2π√

2(√

n1 +√

n1)(√

np +√

np) ∼√

E√

E ∼ E (86)

S = 2π(√

n1 +√

n1)(√

n5 +√

n5)(√

np +√

np) ∼ E3

2 (87)

S = 2π(√

n1 +√

n1)(√

n2 +√

n2)(√

n3 +√

n3)(√

n4 +√

n4) ∼ E2 (88)

S = AN

N∏

i=1

(√

ni +√

ni) ∼ EN2 (89)

ds2 = −dt2 +∑

i

a2i (t)dxidxi (90)

S = 2π(√

n1 +√

n1)(√

n2 +√

n2)(√

n3 +√

n3)(√

n4 +√

n4) (91)

S = 2π(√

n1 +√

n1)(√

n2 +√

n2)(√

n3 +√

n3) (92)

n4 = n4 ! 1 (93)

Smicro = 2π√

2√

n1np = Sbek (94)

Smicro = 2π√

n1n5np = Sbek (95)

Smicro = 2π√

n1n5npnkk = Sbek (96)

Smicro = 2π√

n1n5(√

np +√

np) = Sbek (97)

Smicro = 2π√

n5(√

n1 +√

n1)(√

np +√

np) = Sbek (98)

Smicro = 2π(√

n5 +√

n5)(√

n1 +√

n1)(√

np +√

np) (99)

Smicro = 2π(√

n1 +√

n1)(√

n2 +√

n2)(√

n3 +√

n3)(√

n4 +√

n4) (100)

ni = ni − ni (101)

E =∑

i

(ni + ni) mi (102)

S = CN∏

i=1

(√

ni +√

ni) (103)

Pa =∑

i

(ni + ni) pia (104)

6

S1 → y y : (0, 2πR) (175)

ClV [l] V (176)

N = n1n5 (177)

√N − n

√n + 1 ≈

√N

√n + 1

dn

dt∝ (n + 1) n (178)

ωR =1

R[−l − 2 − mψm + mφn] = ωgravity

R (179)

m = nL + nR + 1, n = nL − nR (180)

|λ − mψn + mφm| = 0, N = 0 (181)

λ = 0, mψ = −l, n = 0, N = 0 (182)

ωI = ωgravityI (183)

|0⟩ |ψ⟩ ⟨0|ψ⟩ ≈ 0 (184)

n1, n2, n3 n4 (185)

1/n1n2n3 (186)

(n1n5)αlp (187)

n1n5

∑

knk = n1n5 n5 (188)

n′p = n1 n′

1 = n5,∑

knk = n′pn

′1 (189)

10

S1 → y y : (0, 2πR) (175)

ClV [l] V (176)

N = n1n5 (177)

√N − n

√n + 1 ≈

√N

√n + 1

dn

dt∝ (n + 1) n (178)

ωR =1

R[−l − 2 − mψm + mφn] = ωgravity

R (179)

m = nL + nR + 1, n = nL − nR (180)

|λ − mψn + mφm| = 0, N = 0 (181)

λ = 0, mψ = −l, n = 0, N = 0 (182)

ωI = ωgravityI (183)

|0⟩ |ψ⟩ ⟨0|ψ⟩ ≈ 0 (184)

n1, n2, n3 n4 (185)

1/n1n2n3 (186)

(n1n5)αlp (187)

n1n5

∑

knk = n1n5 n5 (188)

n′p = n1 n′

1 = n5,∑

knk = n′pn

′1 (189)

10

KK monopole

(Strominger-Vafa 96,Callan-Maldacena 96,Maldacena-Susskind 96)

P, P

In the CFT, all extremal and near extremal states are described by different choices of windings, spins, andleft/right excitations( )

S1 → y y : (0, 2πR) (175)

ClV [l] V (176)

N = n1n5 (177)

√N − n

√n + 1 ≈

√N

√n + 1

dn

dt∝ (n + 1) n (178)

ωR =1

R[−l − 2 − mψm + mφn] = ωgravity

R (179)

m = nL + nR + 1, n = nL − nR (180)

|λ − mψn + mφm| = 0, N = 0 (181)

λ = 0, mψ = −l, n = 0, N = 0 (182)

ωI = ωgravityI (183)

|0⟩ |ψ⟩ ⟨0|ψ⟩ ≈ 0 (184)

n1, n2, n3 n4 (185)

1/n1n2n3 (186)

(n1n5)αlp (187)

n1n5

∑

k mk = n1n5 n5 (188)

n′p = n1 n′

1 = n5,∑

k mk = n′pn

′1 (189)

10

All 2-charge extremal states:

(Lunin+SDM 01,Lunin+Maldacena+Maoz 02,Skenderis+Taylor 07)

All strings equal, one excited by a pair of excitations

Wavefunction for supergravity quantumin “cap”

String in cap (pp-wave limit)

One string highly twisted, excited by several excitations,

(Maldacena+Strominger 98,SDM+Saxena+Srivastava 03,Giusto+Russo 13)

(Lunin+SDM 02)

Ja�n . . . L�m . . . | i2�charge

2-charge extremal + integer moded chiral algebra generators:

Brown-Henneaux: States should lie at boundary of AdS

Sen: Calls these states at neck as “hair”

All states are captured by applying arbitrary fractional mode excitations to different cycles

Ja(i)

� kn

. . . L(j)

� lm

. . . | i2�charge

(SDM+Turton 11, Lunin+SDM+Turton 12)

All strings excited the same way, “filled fermi sea” of fractional excitations

States not in Brown-Henneaux class of “neck excitations”; thus they are give structure at the location where the horizon would have been

Different strings excited differently, integer chiral modes on strings, but not in the combination to give neck modes

⇣J+(1)� 2

nJ+(1)� 4

n. . . J+(1)

� 2kn

⌘⇣J+(2)� 2

nJ+(2)� 4

n. . . J+(2)

� 2kn

⌘. . . | i

(Giusto+Lunin+SDM+Turton 12)

(Bena+Giusto+Russo +Shigemori+Warner 15)

unitary

non-unitary !!�CFT = ⇢thermalL ⇢thermal

R V

= �Hawking

�CFT

= ⇢stateL

⇢stateR

V

= �emission

unitarywhere there is no incoming wave, but we still have an outgoing wave carrying energy out toinfinity. These instability frequencies are given by solutions to the transcendental equation

−e−iνπΓ(1 − ν)

Γ(1 + ν)

(κ

2

)2ν=

Γ(ν)

Γ(−ν)

Γ(12(1 + |ζ| + ξ − ν))Γ(1

2 (1 + |ζ|− ξ − ν))

Γ(12(1 + |ζ| + ξ + ν))Γ(1

2 (1 + |ζ|− ξ + ν))(3.50)

We reproduce the solution to this equation, found in [14], in appendix B. In the large R limit(2.27) the instability frequencies are real to leading order

ω ≃ ωR =1

R(−l − mψm + mφn − |− λ − mψn + mφm|− 2(N + 1)) (3.51)

where N ≥ 0 is an integer. The imaginary part of the frequency is found by iterating to ahigher order; the result is

ωI =1

R

(

2π

[l!]2

[

(ω2 − λ2

R2)Q1Q5

4R2

]l+1l+1+NCl+1

l+1+N+|ζ|Cl+1

)

(3.52)

Note that ωI > 0, so we have an exponentially growing perturbation. Our task will be toreproduce (3.51),(3.52) from the microscopic computation.

4 The Microscopic Model: the D1-D5 CFT

In this section we discuss the CFT duals of the geometries of [13]. Recall that we are workingwith IIB string theory compactified to M4,1×S1×T 4. The S1 is parameterized by a coordinatey with

0 ≤ y < 2πR (4.53)

The T 4 is described by 4 coordinates z1, z2, z3, z4. Let the M4,1 be spanned by t, x1, x2, x3, x4.We have n1 D1 branes on S1, and n5 D5 branes on S1 × T 4. The bound state of these branesis described by a 1+1 dimensional sigma model, with base space (y, t) and target space adeformation of the orbifold (T 4)n1n5/Sn1n5

(the symmetric product of n1n5 copies of T 4). TheCFT has N = 4 supersymmetry, and a moduli space which preserves this supersymmetry. Itis conjectured that in this moduli space we have an ‘orbifold point’ where the target space isjust the orbifold (T 4)n1n5/Sn1n5

[28].The CFT with target space just one copy of T 4 is described by 4 real bosons X1, X2, X3,

X4 (which arise from the 4 directions z1, z2, z3, z4), 4 real left moving fermions ψ1,ψ2,ψ3,ψ4

and 4 real right moving fermions ψ1, ψ2, ψ3, ψ4. The central charge is c = 6. The completetheory with target space (T 4)n1n5/Sn1n5

has n1n5 copies of this c = 6 CFT, with states thatare symmetrized between the n1n5 copies. The orbifolding also generates ‘twist’ sectors, whichare created by twist operators σk. A detailed construction of the twist operators is given in[19, 20], but we summarize here the properties that will be relevant to us.

The twist operator of order k links together k copies of the c = 6 CFT so that the Xi,ψi, ψi

act as free fields living on a circle of length k(2πR). Thus we end up with a c = 6 CFT on acircle of length k(2πR). We term each separate c = 6 CFT a component string. Thus if we arein the completely untwisted sector, then we have n1n5 component strings, each giving a c = 6CFT living on a circle of length 2πR. If we twist k of these component strings together by atwist operator, then they turn into one component string of length k(2πR). In a generic CFTstate there will be component strings of many different twist orders ki with

∑

i ki = n1n5.

10

where there is no incoming wave, but we still have an outgoing wave carrying energy out toinfinity. These instability frequencies are given by solutions to the transcendental equation

−e−iνπΓ(1 − ν)

Γ(1 + ν)

(κ

2

)2ν=

Γ(ν)

Γ(−ν)

Γ(12(1 + |ζ| + ξ − ν))Γ(1

2 (1 + |ζ|− ξ − ν))

Γ(12(1 + |ζ| + ξ + ν))Γ(1

2 (1 + |ζ|− ξ + ν))(3.50)

We reproduce the solution to this equation, found in [14], in appendix B. In the large R limit(2.27) the instability frequencies are real to leading order

ω ≃ ωR =1

R(−l − mψm + mφn − |− λ − mψn + mφm|− 2(N + 1)) (3.51)

where N ≥ 0 is an integer. The imaginary part of the frequency is found by iterating to ahigher order; the result is

ωI =1

R

(

2π

[l!]2

[

(ω2 − λ2

R2)Q1Q5

4R2

]l+1l+1+NCl+1

l+1+N+|ζ|Cl+1

)

(3.52)

Note that ωI > 0, so we have an exponentially growing perturbation. Our task will be toreproduce (3.51),(3.52) from the microscopic computation.

4 The Microscopic Model: the D1-D5 CFT

In this section we discuss the CFT duals of the geometries of [13]. Recall that we are workingwith IIB string theory compactified to M4,1×S1×T 4. The S1 is parameterized by a coordinatey with

0 ≤ y < 2πR (4.53)

The T 4 is described by 4 coordinates z1, z2, z3, z4. Let the M4,1 be spanned by t, x1, x2, x3, x4.We have n1 D1 branes on S1, and n5 D5 branes on S1 × T 4. The bound state of these branesis described by a 1+1 dimensional sigma model, with base space (y, t) and target space adeformation of the orbifold (T 4)n1n5/Sn1n5

(the symmetric product of n1n5 copies of T 4). TheCFT has N = 4 supersymmetry, and a moduli space which preserves this supersymmetry. Itis conjectured that in this moduli space we have an ‘orbifold point’ where the target space isjust the orbifold (T 4)n1n5/Sn1n5

[28].The CFT with target space just one copy of T 4 is described by 4 real bosons X1, X2, X3,

X4 (which arise from the 4 directions z1, z2, z3, z4), 4 real left moving fermions ψ1,ψ2,ψ3,ψ4

and 4 real right moving fermions ψ1, ψ2, ψ3, ψ4. The central charge is c = 6. The completetheory with target space (T 4)n1n5/Sn1n5

has n1n5 copies of this c = 6 CFT, with states thatare symmetrized between the n1n5 copies. The orbifolding also generates ‘twist’ sectors, whichare created by twist operators σk. A detailed construction of the twist operators is given in[19, 20], but we summarize here the properties that will be relevant to us.

The twist operator of order k links together k copies of the c = 6 CFT so that the Xi,ψi, ψi

act as free fields living on a circle of length k(2πR). Thus we end up with a c = 6 CFT on acircle of length k(2πR). We term each separate c = 6 CFT a component string. Thus if we arein the completely untwisted sector, then we have n1n5 component strings, each giving a c = 6CFT living on a circle of length 2πR. If we twist k of these component strings together by atwist operator, then they turn into one component string of length k(2πR). In a generic CFTstate there will be component strings of many different twist orders ki with

∑

i ki = n1n5.

10

ergoregion

(Jejjala,Madden,Ross Titchener 05, Cardoso,Dias,Hobvedo,Myers 05, Chowdhury +SDM 07)

(Bena+Warber 2005 + …)

Large families of states (with perhaps the correct order of the entropy of black holes), have been constructedby Bena+Warner + many others …

These have not been indentified with corresponidng CFT states … it is possible that some of them are liftedslightly (order 1/M ?) by quantum effects …

In that case they contribute to the count of near extremal states rather than extremal states …

(1) (Fuzzball conjecture): No microstate will have a vacuum horizon

(2) Some states will be fuzzballs, but most have a traditional horizon

In principle there are two possibilities:

But if (2) is true, then what is the solution to the information puzzle?

One suggestion was “Large cumulative effect from small corrections”

There can be a small correction (perhaps due to instanton effects), slightly modifying the state at each emission

1p2(|0⇥b|0⇥c + |1⇥b|1⇥c) + ✏k

1p2(|0⇥b|0⇥c � |1⇥b|1⇥c)

1p2(|0�b|0�c + |1�b|1�c)

✏k ⌧ 1( )

not have, until recently, a construction of this hair, but many of them were still not worriedabout Hawking’s paradox. The reason was based on the following misconception. Suppose thehorizon was a place with ‘normal physics’, and let us include a small correction, order � ⇧ 1to the state of each created pair. The number of pairs N is very large, so it might be thatsuitable choices of these small corrections would lead to a situation where Sent does decreasein the manner expected of a normal body.

A priori, it is not wrong to think that small corrections might cause Sent to decrease.Suppose the entangled pair at the first step is 1⇥

2(|0⌃b1 |0⌃c1 + |1⌃b1 |1⌃c1). At the next step we

can have the state

|�⌃ =1

2

�|0⌃b1 |0⌃c1 [(1 + �1)|0⌃b2 |0⌃c2 + (1� �1)|1⌃b2 |1⌃c2 ]

+|1⌃b1 |1⌃c1 [(1 + ��1)|0⌃b2 |0⌃c2 + (1� ��1)|1⌃b2 |1⌃c2 ]⇥

(2.1)

Note that the correction at each step can depend on everything in the hole at all earlier steps;the only requirement is that the correction be small: |�1| < �, |��1| < �. We have ⌅ 2N correctionterms in general after N steps. Since N ⌅ ( M

mp)2 for a 3+1 dimensional black hole, it appears

a priori possible for small corrections to pile up to make Sent decrease after the halfway pointof evaporation.

In [?] it was proved, using strong subadditivity, that such small corrections cannot lead toa decrease in Sent. AMPS invoked this argument in their analysis, so let us outline the steps in[?]. Let {b1, . . . bN} ⇥ {bi} be the quanta radiated in the first N steps, and {ci} their entangledpartners. The entanglement entropy at step N is Sent(N) = S({bi}). The created quanta atthe next step are are bN+1, cN+1. We then have [?]:

(i) By direct computation, one obtains

S(bN+1 + cN+1) < � . (2.2)

(ii) Similarly, by direct computation one obtains

S(cN+1) > ln 2� � . (2.3)

(iii) The unitary evolution of the hole does not a⇥ect quanta already emitted (we haveassumed that nonlocal e⇥ects, if any extend only to distances of order r0, and thus do not a⇥ectquanta that have been emitted from the hole long ago). Thus we have

S({(bi}) = SN . (2.4)

(iv) The strong subadditivity inequality gives

S({bi}+ bN+1) + S(bN+1 + cN+1) ⇤ S(bN+1) + S(cN+1) . (2.5)

Using (i)-(iii) above we find that the entanglement entropy of the radiation after the (N +1)-thtime step, SN+1 ⇥ S({bi}+ bN+1), satisfies

SN+1 > SN + ln 2� 2� . (2.6)

5

First step of emission

not have, until recently, a construction of this hair, but many of them were still not worriedabout Hawking’s paradox. The reason was based on the following misconception. Suppose thehorizon was a place with ‘normal physics’, and let us include a small correction, order � ⇧ 1to the state of each created pair. The number of pairs N is very large, so it might be thatsuitable choices of these small corrections would lead to a situation where Sent does decreasein the manner expected of a normal body.

A priori, it is not wrong to think that small corrections might cause Sent to decrease.Suppose the entangled pair at the first step is 1⇥

2(|0⌃b1 |0⌃c1 + |1⌃b1 |1⌃c1). At the next step we

can have the state

|�⌃ =1

2

�|0⌃b1 |0⌃c1 [(1 + �1)|0⌃b2 |0⌃c2 + (1� �1)|1⌃b2 |1⌃c2 ]

+|1⌃b1 |1⌃c1 [(1 + ��1)|0⌃b2 |0⌃c2 + (1� ��1)|1⌃b2 |1⌃c2 ]⇥

(2.1)

Note that the correction at each step can depend on everything in the hole at all earlier steps;the only requirement is that the correction be small: |�1| < �, |��1| < �. We have ⌅ 2N correctionterms in general after N steps. Since N ⌅ ( M

mp)2 for a 3+1 dimensional black hole, it appears

a priori possible for small corrections to pile up to make Sent decrease after the halfway pointof evaporation.

In [?] it was proved, using strong subadditivity, that such small corrections cannot lead toa decrease in Sent. AMPS invoked this argument in their analysis, so let us outline the steps in[?]. Let {b1, . . . bN} ⇥ {bi} be the quanta radiated in the first N steps, and {ci} their entangledpartners. The entanglement entropy at step N is Sent(N) = S({bi}). The created quanta atthe next step are are bN+1, cN+1. We then have [?]:

(i) By direct computation, one obtains

S(bN+1 + cN+1) < � . (2.2)

(ii) Similarly, by direct computation one obtains

S(cN+1) > ln 2� � . (2.3)

(iii) The unitary evolution of the hole does not a⇥ect quanta already emitted (we haveassumed that nonlocal e⇥ects, if any extend only to distances of order r0, and thus do not a⇥ectquanta that have been emitted from the hole long ago). Thus we have

S({(bi}) = SN . (2.4)

(iv) The strong subadditivity inequality gives

S({bi}+ bN+1) + S(bN+1 + cN+1) ⇤ S(bN+1) + S(cN+1) . (2.5)

Using (i)-(iii) above we find that the entanglement entropy of the radiation after the (N +1)-thtime step, SN+1 ⇥ S({bi}+ bN+1), satisfies

SN+1 > SN + ln 2� 2� . (2.6)

5

Second step of emission

This creates correction terms after stepsN2N

Thus subleading corrections to Hawkings’s leading order computation will resolve the puzzle:

entanglemententanglement(Maldacena 2000Hawking 2004)

??

(SDM 2009)

But using strong subadditivity of entanglement entropy, it can be shown that this is not possible

SN+1 > SN + ln 2 + 2✏

entanglementEntanglement keeps rising monotonically

If we assume

(i) The Niceness conditions N give local Hamiltonian evolution.

(ii) A traditional black hole (i.e. one with an information-free horizon) exists in the theory.

Then formation and evaporation of such a hole will lead to mixed states/remnants.

Hawking’s argument (1974) Hawking’s ‘Theorem’:

Corollary: If the state of Hawking radiation has to be a pure state with no entanglement with the rest of the hole then the evolution of low energy modes at the horizon has to be altered by order unity.

(SDM 2009)

What happens to a collapsing shell ?

What violates the semiclassical approximation?

First consider a toy model:

Consider a shell made of an ingoing gravitational wave with energy M

Let there be a large number of massless scalars in the theory : N

Collapsing shellRadiated quanta

The shell evaporates to a collection of scalar quanta, without forming a horizon

N � (M/mp)2

We do not have a large number of massless scalars in string theory …

But we have massive fuzzball stateseSbek

Emission rate without back reaction:

� ⇠ e�!T ⇠ e�8⇡GM!

Emission rate for a spherical shell with back reaction (Kraus-Wilczek-Parikh)

� ⇠ e�8⇡G(M�!2 )!

This is a special case of the general expression

� ⇠ e��Sbek ⇠ eSbek(M�!)�Sbek(M)

This expression works for both massless and massive shells, and can be used for any !

We set ! = M

� ⇠ f e�Sbek(M) ⇠ eSbeke�Sbek ⇠ 1

Overall rate of tunneling into fuzzballs :

Thus in a time of order unity the collapsing shell transitions to a linear combination of fuzzball states.

� ⇠ e�Sbek(M)

| i !Exp[Sbek]X

i=1

Ci

|Fi

i

The data of the shell is encoded in the set {Ci}

Fuzzball complementarity

How is the AMPS argument against complementarity bypassed ?

Bena: If you fall towards the horizon, you will crack you head and die …

So why did we not say right away that fuzzballsbehave like a “brick wall” or a “firewall” ?

Because there is another possibility :

“fuzzball complementarity”

To see the basic idea, consider AdS/CFT duality…(Maldacena 97, Gubser+Klebanov+Polyakov 98, Witten 98)

This works because the open string excitations are not an arbitrary set of excitations …

The spectrum of excitations along the brane surface reproduces the spectrum of radial infall

In the fuzzball, the infalling graviton cannot go further inside; its energy excites the fuzzball surface

Could this spread of the excitations along the fuzzball surface map, approximately, to the dynamics of radial infall ?

Conjecture of fuzzball complementarity: The dynamics of the fuzzball surface maps approximately to the dynamics of infall into the interior of the traditional hole …

The word “approximately” is crucial here, since different fuzzball microstates have different microstructure, and radiate differently … thus the physics cannot be universal E ⇠ T

But the physics can be a universal “hydrodynamical excitation” of the fuzzball, and this is conjectured to be dual to infall in the traditional hole

E � T

Traditional complementarity :

horizon stretched horizon

In the external observer’s frame,the graviton is absorbed by astretched horizon and its information returned to infinity

Quantum state at the horizon is the vacuum, since we have found no hair in our theory

In the infalling observer’s frame,the graviton just passes through

These two different descriptions are compatible since the two observers cannot communicatebefore the infaller hits the singularity

AMPS rule out traditional complementarity, but fuzzball complementarity is quite different:

With fuzzballs, there are real degrees of freedom where the stretched horizon would have been; these degrees of freedom differentiate microstates

AMPS: You will burn up if you try to fall through the horizon

Fuzzball: But we are not trying to fall through the horizon; there is no horizon to fall through. We are hitting the fuzzball surface and obtaining the infall as a DUAL description of oscillations of the fuzzball surface

AMPS: Our argument against complementarity should still work … you should get burnt by theHawking radiation near the fuzzball surface …

Fuzzball: There is something strange about one of your assumptions. You assume that the stretched horizon does not respond before it is hit. But normal horizons do increase in size before they are hit.

The stretched horizon has one bit of data per planck area. If the infalling bit lands on it, then you have

S =A

4G+ 1 >

A

4G

Fuzzball :

fuzzballs tunnel out to this radius due to the extra energy brought in by the gravitonTemperature is very low at the location where the graviton

is absorbed into the new fuzzball surface, if …

Do not get burnt !

E � T

Fuzzball: When a quantum with falls in, you create a large number of NEW states, and these states are NOT entangled with infinity. Fuzzball complementarity arises only in this limit, and is the dynamics of these new states

AMPS: The black hole is maximally entangled with radiation at infinity after its half-way evaporation point. In this situation, we have argued that you cannot get complementarity

E � T

Nf

Ni=

eS(M+E)

eS(M)=

eS(M)+�S

eS(M)= e�S ⇡ e

ET � 1

(SDM+Turton 2013)

The notion of fuzzball complementarity provides a scenario where the following situation holds:

We fix our theory of the standard model, and also fix the infalling observer. If we now take the black hole mass to be sufficiently large, we will have both of the following

(i) The evaporation is unitary

(ii) The infalling observer will feel no drama in the sense that the number of particles they intercept goes to zero, as does the energy of each particle intercepted (as measured in the infalling frame).

Nevertheless, the density of energy quanta remains much larger than that predicted by the Hawking state.

Caveat (Marolf) : The term “sufficiently large” means , but it needs to be checked if astrophysical holes are large enough to satisfy this condition, in a theory with large gauge hierarchies.

M � mp

Can wormholes be the correct solution ?

Maldacena 2000: Consider the eternal hole in AdS

There are two asymptotically AdS boundaries

Thus the spacetime is dual to two uncoupled CFTs, in an entangled state

?

|�� =P

i e�Ei

2T |Ei�L|Ei�R

?X⌦ =

(Van Raamsdonk 2010)

Hawking radiation is entangled with its partners inside the hole …

Entangled quanta create wormholes …

This can change the physics at the horizon, in such a way that we have the local vacuum, and while being able to get the information out …

1p2(00 + 11)

(Maldacena+Susskind 2013)

wormhole

But there is a problem with such a postulate … (SDM 2014)

(a) Suppose two entangled CFTs are indeed dual to the eternal hole

?

|�� =P

i e�Ei

2T |Ei�L|Ei�R

(b) We can couple the CFTs to external spaces, from which we can throw in or extract particles into the AdS

bc OX

bc OX

(c) We wish to maintain a smooth horizon, which means that we cannot stop particle creation there.

The eternal hole maintains its mass only because we throw in particles at the same rate that they are emitted.

We choose to not throw in these particles, so that both horizons shrink in area as the holes radiates

We do not say how information is recovered, but we assume that the radiation is unitary due to somemechanism …

(d) In the CFT, if we throw in something into the right CFT, we assume that its information will be emitted only on the right (entanglement cannot transfer information) …

|�� =P

i e�Ei

2T |Ei�L|Ei�R

p

(e) In the gravity dual, the particle send in from the right, can cross the central line,though it cannot emerge from the left horizon

We ask: which side will the information emerge ?

p

(f) We may say that left movers send their information out to the right, but we can collide particles and confuse what is right moving and what is left moving …

p

Key point: Even though classical particles cannot traverse from right to left, Hawking radiation“tunnels” out of the horizon, and so we do have to consider what is right and left …

If the dual of two entangled CFTs creates a connected manifold with a “forward wedge”,then there seems to be no obvious way to avoid information transfer from right to left throughthe tunneling process of Hawking radiation

(g) Conclusion: The dual of two entangles CFTs is NOT a connected spacetime …

|�� =P

i e�Ei

2T |Ei�L|Ei�R

NO !

|�� =P

i e�Ei

2T |Ei�L|Ei�R

X⌦

Fuzzballs

(h) But what spacetime do we generate if we start with the wormhole?

wormhole

If we get the eternal hole spacetime, then it is dual to two entangled CFTs, and the problem starts …

r < 2M

r = 2M

Conjecture: If the neck is narrower than then the spacetime slice is unstable to tunneling quickly into an entangled pair of fuzzballs

r = 2M

Thus the eternal hole spacetime does not exist, and we avoid the contradiction with AdS/CFT duality …

But then we also lose the wormhole solution to the information paradox …

Summary

(a) The small corrections theorem

tells us that there are three categories of solutions to the information paradox:

(1) Remnants

(2) Fuzzballs

(3) Wormholes (or other new quantum effects)

SN+1 > SN + ln 2 + 2✏

(b) If we trust AdS/CFT, then we cannot get remnants …

We have also argued that there is a problem with AdS/CFT and the wormhole solution

. . .

p

(c) The fuzzball construction has captured many corners of the space of all states …

The small corrections theorem then suggests that all states are fuzzballs …

(d) The conjecture of fuzzball complementarity says that we can obtain the traditional black hole as an effective geometry for processes … E � T

Conjecture of fuzzball complementarity: The dynamics of the fuzzball surface maps approximately to the dynamics of infall into the interior of the traditional hole …

(e) The AMPS argument has a hole so that it does not rule out fuzzball complementarity …

(1) They do not let the stretched horizon respond before the quantum hits it …

(2) They do not consider the approximation to obtain a complementarity pictureE � T

Nf

Ni=

eS(M+E)

eS(M)=

eS(M)+�S

eS(M)= e�S ⇡ e

ET � 1

(Most of the newly created degrees of freedom are unentangled with infinity)

(f) The underlying theme of black hole physics seems to be that the large number Exp[S] of fuzzball states violates the semiclassical approximation …

The path integral measure is comparable to the classical action …

� ⇠ f e�Sbek(M) ⇠ eSbeke�Sbek ⇠ 1

(SDM+Turton 2013)