The information paradox and the infall problem

Samir D. Mathur

Great Lakes, 2011

There are really three (interrelated) puzzles:

(A) The entropy problem: The entropy of a black hole is given by its surface area A. Is this entropy a count of states, like in usual Statistical mechanics ?

(B) The information paradox: Hawking radiation (E ~ kT) leads to a loss of quantum unitarity. How do we avoid this problem? (SERIOUS PROBLEM !!)

(C) The infall problem: What happens to an object (E >> kT) that falls into the black hole? What does an infalling observer feel ?

S=A/4G

√N − n

√n + 1 ≈

√N

√n + 1

dn

dt∝ (n + 1) n (175)

ωR =1

R[−l − 2 − mψm + mφn] = ωgravity

R (176)

m = nL + nR + 1, n = nL − nR (177)

|λ − mψn + mφm| = 0, N = 0 (178)

λ = 0, mψ = −l, n = 0, N = 0 (179)

ωI = ωgravityI (180)

|0〉 |ψ〉 < 0|ψ〉 ≈ 0 (181)

10

||Λ(1)||2 + ||Λ(2)||2 = 1 (82)

||Λ(2)|| < , 1 (83)

ρp =

Λ(1)|Λ(1) Λ(1)|Λ(2)Λ(2)|Λ(1) Λ(2)|Λ(2)

(84)

|Ψ → ξ(1)Λ(1) + ξ(2)Λ(2) (85)

Sp = (2 − 22) lne

(2 − 22)+O(3) < (86)

|ΨN+1 = |0cn+1 |0bn+1

1√2(Λ(1) + Λ(2)) + |1cn+1 |1bn+1

1√2(Λ(1) − Λ(2)) (87)

M mpl (88)

S ∼ A/G (89)

n1 n5 n1n5 Smicro = 2√2√n1n5 (90)

0|ψ ≈ 1 (91)

5

√N − n

√n + 1 ≈

√N

√n + 1

dn

dt∝ (n + 1) n (175)

ωR =1

R[−l − 2 − mψm + mφn] = ωgravity

R (176)

m = nL + nR + 1, n = nL − nR (177)

|λ − mψn + mφm| = 0, N = 0 (178)

λ = 0, mψ = −l, n = 0, N = 0 (179)

ωI = ωgravityI (180)

|0〉 |ψ〉 〈0|ψ〉 ≈ 0 (181)

10

The information paradox

(Commonly asked questions/confusions on home page: http://www.physics.ohio-state.edu/~mathur )

From (??) we are given that

||ψ2||2 = ψ2|ψ2 ≡ 21 < 2 (65)

|ψ1|ψ2| ≡ 2 < (66)

S(p) = (21 − 22) lne

(21 − 22)+O(3) < (67)

SN+1 > SN + ln 2− 2 (68)

ds2 = − (1− 2M

r)dt2 +

dr2

(1− 2Mr )

+ r2(dθ2 + sin2 θdφ2) (69)

r > 2M r < 2M t = constant r = constant (70)

4

r = 0 r = 2M

lab physics(nothing happensat the horizon)

r=0 horizon

correlated pairs

Older quanta move apart

initial matter⊗ |02|02 + |12|12

⊗ |0n|0n + |1n|1n

. . .

Leading order:Entangled state

⊗ |01|01 + |11|11

ΨM

Entangled state

If the black hole evaporates away,we are left in a configuration which cannot be described by a pure state

(Radiation quanta are entangled, but there is nothing that they are entangled with)

We can get a remnantwith which the radiationis highly entangled

‘lab physics’holds on these slices(Traditional horizon)

E = hν =hc

λ(38)

|ψ|H|ψ − ψ|Hs.c.|ψ| ≤ (39)

lp λ Rs (40)

2 (41)

|ξ1 = |0|0+ |1|1 (42)

|ξ2 = |0|0 − |1|1 (43)

SN+1 = SN + ln 2 (44)

S = Ntotal ln 2 (45)

|ψ → |ψ1|ξ1+ |ψ2|ξ2 (46)

||ψ2|| < (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2 (49)

S(A) = −Tr[ρA ln ρA] (50)

bN+1 cN+1 b c p = cN+1 bN+1 (51)

S(b+ p) > SN − (52)

S(p) < (53)

S(cN+1) > ln 2− (54)

S(b+ bN+1) + S(p) > S(b) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S(b+ bN+1) > SN + ln 2− 2 (57)

S(A+B) ≥ |S(A)− S(B)| (58)

lp Nαlp (59)

SN = S(b) (60)

A = b B = p = bN+1 + cN+1 (61)

A = b B = bN+1 C = cN+1 (62)

S(b+ bN+1) + S(p) ≥ S(b) + S(cN+1) (63)

ρp =

ψ1|ψ1 ψ1|ψ2ψ2|ψ1 ψ2|ψ2

(64)

3

Possibility A

|0k|0k + |1k|1k + δψ

||δψ|| <

State of created pair must be close toHawking’s leading order state

Can depend on anything inside the hole, but not on quanta that have already left

Each emission is only corrected ‘a little’, but these small corrections among N emitted quanta lead to an unentangled state ??

Possibility B

Order unity corrections to low energy evolution near the horizon

i.e., NO ‘lab physics in good slicing’

In this case there is no paradoxsince Hawking’s argument does not work

But we have to find a possible mechanism which can create such order unity corrections ...

Possibility A is removed by the following result:

(SDM arxiv 0909.1038)

Theorem : If we have a traditional horizon

Then no more than a fraction of the entanglement can be removed by the corrections

E = hν =hc

λ(38)

ψ|H|ψ ≈ ψ|Hs.c.|ψ+O() (39)

lp λ Rs (40)

2 (41)

3

E = hν =hc

λ(38)

|ψ|H|ψ − ψ|Hs.c.|ψ| ≤ (39)

lp λ Rs (40)

2 (41)

|ξ1 = |0|0+ |1|1 (42)

|ξ2 = |0|0 − |1|1 (43)

SN+1 = SN + ln 2 (44)

S = Ntotal ln 2 (45)

|ψ → |ψ1|ξ1+ |ψ2|ξ2 (46)

||ψ2|| < (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2 (49)

S(A) = −Tr[ρA ln ρA] (50)

bN+1 cN+1 b c p = cN+1 bN+1 (51)

S(b+ p) > SN − (52)

S(p) < (53)

S(cN+1) > ln 2− (54)

S(b+ bN+1) + S(p) > S(b) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S(b+ bN+1) > SN + ln 2− 2 (57)

S(A+B) ≥ |S(A)− S(B)| (58)

lp Nαlp (59)

SN = S(b) (60)

A = b B = p = bN+1 + cN+1 (61)

A = b B = bN+1 C = cN+1 (62)

S(b+ bN+1) + S(p) ≥ S(b) + S(cN+1) (63)

ρp =

ψ1|ψ1 ψ1|ψ2ψ2|ψ1 ψ2|ψ2

(64)

3

We need to use Strong Subadditivity of quantum entanglement entropy

E = hν =hc

λ(38)

ψ|H|ψ ≈ ψ|Hs.c.|ψ+O() (39)

lp λ Rs (40)

2 (41)

|ξ1 = |0|0+ |1|1 (42)

|ξ2 = |0|0 − |1|1 (43)

SN+1 = SN + ln 2 (44)

S = Ntotal ln 2 (45)

|ψ → |ψ1|ξ1+ |ψ2|ξ2 (46)

||ψ2|| < (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2 (49)

S(A) = −Tr[ρA ln ρA] (50)

bN+1 cN+1 b c p = cN+1 bN+1 (51)

S(b+ p) > SN − (52)

S(p) < (53)

S(cN+1) > ln 2− (54)

S(b+ bN+1) + S(p) > S(b) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S(b+ bN+1) > S) + ln 2− 2 (57)

S(A+B) ≥ |S(A)− S(B)| (58)

3

which does not have a very elementary derivation ...

Possibility B: We can look for alterations to Hawking’s vacuum state at the horizon

horizon

singularity

People wrote down the wave equation for scalars, gauge fields, gravitons ... Looked for solutions with

If find such solutions, then one would expect that the entropy comesfrom horizon fluctuations, and there would be no information problem

But no hair was found ... “no hair theorem”

??

Ylm, l = 1, 2, 3 . . .

’t Hooft: Ingoing quanta distort the outgoing vacuum modes

Susskind: Infalling matter stretches into long strings at the stretched horizon; Thermal emission from this string

Other attempts:

Large commutators around the horizon

Backreaction to Hawking emission

Other attempts at possibility B:

But there was no clear cut evidence for any of these effects ... they are based on large fluctuations in the Schwarzschild frame, but the effects go away in the Kruskal frame ...

Resolution of the problem:

We DO find hair, but

(a) We need all the structure of string theory to get it (do not get the hair in 3+1 dim canonically quantized gravity)

(b) The hair is nonperturbative (not given by linearized wave equation around usual black hole solution)

In string theory we get a new expansion scheme ... an expansion in the COMPLEXITY of the state

Consider the D1D5, D1D5P, D1D5PKK systems, construct microstates one by one (at the coupling where black holesare expected)

Infinite throat

horizon

singularity

Naive structure

Ψ = ψ(x)e−iωt (201)

L =1

2∂µφ∂µφ (202)

τ (203)

|ψ〉1 =1√2

(1.1|0〉b1 ⊗ |0〉c1 + 0.9|1〉b1 ⊗ |1〉c1) (204)

E = mc2 E = mc2 − GMm

rE ∼ 0 r ∼ GM

c2(205)

|Ψ〉 = [|0〉b1|0〉c1 + |1〉b1|1〉c1]⊗ [|0〉b2|0〉c2 + |1〉b2|1〉c2]

. . .

⊗ [|0〉b1|0〉c1 + |1〉b1|1〉c1] (206)

eiθ e−iθ (207)

c, !, G (208)

A

G∼ √

n1np ∼ Smicro (209)

R R + R2 (210)

Sbw =A

2G= 4π

√n1n2 = Smicro (211)

K3 × T 2 (212)

S = 2√

2π√

n1n5 (213)

AdS3 × S3 × T 4 (214)

11

All solutions are ‘capped’ Fuzzball

||Λ(1)||2 + ||Λ(2)||2 = 1 (82)

||Λ(2)|| < , 1 (83)

ρp =

Λ(1)|Λ(1) Λ(1)|Λ(2)Λ(2)|Λ(1) Λ(2)|Λ(2)

(84)

|Ψ → ξ(1)Λ(1) + ξ(2)Λ(2) (85)

Sp = (2 − 22) lne

(2 − 22)+O(3) < (86)

|ΨN+1 = |0cn+1 |0bn+1

1√2(Λ(1) + Λ(2)) + |1cn+1 |1bn+1

1√2(Λ(1) − Λ(2)) (87)

M mpl (88)

S ∼ A/G (89)

5

A common confusion: We cannot write down all quantum corrections to all messy fuzzball states, so have we fully solved the problem?

String theory was incomplete withoutD-branes, so the S-matrix would be nonunitary

But this isNOT the black hole information problem(see notes)

There can be quantum correctionsto the state (and these have been checked for finiteness to first order,Giusto+SDM 05)

Dipole charges KK monopole and and anti-KKmonopole

quanta emitted byergoregion emission

If we take generic CFT states, then CFTemission rate equals Hawking emission rate

Thus Hawking emission arisesby wavefunctions leaking into the throat from the complicated state in the ‘cap’

ΓCFT = Γgravity exactly !! (Chowdhury + SDM 07, 08)

SDM+Lunin (2-charge)Balasubramanian, de Boer, Keski-vakkuri, Ross (maximal 2 charge)Lunin+Maldacena+Maoz (2-charge torus)Skenderis+Taylor (2 charge K3)SDM+Giusto+Saxena (3 charge maximal)Giusto+Saxena+Peet (4 charge)Bena+Warner (3-charge, 4-charge, large body of results)Balasubramanian+Gimon+Levi (4 charge)Denef, de Boer (Branes at angles, halo constructions)Jejjala+Madden+Ross+Titchener (nonextremal 3-charge)Bena+Warner+...Giusto+ ... (4 charge nonextremal)

|ntotal = (J−,total−(2n−2))

n1n5(J−,total−(2n−4))

n1n5 . . . (J−,total−2 )n1n5 |1total (5)

A

4G= S = 2π

√n1n2n3

∆E =1

nR+

1

nR=

2

nR

∆E =2

nR

S = ln(1) = 0 (6)

S = 2√

2π√

n1n2 (7)

S = 2π√

n1n2n3 (8)

S = 2π√

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

e2π√

2√

n1np

1 +Q1

r2

1 +Qp

r2

e2π√

2√

n1n5

w = e−i(t+y)−ikz w(r, θ, φ) (10)

B(2)MN = e−i(t+y)−ikz B(2)

MN(r, θ, φ) , (11)

2

= 2π√

n1n5(2

E

2mp) (54)

S = 2π√

n5(√

n1 +√

n1)(√

np +

np) (55)

= 2π√

n5(E

√m1mp

) (56)

S = 2π√

n1n5npnkk (57)

S = 2π√

n1n5nkk(√

np +

np) (58)

= 2π√

n1n5(E

√mpmkk

) (59)

S = 2π√

n1n5(√

np +

np)(√

nkk +√

nkk) (60)

∼ lp (61)

∼ n16 lp (62)

M9,1 →M4,1 × T 4 × S1 (63)

E/(2mkk) = 0.5 (64)

E/(2mkk) = 1.2 (65)

Lz ∼ [g2α4√n1n5np

V R]13 ∼ Rs (66)

∆S (67)

eS (68)

eS+∆S (69)

S = 2π

n1n5np(1− f) + 2π

n1n5npf(√

nk +√

nk) (70)

nk = nk =1

2

∆E

mk=

1

2Dmk(71)

D ∼ [

√n1n5npg2α4

V Ry]13 ∼ RS (72)

∆S = S − 2π√

n1n5np = 1 (73)

S =A

4G(74)

mk ∼G5

G24

∼ D2

G5(75)

D ∼ G135 (n1n5np)

16 ∼ RS (76)

Nα lp (77)

5

Size of brane bound states in string theory grows with coupling, order horizon size

(SDM 97)

We can now resolve the information paradox by pointing tothe step in Hawking’s argument that goes wrong ...

Traditionally, we took a spherically symmetric ansatz ...

ds2 = −f(r)dr2 + g(r)dr2 + r2dΩ2 + ds2c

getting a unique solution with horizon and singularity

But actual states in string theory have the compact directions fibered over the angular ones, creating ‘dipole charges’ (like KK monopoles and antimonoples) ... the spherically symmetric solution is not realized

Dipole charges heldapart by fluxes

The infall problem

??

We can now use the structure of microstates to conjecture what can happen to a collapsing shell

Consider the amplitude for the shell to tunnel to a fuzzball state

Amplitude to tunnel is very small

But the number of states that one can tunnel to is very large !

Toy model: Small amplitude to tunnel to a neighboring well, but there are a correspondingly large number of adjacent wells

In a time of order unity, the wavefunction in the central well becomes a linear combination of states in all wells (SDM 07)

ttunnel = tdephase tevap

For the black hole

(SDM 07,08)

Is there any role of the standard Penrose diagram ?

We use notions from Israel, Maldacena, Van Raamsdonk(also computations of Arnold+Vaman et al, Skenderis+Van Rees)

Together with the structure of fuzzballs to arrive at a conjecture

(SDM+Plumberg 2011)

Euclidize, 1-loop path integral, saddle point

(1)

(II)

The Minkowski vacuum for a scalar field can be written an an entangled sum of Rindler states

|0M =

k

e−Ek2π |EkL REk|

Interacting, highly excited state of scalar theory, ‘Ending on right side’

For the gravitational field, whatare the corresponding R,L states ?

The fuzzball states we make have exactly this structure ...

|0M =

k

e−Ek2π |EkL REk|

= ⊗

Thus we expect :

ψk ψk

k

e−τhk−τ hk ⊗ =

ψk=

k

e−τhk−τ hk ⊗

=

Actual structure of state whencontinued into the interior ofthe hole

‘Sewing’ process in CFT

Sum over a set of messy entangled geometries equals a smooth geometry

(a) Low energy dynamics (E ∼ kT )

No horizon, radiation from ergoregions, so radiation like that from any warm body

no information loss since radiation depends on choice of microstate

(b) Correlators in high energy infalling frame (E kT )

ψk

ψk

for generic states ψk

⊗

mψm ψm

ψk|O1O2|ψk ≈

m

e−βEmψm|O1O2|ψm

≈

Summary

E = hν =hc

λ(38)

|ψ|H|ψ − ψ|Hs.c.|ψ| ≤ (39)

lp λ Rs (40)

2 (41)

|ξ1 = |0|0+ |1|1 (42)

|ξ2 = |0|0 − |1|1 (43)

SN+1 = SN + ln 2 (44)

S = Ntotal ln 2 (45)

|ψ → |ψ1|ξ1+ |ψ2|ξ2 (46)

||ψ2|| < (47)

SN+1 < SN (48)

SN+1 > SN + ln 2− 2 (49)

S(A) = −Tr[ρA ln ρA] (50)

bN+1 cN+1 b c p = cN+1 bN+1 (51)

S(b+ p) > SN − (52)

S(p) < (53)

S(cN+1) > ln 2− (54)

S(b+ bN+1) + S(p) > S(b) + S(cN+1) (55)

S(A+B) + S(B + C) ≥ S(A) + S(C) (56)

S(b+ bN+1) > SN + ln 2− 2 (57)

S(A+B) ≥ |S(A)− S(B)| (58)

lp Nαlp (59)

SN = S(b) (60)

A = b B = p = bN+1 + cN+1 (61)

A = b B = bN+1 C = cN+1 (62)

S(b+ bN+1) + S(p) ≥ S(b) + S(cN+1) (63)

ρp =

ψ1|ψ1 ψ1|ψ2ψ2|ψ1 ψ2|ψ2

(64)

3

Traditional horizon

Fuzzball

Theorem: If we do not have fuzzballs then we must haveinformation loss/remnants.

We have to examine states in the theory to see if they give the traditional horizon or fuzzballs.

There exists a region around thehorizon where the laboratory limit is achieved (there exists a foliation, low energy Hilbert space)

There is no such region

Corollary: Cannot use AdS/CFT to argue away the paradox

Define gravity to be dual to unitary CFT

(a) No black hole, e.g. matrix model

( tstate ∼ tcrossing )

c = 1

(b) Information leaks slowly from remnant

(e.g. Unitary completion of CGHS modelby Ashtekar et. al)

( tevap M3 )

(c) Order unity corrections at horizon (Fuzzballs): Can get unitary evaporation through Hawking radiation

(E kT ) ψk|O1O2|ψk ≈

m

e−βEmψm|O1O2|ψm

⊗

mψm ψm

≈

(A)

(B)

(E ∼ kT ) Solves information problem

(Infall problem)

How long does this tunneling process take ?

If it takes longer than Hawking evaporation time then it does not help ...

ψL = ψS + ψA

ψ = e−iEStψS + e−iEAtψA

∆E = EA − ES

∆t =π

∆E

1

ψL = ψS + ψA

ψ = e−iEStψS + e−iEAtψA

∆E = EA − ES

∆t =π

∆E

1

The wavefunction tunnels to the other well in a time

ψL = ψS + ψA

ψ = e−iEStψS + e−iEAtψA

∆E = EA − ES

∆t =π

∆E

1

where

Tunneling in the double well:

For the collapsing shell ...

Thus the collapsing shell turns into a linear combination of fuzzball states in a time shortcompared to Hawking evaporation time

Note that the theorem only requires that light modes(E ~ kT) be affected by orderunity

(How microstates differ from each other)

Motion of heavy objects(E >> kT) over the crossing timescale may be effectively given by a traditional black hole geometry (classical correspondence theorem ?)

(How microstates can be effectively similar for statistical processes)(e.g. Balasubramanian, de Boer, Jejjala, Simon 05, 08, SDM 07 )

Informationparadox

Infall problem

Does AdS/CFT solve the information puzzle ?

No !! It is a circular argument (but a very common belief)

AdS-SchwarzschildBlack holewith horizon

Unitary gauge theory

Black hole evaporation must be Unitary ?

String theorist : I have checked AdS/CFT in so many cases. So it is reasonable to assume it is true. Since gravity is dual to a Unitary CFT, information must come out

Wrong ! Quantum mechanics was also checked in so many cases in the lab. My claim is that Unitarity is violated only when you make black holes.

If you write the AdS/Schwarzscild black hole metric, I will use my theorem and prove that you lose Unitarity, and so you also lose string theory, AdS/CFT, etc as a corollary. So you must find the flaw in my theorem !

String theorist : I have another idea: I will define my gravity theory as dual to the CFT. It works for simple processes, so should be a reasonable theory overall

Wont help ! Theory will be Unitary, but can you say which of these will happen?

(i) Black hole does not form (e.g. 1+1 string theory) (ii) Information comes out in radiation: But for this you have to find a flaw in my theorem !

(iii) Information leaks slowly from remnant (e.g. Canonical gravity)