Proof that fuzzballs are essential for Unitarity. (Extending Hawking s computation to a Theorem, and noting its consequences) Samir D.

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1 Proof that fuzzballs are essential for Unitarity (Extending Hawking s computation to a Theorem, and noting its consequences) amir D. Mathur The Ohio tate University

2 The information problem Hawking radiation This process violates Quantum Mechanics!! (Hawking 74)

3 Why are folks not more worried? Common belief: (a) The horizon is just like any other gently curved piece of vacuum spacetime; thus low energy matrix elements here are very close to their semiclassical values (b) Hawking s computation was a leading order computation using this semiclassical evolution. There will always be tiny, delicate corrections to his evolution because of subtle quantum gravity effects (c) The hole emits a very large number of photons. Thus tiny corrections to the leading order evolution can encode the information of the hole as delicate correlations among the photons. In this way information comes out in the Hawking radiation

4 ource of this belief: Burn a piece of paper efactthat r = h implies that Eqs. (5)-(7) give There are subtle correlations between the emitted photons m ln n I n,m = ln m n 1 mn m k=m1 1 n ln m ln n k m. (8) information, along with the average entanglement entropy m,n = plotted versus the subsystem thermodynamic entropy ln m in Fig. 1 mn = 91600, whose 105 integer divisors are taken to be the values of Entanglement entropy between photons and paper first goes up, then comes down (Page 93) Entanglement entropy Information Overall radiation state at end is of course a pure state Why should nt small corrections to Hawking s computation do the same task, and save unitarity? Thermodynamic entropy

5 Question: If small corrections to Hawking s computation could get information out in correlations, why has no one given a conclusive illustration of the mechanism? This talk: The conventional belief is Wrong (DM: arxiv: ) ot similar! Theorem : If we have a traditional horizon Then no more than a fraction ψ H ψψ H s.c. ψ of the information can l p λ R s come out in the Hawking radiation

6 Plan: (a) ummarize Hawking s computation, make assumptions precise (for details, see Hawking 74, DM 08, 09) (b) Allow small corrections to radiation process, Prove `Theorem that such corrections cannot lead to information recovery in Hawking radiation (will focus on the core inequality, see reviews for other details) Proof uses trong ubadditivity theorem from Quantum information theory, so maybe not directly in our intuitive thinking... (c) Discuss how string theory gives order unity corrections at the horizon (fuzzballs), thus resolving the paradox (d) Reason for failure of semiclassical intuition even though the horizon is astronomically large (semiclassical geometry can be an effective description for E>>kT processes)

7 Hawking s derivation of radiation and why it gives information loss

8 Assumption 1 : There must exist a laboratory limit of the full quantum gravity theory, such that : In a bounded region of space and time, under suitable conditions, we can restrict to a finite set of states in describing low energy evolution. This evolution will not be sensitive to quantum gravity effects, and will be described to good accuracy by quantum fields on gently curved space. Lastly, the vacuum is unique; i.e. in a finite volume, energy eigenvalues do not have an accumulation point at E=0 (predictability). local, semiclassical evolution of low energy modes far away quanta have little effect ψ H ψψ H s.c. ψ l p λ R s

9 Assumption : List of suitable conditions (a) Intrinsic curvature of slices should be low (b) Extrinsic curvature of slices should be low (c) 4-curvature should be low (d) Energy and momentum densities on slice should be low, (e) Lapse and shift vectors should change smoothly

10 Hawking: The chwarzshild black hole geometry admits such good slices lab physics limit holds on these slices in this region (Traditional horizon) The infalling matter, and the created pairs, are all at low energy on the slice ψ H ψψ H s.c. ψ l p λ R s

11 tructure of the black hole ds = (1 M r )dt The black hole is described by the chwarzschild metric dr (1 M r ) r (dθ sin θdφ ) Crucial point about the black hole: For r>m the surface t = constant is spacelike For r < M the surface r = constant is spacelike

12 The spacelike slices in a schematic picture t=constant r=constant (no time-independent slicing possible) r=0 horizon

13 The Hawking process Entangled pairs Follow the wavemode from say 1 fm to 1 Km At 1 fm the mode must be in the vacuum state, else there would be a high energy density at the horizon (would violate traditional horizon assumption) At 1 Km we have particle pairs, with wavefunction the Hawking entangled state (Transplanckian physics not needed; bypassed by uniqueness of vacuum assumption)

14 Older quanta move apart correlated pairs Hawking state ξ 1 = (We will use a discretized picture for simplicity; for full state see e.g. Giddings- elson) initial matter light years r=0 horizon

15 Hawking s argument ξ 1 = ξ 1 = ξ 1 = = ln : Entanglement entropy after pairs have been created The radiation state (green quanta) are highly entangled with the infalling members of the Hawking pairs (red quanta) = total ln

16 Entangled state We can get a remnant with which the radiation is highly entangled 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)

17 Allowing small corrections to Hawking s computation: otations and some preliminaries

18 In this finite spacetime volume we can restrict to a finite number of states for O() accuracy Hawking state mall admixture of an orthogonal state ξ 1 = ξ = r=0 horizon

19 At step : (basis for initial matter and inside quanta) χ n (basis for outside quanta) ψ m total state basis change on inner and outer spaces Ψ = C mn ψ m χ n Ψ = i C i ψ i χ i r=0 horizon entanglement entropy = i C i ln C i

20 Leading order evolution (Hawking) χ i χ i ψ i ψ i ξ (1) quanta that have left are not affected, rest changes only by addition of a new Hawking pair Most general evolution to step 1 : ξ (1), ξ () χ i χ i ψ i ψ (1) i ξ (1) ψ () i ξ () ( ) ψ (1) i ψ () i =1 Traditional horizon The admixture of ξ () is small

21 Evolution of the state from timestep to 1: Ψ = C i ψ i χ i C i [ψ (1) i ξ (1) ψ () i ξ () ] χ i i i ξ (1) Λ (1) ξ () Λ () where Λ (1) = C i ψ (1) i χ i, Λ () = i i Λ (1) Λ () = 1 C i ψ () i χ i Having a `traditional horizon means that the evolution of low energy modes must be close to the Hawking evolution, so the coefficient of must be small. Thus we require Λ () <, 1 ξ ()

22 The proof

23 ummary of Goal To get information out, we need 1 < after some point Hawking s leading order result 1 = ln so entanglement keeps increasing We will prove 1 > ln so entanglement keeps increasing even with small corrections to Hawking s calculation Potential obstruction to proof: Each pair may be modified only a little bit, but there are a large number of quanta Perhaps the small correlation between newly created quanta and all the earlier quanta makes entanglement go down? OT TRUE!!

24 chematic notation c 1 p = {c 1 b 1 } {b b 1 From } traditional horizon: tate of new pair is weakly correlated with rest : (p) < {c} {b} Entropy at step : = ({b}) To prove: ({b} b 1 ) > ln

25 Basic tool : trong ubadditivity (Lieb Ruskai 73) A (A) =Tr[ρ A ln ρ A ] B D C Entanglement entropy of A with rest of system (BCD) { } { } (A B)(B C) (A)(C) o elementary proof is known... reasonable relation but not totally obvious...

26 c 1 Our final goal will be to show that ({b}, b ) > ", so that ξ = = ln = total ln ψ ψ1 ξ1 ψ ξ ψ < 1 < 1 > ln {c} (A) = T r[ρa ln ρa ] {b} n1 0(47) λ (44) ψ < = ln (44) = ln ψ < 1 entropy increases lp λ atrsthe timestep small corrections, the entanglement (49) 1 ln ψ H ψ ψ ({b} Hs.c. ψ p) > (39) (48) (40) (45) = total ln 1 < = total lnlp λr1 (45) < s T r[ρa ln ρa ] (50) Lemma L1: If (37) holds, then(p) the entanglement of the pair (cn < Lemma (1) 1: () (46) Λ Λ = 1 the system is bounded as 1 ψ ψ ξ ψ ξ ψ1 ξ1 ψ ξ (46) > ln (49) > ln ξ {c} p ={c1 (51) 1 = 1 b 1 } (41) )> Λ() (c <, 1 1 ln n1, bn1(47) (c )r[ρ trρ ρ(cn1,bn1 ) ln(50) ψ < (A) (47) ψ < (A) (cn1 ) < 1 = T r[ρ ln ρ ] ],b 1 n1 = T ln ρ A A A A (1) (1) (1) () p) > (5) 1 1 ξ1 = 1 1 (4) ξ ({b}) Λb 0 0 Λ Λ, Λ> = 0 0 ({b} ) (p) (c 1 1 ) ρp = (1) () (1) () () () (48) < Λ Λ Λ Λ < (48) Λ Λ =1 1 1 {b} {c}b 1p = {b c(53) }{c} (51) c {b} p = {b c } p) < c 1 = ln 1 Proof: The density for the() system (c b (43) ξ = matrix n1,() n1 ) is (1) (1) () Λ <, (49) 1 ξ Λ (49) ξ! Λ, = > ln 1 > ln Ψ " ln 1 total (1) (1) (1) () ) > ln (54) $Λ Λ Λ!(1) Λ $Λ(1) ΛΛ(44)!(1) Λ( 1 = ln ρ(cn1,bn1(1) = ψ ψ(1) ψ (50) ) () ξ 1 ξ() 1 (1) () ρ Λ =() $Λ p() ln ρ ] A) = T r[ρa ln ρa ] (A) = T r[ρ (50) Λ! $Λ Λ!() Λ( A A Λ = 1 = total ln computation) Λ Λ Λ (45) Proof: (Direct (p) > ({b}) (c 1 ) (55) ψ < () () ( ψ ψ ξ1{b ψ ξ < {c} pthat c 1 }, Ψ (51) 1= Λ From we}are given {b} {c}b 1p =c{b c{b} (51) ξ1(1) Λ(1) ξ(46) Λ 1 1 1(37) 1 < p = {b 1 c (1) () 1 Λ Λ = 1 (1) Λ() ψ < () (1) Λ(1) () () (47) Λ Λ > ln Λ = $Λ 1 Λ! "1 < " ρp = () () (1) 1 < Λ (48) 1 () Λ() Λ<, Λ Λ (A) = T r[ρa ln ρa ] (1) (1) Then by the chwartz (1) () 1inequality > ln (49) Λb 1 Λ c 1 Λ {b} Λ {c} p = {c 1 b ρp = () (1) () () () Λ Λ ΛΛ (A) =inequality T r[ρλ ] (1) (50) A ln ρa $Λ! "p) chwarz < ({b} > " () {c} p ξ=(1) {cλ(1) b ξ } Λ() 1 1 Ψ (p) < ow note that if we have a density matrix ({b} p) > e (c ) > ln b 1 c 1 {b} p = ( (p) < ) ln 1 3 (51) (5) O( ) < 1 ) ( $(p) $σ> ({b}) (53) I) ({b}ρ=b 1 α (c (c 1 ) > ln (A B) (B C) (A) (54) (C) then we can make a unitary transformation to bring it to the form ({b} b 1 ) (p) > ({b}) (c ) b 1 ) > ln (55) 1 ({b}

27 p = {c 1 b 1 } c 1 {b b 1 Lemma : (c 1 ) > ln Λ () <, Proof: (direct computation) { } {c} {b} Ψ 1 = 1 0 cn1 0 bn1 (Λ (1) Λ () 1 ) 1 cn1 1 bn1 (Λ (1) Λ () ) [ ] [ ρ cn1 = ( ( 1 Λ () ) (Λ (1) Λ () ) ) (Λ(1) Λ () ) (Λ (1) Λ () ) ( ) (c n1 ) = =ln [Re( Λ (1) Λ () )] ln ɛ O(ɛ 3 ) > ln ɛ

28 ({b} b 1 ) > ln 3 (A B) (A) (B) 3 ξ = = ln = total ln p = {b 1 c ψ ψ1 ξ1 ψ ξ ψ < 1 < {c} (A) = T r[ρa ln ρa ] {b} 1 > ln 1 e (p) (47) < 3(44) < ln 1 = ln ψ < (44) ψ 1= ) ln (p) = ( O( ({b} b ) > ln 1 ξ ψ ln (49) 1 ψ (A) = T r[ρ 1 1 ξ A ln ρa ] ) < (4 1 > ln > ln ψ 1 ) (1 (c 1)) > > ln (c ln 1 (A (48) = ln=b) T (45) (5 = total ln 1 < (45) (A) (B) < total 1 b c {b} {c} p = {b (A) r[ρ ln ρ ] ψ < T r[ρa ln ρa ] (50) c (A) = T r[ρ ln ρ ] A A A A > ln ({b} b 1 ) (p) > ({b}) (c ) Theorem: (p) 1 1 α ) ({b} b > ({b}) (c ) 1 1(46) l l ψ ψ ξ ψ ξ ψ1 ξ1 ψ ξ (46) > ln (49) p p 1 1 ({b} p) > 1 >(51) p = {b ln 1 c 1 bb {c} (5 b 1 } 1 } c 1 b{b} {c} c p }= {c 1 < 1 {c} p = {c (A 1 B) {b} (B C) (A) (C) (A B) (B C) (p) (A) (C) (47) (5 = ({b}) < ψ < ({b} p) > ψ < (A) (47) ρa ] (A) = > ({b} p)ln >ρln (50) = T r[ρa ln T ] A A (5) p) > 1br[ρ ({b} ) > ln 1 ({b} b 1 ) >(c ln> ln ) (48) (5 (p) < 1 (p) < < < (48) 1 1 {b} {c} c(a) }= b 1 pc=1{b 1 {b} pr[ρ =A{b T (A) ln ρ1 p) 1< c } (53) (A {c} B) (B) A ] c(51) b) (A) (B) ) (p) (c(5 (c ) (A > ln({b} (c B) 1 1ln 1 > > ({b}) (49) > ln > ln 1 (49) α 1 b 1 Proof: c 1 (54) {b}lp {c} αlp p = {c 1 b 1 } ) > ln ({b} b 1 )({b} (p) > ({b}) (c (5 lp(p) >l({b}) 1 p ) b ) (c ) 1 1 (50) A) = T r[ρ ln ρa ] (A) = T r[ρa ln ρa ] ({b} (50) = ({b}) p) > (p) > ({b})a (c ) (55) 1 = ({b}) (A B) (B C) (A) (C) b 1 pc=1{b 1{b} p = {ba (51) {b} {c} c 1{c} } (51) 1=c(p) 1 }< B {b} =p ({b}a = b{b} 1 ) > B ) =lnp A = {b} B1 =) b> C = c 1 1 (c ln 3{b} B) b (A) (B) A =(A B= 1 C = c 1 ({b}bb 1)) (p) (p) ({b} > ({b}) ({b})(c (c11) ) 1 ({b} b 1 ) (p) ({b}) (c 1 ) (A B) (B C) (A) (C)

29 Conclusion: All the time the black hole has a semiclassical horizon, the entanglement entropy (between the radiation and the hole) keeps rising Unlike burning paper, the entanglement does not start going down after the halfway point Thus when the black hole evaporates we have information loss/remnants Theorem : uppose that laboratory physics is obtained under the niceness conditions listed Then if a traditional horizon forms and persists for the duration M m pl then we necessarily have information loss/remnants (DM 08, 09)

30 Why is burning paper different? Let paper consist of atoms, spin up/down Outer atom can burn up to radiation, with same spin Atoms left inside can interact, reshuffle Outermost atom burns, with spin up radiation If 1-d space, no information in radiation -d space of states at interaction: Black hole has 1-d space of possibilities at creation point

31 Fuzzball solutions in string theory

32 Black hole hair horizon singularity People wrote down the wave equation for scalars, gauge fields, gravitons... Looked for solutions with L=1,, 3,... If they had found such solutions, then one would expect that the entropy comes from horizon fluctuations, and there would be no information problem But no hair was found...

33 In string theory we have extra dimensions, which we will take to be compact circles If we want a black hole in 31 d, then we have 6 compact directions If we work PERTURBATIVELY, the extra directions give gauge fields and scalars... A a But we dont get any hair from scalar or vector fields either... We will find that OPERTURBATIVE use of these compact directions, (and the other brane content of string theory), will give the hair...

34 tring theory: How does the size of the brane bound state grow with (a) coupling (b) number of branes?? 3-charge extremal hole: Estimate size of brane bound state Due to fractionation find a size that grows with the number of branes D n1 n 5 n p g α 4 D [ ] 1 3 R V R (DM 97) Infinite throat uggests that we may not get the traditional horizon in string theory horizon singularity

35 Definition: Traditional horizon There exists a region around the horizon where the laboratory limit is achieved (there exists a foliation, low energy Hilbert space) ψ H ψψ H s.c. ψ l p λ R s Fuzzball There is no such region (Fuzzballs are not classical solutions or supergravity solutions; in general they will be very quantum and stringy) The theorem does not prove that our theory has fuzzballs; it only proves that if we do not have fuzzballs then we must have information loss/remnants. We have to examine states in the theory to see if they give the traditional horizon or fuzzballs.

36 Information paradox Infall problem ote that the theorem only requires that light modes (E ~ kt) be affected by order unity (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, imon 05, 08, DM 07 )

37 How do we show more concretely that microstates can be fuzzballs? tring theory gives us a new expansion: an expansion in Complexity -charge extremal D1D5: All states in the brane bound state can be described in terms of different ways of connecting up the strands of an effective string n 1 D1 branes Effective string winding number n 5 D5 branes n 1 n 5 Entropy micro = n 1 n 5

38 Ad 3 3 T 4 (Cvetic,Youm 95 Balasubramanian, de Boer, Keski-Vakkuri, Ross 00 Maldacena, Maoz 00) All solutions are capped (Lunin, DM 01 Lunin, Maldacena Maoz 03 kenderis Taylor et al...) Fuzzball A/G (LuninDM 0)

39 3-charge extremal D1-D5-P (trominger-vafa hole), 4-charge hole... aive geometry Huge class of capped solutions correct order of entropy (Giusto, DM, axena 04) (Bena,Warner et al: Long program Balasubramanian,Gimon,Levi, de Boer..., Denef,)

40 General result of the microstate program : Traditional spherically symmetric solution not realized in string theory, microstates break symmetries Can find a large class of solutions (nonsingular) with same quantum numbers but without the symmetries of the traditional hole. (Limit towards generic solutions with be very quantum, stringy in general) g 0 g nonzero dipole charges appear, their complicated arrangements describe brane bound state 0 ψ 0 ψ 1 0 ψ 0 0 ψ 0

41 The on-extremal Hole : (Jejalla, Madden, Ross Titchener 05)???? D1-D5 CFT has both left and right moving excitations Many other non-bp solution recently (Bena-Warner et al, other groups)

42 ds f = (dt dy ) H1 H5 M H1 H5 (s p dy c p dt) ( r H dr ) 1 H5 (r a 1 )(r a dθ ) Mr ( H1 H5 (a a 1 )( H 1 H ) 5 f) cos θ cos θdψ H1 H5 ( H1 H5 (a a 1 )( H 1 H 5 f) sin θ H1 H5 ) sin θdφ M H1 H5 (a 1 cos θdψ a sin θdφ) M cos θ H1 H5 [(a 1 c 1 c 5 c p a s 1 s 5 s p )dt (a s 1 s 5 c p a 1 c 1 c 5 s p )dy]dψ Q 1 = gα 3 V n 1 Q 5 = gα n 5 Q p = g α 4 V R n p (Jejalla, Madden, Ross Titchener 05) M sin θ H1 H5 [(a c 1 c 5 c p a 1 s 1 s 5 s p )dt (a 1 s 1 s 5 c p a c 1 c 5 s p )dy]dφ H1 H 5 4 i=1 dz i H i = f M sinh δ i, f = r a 1 sin θ a cos θ, Q 1 = M sinh δ 1 cosh δ 1, Q 5 = M sinh δ 5 cosh δ 5, Q p = M sinh δ p cosh δ p

43 Hawking radiation Unitary radiation process in CFT on-unitary radiation from semiclassical gravity Radiation rates agree (pins, greybody factors...) (Callan-Maldacena 96, Dhar-Mandal-Wadia 96, Das-Mathur 96, Maldacena-trominger 96) Can we get UITARY radiation (information carrying) in the GRAVITY description?? Let us start with the simplest microstates...

44 As in any statistical system, each microstate radiates a little differently Γ CF T = V ρ L ρ R Γ CF T = V ρ L ρ R Emission vertex Occupation numbers of left, right excitations Bose, Fermi distributions for generic state Occupation numbers for this particular microstate Emission from the special microstate is peaked at definite frequencies and grows exponentially, like a laser...

45 Hawking radiation from the special microstate Emission grows exponentially because after n de-excited strings have been created, the probability for creating the next one is Bose enhanced by (n1) The emitted frequencies are peaked at ω CF T R = 1 R [l m ψm m φ n] m = n L n R 1, n = n L n R Emission grows as Exp[ω CF T I t]

46 Gravity description of emission : This gravity solution has no horizon, no singularity, but it has an ergoregion (all non-exremal states made so far are either time-dependent or have an ergoregion) ω = ω gravity R iω gravity I (Cardoso, Dias, Jordan, Hovdebo, Myers, 06) egative energy quanta collect in the ergoregion, positive energy quanta radiated to infinity

47 Radiation: The gravity calculation M 9,1 M 4,1 T 4 1 Graviton with indices on the torus is a scalar in 6-d h 1 { Ψ Ψ = 0 M 4,1 t, r, θ, ψ, φ 1 y y : (0, πr) Ψ = exp(iωt iλ y R im ψψ im φ φ)χ(θ)h(r) olve by matching inner and outer region solutions

48 ( ) The real part of the frequency gives the energy of the radiated quanta ω ω R = 1 R (l m ψm m φ n λ m ψ n m φ m ( 1)) ( ) The imaginary part of the frequency gives the exponential growth rate of the perturbation ω I = 1 R ζ λ m ψ n m φ m 0 order; t ( π [l!] ] l1 [(ω λ R )Q 1Q 5 l1 4R C l1 ζ l1 C l1) R radius of 1 l, m φ, m ψ Thus the gravity emission is also characterized by a set of complex frequencies λ angular momenta momentum along 1

49 One finds : ω CF T R ω CF T I = ω gravity R = ω gravity I (Chowdhury DM 07, 08) Thus for a set of (nongeneric) microstates we can explicitly see information carrying radiation which is the Hawking radiation for these microstates

50 pecial and generic states in gravity: conjecture Classical geometry, axial symmetry, standard ergoregion, enhanced emission tar cluster. Different stars have ergoregions with different orientations, so there is no axial symmetry in the emission A generic state is very quantum, with very shallow ergoregions, and quanta leak out slowly as Hawking radiation

51 (D) How does a collapsing shell become fuzzballs? (a genuine question)??

52 How does semiclassical intuition go wrong?

53 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!

54 Toy model: mall 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 (DM 07)

55 (E) How long does this tunneling process take? If it takes longer than Hawking evaporation time then it does not help... Tunneling in the double well: L = A ψ = e iet ψ e ieat ψ A = The wavefunction tunnels to the other well in a time t = π E where E = E A E π

56 For the collapsing shell... Thus the collapsing shell turns into a linear combination of fuzzball states in a time short compared to Hawking evaporation time

57 Large phase space at infinity Large phase space in black hole Wave - function Wavefunction spreads over large phase space (DM 07, de Boer...)

58 ummary

59 (A) Hawking paradox is very nontrivial (Theorem) To get information in Hawking radiation eeds a basic change in our understanding of the structure of the black hole eed O(1) corrections to the evolution of low energy (E~kT) modes at the horizon (B) Lesson from string theory: The extra excitations that complete 31 gravity create a large space of solutions that have no traditional horizon Fuzzball conjecture: In string theory, no energy eigenstate has a traditional horizon

60 Traditional horizon There exists a region around the horizon where the laboratory limit is achieved (there exists a foliation, low energy Hilbert space) ψ H ψψ H s.c. ψ l p λ R s Fuzzball There is no such region (Fuzzballs are not classical solutions or supergravity solutions; in general they will be very quantum and stringy) Theorem: If there exists any eigenstate with a traditional horizon, we will necessarily have information loss/remnants

61 (C) The Hawking paradox arises because we could not find L=1,,3 hair... ince we are resolving a paradox, we do not need to find all solutions with all quantum gravity corrections... we just need to find a flaw in the Hawking evolution process? tring theory microstates show how we get data at the horizon... 0 ψ 0 ψ 1 0 ψ 0 0 ψ 0