PPrepared for submission to JHEP
Quantum tasks require islands on the brane
Alex May a and David Wakeham a a Department of Physics and Astronomy, University of British Columbia 6224 Agricultural Road,Vancouver, B.C., V6T 1W9, Canada
E-mail: [email protected] , [email protected] Abstract:
In recent work, it was argued that quantum computations with inputs andoutputs distributed in spacetime, or quantum tasks, impose constraints on entanglementin holographic theories. The resulting constraint was named the connected wedge theoremand can verified by a direct bulk proof using focusing arguments in general relativity. Inthis article we extend this work to the context of AdS/BCFT, where an end-of-the-worldbrane is present in the bulk. By considering quantum tasks which exploit informationlocalized to the brane, we find a new connected wedge theorem. We apply this theorem tobrane models of black holes, where it relates the formation of islands in the Ryu-Takayanagiformula to causal features of the ambient spacetime. In particular, we find that if the blackhole interior is causally connected to the radiation system through the ambient spacetime,then an island forms. For constant tension branes in pure AdS the converse also holds. a r X i v : . [ h e p - t h ] F e b ontents → → → A Details for AdS calculation 33B Coordinate systems and embedding space 36
In this article, we prove a theorem relating minimal surfaces and causal features of asymp-totically AdS spacetimes which are ended by branes. Such geometries are relevant to theemergence of spacetime [1–3], holographic approaches to cosmology [4–6], and the blackhole information problem [7–14], where they model the formation of islands. In the island– 1 –ontext, our theorem establishes that a causal connection from the black hole interior tothe radiation system implies the existence of an island.Our work is motivated by the operational perspective on AdS/CFT initiated in [15]and elaborated in [16, 17]. In particular, the authors of [15–17] considered a quantum com-putation with inputs given at two boundary spacetime locations and outputs at two otherboundary locations. Considering this computation from a bulk and boundary perspectiveleads to the connected wedge theorem , a relationship between causal features of the bulkgeometry and boundary entanglement. It was then possible to prove this theorem usingtools from general relativity.Here we prove a similar result that applies specifically to the context of AdS spacetimesended by branes. Such spacetimes are described by a manifold with boundary, along witha Lorentzian metric. The metric satisfies Einstein’s equations along with a boundarycondition set at the brane. Holographically, such spacetimes are proposed to be dual toconformal field theories with a boundary [18, 19]. We adapt the theorem of [17] to thissetting. Our theorem is motivated again by an operational perspective on AdS/CFT,but involves considering quantum computations with one input location and two outputlocations. Additional information involved in the computation is localized to the brane.To distinguish our result from the earlier one we refer to it as the 1 → → → C and two ‘output’ regions ˆ R , ˆ R . Note that we use hatted letters to denote regions inthe boundary. We choose ˆ R , ˆ R such that they touch the brane. The theorem is statedin terms of two additional spacetime regions constructed causally from ˆ C , ˆ R , ˆ R .The first region is denoted ˆ V and called the decision region . It is defined byˆ V ≡ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) . (1.1)Here ˆ J ± ( ˆ X ) denotes the future and past of a region ˆ X taken in the boundary geometry.We will restrict our attention to choices of region ˆ C i , ˆ R i such that ˆ C i ⊆ ˆ V i .The second region is called the entanglement scattering region and is defined in thebulk spacetime. To define it, denote the entanglement wedge of a boundary region ˆ X by X , so that X = E W ( ˆ X ). Further, denote the future and past of a region taken in the bulkgeometry by J ± ( X ). Then the entanglement scattering region is defined by J E → ≡ J + ( C ) ∩ J − ( R ) ∩ J − ( R ) ∩ B , (1.2)where B denotes the end-of-the-world brane. This and definition 1.1 are illustrated in figure1. Our main result is as follows. Theorem 1 ( → connected wedge theorem) Consider three boundary regions ˆ C , ˆ R , ˆ R in an asymptotically AdS spacetime with an end-of-the-world brane. Require that ˆ C ⊆ The “region based” connected wedge theorem appearing in [17] is stronger than, and contains as aspecial case, the earlier theorem appearing in [15, 16]. – 2 – V ˆ R ˆ R J E → (a) ˆ V ˆ R ˆ R (b) Figure 1 : Illustration of Theorem 1, shown with a zero tension brane. The input regionis taken to be a point C = c , while the output regions are the light blue half diamondsattached to the edge. The decision region ˆ V is shown in black. a) When a boundary point c and two edge points r , r have a bulk scattering region which intersects the brane, theentanglement wedge of an associated domain of dependence (black shaded region) attachesto the brane. b) When there is no such scattering region, the entanglement wedge neednot be connected.ˆ V , and that ˆ R , ˆ R touch the brane. Then if J E → is non-empty, the entanglement wedgeof ˆ V is attached to the brane. Note that in some cases ˆ V may attach to the brane in the boundary, in which case thetheorem is trivially true. The converse to this theorem does not hold, and we give anexplicit example in the main text.To motivate our theorem, consider the following scenario. Suppose some classicalinformation q is encoded in the brane, either in the choice of boundary state or in brane-localized degrees of freedom. We leave unspecified at this stage in the argument where thiscorresponds to q being localized in the boundary. Alice, an observer, receives a quantumstate H q | b (cid:105) in region ˆ C . H is the Hadamard operator, so if q = 0 this is one of the states | (cid:105) , | (cid:105) and if q = 1 this is one of the states | + (cid:105) , |−(cid:105) . Without knowing q , Alice is notable to measure in the correct basis and learn b . However, Alice’s goal is to bring b to tworegions ˆ R and ˆ R , which will be attached to the CFT edge.– 3 –ausality requires that Alice can succeed in her task only when q is stored in the patchof spacetime formed from the overlap of the past of ˆ R , ˆ R (since she needs to send b to bothoutput regions) and the future of ˆ C (since she needs the input H q | b (cid:105) ). We can considerthis overlap in either the bulk or the boundary perspective. In the boundary we considerthe future or past of the relevant boundary regions, ˆ V = ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ J − ( ˆ R ). Inthe bulk perspective it is appropriate to consider the future or past of the correspondingentanglement wedges, V = J + ( C ) ∩ J − ( R ) ∩ J − ( R ). In either case if the overlap contains q , Alice can complete her task.When the bulk overlap V intersects the brane it contains q . This is the just thestatement that J E → = V ∩ B is non-empty. Then in the bulk picture Alice can completeher goal of bringing b to R and R . Meanwhile, for a holographic BCFT, entanglementwedge reconstruction tells us that the information stored in ˆ V is geometrized as bulkdegrees of freedom in its entanglement wedge. Thus the entanglement wedge of ˆ V shouldinclude the brane whenever J E → is non-empty, which is the claim of the theorem.In earlier work [15, 16], the case where input and output regions consisted of singlepoints was considered. For the applications of our theorem to islands discussed in section6 a point based version of Theorem 1 is appropriate. Partly this is because the statementin terms of regions is equivalent to the point based statement in the setting of pure AdSwith an ETW brane. When not considering pure AdS spacetimes however the region basedstatement is stronger, so we have included the statement and proof of the more generaltheorem.Notice that if we choose ˆ C i such that ˆ C i = ˆ V i the theorem is not useful, in that findingthe entanglement scattering region already involves determining the entanglement wedge ofˆ V i . In this case however we can consider the minimal extremal surface which is not attachedto the brane, call it γ (cid:48)V , and define the region W [ γ (cid:48)V ] whose boundary is γ (cid:48)V ∪ V . Notethat W [ γ (cid:48)V ] is contained within the true entanglement wedge W [ γ V ], which in general maybe larger. In fact it will be larger whenever the minimal extremal surface γ V is attachedto the brane. Then Theorem 1 can be used to conclude that if( J (cid:48) ) E → ≡ J + ( W [ γ (cid:48)V ]) ∩ J − ( R ) ∩ J − ( R ) ∩ B (cid:54) = ∅ (1.3)then W [ γ (cid:48)V ] will not be the full entanglement wedge, and instead it is the brane attachedextremal surface which is minimal. Outline of paper
We now give a summary of this paper.In section 2 we review the AdS/BCFT correspondence, which gives a holographic dualdescription of asymptotically AdS spacetimes with ETW branes.In section 3, we give the quantum information based argument for the 1 → { ˆ C , ˆ R , ˆ R } as the spacetimelocations for inputs and outputs to a quantum computation. We consider a particular such‘distributed quantum computation’, or quantum task , in order to argue for the necessity oflarge correlation between the decision region ˆ V and the brane. In brief, completing the task– 4 –ill be possible in the bulk perspective whenever J E → is non-empty, while completing it inthe boundary will be possible whenever region ˆ V knows information stored on the brane.We argue the task being completed in the bulk implies it is completed in the boundary. Toensure this is possible, a portion of the brane must be in the entanglement wedge of ˆ V .In section 4, we prove the 1 → γ V enclosing the entanglement wedge of ˆ V takes on the brane-detached configuration. Then, we show that this leads to the existence of a surface calledthe null membrane, which connects ˜ γ V to a brane-attached surface of less area, so thatthe candidate surface ˜ γ V cannot have been the correct one.In section 5, we study bulk gravity solutions in 2 + 1 dimensions that have a constanttension brane and are locally pure AdS. We verify the theorem explicitly for those solutionsby comparing calculations of the entanglement entropy with features of bulk null geodesics.We also find that the converse to the theorem holds for these solutions.In section 6, we take up the discussion of islands. We study in particular islandformation in BCFT models of black holes, following the set-up of [7] closely (see also [8–14]). In that context the brane is the black hole, the CFT boundary is the dual quantummechanical description of the black hole, and the CFT is the bath system, into whichinformation from the black hole may escape. We apply the time reversed statement ofTheorem 1 to this setting, and take the input and out regions to be points ˆ C = c ,ˆ R = r , ˆ R = r . In doing so we find that the point c , now at some late time, controlsthe time for which Hawking radiation has been collected from the black hole. Meanwhilethe points r and r define the black hole event horizons. Moving c to gradually latertimes, and so collecting more Hawking quanta, the scattering region J E → opens, whichnow corresponds to c coming into causal contact with the black hole interior. Theorem 1then tells us that this forces the entanglement wedge of the radiation system to include aportion of the black hole interior. We illustrate this in figure 2. In section 7 we conclude with some open questions and remarks.
The 1 → While in general this implication runs in only one direction, for the constant tension solutions of section5 we find that the island forms if and only if the black hole interior is causally connected to the radiationsystem. – 5 – r r (a) c r r (b) Figure 2 : Theorem 1 along with time reversal implies a 2 → r and r as defining the horizons of a black hole.The region ˆ V is then the radiation system. (a) When a light ray reaches ˆ V from theblack hole interior, the entanglement wedge of ˆ V must connect to the brane, so that ˆ V reconstructs a portion of the interior. (b) When the black hole is causally disconnectedfrom the black hole interior, the entanglement wedge of ˆ V may be disconnected from thebrane.AdS region along with an extension of the CFT boundary into the bulk as an end-of-the-world (ETW) brane. To avoid confusion with the bulk-boundary language of the AdS/CFTcorrespondence, we will refer to the CFT boundary as the edge . The bulk spacetime andbrane are described by an action I bulk + I brane = 116 πG N (cid:90) d d +1 x √ g ( R −
2Λ + L matter )+ 18 πG N (cid:90) B d d y √ h ( K + L B matter ) , (2.1)where L matter and L B matter are matter Lagrangians for fields in the bulk and brane respec-tively. As usual, R is the Ricci curvature and Λ the bulk cosmological constant, while K is the trace of the extrinsic curvature of the brane, K ab = ∇ a n b , (2.2)– 6 –or outward normal n j to B , and a, b refer to brane coordinates y a . This action leads toEinstein’s equations in the bulk, along with the boundary condition − πG N ( K ab − Kh ab ) = T B ab . (2.3)In AdS/BCFT, the Ryu-Takayanagi formula [20] and its covariant generalization theHRT formula [21] continue to calculate the entropy of boundary subregions, provided thehomology condition is appropriately adapted [22]. In the context of AdS/CFT, and assum-ing the null energy condition, the HRT formula is equivalent to the maximin formula [23].We will assume this remains the case in AdS/BCFT. The maximin formula states that, toleading order in 1 /G N , S ( A ) = max Σ min γ A (cid:18) Area[ γ A ]4 G N (cid:19) . (2.4)The maximization is over Cauchy surfaces that include A in their boundary, and theminimization is over spacelike codimension 2 surfaces γ A which are homologous to A .We will refer to the surface γ A picked out by such a procedure, whose area computes theentropy, as an entangling surface . In spacetimes with an ETW brane we should understandthe homology constraint as ∂S = γ A ∪ A ∪ b (2.5)for S a spacelike codimension 1 surface in the bulk, and where b is allowed to be any portionof the ETW brane. For a single interval in the CFT, this allows two qualitatively distinctclasses of entangling surface: those which do not include a portion of the brane to satisfythe homology constraint, which we call brane-detached, and those which do, which we callbrane-attached (see figure 1).In the BCFT description there are degrees of freedom which live at the edge and areassociated with the choice of boundary condition. At least for constant, large tensionbranes these edge degrees of freedom are dual in the bulk to degrees of freedom living onthe brane [7, 24]. In this section we give the quantum tasks argument for Theorem 1. Several aspects ofthe argument follow the argument of the 2 → → → A quantum task is a quantum computation which has its inputs and outputs at specifiedspacetime locations. We will consider tasks which have inputs and outputs recorded into– 7 – q | b (cid:105) ˆ V bbxt Figure 3 : The M task, which we employ to argue for the 1 → C the quantum system A is received which holds a state H q | b (cid:105) . For the task to becompleted successfully, b should be produced at both r and r . We show that completingthe task with a high success probability requires the bit q be available in the region ˆ V =ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ).extended spacetime regions. To understand this more precisely, we will say that a quantumsystem A is localized to region X relative to a channel M X if acting on X with M X produces A . If there exists a channel such that quantum system is localized to a region X relative to that channel, we say just that A is localized to X . If it is not possible to learnanything about A from X , we say A is excluded from X . For a review of quantum tasksas they are employed here, see [17].For our particular example, there is one input region ˆ C and two output regions ˆ R , ˆ R .System A is in one of the states H q | b (cid:105) A and is localized to region ˆ C . H is the Hadamardoperator, and b, q ∈ { , } . There is an additional system Q which holds the bit q , andwe leave unspecified for the moment where Q is located in spacetime. To complete thetask the bit b should be localized to ˆ R and ˆ R . We will momentarily leave the channels M ˆ C , M ˆ R , M ˆ R unspecified. This task is illustrated in figure 3, and we refer to it as the M task or “monogamy task”, for reasons that will become apparent.We will need to introduce an equivalent formulation of M that we refer to as purified M . The purified task is modified in two ways: (1) a second system ¯ Q is introduced, andplaced in the maximally entangled state with Q ; and (2) the input qubit H q | b (cid:105) A is replacedwith the A system of a maximally entangled state | Ψ + (cid:105) A ¯ A . We refer to the ¯ Q ¯ A systemas the reference system. Notice that Bob can now perform measurements on the referencesystem to return this to the original task. To do this, Bob first measures the ¯ Q system,and obtains some output q . Then, he measures ¯ A in the computational basis if q = 0,– 8 –nd in the Hadamard basis if q = 1. Bob obtains one bit b of output. Meanwhile, thepost-measurement state on QA is | q (cid:105) Q ⊗ H q | b (cid:105) A , so that the inputs are as in the unpurifiedtask. Alice’s success probability is unaffected whether Bob performs these measurementsbefore or after Alice returns her outputs, since the QA and ¯ Q ¯ A systems never interact.Thus, the purified and unpurified tasks have the same success probability.The three regions ˆ C , ˆ R , ˆ R have a naturally associated spacetime region which welabel ˆ V , defined according toˆ V ≡ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) . (3.1)and which we call the decision region . ˆ V is natural to consider because it is where it ispossible to act on A and reach both of ˆ R and ˆ R . We will in particular be interested intwo situations: (1) the setting where Q is localized to ˆ V and (2) the setting where Q isexcluded from ˆ V .Let us consider first the case where Q is localized to ˆ V . For convenience, take theunpurified task. Then within ˆ V Alice should apply H q to A to obtain ( H q ) | b (cid:105) A = | b (cid:105) A ,measure | b (cid:105) A in the {| (cid:105) , | (cid:105)} basis, and then send the outcome to each of r and r . Doingso, she can complete the task with high probability, say p suc ( M ) = 1 − (cid:15) . We introduce theparameter (cid:15) to account for the effect of any noise present in carrying out this protocol. We can make a stronger statement by introducing a parallel repetition of the monogamytask, which we call M × n . We consider n states { H q i | b i (cid:105)} i being input at ˆ C , with the q i and b i drawn independently and at random. To complete the task, a fraction 1 − δ of the b i should be localized to both ˆ R i . As discussed in the last paragraph, Alice can complete eachof the n runs with a probability p suc ( M ) = 1 − (cid:15) . For (cid:15) < δ , the probability that this leadsto more than a fraction 1 − δ of the runs being successful will be high. For concretenesstake δ = 2 (cid:15) . In this case we have, at large n , p suc ( M × n ) = 1 − (cid:15) n . (3.2)In particular we see that the success probability converges to 1 exponentially in n .Next, consider the case where q is excluded from ˆ V . More precisely, we considerpurified M and state this assumption as I ( ˆ V : ¯ Q ) = 0 . (3.3)Then Alice will be limited in her ability to complete the task, a fact we formalize in thefollowing lemma. Lemma 2
Consider the M task [cf. figure 3] with I ( ˆ V : ¯ Q ) = 0 . Then any strategy forcompleting the task has p suc ( M ) ≤ cos ( π/ . To see why this is true, consider that Alice holds the A subsystem of a maximally entangledstate on A ¯ A in the region ˆ V . After applying a quantum channel to A , she will send part One source of noise may be our assumption that Alice is working in a classical geometry. In theAdS/CFT context, at finite G N , it seems plausible that small errors are inevitable. – 9 –f the output, call it B , to ˆ R and part of the output, call it B , to ˆ R . At best, Alice willlearn Q in the regions ˆ R i . At each of the R i then she can use B i along with q to producea guess for b . This is exactly the guessing game analyzed in [25], known as the monogamyof entanglement game. The stated bound on success probability was proven there.Notice that if B and ¯ A are maximally entangled, Alice can measure in the q basis andproduce an output at ˆ R which is perfectly correlated with Bobs measurement outcome.Similarly if B ¯ A is maximally entangled she can produce the correct output at ˆ R . Themonogamy of entanglement however ensures that there will be a trade-off, and no perfectstrategy will exist. The proof in [25] makes this rigorous.We can also consider the parallel repetition of the task M × n in the case where I ( V :¯ Q ) = 0. Following the reasoning of Lemma 2, this can again be reduced to the guessinggame discussed in [25], who proved that this parallel repetition of the task satisfies thefollowing lemma. Lemma 3
Consider the M × n task with I ( ˆ V : ¯ Q ) = 0 , and require that a fraction − δ ofthe individual M tasks are successful. Then any strategy for completing the task has p suc ≤ (cid:16) h ( δ ) cos (cid:16) π (cid:17)(cid:17) n ≡ (cid:16) h ( δ ) β (cid:17) n (3.4) where h ( δ ) is the binary entropy function h ( δ ) ≡ − δ log δ − (1 − δ ) log(1 − δ ) and the secondequality defines β . For small enough δ we have that 2 h ( δ ) β <
1, so that with zero mutual information thesuccess probability is small. Our next result will be to show that a large success probabilityimplies a large mutual information.In fact, this argument was already completed in [17], albeit in a changed setting. Lemma 4
Suppose that the M × n task is completed with success probability p suc = 1 − (cid:15) n , where we deem the M × n task successful if a fraction − (cid:15) of the individual M tasksare. Then the bound I ( ˆ V : ¯ Q ) ≥ n ( − log 2 h (2 (cid:15) ) β ) − O (( (cid:15)/β ) n ) (3.5) holds. This will be the key technical result in the argument from quantum tasks for the connectedwedge theorem, which we present in the next section.We should highlight an important assumption made in proving Lemma 4. In additionto the region ˆ V , there is also the spacelike complement X = [ J + ( ˆ V ) ∪ J − ( ˆ V )] c . Lemma3, on which Lemma 4 relies, assumes that information from this region is not made use ofin Alice’s protocol. If it were, one could use protocols of the type considered in appendix Bof [16] to perform the M × n task without entanglement between ˆ V and ¯ Q . As discussed in[16], it seems sensible to assume such strategies are not allowed. In particular they requirelarge amounts of GHZ type entanglement in the CFT, which is not expected to exist [26]. This explains our naming convention M for the task. In particular the systems ˆ V and ¯ Q play the role of systems ˆ V and ˆ V discussed in [17]. Our Lemma4 is their Lemma 7 with this replacement made. – 10 – .2 Tasks argument for the → connected wedge theorem With Lemma 4 in hand, we are ready to complete the tasks argument for the 1 → Theorem 1:
Consider three boundary regions ˆ C , ˆ R , ˆ R in an asymptotically AdS spacetime with an end-of-the-world brane. Require that ˆ C ⊆ ˆ V , and that ˆ R , ˆ R touch thebrane. Then if J E → is non-empty, the entanglement wedge of ˆ V is attached to the brane. Argument.
Using our assumption that J E → (cid:54) = ∅ , we have that there exist bulk points c , r , r such that J + ( c ) ∩ J − ( r ) ∩ J − ( r ) ∩ B (cid:54) = ∅ (3.6)with c ∈ C , r ∈ R , r ∈ R , where recall X = E W ( ˆ X ), with ˆ X a boundary region. Wewill consider a M × n task in the bulk such that the input system A = A ....A n is input near c , and each bit b i should be brought near r and r . Further, system Q will be recordedinto the brane degrees of freedom.It is easy to see that the M × n task can be completed in this case with high probability.To see this, note that a simple bulk strategy is to bring A to the brane, learn the q i , anduse them to recover the b i . The b i are then copied and sent to both r and r . Doing so wecan complete each M task with some probability p = 1 − (cid:15) , leading to a success probability p suc = 1 − (cid:15) n for the M × n task. Since the boundary reproduces bulk physics, theboundary must also complete the task with the same probability. Lemma 4 then gives12 I ( ˆ V : ¯ Q ) ≥ n ( − log 2 h (2 (cid:15) ) β ) − O (( (cid:15)/β ) n ) (3.7)so that when the entanglement scattering region is non-empty, we have large mutual infor-mation.This bound on mutual information actually requires the entanglement wedge of ˆ V toattach to the brane. To see this, consider that in the purified M × n task there are n Bellpairs | Ψ + (cid:105) A i ¯ A i with A = A ...A n input at C , and ¯ A = ¯ A ... ¯ A n held by Bob. There are anadditional n Bell pairs | Ψ + (cid:105) Q i ¯ Q i , with Q = Q ...Q n stored on the brane, and ¯ Q = ¯ Q ... ¯ Q n held by Bob. We can choose n to satisfy O (1) < n < O (1 /G N ), so that n grows as G N → /G N .Suppose that E W ( ˆ V ) is not connected to the brane. Then the entropies of the regionˆ V and of system ¯ Q satisfy S ( ˆ V ) = A dis G N + n + O (1) ,S ( ¯ Q ) = n,S ( ˆ V ¯ Q ) = A dis G N + 2 n + O (1) . (3.8)The first statement is just our assumption: the disconnected surface calculates the entropyof ˆ V , and then we add the entropy of the n Bell pairs shared between ˆ V and ¯ A , along with– 11 –ny O (1) contribution. The second statement is due to Q ¯ Q being in the maximally entan-gled state. The third statement follows from the disconnected surface being of minimal areaalong with our choice to take n < O (1 /G N ). This is because the other option, of having theconnected surface calculate the entropy, would imply that the quantum extremal surfacehas moved to enclose the n qubits of Q , which would happen only if n > ( A dis − A con ) /G N .Using these statements about the entropy, the mutual information is I ( ¯ Q : ˆ V ) = S ( ¯ Q ) + S ( ˆ V ) − S ( ˆ V ¯ Q ) = O (1) , (3.9)so that in the disconnected phase the mutual information is O (1). Since 3.7 implies themutual information is O ( n ) > O (1), we find that the entanglement wedge must attach tothe brane.It is interesting to consider this result in the context of entanglement wedge recon-struction. We can observe that when the entanglement wedge connects to the brane Q isreconstructable from ˆ V . This clarifies how the boundary completes the task. Wheneverthe task can be completed in the bulk, the entanglement wedge connects to the brane,which means Q is available in ˆ V . Thus the boundary dynamics can recover the bits q i anduse them to decode the b i , then forward the b i to both output points.We should contrast this boundary picture with the analogous feature of the connectedwedge theorem in AdS/CFT. In that setting there are two decision regions ˆ V and ˆ V , withˆ V associated with the input H q | b (cid:105) and ˆ V associated with the input q . In that case, evenin the connected phase, ˆ V does not reconstruct q . To complete the task then the boundarymust make use of a different strategy. Indeed in [16] the authors argued that the boundarydynamics should be understood as a quantum non-local computation. In this section we prove the 1 → We will briefly review the focusing theorem without an ETW brane present.Consider a null codimension 1 surface N . We assume N is foliated by null geodesicswhich start on a spacelike codimension two surface Σ , and end on another spacelikecodimension two surface Σ . Call the affine parameter along the null geodesics λ , whichwe scale so that λ = 0 on Σ and λ = 1 on Σ . Then A (Σ ) − A (Σ ) = (cid:90) dY √ h λ =0 − (cid:90) dY √ h λ =1 = (cid:90) dλ (cid:90) dY ∂ λ √ h (4.1)– 12 – λ ˆ n Σ Σ B (a) Σ Σ B ∂ λ ˆ n (b) Figure 4 : A portion of the boundary of the past ∂J − ( R i ), showing two cross sections Σ ,Σ , and the end-of-the-world brane B . Null geodesics generating the lightcone are shownin blue. The outward pointing normal to the brane is labelled ˆ n , while the tangent vectorto the null geodesics is labelled ∂ λ . (a) When ˆ n · ∂ λ ≥
0, the brane removes area. (b) Whenˆ n · ∂ λ ≤
0, the brane adds area.where h is the determinant of the induced metric on a constant λ slice of N , and dY = dy ∧ ... ∧ dy d − .Define the expansion, θ , and a d − (cid:15) by θ = 1 √ h ∂ λ √ h (cid:15) = √ h dλ ∧ dY. (4.2)Then the area difference can be written as A (Σ ) − A (Σ ) = (cid:90) (cid:15) θ. (4.3)Expressing the area difference in this way is convenient, since the expansion is constrainedin certain situations if we assume the null energy condition (NEC), k µ k ν T µν ≥ . (4.4)In particular, consider an extremal surface γ . Then the boundary of the future or pastof γ , ∂J ± ( γ ), is generated by a congruence of null geodesics. Assuming the NEC, thiscongruence has non-positive expansion when moving away from γ , as can be shown usingthe Raychaudhuri equation. We will call surfaces wth non-positive expansion light sheets .Considering N to be a portion of either ∂J + ( γ ) or ∂J + ( γ ) then allows us to conclude A (Σ ) ≤ A (Σ ). That is, the area of a cross section of the congruence decreases as wefollow the geodesics.Notice that there are various lightsheets we can define given an extremal surface γ ,in particular the four surfaces ∂J ± in,out ( γ ). To specify it will be more convenient to definelight sheets as the boundary of the future or past of a (codimension 0) entanglement wedge, ∂J ± ( X ). Notice that ∂J ± in ( γ X ) = ∂J ± ( X ), so that defining light sheets in this way choosesthe inward pointing sheets.Next, we consider the focusing theorem in the setting where N intersects the brane.The situation is shown in figure 4. The null surface N is still foliated by a null congruence,– 13 –ut some geodesics end or begin on an additional portion of the boundary, N ∩ B . To provean area theorem in this setting, we will need to assume the NEC holds both for the bulkstress tensor and for the branes stress tensor. This later statement is (cid:96) a (cid:96) b T B ab ≥ , (4.5)where (cid:96) a is a null tangent vector to the brane. This is satisfied with equality for constanttension branes. Using the boundary condition 8 πG N T ab = − ( K ab − Kh ab ) we can alsoexpress this as (cid:96) a (cid:96) b K ab ≤ (cid:82) (cid:15) θ and write this as a boundary integral. A simple application ofthe fundamental theorem of calculus sufficed to derive 4.3, but this was only because thenull geodesics meet Σ and Σ normally. For the additional portion of the boundary weneed to use Stokes theorem in a more general form. To begin, note that (cid:15) θ = ( ∂ λ √ h ) dλ ∧ dY = d ( √ hdY ) ≡ dω (4.6)so (cid:15) θ is closed. The last equality defines ω . Now we will use Stokes theorem in the form (cid:90) M dω = (cid:90) ∂M d d − x √ γn µ V µ (4.7)where γ is the induced metric on the boundary, n µ is the normal vector to the boundary ,and V µ = ( − d − ( ∗ ω ) µ where ∗ denotes the Hodge dual.The one-form V in 4.7 is simple to compute, V = ( − d − ∗ ω = ( − d − dλ . Along Σ we have n µ = − ( ∂ λ ) µ , and along Σ we have n µ = ( ∂ λ ) µ , which recovers the two boundaryterms appearing in 4.3. The boundary B ∩ N returns an additional term, (cid:90) N (cid:15) θ = A (Σ ) − A (Σ ) + (cid:90) Σ ∩N d d − x √ γ n λ . (4.8)We will need this more general statement when we focus backwards in the proof of Theorem1. For 4.8 to relate A (Σ ) and A (Σ ) we would like to fix the sign of n λ . In particular, n λ ≥ θ ≤ A (Σ ) ≥ A (Σ ), recovering the usual area theorem.This is illustrated in figure 4. In fact we can show n λ ≥ N is a portion of ∂J − ( R i ), for R i the entanglement wedge of anedge anchored region. Then we have that at λ = 0, n λ = 0 . (4.9)This holds because the entangling surface γ R i meets the brane normally, which means thenormal vectors of γ R i ∩ B will be tangent to the brane. For spacelike boundaries we should choose the outward pointing normal, while for timelike boundarieswe choose the inward pointing one. More generally we need only that n λ ≥
0. Thus the theorem may still hold in certain cases where ˆ R i is not attached to the edge, though its unclear when this occurs. – 14 – n∂ λ Figure 5 : The lightsheet ∂J − ( R i ) where it meets the brane. For ˆ n · ∂ λ = n λ ≥ n λ ≥ n λ as we move along the brane, (cid:96) µ ∇ µ ( n λ ) = (cid:96) µ ∇ µ ( n σ k σ ) = (cid:96) µ k σ ∇ µ n σ + (cid:96) µ n σ ∇ µ k σ (4.10)Using k σ = n λ n σ + (cid:96) σ , n ν ∇ µ n ν , k µ ∇ µ k ν , (4.11)this becomes (cid:96) µ ∇ µ ( n λ ) = (cid:96) µ (cid:96) σ ∇ µ n σ − n λ n µ n σ ∇ µ k σ . (4.12)Since initially n λ = 0, if we establish that ∇ λ n λ ≥ n λ = 0, we are done. Butwhen n λ = 0 the above is just ∇ λ n λ = (cid:96) µ (cid:96) σ ∇ µ n σ = − (cid:96) a (cid:96) b K ab ≥ K ab is defined using theinward pointing normal vector, whereas the normal vector appearing in Stokes theoremwas outward pointing. The inequality is just the NEC imposed on the brane. How thecurvature in the brane prevents a sign change in n λ is illustrated in figure 5. In this section we prove the 1 → → C Σ γ (cid:48)V Figure 6 : The null membrane. The red surface is the lift L , the blue surfaces make upthe slope. The ridge R , is where the lift meets the brane.The proof relies on three assumptions: (i) that the null energy condition holds in thebulk; (ii) that the null energy condition holds for the branes stress tensor; and (iii) that themaximin procedure [23, 27, 28] for finding HRRT surfaces is correct even in the context ofAdS/BCFT.Given these assumptions, the outline of the proof of Theorem 1 is as follows. Wesuppose, by way of contradiction, that J → (cid:54) = ∅ and the HRRT surface for region ˆ V isbrane-detached. Call this surface γ (cid:48)V . According to the maximin procedure, this surfaceis minimal in some Cauchy slice Σ. We’ll use the focusing theorem and the fact that J E → (cid:54) = ∅ to construct a smaller area surface in Σ which is brane-connected, called the contradiction surface C Σ . This provides a contradiction with γ (cid:48)V having been the HRRTsurface, showing the correct HRRT surface must be brane-attached.To begin, we consider two cases, corresponding to the boundary scattering regionˆ J E → = ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) ∩ B = ˆ V ∩ B (4.14)being empty or non-empty. If it is non-empty, then ˆ V is attached to the brane in theboundary, so its entanglement wedge is immediately brane attached and we are done. If itis empty, we proceed with the proof below.Define the null surface L = ∂J + ( V ) ∩ J − ( R ) ∩ J − ( R ) (4.15)which we call the lift . This is defined by taking the inward pointing null orthogonal vectorsof γ V as generators for a null congruence, and extending those geodesics until they reachthe past of R or R . Additionally, geodesics should not be extended past any causticpoints — defining the lift in terms of ∂J + ( V ) implements this for us, as geodesics leavethe boundary of J + ( V ) after developing a caustic.There are two features of the lift that will be important. The first feature is that thelift has a non-empty intersection with the brane. To see this, recall that by assumption J E → = J + ( C ) ∩ J − ( R ) ∩ J − ( R ) ∩ B (cid:54) = ∅ . (4.16)– 16 –hen, recall that since ˆ C i ⊆ ˆ V i , we have also C i ⊆ V i . Thus we learn J E → ⊆ J + ( V ) ∩ J − ( R ) ∩ J − ( R ) ∩ B (cid:54) = ∅ . (4.17)This gives that J + ( V ) meets the brane while in the past of R and R . In particular thenthe ridge, defined by R ≡ L ∩ B (cid:54) = ∅ (4.18)is non-empty.The second important feature of the lift is that its boundary has a component A along ∂J − ( R ) and a component A along J − ( R ) which are separated by the ridge. Theother possibility would be for the ridge to extend to one or more of the edges. This cannotoccur however, which follows because the ridge is a subregion of the bulk scattering region,which by assumption does not extend to the boundary.Next define a second null sheet which we call the slope , S Σ = ∂ [ J − ( R ) ∩ J − ( R )] ∩ J − [ ∂J + ( V )] ∩ J + (Σ) . (4.19)The slope is generated by past-directed null geodesics beginning as the inward, past directednull normals to γ R and γ R , and extended until they reach Σ. We will be particularlyinterested in C Σ ≡ S Σ ∩ Σ (4.20)which we introduced above as the contradiction surface. The lift, ridge, slope, and contra-diction surface are shown in figure 6.Now, we apply the focusing theorem in the form of equation 4.8 to the lift and to theslope. The lift is a portion of the boundary of the future of an extremal surface, ∂J + ( V ),so focusing applies. We choose a parameterization such that the null generators begin on γ V and end on R ∪ A ∪ A ∪ B L , where B L is any caustics present in the lift. This leadsto area( A ) + area( A ) + 2 area( B L ) + area( R ) − area( γ (cid:48)V ) = (cid:90) (cid:15) θ ≤ . (4.21)Similarly, we can apply 4.8 to the slope, which is a portion of ∂ [ J − ( R ) ∩ J − ( R )]. Choosingthe parameterization such that generators begin on A ∪ A and end on C Σ ∪ B S Σ , where B S Σ is any caustics present in the slope. We have thenarea( C Σ ) + 2 area( B S Σ ) − area( A ) − area( A ) + (cid:90) S Σ ∩B d d − x √ γn λ = (cid:90) (cid:15) θ ≤ . (4.22)Adding these two inequalities and rearranging terms we obtainarea( γ (cid:48)V ) ≥ area( C Σ ) + area( R ) + area( B S Σ ) + area( B L ) + (cid:90) S Σ ∩B d d − x √ γn λ , ≥ area( C Σ ) (4.23)– 17 –here we’ve used that n λ ≥
0, which follows when the NEC applied to the brane matter ten-sor holds, as shown at the end of the last section. This ensures that the brane-disconnectedsurface γ (cid:48)V is not of minimal area in the Cauchy slice Σ, so from the maximin procedurecannot be the correct HRRT surface, completing the proof.We should highlight the modifications made from the similar proof of the 2 → C , C , R and R ,and two decision regions V and V . The lift was formed by a null congruence of geodesicsstarting on γ V ∪ γ V . Points on the ridge corresponded to where a geodesic starting on γ V collided with a geodesic starting on γ V , whereas in our setting the ridge is formed bygenerators from γ (cid:48)V colliding with the brane. Another distinction is the occurrence of theboundary S Σ ∩ B and associated term in 4.22. This is handled in our case by assuming theNEC holds for the brane stress tensor. → connected wedge theoremThe scattering region is inside the entanglement wedge In the context of the 2 → J → isinside of the entanglement wedge of ˆ V ∪ ˆ V . It is straightforward to adapt either of theproofs given there to the 1 → J E → is inside the entanglement wedge of ˆ V . Since J E → lives in the brane, we can be morespecific and say that J E → is inside the island formed by ˆ V ∩ B . Relationship to → theorem and interface branes It is possible to describe ETW brane geometries as a Z identification of an interface branegeometry. In particular, consider a spacetime M described by metric g µν ( x µ ) and satisfyingthe boundary condition K ab − Kh ab = − πG N T B ab (4.24)at the ETW brane. Then we can define a doubled geometry featuring an interface brane,with metric g µν ( x µ + ) on one side of the brane and a copy of that metric g µν ( x µ − ) on theother. At the interface brane Einsteins equations require we satisfy the Israel junctionconditions h + ab = h − ab , (4.25)[ K + ab − K − ab ] − [ K + − K − ] h ab = − πG N T Iab . (4.26)Setting T Iab = 2 T Bab satisfies this condition. Identifying points x + = x − then recovers theETW brane geometry.We can apply the 2 → → C and ˆ C to be mirrorimages across the interface brane. Choose ˆ R and ˆ R to be intervals centered on the twoCFT interfaces. Notice that the brane anchored scattering region J E → is not empty if andonly if the bulk scattering region J E → in the interface geometry is not empty. Further,– 18 –he entanglement wedge of ˆ V ∪ ˆ V will be connected if and only if the entanglement wedgeof ˆ V connects to the brane in the ETW brane geometry. Thus, when the doubled geometrysatisfies the conditions for the 2 → → → → T µν = 0 ,T B ab = − T h ab . (4.27)Then in the interface brane geometry the stress tensor is T µνI = − T h ab e µa e νb δ ( x − x ) (4.28)where the delta function is turned on at the interface. To study the NEC for T Iµν , it’sconvenient to rewrite this using the completeness relation, g µν = n µ n ν + h ab e µa e νb (4.29)so that T µνI (cid:96) µ (cid:96) ν = − T ( g µν − n ν n µ ) (cid:96) µ (cid:96) ν = T ( n µ (cid:96) µ ) (4.30)We see that the NEC is satisfied if and only if T >
0. However, in the ETW brane geometry,the 1 → T <
0. Consequently we find that the 2 → → Counterexample to the converse
We claimed in the introduction that the converse to Theorem 1 is false. In [16, 17], theauthors constructed a counterexample to the converse of the 2 → Z identification of the solution used in their example we can easily construct a counterexampleto the converse of the 1 → The out regions are not entangled
In the 2 → ˆ W = J − ( ˆ R ) ∩ J + ( ˆ C ) ∩ J + ( ˆ C ) , ˆ W = J − ( ˆ R ) ∩ J + ( ˆ C ) ∩ J + ( ˆ C ) . (4.31) We are interested here in the case where ˆ W i ⊆ ˆ R i , analogous to our condition ˆ C i ⊆ ˆ V i on the inputand decision regions. – 19 – V π/ (a) V V π/ (b) V V (c) Figure 7 : A counterexample to the converse of Theorem 1. (a) A constant time sliceof a solution with a T = 0 brane sitting in pure AdS. These solutions are described indetail section 5. We choose a region ˆ V of size π/ T = 0 solution can be viewed as a Z identification of global AdS with theidentification across ρ = 0. (c) In the unfolded geometry, we consider adding a sphericallysymmetric matter distribution (shown in grey). This delays light rays travelling from c tothe brane by some finite amount, closing the scattering region. Due to spherical symmetry,the region ˆ V remains on the phase transition. Increasing its size infinitesimally then keepsthe scattering region closed, while also ensuring the red, brane-attached surface is minimal.In the context of the 1 → x , x as the points where ∂ ˆ J + ( ˆ C ) reaches edge 1 and edge 2, respectively. Thenwe define ˆ W (cid:48) = ˆ J + ( x ) ∩ J − ( ˆ R ) , ˆ W (cid:48) = ˆ J + ( x ) ∩ J − ( ˆ R ) . (4.32)We can ask if ˆ W (cid:48) and ˆ W must also be entangled when the entanglement scattering regionis non-empty.In fact, these regions do not need to be entangled. For an explicit counterexample,begin with the example shown in figure 7a, where ˆ V consists of an interval of size π/ W (cid:48) ∪ ˆ W (cid:48) is on the transition from giving a connected anddisconnected entanglement wedge. Now decrease the tension, moving the brane inward.This shortens the light travel time from ˆ C , so increases the size of the scattering region.Meanwhile, the disconnected surface enclosing ˆ W (cid:48) ∪ ˆ W (cid:48) loses area and becomes dominant,so that there is a non-empty scattering region but only O (1) correlation between the ˆ W i regions. – 20 – → theorem For completeness, we also point out a 1 → C , ˆ R ,both in the AdS boundary, and define the scattering region, J E → = J + ( C ) ∩ J − ( R ) ∩ B , (4.33)and the decision region, ˆ V = ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) . (4.34)By analogy with the 1 → J E → being non-empty implies theentanglement wedge of ˆ V is brane-attached. To verify this, we can give both a tasks andgeometric argument.From tasks, we consider an input H q | b (cid:105) A at C and output b at R , with q recordedinto the brane degrees of freedom. If J E → is non-empty, then one can use a simple bulkstrategy: travel to the brane, learn q , then send q to the output point where it can beused to undo H q and recover b . In the bulk picture knowing q is necessary to successfullyrecover b , so ˆ V must know q , so ˆ V must have the brane in its entanglement wedge. .To understand this from the geometric perspective, note that J + ( C ) ∩ J − ( R ) isinside the entanglement wedge of ˆ V , so J B → non-empty means the brane is inside theentanglement wedge of ˆ V . In this section we give constant tension brane solutions in global AdS , then verify theconnected wedge theorem by explicit calculations in that setting. In this case the converseof Theorem 1 holds, and a bulk scattering region is present if and only if the entanglementwedge is connected. This is similar to the situation for the 2 → [15]. We present a brief overview of the calculation, butrelegate the details to Appendix A. We will consider a simple model where the bulk matter action is set to zero, and the branehas constant tension. This corresponds to a Lagrangian L B matter = − πG N (cid:90) d d y √ hT. (5.1)Extremizing the action 2.1, we obtain the vacuum Einstein’s equations in the bulk and aboundary condition for the brane:( K ab − Kh ab ) = − T h ab . (5.2) A more rigorous argument for this would follow the strategy of section 3. – 21 – a) (b)
Figure 8 : Global AdS with a ETW brane. We’ve shown the T = 0 case for simplicity.Poincar´e patches are shaded in blue. (a) An edge centered choice of Poinar´e patch. Inthe associated Poincar´e solution the ETW brane is flat, described by equation 5.6. (b)A Poincar´e patch centered at σ = 0. In Poincar´e coordinates the brane trajectory is ahyperbola, described by equation 5.9.We can solve this along with Einstein’s equations. The solutions of interest are describedby the metric ds = cosh ρ ds + dρ = cosh ρ (cid:18) − dν + dσ cos σ (cid:19) + dρ , (5.3)with ds the line element for a global 1 + 1 dimensional AdS space. Allowing −∞ < ρ < ∞ , this is global AdS . To add an ETW brane we restrict to ρ < ρ < ∞ , where thebrane is located at ρ = ρ and T = tanh ρ . (5.4)We will call the ( ν, σ, ρ ) coordinates slicing coordinates , since ρ foliates AdS slices toform an spacetime.There are two ways of taking Poincar´e patches of this spacetime that will be of interestto us. First, as shown in figure 8a, we can center our Poincar´e patch on one of the edges, σ = ± π/
2. The associated Poincar´e coordinates are related to slicing coordinates by t = sin ν cos ν − sin σ , x = cos σ tanh ρ cos ν − sin σ , z = cos σ sech ρ cos ν − sin σ . (5.5)– 22 – (a) (b) Figure 9 : (a) Poincar´e-AdS with a constant tension ETW brane, as obtained by takingan edge-centered patch of the global spacetime, as shown in figure 8b. (b) Poincar´e-AdS with a constant tension ETW brane, as obtained by taking a patch as shown in figure 8a.The brane forms a hyperbola, and the two edge trajectories are x = ±√ t . Thehorizons σ = ± ν chosen in the global geometry map to x = ± t , z = (1 − sin Θ) / cos Θ inPoincar´e coordinates, which we’ve shown in red.Under this transformation the boundary becomes the half line x >
0, with one edge locatedat x = 0. The other edge is at x = ∞ . The ETW branes trajectory is xz = sin Θ , (5.6)where Θ is related to the tension T by T = tan Θ. Solutions of this form are shown infigure 9a.Using this planar solution we can relate the bulk parameter T to CFT data. In theCFT, one can calculate the entropy of an interval of size L ending on the CFT-boundary, S ( L ) = c bulk L(cid:15) + log g B . (5.7)The second term is known as the boundary entropy [29], and counts the degrees of freedomlocated at the edge. The Ryu-Takayanagi prescription reproduces this entropy expression inthe simple constant tension model if we relate the tension and boundary entropy accordingto log g B = (cid:96) G N arctanh( T ) . (5.8)The second Poincar´e patch we will be interested in is centered at σ = 0, as shownin figure 8b. This coordinate change is most easily performed using the embedding spaceformalism, see appendix B. The ETW brane in this Poincar´e patch is described by x − t + ( z + tan Θ) = sec Θ (5.9)The edge trajectories are described by t = ±√ x −
1. This solution was studied in [7] inthe context of brane models of black holes and island formation, which we will also takeup in section 6. A solution of this type is shown in figure 9b.– 23 – .2 Null rays and entanglement
In the solutions 5.3, we will check the theorem in the case that the input and output regionsare points, ˆ C = { c } and ˆ R i = { r i } . We will calculate the travel time of null rays in thegeometry (5.3), used to perform the bulk local strategy, and compare to a calculation ofentanglement entropy on the field theory side.We can transform slicing coordinates (5.3) into the following form, as discussed inappendix A: ds = cosh ρ sin θ ( − dν + dθ + sin θ dϕ ) , (5.10)where (cid:96) AdS = 1, θ = σ + π/ ∈ [0 , π ], and ϕ ∈ [ ϕ B , π ] is a warping coordinate for the copiesof AdS , with ϕ = 0 the position of the asymptotic region and ϕ = ϕ B the locationof the brane. While the brane has the geometry of a copy of global AdS , the bulk isconformally equivalent to a patch of R × S enclosed by two lines of longitude. In thesecoordinates, it is easy to trace out light cones. As discussed in more detail in appendix A,if Alice sends a signal from c = (0 , θ ) light rays will arrive at the brane at angle θ at atime cos[ ν ( θ )] = cos θ cos θ + sin θ sin θ cos ϕ B . (5.11)Next we study the von Neumann entropy of subregions of the CFT. This can obtainedusing the replica trick in the BCFT. We start by analytically continuing the Lorentzianmetric (5.10) to Euclidean time τ = iν , and choose a defining factor to obtain the BCFTon M E = R × [0 , π ]. Following the calculation of [22], and as detailed in the appendix,we can calculate the entanglement entropy of the (Euclidean) interval A := [ w , w ], for w j = τ j + iθ j . The phase transition occurs at g /c B = (cid:12)(cid:12)(cid:12)(cid:12) cosh(∆ τ ) − cos(∆ θ )2 sin( θ ) sin( θ ) (cid:12)(cid:12)(cid:12)(cid:12) , (5.12)where ∆ w = w − w = ∆ τ + i ∆ θ , ∆ θ = θ − θ and ∆ τ = τ − τ , c is the central chargeof the CFT, and g B := (cid:104) | B (cid:105) is the boundary entropy.Returning to Lorentzian time, ν = − iτ , (5.12) gives g /c B = (cid:12)(cid:12)(cid:12)(cid:12) sin[(∆ θ + ∆ ν )2] sin[(∆ θ − ∆ ν )2]sin( θ ) sin( θ ) (cid:12)(cid:12)(cid:12)(cid:12) . (5.13)We note that the brane angle ϕ B is related to the boundary entropy by g /c B = tan (cid:16) ϕ B (cid:17) . (5.14)This follows from (5.8) and the relation c = 3 (cid:96) AdS / G N , and is discussed further in ap-pendix A. In the next section, we combine these facts about light rays and entanglement toconfirm the connected wedge theorem for pure AdS ended by constant tension branes.– 24 – .3 A check of the connected wedge theorem Let c = (0 , θ ) be the input point. Without loss of generality, consider output points r , r π on opposite edges. The backward light cones for these points intersect at somepoint x = ( ν , θ ), and hence the decision region is ˆ V = ˆ J + ( c ) ∩ ˆ J − ( x ). If ˆ V intersectsthe edges of the BCFT, then the boundary local strategy can be trivially performed: Alicetravels to the edge, decodes her qubit, and sends the results to r and r π .We will be interested in the case where this strategy cannot be performed, and henceˆ V = ˆ D [ A ] for a boundary interval A with endpoints L = ( θ L , ν L ) = 12 ( θ + θ − ν g , θ − θ + ν g ) , (5.15) R = ( θ R , ν R ) = 12 ( θ + θ + ν g , θ − θ + ν g ) . (5.16)To perform a bulk local strategy, Alice must shoot null rays in the bulk so they intersectthe brane in the past of the point on the brane with boundary coordinates x . This strategymarginally succeeds when the light ray hits x itself. If she sends it from c , (5.11) tellsus it arrives at the brane at a time ν B obeyingcos ν B = cos θ cos θ + sin θ sin θ cos ϕ B . (5.17)Our connected wedge theorem states that when this ray can arrive at x , or ν B ≤ ν , A hasa brane-connected entanglement wedge.Using equations (5.13)–(5.16), the transition to a brane-connected entanglement wedgeoccurs at a time t g obeyingtan (cid:16) ϕ B (cid:17) = (cid:12)(cid:12)(cid:12)(cid:12) cos ν g + cos( θ − θ )cos ν g − cos( θ − θ ) (cid:12)(cid:12)(cid:12)(cid:12) . (5.18)Fixing θ , θ and solving for ν , some algebra shows it obeys (5.17). In other words, thetransition to a connected entanglement wedge occurs precisely when the bulk local strategybecomes possible. This explicitly verifies the connected wedge theorem for vacuum AdS .As a simple illustration take θ = θ = π/
2, corresponding to edge output points atequal times. From (5.14), the phase transition occurs at a time ν g given bytan (cid:16) ϕ B (cid:17) = (cid:12)(cid:12)(cid:12)(cid:12) sin ( ν g / π + ν g ) /
2] sin[( π − ν g ) / (cid:12)(cid:12)(cid:12)(cid:12) = tan (cid:16) ν g (cid:17) , in other words, when ν P = ϕ B . But from (5.11), a light ray from c arrives at the brane attime ν B = ϕ B . So the phase transition occurs precisely when Alice is able to perform thequantum task using the bulk local strategy. In this section, we point out that, in brane models, the 1 → If they are on the same edge, ˆ V intersect the edge and the theorem is trivially true. From (5.10), we note that each AdS slice is conformally equivalent to the flat boundary. Since thisconformal factor is invisible to light rays, the point of intersection on the brane has the same boundarycoordinates as the intersection on the boundary. – 25 –
III IIIV (a)
IIII IIIV (b)
Figure 10 : Choosing an appropriate Poincar´e patch of the global spacetime, we find atwo-sided black hole geometry (on the brane) coupled to two flat regions (wedges of theCFT). The end points of the two flat regions are coupled in the global picture. Note thatwe are most interested in the case where T ≈ T = 0 case however to simplify the diagram.tive of physics on the brane, the RT surface attaching to the brane corresponds to theformation of an island [7–14]. The connected wedge theorem then relates the formation ofthis island to causal features of the higher dimensional AdS geometry. In this section wemake more precise how we can view the brane as a black hole and a portion of the CFTas the radiation system, and finally apply the connected wedge theorem in this context. We will focus on the solutions described in section 5, which have a constant tension braneending a pure, global, AdS spacetime. We are most interested in the case where T ≈ C , ˆ R , ˆ R to be extended regions gives no additional power to the connected wedge theorem in thesesolutions, and consequently for simplicity we will take ˆ C = c , ˆ R = r , ˆ R = r where c , r , r are points on the boundary of AdS, and in particular r , r sit in the edge.For constant tension solutions we have two simplifications that will prove useful in un-derstanding the connected wedge theorems relationship to islands. The first simplificationis that for constant tension branes light rays run tangent to the brane. This allows us todefine horizons in the brane by choosing points r and r on the edge, and considering their– 26 –orward light cones, H = [ ∂J + ( r )] B ,H = [ ∂J + ( r )] B . (6.1)These horizons intersect at p B = ( σ = 0 , ν = 0). From these horizons, define regions I − IV as in figure 10. Region II is the black hole interior, while I and IV are the right and leftexteriors.The second simplification is that the 1 → ds = (cid:96) z ( − dt + dx + dz ) (6.2)with brane located at x − t + ( z + tan Θ) = sec Θ , (6.3)where Θ is related to the tension T according to T = sin Θ. The Poincar´e patch includesonly the − π/ < ν < π/ r and r are mapped to x = t = −∞ and − x = t = −∞ . The details of this coordinate change are given inappendix B.In Poincar´e coordinates the edge trajectory is x = ±√ t . These trajectories asymp-tote to the light rays x = ± t . Mapping the horizons v = ± σ to Poincar´e we find horizons z = 1 − sin Θcos Θ , x = ± t. (6.4)One can also verify directly in the Poincar´e geometry that these are the horizons by studyingnull geodesics in the brane geometry [7].Next we should identify the radiation system. The entire CFT is coupled to the blackhole at the two edges, and information can escape from the black hole into anywhere in theCFT. It seems sensible however to not consider the portion of the CFT which reconstructsthe black hole exterior regions as being part of the radiation system. It is straightforwardto identify the CFT dual to the left and right exterior black hole regions. The interval Y = { σ ∈ ( − π/ , , ν = 0 } has region I inside its entanglement wedge. Similarly theinterval Y = { σ ∈ (0 , π/ , ν = 0 } has region II inside its entanglement wedge. Thisexcludes D ( Y ) and D ( Y ) from the radiation system.– 27 –he remaining portion of the CFT is the future and past of the point x = ( ν = 0 , σ = 0 , ρ = ∞ ) . (6.5)The future of x reconstructs region II of the brane, so we should identify this with theradiation system. To specify that radiation has been collected only up until a certain time,we can choose a second point c and defineˆ R = J + ( x ) ∩ J − ( c ) . (6.6)For c at an early time so that R is small, the entanglement wedge of R will be disconnectedfrom the brane, and R does not see inside of the black hole. At late enough times though, E W ( R ) connects to the brane. Where this transition occurs will be controlled by theconnected wedge theorem. Note also that since the minimal surfaces are at constant σ ,they will in fact lie exactly on the horizons. This is illustrated in figure 2. Finally, we can apply the connected wedge theorem to this black hole on the brane. In fact,we need a time reversed variant of the theorem, which follows immediately from Theorem1 (we also specialize to the case where the input and output regions are points),
Theorem 5 ( → connected wedge theorem) Consider three points r , r , c in anasymptotically AdS spacetime with an end-of-the-world brane, with c in the boundaryand r , r on the edge. Then if J → = J + ( r ) ∩ J + ( r ) ∩ J − ( c ) (6.7) is non-empty, the entanglement wedge of ˆ V = ˆ J + ( r ) ∩ ˆ J + ( r ) ∩ ˆ J − ( c ) (6.8) is attached to the brane. The two input points of the theorem we identify with the points r and r we used aboveto define the black hole horizons H and H . The region ˆ V becomes the subsystem of theradiation which has been collected since J + ( r ) ∩ J + ( r ) = J + ( x ), so ˆ V = ˆ R .Applying Theorem 5 along with its converse (which holds because we are in the con-stant tension solutions) gives a simple condition for when the radiation system reconstructsa portion of the black hole interior: an island forms if and only if there is a causal curvefrom the black hole interior into the radiation system.This causal picture for island formation immediately reveals a set of simple operatorsthat probe behind the black hole horizon. In particular consider an operator O y , whichis localized near a point y , with y in ˆ R and in the future of the black hole interior (suchpoints exist by our theorem). These operators directly probe the black hole interior byvirtue of being in the future of the interior.Notice that the radiation system ˆ R sits outside of the Poincar´e patch we identifiedabove. Thus it sits outside of the black hole spacetime. Ideally, we would understand– 28 – R (a) ˆ R P ˆ R P E W ( R P ) ∩ B (b) Figure 11 : (a) The radiation system R (time-slice in green) picked out by the connectedwedge theorem sits outside the Poincar´e patch. (b) A nearby region ˆ R P inside the patchhas ˆ R inside of its domain of dependence, so that ˆ R P has an island whenever ˆ R does.The entanglement wedge of ˆ R P (shown in light gray) will include a small portion of theblack hole exterior in its entanglement wedge.which subregions of the Poincar´e patch reconstruct the black hole interior. To do this, weneed only note that a nearby subregion ˆ R P of the Poincar´e patch includes ˆ R in its domainof dependence. See figure 11. Evolving the state on this subregion forward using the globalHamiltonian, we can construct the state of the radiation system ˆ R . Notice that ˆ R P isslightly larger than ˆ R and will include a small portion of the black hole exterior in itsentanglement wedge.To write operators which probe behind the black hole horizon in the Hilbert space of V P , we can start with the operators O y which live in ˆ V and time evolve backward using theglobal Hamiltonian. We continue this time evolution until O y is some non-local operator O y,P living on V P .It is interesting that time evolution with the global Hamiltonian, along with localoperators, can be used to probe the black hole interior. We should perhaps be unsurprisedhowever, as the situation is analogous to the traversable wormhole [30]: in both cases wehave a left and right CFT (or in our setting, BCFT), which we couple and then time evolveto find that information from behind the black hole horizon has emerged at the boundary.In the traversable wormhole the coupling is a double trace term which can be understoodperturbatively, while in our setting the coupling is due to time evolution with the global– 29 –amiltonian. In this paper we have proven the 1 → V to knowinformation stored on the brane. Otherwise, the CFT is unable to reproduce bulk physics.The focusing theorem based proof relies on the null membrane, a structure that allowscomparison of the areas of brane-detached extremal surfaces and brane-attached surfaces.When the scattering region is non-empty, we showed there exists a null membrane thatconnects a brane-detached extremal surface to a brane-attached one with less area.Below we make a number of comments. The key technical tool used here to complete the quantum tasks argument for the 1 → I ( ˆ V : ¯ Q ) ≥ n ( − log 2 h (2 (cid:15) ) β ) − O (( (cid:15)/β ) n ) . This bound was first shown in [17]. It is interesting to ask if this bound can be improvedfurther. Supposing a fraction 1 − δ of the M tasks need to be completed successfully forthe M × n task to be declared successful, it is straightforward to achieve12 I ( ˆ V : ¯ Q ) = (1 − δ ) n. (7.1)Thus, (1 − δ ) n is the best lower bound on the mutual information we can hope for. Givensuch a bound, and assuming we can take δ = O (1 /n ), we could directly find that theboundary region ˆ V approximately reconstructs Q [31, 32]. With the existing bound, wecan instead only conclude I ( ˆ V : ¯ Q ) = O (1 /G N ), then use the Ryu-Takayanagi formula toconclude this means the entangling surface are brane-anchored, then use the understandingof entanglement wedge reconstruction to conclude this means ˆ V reconstructs Q . Post-hoc,we can interpret this as being due to ˆ V needing Q to undo H q and complete the task.The better bound presented above would more directly connect the task argument to bulkreconstruction. An un-explored question is the relationship between the quantum tasks considered hereand features of CFT correlation functions. Recall from [33–36] that when there is a bulkpoint p with c , c ≺ p ≺ r , r and p is null separated from each of the four points,there is a perturbative singularity in four point functions (cid:104)O ( c ) O ( c ) O ( r ) O ( r ) (cid:105) . The We thank Henry Lin for pointing out this analogy to us. – 30 – C = x y ˆ V ˆ R ˆ R Figure 12 : View of the boundary of Poincar´e-AdS . The edge is located at x = 0.A region ˆ V is specified, and we are interested in using the connected wedge theorem todetermine if the entanglement wedge of ˆ V is attached to the brane. The figure shows achoice of regions ˆ R , ˆ R and ˆ C which can be used in the theorem. Notice that the inputregion ˆ C is taken to be a point x .appearance of this point p also signals the appearance of a scattering region, and so the2 → → (cid:104)O ( c ) ψ ( r ) ψ ( r ) (cid:105) , where ψ is anedge operator, has a perturbative singularity when the three operator insertion points arenull separated from a single point on the brane. Comparing to the 1 → → theorem in planar brane solutions By using extended input and output regions, [17] applied the 2 → . Here, we have mostly focused on a point basedformulation, and on a class of global solutions with constant tension branes ending pure This is the case for the two point function (cid:104)O ( c ) O ( r ) (cid:105) in a BCFT. For further discussion, anda complementary perspective on causality and spectral properties arising from two-point functions in aBCFT, see the upcoming work [37]. – 31 –dS. However we can also use extended input and output regions and apply the 1 → with a brane.Consider in particular the pure AdS solutions with planar branes discussed in section5. The boundary is the half plane defined by x <
0, and the brane sits at x/z = sin Θ . (7.2)Suppose we are given a region on the boundary ˆ V , and we would like to apply the 1 → x theearliest point on ˆ V , and y the latest point on ˆ V so that ˆ V = ˆ J + ( x ) ∩ ˆ J − ( y ). To applythe theorem non-trivially, choose ˆ C = x ˆ R = D (( y , R = D (( −∞ , y )) (7.3)where by D ( · ) we mean the domain of dependence. This is shown in figure 12. Then the1 → V is connected. The 1 → V touches the brane, and so immediatelyhas a connected entanglement wedge.We have focused on the example of asymptotically global AdS , and discussedPoincar´e-AdS in the last section, where there are many non-trivial configurations. Itwould be interesting to better understand however when the theorem applies non-triviallyin higher dimensions. In section 6 we applied the connected wedge theorem to the static, two sided black holemodel introduced in [7]. While information does escape from this black hole, it is inthermal equilibrium with the radiation system and does not evaporate. The connectedwedge theorem applies much more generally however, including to models of evaporatingblack holes, since the theorem is proven in the context of dynamical spacetimes. It wouldbe interesting to do this explicitly, for example in the dynamical models of [7].
Acknowledgements
We thank Jon Sorce and Geoff Penington for helpful discussions. Mark Van Raamsdonkmade important comments on the connection to islands discussed in section 6. Jamie– 32 –ully was involved in early discussions leading to conjecturing Theorem 1. Jason Pollackprovided feedback on this manuscript. AM is supported by a C-GSM award given by theNational Science and Engineering Research Council of Canada. DW is supported by anInternational Doctoral Fellowship from the University of British Columbia.
A Details for AdS calculation
Light rays
Let’s consider the reflection of bulk light rays in the simplest case, vacuum AdS with abrane of tension T . We can write the global metric (for (cid:96) AdS = 1) in the slicing coordinatesAdS : ds = cosh ρ ds + dρ = cosh ρ (cid:18) − dν + dθ sin θ (cid:19) + dρ , (A.1)where ρ is the position of the brane and ρ = ∞ the boundary, global Lorentzian time is ν ∈ R , and θ = σ + π/ ∈ [0 , π ]. Null rays are simple in conformally flat coordinates, whichwe find by defining a new warping coordinate ϕ = π − (cid:104) tanh (cid:16) ρ (cid:17)(cid:105) = 2 tan − e ρ , dϕ = dρ cosh ρ , (A.2)with ϕ ∈ [ ϕ B , π ] for a brane at ϕ B = ϕ ( ρ ). Then our global metric becomes ds = cosh ρ sin θ ( − dν + dθ + sin θ dϕ ) , (A.3)which is conformally equivalent to patch of S × R enclosed by two lines of longitude. Sincenull rays do not see the conformal factor, our problem reduces to propagating light rayson the sphere. With respect to some affine parameter λ , we have null geodesic equation − ˙ ν + ˙ θ + sin θ ˙ ϕ = 0 . If we set ˙ ν = 1, our problem reduces to finding geodesic lengths on the sphere, with affinetime measuring these lengths.A null ray will start at some initial point θ and with some initial direction θ (cid:48) at theboundary ϕ = π . It travels into the bulk, reflects off the brane at ϕ B , and finally returnsto the boundary at some final position θ . From the cosine rule for spherical trigonometry,the geodesic distance to the brane obeys d = cos − [cos θ cos θ + sin θ sin θ cos( ϕ B )] . (A.4)Thus, the global time it takes a null ray to reach the brane with respect to the parameter-isation ˙ ν = 1 is ν ( θ ) = cos − (cid:2) cos θ cos θ + sin θ sin θ cos( ϕ B ) (cid:3) (A.5)– 33 – ntanglement entropy We now calculate entanglement entropy from the field theory side.First, we analytically continue τ = iν , so that ds = cosh ρ sin θ ( dτ + dθ + sin θ dϕ ) . (A.6)Choosing a defining function to remove the prefactor as we approach the boundary, thedual BCFT is defined on [0 , π ] × R . The first step is to map the strip to the upper half-plane (UHP), z = x + iy for y ≥
0. Let w = τ + iθ = log z . Then correlation functions forprimary operators O i on the strip and UHP are related by (cid:104)O ( w ) · · · O k ( w k ) (cid:105) strip = (cid:89) i | z i | ∆ i (cid:104)O ( z ) · · · O k ( z k ) (cid:105) UHP . (A.7)We define the distances z ij = | z i − z j | and z i ¯ j = | z j − ¯ z j | for future convenience.A twist operator creates an n -fold branched cover of the geometry via boundary con-ditions. The one-point function for a twist in the presence of a boundary is (cid:104) Φ n ( w ) (cid:105) strip = (cid:12)(cid:12)(cid:12)(cid:12) z z (cid:12)(cid:12)(cid:12)(cid:12) d n g − n B , (A.8)where g B := (cid:104) | B (cid:105) is the boundary entropy [29], and the twist scaling dimension for centralcharge c and replica number n is given by [38] d n = c (cid:18) n − n (cid:19) . (A.9)A gap and small OPE coefficients [22] imply the simple form for a correlator of twists: (cid:104) Φ n ( w )Φ − n ( w ) (cid:105) strip = min (cid:40)(cid:12)(cid:12)(cid:12)(cid:12) z z z (cid:12)(cid:12)(cid:12)(cid:12) d n , (cid:12)(cid:12)(cid:12)(cid:12) z z z z (cid:12)(cid:12)(cid:12)(cid:12) d n g − n ) B (cid:41) . (A.10)The entanglement entropy is given by the limit S w w = lim n → + − n log (cid:104) Φ n ( w )Φ − n ( w ) (cid:105) strip = min (cid:26) c (cid:12)(cid:12)(cid:12)(cid:12) z z z (cid:12)(cid:12)(cid:12)(cid:12) , c (cid:12)(cid:12)(cid:12)(cid:12) z z z z (cid:12)(cid:12)(cid:12)(cid:12) + 2 log g B (cid:27) . (A.11)We have neglected the UV regulator, since it cancels when we calculate the transitionbetween expressions. This occurs at g /c B = (cid:12)(cid:12)(cid:12)(cid:12) z z z (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) cosh(∆ τ ) − cos(∆ θ )2 sin( θ ) sin( θ ) (cid:12)(cid:12)(cid:12)(cid:12) (A.12)where w j = τ j + iθ j , ∆ τ = τ − τ , and ∆ θ = θ − θ . Reverting to ν = − iτ , this becomes g /c B = (cid:12)(cid:12)(cid:12)(cid:12) sin[(∆ θ + ∆ ν ) /
2] sin[(∆ θ − ∆ ν ) / θ ) sin( θ ) (cid:12)(cid:12)(cid:12)(cid:12) . (A.13)– 34 – onnected wedge in AdS To relate the location of the brane in different coordinates, first note that ρ = 6 c log g B . Hence, by (A.2), g /c B = tan (cid:16) ϕ B (cid:17) . (A.14)Consider an input point c = ( θ , r = (0 , t ), r = ( π, t ).The backward light cones intersect at coordinates x = 12 ( ν − ν + π, ν + ν − π ) = ( θ , ν g ) . (A.15)Similarly, the forward light cone of c and the backward cone of x intersect at two points, L = ( θ L , ν L ) = 12 ( θ + θ − ν g , θ − θ + ν g ) (A.16) R = ( θ R , ν R ) = 12 ( θ + θ + ν g , θ − θ + ν g ) . (A.17)In order to successfully use a bulk strategy, Alice must send a bulk light ray so thatit hits the brane in the past of the point on the brane with boundary coordinates x . Theextreme case is when her null ray hits x itself. From (A.4), this occurs at a boundary time ν B given by cos( ν B ) = cos θ cos θ + sin θ sin θ cos( ϕ B ) . (A.18)We expect that this is precisely the time at which ( L, R ) experiences a phase transition inentanglement entropy. From (A.13), the transition occurs at g /c = tan (cid:16) ϕ B (cid:17) = (cid:12)(cid:12)(cid:12)(cid:12) sin[( ν g + θ − θ ) /
2] sin[( ν g + θ − θ ) / θ + θ − ν g ) /
2] sin[( θ + θ − ν g ) / (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12) cos ν g + cos( θ − θ )cos ν g − cos( θ − θ ) (cid:12)(cid:12)(cid:12)(cid:12) , (A.19)where we have simplified with trigonometric identities. To verify the connected wedgetheorem, we will show from (A.18) and (A.19) that ν g = ν B . We first use the trigonometricidentity tan (cid:16) ϕ B (cid:17) = 1 − cos( ϕ B )1 + cos( ϕ B ) . (A.20)We can isolate cos( ϕ B ) in (A.18). Substituting this expression into (A.20) yieldstan (cid:16) ϕ B (cid:17) = sin θ sin θ + cos θ cos θ − cos ν B sin θ sin θ − cos θ cos θ + cos ν B = (cid:12)(cid:12)(cid:12)(cid:12) cos ν B + cos( θ − θ )cos ν B − cos( θ − θ ) (cid:12)(cid:12)(cid:12)(cid:12) . (A.21)Comparing to (A.19), we find ν B = ν g as claimed.– 35 – Coordinate systems and embedding space
Here, we briefly discuss the different coordinate systems used for AdS. Rather than ex-plicitly map between coordinates, we use the embedding space formalism, following [39]closely. Recall that we can view AdS d +1 as (the universal cover of) the hyperboloid in R ,d − , given by X + X d +1 − d (cid:88) i =1 X i = (cid:96) . We set (cid:96)
AdS = 1 for convenience. Different choices of coordinates map to different parametriza-tions of this hyperboloid. For instance, consider standard global coordinates on AdS d +1 : ds d +1 = − cosh ˆ ρ d ˆ t + d ˆ ρ + sinh ˆ ρ d Ω d − , where ˆ t is global (Lorentzian) time, and the Ω i are spherical coordinates on S d − . Thiscorresponds to the parametrization X = cosh ˆ ρ cos ˆ tX i = Ω i sinh ˆ ρX d +1 = cosh ˆ ρ sin ˆ t . In this paper, we employ the slicing coordinates ds d +1 = cosh ρ ds d + dρ , with global coordinates ds d on the slices. This arises from the parametrization X = cosh ρ cosh r cos νX a = Ω a cosh ρ sinh rX d = sinh ρX d +1 = cosh ρ cosh r sin ν , where a = 1 , . . . , d − S d − on the AdS d slices.In the global coordinates on AdS d +1 or AdS d , we can always compactify to the ”Einsteinstatic universe” coordinates ˆ σ, σ defined bytan ˆ σ = sinh ˆ ρ, tan σ = sinh ρ, with (for instance) ds d = − dν + dσ + sin θ d Ω d − cos σ . (B.1)Note that for d > ρ >
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