PPrepared for submission to JHEP
Holographic quantum tasks with input and outputregions
Alex May a a Department of Physics and Astronomy, University of British Columbia 6224 Agricultural Road,Vancouver, B.C., V6T 1W9, Canada
E-mail: [email protected]
Abstract:
Quantum tasks are quantum computations with inputs and outputs occurringat specified spacetime locations. In this article we consider tasks where inputs and outputsare encoded into extended spacetime regions, rather than the points previously considered.We show that this leads to stronger constraints relating causal features of the bulk geometryand boundary entanglement than have been derived in the point based setting. In particularwe improve the connected wedge theorem, appearing earlier in [1, 2], by finding a larger bulkregion whose existence implies large boundary correlation. As well, we show how consideringextended input and output regions leads to non-trivial statements in Poincaré-AdS , asetting where the point-based connected wedge theorem is always trivial. a r X i v : . [ h e p - t h ] F e b ontents B × n task 104.2 Connected wedge theorem from quantum tasks 13 A Lower bound on mutual information from success probability 19B Non-trivial configurations in Poincaré-AdS Relativistic quantum tasks are quantum computations with inputs and outputs occurringat specified spacetime locations [3]. In the context of the AdS/CFT correspondence, theframework of holographic quantum tasks [1] considers such computations from a bulk aswell as boundary perspective. By comparing the two perspectives, it has been possible tofind a constraint on boundary entanglement placed by bulk causal features [1, 2]. Thisconstraint is called the connected wedge theorem; it states that a specific entanglementwedge must be connected when a related set of light cones overlap. In this paper we extendthe holographic quantum tasks framework and derive a stronger connected wedge theorem.To derive constraints for AdS/CFT from quantum tasks, we begin by defining a taskin the bulk. This is specified by sets of input and output locations, along with a channel– 1 – C C r c r R R (a) ˆ C ˆ C ˆ C ˆ R ˆ R (b) Figure 1 : (a) Bulk perspective on a particular quantum task. The task has inputs givenat bulk points c and c , and outputs are required at r and r . (b) Boundary view ofthe same task. Now regions ˆ C i whose entanglement wedge C i contains point c i become theinput regions, while the ˆ R i whose entanglement wedge ˆ R i contains the r i become the outputregions. The connected wedge theorem states that when J + ( C ) ∩ J + ( C ) ∩ J − ( R ) ∩ J − ( R ) is non-empty, the entanglement wedge of ˆ C ∪ ˆ C is connected.relating inputs to outputs. We can then discuss protocols for completing this bulk task, anddetermine the probability with which the best such protocol succeeds. Then, we identify acorresponding task in the boundary, and note that because the boundary describes the bulkthe boundary task is completed with the same probability. In some cases we can, beginningwith the success probability, show that there must be large amounts of entanglement in theboundary. Reasoning in this way leads to the connected wedge theorem.In [1, 2] the tasks in the bulk considered always had inputs and outputs taking a specialform. First, the inputs and outputs were given at locations idealized as points, and second,those points were located at asymptotic infinity. This was convenient in that it allowedthe corresponding boundary task to be identified with no additional assumptions: we cannaturally identify a point in AdS at infinity with a point in the CFT, and in this way definethe input and output locations for the boundary task.In this paper we extend this procedure to include bulk tasks with input locations thatare still points, but which are not at asymptotic infinity. In the boundary, entanglementwedge reconstruction implies the corresponding task is one with extended regions as inputlocations, namely regions which have the bulk input point in their entanglement wedge. Asimilar relationship relates bulk output points and boundary output regions. An exampleset-up is shown in figure 1.By considering this broader class of holographic quantum tasks we are led to a strongerversion of the connected wedge theorem than was previously given. A simple adaptation of– 2 –he proof given for the earlier connected wedge theorem in [2] can be used to prove our the-orem. As well, the new connected wedge theorem applies non-trivially to Poincaré-AdS ,where the earlier theorem was always trivial. The application to Poincaré requires consid-ering output regions which consist of two intervals, and so in particular are disconnected.This gives a particularly clear example of how the restriction of input and output locationsto be points failed to capture all useful constructions.Note that this article emphasizes the quantum tasks perspective, however, the readerinterested in the connected wedge theorem as a geometric statement can move directly tothe theorem statement in section 3 and its relativistic proof in section 5.Throughout the article we refer to boundary spacetime regions with hatted script letters ˆ X , ˆ A , ˆ B , ˆ V ... and their bulk entanglement wedges with un-hatted letters E W ( ˆ X ) = X , etc.We will use J ± ( · ) for the causal future or past taken in the bulk geometry, and ˆ J ± ( · ) for thecausal future or past taken in the boundary geometry. We will denote the Ryu-Takayanagisurface of a boundary region ˆ X by γ X . Outline of the article
The outline of this article is as follows.In section 2, we update the quantum tasks framework to consider input and outputspacetime regions. We discuss in detail how to identify bulk and boundary tasks in thiscontext. We emphasize that doing so requires identifying, for a given boundary region,a bulk region which stores the same quantum information. In other work this has beenunderstood to be the entanglement wedge [4–9].In section 3 we state the improved connected wedge theorem, making use of inputand output regions. We explain why this is a stronger theorem than given previously, andcomment on how to choose the inputs and output regions to arrive at the strongest possiblestatement. We also point out that the converse to the theorem does not hold. Finally weexplain how to apply the improved connected wedge theorem to Poincaré-AdS , whichinvolves taking one of the output regions to be disconnected.In section 4 we give the quantum tasks argument for the improved connected wedgetheorem, which exploits the expanded quantum tasks framework developed in section 2.Aside from the generalization to input and output regions, the treatment here also improveson [2] in the way errors in the bulk protocol are handled, in particular we derive a linearlower bound on the mutual information even in this noisy case.In section 5 we prove the stronger connected wedge theorem using the focusing theoremin general relativity. The relativistic proof is a simple modification of the proof of the earliertheorem appearing in [2].We conclude with a brief summary and some comments in section 6.
We will discuss quantum tasks where Alice is given inputs that are initially recorded intoextended spacetime regions, and must be output at extended output regions. To make this– 3 –ore precise, we define a notion of quantum information being localized to a spacetimeregion. Our definitions are adapted from [10].
Definition 1
Suppose one party, Alice, holds system X of a quantum state | Ψ (cid:105) XX (cid:48) . Thenwe say the subsystem X is localized to a spacetime region R if a second party, Bob, forwhom the state is initially unknown can prepare the X system by acting on R with somechannel M R→ X . If system X is localized to region R such that a channel M R→ X recovers X , we will saythat X is localized to R relative to M R→ X . Note that throughout this article by quantumsystem we mean a tensor factor of a Hilbert space, e.g. the X system of H XX (cid:48) = H X ⊗H X (cid:48) . It will also be convenient to say a quantum system X is excluded from a spacetimeregion R if Bob cannot learn anything about X by accessing R . One way to specify thisprecisely is to consider | Ψ (cid:105) XX (cid:48) to be in the maximally entangled state. Then X is excludedfrom R when I ( R : X (cid:48) ) = 0 .We should point out several features of these definitions. First, note that a system X islocalized to R if and only if it is localized to the domain of dependence of R , and similarlyit is excluded from R if and only if it is excluded from the domain of dependence of R .This follows because all the classical and quantum data in the domain of dependence of R is fixed by the data in R by time evolution. Consequently we will identify regions withtheir domains of dependence throughout this paper.Second, note that a quantum system X can be neither localized to nor excluded froma region R if some but not all information about X is available in R . As well, notice thatgiven a Cauchy surface Σ a quantum system can be excluded from both Σ ∩ R and Σ \ R .To do this, encode system X using the one-time pad [11] using a classical key k . This hidesthe state on X , which can only be revealed if k is known. The X register can then be passedthrough Σ ∩ R and k through Σ \ R , and system X will be excluded from both regions. Ofcourse, X will still be localized to the full Cauchy surface Σ .There are many possible ways in which a quantum system can be localized to a givenspacetime region, and a single quantum system can be localized to many different spacetimeregions. Which sets of spacetime regions the same quantum information can be localizedto is restricted however. In particular, if X is localized to a region R then it follows that X is excluded from the spacelike complement R c . This is because otherwise we could actindependently on R and R c to produce copies of X , in violation of the no-cloning theorem.As a simple example of how a single quantum system can be localized to many regions,suppose we have three subregions X , X , X which are all spacelike separated. Then define R = X ∪ X , R = X ∪ X , and R = X ∪ X . To localize a quantum system X toall three regions {R , R , R } , encode X into an error-correcting code on three subsystems S S S that corrects one erasure error. Send system S i to X i . Then each of the R i containtwo of the S i subsystems, and X can be recovered from each of them.Given a quantum system X , we can specify how it is localized in spacetime by specifyingtwo sets of regions, call them {A iX } i and {U iX } i . We specify that X is localized to each of It may also be interesting to repeat our discussion in the more general setting where X is a subalgebra. – 4 –he regions A iX , and excluded from each of the regions U iX . Implicitly, each region A iX comeswith a channel N A iX → X that specifies how X can be recovered from A iX . We summarizethis in the next definition.
Definition 2
A quantum system X is encoded into an access structure S X = ( {A iX } i , {U iX } i ) if X is localized to each of the regions {A iX } i and excluded from each of the regions {U iX } i . The term “access structure” is borrowed from the subject of quantum secret sharing, whichfeatures a closely related object. In [10] the authors characterized which access structuresit is possible to localize a quantum system to.Next, we give a definition of a relativistic quantum task. This builds on the definitionpresented in [1] by allowing for inputs and outputs to be recorded into arbitrary accessstructures.
Definition 3 A relativistic quantum task is defined by a tuple T = ( M , A , S A , B , S B , N A → B ) ,where: • M is the spacetime in which the task occurs, it is described by a manifold equippedwith a (Lorentzian) metric. • A = A ...A n a is the collection of all the input quantum systems, and S A = {S A , ..., S A na } is the set of all access structures for the input systems. • B = B ...B n b is the collection of all the output quantum systems, and S B = {S B , ..., S B nb } is the set of all access structures for the output systems. • N A → B is a quantum channel that maps the input systems A to the output systems B .Bob encodes the input systems A i in such a way that the access structures S A i are satisfied.To complete the task, Alice should apply the channel N A → B and localize each of the systems B i according to the access structure S B i . In order to encode the A i into the appropriate regions, Bob couples the regions A jA i tosome external system which initially hold the A i . To verify Alice has completed the tasksuccessfully, Bob will access one or more of the regions A iB j , U iB j and attempt to recoversystem B j . If Bob is able to produce B j from the authorized region A iB j he declares thetask successful. Similarly if he is unable to produce B j from the unauthorized region U iB j he declares the task successful. The probability that Alice’s outputs pass Bob’s test is hersuccess probability. Alice’s success probability maximized over all possible protocols forcompleting the task is the tasks success probability, p suc ( T ) .Note that if Bob acts on one of the output regions A iB j , U iB j in performing his test, weno longer require Alice have the correct outputs (or exclusions from) regions in the causal In general there could be many such channels, though this will not be important in this article. In quantum secret sharing a system X is recorded into shares X ...X n such that X can be recoveredfrom some subsets of shares, called authorized sets, and no information about X is available in another setof subsets of shares, called the unauthorized sets. – 5 –uture of the accessed region. Similarly, Bob will localize the inputs A j to regions A iA j so long as Alice never interferes. She may choose to access some region A iA j however andobtain A i , in which case Bob is no longer expected to localize A i to regions in the futureof A iA j .In the application considered below, we begin with a spacetime and use tasks as away to probe features of that fixed geometry. Consequently, we have defined quantumtasks to feature a fixed spacetime background. Doing so assumes Alice’s choice of protocoldoes not change the geometry. It is also possible to consider more general tasks, where weallow the spacetime geometry to react to Alice’s protocol, which might for instance involvedistributing large numbers of qubits which then change the geometry. We leave consideringthis to future work. In our definition of a quantum task in the last section, we have used an operational fram-ing. This is only for convenience however, and it is possible to remove this language. Inparticular, the protocol Alice carries out is in fact just a feature of some initial state | Ψ (cid:105) .All the instructions for her protocol are by necessity recorded there, all that happens dur-ing the execution of the protocol is time evolution according to the underlying theory’sHamiltonian. While Alice’s protocol is the internal dynamics of the theory in question,Bob preparing the inputs and collecting outputs correspond to couplings to some externalsystem. Viewing quantum tasks in this way motivates understanding them as probes of theunderlying theory they are defined in.Because tasks probe the underlying theory, if we are given an equivalence betweentwo theories it is natural to try and interpret this equivalence in the language of tasks.In particular we will consider the bulk and boundary theories in AdS/CFT. Within eachtheory, there is a set of tasks that can be defined and associated success probabilities, { ( T i , p suc ( T i )) i } . The equivalence of bulk and boundary theories suggests that for eachtask T defined in the bulk there is some corresponding task ˆT in the boundary, and furtherthat p suc ( ˆT ) = p suc ( T ) . We will make this more precise below.Our first step will be to restrict attention to a bulk described by classical geometryalong with quantum fields living on a curved background (which may be coupled to thegeometry). This means that while the boundary theory completely describes the bulk, theconverse is not true. Consequently we will expect an inequality, p suc ( T ) ≤ p suc ( ˆT ) . Beforeunderstanding this in more detail however, we need to specify how a bulk task should beassociated with a boundary task.Given a task in the bulk T = ( M , A , S A , B , S B , N A → B ) , we should identify theboundary dual of each element of the tuple. Beginning with M , the bulk geometry, wedefine ˆT to be in the geometry ∂ M , the boundary of M . The inputs A , outputs B , andchannel N A → B we may identify trivially across bulk and boundary. This is because whilethe bulk and boundary degrees of freedom look very different, we can record the samequantum states into these different degrees of freedom.Next we need to discuss how to identify an access structure in the bulk with a corre-sponding access structure in the boundary. Note that in principle, because the AdS/CFT– 6 –orrespondence fixes the boundary description given the bulk, the boundary access struc-ture ( { ˆ A jA i } , { ˆ U jA i }}} ) is fixed by the bulk one ( {A jA i } , {U jA i }}} ) . We have not understoodhow to do this in the most general case, but can make some statements which will besufficient for the application discussed here.First, notice that given bulk authorized regions {A jA i } j , we have { ˆ A kA i } k ⊇ (cid:91) j { ˆ X : A jA i ⊆ E W ( ˆ X ) } . (2.1)This is because the entanglement wedge E W ( ˆ X ) is the portion of the bulk which ˆ X canbe used to recover [4–9], so when A jA i ⊆ E W ( ˆ X ) the boundary region ˆ X can be used torecover A i , which implies ˆ X is an authorized region. The other inclusion does not follow ingeneral since there may be some boundary regions ˆ A jA i whose entanglement wedge includesa portion but not all of A jA i and which still construct A i .Given a bulk unauthorized region, we can say that { ˆ U kA i } k ⊇ (cid:91) j { ˆ X : U jA i ⊇ E W ( ˆ X ) } . (2.2)This follows because E W ( ˆ X ) is the largest bulk region whose quantum information can bereconstructed given ˆ X , so U jA i ⊇ E W ( ˆ X ) means ˆ X does not reconstruct A i . Note thatunless one or more of the U jA i are anchored to the boundary, the set { ˆ X : U jA i ⊇ E W ( ˆ X ) } will be empty.We will be interested only in a special case, where the bulk tasks access structures allhave only authorized regions, and where those authorized regions are points. In this casethe inclusion 2.1 becomes an equality, fully specifying the boundary authorized regions fromthe bulk ones. Further, there will be no boundary unauthorized regions. Given a bulk task T and associated boundary task ˆT , we’ve claimed p suc ( T ) ≤ p suc ( ˆT ) .This follows because any protocol that completes the task in the bulk with some proba-bility p will be mapped under the AdS/CFT duality to a protocol in the boundary. Thebulk task’s success probability is determined by the information localized to the regions ( {A jA i } , {U jA i }}} ) . In the boundary description the same information will be available inthe regions ( { ˆ A jA i } , { ˆ U jA i }}} ) , so the boundary protocol will complete the task with prob-ability p as well. Note that we claim only an inequality, rather than an equality, becausemany protocols in the boundary theory will correspond to bulk protocols that change thegeometry M , which we assumed should be fixed and unaffected by the protocol. Worse,some boundary protocols might correspond to leaving a semi-classical description of thebulk altogether. We state the improved connected wedge theorem below. Of course the spacelike complements [ A kA i ] c do not contain any information about A i , but it is notnecessary to designate these as unauthorized, since this is immediate from the A kA i being authorized. – 7 – C ˆ C ˆ C ˆ R ˆ R (a) C C R (b) Figure 2 : (a) A view of the boundary of AdS . Left and right edges of the diagram areidentified. Shown is an example choice of input regions ˆ C i and output regions ˆ R i . Thisparticular choice is maximal for the regions ˆ V i defined by ˆ C i = ˆ V i . (b) Bulk perspectiveon the same choice of regions, showing the entanglement wedges C i and R i . Only one ofthe out regions is shown, to avoid cluttering the diagram. In the bulk there is a non-emptyentanglement scattering region J E → . Theorem 4 (Connected wedge theorem)
Pick four regions ˆ C , ˆ C , ˆ R , ˆ R on the boundaryof AdS. From these, define the decision regions ˆ V ≡ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) , ˆ V ≡ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) . (3.1) Assume that ˆ C i ⊆ ˆ V i . Define the entanglement scattering region J E → ≡ J + ( C ) ∩ J + ( C ) ∩ J − ( R ) ∩ J − ( R ) , (3.2) where C i = E W ( ˆ C i ) and R i = E W ( ˆ R i ) . Then, J E → (cid:54) = ∅ implies that E W ( ˆ V ∪ ˆ V ) isconnected. Notice that this theorem generalizes the earlier one appearing in [2]. In particular choosingthe ˆ C i and ˆ R i to be points we recover the earlier theorem.It is interesting to consider starting with a choice of regions ˆ V and ˆ V , then pick regions ˆ C , ˆ C , ˆ R , ˆ R to understand if ˆ V and ˆ V share a connected entanglement wedge. In AdS there is a unique ‘best’ way to do this, in the sense that one particular choice of regionswill conclude there is a connected wedge whenever any choice of regions does.To find the optimal choice of ˆ C i , ˆ R i , we note first that there is a maximal choice ofregions ˆ C i imposed by the constraint ˆ C i ⊆ ˆ V i : choose ˆ C i = ˆ V i . Further, there is a maximal– 8 – V ˆ V π/ (a) ˆ V ˆ V (b) Figure 3 : A counterexample to the converse of Theorem 4. (a) Vacuum AdS withregions ˆ V and ˆ V chosen antipodally and to each occupy π/ of the boundary. Choosingthe maximal consistent input and output regions, the entanglement scattering region isexactly one point, and the Ryu-Takayanagi surface is on the transition from disconnected(blue) to connected (red). (b) Spherically symmetric matter is added to the bulk. Now theentanglement wedges of ˆ V and ˆ V reach less deeply into the bulk [12], and the light rayssent inward normally from their extremal surfaces are delayed. This closes the entanglementscattering region. By spherical symmetry however the Ryu-Takayanagi surface remains onthe transition. Deforming ˆ V to be larger ensures we are in the connected phase, and forsmall enough deformation ensures the scattering region remains empty. Figure reproducedfrom [2].choice of ˆ R i consistent with a given ˆ V , ˆ V , which is illustrated in figure 2a. Since anyother choice ˆ C (cid:48) i , ˆ R (cid:48) i has ˆ C (cid:48) i ⊆ ˆ C i and ˆ R (cid:48) i ⊆ ˆ R i these maximal choices have J (cid:48)E → ⊆ J E → ,so whenever a non-maximal choice has a non-empty entanglement scattering region themaximal choice will. Thus whenever ˆ C (cid:48) i , ˆ R (cid:48) i can be used to conclude ˆ V ∪ ˆ V has a connectedentanglement wedge, the maximal choice will conclude the same.In general, the converse to Theorem 4 does not hold. This is immediate from the fact,pointed out in [2], that the converse to the point-based case does not hold, which is a specialcase of the theorem presented here. As commented on in the last paragraph however, thepoint based choice is not the strongest choice of regions to understand if the entanglementwedge E W ( ˆ V ∪ ˆ V ) is connected. We can only expect a converse in the case where we takethe optimal choice of regions outlined above. Taking the optimal choice of regions howeverthe theorem still does not have a converse, as we argue in figure 3. The connected wedge theorem applies to any asymptotically AdS spacetime, includingglobal AdS and Poincaré-AdS spacetimes in arbitrary dimensions. To apply the theoremmeaningfully however, we need to find configurations of regions ˆ C , ˆ C , ˆ R , ˆ R such that the– 9 – V ˆ V ˆ R ˆ R ˆ R Figure 4 : A typical choice of regions ˆ C = ˆ V , ˆ C = ˆ V , ˆ R , ˆ R in the boundary of Poincaré-AdS which leads to a non-trivial conclusion in the connected wedge Theorem 4. Region ˆ R consists of two wedges which each extend to infinity.bulk entanglement scattering region is non-empty, while the boundary scattering region isempty.It is not immediately clear how to find non-trivial configurations of input and outputregions in Poincaré AdS . Indeed, at least for pure Poincaré-AdS no such configura-tions exist when the input and output regions are chosen to be points. One way to see thisis to start with non-trivial arrangements of points c , c , r , r in global AdS , and chosea Poincaré patch which includes regions ˆ V and ˆ V . Doing so one always finds that oneof the four points sit outside the patch, and consequently we cannot state the non-trivialinstances of the theorem using points directly in Poincaré AdS .For extended regions it is straightforward to find non-trivial configurations in PoincaréAdS . An example configuration is shown in figure 4. Importantly, region ˆ R consistsof two disconnected parts, where each connected component consists of a half line. Inappendix B we find configurations which are non-trivial in the case where the bulk is pureAdS. Since many of these configurations have extended entanglement scattering regions,and the scattering regions should be deformed only a small amount for small perturbationsto the bulk geometry, there will also be many non-trivial configurations when matter isadded. B × n task Following [1, 2], we discuss the B task. This task has C and C as authorized regionsfor inputs A and A , and R and R as authorized regions for outputs B and B . Alicewill be given a guarantee that A is in one of the states H q | b (cid:105) , and A stores the classicaldata q . Both q and b are bits, q, b ∈ { , } . Alice’s task is to localize b to both R and R . The name of this task comes from its similarity to the BB84 key distribution protocol, itself named forBennet and Brassard [13]. – 10 – C R R J E → (a) | Ψ + (cid:105) ˆ V ˆ V ˆ C ˆ C ˆ R ˆ R (b) Figure 5 : Bulk and boundary perspectives on the ˆB task. (a) Causal features present inthe bulk geometry. Signals from C and C may meet in the scattering region J E → , thentravel to either R or R . The scattering region is a resource, useful for completing quantumtasks. (b) Causal features present in the boundary geometry, which lacks an entanglementscattering region. ˆ C may send signals to ˆ R and ˆ R , and ˆ C may send signals to ˆ R and ˆ R .The boundary replaces the resource of a non-empty scattering region with entanglementbetween ˆ V and ˆ V .We will be interested in two strategies for completing the task: a local strategy and anon-local strategy. The local strategy is one which makes use of the scattering region J + ( C ) ∩ J + ( C ) ∩ J − ( R ) ∩ J − ( R ) (4.1)and so is only available when this region is non-empty. The non-local strategy does not usethis region. The two strategies are shown in figures 5a and 5b respectively. We treat eachbelow. Note that in arguing for Theorem 4, we will be interested in the case where in thebulk the scattering region defined above is available and so the local strategy can be used,while in the boundary the corresponding scattering region is empty, so it is necessary touse a non-local strategy. Local strategy for B × n task The causal features of the task in the local strategy are captured by figure 5a. In this case,there is a protocol which completes the task with high probability. In particular, Aliceshould bring H q | b (cid:105) from C and q from C together inside the above causally defined region.Then, she applies H q to obtain ( H q ) | b (cid:105) = | b (cid:105) , measures in the computational basis to learn b , and then sends b to both R and R . Assuming this can be carried out as describedthis completes the task with probability p suc = 1 . More physically we will allow for the– 11 –resence of noise in carrying out this protocol, and say the success probability satisfies p suc ( ˆB ) ≥ − (cid:15) .Consider repeating the B task n times in parallel. Call this repeated task B × n . Thisrepeated task has inputs A = (cid:78) ni =1 H q i | b i (cid:105) , A = { q i } i , and required outputs { b i } i atboth R and R . The q i and b i are random and independent. We will declare the task tobe completed successfully if a fraction − (cid:15) of the n tasks are completed successfully. Ifeach task is completed with probability p suc = 1 − (cid:15) , then this occurs with probability p suc ( B × n ) = 1 − (cid:15) n . (4.2)This is the success probability for the B × n when a local strategy is available. Non-local strategy for ˆB × n task In the non-local case, where no scattering region is available, we will be interested instrategies of the form shown in figure 5b. These do not use the scattering region butinstead use entanglement shared between the decision regions ˆ V ≡ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) , ˆ V ≡ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) , (4.3)to complete the task. The regions ˆ V and ˆ V are relevant because they are the largest regionswhich contain the inputs A and A but are also in the past of both output regions. Ratherthan discussing specific strategies for completing the task non-locally, we are interestedin lower bounding the amount of entanglement necessary to complete the task with highprobability when using any non-local strategy.We begin by assuming I ( ˆ V : ˆ V ) = 0 , that ˆ C i ⊆ ˆ V i , and considering a single instance ofthe task, ˆB × . We will see that this leads to a success probability bounded strictly below1. Lemma 5
Consider the ˆB × task with I ( ˆ V : ˆ V ) = 0 . Then any strategy for completing thetask has p suc ( ˆB × ) ≤ cos ( π/ . This lemma is proven in [14]. To understand it heuristically we can reason as follows. Inregion ˆ C Alice holds one of the states H q | b (cid:105) for q, b ∈ { , } . If ˆ C also held the basisinformation, Alice could measure in the computational basis {| (cid:105) , | (cid:105)} if q = 0 or theHadamard basis {| + (cid:105) , |−(cid:105)} if q = 1 to determine b . Since I ( ˆ V : ˆ V ) = 0 however, and q is held in ˆ V , Alice must act in a way independent of q . Doing so, she cannot determine b perfectly. Instead her optimal strategy is to measure in an intermediate basis {| ψ (cid:105) , | ψ (cid:105)} where | ψ (cid:105) = cos (cid:16) π (cid:17) | (cid:105) + sin (cid:16) π (cid:17) | (cid:105) , | ψ (cid:105) = cos (cid:18) π (cid:19) | (cid:105) + sin (cid:18) π (cid:19) | (cid:105) . (4.4) This should be contrasted to the condition considered in [1, 2], where B × n was declared successful onlyif all n of the individual B tasks were completed successfully. Taking this more relaxed condition is whateventually leads us to the bound 4.6, which improves on the earlier bound. – 12 –his leads to the cos ( π/ success probability, so that the bound in Lemma 5 is actuallytight. Note that the assumption ˆ C i ⊆ ˆ V i is important for this reasoning to hold. Inparticular, we used that q ∈ ˆ C ⊆ ˆ V in saying I ( ˆ V : ˆ V ) = 0 implies the measurement on A ∈ ˆ C ⊆ ˆ V is independent of q .Next we consider the parallel repetition task, ˆB × n . In [14] the following statement hasbeen proven. Lemma 6
Consider the ˆB × n task with I ( ˆ V : ˆ V ) = 0 , assume that ˆ C i ⊆ ˆ V i and that thescattering region is empty. Require that a fraction − δ of the individual ˆB tasks aresuccessful. Then any strategy for completing the task has p suc ( B × n ) ≤ (cid:16) h ( δ ) cos (cid:16) π (cid:17)(cid:17) n ≡ (cid:16) h ( δ ) β (cid:17) n , (4.5) where h ( δ ) is the binary entropy function h ( δ ) ≡ − δ log δ − (1 − δ ) log (1 − δ ) and thesecond equality defines β . For small enough δ we have that h ( δ ) β < , so this gives a good bound on the successprobability.In the boundary, where the scattering region is empty, the success probability is ex-ponentially small when the mutual information is zero. Comparing to the bulk where thesuccess probability is exponentially close to one suggests the true boundary state containslarge mutual information. This is indeed the case, as we show in the next lemma. Lemma 7
Suppose the ˆB task is completed with probability p suc ≥ − (cid:15) n , and that thescattering region is empty. Then I ( ˆ V : ˆ V ) ≥ n ( − log 2 h (2 (cid:15) ) β ) − O (( (cid:15)/β ) n ) (4.6)This is proven in appendix A. In the next section we employ this bound to argue for theconnected wedge theorem. Note that this improves on the bound presented in [2]. In this section we give the quantum tasks argument for the connected wedge theorem. Wefirst recall the theorem for convenience.
Theorem 4 (Connected wedge theorem)
Pick four regions ˆ C , ˆ C , ˆ R , ˆ R on the boundaryof AdS. From these, define the decision regions ˆ V ≡ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) , ˆ V ≡ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) . (4.7) Assume that ˆ C i ⊆ ˆ V i . Define the entanglement scattering region J E → ≡ J + ( C ) ∩ J + ( C ) ∩ J − ( R ) ∩ J − ( R ) , (4.8)– 13 – here C i = E W ( ˆ C i ) and R i = E W ( ˆ R i ) . Then, J E → (cid:54) = ∅ implies that E W ( ˆ V ∪ ˆ V ) isconnected. Argument.
Consider two cases. First, supposed that ˆ V ∩ ˆ V (cid:54) = ∅ . Then we immediatelyhave that the entanglement wedge of ˆ V ∪ ˆ V is connected, and we are done.Next, assume that ˆ V ∩ ˆ V = ∅ . This is just the statement that the boundary scatteringregion ˆ J → = ˆ J + ( ˆ C ) ∩ ˆ J + ( ˆ C ) ∩ ˆ J − ( ˆ R ) ∩ ˆ J − ( ˆ R ) (4.9)is empty. By assumption however, the bulk entanglement scattering region J E → is non-empty. This implies the existence of four points c , c , r , r such that J + ( c ) ∩ J + ( c ) ∩ J − ( r ) ∩ J − ( r ) (cid:54) = ∅ . (4.10)Choose a B × n task in the bulk with c , c as input points and r , r as output points.Then because the above region is non-empty, we can use the local strategy in the bulk, andwe obtain a high success probability, p suc ( ˆB × n ) ≥ − (cid:15) n . (4.11)Next, we should discuss how large we can take n . The obstruction to taking n arbitrarilylarge is that if we make use of too many qubits, the bulk protocol may change the geometry,deforming the geometry we are attempting to study. To avoid this we choose n to be anyorder in /G N less than linear. Then if each qubit carries some energy ∆ E , Einstein’sequations dictate that the coupling to geometry is G µν = O ( G N ∆ En ) . (4.12)Choosing n < O (1 /G N ) ensures that in the G N → limit we have no backreaction, asneeded to ensure we are studying the intended geometry.Starting with B × n , we label the corresponding boundary task by ˆB × n . Following thediscussion in section 2, we know ˆB × n has the same inputs and outputs as the correspondingbulk task. Further, we have by assumption that c i ∈ E W ( ˆ C i ) ,r i ∈ E W ( ˆ R i ) . (4.13)Thus ˆ C i is an authorized region for A i in the boundary task, and ˆ R i is an authorized regionfor B i . Since ˆ C i ⊆ ˆ V i , and by assumption the boundary scattering region ˆ V ∩ ˆ V is empty,the boundary uses a non-local strategy. Lemma 7 then applies and we can conclude I ( ˆ V : ˆ V ) ≥ n ( − log β ) − O (( (cid:15)/β ) n ) . (4.14)Since n is any order less than O (1 /G N ) , we can conclude that I ( ˆ V : ˆ V ) = O (1 /G N ) , whichoccurs only when E W ( ˆ V ∪ ˆ V ) is connected.– 14 –e should note that there is a gap in this argument, which was noted also in [2]. Inparticular the causal diagram 5b does not include the regions that sit between ˆ V and ˆ V .In general, these can be made use of to complete the ˆB × n task without entanglement. However, such strategies require GHZ correlations in the CFT, which are not expected[15]. As well, it seems possible to rule out such strategies by keeping Alice ignorant ofthe location of the regions ˆ C i before the beginning of the task, in which case she cannotcoordinate actions with the intermediate regions. We leave better understanding this tofuture work, and for now rely on gravitational reasoning to provide a complete proof. The proof of Theorem 4 is nearly identical to the proof of the earlier connected wedgetheorem, which appears already in [2]. For readers familiar with the earlier proof, it sufficesto note that the only change is to replace the causal horizon ∂ [ J − ( r ) ∩ J − ( r )] with the nullsheet ∂ [ J − ( R ) ∩ J − ( R )] . The key point is that ∂ [ J − ( r ) ∩ J − ( r )] and ∂ [ J − ( R ) ∩ J − ( R )] meet the boundary along the same curves, and both surfaces have area theorems for pastdirected null geodesics. This allows the two surfaces to play similar roles in the proof.To be self contained, we also present the proof briefly here. Assume that the null energycondition holds. Work by contradiction by assuming that the entanglement scattering regionis non-empty and the minimal area extremal surface homologous to ˆ V ∪ ˆ V consists of aconnected component γ V homologous to ˆ V and a connected component γ V homologous to ˆ V . Then, the maximin prescription [6] for finding Ryu-Takayanagi surfaces dictates thatthere will exists a Cauchy slice Σ of the bulk in which γ V ∪ γ V is minimal, where by γ X we always mean the Ryu-Takayanagi surface for a boundary region ˆ X . We then constructa codimension surface we call the null membrane N Σ , along with the codimension contradiction surface C Σ . The null membrane facilitates a comparison of the area of γ V ∪ γ V and the contradiction surface. The contradiction surface will turn out to have less area thanthe candidate surface γ V ∪ γ V , which provides the contradiction.The null membrane is illustrated in figure 6. It is defined as the union of two surfaces,called the lift and the slope . The lift is defined by L = ∂J + ( V ∪ V ) ∩ J − ( R ) ∩ J − ( R ) . (5.1)The slope is defined by S Σ = ∂ [ J − ( R ) ∪ J − ( R )] ∩ J − [ ∂J + ( V ∪ V )] ∩ J + (Σ) . (5.2)The contradiction surface is defined by C Σ = S Σ ∩ Σ . Note that ∂J + ( V ∪ V ) is generatedby geodesics starting on the inward, future pointing, null normals to the extremal surface γ V ∪ γ V . Similarly, ∂ [ J − ( R ) ∪ J − ( R )] is generated by geodesics starting on the inward,past directed, null normals to γ R ∪ γ R . See appendix B of [2]. – 15 – C Σ γ V ˆ V ˆ V Figure 6 : The null membrane. The blue surface is the lift L , which is generated by thenull geodesics defined by the inward, future pointing null normals to γ V ∪ γ V , where γ V i is the Ryu-Takayanagi surface for region ˆ V i . The red surfaces make up the slope S Σ , whichis generated by the null geodesics defined by the inward, past directed null normals to γ R ∪ γ R . The ridge R is where null rays from γ V and γ V collide. The contradictionsurface C Σ is where the slope meets a specified Cauchy surface Σ .To repeat the proof of [2] we need to establish various features of the null membrane.The first is that the area of the past directed null geodesics that generate the slope havedecreasing area. This holds because this congruence is defined by beginning with the inward,past directed null normals to γ R ∪ γ R . Since this surface is extremal, the focusing theorem(which assumes the null energy condition) implies that this congruence has decreasing area.Similarly, the null normals that generate the lift also have decreasing area.The second needed feature of the null membrane is that the contradiction surface ishomologous to ˆ V ∪ ˆ V . This follows because the restriction of the contradiction surface tothe boundary is the spacelike boundary of ˆ V ∪ ˆ V .The null membrane can be used to establish that the contradiction surface has less areathan the candidate surface. To see this, consider pushing the candidate surface forwardalong the congruence defined by the lift, removing any generators which collide. Continuepushing the surface forward until reaching the slope. When doing so, any generators whichreach the ridge will be removed, where the ridge is defined by R = ∂J + ( V ) ∩ ∂J + ( V ) ∩ J − ( R ) ∩ J − ( R ) . (5.3)See also figure 6. Assume momentarily that the ridge is non-empty. Then after pushingforward the surface will consist of two disconnected components which sit on the slope.Finally, push the surface backwards along the slope until it reaches Σ , and becomes thecontradiction surface. Because the null congruences defining the lift and the slope begin asnormal vectors to extremal surfaces, moving into the future along the lift and into the pastalong the slope both decrease area. Removing colliding generators also decreases the area.Thus we can conclude the contradiction surface has less area than the candidate surface,as needed.To justify our assumption that the ridge is non-empty, note that this occurs wheneverthe entanglement scattering region J E → is non-empty, since by assumption J E → is– 16 –on-empty and J E → ⊆ J + ( V ) ∩ J + ( V ) ∩ J − ( R ) ∩ J − ( R ) , (5.4)so that the region on the right is non-empty. But this region being non-empty means J + ( V ) and J + ( V ) must meet in the past of R and R , which means the ridge is non-empty. In the context of the connected wedge theorem with input and output regions taken to bepoints, the authors of [2] noted that the scattering region sits inside of the entanglementwedge of ˆ V ∪ ˆ V , at least in the context of bulk dimensions. We can straightforwardlyadapt their argument to our context to see that the larger entanglement scattering regionis also inside of the entanglement wedge, again in bulk dimensions.To see this, define the region X = J + [ ˆ V ∪ ˆ V ] c ∩ [ ˆ V ∪ ˆ V ] c . (5.5)This is the closure of the spacelike complement of ˆ V ∪ ˆ V . In dimensions, thisconsists of the domains of dependence of two intervals which we call ˆ X and ˆ X . Note that ˆ V ∪ ˆ V ∪ ˆ X ∪ ˆ X is a complete Cauchy slice of the boundary, which we extend into the bulkto some Cauchy slice Σ . We will choose this extension such that γ X and γ X are containedin Σ .Next we note that ˆ R is inside the domain of dependence of ˆ V ∪ ˆ X ∪ ˆ V , which welabel ˆ D , while ˆ R sits inside the domain of dependence ˆ V ∪ ˆ X ∪ ˆ V , which we label ˆ D . Because of this, J − ( R ) will be inside J − ( D ) , while J − ( R ) will be inside J − ( D ) .Consequently we learn J − ( R ) ∩ J − ( R ) ⊆ J − ( D ) ∩ J − ( D ) . (5.6)Notice that, assuming the entanglement wedge E W ( ˆ V ∪ ˆ V ) is connected, the future bound-ary of J − ( D ) ∩ J − ( D ) is also the future boundary of E W ( ˆ V ∪ ˆ V ) . This is because theentangling surface for E W ( ˆ V ∪ ˆ V ) consists of two components, one of which is homologousto ˆ X and the other homologous to ˆ X , and so these two components are the entanglingsurfaces for ˆ D and ˆ D respectively. Thus equation 5.6 gives that J − ( R ) ∩ J − ( R ) is inthe past of the future boundary of E W ( ˆ V ∪ ˆ V ) . Since the scattering region is a subregionof J − ( R ) ∩ J − ( R ) , it follows that this also holds for the scattering region.It remains to show that the scattering region is to the future of the past boundary of E W ( ˆ V ∪ ˆ V ) . This is immediate, because the past of γ X and γ X meets the boundary alongthe past boundaries of V and V . Thus any points in the future of V and V must be inthe future of this past boundary. In this article we have expanded the holographic quantum tasks framework to include inputsand outputs encoded into arbitrary access structures. We’ve illustrated the usefulness of thisframework by using this construction to motivate the improved connected wedge theorem,which we could then verify using a geometric proof.– 17 – .1 Non-triviality of the theorem in various spacetimes
We have noted that if the boundary regions ˆ V and ˆ V overlap, then the conclusion ofthe connected wedge theorem is trivial, since in that case ˆ V and ˆ V immediately have aconnected entanglement wedge. Non-trivial configurations in global AdS exist and arediscussed explicitly in [1, 2], while [2] also noted non-trivial configurations exist in the AdSsoliton, and here we have given non-trivial configurations for Poincaré-AdS (which onlyexist when considering the region based statement).We have not explored in detail however if the region based connected wedge theoremapplies non-trivially in higher dimensions. In Poincaré-AdS it seems straightforwardto construct non-trivial configurations by defining the input and output regions to be asdefined in section 3.2, but extended infinitely in the extra transverse direction. In global-AdS it is less clear how to construct such non-trivial configurations, though one plausibleavenue is to begin with the Poincaré configurations and consider their embedding into theglobal spacetime. We leave understanding this in detail to future work. One technical improvement over [2] made in this work is a more robust handling of possiblenoise in the bulk protocol. In particular we proved I ( ˆ V : ˆ V ) ≥ n ( − log β ) − O (( (cid:15)/β ) n ) , (6.1)where (cid:15) was the error in completing a single instance of the B task. Since we argued n can be taken to be any order less than O (1 /G N ) , this bound allows us to directly concludethat the mutual information between two regions ˆ V and ˆ V which have a scattering regionis O (1 /G N ) . In [2], using a weaker bound, it was only possible to prove the mutual infor-mation was O (1 /G N ) by first using the HRT formula to see that the mutual information iseither O (1 /G N ) or O (1) . From a tasks perspective the failure of Theorem 4 to have a converse is tied to the fact thatwe are interested in a fixed bulk geometry. While protocols that take place in that fixedgeometric background have a boundary description, many boundary protocols will deformthe geometry. Because of this, the bulk and boundary success probabilities are relatedby an inequality, p suc ( T ) ≤ p suc ( ˆT ) , so that sufficient entanglement to do the task in theboundary does not imply the task can be done in the bulk fixed geometry, and so does notsignal the appearance of a bulk scattering region.One interesting possibility is that large boundary correlation when measured in someway other than the mutual information will imply the existence of a bulk entanglementscattering region. In particular, this hypothetical measure of correlation should count onlyentanglement that can be made use of by operations that preserve the bulk geometry. Then,its appearance would signal that there should be a bulk protocol in that geometry which– 18 –ompletes the required task, which in turn would imply the existence of the scattering re-gion. Acknowledgements
I thank Kfir Dolev, Jon Sorce, and Jason Pollack for valuable discussions and feedback ondrafts of this article. I am supported by a C-GSM award given by the National Science andEngineering Research Council of Canada.
A Lower bound on mutual information from success probability
In this section we prove the lower bound on mutual information 4.6. Our starting pointis Lemma 6 which bounds the success probability for states with zero mutual information.Our argument will show that states with high probability must be far from these zeroprobability states in terms of trace distance, which we can then translate to a bound onmutual information. Note that the discussion here is a repetition of an argument in [2],but, because various parameters are changed in our context, we’ve included the proof withupdated parameters here.We begin by recalling a continuity bound on success probability for any quantum task,stated earlier in [2].
Lemma 8
Consider a quantum task which takes as input a quantum system A . Then theprobability of completing the task, call it p suc , satisfies the continuity bound | p suc ( ρ A ) − p suc ( σ A ) | ≤ || ρ A − σ A || . (A.1)The proof follows by viewing the task as a procedure for distinguishing ρ from σ . Themaximal success probability of distinguishing states can be written in terms of the tracedistance, which leads to the above inequality. Intuitively, we should understand the lemmaas saying that nearby states produce nearby success probabilities.For the task ˆB × n , we have p suc ( ρ ˆ V ˆ V ) ≥ − (cid:15) n ,p suc ( ρ ˆ V ⊗ ρ ˆ V ) ≤ (2 h ( δ ) β ) n . (A.2)The state ρ ˆ V ˆ V is any boundary state where the bulk scattering region is non-empty, while ρ ˆ V ⊗ ρ ˆ V is the tensor product of its marginals. The second bound follows from Lemma 6.The remainder of the argument consists of relating the trace distance to the relativeentropy. In particular we have that the trace distance and fidelity are related by [16, 17] || ρ − σ || ≤ (cid:112) − F ( ρ, σ ) (A.3)where F ( ρ, σ ) is the fidelity. Additionally, − F ( ρ, σ ) ≤ D ( ρ || σ ) (A.4) We thank Jon Sorce for discussion on these points. – 19 – c ˆ V ˆ V ˆ R ˆ R ˆ R r e q L q R q R q L Figure 7 : A typical choice of regions ˆ C = ˆ V , ˆ C = ˆ V , ˆ R , ˆ R in the boundary of Poincaré-AdS which leads to a non-trivial conclusion in the connected wedge Theorem 4. The regionsare conveniently specified by choosing the four points c , c , r , e .where D ( ρ | σ ) is the relative entropy. The final observation is that D ( ρ AB || ρ A ⊗ ρ B ) = I ( A : B ) ρ (A.5)Combining inequalities A.3 and A.4 and Lemma 8 to lower bound the relative entropy, andhence the mutual information, in terms of success probabilities we find I ( ˆ V : ˆ V ) ρ ≥ − − | p suc ( ρ A ) − p suc ( ρ AB ⊗ ρ B ) | ] . (A.6)Finally using our bounds on success probability A.2 we obtain I ( ˆ V : ˆ V ) ≥ n ( − log 2 h (2 (cid:15) ) β ) − O (( (cid:15)/β ) n ) (A.7)as claimed. B Non-trivial configurations in Poincaré-AdS
Here we explicitly check there are non-trivial configurations for the connected wedge theo-rem in (pure) Poincaré-AdS.Begin by considering points at c = (0 , − (cid:96) ) ,c = (0 , (cid:96) ) ,r = ( T r , ,e = ( T e , . (B.1) Here and throughout the section coordinates are labelled by ( t, x ) . – 20 –he set-up is shown in figure 7. This defines regions, ˆ R = ˆ J + ( e ) ∩ ˆ J − ( r ) , ˆ V = [ ˆ J + ( c ) ∩ ˆ J − ( r )] \ ˆ J + ( e ) , ˆ V = [ ˆ J + ( c ) ∩ ˆ J − ( r )] \ ˆ J + ( e ) . (B.2)We then define ˆ R as the set of points spacelike separated from ˆ R .We are interested in finding choices of points such that the entanglement scatteringregion is non-empty. Because we are in pure AdS, we find J + ( V ) = J + ( c ) ,J + ( V ) = J + ( c ) ,J − ( R ) = J − ( r ) ,J − ( R ) = [ J + ( e )] c . (B.3)Thus a non-empty entanglement scattering region amounts to configurations such that J E → = J + ( c ) ∩ J + ( c ) ∩ J − ( r ) ∩ [ J + ( e )] c (B.4)is non-empty. These light cones are given in Poincaré-AdS by J + ( c ) = { ( t, x, z ) : ( x + (cid:96) ) + z − t ≤ } , (B.5) J + ( c ) = { ( t, x, z ) : ( x − (cid:96) ) + z − t ≤ } , (B.6) J + ( r ) = { ( t, x, z ) : x + z − ( T r − t ) ≤ } , (B.7) [ J + ( e )] c = { ( t, x, z ) : x + z − ( T e − t ) ≥ } . (B.8)By symmetry, if the scattering region is non-empty it will include at least one point with x = 0 . Thus it suffices to find a solution to (cid:96) + z − t ≤ , (B.9) z − ( T r − t ) ≤ , (B.10) z − ( T e − t ) ≥ . (B.11)Eliminating z and t from these inequalities we find one constraint, T e T r − (cid:96) > . (B.12)We need to check this can be satisfied while also having ˆ V ∩ ˆ V = ∅ . In our setup thisoccurs when the rightmost point in ˆ V is to the left of x = 0 and the leftmost point in ˆ V to the right of x = 0 . Via some Minkowski space geometry we can work out the end-points– 21 –f the regions ˆ V , ˆ V , q L = (cid:18) T r − (cid:96) , − T r + (cid:96) (cid:19) ,q R = (cid:18) T e + (cid:96) , T e − (cid:96) (cid:19) ,q R = (cid:18) T r − (cid:96) , T r + (cid:96) (cid:19) ,q L = (cid:18) T e + (cid:96) , (cid:96) − T e (cid:19) , (B.13)where q iL is the left spacelike boundary of ˆ V i , and q iR is the right spacelike boundary of ˆ V i (see figure 7). We see that keeping ˆ V and ˆ V disjoint just requires T e < (cid:96) . Givena choice of T e , (cid:96) such that T e < (cid:96) , we can always choose T r sufficiently large to satisfyB.12, guaranteeing the existence of the scattering region. This establishes that there arenon-trivial configurations for Theorem 4 in Poincaré-AdS .Given that the scattering region exists, Theorem 4 concludes that ˆ V and ˆ V shouldhave a connected entanglement wedge. We can also check this explicitly in the simplesetting considered here. In particular we would like to understand when A min [( q L , q R )] + A min [( q L , q R )] ≥ A min [( q L , q R )] + A min [( q R , q L )] . (B.14)To calculate the minimal area surface for one of the ˆ V i , we need the invariant length of itscross section. This is straightforward to work out from B.13, ∆ x − ∆ t = (cid:96) ( T r − T e ) . (B.15)The intervals ( q L , q R ) and ( q R , q L ) are on a constant t surface, so we can just use theirwidths. The areas of the minimal surfaces are then A min [( q iL , q iR )] = L AdS G N ln (cid:32) (cid:112) (cid:96) ( T r − T e ) (cid:15) (cid:33) , A min [( q L , q R )] = L AdS G N ln (cid:18) (cid:96) − T e (cid:15) (cid:19) , A min [( q R , q L )] = L AdS G N ln (cid:18) T r + (cid:96)(cid:15) (cid:19) . (B.16)The condition B.14 reduces then to just T e T r − (cid:96) ≥ (B.17)which is the same condition we found the existence of the scattering region, verifying thetheorem explicitly in this simple case. References [1] A. May,
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