A supertask is a task that consists in infinitely many componentsteps, but which in some sense is completed in a finite amount oftime. Supertasks were studied by the pre-Socratics and continue to beobjects of interest to modern philosophers, logicians and physicists.The term “super-task” itself was coined by J.F. Thomson(1954).
Here we begin with an overview of the analysis of supertasks and theirmechanics. We then discuss the possibility of supertasks from theperspective of general relativity.
Strange things can happen when one carries out an infinite task.
For example, consider a hotel with a countably infinite number ofrooms. One night when the hotel is completely occupied, a travelershows up and asks for a room. “No problem,” thereceptionist replies, “there’s plenty of space!” The firstoccupant then moves to the second room, the second to the third room,the third to the fourth room, and so on all the way up. The result isa hotel that has gone from being completely occupied to having oneroom free, and the traveler can stay the night after all. Thissupertask was described in a 1924 lecture by David Hilbert, asreported by Gamow (1947).
One might take such unusual results as evidence against thepossibility of supertasks. Alternatively, we might take them to seemstrange because our intuitions are based on experience with finitetasks, and which break down in the analysis of supertasks. For now,let us simply try to come to grips with some of the unusual mechanicalproperties that supertasks can have.
Supertasks often lack a final or initial step. A famous example is thefirst ofZeno’s Paradoxes, the Paradox of the Dichotomy. The runner Achilles begins at thestarting line of a track and runs ½ of the distance to thefinish line. He then runs half of the remaining distance, or ¼of the total. He then runs half the remaining distance again, or⅛ of the total. And he continues in this way ad infinitum,getting ever-closer to the finish line (Figure 1.1.1). But there is nofinal step in this task.

Fig 1.1.1. The Zeno Dichotomy supertask.
There is also a “regressive” version of the Dichotomysupertask that has no initial step. Suppose that Achilles does reachthe finish line. Then he would have had to travel the last ½ ofthe track, and before that ¼ of the track, and before that⅛ of the track, and so on. In this description of the Achillesrace, we imagine winding time backwards and viewing Achilles gettingever-closer to the starting line (Figure 1.1.2). But now there is noinitial step in the task.

Fig 1.1.2. Regressive version of the Zeno Dichotomy supertask.
Zeno, at least as portrayed in Aristotle’sPhysics, arguedthat as a consequence, motion does not exist. Since an infinite numberof steps cannot be completed, Achilles will never reach the finishline (or never have started in the regressive version). However,modern mathematics provides ways of explaining how Achilles cancomplete this supertask. As Salmon (1998) has pointed out, much of themystery of Zeno’s walk is dissolved given the modern definition of alimit. This provides a precise sense in which the following sumconverges:
\[\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \cdots .\]Although it has infinitely many terms, this sum is a geometric seriesthat converges to 1 in the standard topology of the real numbers. Adiscussion of the philosophy underpinning this fact can be found inSalmon (1998), and the mathematics of convergence in any real analysistextbook that deals with infinite series. From this perspective,Achilles actually does complete all of the supertask steps in thelimit as the number of steps goes to infinity. One might only doubtwhether or not the standard topology of the real numbers provides theappropriate notion of convergence in this supertask. A discussion ofthe subtleties of the choice of topology has been given by Mclaughlin(1998).
Max Black (1950) argued that it is nevertheless impossible to completethe Zeno task, since there is no final step in the infinite sequence.The existence of a final step was similarly demanded on a priori termsby Gwiazda (2012). But as Thomson (1954) and Earman and Norton (1996)have pointed out, there is a sense in which this objection equivocateson two different meanings of the word “complete.” On theone hand “complete” can refer to the execution of a finalaction. This sense of completion does not occur in Zeno’s Dichotomy,since for every step in the task there is another step that happenslater. On the other hand, “complete” can refer to carryingout every step in the task, which certainly does occur in Zeno’sDichotomy. From Black’s argument one can see that the Zeno Dichotomycannot be completed in the first sense. But it can be completed in thesecond. The two meanings for the word “complete” happen tobe equivalent for finite tasks, where most of our intuitions abouttasks are developed. But they are not equivalent when it comes tosupertasks.
Hermann Weyl (1949, §2.7) suggested that if one admits that theZeno race is possible, then one should equally admit that it ispossible for a machine to carry out an infinite number of tasks infinite time. However, one difference between the Zeno run and amachine is that the Zeno run is continuous, while the tasks carriedout by a machine are typically discrete. This led Grünbaum (1969)to consider the “staccato” version of the Zeno run, inwhich Achilles pauses for successively shorter times at eachinterval.
Supertasks are often described by sequences that do not converge. J.F. Thomson (1954) introduced one such example now known as Thomson’sLamp, which he thought illustrated a sense in which supertasks trulyare paradoxical.
Suppose we switch off a lamp. After 1 minute we switch it on. After½ a minute more we switch it off again, ¼ on, ⅛off, and so on. Summing each of these times gives rise to an infinitegeometric series that converges to 2 minutes, after which time theentire supertask has been completed. But when 2 minutes is up, is thelamp on or off?

Fig 1.2.1. Thomson’s lamp.
It may seem absurd to claim that it is on: for each moment that thelamp was turned on, there is a later moment at which it was turnedoff. But it would seem equally absurd to claim that it is off: foreach moment that the lamp is turned off, there is a later moment thatit was turned on. This paradox, according to Thomson, suggests thatthe supertask associated with the lamp is impossible.
To analyze the paradox, Thomson suggested we represent the“on” state of the map with the number 1 and the“off” state with 0. The supertask then consists in thesequence of states,
\[0, 1, 0, 1, 0, 1, \ldots .\]This sequence does not converge to any real number in the standardreal topology. However, one might redefine what it means for asequence to converge in response to this. For example, we could defineconvergence in terms of the arithmetic mean. Given a sequence\(x_n\), theCesàro mean is the sequence\(C_1 = x_1\), \(C_2 = (x_1 + x_2)/2\), \(C_3 = (x_1 + x_2 +x_3)/3\), and so on. These numbers describe the average valueof the sequence up to a given term. One says that a sequence\(x_n\)Cesàro converges to a number \(C\) if andonly if \(C_n\) converges (in the ordinary sense) to \(C\). It isthen well-known that the sequence \(0, 1, 0, 1, \ldots\) Cesàroconverges to ½ (see e.g. Bashirov 2014).
Thomson pointed out that this argument is not very helpful without aninterpretation of what lamp-state is represented by ½. We wantto know if the lamp is on or off; saying that its end state isassociated with a convergent arithmetic mean of ½ does littleto answer the question. However, this approach to resolving theparadox has still been pursued, for example by PérezLaraudogoita, Bridger and Alper (2002) and by Dolev (2007).
Are there other consistent ways to describe the final state ofThomson’s lamp in spite of the missing limit?
Benacerraf (1962) pointed out a sense in which the answer is yes. Thedescription of the Thomson lamp only actually specifies what the lampis doing at each finite stage before 2 minutes. It says nothing aboutwhat happens at 2 minutes, especially given the lack of a converginglimit. It may still be possible to “complete” thedescription of Thomson’s lamp in a way that leads it to be either onafter 2 minutes or off after 2 minutes. The price is that the finalstate will not be reached from the previous states by a convergentsequence. But this by itself does not amount to a logicalinconsistency.
Such a completion of Thomson’s description was explicitly constructedby Earman and Norton (1996) using the following example of a bouncingball.
Suppose a metal ball bounces on a conductive plate, bouncing a littlelower each time until it comes to a rest on the plate. Suppose thebounces follow the same geometric pattern as before. Namely, the ballis in the air for 1 minute after the first bounce, ½ minuteafter the second bounce, ¼ minute after the third, ⅛minute after the fourth, and so on. Then the entire infinite sequenceof bounces is a supertask.
Now suppose that the ball completes a circuit when it strikes themetal plate, thereby switching on a lamp. This is a physical systemthat implements Thomson’s lamp. In particular, the lamp is switched onand off infinitely many times over the course of a finite duration of2 minutes.

Fig 1.2.2.Thomson’s lamp implemented by a bouncing ball: contact ofthe bouncing ball with the plate switches the Thomson lamp on. Thesupertask ends with the lamp on.
What is the state of this lamp after 2 minutes? The ball will havecome to rest on the plate, and so the lamp will be on. There is nomystery in this description of Thomson’s lamp.
Alternatively, we could arrange the ball so as to break the circuitwhen it makes contact with the plate. This gives rise to anotherimplementation of Thomson’s lamp, but one that is off after 2 minuteswhen the ball comes to its final resting state.

Fig 1.2.3.Another implementation of Thomson’s lamp: contact of the bouncing ballwith the plate switches the Thomson lamp off. The supertask ends withthe lamp off.
These examples show that is possible to fill in the details ofThomson’s lamp in a way that either renders it definitely on after thesupertask, or definitely off. For this reason, Earman and Nortonconclude with Benacerraf that the Thomson lamp is not a matter ofparadox but of an incomplete description.
As with the Zeno Dichotomy, there is a regressive version of theThomson lamp supertask. Such a lamp has been studied by Uzquiano(2012), although as a set of instructions rather than a set of tasks.Consider a lamp that has been switched on at 2 secondspast the hour, off at 1 second past, on at ½ a second past, offat ¼ a second past, and so on. What is the stateof the lamp on the hour, just before the supertask has begun? Thissupertask can be viewed as incomplete in the same way as the originalThomson lamp. Insofar as the mechanics of bouncing balls and electriccircuits described in Earman and Norton’s lamp are time reversal invariant, it follows that the time-reversed system is a possibility as well, which is spontaneously excited to begin bouncing, providing a physical implementation of the regressive Thomsonlamp. However, whether the reversed Thomson lamp is a physical possibilitydepends on whether or not the system is time reversible. A difficulty is that its initial state will not determine the subsequent history of an infinity of alternations.
Sometimes supertasks require a physical quantity to be discontinuousin time. One example of this, known as Ross’ paradox, was described byJohn Littlewood (1953) as an “infinity paradox” andexpanded upon by Sheldon Ross (1988) in his well-known textbook onprobability. It goes as follows.
Suppose we have a jar—a very large jar—with thecapacity to hold infinitely many balls. We also have a countablyinfinite pile of balls, numbered 1, 2, 3, 4, …. First we drop balls1–10 into the jar, then remove ball 1. (This adds a total of nineballs to the jar.) Then we drop balls 11–20 in the jar, and removeball 2. (This brings the total up to eighteen.) Suppose that wecontinue in this way ad infinitum, and that we do so withever-increasing speed, so that we will have used up our entireinfinite pile of balls in finite time (Figure 1.3.1). How many ballswill be in the jar when this supertask is over?

Fig 1.3.1.The Littlewood-Ross procedure.
Both Littlewood (1953) and Ross (1976) responded that the answer iszero. Their reasoning went as follows.
Ball 1 was removed at the first stage. Ball 2 was removed at thesecond stage. Ball n was removed at the nth stage, and so on adinfinitum. Since each ball has a label n, and since each label n wasremoved at the nth stage of the supertask, there can be only be zeroballs left in the jar at the end after every stage has been completed.One can even identify the moment at which each of them wasremoved.
Some may be tempted to object that, on the contrary, the number ofballs in the jar should be infinite when the supertask is complete.After the first stage there are 9 balls in the jar. After the secondstage there are 18. After the third stage there are 27. In the limitas the number of stages approaches infinity, the total number of ballsin the jar diverges to infinity. If the final state of the jar isdetermined by what the finite-stage states are converging to, then thesupertask should conclude with infinitely many balls in the jar.
If both of these responses are equally reasonable, then we have acontradiction. There cannot be both zero and infinity balls in a jar.It is in this sense that the Littlewood-Ross example might be aparadox.
Allis and Koetsier (1991) argued that only the first response isjustified because of a reasonable “principle ofcontinuity”: that the positions of the balls in space are acontinuous function of time. Without such a principle, the positionsof the balls outside the jar could be allowed to teleportdiscontinuously back into the jar as soon as the supertask iscomplete. But with such a principle in place, one can conclude thatthe jar must be empty at the end of the supertask. This principle hasbeen challenged by Van Bendegum (1994), with a clarifying rejoinder byAllis and Koetsier (1996).
Earman and Norton (1996) follow Allis and Koetsier (and Littlewood andRoss) in demanding that the worldlines of the balls in the jar becontinuous, but point out that there is a different sense ofdiscontinuity that develops as a consequence. (A‘worldline’ is used here to describe the trajectory of aparticle through space and time; it is discussed more below in thesection onTime in Relativistic Spacetime.)Namely, if one views the number of balls in the jar as approximated bya function \(N(t)\) of time, then this “numberfunction” is discontinuous in the Littlewood-Ross supertask,blowing up to an arbitrarily large value over the course of thesupertask before dropping discontinuously to 0 once it is over. Inthis sense, the Littlewood-Ross paradox presents us with a choice, toeither,
but not both. The example thus seems to require a physical quantity tobe discontinuous in time: either in the worldlines of the balls, or inthe number of balls in the jar.
A variation of the Littlewood-Ross example has been posed as a puzzlefor decision theory by Barrett and Arntzenius (1999, 2002). Theypropose a game involving an infinite number of $1 bills, each numberedby a serial number 1, 2, 3, …, and in which a person begins with $0.The person must then choose between the following two options.
At each finite stage of the game it appears to be rational to chooseOption B. For example, at stagen=1 Option B returns $3, whileOption A returns $1. At stagen=2 Option B returns $7 whileOption A returns $1. And so on.
However, suppose that one plays this game as a supertask, so that theentire infinite number of offers is played in finite time. Then howmuch money will the player have? Following exactly the same reasoningas in the Littlewood-Ross paradox, we find that the answer is $0. Foreach bill’s serial number, there is a stage at which that bill wasreturned. So, if we presume the worldlines of the bills must becontinuous, then the infinite game ends with the player winningnothing at all. This is a game in which the rational strategy at eachfinite stage does not provide a winning strategy for the infinitegame.
There are variations on this example that have a more positive yieldfor the players. For example, Earman and Norton (1996) propose thefollowing pyramid marketing scheme. Suppose that an agent sells twoshares of a business for $1,000 each to a pair of agents. Each agentsplits their share in two and sells it for $2,000 to two more agents,thus netting $1,000 while four new agents go into debt for $1,000each. Each of the four new agents then do the same, and so on adinfinitum. How does this game end?
If the pool of agents is only finitely large, then the last agentswill get saddled with the debt while all the previous agents make aprofit. But if the pool is infinitely large, and the pyramid marketingscheme becomes a supertask, then all of the agents will have profitedwhen it is completed. At each stage in which a given agent is in debt,there is a later stage in which the agent sells to shares and makes$1,000. This is thus a game that starts with equal total amount ofprofit and debt, but concludes having converted the debt into pureprofit.
The discussions of supertasks so far suggest that the possibility ofsupertasks is not so much a matter of logical possibility as it is“physical possibility.” But what does “physicalpossibility” mean? One natural interpretation is that it means,“possible according to some laws of physics.” Thus, we canmake the question of whether supertasks are possible more precise byasking, for example, whether supertasks compatible with the laws ofclassical particle mechanics.
Earman and Norton’s (1996) bouncing ball provides one indication thatthe answer is yes. Another particularly simple example was introducedby Pérez Laraudogoita (1996, 1998), which goes as follows.
Suppose an infinite lattice of particles of the same mass are arrangedso that there is a distance of ½ between the first and thesecond, a distance of ¼ between the second and the third, adistance of ⅛ between the third and the fourth, and so on. Nowimagine that a new particle of the same mass collides with the firstparticle in the lattice, as in Figure 1.4.1. If it is a perfectlyelastic collision, then the incoming particle will come to rest andthe velocity will be transferred to the struck particle. Suppose ittakes ½ of a second for the second collision to occur. Then itwill take ¼ of a second for the third to occur, ⅛ of asecond for the fourth, and so on. The entire infinite process willthus be completed after 1 second.

Fig 1.4.1.Jon Pérez Laraudogoita’s ‘Beautiful Supertask’
Earman and Norton (1998) observed several curious facts about thissystem. First, unlike Thomson’s lamp, this supertask does not requireunbounded speeds. The total velocity of the system is never any morethan the velocity of the original moving particle. Second, thissupertask takes place in a bounded region of space. So, there are noboundary conditions “at infinity” that can rule out thesupertask. Third, although energy is conserved in each localcollision, the global energy of this system is not conserved, sinceafter finite time it becomes a lattice of infinitely many particlesall at rest. Finally, the supertask depends crucially on there beingan infinite number of particles, and the width of these particles mustshrink without bound while keeping the mass fixed. This means the massdensity of the particles must grow without bound. The failure ofglobal energy conservation and other curious features of this systemhave been studied by Atkinson (2007, 2008), Atkinson and Johnson(2009, 2010) and by Peijnenburg and Atkinson (2008) and Atkinson andPeijnenburg (2014).
Another kind of classical mechanical supertask was described byPérez Laraudogoita (1997). Consider again the infinite latticeof particles of the same mass, but this time suppose that the firstparticle is motionless, that the second particle is headed towards thefirst with some velocity, and that the velocity of each successiveparticle doubles (Figure 1.4.2). The first collision sets the firstparticle in motion. But a later collision then sets it moving faster,and a later collision even faster, and so on.

Fig 1.4.2.A supertask that relies on unbounded speed.
It is not hard to arrange this situation so that the first collisionhappens after ½ of a second, the second collision after¼ of a second, the third after ⅛ of a second, and so on(Pérez Laraudogoita 1997). So again we have a supertask that iscompleted after one second.
What is the result of this supertask? Their answer is that none of theparticles remain in space. They cannot be anywhere in space, since foreach horizontal position that a given particle can occupy there is atime before 1 second that it is pushed out of that position by acollision. The worldline of any one of the particles from thissupertask can be illustrated using Figure 1.4.3. This is what Malament(2008, 2009) has referred to as a “space evader”trajectory. The time-reversed “space invader” trajectoryis one in which the vacuum is spontaneously populated with particlesafter some fixed time.

Fig 1.4.3. Worldline of the supertask particle.
Earman and Norton (1998) gave some variations on this supertask,including one which occurs in a bounded region in space. Unlike theexample of Pérez Laraudogoita (1996), this supertask alsoessentially requires particles to be accelerated to arbitrarily highspeeds, and in this sense is essentially non-relativistic. SeePérez Laraudogoita (1999) for a rejoinder.
This supertask is modeled on an example of Benardete (1964), whoconsidered a space ship that successively doubles its speed until itescapes to spatial infinity. Supertasks of this kind were also studiedby physicists like Lanford (1975, §4), who identified a system ofparticles colliding elastically that can undergo an infinite number ofcollisions in finite time. Mather and McGehee (1975) pointed out asimilar example. Earman (1986) discussed the curious behavior ofLanford’s example as well, pointing out that such supertasks provideexamples of classical indeterminism, but can be eliminated byrestricting to finitely many particles or by imposing appropriateboundary conditions.
It is possible to carry some of the above considerations of supertasksover from classical to quantum mechanics. The examples of quantummechanical supertasks that have been given so far are somewhat lessstraightforward than the classical supertasks above. However, theyalso bear a more interesting possible relationship to physicalexperiments.
Example 1: Norton’s Lattice
Norton (1999) investigated whether there exists a direct quantummechanical analogue of the kinds of supertasks discussed above. Hebegan by considering the classical scenario shown in Figure 1.5.1 ofan infinite lattice of interacting harmonic oscillators. Assuming thesprings all have the same tension and solving the equation of motionfor this system, Norton found that it can spontaneously excite,producing an infinite succession of oscillations in the lattice in afinite amount of time.

Fig 1.5.1. Norton’s infinite harmonic oscillator system.
Using this example as a model, Norton produced a similar supertask fora quantum lattice of harmonic oscillators. Begin with an infinitelattice of 2-dimensional quantum systems, each with a ground state\(\ket{\phi}\) and an excited state \(\ket{\chi}\). Consider thecollection of vectors,
\[\begin{align}\ket{0} &= \ket{\phi} \otimes \ket{\phi} \otimes \ket{\phi} \otimes \ket{\phi} \otimes \cdots \\\ket{1} &= \ket{\chi} \otimes \ket{\phi} \otimes \ket{\phi} \otimes \ket{\phi} \otimes \cdots \\\ket{2} &= \ket{\phi} \otimes \ket{\chi} \otimes \ket{\phi} \otimes \ket{\phi} \otimes \cdots \\\ket{3} &= \ket{\phi} \otimes \ket{\phi} \otimes \ket{\chi} \otimes \ket{\phi} \otimes \cdots \\\ket{4} &= \ket{\phi} \otimes \ket{\phi} \otimes \ket{\phi} \otimes \ket{\chi} \otimes \cdots \\\;\vdots&\end{align}\]For simplicity, we restrict attention to the possible states of thesystem that are spanned by this set. We posit a Hamiltonian that hasthe effect of leaving |0〉 invariant; of creating |1〉 anddestroying |2〉; of creating |2〉 and destroying |3〉; andso on. Norton then solved the differential form of theSchrödinger equation for this interaction and argued that itadmits solutions in which all of the nodes in the infinite latticestart in their ground state, but all become spontaneously excited infinite time.
Norton’s quantum supertask requires a non-standard quantum systembecause the dynamical evolution he proposes is not unitary, eventhough it obeys a differential equation in wavefunction space thattakes the form of the Schrödinger equation (Norton 1999,§5). Nevertheless, Norton’s quantum supertask has fruitfullyappeared in physical applications, having been found to arisenaturally in a framework for perturbative quantum field theoryproposed by Duncan and Niedermaier (2013, Appendix B).
Example 2: Hepp Measurement
Although quantum systems may sometimes be in a pure superposition ofmeasurable states, we never observe our measurement devices to be insuch states when they interact with quantum systems. On the contrary, our measurement devices always seem to display definite values. Why? Hepp (1972)proposed to explain this by modeling the measurement process using aquantum supertask. This example was popularized by Bell (1987,§6) and proposed as a solution to the measurement problem by Wan(1980) and Bub (1988).
Here is a toy example illustating the idea. Suppose we model anidealised measuring device as consisting in an infinite number offermions. We imagine that the fermions do not interact with eachother, but that a finite number of them will couple to our targetsystem whenever we make a measurement. Then an observablecharacterising the possible outcomes of a given measurement will be aproduct corresponding to some finite number n of observables,
\[A = A_1 \otimes A_2 \otimes \cdots \otimes A_n \otimes I \otimes I \otimes I \otimes \cdots\]Restricting to a finite number of fermions at a time has the effect ofsplitting the Hilbert space of states into special subspaces calledsuperselection sectors, which have the property that when \(\ket{\psi}\)and \(\ket{\phi}\) come from different sectors, any superposition\(a\ket{\psi} + b\ket{\phi}\) with \(|a|^2 + |b|^2 = 1\) will be a mixed state. It turns out inparticular that the space describing the state in which all thefermions are \(z\)-spin-up is in a different superselection sector thanthe space in which they are all spin down. Although this may be puzzlingfor the newcomer, it can be found in any textbook that deals withsuperselection. And it allows us to construct an interesting supertaskdescribing the measurement process. The following simplified versionof it was given by Bell (1987).
Suppose we wish to measure a single fermion. We model this as awavefunction that zips by the locations of each fermion in ourmeasurement device, interacting locally with the individual fermionsin the device as it goes (Figure 1.5.2). The interaction is set up insuch a way that every fermion is passed in finite time, and such thatafter the process is completed, the measurement device indicates whatthe original state of the fermion being measured was. In particular,suppose the single fermion begins in a \(z\)-spin-up state. Then,after it has zipped by each of the infinite fermions, they will all befound in the \(z\)-spin-up state. If the single fermion beginsin a \(z\)-spin-down state, then the infinite collection offermions would all be \(z\)-spin-down. What if the single fermionwas in a superposition? Then the infinite collection of fermions wouldcontain some mixture of \(z\)-spin up and \(z\)-spin downstates.

Fig 1.5.2. Bell’s implementation of the Hepp measurement supertask.
Hepp found that, because of the superselection structure of thissystem, this measurement device admits mixed states that can indicatethe original state of the single fermion, even when the latter beginsin a pure superposition. Suppose we denote the \(z\)-spinobservable for the nth fermion in the measurement device as, \(s_n = I\otimes I \otimes \cdots (n\,times) \cdots \otimes \sigma_z \otimes I\cdots.\) We now construct a new observable, given by,
\[S = \lim_{n\rightarrow\infty} \tfrac{1}{n}(s_1 + s_2 + \cdots + s_n).\]This observable has the property that \(\langle \psi, S\phi\rangle =1\) if \(\ket{\psi}\) and \(\ket{\phi}\) both lie in the samesuperselection sector as the state in which all the fermions in themeasurement device are \(z\)-spin-up. It also has the propertythat \(\langle\psi,S\phi\rangle = -1\) if they lie in the samesuperselection sector as the all-down state. But more interestingly,suppose the target fermion that we want to measure is in a puresuperposition of \(z\)-spin-up and \(z\)-spin-downstates. Then, after it zips by all the fermions in the measurementdevice, that measurement device will be left in a superposition of theform \(a\ket{\uparrow} + b\ket{\downarrow}\), where \(\ket{\uparrow}\)is the state in which all the fermions in the device are spin-up and\(\ket{\downarrow}\) is the state in which they are all spindown. Since \(\ket{\uparrow}\) and \(\ket{\downarrow}\) are indifferent superselection sectors, it follows that their superpositionmust be a mixed state. In other words, this model allows themeasurement device to indicate the pure state of the target fermion,even when that state is a pure superposition, without the deviceitself being in a pure superposition.
The supertask underpinning this model requires an infinite number ofinteractions. As Hepp and Bell described it, the model was unrealisticbecause it required an infinite amount of time. However, a similarsystem was shown by Wan (1980) and Bub (1988) to take place in finitetime. Their approach appears at first glance to be a promising modelof measurement. However, Landsman (1991) pointed out that it isinadequate on one of two levels: either the dynamics is notautomorphic (which is the analogue of unitarity for such systems), orthe task is not completed in finite time. Landsman (1995) has arguedthat neither of these two outcomes is plausible for a realistic localdescription of a quantum system.
Example 3: Continuous Measurement
Another quantum supertask is found in the so-called Quantum ZenoEffect. This literature begins with a question: what would happen ifwe were to continually monitor a quantum system, like an unstableatom? The predicted effect is that the system would not change, evenif it is an unstable atom that would otherwise quickly decay.
Misra and Sudarshan (1977) proposed to make the concept of“continual monitoring” precise using a Zeno-likesupertask. Imagine that an unstable atom is evolving according to somelaw of unitary evolution \(U_t\). Suppose we measure whether ornot the atom has decayed by following that regressive form of Zeno’sDichotomy above. Namely, we measure it at time \(t\), but also at time\(t/2\), and before that at time \(t/4\), and at time \(t/8\), and so on. Let \(E\) bea projection corresponding to the initial undecayed state of theparticle. Finding the atom undecayed at each stage in the supertaskthen corresponds to the sequence,
\[ EU_tE,\; EU_{t/2}E,\; EU_{t/4}E,\; EU_{t/8}E,\ldots.\]Misra and Sudarshan use this sequence as a model for continuousmeasurement, by supposing that the sequence above converges to anoperator \(T(t)=E\), and that it does so for all times \(t\) greater than orequal to zero. The aim is for this to capture the claim that the atomis continually monitored beginning at a fixed time \(t=0\). They provefrom this assumption that, for most reasonable quantum systems, if theinitial state is undecayed in the sense that \(\mathrm{Tr}(\rho E)=1\), then theprobability that the atom will decay in any given time interval \([0,t]\)is equal to zero. That is, continual monitoring implies that the atomwill never decay.
These ideas have given rise to a large literature of responses. Togive a sampling: Ghirardi et al. (1979) and Pati (1996) have objectedthat this Zeno-like model of a quantum measurement runs afoul of otherproperties of quantum theory, such as the time-energy uncertaintyrelations, which they argue should prevent the measurements in thesupertask sequence above from being made with arbitrarily highfrequency. Bokulich (2003) has responded that, nevertheless, such asupertask can still be carried out when the measurement commutes withthe unitary evolution, such as when \(E\) is a projection onto an energyeigenstate.
In Newtonian physics, time passes at the same rate for all observers.If Alice and Bob are both present at Alice’s 20th and 21st birthdayparties, both people will experience an elapsed time of one yearbetween the two events. (This is true no matter what Alice or Bob door where Alice and Bob go in between the two events.) Things aren’t sosimple in relativistic physics. Elapsed time between events isrelative to the path through spacetime a person takes between them. Itturns out that this fact opens up the possibility of a new type ofsupertask. Let’s investigate this possibility in a bit moredetail.
A model of general relativity, a spacetime, is a pair \((M,g)\).It represents a possible universe compatible with the theory. Here, \(M\)is a manifold of events. It gives the shape of the universe. (Lots oftwo-dimensional manifolds are familiar to us: the plane, the sphere,the torus, etc.) Each point on \(M\) represents a localized event in spaceand time. A supernova explosion (properly idealized) is an event. Afirst kiss (properly idealized) is also an event. So is the moonlanding. But July 20, 1969 is not an event. And the moon is not anevent.
Manifolds are great for representing events. But the metric \(g\) dictateshow these events are related. Is it possible for a person to travelfrom this event to that one? If so, how much elapsed time does aperson record between them? The metric \(g\) tells us. At each event, \(g\)assigns a double cone structure. The cone structures can change fromevent to event; we only require that they do so smoothly. Usually, oneworks with models of general relativity in which one can label the twolobes of each double cone as “past” and“future” in a way which involves no discontinuities. Wewill do so in what follows. (See figure 2.1.1.)

Fig 2.1.1. Events in spacetime and the associated double cones.
Intuitively, the double cone structure at an event demarcates thespeed of light. Trajectories through spacetime which thread the insideof the future lobes of these “light cones” are possibleroutes in which travel stays below the speed of light. Such atrajectory is aworldline and, in principle, can be traversedby a person. Now, some events cannot be connected by a worldline. Butif two events can be connected by a worldline, there is aninfinite number of worldlines which connect them.
Each worldline has a “length” as measured by the metric \(g\);this length is the elapsed time along the worldline. Take two eventson a manifold \(M\) which can be connected by a worldline. The elapsedtime between the events might be large along one worldline and smallalong another. Intuitively, if a worldline is such that it stays closeto the boundaries of the cone structures (i.e. if the trajectory stays“close to the speed of light”), then the elapsed time isrelatively small. (See Figure 2.1.2.) In fact, it turns out that iftwo events can be connected by a worldline, then for any number \(t>0\),there is a worldline connecting the events with an elapsed time lessthan \(t\)!

Fig 2.1.2. Elapsed time is worldline dependent.
The fact that, in relativistic physics, elapsed time is relative toworldlines suggests a new type of bifurcated supertask. The idea issimple. (A version of the following idea is given in Pitowsky 1990.)Two people, Alice and Bob, meet at an event \(p\) (the start of thesupertask). Alice then follows a worldline with a finite elapsed timewhich ends at a given event \(q\) (the end of the supertask). On the otherhand, Bob goes another way; he follows a worldline with an infiniteelapsed time. Bob can use this infinite elapsed time to carry out acomputation which need not halt after finitely many steps. Bob mightcheck all possible counterexamples to Goldbach’s conjecture, forexample. (Goldbach’s conjecture is the statement that every evenintegern which is greater than 2 can be expressed as the sumof two primes. It is presently unknown whether the conjecture istrue. One could settle it by sequentially checking to see if eachinstantiated statement is truefor \(n=4\), \(n=6\), \(n=8\), \(n=10\), and so on.) Ifthe computation halts, then Bob sends a signal to Alice at \(q\) saying asmuch. If the computation fails to halt, no such signal is sent. Theupshot is that Alice, after a finite amount of elapsed time, knows theresult of the potentially infinite computation at \(q\).
Let’s work a bit more to make the idea precise. We say that ahalf-curve is a worldline which starts at some event and isextended as far as possible in the future direction. Next, theobservational past of an eventq,OP(q), isthe collection of all eventsx such that there a is a worldlinewhich starts atx and ends atq. Intuitively, a (slowerthan light) signal may be sent from an eventx to aneventq if and only ifx is in thesetOP(q). (See figure 2.2.1.)

Fig 2.2.1. The observational past of an event and a half-curve. A signal can besent to \(q\) from every point in \(OP(q)\). No signalcan be sent to \(q\) from any point on the half-curve \(\gamma\).
We are now ready to define the class of models of general relativitywhich allow for the type of bifurcated supertask mentioned above(Hogarth 1992, 1994).
Definition. A spacetime \((M,g)\) isMalament-Hogarth if there isan event \(q\) in \(M\) and a half-curve \(\gamma\) in \(M\) with infinite elapsedtime such that \(\gamma\) is contained in \(OP(q)\).
One can see how the definition corresponds to the story above. Bobtravels along the half-curve \(\gamma\) and records an infinite elapsedtime. Moreover, at any event on Bob’s worldline, Bob can send a signalto the event \(q\) where Alice finds the result of the computation; thisfollows from the fact that \(\gamma\) is contained in \(OP(q)\). Note thatAlice’s worldline and the starting point \(p\) mentioned in the story didnot make it to the definition; they simply weren’t needed. The halfcurve \(\gamma\) must start at some event – this event is our startingpoint \(p\). Since \(p\) is in \(OP(q)\), there is a worldline from \(p\) to \(q\). Takethis to be Alice’s worldline. One can show that this worldline musthave a finite elapsed time.
Is there a spacetime which satisfies the definition? Yes. Let \(M\) be thetwo-dimensional plane in standard \(t,x\) coordinates. Let the metric \(g\)be such that the light cones are oriented in the \(t\) direction and openup as the absolute value of \(x\) approaches infinity. The resultingspacetime (Anti-de Sitter spacetime) is Malament-Hogarth (see Figure2.2.2).

Fig 2.2.2.Anti-de Sitter Spacetime is Malament-Hogarth. A signal canbe sent to \(q\) from every point on the half-curve \(\gamma\).
In the previous section, we showed the existence of models of generalrelativity which seem to allow for a type of bifurcated supertask.Here, we ask: Are these models “physically reasonable”?Earman and Norton (1993, 1996) and Etesi and Németi (2002) havearticulated a number of potential physical problems concerningMalament-Hogarth spacetimes. First of all, we would like Bob’sworldline to be reasonablly traversable. In the Anti-de Sitter modelabove, the half-curve \(\gamma\) has an infinite total acceleration. Bobwould need an infinite amount of fuel to traverse it! (Malament1985)
Another problem for the Anti-de Sitter spacetime is that a“divergent blueshift” phenomenon occurs. Intuitively, thefrequency of any signal Bob sends to Alice is amplified more and moreas he goes along. Eventually, even the slightest thermal noise will beamplified to such an extent that communication is all but impossible.So, if the counterexample to Goldbach’s conjecture comes late in thegame (or not at all), it is not clear that Alice can ever know this.
One can find Malament-Hogarth spacetimes which can escape both of theproblems mentioned above (Manchak 2010; Andréka et al. 2018). For example, let \(M\) be a two-dimensional plane instandard \(t, x\) coordinates which is then “rolled up” alongthe \(t\) axis. Let the metric \(g\) be such that the light cones are orientedin the \(t\) direction and do not change from point to point. (See Figure2.3.1.)

Fig 2.3.1.An acausal Malament-Hogarth spacetime.
Because worldliness can wrap around and around the cylinder, \(OP(q)=M\)for any event \(q\). This allows for great freedom in choosing Bob’sworldline \(\gamma\). In fact, we can choose it so that the totalacceleration is zero – no fuel is needed to traverse it. Moreover, wecan choose it so that there is also no divergent blueshift phenomenon(see Earman and Norton 1993). But, alas, we have a new problem: thespacetime is acausal. A worldline can start and end at the same eventallowing for a type of “time travel”. It is unclear ifspacetimes allowing for time travel are physically reasonable (seeSmeenk and Wüthrich 2011). It turns out that more complicatedexamples can be constructed which avoid all the potential problemsmentioned so far and more (Manchak 2010). But such examples containspacetime “holes” which may not be physically reasonable(Manchak 2009; Doboszewski 2019). More work is needed to see if such problems canalso be overcome.
We conclude with one final potential problem which threatens to renderall Malament-Hogarth spacetimes physically unreasonable. Penrose(1979) has conjectured that all physically reasonable spacetimes arefree of a certain type of “naked singularities” and thebreakdown of determinism they bring. Whether Penrose’sconjecture is true or not is the subject of much debate (Earman1995). But it turns out that every Malament-Hogarth spacetime harborsthese naked singularities (Hogarth 1992). Even so, the breakdown ofdeterminism guaranteed in such spacetimes does not seem to precludethe construction of Malament-Hogarth “machines” whichcould possibly bring about supertasks from initial data (Earman etal. 2016; Manchak 2018). Stepping back, we find that it is still anopen question whether Malament-Hogarth spacetimes are simply artifactof the formalism of general relativity or if the kind of bifurcatedsupertask they suggest can be implemented in our own universe.
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