# Closed timelike curves make quantum and classical computing equivalent

Scott Aaronson, John Watrous

## Abstract

While closed timelike curves (CTCs) are not known to exist, studying their consequences has led to non-trivial insights into general relativity, quantum information and other areas. In this paper, we show that, if CTCs existed, quantum computers would be no more powerful than classical computers: both would have the (extremely large) power of the complexity class polynomial space (), consisting of all problems solvable by a conventional computer using a polynomial amount of memory. This solves an open problem proposed by one of us in 2005, and gives an essentially complete understanding of computational complexity in the presence of CTCs. Following the work of Deutsch, we treat a CTC as simply a region of spacetime where a ‘causal consistency’ condition is imposed, meaning that nature has to produce a (probabilistic or quantum) fixed point of some evolution operator. Our conclusion is then a consequence of the following theorem: given any quantum circuit (not necessarily unitary), a fixed point of the circuit can be (implicitly) computed in . This theorem might have independent applications in quantum information.

Keywords:

## 1. Introduction

The possibility of closed timelike curves (CTCs) within general relativity and quantum gravity theories has been studied for almost a century (van Stockum 1937; Gödel 1949; Morris et al. 1988). A different line of research has sought to understand the implications of CTCs, supposing they existed, for quantum mechanics, computation and information (Deutsch 1991; Brun 2003; Bacon 2004).

In this paper we contribute to the latter topic, by giving the first complete characterization of the computational power of CTCs. We argue that if CTCs containing polynomially many bits could be created and maintained then both classical and quantum computers would have exactly the power of the complexity class polynomial space (), which consists of all problems solvable on a classical computer with a polynomial amount of memory. To put it differently, CTCs would make polynomial time equivalent to as computational resources, and would also make quantum and classical computers equivalent to each other in their computational power. Our results treat CTCs using the ‘causal consistency’ framework of Deutsch (1991), together with the assumption that a CTC involving polynomially many bits can be maintained using polynomial resources.

It will not be hard to show that classical computers with CTCs can simulate and be simulated in it (though as far as we know, this result is new). The main difficulty will be to show that quantum computers with CTCs can be simulated in . To prove this, we need to give an algorithm for (implicitly) computing fixed points of superoperators in . Our algorithm relies on fast parallel algorithms for linear algebra due to Borodin et al. (1983), and might be of independent interest.

The paper is organized as follows. In §2, we explain needed background about Deutsch's causal consistency framework and computational complexity, and review previous work by Brun (2003), Bacon (2004) and Aaronson (2005). In §3, we show that classical computers with CTCs have exactly the power of . In §4, we extend the analysis of §3 to show that quantum computers with CTCs have exactly the power of . In that section, we make the simplifying assumption that all quantum gates can be applied perfectly and that amplitudes are rational. In §5, we consider what happens when gates are subject to finite error, and extend previous work of Bacon (2004) to show that quantum computers with CTCs can solve problems in a ‘fault-tolerant’ way. We conclude in §6 with some general remarks and open problems.

## 2. Background

### (a) Causal consistency

It was once believed that CTCs would lead inevitably to logical inconsistencies such as the grandfather paradox. But in a ground-breaking paper, Deutsch (1991) argued that this intuition fails, provided the physics of the CTC is quantum mechanical. While Deutsch's resolution of the grandfather paradox is not universally accepted, we will adopt it throughout in this paper, since it leads in a particularly clear and elegant way to a model of computation. Deutsch's insight was that a CTC should simply be regarded as a region of spacetime where nature enforces a requirement of causal consistency: in other words, that the evolution operator within that region should map the state of the initial hypersurface to itself. Given the evolution operator f, nature's ‘task’ is thus to find a fixed point of f, i.e. an input x such that f(x)=x. Of course, not every deterministic evolution operator f has a fixed point: that is just one way of stating the grandfather paradox. On the other hand, it is a basic linear algebra fact that every quantum operation Φ has a fixed point, i.e. a density matrix ρ such that Φ(ρ)=ρ. For any Φ such a ρ can then be used to produce a CTC evolution that satisfies the causal consistency requirement. So, for example, a consistent resolution of the grandfather paradox is that you are born with 1/2 probability, and if you are born you go back in time to kill your grandfather, therefore you are born with 1/2 probability, etc.

Note that Deutsch's resolution works just as well in classical probabilistic theories as in quantum-mechanical ones. For just as every quantum operation has a fixed point, so every Markov chain has a stationary distribution. What matters is simply that the state space and the set of transformations are such that fixed points exist.

It might be thought mysterious that nature ‘finds’ a fixed point ρ of Φ: how, one might ask, does nature do this? Does nature not have to find ρ before the CTC computation starts, so that, in some sense, running the computation is not even necessary? While these issues are admittedly mysterious, to us they are not more mysterious than the starting assumption that CTCs exist! One should keep in mind that any account of a universe with CTCs is going to be strange, so perhaps the most one can hope for is that the account should be mathematically clear and consistent. And at a purely mathematical level, Deutsch's causal consistency account of CTCs is the clearest we have seen.

Although CTCs need not lead to inconsistencies, Deutsch pointed out that they would have striking consequences for the theory of computing. As an example, CTCs could be exploited to solve non-deterministic polynomial time ()-complete and other ‘intractable’ computational problems using only polynomial resources. To see this, suppose some integers x∈{0, 1, …, 2n−1} are ‘solutions’ and others are not, and that our goal is to find a solution in time polynomial in n, assuming solutions exist and can be recognized efficiently. Then we could build a machine M that applied the following transformation to its input x: if x is a solution then M(x)=x, while if x is not a solution then M(x)=(x+1)mod 2n. Now, suppose we use a CTC to feed M its own output as input. Then it is not hard to see that the only way for the evolution to satisfy causal consistency is for M to input, and output, a solution.

In this way, an exponentially hard computational problem could get solved without exponential effort ever being invested to solve it, merely because that is the only way to satisfy causal consistency. A rough analogy would be Shakespeare's plays being written by someone from the present going back in time and dictating the plays to him.

It is sometimes said that, if CTCs existed, one could obviously do computations of unlimited length in an instant, by simply computing the answer, then sending it back in time to before one started. However, this proposal does not work for two reasons. First, it ignores the grandfather paradox: what happens if, on receiving the output, one goes back in time and changes the input? Second, it is perhaps unclear why a computation lasting 101000 years should be considered ‘feasible’, merely because we are able to obtain the solution before performing the computation. It seems that an honest accounting should require the computations performed inside the CTC to be efficient (say, polynomial time), with any computational speedup coming from the requirement of causal consistency.

### (b) Complexity theory

For background on classical computational complexity theory, see for example Arora & Barak (in press); for a recent survey of quantum complexity theory, see Watrous (2008). Here, we briefly describe the main complexity classes that we will consider. is the class of decision problems that are solvable by a classical computer, using an amount of memory that is bounded by a polynomial function of the size of the input n (but possibly an exponential amount of time). An example of such a problem is, given a configuration of an n×n Go board, to decide whether White has a winning strategy using n2 or fewer moves. is the class of decision problems for which every ‘yes’ answer has a polynomial-time checkable, polynomial-size proof or witness. -complete problems are, loosely speaking, the ‘hardest’ problems in , i.e. those problems to which all other problems can be efficiently reduced. An example is, given a graph, to decide whether it has a Hamiltonian cycle (i.e. a cycle that visits each vertex exactly once). contains (thus, in particular, the -complete problems)—since, in , one can simply loop over all possible witnesses and see if any of them are correct. However, is believed to be considerably larger than . So, in saying that computers with CTCs can efficiently solve problems, we are saying something stronger than just that they can solve -complete problems.

Our main result is that computers with polynomial-size CTCs have precisely the power of , and that this is true whether the computers are classical or quantum. Previously, Watrous (1999) showed that bounded-error quantum polynomial space () is equal to , i.e. any problem solvable by a quantum computer with polynomial memory is also solvable by a classical computer with polynomial memory. (By contrast, quantum computers are conjectured to offer an exponential improvement over classical computers in time.) Here, we show that quantum computers are polynomially equivalent to classical computers in the CTC setting as well.

Our results will allow CTCs that contain a polynomial number of bits or qubits, as well as a polynomial number of gate operations. Given our ignorance of the physics of CTCs, it might be wondered whether these choices are justified. Our response is that the choices arise from treating CTCs the same way one would treat any other resource in computational complexity theory: for a ‘polynomial price’ one gets a polynomial amount of the resource (e.g. bits or qubits). Still, it is interesting to consider the effects of other choices, and we discuss some possibilities in §6.

### (c) Related work

Besides Deutsch's (1991) paper, we know of three other works directly relevant to computational complexity in the presence of CTCs. First, Brun (2003) showed that CTCs would allow the efficient solution of -complete problems.1

Second, Bacon (2004) showed that -complete problems can be solved with polynomial resources, even using CTCs that are only ‘one-bit wide’ (i.e. able to transmit a single qubit or probabilistic classical bit back in time).2 Bacon also showed that, using his approach, one can solve not only problems but even # problems, which involve counting solutions rather than just finding one. (The class #—or more formally its decision version —is a subclass of , with the containment believed to be strict.) Finally, Bacon showed that techniques from the theory of quantum fault tolerance could be used to make certain CTC computations, including those used to solve # problems, robust to small errors.

Third, as part of a survey on ‘-complete problems and physical reality’ Aaronson (2005), Aaronson sketched the definitions of CTC and CTC (classical and quantum polynomial time with CTCs) that we adopt in this paper. He also sketched a proof that =CTCCTC. That is, classical computers with polynomial-size CTCs have exactly the power of , while quantum computers with polynomial-size CTCs have at least the power of and at most the power of classical exponential time. The key problem that Aaronson left open was to pin down the power of quantum computers with CTCs precisely. This is the problem we solve in this paper.

## 3. The classical case

To state our results, it is crucial to have a formal model of computation in the presence of CTCs.

We define a deterministic CTC algorithm to be a deterministic polynomial-time algorithm that takes as input a string x=∈{0,1}n, and that produces as output a Boolean circuit C=Cx, consisting of AND, OR and NOT gates. The circuit C acts on bits in two registers: a CTC register CTC and a causality-respecting register CR (figure 1). The registers CTC and CR consist of p(n) and q(n) bits, respectively, for some polynomials p and q depending on . Thus, C can be seen as a Boolean function C:{0,1}p(n)+q(n)→{0,1}p(n)+q(n), which maps an ordered pair 〈y,n〉∈CTC×CR to another ordered pair C(〈y,n〉).

Figure 1

Diagram of a classical CTC computer. A circuit C performs a polynomial-time computation involving ‘CTC bits’ (the register CTC) as well as ‘causality-respecting bits’ (the register CR). Nature must then find a probability distribution over CTC which satisfies Deutsch's causal consistency equation. The final answer is read out from CR.

For convenience, we assume that the causality-respecting register CR is initialized to 0q(n). The CTC register, on the other hand, must be initialized to some probability distribution over p(n)-bit strings that will ensure causal consistency. More formally, let be a probability distribution over CTC×CR and let C() be the distribution over CTC×CR induced by drawing a sample from and then applying C to it. Also, let [.]CTC be an operation that discards the causality-respecting register (i.e. marginalizes it out), leaving only the CTC register. Then we need the initial probability distribution over CTC×CR to satisfy the following two conditions:

1. has support only on pairs of the form 〈y,0q(n)〉 and

2. satisfies the causal consistency equation []CTC=[C()]CTC.

We claim that such a always exists. This is easy to prove: let C′(y)≔[C(〈y,0q(n)〉)]CTC be the induced circuit that acts only on the CTC register. Then it suffices to find a distribution ′ over CTC such that C′(′)=′. To find such a ′, we consider the directed graph representing the function C′≔{0,1}p(n)→{0,1}p(n), find a cycle in that graph (which must exist, since the graph is finite) and let ′ be the uniform distribution over points in the cycle. Finally, we set =〈′,0q(n)〉.

We are now ready to define the complexity class CTC of problems solvable using classical computers with CTCs. We say that a CTC algorithm accepts the input x if, for every distribution satisfying conditions (i) and (ii) above, C() has support only on pairs of the form 〈y,z1〉; i.e. such that the last bit of the causality-respecting register is a 1. (Recall that C=Cx depends on the input x.) Likewise, we say rejects x if, for every satisfying (i) and (ii), C() has support only on pairs of the form 〈y,z0〉. (Of course, it is possible that C() has support on both kinds of pairs, in which case neither accepts nor rejects.) We say decides the language L⊆{0,1}* if accepts every input xL, and rejects every input x∉L. Then CTC is the class of all languages L that are decided by some deterministic CTC algorithm.

Let us make a few remarks about the definition. First, the requirement that some polynomial-time algorithm output the circuit C=Cx is intended to prevent hard-to-compute information from being hard-wired into the circuit. This requirement is standard in complexity theory; it is also used, for example, in the definition of . Second, our definition required C to succeed with certainty, and did not allow C to introduce its own randomness, besides that produced by the causal consistency condition. We could relax these requirements to obtain the complexity class CTC, or bounded-error probabilistic polynomial time with access to a CTC. However, it will turn out that CTC=CTC= anyway.

### (a) Results

We now prove CTC=.

CTC.

Let C be a polynomial-size circuit that maps CTC×CR to itself, as in the definition of CTC. Then our simulation algorithm is as follows. First, let C′=(y)≔[C(〈y,0q(n)〉)]CTC be the induced circuit that acts only on CTC. Then given a string y∈{0,1}p(n), say y is cyclic if C(k)(y)=y for some positive integer k. In other words, y is cyclic if repeated application of C′ takes us from y back to itself. Clearly, every C′ has at least one cyclic string. Furthermore, it is clear from the definition of CTC that if xL then every cyclic string must lead to an output of 1 in the last bit of CR, while if x∉L then every cyclic string must lead to an output of 0. Hence, the problem essentially reduces to finding a cyclic string.

But it is easy to find a cyclic string in : the string y*C(2p(n))(y) will be cyclic for any y. The one remaining step is to compute C(〈y*,0q(n)〉), and then output the last bit of CR. ▪

We are indebted to Lance Fortnow for the following lemma.

CTC.

For some polynomial p, let M be a p(n)-space Turing machine (i.e. every configuration of M takes p(n) bits to describe). We can assume without loss of generality that M includes a ‘clock’, which is incremented at every time step, and which causes M to accept automatically once it reaches its maximum value. This prevents M from ever going into an infinite loop, regardless of its starting configuration.

Let m1, …, mT be the successive configurations of M when run on an input x∈{0,1}n. Then our task is to decide, using a CTC computer, whether mT is an accepting or a rejecting configuration.

Our CTC algorithm will produce a circuit C that acts on two registers: a (p(n)+1)-bit CTC register CTC, and a one-bit causality-respecting register CR. For simplicity, we start by describing the induced circuit C′ that acts on CTC. Given a configuration m of M, let S(m) be the successor of m, i.e. the configuration obtained from m by incrementing the clock and performing one step of computation. Then the circuit C′ acts as follows, on ordered pairs 〈m,b〉 consisting of a configuration m and a ‘control bit’ b.

1. If m is neither an accepting nor a rejecting configuration, then C′ (〈m,b〉)=〈S(m),b〉.

2. If m is an accepting configuration, then C′(〈m,b〉)=〈m1,1〉.

3. If m is a rejecting configuration, then C′(〈m,b〉)=〈m1,0〉.

In other words, if m produces an output then C′ sets the control bit to that output and goes back to the starting configuration; otherwise, C′ increments the computation and leaves the control bit unchanged (figure 2).

Figure 2

To simulate a machine with a CTC, we perform a computation for which the only causally consistent evolution is a loop over all configurations of the machine, with a control bit b set to its final value (in this example, b=1).

Now consider the graph of the function C′≔{0,1}p(n+1)→{0,1}p(n+1). It is not hard to see that the only cycle in this graph is (〈m1,1〉,…,〈mT,1〉) if mT accepts, or (〈m1,0〉,  …,  〈mT,0〉) if mT rejects. Indeed, this is true even if there are configurations that are not reachable from 〈m1,1〉 or 〈m1,0〉, since those configurations will ultimately lead back to either 〈m1,1〉 or 〈m1,0〉 and therefore not produce new cycles. In other words, the only cycle is a loop over m1,  …,  mT, with the control bit b set to M's final output. Therefore, the only probability distribution ′ over {0,1}p(n)+1 that is stationary, in the sense that C′(′)=′, is the uniform distribution over 〈m1,b〉, …, 〈mT,b〉 where b is M's final output.

Finally, the full circuit C simply applies C′ to CTC, and then copies the control bit into the causality-respecting register. ▪

## 4. The quantum case

Let G be a universal set of quantum gates, with amplitudes having rational real and imaginary parts. Then a quantum CTC algorithm is a deterministic polynomial-time algorithm that takes as input a string x∈{0,1}n, and which produces as output an encoding of a unitary quantum circuit Q=Qx with gates from G.

The circuit Q acts on two quantum registers: a q(n)-qubit CTC register CTC and an r(n)-qubit causality-respecting register CR. The causality-respecting register CR is initialized to |0〉r(n) while the CTC register must be initialized to some q(n)-qubit mixed state ρ that will ensure causal consistency. More formally, we require that(4.1)which is equivalent to ρ being a fixed point of the quantum operation defined asDeutsch (1991) proved that every such quantum operation has a fixed point, and an alternative proof of this fact follows from our results in §4c below.

We can now define the complexity class CTC of problems solvable using quantum computers with CTCs. Let be a measurement of the last qubit of CR in the computational basis. Then we say the algorithm accepts x if, for every mixed state ρ satisfying equation (4.1) above, (Q(ρ⊗(|0〉〈0|)r(n))Q) results in output 1 with a probability of at least 2/3. We say rejects x if, for every ρ satisfying the equation, (Q(ρ⊗(|0〉〈0|)r(n))Q) results in output 1 with a probability of at most 1/3. We say decides the language L⊆{0,1}* if accepts every input xL, and rejects every input x∉L. Then CTC is the class of all languages L that are decided by some quantum CTC algorithm.

In what follows, we develop some needed background, and then prove the main result that CTC.

### (a) Matrix representation of superoperators

We will make use of a simple way of representing quantum operations as matrices. This representation begins with a representation of density matrices as vectors by the linear function defined on standard basis states asIf ρ is an N×N density matrix, then vec(ρ) is the N2-dimensional column vector obtained by stacking the rows of ρ on top of one another. For example,

Now, suppose that Φ is a given quantum operation acting on an N-dimensional system, meaning that Φ:N×NN×N is linear, completely positive and trace preserving. Given that the effect of Φ on density matrices is linear, there must exist an N2×N2 matrix K(Φ) that satisfiesfor every possible N×N density matrix ρ. The matrix K(Φ) is called the natural matrix representation of the quantum operation Φ, and is uniquely determined by Φ.

The natural matrix representation can easily be calculated from other standard forms. For example, if an operation Φ is represented in the usual Kraus form asthen it holds thatand, therefore,(Here represents the entry-wise complex conjugate of j.)

In the section that follows, we will make use of the following simple way to calculate the natural matrix representation of a quantum operation that is specified by a quantum circuit. Suppose that is an r-qubit system, is an s-qubit system and U is a unitary operation on r+s qubits. Then, for the quantum operation Φ defined aswe have(4.2)for (In both cases, each identity matrix I acts on , or equivalently is the 2r×2r identity matrix.)

### (b) Space-bounded and depth-bounded computations

When we speak of a family {Cn:n} of Boolean circuits, we assume that each Cn is an acyclic circuit, composed of AND, OR and NOT, and constant-sized fanout gates, with n input bits and an arbitrary number of output bits. Such a family computes a function f:{0,1}*→{0,1}* if, for each n and string x∈{0,1}n, the circuit Cn outputs f(x) when given input x. The depth of a Boolean circuit C is the length of the longest path in C from an input bit to an output bit. The size of C is the sum of the number of input bits, output bits and gates.

For a given function s:, we say that a Boolean circuit family {Cn:n} is space O(s) uniform if there exists a deterministic Turing machine M that runs in space O(s(n)), and that outputs a description of Cn on input 1n for each n.3 As is usual when discussing space-bounded computation, a deterministic Turing machine is assumed to be equipped with a read-only input tape that does not contribute to the space it uses, so it is meaningful to consider sublinear space bounds. Given a space O(s)-uniform family {Cn:n}, the size of Cn can be at most 2O(s(n)).

We say a function f:{0,1}*→{0,1}* is in the class (s) if there exists a space O(s)-uniform family of Boolean circuits {Cn:n} that computes f, and where the depth of Cn is at most s(n)O(1).4 Also, a language L is in (s) if its characteristic function is in (s). We write for (log n), and (poly) for the union of (nc) over all constants c. Borodin (1977) proved that, if s satisfies s(n)=Ω(log n), every function in (s) is computable by a deterministic Turing machine in space s(n)O(1). It follows that (poly)⊆. (The reverse containment also holds, so in fact we have (poly)⊆.)

It is clear that, if f(poly) and g, their composition gf is in (poly), since we can create a circuit for gf by composing the circuits for f and g in the obvious way.

Many functions of matrices are known to be computable in . These include sums and products of matrices, inverses and the trace, determinant and characteristic polynomial, all over a wide range of fields for which computations can be efficiently performed (see von zur Gathen 1993). In particular, we will rely on a fact that follows from a result of Borodin et al. (1983, §4).

The determinant of an n×n matrix whose entries are rational functions in an indeterminate z can be computed in .

### (c) Projecting onto fixed points

In this subsection, we prove a general theorem about efficient construction of quantum operations that project onto the fixed points of other quantum operations. This theorem is the technical core of our CTC result, but it might be of independent interest as well.

Suppose that Φ:N×NN×N is a given quantum operation acting on an N-dimensional system, meaning that it is a completely positive and trace-preserving linear map. We will prove that there exists another quantum operationwhich satisfies the following three properties:

1. for every density matrix σN×N, it holds that ρ=Λ(σ) is a fixed point of Φ,

2. every density matrix ρ that is a fixed point of Φ is also a fixed point of Λ, and

3. Λ can be computed from Φ in .

In essence, Λ is a (non-orthogonal) projection onto fixed points of Φ, so if we want a fixed point of Φ it suffices to compute Λ(σ) for any density matrix σ, and moreover every fixed point ρ of Φ arises in this way from some density matrix σ (which always includes the choice σ=ρ).

The operation Λ is defined as follows. First, for each real number z∈(0,1), we define a superoperator Λz:N×NN×N asHere, Φk represents the k-fold composition of Φ, and Φ0 is the identity operation. Each Φk is obviously completely positive and trace preserving. Given that z(1−z)k∈(0,1) for each choice of z∈(0,1) and k≥0, and that for every z∈(0,1), we have that Λz is a convex combination of completely positive and trace-preserving maps. Thus, Λz is completely positive and trace preserving as well. Finally, we takeWe must of course prove that this limit exists—and in the process, we will prove that Λ can be produced from Φ by an computation, which is an important ingredient of our simulation of CTC in . Once this is done, the required properties of Λ will be easily verified.

We assume that Φ is represented by the N2×N2 complex matrix M=K(Φ) as discussed in §4a. Since Φ is a quantum operation, every eigenvalue of M lies within the unit circle.5 It follows that the matrix I−(1−z) M is invertible for every real z∈(0,1), and moreover there is a convergent series for its inverse,(4.3)Now, for every z∈(0,1) we define an N2×N2 matrix Rz as follows:We note that Rz=K(Λz) for Λz as defined above—and as each Λz is completely positive and trace preserving, each entry of Rz must be bounded in absolute value by 1.

Next, by Cramer's rule, we have(4.4)where (I−(1−z)M)j,i denotes the (N2−1)×(N2−1) matrix obtained by removing the jth row and ith column from I−(1−z)M. It follows that each entry of Rz is given by a rational function in the variable z having degree at most N2. As the entries of Rz are rational functions that are bounded for all z∈(0,1), we have that the limit exists. Defineand note that R=K(Λ). We have therefore proved that the limit exists as claimed.

The fact that R can be computed from M in follows from the above discussion, together with theorem 4.1. In particular, equation (4.4) above expresses the entries of Rz as ratios of polynomials of degree at most N2 in z having coefficients with rational real and imaginary parts. It remains to compute the limit, which is also done symbolically for the real and imaginary parts of each entry. To computefor polynomials and , we perform a binary search on the coefficients of g to find the smallest k for which dk≠0, and then output the ratio ck/dk. Each of the required computations can be done in , and can be applied in parallel for each entry of R to allow R to be computed from M in .

Finally, we verify the required properties of Λ. It is clear that every fixed point ρ of Φ is also a fixed point of Λ, sinceand, therefore, . To prove that ρ=Λ(σ) is a fixed point of Φ for every density matrix σ, it suffices to prove ΦΛ=Λ. For each z∈(0,1) we haveand, therefore,as claimed. ▪

### (d) Proof of containment

We can now complete the proof that quantum computers with CTCs are simulatable in .

CTC.

Let LCTC be given, and assume that is a quantum CTC algorithm for L. As discussed in §4b, it suffices to prove L(poly).

Assume for simplicity that an input x of length n has been fixed. Let Q be the unitary quantum circuit that is output by on input x; then, as in the definition of CTC, define a quantum operationOur goal will be to compute the probability(4.5)for some arbitrary fixed point ρ of Φ. This value can then be compared with 1/2 to decide whether to accept or reject. This computation will be performed in a uniform manner, in (poly), therefore establishing that L(poly).

The first step is to compute the matrix representation M=K(Φ) of the operation Φ. This can be done by a polynomial-space uniform family of Boolean circuits with exponential size and polynomial depth, since M is expressible as in equation (4.2), and Q is expressible as a product of a polynomial number of exponential-size matrices determined by the gates of Q.

Next, we compute the matrix representation R=K(Λ), where Λ is the quantum operation that projects onto fixed points of Φ discussed in §4c. We have argued that R can be computed from M in , and, therefore, by composing this computation with the (poly) computation of M, we have that R can be computed in (poly).

Finally, we compute a fixed point ρ of Φ using R along with an arbitrary choice of a density matrix input for Λ. For instance, we may take ρ=Λ((|0〉〈0|)q(n)), so that vec(ρ)=R|0〉⊗2q(n). The probability (4.5) can then be evaluated in (poly) by performing matrix–vector multiplication. ▪

## 5. Dealing with error

In defining the class CTC, we required the quantum circuits to involve amplitudes with rational real and imaginary parts. However, while this assumption is mathematically convenient, it is also ‘unphysical’. Even in a CTC universe, quantum operations can presumably only be implemented to finite precision. In this section, we consider how to make our upper and lower bounds robust to small errors.

The basic difficulty is that, in a CTC universe, two quantum operations that are arbitrarily close can produce detectably different outcomes. As an example, consider the stochastic matrices,As ϵ→0, these matrices become arbitrarily close to each other and to the identity matrix. Yet, their fixed points remain disjoint: the first has a unique fixed point of (1,0)T, while the second has a unique fixed point of (1,0)T. Hence, were an algorithm to apply one of these matrices inside a CTC, an arbitrarily small error could completely change the outcome of the computation.

However, we will show that, while this ‘pathological’ situation can arise in principle, it does not arise in our simulation of by a CTC computer in lemma 3.2.

Given two probability distributions =(Px) and =(qx), recall that their variation distance is defined asAlso, given two mixed states ρ and σ, their trace distance is defined aswhere the maximum is over all unitary matrices U. Finally, given two superoperators Φ and Φ′, their diamond distance is defined aswhere the maximum is over all mixed states ρ on some larger Hilbert space.

Given a superoperator Φ, call ρ an ϵ-fixed point of Φ if .

Suppose ρ is a fixed point of Φ and . Then, ρ is an ϵ-fixed point of Φ′.

Since ρ=Φ(ρ), we have .

Let Φ be a classical operation mapping a finite set to itself, and let ρ be an ϵ-fixed point of Φ. Then for some state σ supported only on the cycles of Φ.

We prove the contrapositive. Let be the union of all cycles of Φ, and let . Also, for each element x, let px=〈|x〉. Suppose ρ is not δ-close to any state supported only on C, where δ=2||ϵ. Then . Hence, letting be the distribution over obtained by measuring ρ in the standard basis, we haveHere the fourth line follows from the triangle inequality. The fifth line follows from the triangle inequality, together with the fact that, if we fix an maximizing px, and then consider all z's ‘upstream’ from x (i.e. such that Φ(z)=x, or Φ(Φ(z))=x, etc.), we must eventually reach ‘source’ elements, i.e. z's for which there are no y's such that Φ(y)=z, and hence ▪

We can now prove the following.

Every language L is decidable in CTC even if every gate of the CTC circuit is subject to 2q(n) error (for some polynomial q depending on L and the circuit).

Let C′ be the circuit from lemma 3.2 that maps CTC to itself. As part of the proof of lemma 3.2, we showed that the graph of has a unique cycle L, in which every configuration leads to the desired output. Now let C″ be a corrupted version of C′ that satisfies , and let ρ be any fixed point of C″. Then ρ is an ϵ-fixed point of C′ by proposition 5.1. By lemma 5.2, this in turn means that for some state σ supported only on L. So provided , a CTC algorithm that uses C″ in place of C′ will still produce the correct answer with high probability. ▪

Moreover, as pointed out by Bacon (2004), even if every gate in our quantum circuit is subject to (sufficiently small) constant error, we can still use standard results from the theory of quantum error correction (Aharonov & Ben-Or 1997) to ensure that |C′−C″| is exponentially small, where C′ and C″ are the quantum circuits acting on the logical (encoded) qubits. See Bacon (2004) for a detailed version of this argument.

But what about our proof of the CTC upper bound, in §4: is that proof affected by precision issues? It might be thought that we simply need to represent all amplitudes and matrix elements to poly(n) bits of precision. As discussed earlier, however, the trouble is that the set of fixed points of a superoperator Φ can depend sensitively on Φ, so that an arbitrarily small change to Φ produces a large change in the set of fixed points. Indeed, this is precisely the reason why we assumed the amplitudes to be complex rational numbers. Owing to that assumption, we were able to use the algorithm of Borodin et al. (1983) to compute a fixed point symbolically rather than just numerically.

## 6. Discussion and open problems

### (a) CTCs in other computational models

In the proof that CTC we did not actually need the full strength of polynomial-time computation inside the CTC: rather, all we needed was the ability to update the configuration of a machine and increment a counter. Thus, our proof also shows (for example) that , where 0 denotes the class of problems solvable by constant-depth, polynomial-size circuits consisting of AND, OR and NOT gates, and is defined the same way as CTC but with 0 circuits instead of arbitrary polynomial-size circuits.

In the other direction, we could also define CTC the same way as CTC, but with machines in place of polynomial-size circuits. Then it is evident that our proof generalizes to show CTC=.

It would be extremely interesting to study the consequences of Deutsch's causal consistency assumption in other settings besides polynomial-time computation, e.g. communication complexity, branching programmes and finite automata.

### (b) Narrow CTCs

What is the power of classical CTCs with a single bit, or of quantum CTCs with a single qubit (as studied by Bacon (2004))? Let CTC1, CTC1 and CTC1 be the corresponding deterministic, randomized and quantum complexity classes. Then it is not hard to show that CTC1, i.e. a single use of a one-bit CTC, is enough to solve all problems in the class . For we can guess a random string w∈{0,1}p(n), then set the CTC bit b to 1 if w is a yes witness or to 0 if w is a no witness, and leave b unchanged if w is neither. If there exists a yes witness but not a no witness, then the only fixed point of the induced stochastic evolution is b=1, while if there exists a no witness but not a yes witness, then the only fixed point is b=0. Indeed, a simple extension of this idea yields CTC1: we set b=1 if a yes witness w∈{0,1}p(n) was guessed, and set b=0 with some tiny probability independent of the witness. Again, the unique fixed point of the induced stochastic map will fix b=1 with overwhelming probability if there exists a yes witness, or b=0 with certainty if not. Fully understanding the power of ‘bounded-width CTCs’ remains a problem for the future.

### (c) CTC computers with advice

Let CTC/ be defined the same way as CTC except that, instead of being initialized to 0q(n), the chronology-respecting register CR is initialized to a probability distribution n, which depends only on the input length n, but can otherwise be chosen arbitrarily to help the CTC algorithm. Then we claim that CTC/=: in other words, CTC/ contains every computational problem! To see this, let CR be initialized to the uniform distribution over all ordered pairs 〈z,f(z)〉, where z is an n-bit input and f(x)∈{0,1} encodes whether xL. Also, let the CTC register CTC contain a single bit b. Then given an input x, our circuit C acts on b as follows: if z=x then C sets b=f(x); otherwise C leaves b unchanged. It is easy to see that the unique fixed point of the induced stochastic map on CTC fixes b=f(x) with certainty.

While it demonstrates that CTCs combined with randomized advice yield staggering computational power, this result is not quite as surprising as it seems: for it was previously shown by Aaronson (2006) that /=, and by Raz (2005) that (2)/=. In other words, randomized advice has a well-known tendency to yield unlimited computational power when combined with certain other resources.

## Acknowledgments

We thank Lance Fortnow for suggesting the proof of lemma 3.2, and Dave Bacon, Todd Brun and David Deutsch for their helpful discussions. S.A. is supported by MIT CSAIL and the Keck Foundation. J.W. is supported by Canada's NSERC and the Canadian Institute for Advanced Research (CIFAR).

## Footnotes

• He also sketched a possible extension to problems. However, Brun did not specify the model of computation underlying his results, and the most natural interpretation of his ‘CTC algorithms’ would appear to preclude their solving problems. For Brun, a fixed point of a CTC evolution seems to be necessarily deterministic—in which case, finding such a fixed point is an problem (note that is almost universally believed to be smaller than ). Thus, to prove that classical or quantum computers with CTCs give the full power of , it seems essential to adopt Deutsch's causal consistency model.

• On the other hand, Bacon's approach would require a polynomial number of such CTCs, rather than a single CTC as in Deutsch's approach.

• Considering inputs of the form 1n are just a standard trick to force n to be encoded in ‘unary’, i.e. such that it takes n bits to write down.

• stands for ‘Nick's class’; the term is historical. Also, what we call (s) is called (2s) by Borodin et al. (1983).

• This fact is proved in Terhal & DiVincenzo (2000). An alternative proof follows from the fact that ‖Φ=1 and that the diamond norm is submultiplicative (see theorem 5.6.9 of Horn & Johnson (1990)).

• Accepted October 10, 2008.

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