The cost of exactly simulating quantum entanglement with classical communication

Abstract

We investigate the amount of communication that must augment classical local hidden variable models in order to simulate the behaviour of entangled quantum systems. We consider the scenario where a bipartite measurement is given from a set of possibilities and the goal is to obtain exactly the same correlations that arise when the actual quantum system is measured. We show that, in the case of a single pair of qubits in a Bell state, a constant number of bits of communication is always sufficient--regardless of the number of measurements under consideration. We also show that, in the case of a system of n Bell states, a constant times 2^n bits of communication are necessary. Comment: 9 pages, LaTeX, no figures

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arXiv:quant-ph/9901035v1 15 Jan 1999
The cost of exactly simulating quantum entanglement
with classical communication
Gilles Brassard∗
Universit´ e de Montr´ eal†
Richard Cleve‡
University of Calgary§
Alain Tapp‡
Universit´ e de Montr´ eal†
15 January 1999
Abstract
We investigate the amount of communication that must augment classical local hid-
den variable models in order to simulate the behaviour of entangled quantum systems.
We consider the scenario where a bipartite measurement is given from a set of possibil-
ities and the goal is to obtain exactly the same correlations that arise when the actual
quantum system is measured. We show that, in the case of a single pair of qubits in a
Bell state, a constant number of bits of communication is always sufficient—regardless
of the number of measurements under consideration. We also show that, in the case of
a system of n Bell states, a constant times 2nbits of communication are necessary.
1Introduction
Bell’s celebrated theorem [1] shows that certain scenarios involving bipartite quantum mea-
surements result in correlations that are impossible to simulate with a classical system if
the measurement events are space-like separated. If the measurement events are time-like
separated then classical simulation is possible, at the expense of some communication. Our
goal is to quantify the required amount of communication.
The issue that we are addressing is part of the broader question of how quantum informa-
tion affects various resources required to perform tasks in information processing. A two-way
classical communication channel between two separated parties can be regarded as a resource,
and a natural goal is for two parties to produce classical information satisfying a specific
stochastic property. One question is, if the parties have an a priori supply of quantum entan-
glement, can they accomplish such goals with less classical communication than necessary in
∗Supported in part by Canada’s nserc, Qu´ ebec’s fcar and the Canada Council.
†D´ epartement IRO, Universit´ e de Montr´ eal, C.P. 6128, succursale centre-ville, Montr´ eal (Qu´ ebec),
Canada H3C 3J7. Email: {brassard,tappa}@iro.umontreal.ca.
‡Supported in part by Canada’s nserc.
§Department of Computer Science, University of Calgary, Calgary, Alberta, Canada T2N 1N4. Email:
cleve@cpsc.ucalgary.ca.

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the case where their a priori information consists of only classical probabilistic information?
And, if so, by how much? Our question is, to what extent does the fundamental behaviour
of an entangled quantum system itself provide savings, in terms of communication compared
with classical systems?
Imagine a scenario involving two “particles” that may have been “together” (and inter-
acted) at some previous point in time, but are “separated” (in a sense which implies that
they can no longer interact) at the present time. Suppose that a measurement is then ar-
bitrarily selected and performed on each particle (not necessarily the same measurement on
both particles). If the underlying physics governing the behaviour of the system is “classical”
then the behaviour of such a system could be based on correlated random variables (usually
called “local hidden variables”), reflecting the possible results of a previous interaction. If
no communication can occur between the components at the time when the measurements
take place then this imposes restrictions on the possible behaviour of such a system. In fact,
if the underlying physics governing the behaviour of the system is “quantum” (in the sense
that it can be based on entangled quantum states, rather than correlated random variables)
then behaviour can occur that is impossible in the classical case. This is a natural way of
interpreting Bell’s theorem [1, 3]. To formalize—and later generalize—this, we shall define
quantum measurement scenarios and (classical) local hidden variable schemes.
2Definitions and preliminary results
Define a quantum measurement scenario as a triple of the form (|Ψ?AB,MA,MB), where
|Ψ?ABis a bipartite quantum state, MAis a set of measurements on the first component,
and MBis a set of measurements on the second component.
It is convenient to parametrize the simplest von Neumann measurements on individual
qubits by points on the unit circle (more general von Neumann measurements, which involve
complex numbers, are considered later in this paper). Let the parameter x ∈ [0,2π) denote
a measurement with respect to the operator
R(x) =
?cosx
sinx
sinx
−cosx
?
(1)
(whose eigenvectors are cos(x
Consider the case of a pair of qubits in the Bell state |Φ+?AB=
results are written for such states, but can be modified to apply to any of the other Bell
states, including the Einstein-Podolsky-Rosen singlet state |Ψ−?AB=
Let x,y ∈ [0,2π) be the respective measurement parameters of the two components and let
a,b ∈ {0,1} be the respective outcomes. Then the joint probability distribution of these
outcomes is given as:
2)|0? + sin(x
2)|1? and sin(x
2)|0? − cos(x
2)|1?).
1
√2|0?|0?+
1
√2|1?|1?. [Our
1
√2|0?|1? −
1
√2|1?|0?.]
Pr[b = 0] Pr[b = 1]
Pr[a = 0]
1
2cos2(x−y
2)
1
2sin2(x−y
2)
Pr[a = 1]
1
2sin2(x−y
2)
1
2cos2(x−y
2)
2

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Two simple but noteworthy examples of bipartite quantum measurement scenarios with
the Bell state |Φ+?ABare:
Example 1: (|Φ+?AB,MA,MB), where MA= MB= {0,π
2}.
Example 2: (|Φ+?AB,MA,MB), where MA= {−π
8,3π
8} and MB= −MA= {π
8,−3π
8}.
In both examples, each individual outcome is a uniformly distributed bit regardless of the
measurements. In Example 1, if the two measurements are the same then the outcomes
are completely correlated; whereas, if the two measurements are different, the outcomes are
completely independent. In Example 2, the two outcomes are equal with probability sin2(π
if x = −y = +3π
the context of local hidden variable schemes, which are defined next.
Intuitively, we are interested in classical devices that simulate bipartite quantum mea-
surement scenarios to varying degrees, and such devices are naturally explained as local
hidden variable schemes. To define a local hidden variable scheme, it is convenient to view
it as a two-party procedure whose execution occurs in two stages: a preparation stage and
a measurement stage. For ease of reference, call the two parties Alice and Bob. During
the preparation stage, local hidden variables u for Alice and v for Bob are determined by
a classical random process. During this stage, arbitrary communication can occur between
the two parties, so u and v may be arbitrarily correlated. During the measurement stage,
measurements x and y are given to Alice and Bob (respectively), who produce outcomes
a = A(x,u) and b = B(y,v) (respectively). During this stage, no communication is permit-
ted between the parties, which is reflected by the fact that the value of A(x,u) is independent
of the value of y (and vice versa).
A local hidden variable scheme simulates a measurement scenario (|Ψ?AB,MA,MB) if,
for any x ∈ MAand y ∈ MB, the outputs produced by Alice and Bob, (namely, a and b
respectively), have exactly the same bivariate distribution as the outcomes of the quantum
measurement scenario as dictated by the laws of quantum physics.
The measurement scenario in Example 1 is easily simulatable by the following local hidden
variable scheme. Let u and v each consist of a copy of the same uniformly distributed two-bit
string. Then let Alice and Bob each output the first bit of this string if their measurement
is 0 and the second bit if their measurement isπ
scenario of Example 2, it turns out that there does not exist a local hidden variable scheme
that simulates it [3].
Now, we consider a more powerful classical instrument for simulating measurement sce-
narios. Define a local hidden variable scheme augmented by k bits of communication, as
follows. Informally, it is a local hidden variable scheme, except that the prohibition of com-
munication between the parties during the measurement stage is relaxed to a condition that
allows up to k bits of communication (but no more). More formally, a local hidden variable
scheme augmented by k bits of communication, has a preparation stage where random vari-
ables u and v for Alice and Bob are determined and during which arbitrary communication
is permitted between the two parties. Then there is a measurement stage which begins by
measurements x and y being given to Alice and Bob (respectively). Then one party computes
a bit (as a function of his/her measurement and local hidden variables) which is sent to the
8)
8; and with probability cos2(π
8) otherwise. These examples are interesting in
2. On the other hand, for the measurement
3

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other party. This constitutes one round of communication. Then again one party (the same
one or a different one) computes a bit (as a function of his/her measurement, local hidden
variables, and any data communicated from the other party at previous rounds) and sends
it to the other party. And this continues for k rounds, after which Alice and Bob output
bits a and b (respectively).
For example, for the measurement scenario of Example 2, a local hidden variable scheme
augmented with one single bit of communication can simulate it. This is a consequence of
the following more general result, whose easy proof we include for completeness.
Theorem 1. For any quantum measurement scenario (|Ψ?AB,MA,MB), there exists a local
hidden variable scheme augmented with log2(|MA|) bits of communication (from Alice to
Bob) that exactly simulates it.
Proof. First note that, if we allow log2(|MA|) bits of communication from Alice to Bob
and log2(|MB|) bits of communication from Bob to Alice then it is trivial to simulate the
quantum measurement scenario. With this much communication, Alice can obtain y and
Bob can obtain x, which effectively defeats any “nonlocality” in the scenario. More precisely,
during the preparation stage, Alice and Bob can construct |MA|·|MB| random variable pairs,
(a(x,y),b(x,y)), one for each value of x ∈ MAand y ∈ MB. Each such random variable pair
would specify the values of the outcomes of Alice and Bob for the given values of x and y,
with the appropriate correlation. During the measurement stage, after the communication
of x and y between them, Alice and Bob can simply output a(x,y)and b(x,y)(respectively).
To obtain a protocol in which only log2(|MA|) bits of communication from Alice to Bob
occurs, note that the unconditional probability distribution of a(x,y)(the output of Alice
when the measurements are x and y) is independent of the value of y. This is because the
distribution of a(x,y)is completely determined by x and the reduced density matrix of |Ψ?AB
with the second component traced out (TrB(|Ψ?AB)), and this quantity is independent of y.
Therefore, the local hidden variables can be set up as follows. For each x ∈ MA, a(x)is
sampled according to the appropriate probability distribution, and then, for each x ∈ MA
and y ∈ MB, b(x,y)is sampled according the appropriate conditional probability distribution
(conditioned on the value of a(x)) in order to produce the correct bivariate distribution for
(a(x),b(x,y)). Then, during the measurement stage it suffices for Alice to send x to Bob, and
for Alice and Bob to output a(x)and b(x,y)(respectively).
We shall see that in some cases the upper bound of Theorem 1 is asymptotically tight
while in other cases it is not. In the sections that follow, we focus on the case of a single Bell
state and the case of n Bell states, and provide a new upper or lower bound in each case.
3The case of a single Bell state
Consider the case of a single Bell state |Φ+?AB=
of MAand MBmay be arbitrarily large. By Theorem 1, we only obtain an upper bound of
log2(|MA|) bits for the amount of communication necessary for an augmented local hidden
variable scheme to simulate it. In the case where MAand MBare each the entire interval
[0,2π), this communication upper bound would be infinite. If only a finite number, k, bits of
1
√2|0?|0? +
1
√2|1?|1?, but where the sizes
4

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communication are permitted then one alternative that might seem reasonable is for Alice to
send x′, a k-bit approximation of x, to Bob. The protocol for Alice and Bob would be along
the lines of the one in Theorem 1, but using x′in place of x. This would clearly not produce
an exact simulation for a general x ∈ [0,2π), but it would produce an approximation that
improves as k increases. Is this the best that can be done with k bits of communication?
The next theorem demonstrates that it is possible to obtain an exact simulation for any
x,y ∈ [0,2π) with only a constant number of bits of communication.
Theorem 2. For the quantum measurement scenario (|Φ+?AB,MA,MB) with |Φ+?AB=
1
√2|0?|0? +
augmented with four of bits of communication (from Alice to Bob) that exactly simulates it.
1
√2|1?|1? and MA = MB = [0,2π), there exists a local hidden variable scheme
Proof. The local hidden variables are c ∈ {0,1} and θ ∈ [0,3π
distributed.
For j ∈ {0,1,...,9}, define αj=
spaced points on the unit circle. Define the jthα-slot as the interval [αj,α(j+1) mod 10). Also,
define β0= α0+θ, β1= α3+θ, and β2= α6+θ and γ0= α5+θ, γ1= α8+θ, and γ2= α1+θ
(where the addition is understood to be modulo 2π). Define the jthβ-slot as the interval
[βj,β(j+1) mod 3), and the jthγ-slot as the interval [γj,γ(j+1) mod 3).
The protocol starts by Alice sending Bob information specifying the α-slot, β-slot, and γ-
slot in which x is located. Note that these slots partition the unit circle into sixteen intervals,
so Alice can convey this information by sending four bits to Bob. Then Alice outputs the
bit c.
The full procedure for Bob is summarized below, but, in order to explain the idea behind
it, it is helpful to first consider the special case where y is in the 2ndα-slot and the α-
slot number of x is within two of that of y (in other words, the α-slot number of x is in
{0,1,2,3,4}). Note that these conditions depend on the values of x and y only (and not on
the values of the local hidden variables). Also, these conditions imply that |x − y| ≤3π
this case, Bob does the following. If the β-slots of x and y are the same then Bob outputs c.
If the β-slots of x and y are different then exactly one βkis between x and y. Let u = |y−βk|.
Then Bob’s procedure is to output c with probability 1 −3π
To analyse the stochastic behaviour of this procedure (still in the special case), let r =
|x − y| and note that the probability of x and y being in different β-slots is
conditional on x and y being in different β-slots, the probability distribution of the position
of the βkbetween x and y is uniform. Therefore,
5), and both are uniformly
jπ
5. It is useful to view α0,α1,...,α9as ten equally-
5. In
10sin(u).
5r
3π. Also,
Pr[a = b] =
?
1
2(1 + cos(r))
= cos2(r
1 −5r
3π
?
+
?5r
3π
??1
r
??r
0
?
1 −3π
10sin(u)
?
du
=
2),
(2)
which is exactly what is required.
The procedure for Bob in the above special case can be generalized to apply to the other
possible cases by considering various similarities and symmetries among the cases. First
note that the above procedure actually works in all cases where the α-slot number of y is in
{2,3,4,5,6} and the α-slot number of x is within two of that of y. This is because, in these
5