APRIL 2019 | VOL. 62 | NO. 4 | COMMUNICATIONS OF THE ACM 135
holds regardless of the spatial separation between the two
qubits, a conclusion that appears very counterintuitive at first
glance: how does the second qubit “know” the basis in which
the first qubit has been measured, and the outcome obtained?
Such natural puzzlement, however, is an artifact of our
presentation (the same given by Einstein, Podolsky and
Rosen!). Indeed, a simple classical scenario recreates the
same phenomenon. Consider two coins with the same random orientation, so that the normal to their surface makes
the same (random) angle θ in the x − y plane. The two coins
are then spatially separated while leaving their orientation
unchanged and measured. By measurement, we mean pick
an angle ϕ, look at the coin along the angle ϕ (looking from
infinity, towards the coin), and report whether one sees a
heads or tails (0 or 1). Clearly the outcome is uniformly random since the orientation θ was chosen uniformly at random. And if we observe (measure) both coins from the same
angle ϕ, the outcomes will be identical.
The fundamental distinction between the quantum and
classical settings manifests itself when the two qubits are
measured in different bases. If we measure the first qubit in
the standard basis and the second in a ϕ-rotated basis, then
as previously each outcome (0 or 1) is individually random,
but the measurement rule indicates that the two outcomes
will match with probability cos2 ϕ. This differs from the statistics obtained in the classical scenario described above,
in which the chance of obtaining matching outcomes is
| 1 − ϕ/π| for ϕ ∈ [0, π). The difference between these two
numbers, ϕ2 vs. ϕ/π for small ϕ, is the foundation for Bell’s
test of quantumness.
We can now see how this difference plays out in a particu-
larly elegant way in the case of the CHSH game. The quantum
strategy starts with Alice and Bob sharing a Bell state. Each
of them performs a measurement of their qubit in a basis
which depends on their input bit (x or y), and announces the
result: Alice measures her qubit in a θA-rotated basis, with
θA = x, and Bob measures his qubit in a θB-rotated basis,
An observation, or measurement, performed on a qubit
is specified by a choice of basis. The distinguished basis
formed by the x and y-axes is called the standard basis. The
Born rule for the outcomes of quantum measurements
specifies that a measurement of the state |ψθ〉 in the stan-
dard basis yields the outcome 0 with probability cos2 θ, and
1 with probability sin2 θ. Once the measurement has been
performed, the qubit collapses to the post-measurement
state consistent with the outcome obtained, |ψ0〉 = |0〉 or
|ψπ/2〉 = | 1〉 respectively. More generally, for any angle ϕ the
qubit can also be measured in the basis obtained by rotat-
ing the standard basis by an angle of ϕ. Such a measurement
yields the outcome 0 with probability cos2(θ − ϕ), and 1 with
probability sin2(θ − ϕ) (See Figure 1).c The post-measurement
states are |ψϕ〉 or |ψπ/2+ϕ〉 respectively.
Now we turn to a discussion of entanglement. The most
fundamental entangled state is the Bell state, which is a maximally entangled state of two qubits. Using the ket notation the
Bell state is written |φ+〉= |0〉|0〉+ | 1〉| 1〉. To understand
the remarkable properties of this state, let us start by describing the result of measuring both qubits in the same basis
(|ψϕ〉, |ψπ/2+ϕ〉): first note that due to symmetry, the outcome
of the measurement on each qubit is uniformly random;
that is, 0 or 1, with probability 1/2 each. But the symmetry of
the Bell state goes deeper. Note that the Bell state is rotationally invariant and can be written as |φ+〉= |ψϕ〉 |ψϕ〉+
|ψϕ+π/2〉 |ψϕ+π/2〉 (this can be easily verified by direct calculation).
This means that the outcome of a measurement on the two
qubits is that both outcomes are 0, or both outcomes are 1,
each with probability 1/2. Moreover the post-measurement
state is |ψϕ〉|ψϕ〉, or |ψπ/2+ϕ〉|ψπ/2+ϕ〉 respectively. This conclusion
b An example of a no-signaling correlation that is neither local nor quantum is the
family of distributions p(a, b|x, y) = 1/2 if and only if a ⊕ b = xy, for a, b, x, y ∈ {0, 1}.
This family gives a success probability 1 in the CHSH game with probability 1.
c Note here we use the notation “0” to designate the outcome, even though
the associated basis element is the vector |ψϕ〉, not |0〉. This is a standard
convention; labels for outcomes are arbitrary and have no physical meaning.
Figure 2. Our proof of security relies on one of the most fundamental
Bell inequalities, the CHSH inequality,
9 which is pictured here as
a small game. Honest devices measuring a Bell pair in the bases
indicated on the right of the figure will produce outputs such that
Pr(a ⊕ b = xy) = cos2 π/8 whenever x, y OE {0, 1}, and a = b whenever
x = 2 and y = 0.
DA DB
x ∈ {0,
1,
2} y ∈ {0,
1}
a ∈ {0,
1} b ∈ {0,
1}
x =0
DA
x = 1
π/8
y = 1
y =0 x = 2
DB
ΨEPR = 1 √ 2(|0 0+|
1
1)
π/4
Figure 1. Measuring ψ in the standard basis yields the outcome 0 with
probability |α|
2. Measuring it in the basis (u, u⊥ yields the outcome u
with probability cos2 ϕ.
÷0ñ
÷1ñ
÷uñ
÷u^ñ
÷ ψñ = α÷0ñ + β÷1ñ
ϕ