Download Classical vs. Quantum Correlations

Classical vs. Quantum Correlations
Michael Lachenmaier, Marco Michel
June 16, 2016
During this session we will explore the signicance of quantum information theory compared to classical information theory. The dierences will be
visualized with the help of a simple two-player game.
1
Classical Correlations
In this section the aim is to describe correlations using some reasonable
assumptions.
Later we will see that not all quantum correlations can be
described using those assumptions.
1.1 LHV Ansatz
Assumption 1
We assume that quantities have values even before they are measured. They
might be unknown, i.e. hidden, and thus a probabilistic description may be
needed.
Assumption 2
The properties of one subsystem should not immediately depend on events
taking place on very distant subsystems.
Example 1
Let us assume we have a source
observers Alice
A
B
m∈N
and Bob
Both can choose out of
S
that emits pairs of particles to distant
who each perform
±1
measurement devices,
valued measurements.
Ax
and
By
{1, . . . , m}:
±1
Ax
S
1
By
±1
with
x, y ∈
p(a, b | x, y)
We denote by
the probability that
b ∈ {±1}
A
measures outcome
while
B
Here
P
we want to consider the product of their expectations, i.e.
measures outcome
a,b∈{±1}
using devices
Ax
and
By ,
a ∈ {±1}
respectively.
hAx By i :=
ab p(a, b | x, y).
Ansatz 3 (LHV Ansatz)
Let
Ax , By : Ω −→ {±1}
be random variables and
P
a probability measure.
We make the following ansatz for the expectation of the product:
Z
hAx By i =
Ax (ω)By (ω)dP (ω)
(1)
Ω
Here,
ω∈Ω
plays the role of the hidden variable assumed in assumption 1
and locality from assumption 2 is expressed by the fact that
not depend on
y
and
x,
Ax
and
By
do
respectively. This ansatz is called the local hidden
variable (LHV) ansatz.
1.2 Bell Inequalities
Denition 1
Let
C ∈ Rm×m
with entries
Cxy := hAx By i
that are empirically obtained
expectation values for pairs of ±1 valued measurements. We denote by C ⊆
Rm×m the set of all such C for which there exists an LHV description as in
n×n
ansatz 3. An inequality F : R
→ R for C is called Bell inequality if it
holds for all
C ∈ C,
i.e. if
F (C) ≥ 0 ∀C ∈ C
Remark 1
1.
C
2.
|Cxy | ≤ 1
is a closed convex polytope.
are trivial Bell inequalities.
Proposition 1
ρ ∈ B(HA ⊗ HB ) classically correlated (i.e. not entangled) and Mx :
B({±1}) → B(HA ), My0 : B({±1}) → B(HB ) be POVMs for x, y ∈ {1, . . . , m}.
m×m
Let C ∈ R
be dened by
X
Cxy :=
ab tr[ρMx (a) ⊗ My0 (b)] (x, y = 1, . . . , m),
(2)
Let
a,b∈{±1}
then
C ∈ C.
Proof. If we dene
Ax :=
P
a∈{±1}
aMx (a), By :=
−1 ≤ Ax , By ≤ 1.
2
P
b∈{±1}
bMy0 (b),
then
Assume
ρ
Cxy
(ω)
P
(ω)
pω ρA ⊗ ρB . We
h
i h
i
X
(ω)
(ω)
= tr[ρAx ⊗ By ] =
pω tr ρA Ax tr ρB By
has convex combination
ρ=
ω
calculate:
ω
h
i
(ω)
The last term gives a LHV description as in eq. (1) with Ax (ω) = tr ρA Ax
h
i
(ω)
and By (ω) = tr ρB By on some discrete probability space.
Since eq. (2) is just the expected value, the previous proposition yields
that Bell inequalities hold in particular for any unentangled quantum state.
Next we will state and prove a non-trivial Bell inequality.
Theorem 2 (Causer-Horne-Shimony-Holt (CHSH) inequality)
Every LHV theory for the description of
suring devices
A1 , A2 , B1 , B2
±1
valued measurements with mea-
(as in ansatz 3) satises:
|hA1 B1 i + hA1 B2 i + hA2 B1 i − hA2 B2 i| ≤ 2
(3)
Proof. By the denition of the expected values, we can rewrite the left-hand
side as
Z
A1 (ω) (B1 (ω) + B2 (ω)) + A2 (ω) (B1 (ω) − B2 (ω)) dP (ω) .
Ω
If we x
ω ∈ Ω,
we can distinguish two cases:
B1 (ω) − B2 (ω) = 0
B1 (ω) + B2 (ω) ∈ {±2}.
1.
B1 (ω) = B2 (ω),
2.
B1 (ω) 6= B2 (ω): Since there are only values ±1 possible, this implies
B1 (ω) = −B2 (ω), i.e. B1 (ω) + B2 (ω) = 0 and B1 (ω) − B2 (ω) ∈ {±2}.
i.e.
and
Since both A1 (ω) and A2 (ω) are in {±1}, in both cases, the integrand is in
{±2} which yields, using the fact that P is a probability measure, the desired
inequality.
We will later see an example of how entangled states can violate this
inequality.
2
The CHSH game
in this section we will rst look at classical strategies for the CHSH game.
Later we will also consider quantum strategies.
3
qA
A
qB
B
qA
rB
rA
p(rA , rB | qA , qB )
qB
C
rA
rB
Figure 1: structure of a nonlocal game with two players, questions
answers
qA , qB
and
rA , rB
2.1 Denition
Denition 2 (Nonlocal Game with two Players)
A nonlocal game with two players (henceforth only nonlocal game) consists of
players Alice,
A,
and Bob,
B,
and a referee Charlie,
C.
All communication
in the game is only between any player and the referee, however the players
are allowed to choose a strategy beforehand. The referee chooses a question
at random for each of the two players. Every player then responds with an
answer (dependent on the question she got). After receiving both answers, the
referee determines whether both players won or lost the game. It is therefore
not possible for one player to win and for the other one to lose.
See also
g. 1.
Denition 3 (CHSH game)
The CHSH game is a nonlocal game with two players. The referee Charlie
qA , qB for Alice and Bob, respectively, uniformly
{0, 1}. The answers of Alice and Bob, rA , rB , respectively, must
0 or 1. The players win if the following condition is met:
chooses questions
from the
set
be either
rA ⊕ rB = qA ∧ qB .
Here,
⊕
(4)
denotes the exclusive or which is equivalent to addition modulo
The following table shows the winning conditions:
rA ⊕ rB
0
0
0
1
(qA , qB )
(0, 0)
(0, 1)
(1, 0)
(1, 1)
4
2.
2.2 Classical Strategies
Proposition 3
Using a classical strategy, i.e.
Alice and Bob are only allowed to return
classically correlated states, the maximal probability for Alice and Bob to win
the game is
3/4,
i.e.
P (W IN ) ≤ 3/4.
This bound is sharp.
Proof. For the sake of contradiction, let us assume that there is a strategy
P (W IN ) > 3/4. Since there are only 4 possible
question-pairs, this means that P (W IN ) = 1, i.e. they would win every
time. Let rA , rB : {0, 1} −→ {0, 1} be the strategies used by Bob and Alice,
for Alice and Bob with
accordingly. We write the winning conditions as a system of equations:
rA (0) ⊕ rB (0)
rA (0) ⊕ rB (1)
rA (1) ⊕ rB (0)
rA (1) ⊕ rB (1)
2
Adding those four equations modulo
=
=
=
=
0
0
0
1
0 = 1,
gives
a contradiction.
Even
when using probabilistic strategies rather than deterministic ones, the expected probability to win will never be greater than with a deterministic
So far we have proven that there cannot be a strategy P (W IN ) >
3/4, thus P (W IN ) ≤ 3/4 must hold. Next we will give a strategy with
P (W IN ) = 3/4: Suppose, Alice and Bob will always return 0 as an answer.
one.
Then in three out of four cases, they will win the game.
Since the cases
occur with equal probability (are chosen uniformly), this strategy wins with
probability
3/4.
Relation between the CHSH Game and the CHSH Inequality
We
will look into the relation between the CHSH game and the CHSH inequality
from the previous section.
{±1}
rather than
swers
r̃A
and
r̃B
{0, 1},
If we assume that Alice and Bob return values
we get new winning conditions: Call the new an-
for Alice and Bob, respectively. In the case
the winning condition is that the sum of
to saying that both are either
product of
r̃A
and
condition is that
r̃B
rA
is one.
and
rB
0
or
1.
rA
and
(qA , qB ) = (1, 1),
are dierent, i.e.
5
is zero which is equivalent
This is equivalent to saying that the
Similarly, for
minus one. We summarize:
rB
(qA , qB ) 6= (1, 1)
the product of
the winning
r̃A
and
r̃B
is
(qA , qB ) rA ⊕ rB
(0, 0)
0
0
(0, 1)
(1, 0)
0
(1, 1)
1
r̃A r̃B
1
1
1
−1
A0 , A1 for Alice and B0 , B1 for Bob.
Alice uses A0 if qA = 0 and A1 if qA = 1. Similarly for Bob. We are now in
the situation of example 1, i.e. we can make a LHV ansatz. For x, y ∈ {0, 1}
We now consider measuring devices
consider the expectation of the product
X
hAx By i =
a,b∈{±1}
a=b
For
(x, y) 6= (1, 1),
X
p(a, b | x, y) −
p(a, b | x, y)
(5)
a,b∈{±1}
a6=b
eq. (5) gives the probability of winning minus the
probability of losing, while for
(x, y) = (1, 1),
eq. (5) gives the probability
of losing minus the probability of winning. Since all devices are chosen with
equal probability
1/4,
we get the total probability of winning minus the
probability of losing by the following formula:
P (W IN ) − P (LOSE) =
1
[hA0 B0 i + hA0 B1 i + hA1 B0 i − hA1 B1 i]
4
(6)
We can now apply the CHSH inequality to eq. (6) to obtain
1
2
P (W IN ) ≤ 3/4,
P (W IN ) − P (LOSE) ≤
Using
P (W IN ) = 1 − P (LOSE),
we obtain
(7)
the same
bound as in proposition 3.
2.3 Quantum strategies
However, if Alice and Bob share an entangled two-qubit system which is
1
initialized in the Bell state |ψi = √ (|00i + |11i), then they can gain a
2
2
probability of cos (π/8) ≈ 0.85 if they use following strategy:
For a given angle
θ ∈ [0, 2π),
dene:
|φ0 (θ)i = cos(θ) |0i + sin(θ) |1i
(8)
|φ1 (θ)i = − sin(θ) |0i + cos(θ) |1i
(9)
The strategy is now:
6
1. Alice takes the rst qubit and Bob takes the second qubit from the
quantum system.
2. If Alice receives
x=0
then she measures her qubit with respect to the
basis
{|φ0 (0)i , |φ1 (0)i}
(10)
and if she receives the question 1, she will measure her qubit with
respect to the basis
{|φ0 (π/4)i , |φ1 (π/4)i}
(11)
3. Bob uses a similar strategy, except that he measures with respect to
the basis
{|φ0 (π/8)i , |φ1 (π/8)i}
or
{|φ0 (−π/8)i , |φ1 (−π/8)i}
(12)
depending on whether his question was 0 or 1.
4. Alice and Bob output the values obtained as their answers
a
and
b.
Each of these matrices is a rank-one projection matrix, so the measurements Alice and Bob are making are examples of projective measurements.
1
T
Given our particular choice of |ψi, we have hψ| X ⊗ Y |ψi = tr(X Y ) for
2
b
a
arbitrary matrices X and Y . Thus, as each of the matrices Xs and Yt is real
and symmetric, the probability that Alice and Bob answer (s, t) with (a, b) is
1
tr(Xsa Ytb )). Is is now routine to check that in every case, the correct answer
2
2
is given with probability cos (π/8) and the incorrect answer with probability
2
sin (π/8).
3
Tsirelson bound
One might wonder if it is possible to do even better with another quantum
strategy for the CHSH game. The answer is "no" due to Tsirelson's bound:
Theorem 4 (Tsirelson bound for the CHSH inequality)
β̂ := A1 ⊗ (B1 + B2 ) + A2 ⊗ (B1 − B2 ) with Ai , Bj
density operators ρ ∈ B(HA ⊗ HB ) :
√
tr[ρβ̂] ≤ 2 2
For
as above and for all
(13)
√
That is, quantum theory violates the CHSH inequality at most by a factor
2.
2
2
2
Moreover, there exists a pure state on C ⊗C and observables Ax , Ay ∈ B(C )
with eigenvalues
±1,
s.t. equality holds in this equation.
7
i ∈ {1, 2}. Since these are ane
−1 ≤ Ai ≤ 1, the supremum and
inmum is attained for some extreme point for which spec(Ai ) = {±1} and
2
thus Ai = 1. The same holds for Bj , j ∈ {1, 2}. Using this property, direct
Proof. Consider the map
Ai 7→ tr[ρβ̂]
for
functionals over the closed convex set
computations leads to
β̂ 2 = 41 ⊗ 1 + [A2 , A1 ] ⊗ [B1 , B2 ].
(14)
We exploit this via positivity of the variance and obtain
tr[ρβ̂]2 ≤ tr[ρβ̂ 2 ] = 4 + tr[ρ[A2 , A1 ] ⊗ [B1 , B2 ]]
(15)
≤ 4 + ||[A2 , A1 ] ⊗ [B1 , B2 ]|| ≤ 8
(16)
where the last inequality uses that
||[A1 , A2 ]|| ≤ ||A1 A2 || + ||A2 A1√|| ≤ 2||A1 ||||A2 || = 2 and similarly for the B's.
So we end up with tr[ρβ̂] ≤ 2 2 as claimed.
In order to prove that equality can be achieved, assume that ρ = |ψi hψ|
where |ψi is an eigenvector of β̂ with eigenvalue ν . Then equality holds in
(15).
A1 = B1 = σ1 and A2 = B2 = σ2 Pauli matrices, so that β̂ 2 =
41 + 4σ3 ⊗ σ3 has eigenvalues 0 and 8. Hence, ν can be chosen such that
tr[ρβ̂]2 = ν 2 = 8.
Now take
Remark 2
We have previously seen that
β̂
can be interpreted as the total probability of
winning minus the probability of losing for the CHSH game. So for the previously used strategy we archive equality in Tsirelson's bound:
√
P (LOSE)) = 4 × (cos(π/8)2 − sin(π/8)2 ) = 2 2.
4 × (P (W IN ) −
Remark 3
[A2 , A1 ] = 0 or [B1 , B2 ] =
hCHSHi ≤ 2. Clearly, the noncom-
Note that if the local observables are commuting, i.e.,
0,
then we recover the classical bound
mutativity of observables in quantum mechanics plays a crucial role in the
quantum advantage, yielding a winning probability of 85% compared to the
classical 75% in the CHSH game.
Remark 4
The violation of CHSH by a factor of
for the rst time in the early 80's.
√
2
has been veried experimentally
This was done using down conversion
in a non-linear crystal, which produces entangled pairs of photons, whose
polarization degrees of freedom violate CHSH. A more recent experiment [3]
tested the CHSH Bell inequality on photon pairs in maximally entangled states
of polarization in which a value
2.8276 ± 0.00082
8
was observed.
The argumentation can be generalized to more than two observables:
hAx By i =: Cx y, x, y ∈ {1, ..., m} as before, γ ∈ Rm×m and dene
Consider
||γ||LHV
X
γxy ax by := sup a,b∈{±1}m (17)
x,y
||γ||quantum
X
γxy tr[ρAx ⊗ By ]
:= sup ρ,{Ax By }
(18)
x,y
where
ρ
is density operator and
−1 ≤ Ax , By ≤ 1.
The Trirelson bound for
the CHSH inequality then reads:
||γ||quantum √
ν(γ) :=
= 2
||γ||LHV
for
1 1
γ=
1 −1
(19)
Theorem 5 (General Cirelson
bounds)
√
2×2
1.
γ∈R
⇒ ν(γ) ≤
2.
γ ∈ Rm×m , m ∈ N ⇒ ν(γ) ≤ KG <
2
π √
2ln(1+ 2)
≈ 1.782
Remark 5
1. means that the choice of coecients in CHSH is optimal
2. is a non-trivial statement whose proof is based on a deep result of
Grothendieck.
KG
is called Grothendieck's constant, which is unknown
ν(γ) over all γ ∈ Rm×m and all m ∈ N
but equal to the supremum of
References
[1] M. Wolf, Quantum eects, Lecture Notes, 2014.
[2] J. Watrous, Quantum Computation - Lecture 20: Bell inequalities and
nonlocality, Lecture Notes, 2006.
[3] Hou Shun Poh, et al., Approaching Tsirelsons Bound in a Photon Pair
Experiment, Physical Review Letters 115(18), 2015.
9