Bell Inequalities

• I assume familiarity with basic Quantum Information (i.e. Nielsen/Chuang)

How strong are quantum correlations?

• Bell inequalities• as games• geometrically

Alice and Bob win if

CHSH-GameClauser, Horne, Shimony & Holt

a≠b for x=0, y=0a≠b for x=0, y=1a≠b for x=1, y=0a=b for x=1, y=1

Alice and Bob cannot communicate referee supplies questions to Alice and Bob (x and y)(equal probability for all questions)

What is the maximal probability of winning?

Alice Bob

Referee they win if

Mathematical setup

• Alice and Bob have access to correlationsgiven by

• Winning condition • Questions are chosen with probability

conditional probabilty distribution

Alice and Bob win if

Classical Deterministic Strategy

a≠b for x=0, y=0a≠b for x=0, y=1a≠b for x=1, y=0a=b for x=1, y=1

a=f(x)

Average winning probability 75%

100%100%100%0%

b=g(y)

x01

a01

Alicex01

a01

x01

a01

x01

a01

Bob

f(0)=0 f(1)=0 g(0)=1 g(1)=1Example strategy

Optimality

• There is no better strategy• At most three conditions are satisfied

4th3rd 2nd

is in conflict with 1st:

• Winning probability

Alice and Bob win if

Shared Randomness

a≠b for x=0, y=0a≠b for x=0, y=1a≠b for x=1, y=0a=b for x=1, y=1

Bell inequality

Convex combination of classical deterministic

strategies -> fixed deterministic strategy is

best

Can we beat this with shared entanglement?

z

shared randomness(local hidden variable)

Alice Bob

Referee

deterministic strategies

Bell inequality

Quantum StrategyPOVM

best to choose pure state (cf. convex combination)

| ⇥ = 1⇤2(|01⇥ � |10⇥) | ⌅⇤ | =

1

4(1⇥ 1� �

x

⇥ �x

� �y

⇥ �y

� �z

⇥ �z

)

measurement projector

probability of result

post-measurement state is antipodal point

trA

Axa

� 1|⇥⇤⇥⇥| = 12· �

B,x,a

= trA1

2(1+ ⇤r · ⇤�)|⇥⇤⇥⇥| = 1

2

✓1

2(1� ⇤r · ⇤�)

◆

shared quantum state

Alice Bob

Referee

Entangled Qubits

50%

50%

Bob‘s state=antipodal pointfor every measurement

„spooky action at a distance“

Measurement:Madrid or Wellington

Source

Cos2 45°/2= ≈ 85%Cos2 45°/2 ≈ 85%Cos2 45°/2 ≈ 85%1-Cos2 135°/2≈ 85%

a≠b for x=0, y=0a≠b for x=0, y=1a≠b for x=1, y=0a=b for x=1, y=1

x=1a=0

x=1a=1

y=0b=0

y=0b=1

y=1b=1

y=1b=0

x=0a=0

x=0a=1

Violating the CHSH inequality

multiple of 45°

45°

Cirelson‘s bound

1

2(1 +

1p2)

Define

orthogonal projectors

=1/8X

abxy

P

AB|XY (ab|xy)

+ (PAB|XY (01|00) + PAB|XY (10|00))� PAB|XY (00|00)� PAB|XY (11|00))

+ (PAB|XY (01|01) + PAB|XY (10|01))� PAB|XY (00|01)� PAB|XY (11|01))

+ (PAB|XY (01|10) + PAB|XY (10|10))� PAB|XY (00|10)� PAB|XY (11|10))

+ (PAB|XY (00|11) + PAB|XY (11|11))� PAB|XY (01|11)� PAB|XY (10|11))

Cirelson‘s bound

Lemma: Let

Define Note

Then

Proof:

verify with Mathematica

squares are non-negative

qed

a0b0 + a0b1 + a1b0 � a1b1 = 1p2

⇣(a0)2 + (a1)2 + (b0)2 + (b1)2

⌘

�p2� 18

h⇣(p2 + 1)(a0 � b0) + a1 � b1

⌘2

+⇣(p2 + 1)(a0 � b1)� a1 � b0

⌘2

+⇣(p2 + 1)(a1 � b0) + a0 + b1

⌘2

+⇣(p2 + 1)(a1 + b1)� a0 � b0

⌘2i

Cirelson‘s boundCharacterisation of winning probability & Lemma gives Cirelson‘s bound:

Non-signaling distributions

Description of Bob‘s system alone independent of Alice‘s measurement

reduced statePB(b|y) =

X

a

P (ab|0y) =X

a

P (ab|1y)

PA(a|x) =X

b

P (ab|x0) =X

b

P (ab|x1)

Alice Bob

Referee

P (ab|xy)

Only requirement on is that Alice and Bob cannot communicate (no signaling condition)

P (ab|xy)

Popescu-Rohrlich (PR) Box

Alice and Bob win ifa≠b for x=0, y=0a≠b for x=0, y=1a≠b for x=1, y=0a=b for x=1, y=1

P (01|00) = P (10|00) = 12

P (01|01) = P (10|01) = 12

P (01|10) = P (10|10) = 12

P (00|11) = P (11|11) = 12

. They always win!

state is non-signaling:

PB(b|y) =X

a

P (ab|0y) =X

a

P (ab|1y)

PA(a|x) =X

b

P (ab|x0) =X

b

P (ab|x1)

=1

2

=1

2

Alice Bob

Referee

P (ab|xy)

Comparison of correlations

winning probability

0 0.75 0.85.. 1

More parties? More questions? More answers?

active research field

PR box

Prob[ ]

Prob[ ]

Shared randQuantum

Non-sign

Comparison of correlations

Classically local

Quantum

Nonsignaling

Conditional probability distributions

Figure 8.3: The bipartite conditional probability distributions with a fixed and finitenumber of possible inputs and outputs x, y, a, b form a convex polytope. Inthis polytope, the non-signaling, quantum, and classically local distributionsform a sequence of nested subsets. The classically local and the non-signalingsets are both convex polytopes, while the quantum set is convex but not apolytope.

Definition 8.5.1. A conditional probability distribution is called non-signaling if

�

y

PX,Y |a,b(x, y) =�

y

PX,Y |a,b�(x, y), ⇥a, b, b�, ⇥x,

�

x

PX,Y |a,b(x, y) =�

x

PX,Y |a�,b(x, y), ⇥a, a�, b, ⇥y.(8.21)

We can now check that all conditional distributions arising from bipartite quantummeasurements, as in equation (8.6), are non-signaling. By letting �A = trB�AB , and usethe fact that {Bby}y is a POVM, we find

�

y

PX,Y |a,b(x, y) =�

y

tr(Aax �Bby�AB)

=tr(Aax � 1̂B�AB)=tr(Aax�A).

(8.22)

Hence, Alice cannot notice any of Bob’s choices. By instead summing over x, we find theanalogous statement for Bob.

89

Convex sets: non-signaling: polytopequantum: semidefiniteshared rand: polytope

Sharedrandomness

Game is hyperplaneProb[win]