A parallel repetition theorem for entangled
two-player one-round games under product
distributions
Rahul Jain1
1
Attila Pereszlényi2
Penghui Yao2
Centre for Quantum Technologies and Department of Computer Science, National
University of Singapore
2
Centre for Quantum Technologies, National University of Singapore
Quantum Games and Protocols
Workshop, Simons Institute, Berkeley
27th February, 2014
Introduction
Proof of the Main Theorem
Outline
1
Introduction
The Model of Games
Parallel Repetition Theorems
Summary
Introduction
Proof of the Main Theorem
Outline
1
Introduction
The Model of Games
Parallel Repetition Theorems
2
Proof of the Main Theorem
Idea Behind the Proof
Some Details
Summary
Introduction
Proof of the Main Theorem
The Model of Games
Outline
1
Introduction
The Model of Games
Parallel Repetition Theorems
2
Proof of the Main Theorem
Idea Behind the Proof
Some Details
Summary
Introduction
Proof of the Main Theorem
The Model of Games
Two-Player One-Round Games
Game G
Alice and Bob share an
entangled state |ϕi.
|ϕi
Alice
Referee
Bob
Summary
Introduction
Proof of the Main Theorem
The Model of Games
Two-Player One-Round Games
Game G
Alice and Bob share an
entangled state |ϕi.
|ϕi
Alice
Bob
y
x
Referee
Summary
Introduction
Proof of the Main Theorem
The Model of Games
Two-Player One-Round Games
Game G
Alice and Bob share an
entangled state |ϕi.
|ϕi
The referee selects questions
(x, y) ∈ X × Y according to
distribution µ.
Alice
Bob
y
x
Referee
Summary
Introduction
Proof of the Main Theorem
The Model of Games
Two-Player One-Round Games
Game G
Alice and Bob share an
entangled state |ϕi.
|ϕi
The referee selects questions
(x, y) ∈ X × Y according to
distribution µ.
Alice
a
Bob
y
x
b
Referee
Summary
Introduction
Proof of the Main Theorem
Summary
The Model of Games
Two-Player One-Round Games
Game G
Alice and Bob share an
entangled state |ϕi.
|ϕi
The referee selects questions
(x, y) ∈ X × Y according to
distribution µ.
Alice
a
Bob
y
x
b
Referee
Alice and Bob answer a ∈ A and
b ∈ B by performing
measurements on |ϕi.
Introduction
Proof of the Main Theorem
Summary
The Model of Games
Two-Player One-Round Games
Game G
Alice and Bob share an
entangled state |ϕi.
|ϕi
The referee selects questions
(x, y) ∈ X × Y according to
distribution µ.
Alice
a
Bob
y
x
b
V (x, y, a, b)
Referee
Alice and Bob answer a ∈ A and
b ∈ B by performing
measurements on |ϕi.
Introduction
Proof of the Main Theorem
Summary
The Model of Games
Two-Player One-Round Games
Game G
Alice and Bob share an
entangled state |ϕi.
|ϕi
The referee selects questions
(x, y) ∈ X × Y according to
distribution µ.
Alice
a
Bob
y
x
b
Alice and Bob answer a ∈ A and
b ∈ B by performing
measurements on |ϕi.
They win if V (x, y, a, b) = 1.
V (x, y, a, b)
Referee
Introduction
Proof of the Main Theorem
Summary
The Model of Games
Two-Player One-Round Games
Game G
Alice and Bob share an
entangled state |ϕi.
|ϕi
The referee selects questions
(x, y) ∈ X × Y according to
distribution µ.
Alice
a
Bob
y
x
b
Alice and Bob answer a ∈ A and
b ∈ B by performing
measurements on |ϕi.
They win if V (x, y, a, b) = 1.
V (x, y, a, b)
Referee
The value of G, denoted by
ω∗ (G), is the supremum of the
achievable winning probability.
Introduction
Proof of the Main Theorem
Summary
The Model of Games
Parallel Repetition of Games
Game Gk
|ϕi
Gk is the game where k copies
of G are played in parallel.
Alice
a
Bob
y
x
b
V (x, y, a, b)
Referee
Introduction
Proof of the Main Theorem
Summary
The Model of Games
Parallel Repetition of Games
Game Gk
|ϕi
Gk is the game where k copies
of G are played in parallel.
Alice
a
Bob
y
x
b
V (x, y, a, b)
Referee
x = (x1 , x2 , . . . , xk ) ∈ X×k ,
y ∈ Y×k , a ∈ A×k , b ∈ B×k
Introduction
Proof of the Main Theorem
Summary
The Model of Games
Parallel Repetition of Games
Game Gk
|ϕi
Gk is the game where k copies
of G are played in parallel.
Alice
a
Bob
y
x
b
V (x, y, a, b)
Referee
x = (x1 , x2 , . . . , xk ) ∈ X×k ,
y ∈ Y×k , a ∈ A×k , b ∈ B×k
(x, y) is distributed according to
µ⊗k , where µ⊗k denotes k
independent copies of µ.
Introduction
Proof of the Main Theorem
Summary
The Model of Games
Parallel Repetition of Games
Game Gk
|ϕi
Gk is the game where k copies
of G are played in parallel.
Alice
a
Bob
y
x
b
x = (x1 , x2 , . . . , xk ) ∈ X×k ,
y ∈ Y×k , a ∈ A×k , b ∈ B×k
(x, y) is distributed according to
µ⊗k , where µ⊗k denotes k
independent copies of µ.
V (x, y, a, b) = 1 if the players
V (x, y, a, b)
Referee
win all the instances.
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
Outline
1
Introduction
The Model of Games
Parallel Repetition Theorems
2
Proof of the Main Theorem
Idea Behind the Proof
Some Details
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
The Basic Question
Skip this part
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
The Basic Question
How does ω∗ Gk scale with k?
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
The Basic Question
How does ω∗ Gk scale with k?
k
Trivially, ω∗ Gk > ω∗ (G) .
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
The Basic Question
How does ω∗ Gk scale with k?
k
Trivially, ω∗ Gk > ω∗ (G) .
The reverse direction doesn’t hold but we can still hope to show
k
that ω∗ Gk ≈ ω∗ (G) .
Summary
Introduction
Proof of the Main Theorem
Summary
Parallel Repetition Theorems
The Basic Question
How does ω∗ Gk scale with k?
k
Trivially, ω∗ Gk > ω∗ (G) .
The reverse direction doesn’t hold but we can still hope to show
k
that ω∗ Gk ≈ ω∗ (G) .
Analogous result holds for the classical value (denoted by ω(G)):
Theorem ([Raz ’95] and [Holenstein ’07])
∃ constant C s.t.
k
ω G
6 1 − C (1 − ω(G))
3
k
log(|A|·|B|)
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
Parallel Repetition Theorems for the Quantum Value
Parallel repetition theorems for the quantum value were shown for
some classes of games.
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
Parallel Repetition Theorems for the Quantum Value
Parallel repetition theorems for the quantum value were shown for
some classes of games.
Perfect parallel repetition holds for XOR games. [Cleve, Slofstra,
Unger, Upadhyay ’08]
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
Parallel Repetition Theorems for the Quantum Value
Parallel repetition theorems for the quantum value were shown for
some classes of games.
Perfect parallel repetition holds for XOR games. [Cleve, Slofstra,
Unger, Upadhyay ’08]
Parallel repetition holds for the more general class of unique
games [Kempe, Regev, Toner ’10]
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
Parallel Repetition Theorems for the Quantum Value
Parallel repetition theorems for the quantum value were shown for
some classes of games.
Perfect parallel repetition holds for XOR games. [Cleve, Slofstra,
Unger, Upadhyay ’08]
Parallel repetition holds for the more general class of unique
games [Kempe, Regev, Toner ’10]
and the even more general class of projection games [Dinur,
Steurer, Vidick ’13].
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
Parallel Repetition Theorems for the Quantum Value
Parallel repetition theorems for the quantum value were shown for
some classes of games.
Perfect parallel repetition holds for XOR games. [Cleve, Slofstra,
Unger, Upadhyay ’08]
Parallel repetition holds for the more general class of unique
games [Kempe, Regev, Toner ’10]
and the even more general class of projection games [Dinur,
Steurer, Vidick ’13].
For general games, [Kempe and Vidick ’11] showed a theorem
where the rate of decay is inverse-polynomial. (Although not for
Gk .)
Summary
Introduction
Proof of the Main Theorem
Parallel Repetition Theorems
Parallel Repetition Theorems for the Quantum Value
Parallel repetition theorems for the quantum value were shown for
some classes of games.
Perfect parallel repetition holds for XOR games. [Cleve, Slofstra,
Unger, Upadhyay ’08]
Parallel repetition holds for the more general class of unique
games [Kempe, Regev, Toner ’10]
and the even more general class of projection games [Dinur,
Steurer, Vidick ’13].
For general games, [Kempe and Vidick ’11] showed a theorem
where the rate of decay is inverse-polynomial. (Although not for
Gk .)
[Chailloux and Scarpa ’13] showed it for games where the input
distribution is uniform.
Summary
Introduction
Proof of the Main Theorem
Summary
Parallel Repetition Theorems
Our Contribution
Theorem (Main Theorem)
For any game G, where the input distribution µ is product on X × Y, it
holds that
∗
k
ω G
∗
= 1 − (1 − ω (G))
3
Ω
k
log(|A|·|B|)
.
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Outline
1
Introduction
The Model of Games
Parallel Repetition Theorems
2
Proof of the Main Theorem
Idea Behind the Proof
Some Details
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Broad Idea
In Gk , let us condition on success on a set C ⊆ [k] of coordinates.
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Broad Idea
In Gk , let us condition on success on a set C ⊆ [k] of coordinates.
If the success probability is small enough then we are done.
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Broad Idea
In Gk , let us condition on success on a set C ⊆ [k] of coordinates.
If the success probability is small enough then we are done.
Otherwise, we show that there exists a j ∈ C = [k] \ C s.t.
success in the j-th coordinate is bounded away from 1.
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Broad Idea
In Gk , let us condition on success on a set C ⊆ [k] of coordinates.
If the success probability is small enough then we are done.
Otherwise, we show that there exists a j ∈ C = [k] \ C s.t.
success in the j-th coordinate is bounded away from 1.
Doing this Ω(k) times, the success probability will be
exponentially small in k.
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Broad Idea
In Gk , let us condition on success on a set C ⊆ [k] of coordinates.
If the success probability is small enough then we are done.
Otherwise, we show that there exists a j ∈ C = [k] \ C s.t.
success in the j-th coordinate is bounded away from 1.
Doing this Ω(k) times, the success probability will be
exponentially small in k.
Let (x 0 , y 0 ) ∈ X × Y be distributed according to µ.
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Broad Idea
In Gk , let us condition on success on a set C ⊆ [k] of coordinates.
If the success probability is small enough then we are done.
Otherwise, we show that there exists a j ∈ C = [k] \ C s.t.
success in the j-th coordinate is bounded away from 1.
Doing this Ω(k) times, the success probability will be
exponentially small in k.
Let (x 0 , y 0 ) ∈ X × Y be distributed according to µ.
We show that the players can embed (x 0 , y 0 ) into the j-th
coordinate and generate the state of the whole system in Gk .
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Broad Idea
In Gk , let us condition on success on a set C ⊆ [k] of coordinates.
If the success probability is small enough then we are done.
Otherwise, we show that there exists a j ∈ C = [k] \ C s.t.
success in the j-th coordinate is bounded away from 1.
Doing this Ω(k) times, the success probability will be
exponentially small in k.
Let (x 0 , y 0 ) ∈ X × Y be distributed according to µ.
We show that the players can embed (x 0 , y 0 ) into the j-th
coordinate and generate the state of the whole system in Gk .
If they could win the j-th instance with high probability then they
would be able to win G with high probability.
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
Some Simplifications
Without loss of generality:
Let the questions and answers be part of the whole
(classical-)quantum state.
Summary
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
Some Simplifications
Without loss of generality:
Let the questions and answers be part of the whole
(classical-)quantum state.
Upon receiving the questions, the players apply unitaries on their
parts, measure registers A and B, in the standard basis, and
send the outcomes to the referee.
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
Some Simplifications
Without loss of generality:
Let the questions and answers be part of the whole
(classical-)quantum state.
Upon receiving the questions, the players apply unitaries on their
parts, measure registers A and B, in the standard basis, and
send the outcomes to the referee.
The global state, conditioned on success in C, after the unitaries is of
the form
σ=
X
x∈X×k ,y∈Y×k
µ⊗k (x, y) |xyihxy|XY ⊗ |φxy ihφxy |
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
Some Simplifications
Without loss of generality:
Let the questions and answers be part of the whole
(classical-)quantum state.
Upon receiving the questions, the players apply unitaries on their
parts, measure registers A and B, in the standard basis, and
send the outcomes to the referee.
The global state, conditioned on success in C, after the unitaries is of
the form
σ=
X
µ⊗k (x, y) |xyihxy|XY ⊗ |φxy ihφxy |
x∈X×k ,y∈Y×k
where (unnormalized) |φxy i is shared between Alice and Bob.
Introduction
Proof of the Main Theorem
Idea Behind the Proof
How to generate the state in Gk ?
Let us take a strategy for G where the players share
|ϕi =
X
x∈X×k ,y∈Y×k
q
µ⊗k (x, y) |xxyyiXX̃YỸ ⊗ |φxy i .
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
How to generate the state in Gk ?
Let us take a strategy for G where the players share
|ϕi =
X
x∈X×k ,y∈Y×k
TrX̃⊗Ỹ (|ϕihϕ|) = σ.
q
µ⊗k (x, y) |xxyyiXX̃YỸ ⊗ |φxy i .
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
How to generate the state in Gk ?
Let us take a strategy for G where the players share
|ϕi =
X
q
µ⊗k (x, y) |xxyyiXX̃YỸ ⊗ |φxy i .
x∈X×k ,y∈Y×k
TrX̃⊗Ỹ (|ϕihϕ|) = σ.
Alice and Bob get questions (x 0 , y 0 ) ∈ X × Y.
Summary
Introduction
Proof of the Main Theorem
Idea Behind the Proof
How to generate the state in Gk ?
Let us take a strategy for G where the players share
|ϕi =
X
q
µ⊗k (x, y) |xxyyiXX̃YỸ ⊗ |φxy i .
x∈X×k ,y∈Y×k
TrX̃⊗Ỹ (|ϕihϕ|) = σ.
Alice and Bob get questions (x 0 , y 0 ) ∈ X × Y.
Suppose they measure registers Xj and Yj .
Summary
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
How to generate the state in Gk ?
Let us take a strategy for G where the players share
|ϕi =
X
q
µ⊗k (x, y) |xxyyiXX̃YỸ ⊗ |φxy i .
x∈X×k ,y∈Y×k
TrX̃⊗Ỹ (|ϕihϕ|) = σ.
Alice and Bob get questions (x 0 , y 0 ) ∈ X × Y.
Suppose they measure registers Xj and Yj .
If the outcomes of the measurements are x 0 and y 0 then they can
further measure Aj and Bj and reply to the referee.
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
How to generate the state in Gk ?
Let us take a strategy for G where the players share
|ϕi =
X
q
µ⊗k (x, y) |xxyyiXX̃YỸ ⊗ |φxy i .
x∈X×k ,y∈Y×k
TrX̃⊗Ỹ (|ϕihϕ|) = σ.
Alice and Bob get questions (x 0 , y 0 ) ∈ X × Y.
Suppose they measure registers Xj and Yj .
If the outcomes of the measurements are x 0 and y 0 then they can
further measure Aj and Bj and reply to the referee.
This way, we embedded a single instance of G into Gk , with
being conditioned on success in C.
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
How to generate the state in Gk ?
Let us take a strategy for G where the players share
|ϕi =
X
q
µ⊗k (x, y) |xxyyiXX̃YỸ ⊗ |φxy i .
x∈X×k ,y∈Y×k
TrX̃⊗Ỹ (|ϕihϕ|) = σ.
Alice and Bob get questions (x 0 , y 0 ) ∈ X × Y.
Suppose they measure registers Xj and Yj .
If the outcomes of the measurements are x 0 and y 0 then they can
further measure Aj and Bj and reply to the referee.
This way, we embedded a single instance of G into Gk , with
being conditioned on success in C.
If ω∗ (G) is bounded away from 1 then success in the j-th
instance is also bounded away from 1.
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
Generate the state without measurements?
The previous argument doesn’t work because the probability of getting
(x 0 , y 0 ), when measuring Xj and Yj , can be very small.
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
Generate the state without measurements?
The previous argument doesn’t work because the probability of getting
(x 0 , y 0 ), when measuring Xj and Yj , can be very small.
Question
Is there a way to generate the post-measurement state (approximately)
without measurements?
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
Generate the state without measurements?
The previous argument doesn’t work because the probability of getting
(x 0 , y 0 ), when measuring Xj and Yj , can be very small.
Question
Is there a way to generate the post-measurement state (approximately)
without measurements?
Answer
Yes, if the input distribution µ is product.
Introduction
Proof of the Main Theorem
Idea Behind the Proof
How to do it without measurements?
Let ϕx 0 be the resulting state after Alice measures Xj in |ϕi and
j
gets xj0 . States ϕy 0 and ϕx 0 y 0 are defined similarly.
j
j
j
Summary
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
How to do it without measurements?
Let ϕx 0 be the resulting state after Alice measures Xj in |ϕi and
j
gets xj0 . States ϕy 0 and ϕx 0 y 0 are defined similarly.
j
j
j
We show that I Xj : Bob ϕ ≈ 0 and I Yj : Alice ϕ ≈ 0.
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
How to do it without measurements?
Let ϕx 0 be the resulting state after Alice measures Xj in |ϕi and
j
gets xj0 . States ϕy 0 and ϕx 0 y 0 are defined similarly.
j
j
j
We show that I Xj : Bob ϕ ≈ 0 and I Yj : Alice ϕ ≈ 0.
I Xj : Bob ϕ ≈ 0 implies that Bob’s part of ϕx 0 is mostly
independent of xj0 .
j
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
How to do it without measurements?
Let ϕx 0 be the resulting state after Alice measures Xj in |ϕi and
j
gets xj0 . States ϕy 0 and ϕx 0 y 0 are defined similarly.
j
j
j
We show that I Xj : Bob ϕ ≈ 0 and I Yj : Alice ϕ ≈ 0.
I Xj : Bob ϕ ≈ 0 implies that Bob’s part of ϕx 0 is mostly
independent of xj0 .
j
By the unitary equivalence of purifications, ∃ unitary Ux 0 that
j
Alice can apply to get Ux 0 ⊗ 1Bob |ϕi ≈ ϕx 0 .
j
j
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
How to do it without measurements?
Let ϕx 0 be the resulting state after Alice measures Xj in |ϕi and
j
gets xj0 . States ϕy 0 and ϕx 0 y 0 are defined similarly.
j
j
j
We show that I Xj : Bob ϕ ≈ 0 and I Yj : Alice ϕ ≈ 0.
I Xj : Bob ϕ ≈ 0 implies that Bob’s part of ϕx 0 is mostly
independent of xj0 .
j
By the unitary equivalence of purifications, ∃ unitary Ux 0 that
j
Alice can apply to get Ux 0 ⊗ 1Bob |ϕi ≈ ϕx 0 .
j
j
Similarly, ∃Vy 0 s.t. 1Alice ⊗ Vy 0 |ϕi ≈ ϕy 0 .
j
j
j
Introduction
Proof of the Main Theorem
Summary
Idea Behind the Proof
How to do it without measurements?
Let ϕx 0 be the resulting state after Alice measures Xj in |ϕi and
j
gets xj0 . States ϕy 0 and ϕx 0 y 0 are defined similarly.
j
j
j
We show that I Xj : Bob ϕ ≈ 0 and I Yj : Alice ϕ ≈ 0.
I Xj : Bob ϕ ≈ 0 implies that Bob’s part of ϕx 0 is mostly
j
independent of xj0 .
By the unitary equivalence of purifications, ∃ unitary Ux 0 that
j
Alice can apply to get Ux 0 ⊗ 1Bob |ϕi ≈ ϕx 0 .
j
j
Similarly, ∃Vy 0 s.t. 1Alice ⊗ Vy 0 |ϕi ≈ ϕy 0 .
j
j
j
By [Jain, Radhakrishnan, Sen ’08], if the distribution of xj0 , yj0 is
product then
Uxj0 ⊗ Vyj0 |ϕi ≈ ϕxj0 yj0 .
Introduction
Proof of the Main Theorem
Some Details
Outline
1
Introduction
The Model of Games
Parallel Repetition Theorems
2
Proof of the Main Theorem
Idea Behind the Proof
Some Details
Summary
Introduction
Proof of the Main Theorem
Some Details
From Measurements to Unitaries (1 Sided)
Lemma
Let µ be a probability distribution on X. Let
def
|ϕi =
Xp
µ(x) |xxiXX̃ ⊗ |ψx i
x∈X
be shared by Alice and Bob, where X, X̃, and some part of |ψx i are
with Alice and the rest of |ψx i is with Bob.
Summary
Introduction
Proof of the Main Theorem
Summary
Some Details
From Measurements to Unitaries (1 Sided)
Lemma
Let µ be a probability distribution on X. Let
def
|ϕi =
Xp
µ(x) |xxiXX̃ ⊗ |ψx i
x∈X
be shared by Alice and Bob, where X, X̃, and some part of |ψx i are
def
with Alice and the rest of |ψx i is with Bob. Let |ϕx i = |xxi ⊗ |ψx i.
Introduction
Proof of the Main Theorem
Summary
Some Details
From Measurements to Unitaries (1 Sided)
Lemma
Let µ be a probability distribution on X. Let
def
|ϕi =
Xp
µ(x) |xxiXX̃ ⊗ |ψx i
x∈X
be shared by Alice and Bob, where X, X̃, and some part of |ψx i are
def
with Alice and the rest of |ψx i is with Bob. Let |ϕx i = |xxi ⊗ |ψx i. If
I(X : Bob)ϕ 6 ε then there exist unitaries {Ux }x∈X acting on Alice’s
space s.t.
√
E [k|ϕx ihϕx | − (Ux ⊗ 1Bob ) |ϕihϕ| (U∗x ⊗ 1Bob )k1 ] 6 4 ε.
x←µ
Introduction
Proof of the Main Theorem
Summary
Some Details
From Measurements to Unitaries (1 Sided)
Lemma
Let µ be a probability distribution on X. Let
def
|ϕi =
Xp
µ(x) |xxiXX̃ ⊗ |ψx i
x∈X
be shared by Alice and Bob, where X, X̃, and some part of |ψx i are
def
with Alice and the rest of |ψx i is with Bob. Let |ϕx i = |xxi ⊗ |ψx i. If
I(X : Bob)ϕ 6 ε then there exist unitaries {Ux }x∈X acting on Alice’s
space s.t.
√
E [k|ϕx ihϕx | − (Ux ⊗ 1Bob ) |ϕihϕ| (U∗x ⊗ 1Bob )k1 ] 6 4 ε.
x←µ
The proof easily follows from the unitary equivalence of purifications
and Uhlmann’s theorem.
Introduction
Proof of the Main Theorem
Some Details
From Measurements to Unitaries (2 Sided)
Lemma
Let µ be a prob. dist. on X × Y with marginals µX and µY . Let
def
|ϕi =
X
p
µ(x, y) |xxyyiXX̃YỸ ⊗ |ψx,y i
x∈X,y∈Y
be shared by Alice and Bob, where X, X̃, and some part of |ψx i are
with Alice and Y, Ỹ, and the rest of |ψx i are with Bob.
Summary
Introduction
Proof of the Main Theorem
Some Details
From Measurements to Unitaries (2 Sided)
Lemma
Let µ be a prob. dist. on X × Y with marginals µX and µY . Let
def
|ϕi =
X
p
µ(x, y) |xxyyiXX̃YỸ ⊗ |ψx,y i
x∈X,y∈Y
be shared by Alice and Bob, where X, X̃, and some part of |ψx i are
with Alice and Y, Ỹ, and the rest of |ψx i are with Bob. If
I(X : Bob)ϕ 6 ε and I(Y : Alice)ϕ 6 ε then there exist unitaries
{Ux }x∈X on Alice’s space and {Vy }y∈Y on Bob’s space
Summary
Introduction
Proof of the Main Theorem
Some Details
From Measurements to Unitaries (2 Sided)
Lemma
Let µ be a prob. dist. on X × Y with marginals µX and µY . Let
def
|ϕi =
X
p
µ(x, y) |xxyyiXX̃YỸ ⊗ |ψx,y i
x∈X,y∈Y
be shared by Alice and Bob, where X, X̃, and some part of |ψx i are
with Alice and Y, Ỹ, and the rest of |ψx i are with Bob. If
I(X : Bob)ϕ 6 ε and I(Y : Alice)ϕ 6 ε then there exist unitaries
{Ux }x∈X on Alice’s space and {Vy }y∈Y on Bob’s space s.t.
h
i
|ϕx,y ihϕx,y | − (Ux ⊗ Vy ) |ϕihϕ| U∗ ⊗ V ∗ x
y 1
(x,y)←µ
√
6 8 ε + 2 kµ − µX ⊗ µY k1 .
E
Summary
Introduction
Proof of the Main Theorem
Some Details
From Measurements to Unitaries (2 Sided)
Lemma
Let µ be a prob. dist. on X × Y with marginals µX and µY . Let
def
|ϕi =
X
p
µ(x, y) |xxyyiXX̃YỸ ⊗ |ψx,y i
x∈X,y∈Y
be shared by Alice and Bob, where X, X̃, and some part of |ψx i are
with Alice and Y, Ỹ, and the rest of |ψx i are with Bob. If
I(X : Bob)ϕ 6 ε and I(Y : Alice)ϕ 6 ε then there exist unitaries
{Ux }x∈X on Alice’s space and {Vy }y∈Y on Bob’s space s.t.
h
i
|ϕx,y ihϕx,y | − (Ux ⊗ Vy ) |ϕihϕ| U∗ ⊗ V ∗ x
y 1
(x,y)←µ
√
6 8 ε + 2 kµ − µX ⊗ µY k1 .
E
Proved by [Jain, Radhakrishnan, Sen ’08].
Summary
Introduction
Proof of the Main Theorem
Summary
Some Details
Key Lemma
Our main theorem follows from the following lemma.
Lemma (Key Lemma)
Let 1/10 > δ1 , δ2 , δ3 > 0 s.t. δ3 = δ2 + δ1 · log (|A| · |B|). Let
def
k 0 = bδ1 kc.
Show theorem
Introduction
Proof of the Main Theorem
Summary
Some Details
Key Lemma
Our main theorem follows from the following lemma.
Lemma (Key Lemma)
Let 1/10 > δ1 , δ2 , δ3 > 0 s.t. δ3 = δ2 + δ1 · log (|A| · |B|). Let
def
k 0 = bδ1 kc. Given any quantum strategy for Gk , there exists
{i1 , . . . , ik 0 } s.t. for each 1 6 l 6 k 0 − 1, either
h
i
Pr T (l) = 1 6 2−δ2 k
where Ti ∈ {0, 1} indicates success in the i-th repetition and
def
T (l) =
Ql
j=1 Tij .
Show theorem
Introduction
Proof of the Main Theorem
Summary
Some Details
Key Lemma
Our main theorem follows from the following lemma.
Lemma (Key Lemma)
Let 1/10 > δ1 , δ2 , δ3 > 0 s.t. δ3 = δ2 + δ1 · log (|A| · |B|). Let
def
k 0 = bδ1 kc. Given any quantum strategy for Gk , there exists
{i1 , . . . , ik 0 } s.t. for each 1 6 l 6 k 0 − 1, either
h
i
Pr T (l) = 1 6 2−δ2 k
or
h
i
p
Pr Til+1 = 1T (l) = 1 6 ω∗ (G) + 12 10δ3
where Ti ∈ {0, 1} indicates success in the i-th repetition and
def
T (l) =
Ql
j=1 Tij .
Show theorem
Introduction
Proof of the Main Theorem
Some Details
Proof of the Key Lemma
Skip the proof
Summary
Introduction
Proof of the Main Theorem
Some Details
Proof of the Key Lemma
Suppose that we already identified l coordinates and we want to find
the (l + 1)-th coordinate with the given properties.
Summary
Introduction
Proof of the Main Theorem
Some Details
Proof of the Key Lemma
Suppose that we already identified l coordinates and we want to find
the (l + 1)-th coordinate with the given properties.
def
Assume q = Pr T (l) = 1 > 2−δ2 k as otherwise we are done.
Summary
Introduction
Proof of the Main Theorem
Some Details
Proof of the Key Lemma
Suppose that we already identified l coordinates and we want to find
the (l + 1)-th coordinate with the given properties.
def
Assume q = Pr T (l) = 1 > 2−δ2 k as otherwise we are done.
def
Let C = {i1 , . . . , il }.
Summary
Introduction
Proof of the Main Theorem
Some Details
Proof of the Key Lemma
Suppose that we already identified l coordinates and we want to find
the (l + 1)-th coordinate with the given properties.
def
Assume q = Pr T (l) = 1 > 2−δ2 k as otherwise we are done.
def
Let C = {i1 , . . . , il }.
The purified state after the unitaries is
def
|θi =
X
x∈X×k ,y∈Y×k
⊗
q
µ⊗k (x, y) |xxyyiXX̃YỸ
X
aC ∈A×l ,bC ∈B×l
E E
|aC bC iAC BC ⊗ γx,y,aC ,bC A B .
Summary
Introduction
Proof of the Main Theorem
Summary
Some Details
Proof of the Key Lemma
Suppose that we already identified l coordinates and we want to find
the (l + 1)-th coordinate with the given properties.
def
Assume q = Pr T (l) = 1 > 2−δ2 k as otherwise we are done.
def
Let C = {i1 , . . . , il }.
The purified state after the unitaries is
X
def
|θi =
x∈X×k ,y∈Y×k
⊗
q
µ⊗k (x, y) |xxyyiXX̃YỸ
X
E E
|aC bC iAC BC ⊗ γx,y,aC ,bC A B .
aC ∈A×l ,bC ∈B×l
Conditioning on success in C gives us the state
q
1 X
|ϕi = √
µ⊗k (x, y) |xxyyi
def
q
x,y
X
aC ,bC s.t.
Q
i∈C Ti =1
|aC bC i ⊗ γx,y,aC ,bC .
Introduction
Proof of the Main Theorem
Some Details
Lemma about the Relative Entropy
From q > 2−δ2 k , we show the following simple lemma.
Lemma
E
xC ,yC ,aC ,bC ←ϕXC YC AC BC
h i
XYX̃C ỸC EA EB XYX̃C ỸC EA EB
S ϕxC ,yC ,aC ,bC θxC ,yC
6 δ3 k
Summary
Introduction
Proof of the Main Theorem
Some Details
Lemma about the Relative Entropy
From q > 2−δ2 k , we show the following simple lemma.
Lemma
E
xC ,yC ,aC ,bC ←ϕXC YC AC BC
h i
XYX̃C ỸC EA EB XYX̃C ỸC EA EB
S ϕxC ,yC ,aC ,bC θxC ,yC
6 δ3 k
Intuitively,
going from θ to ϕ causes a difference of at most − log q < δ2 k.
Summary
Introduction
Proof of the Main Theorem
Some Details
Lemma about the Relative Entropy
From q > 2−δ2 k , we show the following simple lemma.
Lemma
E
xC ,yC ,aC ,bC ←ϕXC YC AC BC
h i
XYX̃C ỸC EA EB XYX̃C ỸC EA EB
S ϕxC ,yC ,aC ,bC θxC ,yC
6 δ3 k
Intuitively,
going from θ to ϕ causes a difference of at most − log q < δ2 k.
further measuring AC and BC results in a difference of at most
|C| · log (|A| · |B|) 6 δ1 k · log (|A| · |B|).
Summary
Introduction
Proof of the Main Theorem
Some Details
Upper Bound for the Mutual Information
Using the previous lemma, we can show the required upper bound for
Skip
the mutual information.
Summary
Introduction
Proof of the Main Theorem
Some Details
Upper Bound for the Mutual Information
Using the previous lemma, we can show the required upper bound for
the mutual information.
δ3 k >
E
xC ,yC ,aC ,bC ←ϕXC YC AC BC
h i
XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
Summary
Introduction
Proof of the Main Theorem
Some Details
Upper Bound for the Mutual Information
Using the previous lemma, we can show the required upper bound for
the mutual information.
h i
XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
i
h X(Bob)
X(Bob)
> E S ϕxC ,yC ,aC ,bC θxC ,yC
δ3 k >
E
def
where Bob = YỸC EB .
Summary
Introduction
Proof of the Main Theorem
Some Details
Upper Bound for the Mutual Information
Using the previous lemma, we can show the required upper bound for
the mutual information.
h i
XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
i
h X(Bob)
X(Bob)
> E S ϕxC ,yC ,aC ,bC θxC ,yC
h i
X(Bob)
= E S ϕxC ,yC ,aC ,bC θXxC ,yC ⊗ θBob
xC ,yC
δ3 k >
E
def
where Bob = YỸC EB .
Summary
Introduction
Proof of the Main Theorem
Some Details
Upper Bound for the Mutual Information
Using the previous lemma, we can show the required upper bound for
the mutual information.
h i
XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
i
h X(Bob)
X(Bob)
> E S ϕxC ,yC ,aC ,bC θxC ,yC
h i
X(Bob)
= E S ϕxC ,yC ,aC ,bC θXxC ,yC ⊗ θBob
xC ,yC
h i
X(Bob)
> E S ϕxC ,yC ,aC ,bC ϕXxC ,yC ,aC ,bC ⊗ ϕBob
xC ,yC ,aC ,bC
δ3 k >
E
def
where Bob = YỸC EB .
Summary
Introduction
Proof of the Main Theorem
Some Details
Upper Bound for the Mutual Information
Using the previous lemma, we can show the required upper bound for
the mutual information.
h i
XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
i
h X(Bob)
X(Bob)
> E S ϕxC ,yC ,aC ,bC θxC ,yC
h i
X(Bob)
= E S ϕxC ,yC ,aC ,bC θXxC ,yC ⊗ θBob
xC ,yC
h i
X(Bob)
> E S ϕxC ,yC ,aC ,bC ϕXxC ,yC ,aC ,bC ⊗ ϕBob
xC ,yC ,aC ,bC
δ3 k >
E
= I(X : Bob|XC YC AC BC )ϕ
def
where Bob = YỸC EB .
Summary
Introduction
Proof of the Main Theorem
Some Details
Upper Bound for the Mutual Information
Using the previous lemma, we can show the required upper bound for
the mutual information.
h i
XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
i
h X(Bob)
X(Bob)
> E S ϕxC ,yC ,aC ,bC θxC ,yC
h i
X(Bob)
= E S ϕxC ,yC ,aC ,bC θXxC ,yC ⊗ θBob
xC ,yC
h i
X(Bob)
> E S ϕxC ,yC ,aC ,bC ϕXxC ,yC ,aC ,bC ⊗ ϕBob
xC ,yC ,aC ,bC
δ3 k >
E
= I(X : Bob|XC YC AC BC )ϕ
X
=
I Xi : BobXC∪[i−1] YC AC BC ϕ
i∈C
def
where Bob = YỸC EB .
Summary
Introduction
Proof of the Main Theorem
Some Details
Distribution of Questions
Using the same lemma, we show that for most of the coordinates in C
Skip
the distribution of questions is close to µ in |ϕi.
Summary
Introduction
Proof of the Main Theorem
Some Details
Distribution of Questions
Using the same lemma, we show that for most of the coordinates in C
the distribution of questions is close to µ in |ϕi.
δ3 k >
E
xC ,yC ,aC ,bC ←ϕXC YC AC BC
i
h XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
Summary
Introduction
Proof of the Main Theorem
Some Details
Distribution of Questions
Using the same lemma, we show that for most of the coordinates in C
the distribution of questions is close to µ in |ϕi.
i
h XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
XY >
E
S ϕXY
xC ,yC ,aC ,bC θxC ,yC
δ3 k >
E
xC ,yC ,aC ,bC ←ϕXC YC AC BC
Summary
Introduction
Proof of the Main Theorem
Some Details
Distribution of Questions
Using the same lemma, we show that for most of the coordinates in C
the distribution of questions is close to µ in |ϕi.
i
h XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
XY θ
>
E
S ϕXY
xC ,yC
x
,
y
,
a
,
b
C
C
C
C
xC ,yC ,aC ,bC ←ϕXC YC AC BC
h
i
X
Yi
=
E
S ϕrXii Yi θxXiC∪[i−
1] ,yC∪[i−1]
R
δ3 k >
E
i∈C
def
ri ←ϕ
i
where Ri = XC∪[i−1] YC∪[i−1] AC BC .
Summary
Introduction
Proof of the Main Theorem
Some Details
Distribution of Questions
Using the same lemma, we show that for most of the coordinates in C
the distribution of questions is close to µ in |ϕi.
i
h XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
XY θ
>
E
S ϕXY
xC ,yC
x
,
y
,
a
,
b
C
C
C
C
xC ,yC ,aC ,bC ←ϕXC YC AC BC
h
i
X
Yi
=
E
S ϕrXii Yi θxXiC∪[i−
1] ,yC∪[i−1]
R
δ3 k >
E
i∈C
>
ri ←ϕ
i
X
E
i∈C
def
ri ←ϕRi
h
i
ϕXi Yi − µ2
ri
1
where Ri = XC∪[i−1] YC∪[i−1] AC BC .
Summary
Introduction
Proof of the Main Theorem
Some Details
Distribution of Questions
Using the same lemma, we show that for most of the coordinates in C
the distribution of questions is close to µ in |ϕi.
i
h XYX̃ Ỹ E E XYX̃ Ỹ E E
S ϕxC ,yCC ,CaCA,bBC θxC ,yCC C A B
xC ,yC ,aC ,bC ←ϕXC YC AC BC
XY θ
>
E
S ϕXY
xC ,yC
x
,
y
,
a
,
b
C
C
C
C
xC ,yC ,aC ,bC ←ϕXC YC AC BC
h
i
X
Yi
=
E
S ϕrXii Yi θxXiC∪[i−
1] ,yC∪[i−1]
R
δ3 k >
E
i∈C
>
i
X
E
i∈C
>
ri ←ϕ
ri ←ϕRi
h
i
ϕXi Yi − µ2
ri
1
X
E
i∈C
def
ri ←ϕRi
h
i 2
X
Y
i
i
ϕ
ri − µ 1
where Ri = XC∪[i−1] YC∪[i−1] AC BC .
Summary
Introduction
Proof of the Main Theorem
Some Details
Final Upper Bounds
By Markov’s inequality, there exists a j ∈ C s.t.
Summary
Introduction
Proof of the Main Theorem
Some Details
Final Upper Bounds
By Markov’s inequality, there exists a j ∈ C s.t.
I Xj : BobRj ϕ 6 10δ3
where Rj = XC∪[j−1] YC∪[j−1] AC BC .
Summary
Introduction
Proof of the Main Theorem
Some Details
Final Upper Bounds
By Markov’s inequality, there exists a j ∈ C s.t.
I Xj : BobRj ϕ 6 10δ3
I Yj : AliceRj ϕ 6 10δ3
where Rj = XC∪[j−1] YC∪[j−1] AC BC .
Summary
Introduction
Proof of the Main Theorem
Summary
Some Details
Final Upper Bounds
By Markov’s inequality, there exists a j ∈ C s.t.
XY
ϕ j j − µ 6
1
I Xj : BobRj ϕ 6 10δ3
I Yj : AliceRj ϕ 6 10δ3
i p
h
Xj Yj
E
ϕrj − µ 6 10δ3
Rj
rj ←ϕ
where Rj = XC∪[j−1] YC∪[j−1] AC BC .
1
Introduction
Proof of the Main Theorem
Summary
Some Details
Final Upper Bounds
By Markov’s inequality, there exists a j ∈ C s.t.
XY
ϕ j j − µ 6
1
I Xj : BobRj ϕ 6 10δ3
I Yj : AliceRj ϕ 6 10δ3
i p
h
Xj Yj
E
ϕrj − µ 6 10δ3
Rj
rj ←ϕ
1
where Rj = XC∪[j−1] YC∪[j−1] AC BC . By similar arguments as in the
previous slide, we also have
E
Rj
rj ←ϕ
i p
h
X
Y Xj Yj
ϕrj − ϕrjj ⊗ ϕrjj 6 10δ3
1
Introduction
Proof of the Main Theorem
Summary
Some Details
Final Upper Bounds
By Markov’s inequality, there exists a j ∈ C s.t.
XY
ϕ j j − µ 6
1
I Xj : BobRj ϕ 6 10δ3
I Yj : AliceRj ϕ 6 10δ3
i p
h
Xj Yj
E
ϕrj − µ 6 10δ3
Rj
rj ←ϕ
1
where Rj = XC∪[j−1] YC∪[j−1] AC BC . By similar arguments as in the
previous slide, we also have
E
Rj
rj ←ϕ
i p
h
X
Y Xj Yj
ϕrj − ϕrjj ⊗ ϕrjj 6 10δ3
E
1
Xj Yj
xj ,yj ←ϕ
i p
h
Rj
Rj ϕxj ,yj − ϕ 6 10δ3 .
1
Introduction
Proof of the Main Theorem
Summary
Some Details
Final Upper Bounds
By Markov’s inequality, there exists a j ∈ C s.t.
XY
ϕ j j − µ 6
1
I Xj : BobRj ϕ 6 10δ3
I Yj : AliceRj ϕ 6 10δ3
i p
h
Xj Yj
E
ϕrj − µ 6 10δ3
Rj
rj ←ϕ
1
where Rj = XC∪[j−1] YC∪[j−1] AC BC . By similar arguments as in the
previous slide, we also have
E
Rj
rj ←ϕ
i p
h
X
Y Xj Yj
ϕrj − ϕrjj ⊗ ϕrjj 6 10δ3
E
1
Xj Yj
xj ,yj ←ϕ
i p
h
Rj
Rj ϕxj ,yj − ϕ 6 10δ3 .
1
With these, and by treating Rj as public coins, it’s easy to show that
we can embed G into Gk .
Introduction
Proof of the Main Theorem
Summary
Summary
We proved the following parallel repetition theorem.
Theorem (Main Theorem)
For any game G, where the input distribution µ is product on X × Y, it
holds that
∗
k
ω G
∗
= 1 − (1 − ω (G))
3
Ω
k
log(|A|·|B|)
.
Introduction
Proof of the Main Theorem
Summary
Summary
We proved the following parallel repetition theorem.
Theorem (Main Theorem)
For any game G, where the input distribution µ is product on X × Y, it
holds that
∗
k
ω G
∗
= 1 − (1 − ω (G))
3
Ω
k
log(|A|·|B|)
.
It improves upon the result of [Chailloux and Scarpa ’13] by
generalizing it from uniform to product distributions
Introduction
Proof of the Main Theorem
Summary
Summary
We proved the following parallel repetition theorem.
Theorem (Main Theorem)
For any game G, where the input distribution µ is product on X × Y, it
holds that
∗
k
ω G
∗
= 1 − (1 − ω (G))
3
Ω
k
log(|A|·|B|)
.
It improves upon the result of [Chailloux and Scarpa ’13] by
generalizing it from uniform to product distributions and by
removing the dependence on |X| and |Y|.
Introduction
Proof of the Main Theorem
Summary
Summary
We proved the following parallel repetition theorem.
Theorem (Main Theorem)
For any game G, where the input distribution µ is product on X × Y, it
holds that
∗
k
ω G
∗
= 1 − (1 − ω (G))
3
Ω
k
log(|A|·|B|)
.
It improves upon the result of [Chailloux and Scarpa ’13] by
generalizing it from uniform to product distributions and by
removing the dependence on |X| and |Y|.
A parallel repetition theorem for arbitrary games where the exponent
only depends on k and |A| · |B| is still unknown.
Introduction
Proof of the Main Theorem
Thank you for your attention!
The manuscript is available at arXiv:1311.6309.
Summary