Squashed Entanglement Nicholas LeCompte based on Squashed - - PowerPoint PPT Presentation

Entanglement measures

Squashed Entanglement

Nicholas LeCompte based on ”Squashed Entanglement” - An Additive Entanglement Measure (M. Christandl, A. Winter, quant-ph/0308088), and A paradigm for entanglement theory based on quantum communication (J. Oppenheim arXiv:0801.0458)

Nicholas LeCompte Squashed entanglement

Entanglement measures

(Classical) intrinsic information

◮ The intrinsic information between two random variables X, Y ,

given a third r.v. Z, is given by I(X; Y ↓ Z) := inf

¯ Z (I(X; Y |¯

Z)) with ¯ Z governed by a conditional probability distribution P¯

Z|Z. ◮ Idea: two parties have access to X and Y , and want to

generate a secret key even if an adversary has access to some additional knowledge (Z).

Nicholas LeCompte Squashed entanglement

Entanglement measures

(Classical) intrinsic information

◮ The intrinsic information between two random variables X, Y ,

given a third r.v. Z, is given by I(X; Y ↓ Z) := inf

¯ Z (I(X; Y |¯

Z)) with ¯ Z governed by a conditional probability distribution P¯

Z|Z. ◮ Idea: two parties have access to X and Y , and want to

generate a secret key even if an adversary has access to some additional knowledge (Z).

◮ Mauer, Wolf 1999: The intrinsic information upper-bounds

the rate at which the key can be extracted.

Nicholas LeCompte Squashed entanglement

Entanglement measures

A quantum analogue (Christandl, Winter, quant-ph/0308088)

◮ Replace classical mutual conditional infos with quantum, P¯ Z|Z

with quantum channel: Esq(ρAB) = inf

Λ:B(HC )→B(HE )

1 2I(A; B|E) : ρABE = (id ⊗ Λ)|ΨΨ|ABC

• ◮ Here, |ΨABC is a purification of ρAB, and ρABE is simply a

tripartite extension; ρAB = TrE(ρABE).

Nicholas LeCompte Squashed entanglement

Entanglement measures

A quantum analogue (Christandl, Winter, quant-ph/0308088)

◮ Replace classical mutual conditional infos with quantum, P¯ Z|Z

with quantum channel: Esq(ρAB) = inf

Λ:B(HC )→B(HE )

1 2I(A; B|E) : ρABE = (id ⊗ Λ)|ΨΨ|ABC

• ◮ Here, |ΨABC is a purification of ρAB, and ρABE is simply a

tripartite extension; ρAB = TrE(ρABE).

◮ Esq(ρAB) is the squashed entanglement.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Squashed entanglement

◮ Equivalently,

Esq(ρAB) = inf

ρABE

1 2I(A; B|E) : TrE(ρABE) = ρAB

• ◮ (equivalence follows by the equivalence of purifications up to a

unitary map)

Nicholas LeCompte Squashed entanglement

Entanglement measures

Squashed entanglement

◮ Equivalently,

Esq(ρAB) = inf

ρABE

1 2I(A; B|E) : TrE(ρABE) = ρAB

• ◮ (equivalence follows by the equivalence of purifications up to a

unitary map)

◮ Sanity check: If ρAB = |ψψ|AB, all tripartite extensions are

given by ρAB ⊗ ρE, so Esq(ρAB) = S(ρA) = E(|ψ).

Nicholas LeCompte Squashed entanglement

Entanglement measures

Some basic properties

◮ Esq has basic properties we want in an entanglement measure,

e.g.:

Nicholas LeCompte Squashed entanglement

Entanglement measures

Some basic properties

◮ Esq has basic properties we want in an entanglement measure,

e.g.:

◮ does not increase under LOCC ◮ is convex ◮ is continuous (by Alicki-Fannes theorem, quant-ph/0312081

[or Lecture 10])

Nicholas LeCompte Squashed entanglement

Entanglement measures

Some basic properties

◮ Esq has basic properties we want in an entanglement measure,

e.g.:

◮ does not increase under LOCC ◮ is convex ◮ is continuous (by Alicki-Fannes theorem, quant-ph/0312081

[or Lecture 10])

◮ It’s also superadditive, and additive on tensor products.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Recall that the intrinsic information upper-bounds the rate at

which the key can be extracted given an adversary has access to some information.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Recall that the intrinsic information upper-bounds the rate at

which the key can be extracted given an adversary has access to some information.

◮ Prop. ED(ρAB) ≤ Esq(ρAB), where ED is the distillable

entanglement.

◮ (Recall ED(ρ) is the limiting ratio n/m of n Bell pairs created

from m copies of ρ with LOCC.)

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Recall that the intrinsic information upper-bounds the rate at

which the key can be extracted given an adversary has access to some information.

◮ Prop. ED(ρAB) ≤ Esq(ρAB), where ED is the distillable

entanglement.

◮ (Recall ED(ρ) is the limiting ratio n/m of n Bell pairs created

from m copies of ρ with LOCC.)

◮ The squashed entanglement is an analogous upper-bound

(viewing maximally entangled states as secret quantum correlations).

Nicholas LeCompte Squashed entanglement

Entanglement measures

Proof sketch: ED(ρAB) ≤ Esq(ρAB)

◮ Start with an arbitrary distillation protocol, taking

(ρAB)⊗n →LOCC ΨAB such that ||ΨAB − |kk|AB||1 ≤ ε, where |k is a rank-k maximally entangled state.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Proof sketch: ED(ρAB) ≤ Esq(ρAB)

◮ Start with an arbitrary distillation protocol, taking

(ρAB)⊗n →LOCC ΨAB such that ||ΨAB − |kk|AB||1 ≤ ε, where |k is a rank-k maximally entangled state.

◮ By superadditivity and monotonicity under LOCC,

nEsq(ρAB) ≥ Esq(ΨAB).

Nicholas LeCompte Squashed entanglement

Entanglement measures

Proof sketch: ED(ρAB) ≤ Esq(ρAB)

◮ Start with an arbitrary distillation protocol, taking

(ρAB)⊗n →LOCC ΨAB such that ||ΨAB − |kk|AB||1 ≤ ε, where |k is a rank-k maximally entangled state.

◮ By superadditivity and monotonicity under LOCC,

nEsq(ρAB) ≥ Esq(ΨAB).

◮ Note Esq(|kk|) = log k (since |k is maximally entangled),

and using Fannes’ inequality, one can show that for any extension ΨABE, 1

2I(A; B|E) ≥ log k − f (ǫ) log(k) for some

function f with limǫ→0 f (ǫ) = 0.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Proof sketch: ED(ρAB) ≤ Esq(ρAB)

◮ Start with an arbitrary distillation protocol, taking

(ρAB)⊗n →LOCC ΨAB such that ||ΨAB − |kk|AB||1 ≤ ε, where |k is a rank-k maximally entangled state.

◮ By superadditivity and monotonicity under LOCC,

nEsq(ρAB) ≥ Esq(ΨAB).

◮ Note Esq(|kk|) = log k (since |k is maximally entangled),

and using Fannes’ inequality, one can show that for any extension ΨABE, 1

2I(A; B|E) ≥ log k − f (ǫ) log(k) for some

function f with limǫ→0 f (ǫ) = 0.

◮ Hence Esq(ρAB) ≥ 1 n(1 − f (ǫ)) log k.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Recall that the entanglement of formation is given by

EF(ρAB) = inf

ρ=

i pi|ψiψi|

• i

piE(|ψi) = inf 1 2

• i

piI(|ψiA; |ψiB)

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Recall that the entanglement of formation is given by

EF(ρAB) = inf

ρ=

i pi|ψiψi|

• i

piE(|ψi) = inf 1 2

• i

piI(|ψiA; |ψiB)

◮ Consider the following extension

ρABE =

• i

pi|ψiψi|AB ⊗ |ii|E

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Recall that the entanglement of formation is given by

EF(ρAB) = inf

ρ=

i pi|ψiψi|

• i

piE(|ψi) = inf 1 2

• i

piI(|ψiA; |ψiB)

◮ Consider the following extension

ρABE =

• i

pi|ψiψi|AB ⊗ |ii|E

◮ Then

1 2

• i

piI(|ψiA; |ψiB) = 1 2I(A; B|E).

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Hence the entanglement of formation can be thought of as

the infinum of I(A; B|E) over a certain type of tripartite extensions.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Hence the entanglement of formation can be thought of as

the infinum of I(A; B|E) over a certain type of tripartite extensions.

◮ It follows that Esq(ρAB) ≤ EF(ρAB).

Nicholas LeCompte Squashed entanglement

Entanglement measures

Relations to other entanglement measures

◮ Hence the entanglement of formation can be thought of as

the infinum of I(A; B|E) over a certain type of tripartite extensions.

◮ It follows that Esq(ρAB) ≤ EF(ρAB). ◮ Invoking additivity, the entanglement cost

EC(ρAB) = lim

n→∞

1 nEF((ρAB)⊗n) is also bounded below by Esq(ρAB).

Nicholas LeCompte Squashed entanglement

Entanglement measures

Another interpretation of squashed entanglement (Oppenheim, arXiv:0801.0458)

◮ Suppose Alice and Bob share a state ρAB, and Alice wants to

send her part to Eve. How many qubits does she need?

◮ We allow Alice and Eve to have as much side information

(e.g., ancillas) as they like.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Another interpretation of squashed entanglement (Oppenheim, arXiv:0801.0458)

◮ Suppose Alice and Bob share a state ρAB, and Alice wants to

send her part to Eve. How many qubits does she need?

◮ We allow Alice and Eve to have as much side information

(e.g., ancillas) as they like.

◮ However, the amount of useful information is limited, since

Alice’s share of the state may be entangled with Bob’s.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Another interpretation of squashed entanglement (Oppenheim, arXiv:0801.0458)

◮ Suppose Alice and Bob share a state ρAB, and Alice wants to

send her part to Eve. How many qubits does she need?

◮ We allow Alice and Eve to have as much side information

(e.g., ancillas) as they like.

◮ However, the amount of useful information is limited, since

Alice’s share of the state may be entangled with Bob’s.

◮ Let QA′E(ρAB) be the rate of communication required to send

ρA to a receiver who holds ρC, with ρA′ held by the sender.

Nicholas LeCompte Squashed entanglement

Entanglement measures

State redistribution and merging (Oppenheim, arXiv:0805.1065, and Lecture 16)

◮ Using state merging, the minimum amount of quantum

communication required is QA′C = 1 2I(A : B|C) E (entanglement) = 1 2I(A : A′) − 1 2I(A : C)

Nicholas LeCompte Squashed entanglement

Entanglement measures

State redistribution and merging (Oppenheim, arXiv:0805.1065, and Lecture 16)

◮ Using state merging, the minimum amount of quantum

communication required is QA′C = 1 2I(A : B|C) E (entanglement) = 1 2I(A : A′) − 1 2I(A : C)

◮ To clear up some confusion: in terms of the merging mother

protocol, Bob is the reference here.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Back to squashed entanglement

◮ Allowing arbitrary side information, the entanglement E is

certainly attainable.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Back to squashed entanglement

◮ Allowing arbitrary side information, the entanglement E is

certainly attainable.

◮ So, let ρS be the side information shared between Alice and

Eve (so that ρABS is pure), with Alice’s share being ρA′ and Eve’s ρC.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Back to squashed entanglement

◮ Allowing arbitrary side information, the entanglement E is

certainly attainable.

◮ So, let ρS be the side information shared between Alice and

Eve (so that ρABS is pure), with Alice’s share being ρA′ and Eve’s ρC.

◮ Then

inf

ρA′C

QA′C = inf

ρA′C

1 2I(A : B|C) = inf

Λ:B(S)→B(C)

1 2I(A : B|Λ(S)) = Esq(ρAB)

Nicholas LeCompte Squashed entanglement

Entanglement measures

Back to squashed entanglement

◮ Allowing arbitrary side information, the entanglement E is

certainly attainable.

◮ So, let ρS be the side information shared between Alice and

Eve (so that ρABS is pure), with Alice’s share being ρA′ and Eve’s ρC.

◮ Then

inf

ρA′C

QA′C = inf

ρA′C

1 2I(A : B|C) = inf

Λ:B(S)→B(C)

1 2I(A : B|Λ(S)) = Esq(ρAB)

◮ So the squashed entanglement is the fastest rate Alice can

send a state to Eve given arbitrary side information.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Sanity check

◮ If ρAB = |ψψ|, Esq(ρ(AB)) = E(|ψ). ◮ Likewise, any pure state extension ρABS must have S

completely uncorrelated with AB.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Sanity check

◮ If ρAB = |ψψ|, Esq(ρ(AB)) = E(|ψ). ◮ Likewise, any pure state extension ρABS must have S

completely uncorrelated with AB.

◮ If Alice and Bob share a separable state ρAB, written as a

convex combination of separable pure states, Esq(ρAB) = 0:

Nicholas LeCompte Squashed entanglement

Entanglement measures

Sanity check

◮ If ρAB = |ψψ|, Esq(ρ(AB)) = E(|ψ). ◮ Likewise, any pure state extension ρABS must have S

completely uncorrelated with AB.

◮ If Alice and Bob share a separable state ρAB, written as a

convex combination of separable pure states, Esq(ρAB) = 0:

◮ The extension

ρABE =

• i

pi|ψiψi|A ⊗ |φiφi|B ⊗ |ii|E has zero quantum conditional mutual information.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Sanity check

◮ If ρAB = |ψψ|, Esq(ρ(AB)) = E(|ψ). ◮ Likewise, any pure state extension ρABS must have S

completely uncorrelated with AB.

◮ If Alice and Bob share a separable state ρAB, written as a

convex combination of separable pure states, Esq(ρAB) = 0:

◮ The extension

ρABE =

• i

pi|ψiψi|A ⊗ |φiφi|B ⊗ |ii|E has zero quantum conditional mutual information.

◮ Of course, the rate also vanishes - teleportation does the trick.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Why “squashed” entanglement?

Nicholas LeCompte Squashed entanglement

Entanglement measures

Why “squashed” entanglement?

◮ Christandl and Winter:

“Our name for this functional comes from the idea that the right choice of a conditioning system reduces the quantum mutual information between A and B, thus “squashing out” the nonquantum correlations.”

Nicholas LeCompte Squashed entanglement

Entanglement measures

Why “squashed” entanglement?

◮ Christandl and Winter:

“Our name for this functional comes from the idea that the right choice of a conditioning system reduces the quantum mutual information between A and B, thus “squashing out” the nonquantum correlations.”

◮ Oppenheim defines the “puffed entanglement” as the

supremum of I(A : B|E) over all extensions E, with the

• perational interpretation of the rate required given the side

information is as “useless” as possible.

Nicholas LeCompte Squashed entanglement

Entanglement measures

Thank you!

Questions?

Nicholas LeCompte Squashed entanglement