[PPT] - Polynomial optimization on the sphere and quantum entanglement PowerPoint Presentation

SLIDE 1

Kun Fang

Presented at ICCOPT 2019, Berlin

Polynomial optimization on the sphere and quantum entanglement testing

Joint work with Hamza Fawzi

SLIDE 2

◎Polynomial Optimization and SOS Hierarchy ◎An improved Convergence Rate

Main Result and Proof Strategy

◎Relation to Entanglement Testing

SOS Hierarchy (polynomial) v.s. DPS Hierarchy (quantum)

◎Summary and Discussions

Talk Outline

SLIDE 3

Polynomial Optimization

n the Sphere

SLIDE 4

Polynomial Optimization on the Sphere

Applications:

the largest stable/independent set of a graph
Degree 3 polynomial opt. on the sphere (e.g. [Nesterov’03, De Klerk’08])
2 → 4 norm of a matrix A,
Degree 4 polynomial opt. on the sphere (e.g. [Barak et al.’12])
Best Separable State problem in quantum information theory
Degree 4 polynomial opt. on the product of spheres (e.g. [Barak-Kothari-Steurer’17])
…

p(x) = kAxk4

4

Given a multivariate polynomial with

pmax = max

x∈Sd−1 p(x)

p(x)

x = (x1, · · · , xd)

Q

Sd−1 =

x ∈ Rd : x2

1 + · · · + x2 d = 1

Computing the maximal value Over the unit sphere

SLIDE 5

Polynomial Optimization on the Sphere

Difficulty:

Degree = 2, efficiently solved as an eigenvalue problem;
Degree > 2, NP-hard in general!
Sum-of-square (SOS) hierarchy [Parrilo’00; Lasserre’01]

where each level is efficiently computable by semidefinite program

Solution:

Q

Given a multivariate polynomial with

pmax = max

x∈Sd−1 p(x)

p(x)

x = (x1, · · · , xd)

Sd−1 =

x ∈ Rd : x2

1 + · · · + x2 d = 1

Computing the maximal value Over the unit sphere

SLIDE 6

Sum-of-Square (SOS) Hierarchy

pmax = max

x∈Sd−1 p(x)

[ SDP of size ] dO(`)

p1

p`+1

pmax

R

p`

Relation with polynomial optimization:

= p`

Approaching from above:

pmax

NN 1-SOS 2-SOS … `-SOS

…

restriction

≤ min

∈ R : − p ∈ `-SOS on Sd−1

`-SOS = n p(x) = X

i

qi(x)2 on Sd−1

s.t. deg(qi) ≤ `

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NN =

p(x) : p(x) ≥ 0, ∀x ∈ Sd−1

polynomial polynomial

= min{γ ∈ R : γ − p ∈ NN on Sd−1}

SLIDE 7

A: Positive answer in this work, convergence rate at least Q: How fast does converge to ?

Main Result: improved Convergence Rate

pmax

A: [Reznick’95; Doherty-Wehner’12], convergence rate at least O(d/`)

Q: Can we further sharpen the convergence rate? (see recent works by de Klerk & Laurent 1811.05439 & 1904.08828)

pmin = min

x∈Sd−1 p(x)

` ≥ Cnd

1 ≤ p` − pmin pmax − pmin ≤ 1 + ✓ Cn · d ` ◆2

for all

p1

p`+1

pmax

R

p`

p`

reference point A constant depends only on n.

Q u a d r a t i c i m p r

v

e m e n t

Suppose is a homo. poly. of degree in variables with ,

p(x1, · · · , xd)

2n

d

n ≤ d

Main Result (technical statement)

O((d/`)2)

SLIDE 8

For any homo. matrix-valued poly.

f degree

in variables with 2n

d

n ≤ d

` ≥ Cnd

for all

0 ≤ F(x) ≤ I

F(x) ∈ Sk[x]

for all , x ∈ Sd−1

and

A Stronger Result [take home message]

Matrix-valued polynomials: : k by k matrix with polynomial entries, symmetric for any . : positive semidefinite matrix for any .

x

F(x) ≥ 0

F(x) ∈ `-SOS

x

F(x) = X

j

Uj(x)Uj(x)T, deg(Uj) ≤ `

if

These definitions reduce to (scalar-valued) polynomial if k = 1;

A small perturbation δ

F + C0

n

✓d ` ◆2 · I ∈ `-SOS on Sd1

A remarkable fact: the result is totally independent on the size of the matrix F(x).

F(x) ∈ Sk[x]

x = (x1, · · · , xd)

Identity

Remark:

But the results cannot be trivially extended. (e.g. Nonnegative quadratic

polynomial is necessarily a SOS. But not true for matrix-valued case.)

SLIDE 9

For any homo. matrix-valued poly.

f degree

in variables with 2n

d

n ≤ d

` ≥ Cnd

for all

0 ≤ F(x) ≤ I

F(x) ∈ Sk[x]

for all , x ∈ Sd−1

and

A Stronger Result

A small perturbation δ

F + C0

n

✓d ` ◆2 · I ∈ `-SOS on Sd1

A remarkable fact: the result is totally independent on the size of the matrix F(x).

Identity

Suppose is a homo. poly. of degree in variables with ,

` ≥ Cnd

1 ≤ p` − pmin pmax − pmin ≤ 1 + ✓ Cn · d ` ◆2

for all

p(x1, · · · , xd)

2n

d

n ≤ d

Main Result (technical statement) F = pmax − p pmax − pmin

SLIDE 10

Convergence Rate Proof Outline

SLIDE 11

For any polynomial:

f

, consider

Proof Outline

Key observation: (Kh)(x) = Z

y∈Sd−1 K(x, y)h(y)dσ(y)

∀x ∈ Sd−1

K(x, y) = q(hx, yi)2

q(t) : [−1, 1] → R

h(y) ≥ 0 = ⇒ Kh ∈ `-SOS

e F = K(K−1 e F)

e F ∈ `-SOS

e F = F + δ ≥ δ

kK−1 e F e Fk∞  δ

K−1 e F ≥ 0

= Z

y∈Sd−1 q(hx, yi)2 h(y) dσ(y)

8x 2 Sd−1

deg(q) = `

Goal: Given poly. . Find such that

δ > 0

e F = F + is `-SOS

0 ≤ F ≤ 1

How to construct SOS:

[Reznick’95; Doherty-Wehner’12 ;Parrilo’13] nonnegative weight Poly^2 sum

SLIDE 12

Harmonic Decomposition

Since

, its harmonic components won’t be too large

F = F0 + F2 + · · · F2n

e F = (F0 + δ) + F2 + · · · F2n

φ = q2 = λ0C0 + λ1C1 + · · · λ2`C2`

Kernel Decomposition

K−1 e F = λ−1

0 (F0 + δ) + λ−1 2 F2 + · · · λ−1 2n F2n

Funk-Hecke formula

kK−1( e F) e Fk∞ =

n

X

k=1

✓ 1 λ2k 1 ◆ F2k

∞



n

X

k=1

1

λ2k 1

kF2kk∞

0 ≤ F ≤ 1

kF2kk∞  B2nkFk∞

n

X

k=1

1

λ2k − 1

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` ≥ 2nd

Estimate

n

X

k=1

1

λ2k − 1

≤ 2

n

X

k=1

(1 − λ2k),

kK−1 e F e Fk∞  δ

e F = F + is `-SOS

Estimate kK−1 e

F e Fk∞  δ

K(x, y) = q(hx, yi)2

Rotation invariant kernel

K e F = λ0(F0 + δ) + λ2F2 + · · · λ2nF2n

Some well-studied results

[ Gegenbauer ]

Ck

SLIDE 13

If we choose polynomial

, each coefficient can be computed explicitly. Observe that scales as . Recover results by [Reznick’95; Doherty-Wehner’12].

φ = q2 = λ0C0 + λ1C1 + · · · λ2`C2`

φ(t) = [q(t)]2 = " ` X

i=0

ei Ci(t) p Ci(1) #2

λi = ωd−1 ωd Z 1

−1

φ(t) Ci(t) Ci(1)(1 − t2)

d−3 2 dt

λi

q(t) ∝ t`

O(d/`)

To obtain a better result, we do not choose specific

at this moment. q(t)

e = [e0 e1 · · · e`]T

λ2k = eTT [C2k/C2k(1)]e

Generalized Toeplitz matrix

T [g]i,j = ωd−1 ωd Z 1

−1

Ci(t) p Ci(1) Cj(t) p Cj(1) g(t)(1 − t2)

d−3 2 dt

1 − λmax(T [h]) ≤ δ

n

X

k=1

(1 − λ2k) = n

1 − eTT [h] e
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h = 1 n

n

X

k=1

C2k C2k(1)

kK−1 e F e Fk∞  δ

e F = F + is `-SOS

n

X

k=1

(1 − λ2k) ≤ δ

Estimate

n

X

k=1

(1 − λ2k) ≤ δ

SLIDE 14

For linear polynomial , we have a good understanding of the eigenvalues of .

kK−1 e F e Fk∞  δ

e F = F + is `-SOS

n

X

k=1

(1 − λ2k) ≤ δ

1 − λmax(T [h]) ≤ δ

Estimate 1 − λmax(T [h]) ≤ δ

h = 1 n

n

X

k=1

C2k C2k(1)

f

T [f]

1.0
0.5

0.5 1.0

0.2

0.2 0.4 0.6 0.8 1.0

¯ h(t) = h0(1)(t − 1) + h(1)

h is non-linear -> Consider the tangent line at t = 1,

max(T [h]) ≥ max(T [¯ h]) = ¯ h(x`+1,`+1) ≥ 1 − 7n 12 d2 `2

x`+1,`+1 ≥ 1 − 1 4 d2 `2

The largest root of ,

C`+1

d = 4, n = 2

Remark: The estimation of is tight.

λmax(T [h])

This completes the proof.

h

¯ h

δ

SLIDE 15

Some Remarks

Recall the result Some Remarks:

We can estimate the convergence for all values of level, not just

.

` ≥ Cnd

The proof works for polynomials on the complex sphere.
The proof works for matrix-valued polynomials.

2n

d

n ≤ d

` ≥ Cnd

for all

0 ≤ F(x) ≤ I

F(x) ∈ Sk[x]

for all x ∈ Sd−1

and

F + C0

n

✓d ` ◆2 · I ∈ `-SOS on Sd1

For any homo. matrix-valued poly.

f degree

in variables with

SLIDE 16

Entanglement Testing

SLIDE 17

Quantum States

Quantum state Separable state Entangled state Any quantum state that is not separable

SEP(HA ⊗ HB) = nX

i pi(xix† i) ⊗ (yiy† i ) : pi ≥ 0, xi ∈ HA, yi ∈ HB

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S(HA) = nX

i pixix† i : pi ≥ 0, xi ∈ HA

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Q: Whether a given quantum state is entangled or not? A: Doherty-Parrilo-Spedalieri (DPS) hierarchy

ρAB

1. Reduction under partial trace:

TrB2···B`[ρAB1B2···B`] = ρAB

2. Symmetry on B systems:

(I ⊗ ΠB1···B`)ρAB1···B`(I ⊗ ΠB1···B`) = ρAB1···B`

3. Positive partial transpose (PPT):

(IA ⊗ TB1 ⊗ · · · TBs ⊗ IBi+1 ⊗ IB`)(ρAB1···B`) ≥ 0

DPS`(HA ⊗ HB) = {ρAB : ∃ρAB1···B` s.t. (1, 2, 3) holds}

… …

SEP

DPS1

DPS`

DPS2

Form a complete hierarchy [Doherty-Parrilo-Spedalieri’02&04]

SLIDE 18

Quantum States

Quantum state Separable state Entangled state Any quantum state that is not separable

SEP(HA ⊗ HB) = nX

i pi(xix† i) ⊗ (yiy† i ) : pi ≥ 0, xi ∈ HA, yi ∈ HB

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S(HA) = nX

i pixix† i : pi ≥ 0, xi ∈ HA

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A: Doherty-Parrilo-Spedalieri (DPS) hierarchy

1. Reduction under partial trace:

TrB2···B`[ρAB1B2···B`] = ρAB

2. Symmetry on B systems:

(I ⊗ ΠB1···B`)ρAB1···B`(I ⊗ ΠB1···B`) = ρAB1···B`

3. Positive partial transpose (PPT):

(IA ⊗ TB1 ⊗ · · · TBs ⊗ IBi+1 ⊗ IB`)(ρAB1···B`) ≥ 0

DPS`(HA ⊗ HB) = {ρAB : ∃ρAB1···B` s.t. (1, 2, 3) holds}

… …

SEP

DPS1

DPS`

DPS2

EXT `(HA ⊗ HB) = {ρAB : ∃ρAB1···B` s.t. (1, 2) holds}

This is a weaker hierarchy but it is still complete. Without PPT conditions

Q: Whether a given quantum state is entangled or not?

ρAB

SLIDE 19

Duality Relation

… …

SEP

DPS1

DPS`

DPS2

1-SOS 2-SOS … `-SOS

NN

…

SOS hierarchy (polynomials) DPS hierarchy (quantum states) For any Hermitian operator M on , define its associated Hermitian polynomial

pM(x, ¯ x, y, ¯ y) = (x ⊗ y)†M(x ⊗ y) = X

i,j,k,l

Mij,klxi¯ xkyj ¯ y` ∀x ∈ CdA, y ∈ CdB

Duality relation

SEP∗ = {M ∈ Herm(HA ⊗ HB) : pM is nonnegative}

dual ?

HA ⊗ HB

DPS∗

` =

n M 2 Herm(HA ⌦ HB) : kyk2(`−1)pM is rSOS

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EXT ∗

` =

n M 2 Herm(HA ⌦ HB) : kyk2(`−1)pM is cSOS

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X

i qi(x, ¯

x, y, ¯ y)2

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i |gi(x, y)|2

PPT conditions determine the choice of monomials in the SOS decomposition.

SLIDE 20

Relation with [NOP’09]

[Navascues-Owari-Plenio’09] For any quantum state with reduced state

ρAB ∈ DPS`

ρA = TrB[ρAB]

(1 − t)ρAB + tρA ⊗ IB dB is seperable

t = O ✓d2

B

`2 ◆

with 2n

d

n ≤ d

` ≥ Cnd

for all

0 ≤ F(x) ≤ I

F(x) ∈ Sk[x]

for all x ∈ Sd−1

and For any homo. matrix-valued poly.

f degree

in variables with

F + C0

n

✓d ` ◆2 · I ∈ `-SOS on Sd1

Utilizing the duality between DPS and SOS, this is equivalent to our result of matrix-valued polynomial with degree 2.

A small perturbation A small perturbation

Recall our result in this work

SLIDE 21

Summary & Discussions

SLIDE 22

Summary

An quadratic improvement of convergence rate of the SOS hierarchy
Works for matrix-valued polynomials
Works for complex variables
Works for all values of the level
Exact Duality relation between SOS and DPS hierarchies
Connection with [Navascues-Owari-Plenio’09] from quantum community

p1

p`+1

pmax

R

p`

SOS hierarchy appr. from above

Lasserre hierarchy appr. from below

[de Klerk-Laurent 1904.08828] [Fang-Fawzi-This work]

Question: Further sharpening the convergence rate? New techniques are required.

Empirically much faster

Other related works:

Analysis of the SOS hierarchy from the computer science community

(e.g. [Bhattiprolu et al.’17; Barak-Kothari-Steurer’17])

SLIDE 23

Kun Fang

Polynomial optimization on the sphere and quantum entanglement testing

◎Polynomial Optimization and SOS Hierarchy ◎An improved Convergence Rate

Main Result and Proof Strategy

◎Relation to Entanglement Testing

SOS Hierarchy (polynomial) v.s. DPS Hierarchy (quantum)

◎Summary and Discussions

Talk Outline

Polynomial Optimization

Polynomial Optimization on the Sphere

Q

Polynomial Optimization on the Sphere

Q

Sum-of-Square (SOS) Hierarchy

…

Main Result: improved Convergence Rate

A Stronger Result [take home message]

x

x

A Stronger Result

Convergence Rate Proof Outline

Proof Outline

Estimate kK−1 e

Estimate

Estimate 1 − λmax(T [h]) ≤ δ

h

¯ h

δ

Some Remarks

Entanglement Testing

Quantum States

Q: Whether a given quantum state is entangled or not? A: Doherty-Parrilo-Spedalieri (DPS) hierarchy

… …

Quantum States

A: Doherty-Parrilo-Spedalieri (DPS) hierarchy

… …

Q: Whether a given quantum state is entangled or not?

Duality Relation

… …

…

Relation with [NOP’09]

Summary & Discussions

Summary

Thanks for your attention!

Full paper will be online soon.