Two-prover entangled games are NP hard Anand Natarajan (MIT) - PowerPoint PPT Presentation

Two-prover entangled games are NP hard Anand Natarajan (MIT) Thomas Vidick (Caltech) arXiv:1710.03062

Interactive proofs MIP IP Completeness: If YES, prover can convince verifier with Pr ≥ c (e.g. 2/3) Soundness : If NO, prover cannot convince verifier with Pr ≥ s (e.g. 1/3) = NEXP = PSPACE [BFL ‘91] [Shamir ‘90]

Interactive proofs IP MIP (log n bit messages) Completeness: If YES, prover can convince verifier with Pr ≥ c (e.g. 2/3) Soundness : If NO, prover cannot convince verifier with Pr ≥ s (e.g. 1/3) = PSPACE = NP [Shamir ‘90] [ALMSS’98,AS’98]

Quantum interactive proofs What about a quantum verifier, or quantum messages? MIP* QIP: (c. verifier, c. messages, q. provers) (q. verifier, q. messages) | Ψ ⟩ How powerful is it? Still = PSPACE ! [JJUW’09]

Entangled provers • Entanglement makes provers more powerful • [Bell64, CHSH68]: T wo-prover, one round protocol – Best classical p success = 0.75 – With entanglement (1 EPR pair) p success ≈ 0.85 • This can hurt soundness! • [CHTW04]: ⊕ MIP*(2) ⊆ PSPACE – ⊕ MIP(2) = MIP = NEXP [Håstad’01] – Based on SDP characterization of quantum strategies due to Tsirelson

Quantum interactive proofs What about a quantum verifier, or quantum messages? MIP* QIP: (c. verifier, c. messages, q. provers) (q. verifier, q. messages) | Ψ ⟩ Contains NEXP [IV’12] Still = PSPACE ! ⊕ MIP*(3) contains NEXP [JJUW’09] [Vidick’13]

Two vs three provers • Classically: reduce any MIP protocol to 2 provers using oracularization • Quantumly: no general prover reduction known – Monogamy of entanglement: if Alice is maximally entangled with Bob, must share zero entanglement with Charlie! – [T oner’09] shows 3-prover versions of CHSH, odd-cycle game have lower entangled p success

Results What about a quantum verifier, or quantum messages? MIP* QIP: (c. verifier, c. messages, q. provers) (q. verifier, q. messages) | Ψ ⟩ This work: Contains NEXP • Still = PSPACE ! • For log(n)-bit [JJUW’09] messages, contains NP

Low-degree test • Encode an NP-proof as a low-degree m → ! q polynomial g: ! q x – q m = Θ (n) m be random point, • Let x ∈ ! q m random plane containing x s ⊆ ! q – Ask Alice for g(x) – Ask Bob for g| s as polynomial – Check that g| s (x) = g(x) • Thm [RS’97]: If pass with prob ≥ 1- ε , agree with some low-degree g on s ≥ 1 –O( ε c ) fraction of points

Soundness against entangled provers • Entangled strategy: x – Shared state | Ψ ⟩ – Points measurements: a ≥ 0, Σ a A x a = Id A x – Planes measurements: h ≥ 0, Σ h B s h = Id B s • Thm: If pass with prob ≥ 1- ε , exists measurement {M g } s.t. s ⟨ Ψ | M g ⊗ A x g(x) | Ψ ⟩ ≥ 1- O( ε c )

Induction m-1 with • Hypothesis: If pass test on ! q prob ≥ 1- ε , exists measurement {M g } with m-1 → ! q s.t. ⟨ Ψ | M g ⊗ A x g: ! q g(x) | Ψ ⟩ ≥ 1- δ m with prob ≥ 1- ε , • Goal: If pass test on ! q exists measurement M g with g: ! q m → ! q s.t. ⟨ Ψ | M g ⊗ A x g(x) | Ψ ⟩ ≥ 1- δ

Induction m-1 with prob ≥ • Hypothesis: If pass test on ! q m-1 → 1- ε , exists measurement {M g } with g: ! q ! q s.t. ⟨ Ψ | M g ⊗ A x g(x) | Ψ ⟩ ≥ 1- δ • Implies: for most dim-(m-1) subspaces s, exists {M s g } that is δ ’- consistent with {A x a } m with prob ≥ 1- ε , • Goal: If pass test on ! q exists measurement M g with g: ! q m → ! q s.t. ⟨ Ψ | M g ⊗ A x g(x) | Ψ ⟩ ≥ 1- δ ’’ • Problem: δ ’’ > δ ’ > δ !

Induction with self- improvement m-1 with • Hypothesis: If pass test on ! q prob ≥ 1- ε , exists measurement {M g } with m-1 → ! q s.t. ⟨ Ψ | M g ⊗ A x g: ! q g(x) | Ψ ⟩ ≥ 1- δ • Implies: for most dim-(m-1) subspaces s ⊆ ! q m , exists {M s g } that is δ -consistent with {A x a } m with prob ≥ 1- ε , • Goal: If pass test on ! q exists measurement M g with g: ! q m → ! q s.t. ⟨ Ψ | M g ⊗ A x g(x) | Ψ ⟩ ≥ 1- δ

Self-improvement {A x a } • Given: – {A x a } ε -”globally consistent” {A y a } x – {Q g } η -consistent with {A x a } E x Σ g Σ a ≠ g(x) ⟨ Ψ | Q g ⊗ A x a | Ψ ⟩ ≤ η • Construct improved {S g } y – ε 1/2 -consistent with {A x a } E x Σ g Σ a ≠ g(x) ⟨ Ψ | S g ⊗ A x a | Ψ ⟩ ≤ ε 1/2 – η ’ := 1 – ε 1/2 – η complete {S g } Σ g ⟨ Ψ | S g ⊗ Id | Ψ ⟩ ≥ η ’

Self-improvement: SDP • A g := E x [A x g(x) ] – Note: Σ g A g may not be ≤ Id {A x a } • SDP (“majority vote”) {A y a } x max Σ g Tr[T g A g ] s.t.T g ≥ 0, Σ g T g ≥ Id y • Q g is primal feasible, so SDP value ≥ 1 - η {S g } • Define S g := E x A x g(x) T g A x g(x)

Self-improvement: consistency • S g := E x A x g(x) T g A x g(x) • E x Σ g Σ a ≠ g(x) ⟨ Ψ | S g ⊗ A x a | Ψ ⟩ g(y) ⊗ A x g(y) T g A y = E x E y Σ g Σ a ≠ g(x) ⟨ Ψ | A y a | Ψ ⟩ ≤ O( ε 1/2 )

Self-improvement: completeness • A g := E x [A x g(x) ] S g = E x A x g(x) T g A x Σ g ⟨ Ψ | S g ⊗ Id| Ψ ⟩ g(x) ≈ Σ g ⟨ Ψ | T g A g ⊗ Id | Ψ ⟩ • Dual SDP = Σ g ⟨ Ψ | T g Z ⊗ Id | Ψ ⟩ min Tr[Z] = ⟨ Ψ | Z ⊗ Id | Ψ ⟩ s.t. Z ≥ A g , Z ≥ 0 ≥ 1- η ’ • Complementary slackness: T g A g = T g Z

Applications and future work • Superclassical hardness results – In [NV’18] improved to QMA-hardness – [Ji’17,FJVY’18] for small soundness • Entanglement-sound tests for locally testable codes? • Non-signaling soundness? • Prover reduction for MIP*?

arXiv:1710.03062 THE END