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Quantum Lovasz local lemma Andris Ambainis (Latvia), Julia Kempe (Tel Aviv), Or Sattath (Hebrew U.) arXiv:

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Part 1 (classical) Lovasz local lemma

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The setting Bad events A 1,..., A m. Bad events A 1,..., A m. Pr [A i ]. Pr [A i ]. When can we say that When can we say that Pr [none of A i ] > 0?

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Obvious results A i independent: A i independent: Pr [not A i ] 1-.Pr [not A i ] 1-. Pr [none of A i ] (1- ) m >0.Pr [none of A i ] (1- ) m >0. No assumptions about A i : No assumptions about A i : Pr [A i ].Pr [A i ]. Pr [some A i occurs] m.Pr [some A i occurs] m. If m 0.If m 0.

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Limited independence Each A i is independent of all but at most d other events A j. Each A i is independent of all but at most d other events A j. [Erdös, Lovász, 1975] If Pr[A i ] and e(d+1) < 1, then [Erdös, Lovász, 1975] If Pr[A i ] and e(d+1) < 1, then Pr [none of A i ] > 0. < 1 enough; Full independence: m < 1 enough; Limited independence: e ( d +1) < 1. Limited independence: e ( d +1) < 1.

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Application 1: k-SAT k-SAT formula F, k-SAT formula F, F = F 1 F 2... F m ;F = F 1 F 2... F m ; F i = y i,1 y i,2... y i,k ;F i = y i,1 y i,2... y i,k ; y i,l = x j or x j.y i,l = x j or x j. Theorem If each F i has common variables with at most 2 k /e-1 other clauses F j, thenx 1,..., x n : F(x 1,..., x n )=TRUE. Theorem If each F i has common variables with at most 2 k /e-1 other clauses F j, thenx 1,..., x n : F(x 1,..., x n )=TRUE.

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Proof Pick x i at random: Pick x i at random: F i = x 1 x 2... x k

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Proof Bad events - Bad events - F i and F j independent = F i and F j have no common variables. F i and F j independent = F i and F j have no common variables. Each F i has common variables with at most m=2 k /e-1 other F j. Each F i has common variables with at most m=2 k /e-1 other F j. Lovasz local lemma applies

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Application 2: Ramsey graphs Complete graph K n. Complete graph K n. Colour edges in two colours so that no K k has all edges in one colour. Colour edges in two colours so that no K k has all edges in one colour.

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Solution Colour edges randomly. Colour edges randomly. Events A i – a fixed k-vertex subgraph does not have all edges in the same colour. Events A i – a fixed k-vertex subgraph does not have all edges in the same colour. Independent for subgraphs with no common edges. Independent for subgraphs with no common edges.

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Result Theorem If, then edges of Theorem If, then edges of K m can be coloured with two colours so that K m can be coloured with two colours so that no there is no k vertices with all edges among them in one colour.

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Other applications Coverings of R 3 by unit balls; Coverings of R 3 by unit balls; Linear arboricity (partitioning edges of a graph into linear forests). Linear arboricity (partitioning edges of a graph into linear forests).

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Quantum Lovasz lemma

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Events subspaces Finite-dimensional Hilbert space H. Finite-dimensional Hilbert space H. Events A i bad subspaces S i. Events A i bad subspaces S i. Event does not occur a state | is orthogonal to S i. Event does not occur a state | is orthogonal to S i. Goal: a state |, | S i for all i. Goal: a state |, | S i for all i.

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Hamiltonian version Hamiltonian H = i P i. Hamiltonian H = i P i. Terms H i subspaces S i. Terms H i subspaces S i. H i |=0 | S i. H i |=0 | S i. Is there a state | with H | = 0? Is there a state | with H | = 0?

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Probability dimension Relative dimension Relative dimension Pr [A i ] d(H i ). Pr [A i ] d(H i ). When are two subspaces independent?

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Independence: definition #1 Bipartite system H AH B. Bipartite system H AH B. Subspaces H 1H B and H AH 2 are independent. Subspaces H 1H B and H AH 2 are independent.

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Definition #2 Classically, A 1, A 2 independent if Classically, A 1, A 2 independent if Pr[A 1 A 2 ] = Pr[A 1 ] Pr[A 2 ]. Quantumly, H 1, H 2 – independent if Quantumly, H 1, H 2 – independent if d(H 1 H 2 ) = d(H 1 ) d(H 2 ); d(H 1 H 2 ) = d(H 1 ) d(H 2 ).

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More than 2 subspaces H is independent of H 1,..., H m if H is independent of any combination (union, intersection, complement) of H 1,..., H m. H is independent of H 1,..., H m if H is independent of any combination (union, intersection, complement) of H 1,..., H m.

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Quantum LLL Theorem Let H 1,..., H m be subspaces with: Theorem Let H 1,..., H m be subspaces with: d(H i ) ; d(H i ) ; Each H i independent of all but at most d other H j.Each H i independent of all but at most d other H j. e(d+1) <1.e(d+1) <1. Then, there is |, | H i for all i.

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Proof of quantum LLL

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Our goal Need to show: there exists |, | H i for all i. Need to show: there exists |, | H i for all i. Equivalently, | H i for all i. Equivalently, | H i for all i.

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Main lemma Corollary: Then for any other H i.

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Application: quantum k-SAT

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Quantum SAT k-SAT: k-SAT: variables x 1,..., x N.variables x 1,..., x N. F = F 1... F m ;F = F 1... F m ; F i = y i,1... y i,k ;F i = y i,1... y i,k ; y i,l = x j or x j.y i,l = x j or x j. Goal: F = true. Goal: F = true. k-QSAT k-QSAT N qubits; H = H H m ; Each H i involves k qubits; Each H i – projector to 1 of 2 k dimensions. Goal: H = 0. Goal: H = 0.

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Theorem Assume that Assume that H = H H m, etc.H = H H m, etc. each H i has common qubits with at most d = 2 k /e-1 other H j.each H i has common qubits with at most d = 2 k /e-1 other H j. Then there exists :H = 0. Then there exists :H = 0.

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Proof Each H i is a projector on S i : Each H i is a projector on S i : QLLL: QLLL:

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Random k-SAT and k- QSAT

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Random k-SAT F = F 1... F m ; F = F 1... F m ; Each F i – random k-clause. Each F i – random k-clause. What should m be so that F is satisfiable w.h.p.? What should m be so that F is satisfiable w.h.p.? Ratio m/n.

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Random k-SAT Threshold c k, for large n: Threshold c k, for large n: If m<(c k -) n, then FSAT w.h.p.If m<(c k -) n, then FSAT w.h.p. If m>(c k +) n, then FSAT w.h.p.If m>(c k +) n, then FSAT w.h.p < c 3 < < c 3 < Large k: Large k: 2 k ln 2 – O(k) c k 2 k ln 2.

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Random k-QSAT [Bravyi, 06] H = H H m. H = H H m. Each H i – random projector to 1 of 2 k dimensions for random k qubits. Each H i – random projector to 1 of 2 k dimensions for random k qubits. Do we have :H = 0? Do we have :H = 0? Ratio q k =m/n.

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Results on quantum k-SAT [Laumann et al., 09] q 2 =1/2. q 2 =1/2. For large k, 1- < q k < k. For large k, 1- < q k < k. 2 k = k. Classically, c k < ln 2 2 k = k. Huge gap between upper and lower bounds.

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Our result Theorem Theorem Since each H i involves k qubits, this corresponds to each H i having common qubits withother H j, on average. Since each H i involves k qubits, this corresponds to each H i having common qubits withother H j, on average. QLLL:, worst case.

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Solution Divide qubits into two sets: Divide qubits into two sets: high-degree: includes all qubits that are contained in many H j and those that are in H i with such qubits.high-degree: includes all qubits that are contained in many H j and those that are in H i with such qubits. low-degree.low-degree. Use QLLL on low-degree set, another approach on high-degree set. Use QLLL on low-degree set, another approach on high-degree set. Combine the two solutions. Combine the two solutions.

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[Laumann, et al., 09] H = H H m. H = H H m. Theorem If f:{1,..., m} qubits: Theorem If f:{1,..., m} qubits: f(i) – qubit that is involved in H i ;f(i) – qubit that is involved in H i ; f(i) f(j),f(i) f(j), there exists :H = 0. there exists :H = 0.

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