# Quantum Lovasz local lemma Andris Ambainis (Latvia), Julia Kempe (Tel Aviv), Or Sattath (Hebrew U.) arXiv:0911.1696.

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Quantum Lovasz local lemma Andris Ambainis (Latvia), Julia Kempe (Tel Aviv), Or Sattath (Hebrew U.) arXiv:0911.1696

Part 1 (classical) Lovasz local lemma

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?

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.

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.

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.

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

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

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.

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.

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.

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).

Quantum Lovasz lemma

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.

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?

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

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.

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 ).

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.

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.

Proof of quantum LLL

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.

Main lemma Corollary: Then for any other H i.

Application: quantum k-SAT

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 1 +... + H m ; Each H i involves k qubits; Each H i – projector to 1 of 2 k dimensions. Goal: H = 0. Goal: H = 0.

Theorem Assume that Assume that H = H 1 +... + H m, etc.H = H 1 +... + 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.

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

Random k-SAT and k- QSAT

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.

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. 3.52 < c 3 < 4.49. 3.52 < c 3 < 4.49. Large k: Large k: 2 k ln 2 – O(k) c k 2 k ln 2.

Random k-QSAT [Bravyi, 06] H = H 1 +...+H m. H = H 1 +...+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.

Results on quantum k-SAT [Laumann et al., 09] q 2 =1/2. q 2 =1/2. For large k, 1- < q k < 0.5742 k. For large k, 1- < q k < 0.5742 k. 2 k = 0.69 2 k. Classically, c k < ln 2 2 k = 0.69 2 k. Huge gap between upper and lower bounds.

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.

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.

[Laumann, et al., 09] H = H 1 +...+H m. H = H 1 +...+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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