A Quantum Local Lemma

1. A Quantum Local Lemma Or Sattath TAU and HUJI Joint work with AndrisAmbainis Julia Kempe

2. Outline • The classical Lovász Local Lemma • Motivation in the quantum case • A quantum local lemma • Application to random QSAT

3. The Lovász Local Lemma If a large number of events are all independent, then there is a positive (small) probability that none of them occurs. I.e.: If each of m events occurs with probability at most p<1 then Pr[no events occur] ≥ (1-p)m >0. But what if the events are “weakly” dependent?

4. Example: sparse k-SAT Given a k-SAT formula  where each of the m clauses shares a variable with at most d other clauses.

5. really stupid Example: sparse k-SAT Given a k-SAT formula  where each of the m clauses shares a variable with at most d other clause. In a random assignment each clause is violated with probability p=2-k. These events are independent. A random assignment satisfies  with probability (1-2-k)m >0.  is satisfiable. any • no

6. Example: sparse k-SAT Given a k-SAT formula  where each of the clauses shares a variable with at most d other clauses. In a random assignment each clause is violated with probability p=2-k. However, these events are not independent. Corollary of LLL [ErdősLovász75]: If then a random assignment satisfies  with probability >0.  is satisfiable.

7. Example: sparse k-SAT Corollary: A k-SAT formula where each variable appears in at most 2k/(ek)clauses is always satisfiable.

8. The Lovász Local Lemma LLL [ErdősLovász75]: Let B1,…,Bn be events with Pr[Bi] ≤ p and s.t. each event is independent of all but d of the others. If then there is a non-zero probability that none of them occur.

9. Outline • The classical Lovász Local Lemma • Motivation in the quantum case • A quantum local lemma • Application to random QSAT

10. Quantum bit A classical bit can be either “0” or “1”. A quantum bit (qubit), can be in either |0 or |1, or a linear combination: A qubit is a vector in a 2 dimensional vector space. |0 and |1 form an orthonormal basis for this vector space.

11. n qubits n classicals bits can be either “00…0”, “00…1”,… or “11…1”. n qubit state can be in either |00…0, |00…1,…, |11…1, or some linear combination: |v=a00…0|00…0+a00…1|00…1+…+a11…1|11…1. n qubit state is a vector in a 2n dimensional vector space. |0…0,…,|1…1 form an orthonormal basis for this vector space.

12. Quantum SAT k-SAT: each clause excludes 1 configuration out of the 2k possible configurations. k-QSAT[Bravyi06]: each quantum clause excludes one dimensional subspace out of 2k dimensions of the involved qubits. Clauses  Rank-1 Projectors Satisfying State Excluded 2n-k dimensional subspace

13. QSAT example QSAT generalizes SAT: is satisfiableiff is satisfiable. The state |0011 is a satisfying state: |0011=0, |0011=0

14. Quantum SAT Formal Def: (k-QSAT) Given a collection of k-local rank-1 projectors on n qubits, Is there a state |s inside the allowed subspace of 0 for i=1..m. Importance: known to be QMA1-Complete (quantum analogue of NP) , for k≥4[Bravyi06].

15. Quantum SAT If each projector acts on a set of mutually disjoint qubits, then |s= |s1…|sm is a satisfying state. But what if each qubit appears in a few projections?

16. A statement we would like If each projector excludes a p-fraction of the space and shares a qubit with at most d other projectors, then the k-QSAT instance is satisfiable as long as Or: Let I be an instance of k-QSAT. If each qubit appears in at most 2k/(ek) projectors, then I is satisfiable.

17. Events and independence Correspondences: Probability space: Vector space V Events: Subspaces X  V Probabilities Pr: relative dimension R Conditional Probability Pr(X|Y): Independence: X, Y are R-independent if R(X|Y)=R(X) (equivalently R(XY)=R(X)R(Y) )

18. Properties of relative dim Properties or R: • 0 ≤R(X)≤1 • XY R(X)≤R(Y) • Chain rule: • “Inclusion/Exclusion”: Let X+Y={x+y|xX,yY} So far complete analogy to classical probability.

19. The complement Properties of classical Pr: Let Xc be the complement of X. Then Pr(X)+Pr(Xc)=1 and Pr(X|Y)+Pr(Xc|Y)=1 (needed in proof of LLL). This is not true for R. We can define a “complement”: Xc=X=subspace orthogonal to X R(X)+R(X)=1 but we only have R(X|Y)+R(X|Y)≤1 !

20. The complement Example: R(X|Y)+R(X|Y)<1 Xc Y X R(X|Y)=R(X|Y)=0

21. The complement Classically, if X and Y are independent, then Xc and Y are also independent. For relative dimension this is wrong! Care needed in the formulation of the Local Lemma.

22. A quantum local lemma QLLL: Let X1,…,Xmbe subspaces of V with R(Xi)≥1-p and such that Xi is R-independent of all but at most d of the others. If then In particular Proof: Use properties of R, especially chain-rule and inclusion-exclusion. Induction.

23. Sparse QSAT Corollary of QLLL: Let 1,…,m be k-local projectors on n qubits s.t. each qubit appears in at most 2k/(ek) projectors. Then there is a state satisfying all i. We show: If i and j do not share a qubit, then their satisfying subspaces Xi and Xj are R-independent.

24. Sparse QSAT Corollary of QLLL: Let 1,…,m be k-local projectors on n qubits s.t. each qubit appears in at most 2k/(ek) projectors. Then there is a state satisfying all i. Proof: Xi=satisfying subspace for i. Then R(Xi)=1-2-k, i.e. p=2-k. Each Xi is R-dependent only on d=2k/e-1 others (d+1)pe1.

25. Outline • The classical Lovász Local Lemma • Motivation in the quantum case • A quantum local lemma • Application to random QSAT

26. Random SAT Classically: Properties of random k-SAT formulas have been studied in order to understand easy and hard instances as a function of clause density  ( = #clauses/#variables). Generating random k-SAT on n variables: For i=1,…,m=n • Pick a random set of k variables (random hyperedge – Gk(n,m) model ) • Negate each variable with probability ½.

27. Random-k-SAT Threshold Threshold phenomenon[Friedgut99]: For every k, there exists c(k) such that

28. Random SAT and QSAT Results: • Various properties [KS94,MPZ02,MMZ05]. • c(2)=1 [CR92,Goe92] • 3.52 ≤ c(3)≤ 4.49 [KKL03,HS03] • 2kln2-O(k) ≤ c(k)≤ 2kln2 [AP04] What about k-QSAT?

29. Random k-QSAT A random k-QSAT on n qubits is constructed as follows: For i=1,…,m =n : • Pick a random set of k qubits (random hyperedge – Gk(n,m) model) • Pick a uniformly random k-qubit state |vi on those k qubits and exclude it.

30. The case k=2 [LaumannMSS09,Bravyi07]:Threshold at density ½ The satisfying states in the satisfiable phase are tensor product states. 2-QSAT is fully understood.

31. Random k-QSAT at k≥3 Lower bound [LaumannLMSS09] : “Matching condition”: if there is a matching between clauses and qubits contained in a clause, there is a satisfying product state -clauses (projectors) -qubits

32. Random k-QSAT at k≥3 1 2 1 3 2 4 3 4 Matching condition  there is a left matching in G. True w.h.p for random graphs if m≤rkn , i.e. density ≤ rk 1 5 5 6 Random left-k-regular 1 1 2 2 3 3 4 4 5 5 6 ClausesProjectors

33. Random k-QSAT at k≥3 Lower bound: [LaumannLMSS09] As long as density <1 there is a satisfiable product state. Nothing was known about non-product states or above density 1. Upper bound: [BravyiMooreRussell09] For k=3: critical density <3.549… For k≥4: critical density <2k0.573... Large gap between lower bound 1 and upper bounds.

34. Random k-QSAT at k≥3 Remark: Let Gk(n,d) be a random k-uniform hypergraph of fixed degree d. Matching  dk.By [LaumannLMSS09], d≤ksatisfiable product state. Nothing was known for d>k. deg =k deg d

35. Random k-QSAT at k≥3 Remark: Let Gk(n,d) be a random k-uniform hypergraph of fixed degree d. Matching  dk.By [LaumannLMSS09], d≤ksatisfiable product state. Corollary of QLLL: If d ≤2k/(ek) there is a satisfying state. [LaumannLMSS09] conjecture that there is no satisfying product state above degree d=k. Would show that QLLL can deal with entangled satisfying states.

36. Random k-QSAT and QLLL What about Gk(n,m) (random hypergraph with n vertices and m hyperedges)? Problem: QLLL can deal with degree up to 2k/(ek). But Gk(n,m) of average degree 2k/(ek) will have some vertices with much higher degree. Degrees are Poisson distributed.

37. Random k-QSAT and QLLL Theorem [using QLLL]: Gk(n,m) of density c2k/k2 has a satisfying groundstate with high probability.

38. Random k-QSAT and QLLL unsatisfiable satisfiable classical threshold for large k QLLL Product states ln2  2k 1 c2k/k2 0.573  2k clause density [BravyiMooreRussell09] [LaumannLMSS09] Entangled states suspected

39. Random k-QSAT and QLLL L Theorem [using QLLL]: Gk(n,m) of density c2k/k2 has a satisfying groundstate with high probability. Idea: hybrid approach – split the graph into two parts: a high degree part H and a low degree part L. H

40. Gluing Lemma Gluing Lemma: If the vertices of the hypergraph G can be partitioned into H and L s.t.:1) All vertices in L have a degree somewhat below the QLLL threshold.2) All edges that involve only H can be satisfied.3) All edges that involve both H and L have the form: . Then there is a satisfying assignment for G. H L

41. Random QSAT and QLLL Proof sketch: We know there is a satisfying state for all edges that involve H. If edge e involves both L and H: Define two new projectors on qubits 2,3,4. Any state on qubits 2,3,4 orthogonal to |0 and |1 will be orthogonal to |  effectively decoupled H and L Apply QLLL to qubits 2,3,4 with 2 new constraints. 1 3 2 4 constraint ||

42. Random QSAT and QLLL L Constructing the partition H and L: H H H • We show w.h.p. |H| is small. • This is enoguh. • Intuition: Smaller sets have smaller density • H density becomes much smaller than 1. • w.h.p. it has a matching • H is satisfiable.

43. Notable points • Lovász Local Lemma generalizes to the geometric/quantum setting. • Allows making statements about satisfiability of (sparse) QSAT. We avoid the “tensor product structure”, by using the probabilistic method! • Allows to improve lower bounds on threshold for random k-QSAT and to deal with entangled satisfying states.

44. Open Questions QLLL is a geometric statement about subspaces: are there any other applications? Finding the satisfying state? Recent breakthrough by [Moser09] gives efficient algorithm to find it classically. Essentially Walk-SAT. Is there a generalization of Moser’s algorithm to the quantum case?

45. Thank you! Blahfdsfdsfds