1. A Quantum Local Lemma Or Sattath TAU and HUJI Andris Ambainis
Joint work with
Andris Ambainis
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. no
any

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. Excluded 2n-k dimensional subspace
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 satisfiable iff 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 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,…,Xm be 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. 2-QSAT is fully understood.
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 left-k-regular
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
Clauses Projectors

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

39. Random k-QSAT and QLLL H 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. 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 2 3 4
constraint ||

42. Random QSAT and QLLL L H H H Constructing the partition H and L:
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!
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