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A Quantum Local Lemma. Or Sattath TAU and HUJI. Joint work with Andris Ambainis Julia Kempe. Outline. The classical Lovász Local Lemma Motivation in the quantum case A quantum local lemma Application to random QSAT. The Lovász Local Lemma. - PowerPoint PPT Presentation
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A Quantum Local Lemma
Or SattathTAU and HUJI
Joint work with Andris Ambainis
Julia Kempe
• The classical Lovász Local Lemma
• Motivation in the quantum case
• A quantum local lemma
• Application to random QSAT
Outline
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.
The Lovász Local Lemma
But what if the events are “weakly” dependent?
Given a k-SAT formula where each of the m clauses shares a variable with at most d other clauses.
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.
Example: sparse k-SATreally stupid
anyno
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.
Example: sparse k-SAT
Corollary: A k-SAT formula where each variable appears in at most 2k/(ek)
clauses is always satisfiable.
Example: sparse k-SAT
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.
The Lovász Local Lemma
• The classical Lovász Local Lemma
• Motivation in the quantum case
• A quantum local lemma
• Application to random QSAT
Outline
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.
Quantum bit
ba
1b0av
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.
n qubits
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
Quantum SAT
0, vvv vv
Satisfying State
Excluded 2n-k dimensional subspace
QSAT generalizes SAT:
is satisfiable iff is satisfiable.
The state |0011 is a satisfying state:
|0011=0, |0011=0
QSAT example
)()( 432321 xxxxxx
€
000123 ,Π 111
234{ }
123000 234
111
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].
Quantum SAT
€
v1,...,Π vm
€
v i
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?
Quantum SAT
€
∀i Π visi = 0
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/(ek)
projectors, then I is satisfiable.
A statement we would like
Correspondences:Probability space: Vector space VEvents: Subspaces X VProbabilities Pr: relative dimension R
Conditional Probability Pr(X|Y):
Independence: X, Y are R-independent if R(X|Y)=R(X) (equivalently R(XY)=R(X)R(Y) )
Events and independence
Properties or R:1) 0 ≤R(X)≤12) XY R(X)≤R(Y)3) Chain rule:
4) “Inclusion/Exclusion”: Let X+Y={x+y|xX,yY}
So far complete analogy to classical probability.
Properties of relative dim
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 !
The complement
Example: R(X|Y)+R(X|Y)<1
The complement
XY Xc
R(X|Y)=R(X|Y)=0
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.
The complement
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.
A quantum local lemma
Corollary of QLLL: Let 1,…,m be k-local projectors on n qubits s.t. each qubit appears in at most 2k/(ek) 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.
Sparse QSAT
Corollary of QLLL: Let 1,…,m be k-local projectors on n qubits s.t. each qubit appears in at most 2k/(ek) 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)pe1.
Sparse QSAT
• The classical Lovász Local Lemma
• Motivation in the quantum case
• A quantum local lemma
• Application to random QSAT
Outline
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.
Random SAT
Threshold phenomenon[Friedgut99]: For every k, there exists c(k) such that
Random-k-SAT Threshold
)(if0
)(if1)esatisfiablisPr(
kk
c
c
n
Results:• Various properties [KS94,MPZ02,MMZ05].• c(2)=1 [CR92,Goe92]• 3.52 ≤ c(3)≤ 4.49 [KKL03,HS03]• 2kln2-O(k) ≤ c(k)≤ 2kln2 [AP04]
What about k-QSAT?
Random SAT and 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.
Random k-QSAT
[LaumannMSS09,Bravyi07]:Threshold at density ½
The satisfying states in the satisfiable phase are tensor product states.
2-QSAT is fully understood.
The case k=2
Lower bound [LaumannLMSS09] :“Matching condition”: if there is a
matching between clauses and qubits contained in a clause, there is a satisfying product state
Random k-QSAT at k≥3
-clauses (projectors)-qubits
Random k-QSAT at k≥31
2 34
5
1
23 4
5 6
Random left-k-regular
Matching condition there is a left matching in G.
True w.h.p for random graphs if m≤rkn , i.e. density ≤ rk 1
1
234
5
1
2
3
4
5
6Clauses Projectors
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 <2k0.573... Large gap between lower bound 1 and
upper bounds.
Random k-QSAT at k≥3
Remark: Let Gk(n,d) be a random k-uniform hypergraph of fixed degree d.
Matching dk. By [LaumannLMSS09], d≤k satisfiable product state. Nothing was known for d>k.
Random k-QSAT at k≥3
deg ddeg =k
Remark: Let Gk(n,d) be a random k-uniform hypergraph of fixed degree d.
Matching dk. By [LaumannLMSS09], d≤k satisfiable product state. Corollary of QLLL: If d ≤2k/(ek) 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.
Random k-QSAT at k≥3
What about Gk(n,m) (random hypergraph with n vertices and m hyperedges)?
Problem: QLLL can deal with degree up to 2k/(ek). But Gk(n,m) of average degree 2k/(ek) will have some vertices with much higher degree.
Degrees are Poisson distributed.
Random k-QSAT and QLLL
Theorem [using QLLL]: Gk(n,m) of density c2k/k2 has a satisfying groundstate with high probability.
Random k-QSAT and QLLL
satisfiable unsatisfiable
c2k/k2
QLLL
ln2 2k
classical thresholdfor large k
1
[LaumannLMSS09]
Product states 0.573 2k
[BravyiMooreRussell09]
Entangled states suspected
clause density
Random k-QSAT and QLLL
Theorem [using QLLL]: Gk(n,m) of density c2k/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.
Random k-QSAT and QLLL
L
H
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.
Gluing Lemma
H
L
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.
Random QSAT and QLLL
constraint ||
12 34
10,
L
HH
Constructing the partition H and L:Random QSAT and QLLL
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.Þ w.h.p. it has a matchingÞ H is satisfiable.
• 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.
Notable points
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?
Open Questions
Thank you!
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