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
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?
Example: sparse k-SAT
Given a k-SAT formula  where each of the
m clauses shares a variable with at most d
other clauses.
really stupid
Example: sparse k-SAT
Given a k-SAT formula  where each of the
m clauses shares a variable with at most d
any
no
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-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.
Example: sparse k-SAT
Corollary: A k-SAT formula where each
variable appears in at most 2k/(ek) clauses
is always satisfiable.
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 nonzero probability that none of them occur.
Outline
• The classical Lovász Local Lemma
• Motivation in the quantum case
• A quantum local lemma
• Application to random QSAT
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 
v  a 0  b 1   
b
A qubit is a vector in a 2 dimensional vector
space. |0 and |1 form an orthonormal basis
for this vector space.
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.
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.
Satisfying
State
Clauses  Rank-1 Projectors
v v  v , v v

0
Excluded 2n-k dimensional subspace
QSAT example
QSAT generalizes SAT:
  ( x1  x2  x3 )  (x2  x3  x4 )
is satisfiable iff


123
000
,
234
111

is satisfiable.
The state |0011 is a satisfying state:

123
|0011=0,
000

234
111 |0011=0
Quantum SAT
Formal Def: (k-QSAT)
Given a collection of k-local rank-1 projectors on
n qubits,  v ,..., v
1
m
Is there a state |s inside the allowed subspace
of 
0
for i=1..m.
vi

Importance: known to be QMA1-Complete
(quantum analogue of NP) , for k≥4[Bravyi06].
Quantum SAT
If each projector acts on a set of mutually
disjoint qubits,
i  vi si  0
then |s= |s1…|sm is a satisfying state.
But what if each qubit appears in a few
projections?
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.
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) )
Properties of relative dim
Properties or R:
1) 0 ≤R(X)≤1
2) XY R(X)≤R(Y)
3) Chain rule:
4) “Inclusion/Exclusion”: Let X+Y={x+y|xX,yY}
So far complete analogy to classical probability.
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 !
The complement
Example: R(X|Y)+R(X|Y)<1
Y
Xc
X
R(X|Y)=R(X|Y)=0
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.
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 chainrule and inclusion-exclusion. Induction.
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.
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.
Outline
• The classical Lovász Local Lemma
• Motivation in the quantum case
• A quantum local lemma
• Application to random QSAT
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.
Random-k-SAT Threshold
Threshold phenomenon[Friedgut99]:
For every k, there exists c(k) such that
1 if    c (k )
Pr( is satisfiabl e)  
n 0 if    (k )
c

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?
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.
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.
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
Random k-QSAT at k≥3
1
2
3
1
2
4
3
4
5
5
6
Random left-k-regular
1
1
2
2
3
3
4
4
5
5
6
Clauses Projectors
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
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.
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
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.
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.
Random k-QSAT and QLLL
Theorem [using QLLL]: Gk(n,m) of density
c2k/k2 has a satisfying groundstate with
high probability.
Random k-QSAT and QLLL
unsatisfiable
satisfiable
classical
threshold
for large k
QLLL
Product states 1
[LaumannLMSS09]
c2k/k2
0.573  2k
ln2  2k
[BravyiMooreRussell09]
Entangled states suspected
clause
density
Random k-QSAT and QLLL
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
L
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
H
the form:
.
L
Then there is a satisfying assignment for G.
Random QSAT and QLLL
Proof sketch: We know there is a satisfying state
for all edges that involve H.
1
If edge e involves both L and H:
3
2
4
constraint ||
Define two new projectors 
,  1 on qubits
0
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
Constructing the partition H and L:
HH
We show w.h.p. |H| is small.
L
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.
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.
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?
Thank you!