quantum lovasz local lemma andris ambainis (latvia), julia kempe (tel aviv), or sattath (hebrew u.)...

Quantum Lovasz local Quantum Lovasz local lemmalemma

Andris Ambainis (Latvia), Julia Andris Ambainis (Latvia), Julia Kempe (Tel Aviv), Or Sattath Kempe (Tel Aviv), Or Sattath

(Hebrew U.)(Hebrew U.)

arXiv:0911.1696

Part 1Part 1

(classical) Lovasz local lemma(classical) Lovasz local lemma

The settingThe setting

““Bad” events ABad” events A11, ..., A, ..., Amm. . Pr [APr [Aii] ] .. When can we say that When can we say that

Pr [none of APr [none of Aii] > 0?] > 0?

Obvious resultsObvious results

AAii independent: independent:

• Pr [not APr [not Aii] ] 1- 1- ..

• Pr [none of APr [none of Aii] ] (1- (1- ))mm>0.>0.

No assumptions about ANo assumptions about Aii::

• Pr [APr [Aii] ] ..

• Pr [some APr [some Aii occurs] occurs] m m..

• If mIf m < 1, then Pr [none of A < 1, then Pr [none of Aii] >0.] >0.

Limited independenceLimited independence

Each AEach Aii is independent of all but at is independent of all but at most d other events Amost d other events Ajj. .

[Erd[Erdöös, Lovs, Lováász, 1975] If Pr[Asz, 1975] If Pr[Aii] ] and and e(d+1) e(d+1) < 1, then < 1, then

Pr [none of APr [none of Aii] > 0.] > 0.

Full independence: m < 1 enough; < 1 enough; Limited independence: e (Limited independence: e (dd+1) +1) < 1. < 1.

Application 1: k-SATApplication 1: k-SAT

k-SAT formula F,k-SAT formula F,• F = FF = F11 F F22 ... ... F Fmm;;

• FFii = y = yi,1i,1 y yi,2i,2 ... ... y yi,ki,k;;

• yyi,li,l = x = xjj or or xxjj..

TheoremTheorem If each F If each Fii has common variables has common variables with at most 2with at most 2kk/e-1 other clauses F/e-1 other clauses Fjj, then , then xx11, ..., x, ..., xnn: F(x: F(x11, ..., x, ..., xnn)=TRUE.)=TRUE.

ProofProof

Pick xPick xii at random: at random:

FFii = x = x11 x x22 ... ... xxkk

ProofProof

““Bad events” -Bad events” -

FFii and F and Fjj independent = F independent = Fii and F and Fjj have no have no common variables. common variables.

Each FEach Fii has common variables with at most has common variables with at most m=2m=2kk/e-1 other F/e-1 other Fjj..

Lovasz local lemma applies

Application 2: Ramsey graphsApplication 2: Ramsey graphs

Complete graph Complete graph KKnn..

Colour edges in Colour edges in two colours so two colours so that no Kthat no Kkk has all has all edges in one edges in one colour.colour.

SolutionSolution

Colour edges randomly.Colour edges randomly. Events AEvents Aii – a fixed k- – a fixed k-

vertex subgraph does not vertex subgraph does not have all edges in the have all edges in the same colour.same colour.

Independent for Independent for subgraphs with no subgraphs with no common edges.common edges.

ResultResult

TheoremTheorem If If , then edges of, then edges of

KKmm can be coloured with two colours so that can be coloured with two colours so that

no there is no k vertices with all edges no there is no k vertices with all edges among them in one colour.among them in one colour.

Other applicationsOther applications

Coverings of RCoverings of R33 by unit balls; by unit balls; Linear arboricity (partitioning edges Linear arboricity (partitioning edges

of a graph into linear forests).of a graph into linear forests).

Quantum Lovasz lemmaQuantum Lovasz lemma

Events Events subspaces subspaces

Finite-dimensional Hilbert space H.Finite-dimensional Hilbert space H. Events AEvents Aii “bad subspaces” S “bad subspaces” Sii.. Event does not occur Event does not occur a state | a state | is is

orthogonal to Sorthogonal to Sii. . Goal: a state |Goal: a state |, |, | S Sii for all i. for all i.

Hamiltonian versionHamiltonian version

Hamiltonian H = Hamiltonian H = i i PPii.. Terms HTerms Hii subspaces S subspaces Sii.. HHii||=0 =0 || S Sii. . Is there a state |Is there a state | with H | with H | = 0? = 0?

Probability Probability dimension dimension

Relative dimensionRelative dimension

Pr [APr [Aii] ] d(H d(Hii) ) ..

When are two subspaces independent?

Independence: definition #1Independence: definition #1

Bipartite system HBipartite system HAAHHBB.. Subspaces HSubspaces H11HHB B and Hand HAAHH22 are are

independent.independent.

Definition #2Definition #2

Classically, AClassically, A11, A, A22 independent if independent if

Pr[APr[A11 A A22] = Pr[A] = Pr[A11] Pr[A] Pr[A22]. ].

Quantumly, HQuantumly, H11, H, H22 – independent if – independent if d(Hd(H11 H H22) = d(H) = d(H11) d(H) d(H22););

d(Hd(H11 H H22) = d(H) = d(H11) d(H) d(H22

);); d(Hd(H11

H H22) = d(H) = d(H11) d(H) d(H22););

d(Hd(H11 H H22

) = d(H) = d(H11) d(H) d(H22

).).

More than 2 subspacesMore than 2 subspaces

H is independent of HH is independent of H11, ..., H, ..., Hmm if H is if H is independent of any combination independent of any combination (union, intersection, complement) of (union, intersection, complement) of HH11, ..., H, ..., Hmm..

Quantum LLLQuantum LLL

TheoremTheorem Let H Let H11, ..., H, ..., Hmm be subspaces be subspaces with:with:• d(Hd(Hii) ) ;;

• Each HEach Hii independent of all but at most d independent of all but at most d other Hother Hjj. .

• e(d+1) e(d+1) <1. <1.

Then, there is |Then, there is |, |, | H Hii for all i. for all i.

Proof of quantum LLLProof of quantum LLL

Our goalOur goal

Need to show: there exists Need to show: there exists ||, |, | H Hii for all i. for all i.

Equivalently, |Equivalently, | H Hii for all i. for all i.

Main lemmaMain lemma

Corollary:

Then

for any other Hi.

Application: Application: quantum k-SATquantum k-SAT

Quantum SATQuantum SAT

k-SAT:k-SAT:• variables xvariables x11, ..., x, ..., xNN..

• F = FF = F11 ... ... F Fmm;;

• FFii = y = yi,1i,1 ... ... y yi,ki,k;;

• yyi,li,l = x = xjj or or xxjj..

Goal: F = true.Goal: F = true.

k-QSATk-QSAT• N qubits;N qubits;

• H = HH = H11 + ... + H + ... + Hmm;;

• Each HEach Hii involves k involves k qubits;qubits;

• Each HEach Hii – projector to 1 – projector to 1 of 2of 2kk dimensions. dimensions.

Goal: HGoal: H = 0. = 0.

TheoremTheorem

Assume that Assume that • H = HH = H11 + ... + H + ... + Hmm, etc. , etc.

• each Heach Hii has common qubits with at most has common qubits with at most d = 2d = 2kk/e-1 other H/e-1 other Hjj. .

Then there exists Then there exists :H :H = 0. = 0.

ProofProof

Each HEach Hii is a projector on S is a projector on Sii: :

QLLL: QLLL:

Random k-SAT and k-Random k-SAT and k-QSATQSAT

Random k-SATRandom k-SAT

F = FF = F11 ... ... F Fmm;; Each FEach Fii – random k-clause. – random k-clause. What should m be so that F is What should m be so that F is

satisfiable w.h.p.?satisfiable w.h.p.?

Ratio m/n.

Random k-SATRandom k-SAT

Threshold cThreshold ckk, for large n:, for large n:

• If m<(cIf m<(ckk--) n, then F) n, then FSAT w.h.p.SAT w.h.p.

• If m>(cIf m>(ckk++) n, then F) n, then FSAT w.h.p.SAT w.h.p.

3.52 < c3.52 < c33 < 4.49. < 4.49. Large k: Large k:

22kk ln 2 – O(k) ln 2 – O(k) c ckk 2 2kk ln 2. ln 2.

Random k-QSAT [Bravyi, 06]Random k-QSAT [Bravyi, 06]

H = HH = H11+...+H+...+Hmm.. Each HEach Hii – random projector to 1 of 2 – random projector to 1 of 2kk

dimensions for random k qubits. dimensions for random k qubits. Do we have Do we have :H:H = 0? = 0?

Ratio qk=m/n.

Results on quantum k-SAT Results on quantum k-SAT [Laumann et al., 09][Laumann et al., 09]

qq22=1/2.=1/2. For large k, 1-For large k, 1- < q < qkk < 0.574 < 0.57422kk..

Classically, ck < ln 2 22k k = 0.69 = 0.69 22kk..

Huge gap between upper and lower bounds.

Our resultOur result

TheoremTheorem

Since each HSince each Hii involves k qubits, this involves k qubits, this corresponds to each Hcorresponds to each Hii having having common qubits withcommon qubits with other Hother Hjj, , on averageon average. .

QLLL: , worst case.

SolutionSolution

Divide qubits into two sets:Divide qubits into two sets:• ““high-degree”: includes all qubits that high-degree”: includes all qubits that

are contained in many Hare contained in many Hjj and those that and those that are in Hare in Hii with such qubits. with such qubits.

• ““low-degree”.low-degree”. Use QLLL on “low-degree” set, Use QLLL on “low-degree” set,

another approach on “high-degree” another approach on “high-degree” set.set.

Combine the two solutions.Combine the two solutions.

[Laumann, et al., 09][Laumann, et al., 09]

H = HH = H11+...+H+...+Hmm.. Theorem If f:{1, ..., m} Theorem If f:{1, ..., m} qubits: qubits:

• f(i) – qubit that is involved in Hf(i) – qubit that is involved in Hii;;

• f(i) f(i) f(j), f(j),

there exists there exists :H:H = 0. = 0.