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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.