Boris Altshuler Columbia University

Anderson Localization against Adiabatic Quantum Computation Hari Krovi, Jérémie Roland NEC Laboratories America Boris AltshulerColumbia University

Computational Complexity Complexity Classes P Solution can be found in a polynomial time, e.g. multiplication Polynomial time is the size of the problem Solution can be checked in a polynomial time, e.g. factorization NP Nondeterministic polynomial time NP complete Every NP problem can be reduced to this problem in a polynomial time

NP complete Every NP problem can be reduced to this problem in a polynomial time Cook – Levin theorem (1971): SAT problem is NP complete Now: ~ 3000 known NP complete problems

? P = NP

1 in 3 SAT problem M bits laterals Ising spins clauses N Definition Clause c is satisfied if one of the three spins is down and other two are up or or Otherwise the clause is not satisfied Task: to satisfy allM clauses

1 in 3 SAT problem bits laterals Ising spins M N clauses Clause c is satisfied if one of the three spins is down and other two are up. Otherwise the clause is not satisfied Task: to satisfy allM clauses Size of the problem: Many solutions Few solutions No solutions

bits laterals Ising spins M N clauses Many solutions Few solutions No solutions clustering threshold satisfiability threshold

Otherwise Solutions and only solutions are zero energy ground states of the Hamiltonian Bi – number of clauses, which involve spin i Jij – number of clauses, where both i and jparticipate

Adiabatic Quantum Computation E. Farhi, J. Goldstone, S. Gutmann, J. Lapan, A. Lundgren, and D. Preda, Science 292, 472(2001) • Assume that • Solution can be coded by some assignment • of bits (Ising spins) • 2. It is a ground state of a Hamiltonian • 3. We have a system of qubits and can initialize it in the • ground state of another Hamiltonian Recipe: 1.Construct the Hamiltonian 2.Slowly change adiabatic parameter s from 0 to 1

Adiabatic Quantum Computation E. Farhi, J. Goldstone, S. Gutmann, J. Lapan, A. Lundgren, and D. Preda, Science 292, 472(2001) Recipe: 1.Construct the Hamiltonian 2.Slowly change adiabatic parameter s from 0 to 1 Adiabatic theorem: Quantum system initialized in a ground state remains in the ground state at any moment of time if the time evolution of its Hamiltonian is slow enough

Adiabatic Quantum Algorithm for 1 in 3 SAT Recipe: 1.Construct the Hamiltonian 2.Slowly change adiabatic parameter s from 0 to 1

Adiabatic Quantum Algorithm for 1 in 3 SAT Ising model (determined on a graph ) in a perpendicular field l Another way of thinking: determines a site of N-dimensional cube onsite energy hoping between nearest neighbors

Anderson Model • Lattice - tight binding model • Onsite energies ei- random • Hopping matrix elements Iij i j Iij { I i andjare nearest neighbors 0otherwise Iij= -W < ei <Wuniformly distributed Anderson Transition I < Ic I > Ic Metal There appear states extended all over the whole system Insulator All eigenstates are localized Localization length zloc

Adiabatic Quantum Algorithm for 1 in 3 SAT Another way of thinking: determines a site of N-dimensional cube onsite energy hoping between nearest neighbors Anderson Model on N-dimensional cube

Adiabatic Quantum Algorithm for 1 in 3 SAT Anderson Model on N-dimensional cube Usually: # of dimensions system linear size Here: # of dimensions system linear size

Adiabatic Quantum Algorithm for 1 in 3 SAT Adiabatic theorem: Quantum system initialized in a ground state remains in the ground state at any moment of time if the time evolution of its Hamiltonian is slow enough anticrossing g.s. Calculation time is

Calculation time is ! System needs time to tunnel barrier Tunneling matrix element Minimal gap Exponentially long tunneling times Exponentially small anticrossing gaps Localized states

When the gaps decrease even quicker than exponentially • Hamiltonian is integrable: it commutes with all . Its states thus can be degenerated. These degeneracies should split at finite l since is non-integrable 2 1 2. For ais close to asthere typically are several solutions separated by distances . Consider two.

When the gaps decrease even quicker than exponentially • Hamiltonian is integrable: it commutes with all . Its states thus can be degenerated. These degeneracies should split at finite l since is non-integrable 2 1 2. For ais close to asthere typically are several solutions separated by distances . Consider two. 3. Let us add one more clause, which is satisfied by but not by

When the gaps decrease even quicker than exponentially • Hamiltonian is integrable: it commutes with all . Its states thus can be degenerated. These degeneracies should be split by finite l in non-integrable 2 1 2. For ais close to asthere typically are several solutions separated by distances . Consider two. 3. Let us add one more clause, which is satisfied by but not by

2 2 1 1 Q1: ? Is the splitting big enough for to remain the ground state at large Q2: ? How big would be the anticrossing gap

Q1: ? Is the splitting big enough for to remain the ground state at large Perturbation theory inl } Cluster expansion: ~N terms of order 1 1. is exactly the same for all states, i.e. for all solutions. In the leading order 2. In each order of the perturbation theory a sum of terms with random signs. In the leading order in l

Is the splitting big enough for to remain the ground state at finite ? Q1: In the leading order in l Q1.1: ? How big is the interval in , where perturbation theory is valid A1.1: It works as long as -Anderson localization !Important: (?) when

Q2: ? How big is the anticrossing gap Two solutions. Spins: common -1 common 1 different Clause, which involves spins different in the two solutions • Spins that distinguish the two solutions form a graph • This graph is connected • Both solutions correspond to minimal energy • : energy is 1 if one of the spins if flipped and 0 otherwise Ising model in field on the graph. • 4. The field forms symmetric and antisymmetric linear combinations of the two ground states. • 5. The anticrossing gap is the difference between the ground state energies of the two “vacuums”.

Ising model in perp.field on the graph. The anticrossing gap is the difference between the ground state energies of the two “vacuums”. Q2: ? How big is the anticrossing gap Conventional case –number of different spins Tree

Q2: ? How big is the anticrossing gap Adiabatic quantum computer badly fails at large enoughN Existing classical algorithms for solving 1 in 3 SAT problem work for

Conclusion Original idea of adiabatic quantum computation will not work Hope Maybe the delocalized ground state at finite l contains information that can speed up the classical algorithm ?

Boris Altshuler Columbia University