1. Boris Altshuler Columbia University Anderson Localization against Adiabatic Quantum Computation Hari Krovi, Jérémie Roland NEC Laboratories America

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

3. NPcomplete Cook – Levin theorem (1971): SAT problem is NP complete Now: ~ 3000 known NP complete problems

4. P = NP ?

5. 1 in 3 SAT 1 in 3 SAT problem bits laterals Ising spins clauses N M Clause c is satisfied if one of the three spins is down and other two are up or Otherwise the clause is not satisfied Task: Task: to satisfy all M clauses Definition

6. 1 in 3 SAT 1 in 3 SAT problem bits laterals Ising spins clauses N M 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 all M clauses Size of the problem: No solutions Few solutions Many solutions

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

8. Otherwise Solutions and only solutions are zero energy ground states of the Hamiltonian B i – number of clauses, which involve spin i J ij – number of clauses, where both i and j participate

9. Recipe: 1. Construct the Hamiltonian 2.Slowly change adiabatic parameter s from 0 to 1 Adiabatic Quantum Computation Assume that 1.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 E. Farhi, J. Goldstone, S. Gutmann, J. Lapan, A. Lundgren, and D. Preda, Science 292, 472(2001)

10. 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) 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

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

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

14. Anderson Model Lattice - tight binding model Onsite energies  i - random Hopping matrix elements I ij j i I ij I ij = { I i and j are nearest neighbors 0 otherwise -W <  i <W uniformly distributed I < I c I > I c Insulator All eigenstates are localized Localization length  loc Metal There appear states extended all over the whole system Anderson Transition

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

16. 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

17. Adiabatic theorem: slow enough 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 g.s. Calculation time is anticrossing

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

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

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

21. 2 1 3. Let us add one more clause, which is satisfied by but not by When the gaps decrease even quicker than exponentially 2. For   is close to  s  there typically are several solutions separated by distances. Consider two. 1.Hamiltonian is integrable: it commutes with all. Its states thus can be degenerated. These degeneracies should split at finite since is non-integrable

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

23. 2 1 2 1 Q1: Is the splitting big enough for to remain the ground state at large ? Q2: How big would be the anticrossing gap ?

24. Q1: Is the splitting big enough for to remain the ground state at large ? Perturbation theory in  } 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

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

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

28. Clause, which involves spins different in the two solutions 1.Spins that distinguish the two solutions form a graph 2.This graph is connected 3.Both solutions correspond to minimal energy on the graph. : 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”. Q2: How big is the anticrossing gap ? common - 1 common 1 different Two solutions. Spins:

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

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

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