Anderson Localization against Adiabatic Quantum Computation Boris Altshuler Hari Krovi, Jérémie Roland Columbia University NEC Laboratories America Computational Complexity Complexity Classes P NP NP complete N is the size of Solution can be found in a polynomial time, Nondeterministic polynomial time Solution can be checked in a polynomial time, Polynomial time N p the problem e.g. multiplication e.g. factorization 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 1 bits i laterals i 1, 2, ..., N Ising spins N Definition Clause c is satisfied if one of the three spins is down and other two are up M clauses ic , jc , kc c 1, 2, ..., M 1 ic , jc , kc N i 1, j 1, k 1 or i 1, j 1, k 1 or i 1, j 1, k 1 c c c c c c c c c Otherwise the clause is not satisfied Task: to satisfy all M clauses 1 in 3 SAT problem i 1 i , j , k c c c clauses M bits laterals i 1, 2, ..., N Ising spins N c 1, 2, ..., M Clause c is satisfied if one of the three spins is down and other two are up. Otherwise the clause is not satisfied Size of the problem: Many solutions 0 c Task: to satisfy all M clauses M N , M , N Few solutions s No solutions 1 bits i laterals i 1, 2, ..., N Ising spins N i , j , k c c c clauses M c 1, 2, ..., M M N , M , N Many solutions 0 clustering c threshold c Few solutions s No solutions s satisfiability 0.546 < s 0.644 threshold ic 1, jc 1, kc 1 ic 1, jc 1, kc 1 ic 1, jc 1, kc 1 ic 2 jc kc 1 0 2 Otherwise ic jc kc 1 0 Solutions i and only solutions are zero energy ground states of the Hamiltonian M N H p ic jc kc 1 Bi i J ij i j c 1 2 i 1 i j Bi – number of clauses, which involve spin i Jij – number of clauses, where both i and j participate E. Farhi, J. Goldstone, S. Gutmann, J. Lapan, A. Lundgren, and D. Preda, Science 292, 472(2001) Adiabatic Quantum Computation Assume that 1.Solution can be coded by some assignment i of bits (Ising spins) 2. It is a ground state of a Hamiltonian H p i x z 3. We have a system of qubits ˆ i ˆ i , ˆ i and can initialize it in the ground state of another Hamiltonian Hˆ 0 ˆ i Recipe: 1.Construct the Hamiltonian Hˆ ˆ sHˆ ˆ 1 s H ˆ i s p z i 2.Slowly change adiabatic parameter 0 s i 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 Hˆ s ˆ i sHˆ p ˆ iz 1 s H 0 ˆ i 2.Slowly change adiabatic parameter Adiabatic theorem: s from 0 to 1 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 Hˆ s ˆ i sHˆ p ˆ iz 1 s H 0 ˆ i 2.Slowly change adiabatic parameter M s from 0 to 1 N Hˆ p ˆ izc ˆ zjc ˆ kzc 1 Biˆ iz J ijˆ izˆ zj c 1 Hˆ 0 ˆ i i 1 i j N x ˆ i Hˆ ˆ i 2 i 1 H Hˆ p ˆ iz 0 ˆ i 1 s ; s Adiabatic Quantum Algorithm for 1 in 3 SAT M N Hˆ p ˆ izc ˆ zjc ˆ kzc 1 Biˆ iz J ijˆ izˆ zj ; c 1 Hˆ ˆ i 2 i 1 i j H z ˆ H p ˆ i Hˆ 0 ˆ i ˆ 0 i N ˆ ix i 1 1 s ; s Ising model (determined on a graph ) in a perpendicular field Another way of thinking: H p i onsite energy determines a site i of N-dimensional cube Hˆ 0 ˆ i N ˆ ix i 1 ˆ x ˆ ˆ hoping between nearest neighbors • Lattice - tight binding model Anderson Model • Onsite energies j i Iij -W < ei <W uniformly distributed ei - random • Hopping matrix elements Iij { Iij = I i and j are nearest neighbors 0 otherwise Anderson Transition I < Ic Insulator All eigenstates are localized Localization length zloc I > Ic Metal There appear states extended all over the whole system Adiabatic Quantum Algorithm for 1 in 3 SAT M N Hˆ p ˆ izc ˆ zjc ˆ kzc 1 Biˆ iz J ijˆ izˆ zj ; c 1 Hˆ ˆ i i 1 onsite energy Hˆ 0 ˆ i i j H z ˆ H p ˆ i Another way of thinking: H p i 2 ˆ 0 i N ˆ ix i 1 1 s ; s determines a site i of N-dimensional cube Hˆ 0 ˆ i N x ˆ i x ˆ ˆ ˆ i 1 hoping between nearest neighbors Anderson Model on N-dimensional cube Adiabatic Quantum Algorithm for 1 in 3 SAT M N Hˆ p ˆ izc ˆ zjc ˆ kzc 1 Biˆ iz J ijˆ izˆ zj ; c 1 Hˆ ˆ i 2 i 1 i j H z ˆ H p ˆ i 0 ˆ i Hˆ 0 ˆ i N ˆ ix i 1 1 s ; s Anderson Model on N-dimensional cube Usually: # of dimensions d const system linear size L Here: # of dimensions d N system linear size L 1 Adiabatic Quantum Algorithm for 1 in 3 SAT M N Hˆ p ˆ izc ˆ zjc ˆ kzc 1 Biˆ iz J ijˆ izˆ zj ; c 1 Hˆ ˆ i 2 i 1 H 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 t i j z ˆ H p ˆ i Adiabatic theorem: 2 Hˆ 0 ˆ i 0 E ˆ i N ˆ ix i 1 1 s ; s Ĥ E g.s. Calculation time is anticrossing 2 O min t 2 Calculation time is barrier System needs time to tunnel Minimal gap Localized states Exponentially long tunneling times 2 O min ! Tunneling matrix element Exponentially small anticrossing gaps t 2 Calculation time is barrier System needs time to tunnel Minimal gap Localized states Exponentially long tunneling times 2 O min ! Tunneling matrix element Exponentially small anticrossing gaps When N the gaps decrease even quicker than exponentially Hˆ ˆ i H ˆ Hˆ p ˆ iz ˆ 0 i z ˆ ˆ H 1. Hamiltonian p is integrable: i z it commutes with all i . Its states thus can be degenerated. These degeneracies should split at finite since Hˆ ˆ i is non-integrable 2. For is close to s there typically are several solutions separated by distances N . Consider two. E 2 1 1 E 0 When N the gaps decrease even quicker than exponentially Hˆ ˆ i H ˆ Hˆ p ˆ iz ˆ 0 i z ˆ ˆ H 1. Hamiltonian p is integrable: i z it commutes with all i . Its states thus can be degenerated. These degeneracies should split at finite since Hˆ ˆ i is non-integrable 2. For is close to s there typically are several solutions separated by distances N . Consider two. 3. Let us add one more clause, which is satisfied by 1 but not by 0 E 2 1 1 E 0 When N the gaps decrease even quicker than exponentially Hˆ ˆ i H ˆ Hˆ p ˆ iz 0 1. Hamiltonian Hˆ p ˆ iz is integrable: z it commutes with all ˆ i . Its states thus can be degenerated. These 2 degeneracies should be split by 1 finite in non-integrable Hˆ ˆ i 2. For is close to s there typically are several solutions separated by distances N . Consider two. 3. Let us add one more clause, which is satisfied by 1 but not by 0 i E 1 0 0 1 E E 2 1 2 1 1 0 1 E 0 1 0 Q1: Is the splitting E big enough for remain the ground state at large Q2: How big would be the anticrossing gap 0 to ? ? Q1: Is the splitting E big enough for remain the ground state at large Perturbation theory in N M const N } 0 to ? E N Ck 2k k E N Cluster expansion: ~N terms of order 1 1.C1 is exactly the same for all E 0 0 states, 4 i.e. for all solutions. In the leading order E 2. In each order of the perturbation theory E a sum N terms with random signs. of In the leading order in E 4 N E , N 4 4 6 6 ... 4 , 6 ,... N 2 4 6 Q1: Is the splitting E big enough for 0 to remain the ground state at finite In the leading order in E 4 N Q1.1: A1.1: ? Hˆ ˆ i H ˆ Hˆ p ˆ iz 0 i E 1 N How big is the interval in , where perturbation theory is valid 1 8 ? It works as long as c -Anderson localization ! Important: c const 0.3 (?) when N Q2: How big is the anticrossing gap Two solutions. Spins: common -1 1. 2. 3. 4. 5. ? Clause, which involves spins different in the two solutions common 1 different 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. The field forms symmetric and antisymmetric linear combinations of the two ground states. The anticrossing gap is the difference between the ground state energies of the two “vacuums”. 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 n E 2n N–number of different spins Tree E n Q2: How big is the anticrossing gap ? E E exp # N ln N 1 8 # N #N e Adiabatic quantum computer badly fails at large enough N c NN 10 4 Existing classical algorithms for solving 1 3 in 3 SAT problem work for N 3 4 10 # N 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 ?