Boris Altshuler Columbia University

Anderson Localization Anderson Localization againstagainst

Adiabatic Quantum ComputationAdiabatic Quantum Computation

Hari Krovi, Jérémie Roland NEC Laboratories America

Nondeterministic polynomial

timeNPNP

Solution can be checked in a polynomial

time, e.g. factorization

Computational ComplexityComplexity 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

NPNPcompletecomplete

pNNPP

Every NP problem can be reduced to this

problem in a polynomial time

NPNPcompletecomplete

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

problems

PP = NP= NP?

1 in 3 SAT 1 in 3 SAT problem

1,2,...,i N

1, 1, 1c c ci j k

1i bitslaterals

Ising spins 1,2,...,

1 , ,c c c

c M

i j k N

, ,c c ci j kclausesN M

Clause c is satisfied if one of the three spins is down and other two are up

or

1, 1, 1c c ci j k or

1, 1, 1c c ci j k

Otherwise the clause is not satisfied

Task:Task: to satisfy all M clauses

DefinitionDefinition

1 in 3 SAT 1 in 3 SAT problem

1,2,...,i N1i bits

lateralsIsing spins 1,2,...,c M

, ,c c ci j kclausesN 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:Task: to satisfy all M clauses

Size of the problem:

, ,M

N MN

No

solutionsFew

solutionsMany

solutions

c s0

1,2,...,i N1i bits

lateralsIsing spins 1,2,...,c M

, ,c c ci j kclausesN M

, ,M

N MN

No

solutionsFew

solutionsMany

solutions

c s0clustering threshold

satisfiability

threshold

c s 0.546 < 0.644s

21, 1, 1

1, 1, 1 1 0

1, 1, 1

c c c

c c c c c c

c c c

i j k

i j k i j k

i j k

21 1

1c c c

M N

p i j k i i ij i jc i i j

H B J

Otherwise

21 0c c ci j k

Solutions and only solutions are zero energy ground states of the Hamiltonian

i

Bi – number of clauses, which involve spin iJij – number of clauses, where both i and j participate

Recipe: 1.Construct the Hamiltonian

2.Slowly change adiabatic parameter s from 0 to 1

Adiabatic Adiabatic Quantum Quantum ComputationComputationAssume 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

i p iH

ˆ ˆ ˆ,x zi i i

0ˆˆiH

0ˆ ˆˆ ˆ ˆ 1zi p i i

sH sH s H

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 Adiabatic Quantum Quantum ComputationComputation

0ˆ ˆˆ ˆ ˆ 1z

s i p i iH sH s H

E. Farhi, J. Goldstone, S. Gutmann, J. Lapan, A. Lundgren, and D. Preda, Science 292, 472(2001)

Adiabatic theorem: 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

2

1 1

01

ˆ ˆ ˆ ˆ ˆ ˆ ˆ1

ˆˆ ˆ

c c c

M Nz z z z z z

p i j k i i ij i jc i i j

Nx

i ii

H B J

H

Adiabatic Quantum Algorithm for 1 in 3 SAT Adiabatic Quantum Algorithm for 1 in 3 SAT Recipe: 1.Construct the Hamiltonian

2.Slowly change adiabatic parameter s from 0 to 1

0ˆ ˆˆ ˆ ˆ 1z

s i p i iH sH s H

0

1ˆ ˆˆ ˆ ˆ ;zi p i i

sH H H

s

determines a site of N-dimensional cube

2

01 1 1

ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ1 ;c c c

M N Nz z z z z z x

p i j k i i ij i j i ic i i j i

H B J H

Adiabatic Quantum Algorithm for 1 in 3 SAT Adiabatic Quantum Algorithm for 1 in 3 SAT

0

1ˆ ˆˆ ˆ ˆ ;zi p i i

sH H H

s

Ising model (determined on a graph ) in a perpendicular field

Another way of thinking:

i

p iH onsite energy

01

ˆˆ ˆ ˆ ˆ ˆN

x xi i

i

H

hoping between nearest neighbors

Anderson

Model

• Lattice - tight binding model

• Onsite energies i - random

• Hopping matrix elements Iij j iIij

Iij ={ I i and j are nearest neighbors

0 otherwise

-W < i <W uniformly distributed

I < Ic I > IcInsulator

All eigenstates are localized

Localization length loc

MetalThere appear states extended

all over the whole system

Anderson Anderson TransitionTransition

determines a site of N-dimensional cube

2

01 1 1

ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ1 ;c c c

M N Nz z z z z z x

p i j k i i ij i j i ic i i j i

H B J H

Adiabatic Quantum Algorithm for 1 in 3 SAT Adiabatic Quantum Algorithm for 1 in 3 SAT

0

1ˆ ˆˆ ˆ ˆ ;zi p i i

sH H H

s

Another way of thinking:

i

p iH onsite energy

01

ˆˆ ˆ ˆ ˆ ˆN

x xi i

i

H

hoping between nearest neighbors

Anderson Model on Anderson Model on NN--dimensional cubedimensional cube

2

01 1 1

ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ1 ;c c c

M N Nz z z z z z x

p i j k i i ij i j i ic i i j i

H B J H

Adiabatic Quantum Algorithm for 1 in 3 SAT Adiabatic Quantum Algorithm for 1 in 3 SAT

0

1ˆ ˆˆ ˆ ˆ ;zi p i i

sH H H

s

Anderson Model on Anderson Model on NN--dimensional cubedimensional cubeUsually:# of dimensions system linear size

d constL

Here:# of dimensions system linear size

d N 1L

Adiabatic Adiabatic theorem: 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 slow enoughenough

Adiabatic Quantum Algorithm for 1 in 3 SAT Adiabatic Quantum Algorithm for 1 in 3 SAT

0

1ˆ ˆˆ ˆ ˆ ;zi p i i

sH H H

s

E

g.s.

H E

2t Calculation time Calculation time isis

2minO

2

01 1 1

ˆˆ ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ1 ;c c c

M N Nz z z z z z x

p i j k i i ij i j i ic i i j i

H B J H

anticrossing

2t Calculation time Calculation time isis

2minO

barrierSystem needs time to tunnel

!

Localized states

Exponentially long

tunneling times

Exponentially small

anticrossing gaps

Minimal gapTunneling

matrix element

2t Calculation time Calculation time isis

2minO

barrierSystem needs time to tunnel

!

Localized states

Exponentially long

tunneling times

Exponentially small

anticrossing gaps

Minimal gapTunneling

matrix element

1. Hamiltonian is integrable: it commutes with all . Its states thus can be degenerated. These degeneracies should split at finite since is non-integrable

21

E1

0

E

2. For is close to sthere

typically are several solutions separated by distances . Consider two.

N

ˆ ˆ zp iH ˆ zi

ˆˆiH

0ˆ ˆˆ ˆ ˆ zi p i iH H H

When the gaps decrease even quicker than exponentially

N

0ˆ ˆˆ ˆ ˆ zi p i iH H H

21

E1

0

E

3. Let us add one more clause, which is satisfied by but not by

1 0

When the gaps decrease even quicker than exponentially

N

2. For is close to sthere

typically are several solutions separated by distances . Consider two.

N

1. Hamiltonian is integrable: it commutes with all . Its states thus can be degenerated. These degeneracies should split at finite since is non-integrable

ˆ ˆ zp iH ˆ zi

ˆˆiH

1

0

E

0

1

0ˆ ˆˆ ˆ ˆ zi p i iH H H

When the gaps decrease even quicker than exponentially

N

3. Let us add one more clause, which is satisfied by but not by

1 0

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 sthere

typically are several solutions separated by distances . Consider two.

N

ˆ ˆ zp iH ˆ zi

ˆˆiH

21

21

1

0

E

0

1

21

E1

0

E

Q1: Is the splitting big enough for to remain the ground state at large

E 0 ?

Q2: How big would be the anticrossing gap

?

2kk

k

E N C Q1: Is the splitting big enough for

to remain the ground state at largeE 0

?Perturbation theory inN

Mconst

N

} E N Cluster expansion: ~N terms of order

1 1. is exactly the same for all

states, i.e. for all solutions. In the leading order

1C 0 0E

4E 2. In each order of the perturbation theory a

sum of terms with random signs.E

NIn the leading order in

4E N

2 4 64 6

4 6

, ...

, ,...

E N

N

4 6

1 81E N

In the leading order in

4E N

Q1:Is the splitting big enough for to remain the ground state at finite

E0

?

Q1.1: How big is the interval in , where perturbation theory is valid

?A1.1:

It works as long as -Anderson localization ! Important: (?) when

c 0.3c const N

0

ˆˆ ˆ ˆ

ˆ

zi p i

i

H H

H

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 : energy is 1 if one of the spins if flipped and

0 otherwise Ising model in field on the on the graph.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 1different

Two solutions.Spins:

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

2nE n N

TreenE

Q2:How big is the anticrossing gap

?

#

#

1 8exp # ln

#

NNE

E N N eN

Adiabatic quantum computer badly fails at large enough N

410N N c

Existing classical algorithms for solving 1 in 3 SAT problem work for 33 4 10N

ConclusionConclusionOriginal idea of adiabatic quantum computation will not work

HopeHopeMaybe the delocalized ground state at

finite contains information that can speed up the classical algorithm ?