Boris Altshuler Columbia University

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# Boris Altshuler Columbia University - PowerPoint PPT Presentation

Anderson Localization against Adiabatic Quantum Computation. Hari Krovi, Jérémie Roland NEC Laboratories America. Boris Altshuler Columbia University . Computational Complexity Complexity Classes . P. Solution can be found in a polynomial time, e.g. multiplication. Polynomial time

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## Boris Altshuler Columbia University

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Anderson Localization

against

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

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

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

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

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

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

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

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