boris altshuler columbia university anderson localization against adiabatic quantum computation hari...
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Boris Altshuler Columbia University
Anderson Localization Anderson Localization againstagainst
Adiabatic Quantum ComputationAdiabatic Quantum Computation
Hari Krovi, Jérémie Roland NEC Laboratories America
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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
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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
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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
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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
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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
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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
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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)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
?
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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
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2 4 64 6
4 6
, ...
, ,...
E N
N
4 6
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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
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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:
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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
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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
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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 ?