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   Biˆ iz   J ijˆ izˆ 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   Biˆ iz   J ijˆ izˆ 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   Biˆ iz   J ijˆ izˆ 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   Biˆ iz   J ijˆ izˆ 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   Biˆ iz   J ijˆ izˆ 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
NN

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 ?