Quantum Adiabatic Evolution Algorithms versus Simulated Annealing

16 Pages·0.22 MB·English

by  Edward Farhi

Quantum Adiabatic Evolution Algorithms
versus Simulated Annealing
Edward Farhi, Jeffrey Goldstone∗
Center for Theoretical Physics, Massachusetts Institute of Technology, Cambridge, MA 02139
2
Sam Gutmann†
0
0 Department of Mathematics, Northeastern University, Boston, MA 02115
2
MIT-CTP #3228 quant-ph/0201031
n
a
J Abstract

8
We explain why quantum adiabatic evolution and simulated annealing perform simi-

1
larly in certain examples of searching for the minimum of a cost function of n bits. In
v
1 these examples each bit is treated symmetrically so the cost function depends only on
3 the Hamming weight of the n bits. We also give two examples, closely related to these,
0
where the similarity breaks down in that the quantum adiabatic algorithm succeeds in
1
0 polynomial time whereas simulated annealing requires exponential time.
2
0
/ 1 Introduction
h
p
- Quantum adiabatic evolution algorithms [1] are designed to minimize a (classical) cost func-
t
n tion whose domain is the 2n values taken by n bits. To test the algorithm it is natural to
a
u look at problems where classical local search algorithms, such as simulated annealing, have
q difficulty. It is easy to construct examples where both classical local search and quantum
:
v adiabatic evolution require time exponential in n. An example is the Grover problem, where
i
X the function takes the value 0 at a single input and is 1 on the 2n−1 other inputs. A variant
r of this Grover example can be found in [2].
a
More interesting examples arise when the cost function is local in the sense that it can be
written as a sum of terms each of which involves only a few bits. A 3sat example where both
the classical local search and “standard” quantum adiabatic algorithms are exponentially
slow has recently been given in [3]. In Section 4 below we analyze an example that is based
on [3] and give a quantum mechanical argument for why quantum adiabatic algorithms have
trouble with these kinds of examples.
In Section 5 we give a nonlocal example where simulated annealing requires exponential
time but quantum adiabatic evolution succeeds in polynomial time. In Section 6 we give a
detailed analysis of a local (2sat) example that first appeared in [1]. Here again a quan-
tum adiabatic algorithm succeeds even though simulated annealing fails in polynomial time.
These examples show that there is no general theorem that quantum adiabatic algorithms
must fail if simulated annealing fails.
∗[email protected], [email protected]
†[email protected]
2 E. Farhi, J. Goldstone, S. Gutmann
2 Quantum adiabatic algorithms
Quantum adiabatic evolution algorithms offer a general approach to solving combinatorial
search problems on a quantum computer. A local combinatorial search problem can be cast
as the classical problem of finding the minimum of cost function h(z ,... ,z ) where each
1 n
z = 0,1 and
i
h = h (1)
C
C
X
and each h is a nonnegative integer-valued function that depends only on a few z . A
C i
quantum adiabatic algorithm converts this into the problem of producing the ground state
of a quantum Hamiltonian H given by
P
H z ,... ,z = h(z ,... ,z ) z ,... ,z (2)
P 1 n 1 n 1 n
where{ z ,... ,z }isabasi(cid:12)sforthe2n(cid:11)-dimensionalHil(cid:12)bertspace(cid:11)ofthequantumcomputer.
1 n (cid:12) (cid:12)
Specifying the algorithm consists of choosing a smoothly varying Hamiltonian H(s) for 0 ≤
(cid:12) (cid:11)
(cid:12)
s ≤ 1 such that H(1) = H and where H(0) has a ground state ψ that is known and
P 0
easily constructed. A run time T must also be specified. The Hamiltonian thategoverns the
(cid:12) (cid:11)
evolution is giveneby e (cid:12)
H(t) = H(t/T) (3)
and the state of the quantum computer evolves according to the Schr¨odinger equation
e
d
i ψ(t) = H(t) ψ(t) (4)
dt
with the state at time t = 0 given by(cid:12) (cid:11) (cid:12) (cid:11)
(cid:12) (cid:12)
ψ(0) = ψ . (5)
0
Under conditions that generally apply(cid:12)in th(cid:11)e ca(cid:12)ses(cid:11)of interest here, the quantum adiabatic
(cid:12) (cid:12)
theorem guarantees that for T large enough the state of the quantum computer, ψ(t) for
0 ≤ t ≤ T, will be close to the(instantaneous) ground state ofH(t), and inparticular ψ(T)
(cid:12) (cid:11)
will be close to the ground state of H , encoding the solution to the problem at han(cid:12)d. When
P (cid:12) (cid:11)
applied to a particular combinatorial search problem, the quantum algorithm is cons(cid:12)idered
to be successful if the required running time grows polynomially in the number of bits n.
The required running time can be related to the spectrum of H(s), in particular the
difference between the two lowest eigenvalues E (s) and E (s). The required running time T
0 1
must obey e
E
T ≫ (6)
gap2
where
gap = min E (s)−E (s) (7)
1 0
0≤s≤1
(cid:0) (cid:1)
andE islessthanthelargesteigenvalueofH −H(0),alwayspolynomialinnintheexamples
P
here.
e
Quantum Adiabatic Evolution Algorithms versus Simulated Annealing 3
3 Simulated annealing
Simulated annealing is a classical local search strategy that can be used to find the global
minimum of a function h(z ,... ,z ) of the form (1). For any temperature τ, 0 ≤ τ ≤ ∞,
1 n
define the Boltzmann distribution by
P (z ,... ,z ) = c(τ)exp −h(z ,... ,z )/τ (8)
τ 1 n 1 n
where c(τ) is the normalizing constant such that(cid:2) (cid:3)
P (z ,... ,z ) = 1 . (9)
τ 1 n
z1X,...,zn
For τ = ∞, all 2n strings z ,... ,z are equally likely. For τ = 0, the Boltzmann distribution
1 n
concentrates on the global minimum (or minima) of n.
The idea is to construct a Markov chain that starts in the τ = ∞ Boltzmann distribution
and gradually moves through the Boltzmann distributions for decreasing τ down to τ near 0.
If the process succeeds, the final distribution will be close to the zero temperature Boltz-
mann distribution, which means that the global minimum of h has been found with high
probability. More specifically, choose a sequence of temperatures ∞ ≥ τ ≥ τ ≥ ···τ ≥ 0
1 2 M
at which bit values may change. The τ can be chosen deterministically or randomly. Also
i
choose a sequence of bits i ,i ,... ,i . Bit i is the bit that might be changed (according
1 2 M k
to an acceptance rule to be given below in (10)) at step k. The sequence i ,i ,... ,i can
1 2 M
be deterministic or random.
The initial string is picked uniformly, i.e., all 2n strings z ,... ,z have probability
1 n
2−n. At step k, if the current string is z ,... ,z ,... ,z we flip bit i and move to string
1 ik n k
z ,... ,(1−z ),... ,z with probability
1 ik n
min 1,exp −∆h/τ
k
where ∆h = h(z ,... ,(1−z ),... ,z )−h(z ,... ,z ,... ,z ). (10)
1 (cid:8) ik (cid:0) n (cid:1)(cid:9) 1 ik n
(This is the Metropolis rule.) Otherwise the string remains unchanged. Note that bit i is
k
always flipped if the change decreases h, and is flipped with only small probability if the
change would increase h by a large amount relative to the current temperature.
The acceptance rule (10) guarantees that if many transitions are made at a single tem-
perature τ, the distribution of the string will converge to the Boltzmann distribution at
temperature τ. If the sequence τ ,τ ,... ,τ decreases slowly enough, the distribution,
1 2 M
which starts in the τ = ∞ Boltzmann distribution, will at step k be close to the Boltzmann
distribution for temperatureτ . The question ishow many transitions must bemade inorder
k
to stay near the Boltzmann distributions as the temperature is lowered down to near 0. If
the totalnumber ofsteps required grows only as a polynomial inn, we say that the simulated
annealing algorithm succeeds.
4 Symmetrized cost functions
In this section we discuss cost functions of the form (1) that are functions only of the
Hamming weight w = z +z +···+z as a result of all bits being treated symmetrically.
1 2 n
4 E. Farhi, J. Goldstone, S. Gutmann
As an example, let h depend on three bits z,z′, and z′′ with the form
3
0 z +z′ +z′′ = 0
q z +z′ +z′′ = 1
h (z,z′,z′′) =  (11)
3 1 z +z′ +z′′ = 2


1 z +z′ +z′′ = 3



where q is an integer greater than or equalto 3. Now let
h(z ,... ,z ) = h (z ,z ,z ) (12)
1 n 3 i j k
i<j<k
X
which gives
h(w) ≡ h(z ,... ,z )
1 n
= qw(n−w)(n−w −1)+ 1w(w−1)(n−w)+ 1w(w−1)(w−2) . (13)
2 2 6
We can also write
n 3
h(w) = g(w/n)+O(n2) (14)
2
(cid:16) (cid:17)
where
g(u) = 4qu(1−u)2 +4u2(1−u)+ 4u3 . (15)
3
Fornlargetheformofg(u)determines thebehavior oftheclassical andquantum algorithms.
In Figure 1 we plot g(u) for q = 3. Note that the global minimum of g is at u = 0
corresponding to the unique input z ,... ,z = 0,...,0. The derivative g′(1/2) is negative
1 n
and there is a local minimum at u = 1 corresponding to z ,... ,z = 1,...,1.
1 n
We now look at the quantum algorithm applied to this problem. The cost function (13)
corresponds to the Hamiltonian
q n n n
H = −S +S +S −1
P z z z
2 2 2 2
(cid:16) (cid:17)(cid:16) (cid:17)(cid:16)1 n (cid:17)n n
+ −S −S −1 +S
z z z
2 2 2 2
(cid:16) (cid:17)(cid:16) 1 n (cid:17)(cid:16) n (cid:17) n
+ −S −S −1 −S −2 (16)
z z z
6 2 2 2
(cid:16) (cid:17)(cid:16) (cid:17)(cid:16) (cid:17)
where
n
S z ,... ,z = −w z ,... ,z . (17)
z 1 n 1 n
2
(cid:16) (cid:17)
(cid:12) (cid:11) (cid:12) (cid:11)
We choose for the initial Ham(cid:12) iltonian (following [1],(cid:12)formula 2.28)
n−1 n
H(0) = −S (18)
2 2 x
(cid:16) (cid:17)(cid:16) (cid:17)
e
Quantum Adiabatic Evolution Algorithms versus Simulated Annealing 5
2.5
2
1.5
u)
g(
1
0.5
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
u
Figure 1: Theleading term in the scaled cost as a function of the scaled Hamming weight u= w/n.
where
n
S = 1 σ(i) . (19)
x 2 x
i=1
X
Choosing a linear interpolation between H(0) and H gives
P
H(s) = (1−s)H(e0)+sHP for 0 ≤ s ≤ 1. (20)
Since H(s) depends onlyeonthetwo opereatorsS andS , we startour analysis by considering
x z
the states θ defined by
e
(cid:12) (cid:11) n
(cid:12) (sinθS +cosθS ) θ = θ . (21)
x z
2
(cid:12) (cid:11) (cid:12) (cid:11)
Note that (cid:12) (cid:12)
n
θ S θ = sinθ (22a)
x
2
(cid:10) (cid:12) (cid:12) (cid:11) n 2
θ(cid:12)S2(cid:12)θ = sinθ +O(n) (22b)
x 2
(cid:16)n (cid:17)
(cid:10)θ(cid:12)S (cid:12)θ(cid:11) = cosθ (22c)
(cid:12) z(cid:12)
2
(cid:10) (cid:12) (cid:12) (cid:11) n 2
θ(cid:12)S2(cid:12)θ = cosθ +O(n) (22d)
z 2
(cid:10) (cid:12) (cid:12) (cid:11) (cid:16)n (cid:17)3
θ(cid:12)S3(cid:12)θ = cosθ +O(n2) (22e)
z 2
(cid:16) (cid:17)
(cid:10) (cid:12) (cid:12) (cid:11)
Now define (cid:12) (cid:12)
V(θ,s) = 2(1−s)(1−sinθ)+sg 1(1−cosθ) (23)
2
(cid:0) (cid:1)
6 E. Farhi, J. Goldstone, S. Gutmann
where g is given in (15). Using (16), (18), (20), and (22) we have
2 3
V(θ,s) = θ H(s) θ +O(1/n) (24)
n
(cid:16) (cid:17)
(cid:10) (cid:12) (cid:12) (cid:11)
where we have also used the operator relation(cid:12)we =(cid:12)n −S . The function V(θ,s), for each s,
2 z
represents a large-n “effective potential”. For n large, for each s, the ground state of H(s)
is well approximated by θ where θ minimizes V(θ,s).
m m
e
(cid:12) (cid:11)
(cid:12) s=0 s=.2 s=.4
2 2 1.8
1.6
1.5 1.5
e
v
cti 1.4
e 1 1
eff 1.2
V
0.5 0.5
1
0 0 0.8
0 1 2 3 0 1 2 3 0 1 2 3
theta
s=.4341 s=.46 s=1
1.8 1.8 2.5
1.6 1.6 2
e
v
cti1.4 1.4 1.5
e
eff1.2 1.2 1
V
1 1 0.5
0.8 0.8 0
0 1 2 3 0 1 2 3 0 1 2 3
theta
Figure 2: The effective potential versus θ changes shape as s increases.
In Figure 2 we plot V(θ,s) versus θ for s = 0,0.2,0.4,0.4341,0.46, and 1. At s = 0 the
minimum is at θ = π/2 corresponding to Hamming weight n/2. Because g′(1/2) < 0, this
minimum moves to θ bigger than π/2 for small s as can be seen in the s = 0.2 panel. As s
increases, a local minimum forms at values of θ less than π/2. There is a critical value of s,
s∗ = 0.4341, at which the two minima occurring at θ < π/2 and at θ > π/2 are degenerate;
1 2
that is, V(θ ,s∗) = V(θ ,s∗). For s > s∗ the global minimum is always at a value of θ less
1 2
than π/2. For s = 1 this global minimum is at θ = 0.
1
For the quantum adiabatic evolution algorithm to work, the quantum state of the system
should remain in (or very near) the instantaneous ground state of H(s) with s = t/T.
For s < s∗ the ground state is smoothly varying. However, at s = s∗ the ground state
changes from ≈ θ to ≈ θ . For the quantum system to go from eθ to θ requires
2 1 2 1
quantum tunneling through the barrier separating the minima at θ and θ . This takes a
(cid:12) (cid:11) (cid:12) (cid:11) 1 (cid:12) (cid:11) 2 (cid:12) (cid:11)
time exponentia(cid:12)l in n and(cid:12) implies that the gap of H(s∗) is exponent(cid:12)ially sm(cid:12)all. Using a
combination of standard large spin and instanton methods [4] we can estimate the dominant
exponentially small term. We obtain e
gap ∼ poly(n)e−S0n (25)
Quantum Adiabatic Evolution Algorithms versus Simulated Annealing 7
-3.5
-4
-4.5
p
a
g -5
e
h
of t
g -5.5
o
l
-6
-6.5
-7
60 65 70 75 80 85 90 95 100
number of bits
Figure 3: The log of the gap plotted versus n. The straight line behavior indicates that the gap is
decreasing exponentially in n.
where
θ2
S = 1 ϕ(θ)sinθdθ (26a)
0 2
Zθ1
and
V(θ,s∗)−V(θ ,s∗)
ϕ(θ) = cosh−1 1+ 1 . (26b)
2(1−s∗)sinθ
h i
Forq = 3, we find s∗ = 0.4341 and S = 0.2015. We can also determine the minimum gap
0
of H(s) by numerical diagonalization. Because of the symmetry of H and the initial state,
although the full Hilbert space is 2n dimensional, the evolution takes place in the (n+ 1)-
dimeensional subspace of total spin n/2. In Figure 3 we show the logeof the gap versus n for
n between 60 and 100. At n = 100 the minimum gap occurs at s = 0.4338. Fitting the gap
to the functional form Anpe−cn and finding the best values of A, p, and c gives c = 0.2021,
which certainly supports our approach. For q = 5 we have S = 0.4946 and c = 0.4966, and
0
for q = 7 we have S = 0.7008 and c = 0.6930.
0
Formula (25) is valid for any example where the cost function depends only on the
Hamming weight w and we can write
h(w) = nrg(w/n)+O(nr−1) (27)
where the global minimum of g(u) occurs at u < 1/2 and g′(1/2) < 0. This guarantees that
there is a value s∗ such that V(θ,s∗) given by (23) has degenerate minima at θ < π/2 and
1
θ > π/2. Then (26) can be used to calculate the coefficient of n in the exponent of (25).
2
8 E. Farhi, J. Goldstone, S. Gutmann
We now study the performance of simulated annealing for this kind of problem. Again
we start with a cost function h(z ,... ,z ), which is a sum of local terms but because of
1 n
symmetry only depends on the Hamming weight, so we can write the cost function as h(w).
Now for any temperature τ, the Boltzmann distribution (8) depends only on the Hamming
weight of the string w = z +z +···+z . The number of strings with Hamming weight w
1 2 n
is (n). We can define the (unnormalized) probability
w
n
p (w) = exp −h(w)/τ (28)
τ w
(cid:16) (cid:17)
(cid:0) (cid:1)
which gives the probability of finding a string with Hamming weight w at temperature τ in
the Boltzmann distribution.
We also define the “Free energy”, F(w,τ), by
n
exp −F(w,τ)/τ = exp −h(w)/τ . (29)
w
(cid:16) (cid:17)
(cid:0) (cid:1) (cid:0) (cid:1)
Note that for any τ, the most likely value of the Hamming weight occurs at the value of w
that minimizes F(w,τ). Now we further specialize to the case (13) so that (14) and (15)
apply. If we rescale the temperature and let a = 8τ/n2 then we can write
n 3
F(w,an2/8) = F(u,a)+O(n2) (30)
2
(cid:16) (cid:17)
where
F(u,a) = a (1−u)log(1−u)+ulogu +g(u) (31)
and again u = w/n. (cid:2) (cid:3)
For n large, for any value of the temperature τ = an2/8, the Boltzmann distribution
is peaked on those strings with Hamming weight nu where u minimizes (31). At infinite
temperature, the first term in (31) dominates and is minimized at u = 1/2. As a is decreased
the minimum moves to a value of u above 1/2. In Figure 4 we show F(u,a)/a versus u for
a = 100,10,3.4,3.0574,2.8, and 0.01. The similarity with Figure 2 is apparent. The critical
value of a, a∗ = 3.0574, is where the global minimum switches from a value of u above 1/2
to a value below. To follow the Boltzmann distribution, in the simulated annealing Markov
chain, the Hamming weight, nu, must change from the value corresponding to the larger
local minimum to the smaller while τ is approximately a∗n2/8.
This order-n change in the Hamming weight requires exponentially many steps: The sign
of the derivative ∂ F(u,a∗) determines whether the Markov chain is more likely to move left
∂u
or right. For large n, if in the steps of the annealing process the bits are chosen uniformly
at random, w is more likely to move to w +1 than to w −1 if ∂ F(u,a∗) < 0. This means
∂u
that climbing the hill separating the local minima is exponentially unlikely in polynomially
many steps.
As long as g(u) has a global minimum at u < 1/2 and g′(1/2) < 0 simulated annealing,
likequantumadiabaticevolution,willrequireatimeexponentialinntosucceed. Themethod
used in this section, tracking the local minima of the effective potential and the free energy,
can be used to show the similarity of the performance of quantum adiabatic evolution and
simulated annealing for many examples with a cost that depends on the Hamming weight
alone.
Quantum Adiabatic Evolution Algorithms versus Simulated Annealing 9
a=100 a=10 a=3.4
0 0.4
-0.1 0.1
a 0.3
y / -0.2 0
erg-0.3 -0.1 0.2
n
e E-0.4 -0.2 0.1
e-0.5 -0.3
Fr 0
-0.6 -0.4
-0.5 -0.1
0 0.5 1 0 0.5 1 0 0.5 1
u
a=3.0574 a=2.8 a=.01
0.5 0.6 250
0.4 0.5
a 200
y / 0.3 0.4
erg 0.3 150
n 0.2
e E 0.1 0.2 100
e 0.1
Fr 50
0 0
-0.1 -0.1 0
0 0.5 1 0 0.5 1 0 0.5 1
u
Figure 4: The scaled free energy F/a plotted as a function of the scaled Hamming weight changes
shape as the temperature decreases. Compare to Figure 2.
5 The Hamming weight with a spike
Here we consider the cost function
w, w 6= n/4
h(w) = (32)
(n, w = n/4 .
where again w is the Hamming weight of n bits and also n is taken to be a multiple of 4.
Note that this cost function cannot be written as a sum of terms each of which involves
only a few bits. The global minimum of h(w) is at w = 0 and there is a local minimum at
w = (n/4)+1.
It is easy to see that simulated annealing fails to find the global minimum in polynomial
time. For τ > 1/log3, the Boltzmann distribution concentrates at Hamming weights larger
than n/4. For τ < 1/log3 the Boltzmann distribution concentrates at Hamming weights
smaller than n/4. At temperatures ∼ 1/log3, according to the acceptance rule (10), a string
with Hamming weight (n/4)+1 will only flip to a string with Hamming weight n/4 with a
probability that is exponentially small in n. Thus a simulated annealing algorithm running
for only polynomial time gets hung up in the false minimum of h.
In quantum mechanics, it is well known that a low-energy particle can penetrate a high
potential barrier. If the barrier is thin, the penetration probability need not be small. This
turns out to be the case here. As in the examples of Section 4, since the cost is a function
only of the Hamming weight, the quantum evolution takes place in an (n+1)-dimensional
subspace. The gap can be calculated by explicit diagonalization (using a technique similar
10 E. Farhi, J. Goldstone, S. Gutmann
–2.8(cid:13)
–2.85(cid:13)(cid:13)
–2.9(cid:13)
n))(cid:13)–2.95(cid:13)(cid:13)
p(
a
g
g(
o –3(cid:13)
l
–3.05(cid:13)(cid:13)
–3.1(cid:13)
6.1(cid:13)(cid:13) 6.2(cid:13)(cid:13) 6.3(cid:13)(cid:13) 6.4(cid:13)(cid:13) 6.5(cid:13)(cid:13) 6.6(cid:13)(cid:13) 6.7(cid:13)(cid:13) 6.8(cid:13)(cid:13)
log(n)(cid:13)
Figure5: TheHammingweightwithaspike. Thestraightlinebehavioronthelog-logplotindicates
that the gap decreases polynomially in n.
to the one in [1], section 4.2). The result is that
gap(n) ≈ 1.35n−1/2 (33)
and, for n large, occurs at s = 0.366. We can also find the gap by numerical diagonalization;
see Figure 5, which plots loggap(n) versus log(n) for 500 ≤ n ≤ 848. The straight line fit
has slope −0.474, consistent with the predicted −0.5 in (33). For n = 848, the gap occurs
at 0.3675, close to the predicted 0.366.
6 The bush of implications
In the previous section we discussed an example with a cost function that treats all bits
symmetrically and depends only on the Hamming weight of the n bits. Here we describe an
example where the cost function depends on the Hamming weight of n bits, but also on the
value of a single additional bit. As in Section 4, we consider the effective potential and free
energy. In the example in this section, quantum adiabatic evolution succeeds in polynomial
time while simulated annealing does not.
We label the n+1 bits 0,1,2,... ,n. The cost function is given by
n
h(z ,z ,... ,z ) = z (1−z )+(1−z ) . (34)
0 1 n 0 i 0
i=1
X
The terms z (1 − z ) can be viewed as “imply” clauses, since they are 0 unless z = 1
0 i 0
and z = 0. The unique string with h = 0 has z = 1 and all z = 1. If z = 0, then
i 0 i 0
h = 1 no matter what the values of z ,... ,z are. If z = 1, then h equals n − w where
1 n 0