Barry C. SandersInstitute for Quantum Information Science, University of

Calgarywith G Ahokas (Calgary), D W Berry (Macquarie), R Cleve

(Waterloo),P Hoyer (Calgary), N Wiebe (Calgary)

Efficiently algorithm for universal quantum simulation

Quantum Information and Many Body Physics Workshop

University of British Columbia, 1 December 2007

Comm. Math. Phys. 270(2): 359 - 371 (March 2007) + New Work.

Simulating evolution: quantum state generation

ClassicalPreprocessor

ClassicalPreprocessor

Λ ≥sup H , &H , &&H3 ,K{ },dimH ,

ε ,T ,d = sparseness H( )

n (including ancillae) , %ti{ }

Background

t3/2

(d+1)2 n6

Lie-TrotterLie-Trotter

Graph ColouringGraph Colouring

t3/2

d2 n2

Lie-TrotterLie-Trotter

Graph ColouringGraph Colouring

t1+1/2k

d2 log*n

Lie-Trotter-Suzuki(kth order)

Lie-Trotter-Suzuki(kth order)

Deterministic Coin Tossing

Deterministic Coin Tossing

ATS 2003ATS 2003

Childs 2004Childs 2004

Our ResultsOur Results

Feynman 1982: Quantum Computer would efficientlysimulate dynamics of quantum systems.

Lloyd 1996: Formalized conjecture, assumed tensorproduct structure, showed efficient algorithm.

Optimal in t; nearly constant in n

t1+1/2k

d2 log*n

Lie-Trotter-Suzuki(kth order)

Lie-Trotter-Suzuki(kth order)

Deterministic Coin Tossing

Deterministic Coin Tossing

Our ResultsOur Results

log*n is the height of the smallesttower of powers of 2 that exceeds n:

%ψ

%ψ '

%U = %U jN

L %U j2%U j1

j1 j2 j3 j4

0L 0 0L 0

%U = %U jN

L %U j2%U j1

ψ = cl l∑

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)',max( ll=x

( )',min ll=y

%H x,y , colour(l , l ') = j

0 ,colour(l , l ') ≠ j

⎧⎨⎪

⎩⎪

x

y

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Hamiltonian H generates unitary: break up

H: sum of local Hamiltonians

Trotter (m=2): eiHt(eiH1t/2r eiH2t/r eiH1t/2r)r, HH1+H2.

Number of steps for quantum computer N t3/2.

Suzuki generalization of Trotter formula:

Suzuki proves for small :

H = Hii=1

m

∑

, pk = 4 −41/ 2k−1( )( )−1

5 terms5 terms

Lemma: Strict bound for Lie-Trotter-Suzuki

qk = 1−4pk'( )k'=2

k

∏

exp −it Hii=1

m

∑⎛⎝⎜

⎞⎠⎟− S2k −i

tr

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥

r

≤22m5k−1qkτ( )

2k+1

2k+1( )!r2k

12m5k−1qkτ / r ≤1,

32

2m5k−1qkτ( )2k+1

2k+1( )!r2k ≤1.

τ =t × max H j

ε ≤1 ≤4m5k −1qkτ

2k +1( )!62k

Theorem: Simulation cost nearly linear in time

Theorem:

Optimal choice of k:

Then

N ≤m52k mqkτ( )

1+1/2k

2 2k+1( )!ε⎡⎣ ⎤⎦1/2k

k ≈12

log5

mτε

⎛⎝⎜

⎞⎠⎟

N ≤4m2τ exp 2 log5 mτ / ε( )⎡⎣

⎤⎦

2

t=0

4

34

Xj = 0 1 1 0 1 0 0 1

Simulation time cannot be sublinear in t

Lemma (decomposition of H unknown)

€

∃ decomposition H = H j

j=1

m

∑ , with each H j 1- sparse,

such that m = 6d2, and each query to any H j can be

simulated by O log* n( ) queries to H.

Graph associated with Graph associated with HH

2

2

2

2

2

1

11

2

1

1

1

1

1

1

1

1

1

1

1

2

2

2

2

22

1

1

1

1 1

1

3

3

3

3

3

3

33

3

3

3

3

21

2

22

2

2

2

3

3

3

3

3

3

y1

yd

x :α1

αd

Connect x to yk (x) with an

edge of weight αk (x)

SymmetricallySymmetrically labeled graphs labeled graphs

2

1

2

1

3

1

31

1

2

1

2

1

3

2

1

1

2

1

1

3

2

1

3

13

1

3

3

3 2

3

2

1

2

3

3

1

32

3

3

2

2

22

2

31

2

3

2

3

1

1

1

2

2

Non-symmetric caseNon-symmetric caseModify labeling to be symmetric (with an overhead cost)

(a, b)We now have d

2 labels

instead of d labels, but a symmetric labeling

a bx y with x < y

x y

(1, 3)with z < y

with y < w

(1, 2)

(1, 3)

x y

z

w

1 32

1

1

3

Example:

Problem!

(a, b)

(1, 3)

(1, 2)

(1, 3)

Graph with monochromatic pathsGraph with monochromatic paths1

2

1

3

1

1

33

1

2

1

3

1

3

3

3

1

1

3

1

3

2

3

3

33

2

1

1

3 2

3

2

2

1

3

3

2

21

1

1

1

2

21

1

21

2

1

1

3

2

2

1

1

2

To break up the paths, we increase the number of colours

x

y

z

w

(a,b, x

(a,b, y

(a,b, z

(a,b, w

n bits

x

y

z

w

x′

y′

z′

w′

d 2 2n

colourslog(n)+1 bits

y′ (i, yi), where i = min{ j : yj zj}

Then y′ = (010,1)

Example: y = 01100101

z = 01001101

010

x < y < z < w

Note: still a valid coloring!x′ y′ & y′ z′ & z′ w′

“Deterministic coin-tossing” [Cole & Vishkin ’86]

Breaking up the paths IIBreaking up the paths II

x

y

z

w

(a,b, x

(a,b, y

(a,b, z

(a,b, w

n bits

x

y

z

w

x′

y′

z′

w′

x

y

z

w

x′′

y′′

z′′

w′′

d 2 2n

colorslog(n)+1 bits

log(log(n)+1)+1 bits

x

y

z

w

x′′′

y′′′

z′′′

w′′′

6 elements

...

...

...

...

O(log*(n)) iterations

Just 5 iterations for n 101037

Sketch of Proof:

# of Hj’s is m = 6d2. Need to call the black-box O(log*n) times for each Hj.

Substituting into theorem for upper bound on Nexp gives result.

Further Reading S. Lloyd, Science 273, 1073 (1996). R. P. Feynman, Int. J. Th.. Phys. 21, 467 (1982). D. Aharonov and A. Ta-Shma, Proc. ACM STOC, 20 (2003). M. Suzuki, Phys. Lett. A 146, 319 (1990); JMP 32, 400

(1991). A. Childs, Ph.D. Thesis, MIT (2004). R. Cole and U. Vishkin, Inform. and Control 70, 32 (1986). N. Linial, SIAM J. Comp. 21, 193 (1992). A. Childs, R. Cleve, E. Deotto, E. Farhi, S. Guttman, and D.

Spielman, Proc. ACM STOC, 59 (2003). R. Beals, H. Buhrman, R. Cleve, M. Mosca, and R. de Wolf, J.

ACM 48, 778 (2001). G. Ahokas, D. W. Berry, R. Cleve, and B. C. Sanders, Comm.

Math. Phys. 270(2): 359 - 371 (March 2007); quant-ph/0508139.