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quantum

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· | i | i ··· · | i | i ···
·( |0 i+e2πi0.j1j2 ···jn |1 i). ·( |0 i+e2πi0.ϕ1ϕ1 ···ϕb |1 i),
It can be easily checked from the product represen- whichisthequantumFouriertransformoftheprod-
tation that with quantum parallelism the quantum uct state ϕ ϕ ϕ .
1 2 b
| ··· i
Fourier transform can be realized as a quantum cir- The second stage of phase estimation is to take
cuitwithonlyO(n2)operations,whileclassicallythe the inverse quantum Fourier transform on the first
fast Fourier transform requires O(n2n) operations register. For ϕ=0.ϕ ϕ , the output state from
1 b
···
for processing 2n data, which indicates an exponen- thesecondstage is ϕ ϕ ϕ ,andameasurement
1 2 b
| ··· i
tial speed-up (Nielsen and Chuang (2000)). Realiz- in the computational basis yields ϕ ϕ and di-
1 b
···
ing such an exponential saving accommodated by viding the measurement by 2b gives ϕ ϕ /2b =
1 b
···
quantum parallelism requires clever measurement 0.ϕ ϕ =ϕ. We obtain a perfect estimate of ϕ.
1 b
···
schemes.Successfulexamplesincludequantumphase Now we consider the case that ϕ cannot be ex-
estimation and Shor’s algorithms for factoring and pressed exactly with a b bit binary expansion. Take
discrete logarithm. 0 η < 2b to be the integer that its binary frac-
≤
tion η/2b =0.η η η is the first b bit represen-
4.2 Phase Estimation 1 2 ··· b
tation in the binary expansion of ϕ, which satisfies
Quantum algorithms are of random nature in the 0 ϕ η/2b 2 b.
−
≤ − ≤
sense that they are able to produce correct answers Perform the inverse quantum Fourier transform
only with some probabilities. Consider quantum on the first register given by (11), which is obtained
phase estimation which provides the key to many in the first stage results, and get
quantum algorithms. Assume that a unitary opera-
Tto hr eU phh aa ss ea ϕn oe fig te hn eve ec igto er ax uewi it sh ue ni kg nen ov wa nlu ae ne d2π ti hϕ e. 21 2b −1 i=2b −1
nv| li e −2πikℓ/2b e2πiϕk |ℓ β |ℓ i,
b ℓ
kX,ℓ=0 Xℓ=0
goal of the phase estimation algorithm is to esti-
mate ϕ based on the assumption that the state x where amplitudes of (η+ℓ)(mod2b) are
controlled-U2j operatio| nsi | i
can be prepared and the
2b 1
[see Section 3.2 for control gate] can be performed β = 1 − e2πi[ϕ (η+ℓ)/2b] k
for suitable nonnegative integers j. ℓ 2b { − }
Xk=0
The registers are used in phase estimation. The
firstregister consists of b qubitsinitially in the state 1 1 e2πi(2bϕ η ℓ)
− −
= − .
0 . The second register starts in the state x and 2b(cid:18)1 e2πi(ϕ η/2b ℓ/2b)(cid:19)
| i | i − −
−
13
QUANTUMCOMPUTATIONAND QUANTUMINFORMATION
Assume that the result of the final measurement are correct with probability 1 ǫ and incorrect with
−
fromphaseestimationisη˜anddividingtheresultby probability ǫ.
2b gives ϕ˜=η˜/2b.Letζ bethespecifiedaccuracyfor If the outcome result of the algorithm is verifi-
the phase estimation procedure. By adding up β 2 able to be a correct answer or not [as in the case
ℓ
| |
with ℓ being within ζ2b, we bound the probability of Shor’s algorithms for factoring and order-finding
that the obtained ϕ˜ is within ζ from ϕ: in Section 4.4 below], the obtained result from each
P(ϕ˜ ϕ ζ) P(η˜ η ζ2b 1) run is checked to be a correct answer or not. Then
| − |≤ ≥ | − |≤ − the number of times required to run the algorithm
1
for obtaining a correct answer follows a geometric
1 .
≥ − 2(ζ2b 2) distribution. Thus the probability that we obtain
−
For ǫ>0, set a correct answer in n repetitions is equal to
1 1 P(obtain a correct answer in n trials)
(13) b= log + log 2+ .
(cid:20) 2(cid:18)ζ(cid:19)(cid:21) (cid:20) 2(cid:18) 2ǫ(cid:19)(cid:21) =1 P(no success in the n trials)=1 ǫn.
− −
ThenP(ϕ˜ ϕ ζ) 1 ǫ,thatis,withprobability Since ǫn goes to zero geometrically fast, we may
| − |≤ ≥ −
at least 1 ǫ the phase estimation procedure can
choose a moderate ǫ with fewer qubits to achieve
−
successfully produce ϕ˜ within ζ from the true ϕ.
veryhighprobabilityofsuccessfullyobtainingacor-
See Nielsen and Chuang (2000).
rect answer by repeatedly running the algorithm
4.3 Statistical Analysis enough times.
On the other hand, if the outcome result is not
The phase estimation algorithm requires b qubits
verifiable to be a correct answer or not [as in the
forthefirstregistertoachieve [ log ζ]bitaccuracy
− 2 case of phase estimation], careful analysis is needed
and success probability 1 ǫ. With accuracy fixed,
to design ways for obtaining a correct answer with
−
to increase the success probability the required qu-
veryhighprobability.Aswronganswersarefaraway
bits