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Introduction to Introduction to Quantum Information ProcessingQuantum Information Processing

CS 467 / CS 667CS 467 / CS 667Phys 467 / Phys 767Phys 467 / Phys 767C&O 481 / C&O 681C&O 481 / C&O 681

Richard Cleve DC 3524cleve@cs.uwaterloo.ca

Course web site at: http://www.cs.uwaterloo.ca/~cleve/courses/cs467

Lecture 9 (2005)

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ContentsContents

• Loose ends in discrete log algorithm

• Universal sets of quantum gates

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• Loose ends in discrete log algorithm

• Universal sets of quantum gates

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Discrete log algorithm (I)Discrete log algorithm (I)Input: p (n-bit prime), g (generator of Z*p), a Z*p

Output: r Zp1 such that g r mod p = a

Example: p = 7, Z*7 = {1, 2, 3, 4, 5, 6} = {30, 32, 31, 34, 35, 33} (hence 3 is a generator of Z*7)

Define f : Zp1 Zp1 Z*p as f (x , y) = gx ay mod p

Then f (x1 , y1) = f (x2, y2) iff (x1, y1) (x2, y2) k (r, 1) (mod p 1)

fF

0

0F0

0

F†

F†

produces a random (s , t) such that

(s , t)(r, 1) 0 (mod p 1)

sr + t 0 (mod p 1)

(for some k)

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Discrete log algorithm (II)Discrete log algorithm (II)

fF

0

0F0

0

F†

F† produces a random (s , t) such that

sr + t 0 (mod p 1)

If gcd(s, p 1) = 1 then r can be computed as r = ts1 mod p 1

The probability that this occurs is (p 1)/ (p 1), where is Euler’s totient function

It is known that (N) = (N/ loglogN), which implies that the

above probability is at least (1/ loglog p) = (1/ log n)

Therefore, O( log n) repetitions are sufficient

… this is not bad—but things are actually better than that …

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Discrete log algorithm (III)Discrete log algorithm (III)We obtain a random (s , t) such that sr + t 0 (mod p 1)

Therefore, a constant number of repetitions suffices

Note that each s {0,…, p 2} occurs with equal probability

Therefore, if we run the algorithm twice: we obtain two independent samples s1, s2 {0,…,p 2}

Question: what is the probability that gcd(s1, s2) = 1?

prime

22prime

1 5401

1PrPr1qq

.q

s|qs|q

If it happens that gcd(s1, s2) = 1 then (by Euclid) there exist

integers a and b such that as1 + bs2 = 1 r = (at1 + bt2)

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Discrete log algorithm (IV)Discrete log algorithm (IV)Another loose end: our algorithm uses QFTs modulo p 1,

whereas we have only seen how to compute QFTs modulo 2n

A variation of our QFT algorithm would work for moduli of the form 3n, and, more generally, all smooth numbers (those that are products of “small” primes)

2113121

13963

12642

132

ωωωω1

ωωωω1

ωωωω1

ωωωω1

11111

1

)N()N()N(N

)N(

)N(

N

N

H4

8 4

816 4

32 16 8 4

H

HH

H

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Discrete log algorithm (V)Discrete log algorithm (V)

Shor just used a modulus close to p 1, and, using careful error-analysis, showed that this was good enough ...

In fact, for the case where p 1 is smooth, there already exist polynomial-time classical algorithms for discrete log!

It’s only the case where p 1 is not smooth that is interesting

There are also ways of attaining good approximations of QFTs for arbitrary moduli (which we won’t consider now)

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• Loose ends in discrete log algorithm

• Universal sets of quantum gates

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A universal set of gates (I)A universal set of gates (I)Main Theorem: any unitary operation U acting on k qubits

can be decomposed into O(4k) CNOT and one-qubit gates

Proof sketch (for a slightly worse bound of O(k24k)) :

We first show how to simulate a controlled-U, for any one-qubit unitary U

2β

2β

2α

2α

δ

δ

0

0

2θcos2θsin

2θsin2θcos

0

0

0

0/i

/i

/i

/i

i

i

e

e

//

//

e

e

e

e

Straightforward to show: every one-qubit unitary matrix can be expressed as a product of the form

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A universal set of gates (II)A universal set of gates (II)

• A B C = I• ei A X B X C = U, where

This can be used to show that, for every one-qubit unitary U,

there exist A, B, C, and , such that:

01

10X

The fact implies that

U A B C

Pwhere

λ0

01ie

P

Exercise: show how this follows

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A universal set of gates (III)A universal set of gates (III)

Controlled-U gates can also simulate controlled-controlled-V gates, for an arbitrary unitary one-qubit unitary V:

UV U U †

where V = U 2

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A universal set of gates (IV)A universal set of gates (IV)

Example: Toffoli gate“controlled-controlled-NOT”

In this case, the one-qubit gates can be:

4π0

01/ie

T

c (ab)

b

aa

b

c

11

11

2

1H

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A universal set of gates (V)A universal set of gates (V)

From the Toffoli gate, generalized Toffoli gates (which are controlled-controlled- ... -NOT gates) can be constructed:

a1a2a3

:akb1

:

c

c (a1 a2 ... ak)

bk2

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A universal set of gates (VI)A universal set of gates (VI)From generalized Toffoli gates, generalized controlled-U

gates (controlled-controlled- ... -U) can be constructed:

U0

U

0

1110

0100

000000

000000

00100000

00010000

00001000

00000100

00000010

00000001

UU

UU

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A universal set of gates (VII)A universal set of gates (VII)

10000000

01000000

00100000

000000

00001000

00000100

000000

00000001

1110

0100

UU

UU

The approach essentially enables any k-qubit operation of the simple form

to be computed with O(k2) CNOT and one-qubit gates

In a spirit similar to Gaussian elimination, any 2k2k unitary

matrix can be decomposed into a product of O(4k) of these

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A universal set of gates (VIII)A universal set of gates (VIII)

Thus, the set of all one-qubit gates and the CNOT gate are universal in that they can simulate any other gate set

This completes the proof sketch*

Question: is there a finite set of gates that is universal?

Answer 1: strictly speaking, no, because this results in only countably many quantum circuits, whereas there are uncountably many unitary operations on k qubits (for any k)

Actually we proved a slightly worse bound of O(k24k)

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ApproximatelyApproximately universal gate sets universal gate sets

Answer 2: yes, for universality in an approximate sense ...

To be continued ...

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