1 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 467 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 3524 Course web site at: Lecture 7 (2005)
2 Contents Recap of phase estimation problem/algorithm How the algorithm works for general phases Recap of the order-finding problem/algorithm How to bypass the need for an eigenstate
3 Recap of phase estimation problem/algorithm How the algorithm works for general phases Recap of the order-finding problem/algorithm How to bypass the need for an eigenstate
4 Phase estimation problem U is a unitary operation on n qubits is an eigenvector of U, with eigenvalue e 2 i ( 0 ≤ < 1 ) Output: ( m -bit approximation) Input: black-box for U n qubits m qubits and a copy of
5 Algorithm for phase estimation U H H H 00 00 00 When = 0. a 1 a 2 …a m : Therefore, applying the inverse of F M yields the digits of
6 Recap of phase estimation problem/algorithm How the algorithm works for general phases Recap of the order-finding problem/algorithm How to bypass the need for an eigenstate
7 Arbitrary phases (I) What if is not of the nice form = 0. a 1 a 2 …a m ? Example: = ⅓ = … Let’s calculate what the previously-described procedure does: Let a / 2 m = 0. a 1 a 2 …a m be an m -bit approximation of , in the sense that = a / 2 m + , where | | ≤ 1 / 2 m+1 What is the amplitude of a 1 a 2 …a m ?
8 Arbitrary phases (II) geometric series! The amplitude of y , for y = a is State is: Numerator: 1 e2ie2i lower bounded by 2 i 2 m (2/ ) > 4 2 m Denominator: 1 e2i2me2i2m upper bounded by 2 Therefore, the absolute value of the amplitude of y is at least the quotient of (1/2 m ) (numerator/denominator), which is 2/
9 Arbitrary phases (III) Therefore, the probability of measuring an m -bit approximation of is always at least 4/ 2 0.4 For example, when = ⅓ = …, the outcome probabilities look roughly like this: 424
10 Recap of phase estimation problem/algorithm How the algorithm works for general phases Recap of the order-finding problem/algorithm How to bypass the need for an eigenstate
11 Order-finding problem Let M be an m -bit integer Def: Z M * = {x {1,2,…, M 1} : gcd( x,M ) = 1 } (a group) Def: ord M ( a ) is the minimum r > 0 such that a r = 1 ( mod M ) Order-finding problem: given a and M, find ord M ( a ) Example: Z 21 * = { 1,2,4,5,8,10,11,13,16,17,19,20 } The powers of 10 are: 1, 10, 16, 13, 4, 19, 1, 10, 16, … Therefore, ord 21 (10) = 6
12 Order-finding algorithm (I) Define: U (an operation on m qubits) as: U y = a y mod M Define: Then
13 Order-finding algorithm (II) U n qubits 2n qubits corresponds to the mapping: x y x a x y mod M Moreover, this mapping can be implemented with roughly O( n 2 ) gates The phase estimation algorithm yields a 2 n - bit estimate of 1/ r From this, a good estimate of r can be calculated by taking the reciprocal, and rounding off to the nearest integer Problem: how do we construct state 1 to begin with? Exercise: why are 2 n bits necessary and sufficient for this?
14 Recap of phase estimation problem/algorithm How the algorithm works for general phases Recap of the order-finding problem/algorithm How to bypass the need for an eigenstate
15 Bypassing the need for 1 (I) Let Any one of these could be used in the previous procedure, to yield an estimate of k/r, from which r can be extracted What if k is chosen randomly and kept secret? Can still uniquely determine k and r, provided they have no common factors (and O(log n ) trials suffice for this)
16 Bypassing the need for 1 (II) Returning to the phase estimation problem, suppose that 1 and 2 have respective eigenvalues e 2 i 1 and e 2 i 2, and that 1 1 + 2 2 is used in place of an eigenvalue: U 11 + 2211 + 22 H H H 00 00 00 F†MF†M What will the outcome be? It will be an estimate of 1 with probability | 1 | 2 2 with probability | 2 | 2
17 Bypassing the need for 1 (III) Using the state Is it hard to construct the state ? yields results equivalent to choosing a k at random In fact, it’s easy, since This is how the previous requirement for 1 is bypassed
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