1. 1 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 467 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 3524 cleve@cs.uwaterloo.ca Course web site at: http://www.cs.uwaterloo.ca/~cleve Lecture 7 (2005)

2. 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. 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. 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. 5 Algorithm for phase estimation U  H H H 00 00 00  When  = 0. a 1 a 2 …a m : Therefore, applying the inverse of F M yields the digits of 

6. 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. 7 Arbitrary phases (I) What if  is not of the nice form  = 0. a 1 a 2 …a m ? Example:  = ⅓ = 0.0101010101010101… 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. 8 Arbitrary phases (II) geometric series! The amplitude of  y , for y = a is State is: Numerator: 1 e2ie2i lower bounded by 2  i  2 m (2/  ) > 4  2 m Denominator: 1 e2i2me2i2m 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. 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 = 0.01010101010101…, the outcome probabilities look roughly like this: 4242 0000000100100011010001010110011110001001101010111000110111101111 

10. 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. 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. 12 Order-finding algorithm (I) Define: U (an operation on m qubits) as: U  y  =  a y mod M  Define: Then

13. 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. 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. 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. 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 11 + 2211 + 22 H H H 00 00 00 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. 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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