1. 1 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 10 (2008)

2. 2 Order-finding via eigenvalue estimation

3. 3 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 Note: no classical polynomial-time algorithm is known for this problem

4. 4 Order-finding algorithm (1) Define: U (an operation on m qubits) as: U  y  =  a y mod M  Define: Then

5. 5 Order-finding algorithm (2) 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?

6. 6 Bypassing the need for   1  (1) 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)

7. 7 Bypassing the need for   1  (2) 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

8. 8 Bypassing the need for   1  (3) 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

9. 9 Quantum algorithm for order-finding U a,M H H H 00 00 00 00 00 11 H H 44 44 88 H measure these qubits and apply continued fractions* algorithm to determine a quotient, whose denominator divides r U a,M  y  =  a y mod M  inverse QFT * For a discussion of the continued fractions algorithm, please see Appendix A4.4 in [Nielsen & Chuang] Number of gates for a constant success probability is: O(n 2 log n loglog n)

10. 10 Reduction from factoring to order-finding

11. 11 The integer factorization problem Input: M ( n -bit integer; we can assume it is composite) Output: p, q (each greater than 1 ) such that pq = N Note 2: given any efficient algorithm for the above, we can recursively apply it to fully factor M into primes* efficiently * A polynomial-time classical algorithm for primality testing exists Note 1: no efficient (polynomial-time) classical algorithm is known for this problem

12. 12 Factoring prime-powers There is a straightforward classical algorithm for factoring numbers of the form M = p k, for some prime p What is this algorithm? Therefore, the interesting remaining case is where M has at least two distinct prime factors

13. 13 Proposed quantum algorithm (repeatedly do): 1.randomly choose a  {2, 3, …, M  1} 2.compute g = gcd (a,M ) 3. if g > 1 then output g, M/g else compute r = ord M ( a ) (quantum part) if r is even then compute x = a r /2  1 mod M compute h = gcd (x,M ) if h > 1 then output h, M /h Numbers other than prime-powers so M | (a r /2 +1 )(a r /2  1 ) we have M | a r  1 thus, either M | a r /2 +1 or gcd (a r /2 +1,M ) is a nontrivial factor of M latter event occurs with probability  ½ Analysis:

14. 14 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 11 (2008)

15. 15 Universal sets of gates

16. 16 A universal set of gates (1) Main Theorem: any unitary operation U acting on k qubits can be decomposed into O( 4 k ) CNOT and one-qubit gates Proof sketch (for a slightly worse bound of O(k 2 4 k ) ) : We first show how to simulate a controlled- U, for any one- qubit unitary U Straightforward to show: every one-qubit unitary matrix can be expressed as a product of the form

17. 17 A universal set of gates (2) A B C = I e i 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: The fact implies that UCBA P where  Exercise: show how this follows

18. 18 A universal set of gates (3) Controlled- U gates can also simulate controlled-controlled- V gates, for an arbitrary unitary one-qubit unitary V : UVUU †U †  where V = U 2

19. 19 A universal set of gates (4) Example: Toffoli gate “controlled-controlled-NOT” In this case, the one-qubit gates can be:  c  (a  b)  bb aaaa bb cc

20. 20 A universal set of gates (5) From the Toffoli gate, generalized Toffoli gates (which are controlled-controlled-... -NOT gates) can be constructed:  a1a1 a2a2 a3a3 : akak b1b1 : cc  c  ( a 1  a 2 ...  a k )  bk2bk2

21. 21 A universal set of gates (6) From generalized Toffoli gates, generalized controlled- U gates (controlled-controlled-... - U ) can be constructed: U 00 U  00

22. 22 A universal set of gates (7) The approach essentially enables any k -qubit operation of the simple form to be computed with O(k 2 ) CNOT and one-qubit gates In a spirit similar to Gaussian elimination, any 2 k  2 k unitary matrix can be decomposed into a product of O(4 k ) of these

23. 23 A universal set of gates (8) 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(k 2 4 k )

24. 24 Approximately universal gate sets Answer 2: yes, for universality in an approximate sense... To be continued...

25. 25 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 12 (2008)

26. 26 Approximately universal sets of gates

27. 27 Universal gate sets The set of all one-qubit gates and the CNOT gate are universal in that they can simulate any other gate set 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 ) Quantitatively, any unitary operation U acting on k qubits can be decomposed into O( 4 k ) CNOT and one-qubit gates

28. 28 Approximately universal gate sets Answer 2: yes, for universality in an approximate sense As an illustrative example, any rotation can be approximated within any precision by repeatedly applying some number of times In this sense, R is approximately universal for the set of all one-qubit rotations: any rotation S can be approximated within precision  by applying R a suitable number of times It turns out that O ( ( 1 /  ) c ) times suffices (for a constant c )

29. 29 Approximately universal gate sets Theorem 2: the gates CNOT, H, and are approximately universal, in that any unitary operation on k qubits can be simulated within precision  by applying O ( 4 k log c (1 /  ) ) of them ( c is a constant) [Solovay, 1996][Kitaev, 1997] In three or more dimensions, the rate of convergence with respect to  can be exponentially faster

30. 30 Complexity classes

31. 31 Complexity classes P (polynomial time): problems solved by O( n c ) -size classical circuits (decision problems and uniform circuit families) BPP (bounded error probabilistic polynomial time): problems solved by O( n c ) -size probabilistic circuits that err with probability  ¼ BQP (bounded error quantum polynomial time): problems solved by O( n c ) -size quantum circuits that err with probability  ¼ PSPACE (polynomial space): problems solved by algorithms that use O ( n c ) memory. Recall:

32. 32 Summary of previous containments P  BPP  BQP  PSPACE  EXP P BPP BQP PSPACE EXP We now consider further structure between P and PSPACE Technically, we will restrict our attention to languages (i.e. {0,1} -valued problems) Many problems of interest can be cast in terms of languages For example, we could define FACTORING = { ( x, y ) :  2  z  y, such that z divides x}

33. 33 NP Define NP (non-deterministic polynomial time) as the class of languages whose positive instances have “witnesses” that can be verified in polynomial time Example: Let 3-CNF-SAT be the language consisting of all 3-CNF formulas that are satisfiable is satisfiable iff there exists such that But poly-time verifiable witnesses exist (namely, b 1,..., b n ) 3-CNF formula: No sub-exponential-time algorithm is known for 3-CNF-SAT

34. 34 Other “logic” problems in NP k -DNF-SAT: CIRCUIT-SAT:      0 0 1 1 1  Λ Λ  Λ Λ Λ  Λ Λ  Λ Λ Λ Λ Λ Λ Λ  output bit  But, unlike with k -CNF-SAT, this one is known to be in P  All known algorithms exponential- time

35. 35 “Graph theory” problems in NP k -COLOR: does G have a k -coloring ? k -CLIQUE: does G have a clique of size k ? HAM-PATH: does G have a Hamiltonian path? EUL-PATH: does G have an Eulerian path?

36. 36 “Arithmetic” problems in NP FACTORING = { ( x, y ) :  2  z  y, such that z divides x} SUBSET-SUM: given integers x 1, x 2,..., x n, y, do there exist i 1, i 2,..., i k  { 1, 2,..., n} such that x i 1 + x i 2 +... + x i k = y ? INTEGER-LINEAR-PROGRAMMING: linear programming where one seeks an integer-valued solution (its existence)

37. 37 P vs. NP If a polynomial-time algorithm is discovered for 3-CNF-SAT then a polynomial-time algorithm for 3-COLOR easily follows All of the aforementioned problems have the property that they reduce to 3-CNF-SAT, in the sense that a polynomial- time algorithm for 3-CNF-SAT can be converted into a poly- time algorithm for the problem algorithm for 3-CNF-SAT algorithm for 3-COLOR Example: In fact, this holds for any problem X  NP, hence 3-CNF-SAT is NP-hard...

38. 38 P vs. NP If a polynomial-time algorithm is discovered for 3-CNF-SAT then a polynomial-time algorithm for 3-COLOR easily follows All of the aforementioned problems have the property that they reduce to 3-CNF-SAT, in the sense that a polynomial- time algorithm for 3-CNF-SAT can be converted into a poly- time algorithm for the problem algorithm for 3-CNF-SAT algorithm for X Example: In fact, this holds for any problem X  NP, hence 3-CNF-SAT is NP-hard... Also NP-hard: CIRCUIT-SAT, k -COLOR,...

39. 39 FACTORING vs. NP 3-CNF-SAT FACTORING P NP PSPACE co-NP Is FACTORING NP-hard too? But FACTORING has not been shown to be NP-hard Moreover, there is “evidence” that it is not NP-hard: FACTORING  NP  co-NP If so, then every problem in NP is solvable by a poly-time quantum algorithm! If FACTORING is NP-hard then NP = co-NP

40. 40 FACTORING vs.NP FACTORING vs. co-NP P NP PSPACE co-NP FACTORING FACTORING = { ( x, y ) :  2  z  y, s.t. z divides x} Question: what is a good witness for the negative instances? co-NP: languages whose negative instances have “witnesses” that can be verified in poly-time Answer: the prime factorization p 1, p 2,..., p m of x will work Can verify primality and compare p 1, p 2,..., p m with y, all in poly-time

41. 41 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 13 (2008)

42. 42 More state distinguishing problems

43. 43 More state distinguishing problems Which of these states are distinguishable? Divide them into equivalence classes: 1.  0  +  1  2. −  0  −  1  3.  0  with prob. ½  1  with prob. ½ 4.  0  +  1  with prob. ½  0  −  1  with prob. ½ 5.  0  with prob. ½  0  +  1  with prob. ½ 6.  0  with prob. ¼  1  with prob. ¼  0  +  1  with prob. ¼  0  −  1  with prob. ¼ 7. The first qubit of  01  −  10  This is a probabilistic mixed state Answers later on...

44. 44 Density matrix formalism

45. 45 Density matrices (1) Until now, we’ve represented quantum states as vectors (e.g.  ψ , and all such states are called pure states) An alternative way of representing quantum states is in terms of density matrices (a.k.a. density operators) The density matrix of a pure state  ψ  is the matrix  =  ψ   ψ  Example: the density matrix of  0  +  1  is

46. 46 Density matrices (2) Effect of a unitary operation on a density matrix: applying U to  yields U  U † Effect of a measurement on a density matrix: measuring state  with respect to the basis  1 ,  2 ,...,  d , yields the k th outcome with probability  k   k  How do quantum operations work using density matrices? ( this is because the modified state is U  ψ   ψ  U † ) ( this is because  k   k  =  k  ψ   ψ  k  =   k  ψ   2 ) —and the state collapses to  k   k 

47. 47 Density matrices (3) A probability distribution on pure states is called a mixed state: ( (  ψ 1 , p 1 ), (  ψ 2 , p 2 ), …, (  ψ d , p d ) ) The density matrix associated with such a mixed state is: Example: the density matrix for ( (  0 , ½ ), (  1 , ½ ) ) is: Question: what is the density matrix of ( (  0  +  1 , ½ ), (  0  −  1 , ½ ) ) ?

48. 48 Density matrices (4) Effect of a unitary operation on a density matrix: applying U to  still yields U  U † How do quantum operations work for these mixed states? This is because the modified state is: Effect of a measurement on a density matrix: measuring state  with respect to the basis  1 ,  2 ,...,  d , still yields the k th outcome with probability  k   k  Why?

49. 49 Recap: density matrices Applying U to  yields U  U † Measuring state  with respect to the basis  1 ,  2 ,...,  d , yields: k th outcome with probability  k   k  —and causes the state to collapse to  k   k  Quantum operations in terms of density matrices: Since these are expressible in terms of density matrices alone (independent of any specific probabilistic mixtures), states with identical density matrices are operationally indistinguishable

50. 50 Return to state distinguishing problems …

51. 51 State distinguishing problems (1) The density matrix of the mixed state ( (  ψ 1 , p 1 ), (  ψ 2 , p 2 ), …, (  ψ d , p d ) ) is: 1. & 2.  0  +  1  and −  0  −  1  both have 3.  0  with prob. ½  1  with prob. ½ 4.  0  +  1  with prob. ½  0  −  1  with prob. ½ 6.  0  with prob. ¼  1  with prob. ¼  0  +  1  with prob. ¼  0  −  1  with prob. ¼ Examples (from beginning of lecture):

52. 52 State distinguishing problems (2) 5.  0  with prob. ½  0  +  1  with prob. ½ 7. The first qubit of  01  −  10  Examples (continued): has:... ? (later)

53. 53 Characterizing density matrices Three properties of  : Tr  = 1 ( Tr M = M 11 + M 22 +... + M dd )  =  † ( i.e.  is Hermitian)    0, for all states  Moreover, for any matrix  satisfying the above properties, there exists a probabilistic mixture whose density matrix is  Exercise: show this

54. 54 Taxonomy of various normal matrices

55. 55 Normal matrices Definition: A matrix M is normal if M † M = MM † Theorem: M is normal iff there exists a unitary U such that M = U † DU, where D is diagonal (i.e. unitarily diagonalizable) Examples of abnormal matrices: is not even diagonalizable is diagonalizable, but not unitarily eigenvectors:

56. 56 Unitary and Hermitian matrices with respect to some orthonormal basis Normal: Unitary: M † M = I which implies | k | 2 = 1, for all k Hermitian: M = M † which implies k  R, for all k Question: which matrices are both unitary and Hermitian? Answer: reflections ( k  {+1,  1}, for all k )

57. 57 Positive semidefinite Positive semidefinite: Hermitian and k  0, for all k Theorem: M is positive semidefinite iff M is Hermitian and, for all ,  M   0 ( Positive definite: k > 0, for all k )

58. 58 Projectors and density matrices Projector: Hermitian and M 2 = M, which implies that M is positive semidefinite and k  {0,1}, for all k Density matrix: positive semidefinite and Tr M = 1, so Question: which matrices are both projectors and density matrices? Answer: rank-1 projectors ( k = 1 if k = j ; otherwise k = 0 )

59. 59 Taxonomy of normal matrices normal unitaryHermitian reflection positive semidefinite projector density matrix rank one projector

60. 60 Introduction to Quantum Information Processing CS 467 / CS 667 Phys 667 / Phys 767 C&O 481 / C&O 681 Richard Cleve DC 2117 cleve@cs.uwaterloo.ca Lecture 14 (2008)

61. 61 Bloch sphere for qubits

62. 62 Bloch sphere for qubits (1) Consider the set of all 2x2 density matrices  Note that the coefficient of I is ½, since X, Y, Y are traceless They have a nice representation in terms of the Pauli matrices: Note that these matrices—combined with I —form a basis for the vector space of all 2x2 matrices We will express density matrices  in this basis

63. 63 Bloch sphere for qubits (2) We will express First consider the case of pure states  , where, without loss of generality,  = cos (  )  0  + e 2i  sin (  )  1  ( ,   R ) Therefore c z = cos (2  ), c x = cos (2  ) sin (2  ), c y = sin (2  ) sin (2  ) These are polar coordinates of a unit vector ( c x, c y, c z )  R 3

64. 64 Bloch sphere for qubits (3) ++ 00 11 –– +i+i –i–i +i = 0 + i1+i = 0 + i1 –i = 0 – i1–i = 0 – i1 – = 0 – 1– = 0 – 1 + = 0 +1+ = 0 +1 Pure states are on the surface, and mixed states are inside (being weighted averages of pure states) Note that orthogonal corresponds to antipodal here

65. 65 Basic properties of the trace

66. 66 Basic properties of the trace The trace of a square matrix is defined as It is easy to check that The second property implies and Calculation maneuvers worth remembering are: and Also, keep in mind that, in general,

67. 67 Partial trace (1) In such circumstances, if the second register (say) is discarded then the state of the first register remains  Two quantum registers (e.g. two qubits) in states  and  (respectively) are independent if then the combined system is in state  =    In general, the state of a two-register system may not be of the form    (it may contain entanglement or correlations) We can define the partial trace, Tr 2 , as the unique linear operator satisfying the identity Tr 2 (    ) =  For example, it turns out that Tr 2 () =) = index means 2 nd system traced out

68. 68 Partial trace (2) Example: discarding the second of two qubits Let A 0 = I   0  and A 1 = I   1  For the resulting quantum operation, state    becomes  For d -dimensional registers, the operators are A k = I   k , where  0 ,  1 , …,  d  1  are an orthonormal basis

69. 69 Partial trace (3) For 2-qubit systems, the partial trace is explicitly and

70. 70 POVMs (Positive Operator Valued Measurements)

71. 71 POVMs (1) Positive operator valued measurement (POVM): Let A 1, A 2, …, A m be matrices satisfying Then the corresponding POVM is a stochastic operation on  that, with probability produces the outcome: j (classical information) (the collapsed quantum state) Example 1: A j =  j  j  (orthogonal projectors) This reduces to our previously defined measurements …

72. 72 POVMs (2) Moreover, When A j =  j  j  are orthogonal projectors and  = , = Tr  j  j  j  j  =  j  j  j  j  =  j  2

73. 73 POVMs (3) Example 3 (trine state “measurent”): Let  0  =  0 ,  1  =  1/2  0  +  3/2  1 ,  2  =  1/2  0    3/2  1  Then If the input itself is an unknown trine state,  k  k , then the probability that classical outcome is k is 2/3 = 0.6666… Define A 0 =  2/3  0  0  A 1 =  2/3  1  1  A 2 =  2/3  2  2 

74. 74 General quantum operations

75. 75 General quantum operations (1) Example 1 (unitary op): applying U to  yields U  U † General quantum operations (a.k.a. “completely positive trace preserving maps”, “admissible operations” ): Let A 1, A 2, …, A m be matrices satisfying Then the mappingis a general quantum op

76. 76 General quantum operations (2) Example 2 (decoherence): let A 0 =  0  0  and A 1 =  1  1  This quantum op maps  to  0  0  0  0  +  1  1  1  1  Corresponds to measuring  “ without looking at the outcome” For  ψ  =  0  +  1 , After looking at the outcome,  becomes  0  0  with prob. |  | 2  1  1  with prob. |  | 2

77. 77 General quantum operations (3) Example 4 (discarding the second of two qubits): Let A 0 = I   0  and A 1 = I   1  State    becomes  State becomes Note 1: it’s the same density matrix as for ( (½,  0  ), (½,  1  ) ) Note 2: the operation is the partial trace Tr 2 

78. 78 Distinguishing mixed states

79. 79 Distinguishing mixed states (1)  0  with prob. ½  0  +  1  with prob. ½  0  with prob. ½  1  with prob. ½  0  with prob. cos 2 (  /8)  1  with prob. sin 2 (  /8) 00 ++ 00 11  0  with prob. ½  1  with prob. ½ What’s the best distinguishing strategy between these two mixed states?  1 also arises from this orthogonal mixture: … as does  2 from:

80. 80 Distinguishing mixed states (2) 00 ++ 00 11 We’ve effectively found an orthonormal basis   0 ,   1  in which both density matrices are diagonal: Rotating   0 ,   1  to  0 ,  1  the scenario can now be examined using classical probability theory: Question: what do we do if we aren’t so lucky to get two density matrices that are simultaneously diagonalizable? Distinguish between two classical coins, whose probabilities of “heads” are cos 2 (  /8) and ½ respectively (details: exercise) 11

81. 81 Simulations among operations

82. 82 Simulations among operations (1) Fact 1: any general quantum operation can be simulated by applying a unitary operation on a larger quantum system: U 00 00 00   Example: decoherence 00  0  +  1  output discard input

83. 83 Simulations among operations (2) Fact 2: any POVM can also be simulated by applying a unitary operation on a larger quantum system and then measuring: U 00 00 00   quantum output input classical output j

84. 84 Separable states

85. 85 Separable states product state if  =  separable state if A bipartite (i.e. two register) state  is a: Question: which of the following states are separable? (i.e. a probabilistic mixture of product states) ( p 1,…, p m  0 )