1 Introduction to Quantum Information Processing CS 667 / Phys 767 / C&O 681 Richard Cleve DC 2117 Lecture 15 (2008)

2 Continuous-time evolution

3 Although we’ve expressed quantum operations in discrete terms, in real physical systems, the evolution is continuous 00 11 Let H be any Hermitian matrix and t  R Then e iHt is unitary – why? H = U † DU, where Therefore e iHt = U † e iDt U = (unitary)

4 Grover’s quantum search algorithm

5 Quantum search problem Given: a black box computing f : {0,1} n  {0,1} Goal: determine if f is satisfiable (if  x  {0,1} n s.t. f(x) = 1 ) In positive instances, it makes sense to also find such a satisfying assignment x Classically, using probabilistic procedures, order 2 n queries are necessary to succeed—even with probability ¾ (say) x1x1 xnxn yy xnxn x1x1  y  f ( x 1,...,x n )  UfUf Grover’s quantum algorithm that makes only O(  2 n ) queries Query: [Grover ’96]

6 Applications of quantum search The function f could be realized as a 3-CNF formula: Alternatively, the search could be for a certificate for any problem in NP 3-CNF-SAT FACTORING P NP PSPACE co-NP The resulting quantum algorithms appear to be quadratically more efficient than the best classical algorithms known

7 Prelude to Grover’s algorithm: two reflections = a rotation  11 22 11 22 Consider two lines with intersection angle  : reflection 1 reflection 2 Net effect: rotation by angle 2 , regardless of starting vector

8 Grover’s algorithm: description I x1x1 xnxn yy xnxn x1x1  y  [ x = ]  U0U0 x1x1 xnxn yy xnxn x1x1  y  f ( x 1,...,x n )  UfUf U f  x  = (  1) f(x)  x  H H H H X X X X X X Hadamard Basic operations used: Implementation? U 0  x  = (  1) [ x = 0...0]  x 

9 Grover’s algorithm: description II 1.construct state H   2.repeat k times: apply  H U 0 H U f to state 3. measure state, to get x  {0,1} n, and check if f (x) =1 00 00   UfUf HHU0U0 HUfUf HU0U0 H iteration 1 iteration 2... (The setting of k will be determined later)

10 Grover’s algorithm: analysis I and N = 2 n and a = | A | and b = | B | Let A = { x  {0,1} n : f (x) = 1 } and B = { x  {0,1} n : f (x) = 0 } Letand BB AA H   Consider the space spanned by  A  and  B  Interesting case: a << N  goal is to get close to this state

11 Grover’s algorithm: analysis II Algorithm: (  H U 0 H U f ) k H   BB AA H   Observation: U f is a reflection about  B  : U f  A  =   A  and U f  B  =  B  Question: what is  H U 0 H ? Partial proof:  H U 0 H H   =  H U 0   =  H (    ) = H   U 0 is a reflection about H  

12 Grover’s algorithm: analysis III Algorithm: (  H U 0 H U f ) k H   BB AA H    22 22 22 22 Since  H U 0 H U f is a composition of two reflections, it is a rotation by 2 , where sin (  )=  a/N   a/N When a = 1, we want (2k+1)(1/  N)   / 2, so k  (  / 4)  N More generally, it suffices to set k  (  / 4)  N/a Question: what if a is not known in advance?

13 Introduction to Quantum Information Processing CS 667 / Phys 767 / C&O 681 Richard Cleve DC 2117 Lecture 17 (2008)

14 Optimality of Grover’s algorithm

15 Optimality of Grover’s algorithm Proof (of a slightly simplified version): f U0U0 U1U1 U2U2 U3U3 UkUk fff and that a k -query algorithm is of the form where U 0, U 1, U 2,..., U k, are arbitrary unitary operations Theorem: any quantum search algorithm for f : {0,1} n  {0,1} must make  (  2 n ) queries to f (if f is used as a black-box) f |x|x (  1) f(x) | x  Assume queries are of the form | 

16 Optimality of Grover’s algorithm Define f r : {0,1} n  {0,1} as f r ( x ) = 1 iff x = r frfr U0U0 U1U1 U2U2 U3U3 UkUk frfr frfr frfr |0|0 | ψ r,k  Consider versus I U0U0 U1U1 U2U2 U3U3 UkUk III |0|0 | ψ r, 0  We’ll show that, averaging over all r  {0,1} n, || | ψ r,k   | ψ r, 0  ||  2 k /  2 n

17 Optimality of Grover’s algorithm Consider I U0U0 U1U1 U2U2 U3U3 UkUk Ifrfr frfr |0|0 | ψ r,i  k  ik  i i | ψ r,k   | ψ r, 0  = ( | ψ r,k   | ψ r,k  1  ) + ( | ψ r,k  1   | ψ r,k  2  ) ( | ψ r,1   | ψ r,0  ) Note that || | ψ r,k   | ψ r, 0  ||  || | ψ r,k   | ψ r,k  1  || || | ψ r,1   | ψ r,0  || which implies

18 Optimality of Grover’s algorithm I U0U0 U1U1 U2U2 U3U3 UkUk Ifrfr frfr |0|0 | ψ r,i  query i query i +1 || | ψ r,i   | ψ r,i -1  || = | 2  i,r |, since query only negates | r  I U0U0 U1U1 U2U2 U3U3 UkUk IIfrfr |0|0 query i query i +1 | ψ r,i-1  Therefore, || | ψ r,k   | ψ r, 0  || 

19 Optimality of Grover’s algorithm Therefore, for some r  {0,1} n, the number of queries k must be  (  2 n ), in order to distinguish f r from the all-zero function Now, averaging over all r  {0,1} n, (By Cauchy-Schwarz) This completes the proof

20 Introduction to Quantum Information Processing CS 667 / Phys 767 / C&O 681 Richard Cleve DC 2117 Lecture 18 (2008)

21 Lab tour (instead of a regular lecture)

22 Introduction to Quantum Information Processing CS 667 / Phys 767 / C&O 681 Richard Cleve DC 2117 Lecture 19 (2008)

23 Preliminary remarks about quantum communication

24 How does quantum information affect the communication costs of information processing tasks? Quantum information can apparently be used to substantially reduce computation costs for a number of interesting problems We explore this issue...

25 Entanglement and signaling Recall that Entangled states, such as, Any operation performed on one system has no affect on the state of the other system (its reduced density matrix) qubit can be used to perform some intriguing feats, such as teleportation and superdense coding —but they cannot be used to “signal instantaneously”

26 AliceBob Basic communication scenario Resources x 1 x 2  x n Goal: convey n bits from Alice to Bob x 1 x 2  x n

27 Basic communication scenario Bit communication: Cost: n Qubit communication: Cost: n [Holevo’s Theorem, 1973] Bit communication & prior entanglement: Cost: n (can be deduced)Cost: n / 2 superdense coding [Bennett & Wiesner, 1992] Qubit communication & prior entanglement:

28 The GHZ “paradox”

29 GHZ scenario AliceBobCarol Input: r ts Output: acb Rules of the game: 1.It is promised that r  s  t = 0 2.No communication after inputs received 3.They win if a  b  c = r  s  t rstabcabc ← r ← ¬s← ¬s ← 1← 1 abc [Greenberger, Horne, Zeilinger, 1980]

30 No perfect strategy for GHZ Input: r ts Output: acb rstabcabc General deterministic strategy: a 0, a 1, b 0, b 1, c 0, c 1 Winning conditions: a 0  b 0  c 0 = 0 a 0  b 1  c 1 = 1 a 1  b 0  c 1 = 1 a 1  b 1  c 0 = 1 Has no solution, thus no perfect strategy exists

31 GHZ : preventing communication Input: r ts Output: acb Input and output events can be space-like separated: so signals at the speed of light are not fast enough for cheating What if Alice, Bob, and Carol still keep on winning?

32 “ GHZ Paradox” explained r ts acb Prior entanglement:    =  000  –  011  –  101  –  110  Alice’s strategy: 1. if r = 1 then apply H to qubit 2. measure qubit and set a to result Bob’s & Carol’s strategies: similar Case 1 ( rst = 000 ): state is measured directly … Case 2 ( rst = 011 ): new state  001  +  010  –  100  +  111  Cases 3 & 4 ( rst = 101 & 110 ): similar by symmetry

33 GHZ : conclusions For the GHZ game, any classical team succeeds with probability at most ¾ Allowing the players to communicate would enable them to succeed with probability 1 Entanglement cannot be used to communicate Nevertheless, allowing the players to have entanglement enables them to succeed with probability 1 Thus, entanglement is a useful resource for the task of winning the GHZ game

34 The Bell inequality and its violation – Physicist’s perspective

35 Bell’s Inequality and its violation Part I: physicist’s view: Can a quantum state have pre-determined outcomes for each possible measurement that can be applied to it? if {  0 ,  1  } measurement then output 0 if {  + ,  −  } measurement then output 1 if... (etc) qubit: where the “manuscript” is something like this: called hidden variables [Bell, 1964] [Clauser, Horne, Shimony, Holt, 1969] table could be implicitly given by some formula

36 Bell Inequality Imagine a two-qubit system, where one of two measurements, called M 0 and M 1, will be applied to each qubit: M 0 : a 0 M 1 : a 1 M 0 : b 0 M 1 : b 1 Define: A 0 = (  1) a 0 A 1 = (  1) a 1 B 0 = (  1) b 0 B 1 = (  1) b 1 Claim: A 0 B 0 + A 0 B 1 + A 1 B 0  A 1 B 1  2 Proof: A 0 (B 0 + B 1 ) + A 1 (B 0  B 1 )  2 one is  2 and the other is 0 space-like separated, so no cross-coordination

37 Bell Inequality Question: could one, in principle, design an experiment to check if this Bell Inequality holds for a particular system? Answer 1: no, not directly, because A 0, A 1, B 0, B 1 cannot all be measured (only one A s B t term can be measured) Answer 2: yes, indirectly, by making many runs of this experiment: pick a random st  {00, 01, 10, 11} and then measure with M s and M t to get the value of A s B t The average of A 0 B 0, A 0 B 1, A 1 B 0,  A 1 B 1 should be  ½ A 0 B 0 + A 0 B 1 + A 1 B 0  A 1 B 1  2 is called a Bell Inequality* * also called CHSH Inequality

38 Introduction to Quantum Information Processing CS 667 / Phys 767 / C&O 681 Richard Cleve DC 2117 Lecture 20 (2008)

39 Violating the Bell Inequality Two-qubit system in state    =  00  –  11  Define M 0 : rotate by   / 16 then measure M 1 : rotate by +3  / 16 then measure st = 01 or 10  /8 3  /8 -  /8 st = 11 st = 00 Applying rotations  A and  B yields: cos (  A +  B ) (  00  –  11  ) + sin (  A +  B ) (  01  +  10  ) A B = +1 A B =  1 Then A 0 B 0, A 0 B 1, A 1 B 0,  A 1 B 1 all have expected value ½√2, which contradicts the upper bound of ½ cos 2 (  / 8) = ½ + ¼√2

40 Bell Inequality violation: summary Assuming that quantum systems are governed by local hidden variables leads to the Bell inequality A 0 B 0 + A 0 B 1 + A 1 B 0  A 1 B 1  2 But this is violated in the case of Bell states (by a factor of √2 ) Therefore, no such hidden variables exist This is, in principle, experimentally verifiable, and experiments along these lines have actually been conducted

41 The Bell inequality and its violation – Computer Scientist’s perspective

42 Bell’s Inequality and its violation b st a input: output: With classical resources, Pr[ a  b = s  t ] ≤ 0.75 But, with prior entanglement state  00  –  11 , Pr[ a  b = s  t ] = cos 2 (  / 8) = ½ + ¼√2 = 0.853… Rules:1.No communication after inputs received 2.They win if a  b = s  t stabab Part II: computer scientist’s view:

43 The quantum strategy Alice and Bob start with entanglement    =  00  –  11  Alice: if s = 0 then rotate by  A =   / 16 else rotate by  A = + 3  / 16 and measure Bob: if t = 0 then rotate by  B =   / 16 else rotate by  B = + 3  / 16 and measure st = 01 or 10  /8 3  /8 -  /8 st = 11 st = 00 cos (  A –  B ) (  00  –  11  ) + sin (  A –  B ) (  01  +  10  ) Success probability: Pr[ a  b = s  t ] = cos 2 (  / 8) = ½ + ¼√2 = 0.853…

44 Nonlocality in operational terms information processing task quantum entanglement ! classically, communication is needed

45 The magic square game

46 Magic square game a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 Problem: fill in the matrix with bits such that each row has even parity and each column has odd parity even odd even IMPOSSIBLE Game: ask Alice to fill in one row and Bob to fill in one column They win iff parities are correct and bits agree at intersection Success probabilities: classical and 1 quantum 8/98/9 [Aravind, 2002] (details omitted here)

47 Preview of communication complexity

48 Classical Communication Complexity f (x,y) x 1 x 2  x n y 1 y 2  y n E.g. equality function: f (x,y) = 1 if x = y, and 0 if x  y Any deterministic protocol requires n bits communication Probabilistic protocols can solve with only O(log(n /  )) bits communication (error probability  ) [Yao, 1979]

49 Quantum Communication Complexity Qubit communication Prior entanglement f (x,y) x 1 x 2  x n y 1 y 2  y n qubits f (x,y) x 1 x 2  x n y 1 y 2  y n  entangled qubits bits Question: can quantum beast classical in this context?

50 Appointment scheduling i ( x i = y i = 1 ) Classically,  (n) bits necessary to succeed with prob.  3/4 For all  > 0, O(n 1 / 2 log n) qubits sufficient for error prob. <  … n … n x =x = y =y = [KS ‘87] [BCW ‘98]

51 Search problem … n x =x = Given:accessible via queries ii  b  x i  ii bb Goal: find i  {1, 2, …, n } such that x i = 1 Classically:  (n) queries are necessary Quantum mechanically: O(n 1 / 2 ) queries are sufficient i b  x i i b log n 1 x [Grover, 1996]

… n x =x = … 1 y =y = … 0 xy =xy = Alice Bob ii 00 00 bb xy xy  ii 00 00 bb y y Alice xx Communication per x  y -query: 2 ( log n + 3 ) = O ( log n)

53 Appointment scheduling: epilogue Bit communication: Cost: θ( n ) Qubit communication: Cost: θ( n 1 / 2 ) (with refinements) Bit communication & prior entanglement: Cost: θ( n 1 / 2 ) Qubit communication & prior entanglement: Cost: θ( n 1 / 2 ) [R ‘02] [AA ‘03]

54 Are exponential savings possible?

55 Restricted version of equality Precondition (i.e. promise): either x = y or  ( x, y ) = n / 2 Hamming distance Classically,  (n) bits communication are necessary for an exact solution Quantum mechanically, O(log n) qubits communication are sufficient for an exact solution [BCW ‘98] (Distributed variant of “constant” vs. “balanced”)

56 Classical lower bound Theorem: If S  {0,1} n has the property that, for all x, x ′  S, their intersection size is not n/ 4 then  S  < 1.99 n [Frankl and Rödl, 1987] Let some protocol solve restricted equality with k bits comm. ● approximately 2 n /  n input pairs ( x, x ), where Δ ( x ) = n/ 2 Define S = { x : Δ ( x ) = n/ 2 and ( x, x ) yields conv. C } Therefore, 2 n / 2 k  n input pairs ( x, x ) that yield same conv. C ● 2 k conversations of length k For any x, x ′  S, input pair ( x, x ′ ) also yields conversation C Therefore, Δ ( x, x ′)  n/ 2, implying intersection size is not n/ 4 Theorem implies 2 n / 2 k  n n

57 Quantum protocol For each x  {0,1} n, define Protocol: 1.Alice sends  x  to Bob ( log( n ) qubits) 2.Bob measures state in a basis that includes  y  If x = y then Bob’s result is definitely  y  If  ( x, y ) = n / 2 then  x  y  = 0, so result is definitely not  y  Question: How much communication if error ¼ is permitted? Answer: just 2 bits are sufficient! Correctness of protocol:

58 Exponential quantum vs. classical separation in bounded-error models O(log n) quantum vs.  (n 1 / 4 / log n) classical Output: result of applying M to U   : a log( n ) -qubit state (described classically) M : two-outcome measurement U : unitary operation on log( n ) qubits [Raz, 1999]

59 Lower bound for the inner product problem

60 Inner product IP(x, y) = x 1 y 1 + x 2 y 2 +  + x n y n mod 2 Classically,  (n) bits of communication are required, even for bounded-error protocols Quantum protocols also require  (n) communication [KY ‘95] [CNDT ‘98] [NS ‘02]

61 The BV black-box problem Let f ( x 1, x 2, …, x n ) = a 1 x 1 + a 2 x 2 +  + a n x n mod 2 Given: f bb x1x1 xnxn x2x2  x2x2  b  f ( x 1, x 2, …, x n )  xnxn x1x1  H H H H H H H H H H 11 00 00 00  11 a1a1 anan a2a2  Goal: determine a 1, a 2, …, a n Classically, n queries are necessary Quantum mechanically, 1 query is sufficient Bernstein & Vazirani

62 Lower bound for inner product IP(x, y) = x 1 y 1 + x 2 y 2 +  + x n y n mod 2 y1y1ynyny2y2 Alice and Bob’s IP protocol x2x2x1x1xnxn  z  IP(x, y)  Alice and Bob’s IP protocol inverted y1y1y2y2ynynx1x1x2x2xnxn zz Proof:

63 Lower bound for inner product IP(x, y) = x 1 y 1 + x 2 y 2 +  + x n y n mod 2 Since n bits are conveyed from Alice to Bob, n qubits communication necessary (by Holevo’s Theorem) Alice and Bob’s IP protocol x2x2x1x1xnxn Alice and Bob’s IP protocol inverted x1x1x2x2xnxn x1x1x2x2xnxn HHHHHH 00110000 11 HH Proof:

64 Simultaneous message passing and fingerprinting

65 Equality revisited in simultaneous message model x 1 x 2  x n y 1 y 2  y n f (x,y) Exact protocols: require 2 n bits communication Equality function: f (x,y) = 1 if x = y 0 if x  y

66 Equality revisited in simultaneous message model x 1 x 2  x n y 1 y 2  y n f (x,y) Bounded-error protocols with a shared random key: require only O(1) bits communication Error-correcting code: e( x ) = e( y ) = Pr[ 00 ] = Pr[ 11 ] = ½ random k classical

67 Equality revisited in simultaneous message model x 1 x 2  x n y 1 y 2  y n f (x,y) Bounded-error protocols without a shared key: Classical: θ(n 1 / 2 ) Quantum: θ(log n) [A ‘96] [NS ‘96] [BCWW ‘01]

68 Quantum fingerprints Question 1: how many orthogonal states in m qubits? Answer: 2 m Answer: 2 2 am, for some constant a > 0 Let  be an arbitrarily small positive constant Question 2: how many almost orthogonal* states in m qubits? (* where |  x  y  | ≤  ) The states can be constructed via a suitable (classical) error- correcting code, which is a function e : {0,1} n  {0,1} cn where, for all x ≠ y, dcn ≤ Δ( e ( x ), e ( y )) ≤ (1− d ) cn ( c, d are constants)

69 Construction of almost orthogonal states Since dcn ≤ Δ( e ( x ), e ( y )) ≤ (1− d ) cn, we have |  x  y  | ≤ 1− 2d Set  x  for each x  {0,1} n ( log( cn ) qubits) Then  x  y  By duplicating each state,  x  x  …  x , the pairwise inner products can be made arbitrarily small: (1− 2d ) r ≤  Result: m = r log( cn ) qubits storing 2 n = 2 ( 1/c ) 2 m/r different states

70 Quantum fingerprints if x = y, Pr[ output = 0] = 1 if x ≠ y, Pr[ output = 0] = (1+  2 ) / 2 Given  x  y , one can check if x = y or x ≠ y as follows: Let  000 ,  001 , …,  111  be 2 n states on O(log n ) qubits such that |  x  y  | ≤  for all x ≠ y H SWAPSWAP H xx yy 00 Intuition:  0  x  y  +  1  y  x  Note: error probability can be reduced to ( (1+  2 ) / 2 ) r

71 Equality revisited in simultaneous message model x 1 x 2  x n y 1 y 2  y n f (x,y) Bounded-error protocols without a shared key: Classical: θ(n 1 / 2 ) Quantum: θ(log n) [A ‘96] [NS ‘96] [BCWW ‘01]

72 Quantum protocol for equality in simultaneous message model x 1 x 2  x n y 1 y 2  y n xx yy Orthogonality test xxyy Recall that, with a shared key, the problem is easy classically...

73 Hidden matching problem

74 Hidden matching problem For this problem, a quantum protocol is exponentially more efficient than any classical protocol—even with a shared key x  {0,1} n matching on {1, 2, …, n } Inputs: M =M = [Bar-Yossef, Jayram, Kerenidis, 2004] (i, j, x i  x j ), such that ( i, j )  M Output: Only one-way communication (Alice to Bob) is permitted

75 The hidden matching problem x  {0,1} n matching on {1,2, …, n } Inputs: Output: (i, j, x i  x j ), ( i, j )  M M =M = Rough intuition: Alice doesn’t know which edges are in M, so she apparently has to send  (  n ) bits of the form x i  x j … Classically, one-way communication is  (  n ), even with a shared classical key (the proof is omitted here)

76 The hidden matching problem x  {0,1} n matching on {1,2, …, n } Inputs: M =M = Output: (i, j, x i  x j ), ( i, j )  M Quantum protocol: Alice sends ( log n qubits) Bob measures in  i    j  basis, ( i, j )  M, and uses the outcome’s relative phase to determine x i  x j

77 Nonlocality revisited

78 Restricted-equality nonlocality b xy a inputs: outputs: ( n bits) ( log n bits) ( n bits) ( log n bits) With classical resources,  ( n ) bits of communication needed for an exact solution* With (  00  +  11  )  log n prior entanglement, no communication is needed at all* Precondition: either x = y or  ( x, y ) = n / 2 Required postcondition: a = b iff x = y [BCT ‘99]  Technical details similar to restricted equality of Lecture 17

79 Restricted-equality nonlocality Restricted-equality nonlocality Bit communication: Cost: θ( n ) Qubit communication: Cost: log n Bit communication & prior entanglement: Cost: zero Qubit communication & prior entanglement:

80 Nonlocality and communication complexity conclusions Quantum information affects communication complexity in interesting ways There is a rich interplay between quantum communication complexity and: –quantum algorithms –quantum information theory –other notions of complexity theory …

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