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 17 (2005)

2 Contents The Bell inequality and its violation –Physicist’s perspective –Computer Scientist’s perspective The magic square game Communication complexity –Equality checking –Appointment scheduling (quadratic savings) –Are exponential savings possible? –Lower bound for the inner product problem Simultaneous message passing and fingerprinting

3 The Bell inequality and its violation –Physicist’s perspective –Computer Scientist’s perspective The magic square game Communication complexity –Equality checking –Appointment scheduling (quadratic savings) –Are exponential savings possible? –Lower bound for the inner product problem Simultaneous message passing and fingerprinting

4 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:

5 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…

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

7 The Bell inequality and its violation –Physicist’s perspective –Computer Scientist’s perspective The magic square game Communication complexity –Equality checking –Appointment scheduling (quadratic savings) –Are exponential savings possible? –Lower bound for the inner product problem Simultaneous message passing and fingerprinting

8 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)

9 The Bell inequality and its violation –Physicist’s perspective –Computer Scientist’s perspective The magic square game Communication complexity –Equality checking –Appointment scheduling (quadratic savings) –Are exponential savings possible? –Lower bound for the inner product problem Simultaneous message passing and fingerprinting

10 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  ), via random hashing [Yao, 1979]

11 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

12 The Bell inequality and its violation –Physicist’s perspective –Computer Scientist’s perspective The magic square game Communication complexity –Equality checking –Appointment scheduling (quadratic savings) –Are exponential savings possible? –Lower bound for the inner product problem Simultaneous message passing and fingerprinting

13 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]

14 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)

16 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]

17 The Bell inequality and its violation –Physicist’s perspective –Computer Scientist’s perspective The magic square game Communication complexity –Equality checking –Appointment scheduling (quadratic savings) –Are exponential savings possible? –Lower bound for the inner product problem Simultaneous message passing and fingerprinting

18 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”)

19 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

20 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:

21 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]

22 The Bell inequality and its violation –Physicist’s perspective –Computer Scientist’s perspective The magic square game Communication complexity –Equality checking –Appointment scheduling (quadratic savings) –Are exponential savings possible? –Lower bound for the inner product problem Simultaneous message passing and fingerprinting

23 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]

24 Recall the BV 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

25 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:

26 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:

27 The Bell inequality and its violation –Physicist’s perspective –Computer Scientist’s perspective The magic square game Communication complexity –Equality checking –Appointment scheduling (quadratic savings) –Are exponential savings possible? –Lower bound for the inner product problem Simultaneous message passing and fingerprinting

28 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

29 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

30 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]

31 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)

32 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

33 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

34 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]

35 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...

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