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Introduction to Quantum Information Processing CS 467 / CS 667 Phys 467 / Phys 767 C&O 481 / C&O 681 Lecture 17 (2005) Richard Cleve DC 3524 cleve@cs.uwaterloo.ca Course web site at: http://www.cs.uwaterloo.ca/~cleve/courses/cs467 1 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 2 • 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 Bell’s Inequality and its violation Part II: computer scientist’s view: input: s t output: a b Rules: 1. No communication after inputs received st ab 2. They win if ab = st 00 0 01 0 With classical resources, Pr[ab = st] ≤ 0.75 10 0 11 1 But, with prior entanglement state 00 – 11, Pr[ab = st] = cos2(/8) = ½ + ¼√2 = 0.853… 4 The quantum strategy • Alice and Bob start with entanglement = 00 – 11 • Alice: if s = 0 then rotate by A = /16 st = 11 else rotate by A = + 3/16 and measure 3/8 st = 01 or 10 /8 -/8 • Bob: if t = 0 then rotate by B = /16 st = 00 else rotate by B = + 3/16 and measure cos(A – B ) (00 – 11) + sin(A – B ) (01 + 10) Success probability: Pr[ab = st] = cos2(/8) = ½ + ¼√2 = 0.853… 5 Nonlocality in operational terms information processing task ! classically, quantum communication entanglement is needed 6 • 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 7 Magic square game Problem: fill in the matrix with bits such that each row has even parity and each column has odd parity a11 a12 a13 even a21 a22 a23 even a31 a32 a33 even odd odd odd 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: 8/9 classical and 1 quantum [Aravind, 2002] (details omitted here) 8 • 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 9 Classical communication complexity [Yao, 1979] x1x2 xn y1y2 yn f (x,y) 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 10 Quantum communication complexity x1 x2 xn y1 y2 yn Qubit communication qubits f (x,y) entangled qubits x1 x2 xn y1 y2 yn Prior entanglement bits f (x,y) 11 • 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 12 Appointment scheduling 1 2 3 4 5 ... n 1 2 3 4 5 ... n x= 01101…0 y= 10011…1 i (xi = yi = 1) Classically, (n) bits necessary to succeed with prob. 3/4 For all > 0, O(n1/2 log n) qubits sufficient for error prob. < [KS „87] [BCW „98] 13 Search problem 1 2 3 4 5 6 ... n Given: x= 000010…1 accessible via queries log n i x i 1 b b b xi b Goal: find i{1, 2, …, n} such that xi = 1 Classically: (n) queries are necessary Quantum mechanically: O(n1/2) queries are sufficient [Grover, 1996] 14 1 2 3 4 5 6 ... n Alice x= 011010…0 Bob y= 100110…1 xy = 0 0 0 0 1 0 … 0 i xy i y x x y 0 0 0 0 b b Bob Alice Bob Communication per xy-query: 2(log n + 3) = O(log n) 15 Appointment scheduling: epilogue Bit communication: Qubit communication: Cost: θ( n) Cost: θ( n1/2 ) (with refinements) Bit communication Qubit communication & prior entanglement: & prior entanglement: Cost: θ( n1/2) Cost: θ( n1/2) [R „02] [AA „03] 16 • 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 17 Restricted version of equality Precondition (i.e. promise): either x = y or (x,y) = n/2 Hamming distance (Distributed variant of “constant” vs. “balanced”) 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] 18 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 < 1.99n Let some protocol solve restricted equality with k bits comm. ● 2k conversations of length k ● approximately 2n/n input pairs (x, x), where Δ(x) = n/2 Therefore, 2n/2kn input pairs (x, x) that yield same conv. C Define S = {x : Δ(x) = n/2 and (x, x) yields conv. C } 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 2n/2kn < 1.99n , so k > 0.007n 19 [Frankl and Rödl, 1987] Quantum protocol n define ψ x (1) xj For each x {0,1}n, j j 1 Protocol: 1. Alice sends x to Bob (log(n) qubits) 2. Bob measures state in a basis that includes y Correctness of protocol: If x = y then Bob‟s result is definitely y If (x,y) = n/2 then xy = 0, so result is definitely not y Question: How much communication if error ¼ is permitted? Answer: just 2 bits are sufficient! 20 Exponential quantum vs. classical separation in bounded-error models O(log n) quantum vs. (n1/4 / log n) classical : a log(n)-qubit state U: unitary operation (described classically) on log(n) qubits M: two-outcome measurement Output: result of applying M to U [Raz, 1999] 21 • 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 22 Inner product IP(x, y) = x1 y1 + x2 y2 + + xn yn 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] 23 Recall the BV problem Let f(x1, x2, …, xn) = a1 x1 + a2 x2 + + an xn mod 2 Given: 0 x1H x H 1 a1 0 x2H x H 2 a2 H f H 0 xnH x H n an 1 bH b f H 1(x1, x2, …, xn) Goal: determine a1, a2 , …, an Classically, n queries are necessary Quantum mechanically, 1 query is sufficient 24 Lower bound for inner product IP(x, y) = x1 y1 + x2 y2 + + xn yn mod 2 Proof: x1 x2 xn y1 y2 yn z Alice and Bob‟s IP protocol Alice and Bob‟s IP protocol inverted x1 x2 xn y1 y2 yn zIP(x, y) 25 Lower bound for inner product IP(x, y) = x1 y1 + x2 y2 + + xn yn mod 2 0 0 0 1 Proof: x1 x2 xn H H H H Alice and Bob‟s IP protocol Alice and Bob‟s IP protocol inverted H H H H x1 x2 xn x1 x2 xn 1 Since n bits are conveyed from Alice to Bob, n qubits communication necessary (by Holevo‟s Theorem) 26 • 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 27 Equality revisited in simultaneous message model x1x2 xn y1y2 yn Equality function: f (x,y) f (x,y) = 1 if x = y 0 if x y Exact protocols: require 2n bits communication 28 Equality revisited in simultaneous message model x1x2 xn y1y2 yn classical classical f (x,y) Pr[00] = Pr[11] = ½ Bounded-error protocols with a shared random key: require only O(1) bits communication Error-correcting code: e(x) = 1 0 1 1 1 1 0 1 0 1 1 0 0 1 1 0 0 1 e(y) = 0 1 1 0 1 0 0 1 0 0 1 1 0 0 1 0 1 0 29 random k Equality revisited in simultaneous message model x1x2 xn y1y2 yn Bounded-error protocols without a shared key: f (x,y) Classical: θ(n1/2) Quantum: θ(log n) [A „96] [NS „96] [BCWW „01] 30 Quantum fingerprints Question 1: how many orthogonal states in m qubits? Answer: 2m Let be an arbitrarily small positive constant Question 2: how many almost orthogonal* states in m qubits? (* where |xy| ≤ ) am Answer: 22 , for some constant a > 0 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) 31 Construction of almost orthogonal states cn 1 Set x (1) e ( x )k k for each x{0,1}n (log(cn) qubits) cn k 1 Then xy 1 cn [ e ( x )e ( y )]k 2e( x), e( y ) cn (1) k 1 k 1 cn Since dcn ≤ Δ(e(x),e(y)) ≤ (1− d )cn, we have |xy| ≤ 1− 2d By duplicating each state, xx … x, the pairwise inner products can be made arbitrarily small: (1− 2d )r ≤ m/r Result: m = r log(cn) qubits storing 2n = 2(1/c)2 different states 32 Quantum fingerprints Let 000, 001, …, 111 be 2n states on O(log n) qubits such that |xy| ≤ for all x ≠ y Given xy, one can check if x = y or x ≠ y as follows: if x = y, Pr[output = 0] = 1 0 H H if x ≠ y, Pr[output = 0] = (1+ 2)/2 x S W y A P Note: error probability can Intuition: 0xy + 1yx be reduced to ((1+ 2)/2)r 33 Equality revisited in simultaneous message model x1x2 xn y1y2 yn Bounded-error protocols without a shared key: f (x,y) Classical: θ(n1/2) Quantum: θ(log n) [A „96] [NS „96] [BCWW „01] 34 Quantum protocol for equality in simultaneous message model x1x2 xn y1y2 yn x y x y Recall that, with a Orthogonality shared key, the test problem is easy classically ... 35 36

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posted: | 8/28/2010 |

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