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
[email protected]
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:
s
input:
output:
a
t
b
Rules: 1. No communication after inputs received
2. They win if ab = st
With classical resources, Pr[ab = st] ≤ 0.75
But, with prior entanglement state 00 – 11,
Pr[ab = st] = cos2(/8) = ½ + ¼√2 = 0.853…
st
ab
00
0
01
0
10
0
11
1
4
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 = 11
3/8
st = 01 or 10
/8
-/8
st = 00
cos(A – B ) (00 – 11) + sin(A – B ) (01 + 10)
Success probability:
Pr[ab = st] = cos2(/8) = ½ + ¼√2 = 0.853…
5
Nonlocality in operational terms
information
processing
task
!
classically,
communication
is needed
quantum
entanglement
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
a21 a22 a23
even
a31 a32 a33
even
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:
[Aravind, 2002]
8/9
classical and 1 quantum
(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

Prior entanglement
entangled qubits
x1 x2  xn
f (x,y)

y1 y2  yn
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
x= 01101…0
1
2
3
4
5
...
n
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
Given:
2
3
4
5
6
...
n
x= 000010…1
log n
i 
1
b
b
x
accessible via queries
i 
b
b  xi 
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
xy = 0 0 0 0 1 0 … 0
i xy
0
0
b

i y
x
x
y
0
0
b
Bob
Alice
Bob
Communication per xy-query: 2(log n + 3) = O(log n)
15
Appointment scheduling: epilogue
Bit communication:
Qubit communication:
Cost: θ(
Cost: θ(
n)
n
1/2
)
(with refinements)
Bit communication
& prior entanglement:
Qubit communication
& prior entanglement:
Cost: θ(
Cost: θ(
n1/2)
[R ‘02] [AA ‘03]
n1/2)
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/2kn 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/2kn < 1.99n , so k > 0.007n
[Frankl and Rödl, 1987]
19
Quantum protocol
For each x 
n
{0,1}n,
define ψ x   (1)
xj
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 xy = 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
(described classically)
M: two-outcome measurement
U: unitary operation
on log(n) qubits
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

0 x2H

 H
0 xnH

1 bH
f
Hx1 a1
Hx2 a2
H 
Hxn an
Hb 1f(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 zIP(x, y)
25
Lower bound for inner product
IP(x, y) = x1 y1 + x2 y2 +  + xn yn mod 2
Proof: x1 x2 xn
0 0 0
1
H
H
H
H
H
x1 x2 xn
1
H
Alice and Bob’s IP protocol
Alice and Bob’s IP protocol inverted
x1 x2 xn
H
H
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
f (x,y)
Equality function:
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
random k
29
Equality revisited
in simultaneous message model
x1x2  xn
Bounded-error
protocols without
a shared key:
y1y2  yn
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 |xy| ≤  )
am
Answer: 22
, for some constant a > 0
The states can be constructed via a suitable (classical) errorcorrecting 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
Set x 
1
cn
cn
 (1)
e ( x )k
k
for each x{0,1}n (log(cn) qubits)
k 1
Then xy 
1
cn
[ e ( x )e ( y )]k
(1)

cn
k 1
2e( x), e( y )
k  1
cn
Since dcn ≤ Δ(e(x),e(y)) ≤ (1− d )cn, we have |xy| ≤ 1− 2d
By duplicating each state, xx … 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 |xy| ≤  for all x ≠ y
Given xy, one can check if x = y or x ≠ y as follows:
0
x
y
H
H
if x = y, Pr[output = 0] = 1
if x ≠ y, Pr[output = 0] = (1+ 2)/2
S
W
A
P
Intuition: 0xy + 1yx
Note: error probability can
be reduced to ((1+ 2)/2)r
33
Equality revisited
in simultaneous message model
x1x2  xn
Bounded-error
protocols without
a shared key:
y1y2  yn
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
Orthogonality
test
Recall that, with a
shared key, the
problem is easy
classically ...
35
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