Introduction to
Quantum Information Processing
CS 667 / Phys 767 / C&O 681
Lecture 15 (2008)
Richard Cleve
DC 2117
[email protected]
1
Continuous-time evolution
2
Continuous-time evolution
Although we’ve expressed quantum operations in discrete
terms, in real physical systems, the evolution is continuous
Let H be any Hermitian matrix and t  R
1
Then eiHt is unitary – why?
 λ1



†
D


H = U DU, where



λd 
e iλ1t
t
† iDt
iHt
Therefore e = U e U = U 


0


(unitary)

U
e iλd t 
3
Grover’s quantum
search algorithm
4
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 2n queries
are necessary to succeed—even with probability ¾ (say)
Grover’s quantum algorithm that makes only O(2n) queries
Query: x1
[Grover ’96]
xn
 y
 x1 
Uf
xn
y  f(x1,...,xn)
5
Applications of quantum search
The function f could be realized as a 3-CNF formula:
f x1 ,..., xn   x1  x3  x4   x2  x3  x5     x1  x5  xn 
PSPACE
Alternatively, the search
could be for a certificate
for any problem in NP
The resulting quantum
algorithms appear to be
quadratically more
efficient than the best
classical algorithms
known
3-CNF-SAT
NP
co-NP
FACTORING
P
6
Prelude to Grover’s algorithm:
two reflections = a rotation
Consider two lines with intersection angle :
reflection 2
2

2
1
1
reflection 1
Net effect: rotation by angle 2, regardless of starting vector
7
Grover’s algorithm: description I
Basic operations used:
 x1 
 x n
 y
 x1 
Uf
 x1 
 x n
 y
xn
y  f(x1,...,xn)
 x1 
U0
H
Uf x = (1) f(x) x
xn
y  [x = 0...0]
Hadamard
Implementation?
X
X
X
X
X
X
U0 x = (1) [x = 0...0]x
H
H
H
8
Grover’s algorithm: description II
iteration 1
iteration 2
...
H
H
 0
0
 
H
Uf
H
U0
H
Uf
U0
1. construct state H 0...0
2. repeat k times:
apply H U0 HUf to state
3. measure state, to get x{0,1}n, and check if f(x) =1
(The setting of k will be determined later)
9
Grover’s algorithm: analysis I
Let A = {x {0,1}n : f (x) = 1} and B = {x {0,1}n : f (x) = 0}
and N = 2n and a = |A| and b = |B|
Let
A 

a
1
x
x

b
B 
and
1
xA
xB
Consider the space spanned by A and B
A  goal is to get close to this state
H0...0 
B
1
N
x

a
N
A 
b
N
B
x{ 0 ,1} n
Interesting case: a << N
10
Grover’s algorithm: analysis II
A
Algorithm: (HU0 HUf )k H 0...0
H0...0
B
Observation:
Uf is a reflection about B: Uf A = A and Uf B = B
Question: what is HU0 H ? U0 is a reflection about H0...0
Partial proof:
H U0 HH0...0 = H U0 0...0 = H( 0...0) = H 0...0
11
Grover’s algorithm: analysis III
A
2
Algorithm: (HU0 HUf )k H 0...0
2
2
2

H0...0
B
Since H U0 HUf 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?
12
Introduction to
Quantum Information Processing
CS 667 / Phys 767 / C&O 681
Lecture 17 (2008)
Richard Cleve
DC 2117
[email protected]
13
Optimality of
Grover’s algorithm
14
Optimality of Grover’s algorithm
Theorem: any quantum search algorithm for f : {0,1}n  {0,1}
must make (2n) queries to f (if f is used as a black-box)
Proof (of a slightly simplified version):
Assume queries are of the form
|x
(1) f(x)|x
f
and that a k-query algorithm is of the form
|0...0 U0
f
U1
f
U2
f
U3
f
where U0, U1, U2, ..., Uk, are arbitrary unitary operations
Uk
15
Optimality of Grover’s algorithm
Define fr : {0,1}n  {0,1} as fr (x) = 1 iff x = r
Consider
|0
U0
fr
U1
fr
U2
fr
U3
fr
Uk |ψr,k
I
U1
I
U2
I
U3
I
Uk |ψr,0
versus
|0
U0
We’ll show that, averaging over all r  {0,1}n,
|| |ψr,k  |ψr,0 ||  2k / 2n
16
Optimality of Grover’s algorithm
Consider
|0
U0
I
U1
ki
I
U2
fr
U3
fr
Uk |ψr,i
i
Note that
|ψr,k  |ψr,0 = (|ψr,k  |ψr,k1) + (|ψr,k1  |ψr,k2) + ... + (|ψr,1  |ψr,0)
which implies
|| |ψr,k  |ψr,0 ||  || |ψr,k  |ψr,k1 || + ... + || |ψr,1  |ψr,0 ||
17
Optimality of Grover’s algorithm
query
|0
U0
I
U1
U0
I
U1
query i+1
I
query
|0
i
fr
U2
i
U3
|ψr,i
fr
Uk
fr
Uk |ψr,i-1
query i+1
I
I
U2
α
i ,x
U3
x
x
|| |ψr,i  |ψr,i-1 || = |2i,r|, since query only negates |r
Therefore, || |ψr,k  |ψr,0 || 
k 1
2α
i 0
i ,r
18
Optimality of Grover’s algorithm
Now, averaging over all r  {0,1}n,
1
2n

r
ψ r ,k  ψ r ,0
1
 n
2
 k 1

2 α i ,r 
r  
i 0

1
 n
2


2  α i ,r 

i 0  r

1
 n
2

k 1
k 1
 
n
2
2

i 0
(By Cauchy-Schwarz)
2k
2n
Therefore, for some r  {0,1}n, the number of queries k must
be (2n), in order to distinguish fr from the all-zero function
This completes the proof
19
Introduction to
Quantum Information Processing
CS 667 / Phys 767 / C&O 681
Lecture 18 (2008)
Richard Cleve
DC 2117
[email protected]
20
Lab tour
(instead of a regular lecture)
21
Introduction to
Quantum Information Processing
CS 667 / Phys 767 / C&O 681
Lecture 19 (2008)
Richard Cleve
DC 2117
[email protected]
22
Preliminary remarks about
quantum communication
23
Quantum information can apparently be
used to substantially reduce computation
costs for a number of interesting problems
How does quantum information affect the
communication costs of information
processing tasks?
We explore this issue ...
24
Entanglement and signaling
Recall that Entangled states, such as
qubit
1
2
00 
1
2
11 ,
qubit
can be used to perform some intriguing feats, such as
teleportation and superdense coding
—but they cannot be used to “signal instantaneously”
Any operation performed on one system has no affect on
the state of the other system (its reduced density matrix)
25
Basic communication scenario
Goal: convey n bits from Alice to Bob
x1x2  xn
Alice
Resources
Bob
x1x2  xn
26
Basic communication scenario
Bit communication:
Cost:
n
Cost:
Bit communication
& prior entanglement:
Cost:
n
Qubit communication:
(can be deduced)
n
[Holevo’s Theorem, 1973]
Qubit communication
& prior entanglement:
Cost:
n/2
superdense coding
[Bennett & Wiesner, 1992]
27
The GHZ “paradox”
28
GHZ scenario
[Greenberger, Horne, Zeilinger, 1980]
Input:
Output:
r
s
t
a ←r
b ← ¬s
c ←1
Alice
Rules of the game:
1. It is promised that
Bob
r s t = 0
Carol
rst abc abc
2. No communication after inputs received
000
0
011
3. They win if abc = rst
011
1
001
101
1
111
110
1
101
29
No perfect strategy for GHZ
Input:
r
s
t
Output:
a
b
c
rst abc
000
0
011
1
101
1
110
1
General deterministic strategy:
a0, a1, b0, b1, c0, c1
Winning conditions:
a0  b0  c0 = 0
Has no solution,
a0  b1  c1 = 1
thus no perfect
a1  b0  c1 = 1
strategy exists
a1  b1  c0 = 1
30
GHZ: preventing communication
Input:
r
s
t
Output:
a
b
c
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?
31
“GHZ Paradox” explained
Prior entanglement:  = 000 – 011 – 101 – 110
r
s
t
a
b
c
Alice’s strategy:
1. if r = 1 then apply H to qubit
2. measure qubit and set a to result
1 1
H 

2 1  1
1
Bob’s & Carol’s strategies: similar
Case 2
1 (rst = 011):
000): new
statestate
is measured
001 + 010
directly
– 100
… + 111
Cases 3 & 4 (rst = 101 & 110): similar by symmetry
32
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
33
The Bell inequality and its violation
– Physicist’s perspective
34
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?
qubit:
where the “manuscript”
is something like this:
if {0,1} measurement
then output 0
if {+,−} measurement
then output 1
if ... (etc)
called hidden variables
[Bell, 1964]
[Clauser, Horne, Shimony, Holt, 1969]
table could be implicitly
given by some formula
35
Bell Inequality
Imagine a two-qubit system, where one of two measurements,
called M0 and M1, will be applied to each qubit:
M0 : a0
M1 : a1
Define:
A0 = (1)a0
A1 = (1)a1
B0 = (1)b0
B1 = (1)b1
space-like separated, so
no cross-coordination
M0 : b0
M1 : b1
Claim: A0 B0 + A0 B1 + A1B0  A1 B1  2
Proof: A0 (B0 + B1) + A1 (B0  B1)  2
one is  2 and the other is 0
36
Bell Inequality
A0 B0 + A0 B1 + A1B0  A1 B1  2 is called a 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 A0, A1, B0, B1 cannot
all be measured (only one As Bt 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 Ms and Mt to get the value of As Bt
The average of A0 B0, A0 B1, A1B0, A1 B1 should be  ½
* also called CHSH Inequality
37
Introduction to
Quantum Information Processing
CS 667 / Phys 767 / C&O 681
Lecture 20 (2008)
Richard Cleve
DC 2117
[email protected]
38
Violating the Bell Inequality
Two-qubit system in state
 = 00 – 11
Applying rotations A and B yields:
cos(A + B ) (00 – 11) + sin(A + B ) (01 + 10)
A B = +1
Define
M0: rotate by  /16 then measure
M1: rotate by +3/16 then measure
Then A0 B0, A0 B1, A1B0, A1 B1 all have
expected value ½√2, which contradicts
the upper bound of ½
A B = 1
st = 11
3/8
st = 01 or 10
/8
-/8
st = 00
cos2(/8) = ½ + ¼√2
39
Bell Inequality violation: summary
Assuming that quantum systems are
governed by local hidden variables
leads to the Bell inequality
A0 B0 + A0 B1 + A1B0  A1 B1  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
40
The Bell inequality and its violation
– Computer Scientist’s perspective
41
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
42
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…
43
Nonlocality in operational terms
information
processing
task
!
classically,
communication
is needed
quantum
entanglement
44
The magic square game
45
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)
46
Preview of
communication
complexity
47
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 )
48
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)
Question: can quantum beast classical in this context?
49
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]
50
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]
51
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)
52
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)
53
Are exponential
savings possible?
54
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]
55
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]
56
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!
57
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]
58
Lower bound for the
inner product problem
59
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]
60
The BV black-box problem
Bernstein & Vazirani
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
61
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)
62
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)
63
Simultaneous message
passing and fingerprinting
64
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
65
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
66
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]
67
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)
68
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
69
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
70
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]
71
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 ...
72
Hidden matching problem
73
Hidden matching problem
For this problem, a quantum protocol is exponentially more
efficient than any classical protocol—even with a shared key
Inputs:
x  {0,1}n
M=
matching on
{1, 2, …, n}
Output: (i, j, xixj), such that
( i, j)  M
Only one-way communication (Alice to Bob) is permitted
[Bar-Yossef, Jayram, Kerenidis, 2004]
74
The hidden matching problem
Inputs:
x  {0,1}n
M=
matching on
{1,2, …, n}
Output: (i, j, xixj), (i, j)  M
Classically, one-way communication is (n), even with a
shared classical key (the proof is omitted here)
Rough intuition: Alice doesn’t know which edges are in M,
so she apparently has to send (n) bits of the form xixj …
75
The hidden matching problem
Inputs:
matching on
{1,2, …, n}
M=
x  {0,1}n
Output: (i, j, xixj), (i, j)  M
Quantum protocol: Alice sends
1
n
n
 (1) k k
x
(log n qubits)
k 1
Bob measures in i  j basis, (i, j)  M,
and uses the outcome’s relative phase to
determine xixj
76
Nonlocality revisited
77
Restricted-equality nonlocality
inputs:
x
(n bits)
y
(n bits)
outputs:
a
(log n bits)
b
(log n bits)
Precondition: either x = y or (x,y) = n/2
Required postcondition: a = b iff x = y
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*
 Technical details similar to restricted equality of Lecture 17
[BCT ‘99]
78
Restricted-equality nonlocality
Bit communication:
Qubit communication:
Cost: θ(
Cost: log
n)
Bit communication
& prior entanglement:
Cost:
zero
n
Qubit communication
& prior entanglement:
Cost:
zero
79
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 …
80
81