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Quantum Network Coding
Harumichi Nishimura
西村治道
Graduate School of Science, Osaka Prefecture University
大阪府立大学理学系研究科
March 30, 2011, Institute of Network Coding
Why Quantum Network Coding?
 Power of quantum information processing
– Fast algorithms
• Factoring large integers
• Database search
– Secure cryptosystems
• Unconditionally secure key distribution
• Public key cryptosystems
– Communication-efficient multi-party protocols
• Communication complexity
• Leader election
 But quantum channel is "expensive"
Q. Can we reduce the amount of quantum
communication using the idea of network coding?
Outline of This Talk
• Basic of quantum information
– States
– Measurements
– Transformations
• Quantum network coding
– Negative results
– Positive results
Basic of Quantum Information
Mathematical Representation of
Quantum Mechanics
Quantum mechanics
quantum state
state space
measurement
evolution
Representations
vector
vector space
projection
unitary
Qubit
• Bit：＝Basic unit of information and
computation
• Quantum bit (qubit)：＝Basic unit of
"quantum" information and computation
– allows a superposed state of the 2 basis
states which corresponds "0'' and "1"
Implemented by various micro systems:
|0>+|1>
•
Nuclear spin of an atom ("up" or "down")
•
Polarization of the light （"vertical" or "horizontal"）
Qubit
Represented by a unit vector of the two dim. complex-valued
vector space
(probability) amplitude
orthogonal
basis state
1
0 :  
0
a
  a 0  b 1 :  
b
 0
1 :  
1
ket
ei 
and

represent the same state
0

b
If a, b are complex, it
is identified as a vector
1
0  1 
2
on the unit ball, called
"Bloch sphere"
a
If a, b are real, a state is represented in 2D plane
1
0 i1 
2

1
To Get Classical Information
Measurement is represented by a set of projections
  a 0 b1
Measurement
E0 , E1
projection on
span 0 
After the measurement
0
prob.
a (: E0 
2
The state becomes
1
prob.

b
2
b
Measurement
a
0
a
)
0
b (: E1 
The state becomes
2
1
2
)
Multiple Qubits
A m-qubit state is in the 2m dim. vector space that is the tensor product of
m 2-dim. spaces.
1
 
1 1  0
00 :        
 0  0  0
 0
 
1  0
01 :     
 0 1
 1  0 : 1 0
2-qubit state
 0 1
10 :     
1  0
 :
a 00  b 01  c 10  d 11
 0  0
11 :     
1 1
After the measurement
 
 x x
x0,1n
Measurement
Ex x0,1
x
prob.
x
n
The state becomes
normalization
x 
Ex 
Ex 
2
Entanglement
Entanglement is one of important key words in the quantum world,
which is a quantum correlation between two (or more) qubit states.
Mathematically, a two-qubit state is called entangled if it cannot be
"represented" by any tensor product of two single qubits
Ex.
1
 00  11   a 0  b 1 c 0  d 1 
2
A quantum state on two quantum registers (=quantum
systems) R and Q are called entangled if

R
Q
PQ
 
P

Q
Partial Measurement
1
( 0  1 ) 0
( I  E0 ) 
1
3


( 0  1 ) 0
State after getting 0: ＝
( I  E0 ) 
2
2
3
2-qubit state
1
 
( 00  10  11 )
3
1

( 0  1 ) 0
3
1

1 1
3
1
( 00  10 )
2
Result 0 (prob. 2/3)
Measure 2nd
qubit
a     b     (a   b  )  
11
Result 1 (prob. 1/3)
To Transform Quantum States
A measurement collapses a superposition
  a 0 b1
Measurement
E0 , E1
0
On the contrary, the quantum mechanics allows us to transform a
superposed state into another state, which is mathematically
described as a unitary transformation.
Important quantum gates

Quantum gate
U
U
•
Hadamard transformation
•
Controlled NOT
•
(Controlled) Phase-shift
Hadamard Transformation
matrix representation
Hadamard gate

1
x
H x 
0   1 1
2
1
0 :  
0
H
 0
1 :  
1
H

1 1 1 


H
2 1  1
1
1 1
 
( 0  1 ) :
2
2 1
1
1 1
 
( 0  1 ) :
2
2   1
Hadamard gate is essential to prepare the uniform superposed state of all n classical bits
0 0 0
Apply H for each qubit
n qubits
1
 0  1  1  0  1  1  0  1 
2
2
2

1
2
n
x
x{0,1}n
Controlled NOT
Controlled NOT gate
CN x y  x x  y
control part
target part
matrix representation
1

0
CN  
0

0

0
1
0
0
0
0
0
1
0

0
1

0 
Notice: CN is essentially "classical" transformation since it flips the target part
if the control part is 1. But the fact that CN can apply to any superposed
state is very important in quantum information processing.
Toffoli gate (Controlled-Controlled NOT)
T x y z  x y xy  z
Basic Fact
Any function f computed efficiently by a classical computer can be computed
efficiently using only Toffoli gates, which means that the transformation
x y 0  x y f ( x, y)
can be efficiently implemented by quantum computing
Phase-Shift
Phase-shift gate
Z x   1 x
x
matrix representation
1 0 

Z  
 0  1
1/√2
It does not change classical states
but change superposed states!
1
0  1 
2
1
 00  01  10  11 
2
Z
Apply Z at 2nd
qubit
1/√2
1
0  1 
2
1
 00  01  10  11 
2
How to Process Quantum Information
 Prepare quantum bits as much as you need
 Choose a few qubits and apply
 unitary transformation
 measurement
 You can introduce a "fresh" qubit if you want
 You can use previous measurement results for
choosing your operations
 The final result may be classical or quantum
(depending on your task)
Ex.1: Controlled Phase-Shift
Controlled phase-shift gate
matrix representation
CZ x y   1 x y
1

0
CN  
0

0

xy
control part
target part
0
1
0
0
0 0

0 0
1 0

0  1
CZ can be implemented using Toffoli and Hadamard.
x y 1
input
x y
1
0  1 
2
H on the 3rd qubit
control part
target part
1
1
xy
 xy  1  xy    1 x y  0  1 
x y
2
2
Toffoli
output
Similarly,
x y   1
f ( x, y )
x y
is efficiently implementable
Ex.2: Quantum Teleportation
Quantum teleportation is sending an unknown state
from A to B under only local operations and
classical communication with the assistance of
pre-shared state between A and B
1
(0
2
A2
0
B
A

A1
a0
1
A2
pre-shared state (prior "entanglement")
1 B)
B
A1
b1
A
B

A1
input
B
output
only local operation
& classical communication
Quantum Teleportation: A's Local Transformation
1
(0
2
A2
0
B
1
A2
1 B)
A

A1
pre-shared
B
a0
A1
b1
A1
& classical communication
A1
 a 00
b1
A
0
B
A1
 0
A2
0
B
target part
A
A2
 a 01 A 1 B  b 11
1
A
B

0
B
 b 10
A
1
B
A applies CN
control part
a 00
1
0 B  a 01 A 1 B  b 10
A
0 B  b 11 A 1 B
A applies Hadamard on 1st qubit
a 00
A
 b 01
0
A
B
0
 a 10
B
A
 b 11
0
A
B
0
 a 01 A 1 B  a 11 A 1
B
 b 00
A
1 B  b 10
B

local operation
input
a 0
A
A
B
1
B
B
output
Quantum Teleportation: A's Measurement
1
(0
2
A2
0
B
1
A2
1 B)
A

A1
pre-shared
B
a0
A1
b1
A
A1
a 00
& classical communication
A
 b 01
 00
 10

local operation
input
0
A
B
0
 a 10
B
a 0
a 0
 b 11
A
B
A
B
cd
A
A
b1
b1
a d
0
A
B
B
B
0
 a 01 A 1 B  a 11 A 1
B
 b 00
1 B  b 10
A
  01 a 1
 11 a 1
A
A
b 0
B
B
b 0


A
B
1
B
B
A measures the two qubits
 b 1 1  d
B
c
B

B
B
B
output
Quantum Teleportation: Classical
Communication + B's Local Transformation
1
(0
2
A2
0
B
1
A2
A

A1
a0
1 B)
pre-shared
B
b1
A1
A1
A
B

local operation
input
output
& classical communication
cd
cd
A
A
a d
a d
 b 1 1  d
B
c
B

A sends 2bits c and d
 b 1 1  d
B
c
B
 cd
target part
cd
A
a 0
B
control part
B applies CN
B

 b 1 1 B cd
c
target part
cd
A
a 0
B

 b 1 B cd
B
output
B applies Controlled phase-shift
B
control part
B
Quantum Network Coding
Network Coding Problems in This Talk
 We consider the "solvability."
 We consider the multiple unicast problem.
Instance:
– a directed acyclic graph G=(V,E), where each edge
has a unit capacity
– k source-target pairs (s1,t1),...,(sk,tk) where each sj has
a message xj
An instance is solvable if there is a network
coding protocol which sends xj from sj to tj for
every j.
 For simplicity, we assume that the alphabet is binary
Revisiting Butterfly
[Ahlswede-Li-Cai-Yeung 2000]
y
x
s1
POINT 1
Information can
be copied
y
x
x
s2
s0
y
x⊕y
x⊕y
t2
y=x⊕x⊕y
t0
POINT 2
Information can be encoded
x⊕y
t1
x=y⊕x⊕y
Quantum Butterfly
|Ψ2>
|Ψ1>
s1
s2
s0
• Information to be sent is
quantum states
• Quantum operation is possible
at each node
POINT 1
Quantum information
cannot be copied
• Every channel is quantum
t0
t2
• Source nodes may have an
POINT state
2
entangled
How quantum information is encoded?
t1
Q. If an instance is classically solvable, is quantum also?
Quantum Information cannot be Copied
(No-cloning theorem: Wootters-Zurek)
An "unknown" quantum state cannot be cloned.
X
0 a  00  ,
U
U
1 a 
11 
U
( 0   1 ) a 
 00    11 
 ( 0   1 )( 0   1 ) 
Multiple Unicast is a Natural Target
Multicast
Multiple Unicast
|φ1>, |φ2>, |φ3>
|φ1>
|φ2>
|φ1>, |φ2>, |φ3>
|φ1>, |φ2>, |φ3>
|φ1>, |φ2>, |φ3>
c.f. Shi-Soljanin 2006, Kobayashi et al. 2010
|φ2>
|φ1>
How Quantum Information is Encoded?
We cannot whether b=0 or b=1 when U|Ψ1>＝V |Ψ1>
|Ψ1>
s1
b
|Ψ1>
s0
b
s2
Apply U if b=0,
V if b=1
Sending |Ψ1> and b
simultaneously seems
to be impossible
t0
t2
t1
Negative Results
１qubit
１qubit
|φ1>
|φ2>
m qubits
|φ1>
one shot
|φ2>
m qubits
|φ2>
Use of network n times
|φ1>
|φ2>
|φ1>
Unsolvable under a single use
Unsolvable even asymptotically, i.e., under
of the network (one shot)
the condition m/n goes to 1 asymptotically
[Hayashi-Iwama-N-Raymond-Yamashita07]
[Hayashi07, Leung-Oppenheim-Winter10]
Under Additional Resources
• Entanglement
– among sources [Hayashi07]
– among neighboring nodes [Leung-Oppenheim-Winter10]
• Classical channel [LOW10, Kobayashi-Le Gall-N-Roetteler09 & 11]
– Much cheaper than quantum: LOCC (Local Operation &
Classical Communication) is easier than quantum
communication.
Q. If an instance is classically solvable, then is quantum
also under free classical communication?
Our Question
Q. If an instance is classically solvable, then is quantum
also under free classical communication?
Quantum
Classical
x1
xk
x2

x1 ,.., xk
x1 ,..., xk
x1 ,.., xk
sources
sources
quantum channel
classical channel
solvable!
?
message is classical
x1
x2
message is quantum
sinks
xk
solvable!
free classical communication

x1 ,.., xk
sinks
x1 ,.., xk
x1 ,..., xk
Our Result
If there is a classical coding protocol for an instance,
then there is also a quantum coding protocol for the
corresponding quantum instance.
[Kobayashi-Le Gall-N-Roetteler 2011]
Our previous results was:
If there is a classical linear coding protocol for an instance,
then there is also a quantum coding protocol for the
corresponding quantum instance.
[Kobayashi-Le Gall-N-Roetteler 2009]
Idea of Our Protocol
Our protocol consists of three stages
1. Node-by-node simulation
– 1 qubit for each edge
2. Removal of internal registers
– 1 bit backward for each edge
3. Removal of initial registers
– 1bit forward for each edge
Stage1: Node-by-node Simulation
Classical
Quantum
um
u2
y2
u1
1 qubit for each edge
u2
um
u1
P2
ym
y1
v
P1
Pm
y1
v
R
R
y1
f : f ( y1 ,, ym )
P1
P1
 ym
 ym
Pm
Pm
f
w
w
1-1. Receive registers P1,...,Pm
1-2. Introduce fresh register R and apply the
unitary transformation
y1 ,..., ym 0  y1 ,..., ym f
on registers P1, P2,..., Pm and R.
1-3. Send R to w.
R
Stage1: Node-by-node Simulation

   x, y x
P1, P 2
P1
P1
y
x
R2
P2
x, y
y
R1
R4
x
R2 R4
COPY : x 0 0  x x x
x y
x
R4
R6
x, y
R1
XOR : x y 0  x y x  y
x
P1
R1
x
R2
y
P2
y
R3
R3
y
R5

x P1 x R1 x R 2 y P 2

x
,
y
x, y x, y x P1 x R1 x R 2 y P 2 y R3 y
  x, y x
y
R4
x y
Q2
R6
y
R5
R7
x y
x  y R7
R3
x
Q1
y
R5
x, y

x
x, y
P1
x
R1
x
R2
y
P2
y
R3
y
R4
x y
R5
x y
R6
x y
R7
x, y

x, y
x, y
x
P1
x
R1
x
R2
y
P2
y
R3
P2
y
R4
x y
R5
x y
R6
x y
R7
x
Q1
y
Q2
Stage2: Removal of Internal Registers
1 bit backward for each edge
2. Do the following for internal registers in the
inverse topological order
2-1. Apply the Hadamard transformation on R,
and measure it.
v
H
zy
y 
(

1
)
z

P1, P2,..., Pm
a
z
measure
  1 a
ay
R
w
y1 2-1
 ym
P1
After
 1af ( y
1,, y m )
f
Pm
y1
P1
2-2. Send the measurement value backward
R
 ym
Pm
a
2-3. Erase ''phase error" using phase-shift
transformation
R
y1
 ym
P1
  1
Pm
af ( y1 ,, ym )
After 2-3
y1
P1
 ym
Ignore R
Pm
since no correlation with other registers!!
y1
P1
 ym
Pm
Stage2: Removal of Internal Registers
2-1. Apply the Hadamard transformation on R7,
and measure it
P1
x
y
x
2-2. Send the measurement value backward.
R2 R4
P2
y
2-3. Erase phase error by
  1
a( x y )
x y
R5
x y
x y
x
y
R5
R5
a
R1
Q2
 x
 1
After
2-1
x, y
x, y
P1
x
x
R1
a( x y )
x, y
x
y
R2
P1
x
P2
R1
x
y
R3
R2
y
y
R4
P2
y
x y
R3
y
R5
R4
R6
x y
x  y R7
R3
x
Q1
y
x y
x y
R6
R5
x y
x y
x
R7
R6
a
Q1
R7
y
x
Q2
Q1
y
x, y
After 2-3

x, y
x, y
x
P1
x
R1
x
R2
y
P2
y
R3
y
R4
x y
R5
x y
R6
x ignore
y this
Q1
Q2
Q2
Stage2: Removal of Internal Registers
2-1. Apply the Hadamard transformation on R6,
and measure it
P1
x
y
x
2-2. Send the measurement value backward.
R2 R4
P2
y
2-3. Erase phase error by
x y
  1
b( x y )
R5
x y
x y
x
y
R5
R5
b
R1
Q2
 x
 1
After 2-3  
After 2-1
x, y
x, y
P1
x
b( x y )
x, y
R1
x
x
R2
P1
x
y
R1
P2
x
y
R2
R3
y
y
P2
R4
y
x y
R3
y
x, y
x, y
x, y
x
P1
x
R1
x
R2
y
P2
y
R3
y
R4
R5
R4
R6
x y
x  y R7
R3
x
Q1
y
x y
x y
x y
R5
R6
R5
x
x
b
Q1
Q1
R6
y
y
x
Q2
Q2
Q1
y
ignore this
Q2
Stage2: Removal of Internal Registers
2-1. Apply the Hadamard transformation on R6,
and measure it
P1
x
y
x
2-2. Send the measurement value backward.
R2 R4
P2
y
2-3. Erase phase error by
x
R2
y
  1
c( x y )
R4
x
x
y
R2
Q2
x, y
x, y
x, y
x, y
x, y
P1
x
R1
x
R2
y
P2
y
R3
y
R4
R6
x y
R5
x
Q1
y
x xx xx x y y y y y y x c y x
x, y
x, y
x
P1
y
P2
x
Q1
y
Q2
R3
x
Q1
Q2
c( x y )
x, y
P1 P1 R1 R1 R 2 R 2 P 2 P 2 R 3 R 3 R 4 R 4 Q1 R 5 Q 2 Q1

x  y R7
y
x y
By continuing these, we have
y
R5
R4
R1
 x
After
After 2-1
2-3
 1
x y c
y
Q2
Stage3: Removal of Initial Registers
1 bit forward for each edge
After Stage 2

P1
x1 ,.., xk
x1
 xk
P1
Pk
x1
 xk
Q1
P2
Pk
sources
Qk
x1 ,.., xk
3-1. Apply the Hadamard transformations
on the initial registers P1,....,Pk, and
measure them.
sinks
After Stage 3.1
Q1
a x  a x




1
a1
 x ,.., x
1 1
1
k
k k
P1
 ak
Pk
x1
Q1
 xk
Q2
Qk
x1 ,.., xk
ignore these
3-2. Send the measurement values using the classical network coding protocol.
3-3. Erase ''phase error" using phase-shift transformation.
Qk
Stage3: Removal of Initial Registers
After Stage 2
 x, y x
y
P1
x
P2
Q1
P1
y
a
b
Q2
x, y
a
b
3-1. Apply the Hadamard transformations
on P1,P2, and measure them.
 1
axby
x, y
a
P1
b
P2
a
x
P2
Q1
y
a b
b
Q2
x, y
a b
ignore
3-2. Send the measurement values using
Q2
a b
a
b
the classical network coding protocol.
Finally, we obtain
3-3. Erase ''phase error" by
x
y
  1
Q1
ax
  1
by
Q2
x
Q1
y
Q2

x, y
x, y
x
Q1
y
Q2
Q1
Comments for Free Classical
Communication
• KLNR11 reduces the amount of classical
communication compared to KLNR09
– k*m*#(node) where m:=max fan-in of all nodes
[KLNR09]
– 1 bit forward +1 bit backward for each edge = total
2*#(edge) [KLNR11]
• Sending classical bits backward is necessary
– Quantum butterfly is not solvable even for the case
where free classical communication is allowed in the
direction of edges. [Leung-Oppenheim-Winter 2010]
Summary
• No additional assistance
– Butterfly is not solvable [HINRY07, LOW10, H07]
– Routing is optimal for a few cases [LOW10]
• 2 source-sink pairs
• shallow networks (including butterfly)
• Under free classical communication
– If an instance is classically solvable, then the
corresponding instance is also quantumly solvable [KLNR11]
– Additional classical communication is efficient
– Outer/inner bound for a few cases [LOW10]
Future Work
Generally;
• Advantage from classical, say, for security or complexity
• Lossy quantum channels
• Application (such as wireless communication in classical case), etc.
[No additional resources]
Q. If an instance is quantumly solvable with network coding,
then is it solvable with routing?
A. Unknown. Yes under only a few special cases
[LOW10]
[Under free classical communication]
[KLNR11]
An instance is classically
solvable
?
The corresponding quantum
instance is quantumly solvable
Future Work
In case where underlying graphs are directed
[Under free classical communication]
[KLNR11]
An instance is classically
solvable
X
The corresponding quantum
instance is quantumly solvable

x
s
s
t
t
quantumly solvable since sending two
bits backward enables us to
reverse the direction of the edge
by quantum teleportation!!
classically not solvable
In case where underlying graphs are undirected
[Under free classical communication]
[KLNR11]
An instance is classically
solvable
?
The corresponding quantum
instance is quantumly solvable