Download Hadamard Gates - UW

Alice and Bob’s
Excellent Adventure
Presented by:
Lacie Zimmerman
Adam Serdar
Jacquie Otto
Paul Weiss
What’s to Come…
• Brief Review of Quantum Mechanics
• Quantum Circuits/Gates
• No-Cloning
• Distinguishability of Quantum States
• Superdense Coding
• Quantum Teleportation
Dirac Bra-Ket Notation
Notation
Inner Products
Outer Products
Bra-Ket Notation Involves
Vector Xn can be represented two
ways
Ket
Bra
|n>
<n| = |n>t
v 
 
 w
x 
 
y
z 
 
v w x y z 
*
*
* *
* *
m is the complex conjugate of m.
Inner Products
An Inner Product is a Bra multiplied by a Ket
<a| |b> can be simplified to <a|b>
<a|b> =
v
*
w* x* y* z *

l 
 
 m
 n  = lv *  mw*  nx*  oy*  pz*
 
o 
 p
 
Outer Products
An Outer Product is a Ket multiplied by a Bra
l 
 lv * lw*
 
 * *
 m
 mv mw
n  *

*
*
*
*
*
*
|a><b| =
=
nv
nw
  v w x y z

o 
 ov* ow*
 * *
 p
 
 pv pw

By Definition

 a b c
 bc a
lx * ly * lz * 

*
*
*
mx my mz 
*
*
*
nx ny nz 
*
*
*
ox oy oz 
*
*
*
px py pz 
• State Space: The inner product space V  C

associated with an isolated quantum system.
n
•The system at any given time is described by a
“state”, which is a unit vector in V.
 (Qubit)
• Simplest state space - V  C
If | 0 and | 1 form a basis for V,
then an arbitrary qubit state has the form
| x  a | 0  b | 1 , where a and b in C

2
2
have | a |  | b |  1 .
2
• Qubit state differs from a bit because
“superpositions” of a qubit state are possible.
The evolution of an isolated quantum system is
described by a unitary operator on its state
space.
The state |  (t1 ) is related to the state |  (t2 ) by a
unitary operator U t1,t2 .
i.e., |  (t2 )  U t ,t |  (t1 )
1 2
Quantum measurements are described by a
finite set of projections, {Pm}, acting on the
state space of the system being measured.
• If |   is the state of the system immediately
before the measurement.
•Then the probability that the result m occurs is
given by p(m)   | Pm |   .
• If the result m occurs, then the state of the
system immediately after the measurement is
Pm |  
Pm |  

1/ 2
 | Pm |  
p(m)
• The state space of a composite quantum system
is the tensor product of the state of its
components.
• If the systems numbered 1 through n are
prepared in states |  (ti ), i = 1,…, n, then
the joint state of the total composite system
is |  1      |  n .
Quantum Uncertainty and
Quantum Circuits
Classical Circuits vs. Quantum Circuits
Hadamard Gates
C-not Gates
Bell States
Other Important Quantum Circuit Items
Classical Circuits
vs.
Quantum Circuits
Classical Circuits based upon bits, which are
represented with on and off states. These states
are usually alternatively represented by 1 and 0
respectively.
The medium of transportation of a bit is a
conductive material, usually a copper wire or
something similar. The 1 or 0 is represented with
2 different levels of current through the wire.
Circuits Continued…
Quantum circuits use electron “spin” to hold
their information, instead of the conductor
that a classical circuit uses.
While a classical circuit uses transistors to
perform logic, quantum circuits use
“quantum gates” such as the Hadamard
Gates.
Hadamard Gates
Hadamard Gates can perform logic and are
usually used to initialize states and to add
random information to a circuit.
Hadamard Gates are represented
mathematically by the Hadamard Matrix which
is below.
1
H
2
1 1 
1  1


Circuit Diagram of a
Hadamard Gate
When represented in a Quantum Circuit
Diagram, a Hadamard Gate looks like this:
x
H
y
Where the x is the input qubit and the y is
the output qubit.
C-Not Gates
C-not Gates are one of the basic 2-qubit gates in
quantum computing. C-not is short for controlled
not, which means that one qubit (target qubit) is
flipped if the other qubit (control qubit) is |1>,
otherwise the target qubit is left alone.
The mathematical representation of a C-Not Gate
is below.
1 0 0 0
U CN
0 1 0 0 


0 0 0 1 


0 0 1 0 
Circuit Diagram of a C-Not Gate
When represented in a Quantum Circuit
Diagram, a C-Not Gate looks like this:
x
x
y
x y
Where x is the control qubit and y is the
target qubit.
Bell States
Bell States are sets of qubits that are entangled.
They can be created with the following Quantum
Circuit called a Bell State Generator:
x
H
 xy
y
With H being a Hadamard Gate and x and y
being the input qubits.  is the Bell State.
Bell State Equations
The following equations map the previous Bell State
Generator:
1
 00  00   1  00  11    00
2
2
1
 01  11   1  01  10    01
01 
2
2
1
 00  10   1  00  11   10
10 
2
2
00 
11 
1
 01  11   1  01  10   11
2
2
So we can write:
0 y   1 1y
x
 xy 
2
Other Important Quantum
Circuit Items
• Controlled U-Gates
• Measurement Devices
Controlled U-Gate
A Controlled U-Gate is an extension of a C-Not
Gate. Where a C-Not Gate works on one qubit
based upon a control qubit, a U-Gate works on
many qubits based upon a control qubit.
A Controlled U-Gate can be represented with the
following diagram:
n
U
n
Where n is the number of qubits the gate is
acting on.
Measurement Devices
These devices convert a qubit state into a
probabilistic classical bit.
It can be represented in a diagram with the
following:

M
x
A measurement with x possible outcomes has x
wires coming from the device that measures it.
Cloning of a
Quantum
State
Cloning
Can copying of an unknown qubit
state really happen?
By copy we mean:
1. Take a quantum state Z
2. Perform an operation
3. End with an exact copy of Z
Using a Classical Idea
• A classical CNOT gate can be used for an
unknown bit x
• Let x be the control bit and 0 be the target
• Send x0  xx where  is a CNOT gate
• Yields an exact copy of x in the classical
setting
Move the Logic to Quantum
States
• Given a qubit in an unknown quantum state
Z such that Z  a 0  b 1
• Through a CNOT gate we take Z 0  Z Z
such that (a 0  b 1 ) 0  a 00  b 10
• Note if indeed we copied Z we would thus
end up with Z Z which would equal
a² 00  ab 01  ab 10  b² 11
Limits on Copying
Note that:
a 00  b 10  a² 00  ab 01  ab 10  b² 11
only at ab=0 and for a and b being 0 or 1
Proving the difficulty of cloning
• Suppose there was a copying machine
• Such that Z can be copied with a standard
state S
• This gives an initial state Z S which when
the unitary operation U is applied yields
U Z S  Z Z
…difficulty cloning
• Let U( z s )  z z & U( y s )  y y
• By taking inner products of both sides:
zy  zy²
• From this we can see that: z y = 0 or 1
• Therefore this must be true: z  y or z  y
• Thus if the machine can successfully copy z it is
highly unlikely that the machine will copy an
arbitrary unknown state y unless y is
orthogonal to z
Final cloning summary
• Cloning is improbable.
• Basically all that can be accomplished is
what we know as a cut-n-paste.
• Original data is lost.
• The process of this will be shown in the
teleportation section soon to follow.
Distinguishability
• To determine the state of an element in the
set: y1 , y 2 ,..., y n
• This must be true:
- y1  y2  ...  yn
• Finding the probability of observing a
specific state ym , let Pm be the measurement
such that Pm  y m y m
Distinguishability cont.
• Then the probability that m will be observed is:
- P(m)  y m | Pm | y m
• Which yields P(m)  ym ym ym ym
• Because the set is orthogonal
-
P(m)  11  1
• If the set was not orthogonal we couldn’t know for
certain that m will be observed.
Cloning and
Distinguishability
• Take some quantum information
• Send it from one place to another
• Original is destroyed because it can’t just be
cloned (copied)
• Basically it must be combined with some
orthogonal group or distinguishing the quantum
state with absolute certainty is impossible.
• Pauli Matrices
• Alice & Bob
• The Conditions
• How it Works
0 1 
X 

1 0
0  i 
Y 

i 0 
1 0 
Z 

0  1
THE CONDITIONS…
• Alice and Bob are a long way from one
another.
• Alice wants to transmit some classical
information in the form of a 2-bit to Bob.
HOW IT WORKS…
• Alice and Bob initially share a 2-qubit in the
entangled Bell state


00

 11 
2
which is just a pair of quantum particles.
HOW IT WORKS…
•  is a fixed state and it is not necessary for
Alice to send any qubits to Bob to prepare
this state.
• For example, a third party may prepare the
entangled state ahead of time, sending one of
the qubits to Alice and the other to Bob.
HOW IT WORKS…
1) Alice keeps the first qubit (particle).
2) Bob keeps the second qubit (particle).
3) Bob moves far away from Alice.
HOW IT WORKS…
• The 2-bit that Alice wishes to send to Bob
determines what quantum gate she must
apply to her qubit before she sends it to
Bob.
The four resulting states are:
00 :          00 ,
01 : (    )    01 ,
10 :       10 ,
11 : (i  )   11 .
HOW IT WORKS…
• Since Bob is in possession of both qubits,
he can perform a measurement on this Bell
basis and reliably determine which of the
four possible 2-bits Alice sent.
Teleportation
What is it used for?
Teleportation Circuit
Teleportation
•Teleportation is sending unknown quantum
information not classical information.
•Teleportation starts just like Superdense coding.
•Alice and Bob each take half of the 2-qubit Bell
state:
 00   00  11  / 2
•Alice takes the first qubit (particle) and Bob
moves with the other particle to another location.
Teleportation
•Alice wants to teleport  to Bob:
•She combines the qubit  with her half of the
Bell state and sends the resulting 3-qubit (the 2
qubits-Alice & 1 qubit-Bob) through the
Teleportation circuit (shown on the next slide):
Teleportation
Circuit
Single line denotes
quantum information
being transmitted
Double line denotes
classical info being
transmitted

 00
H

{

M1

M2
Xy
0
1
2

Zx
3
Top 2 wires represent Alice's system
Bottom wire represents Bob’s system
4
Teleportation
Circuit

 00
H
{

0

M1

M2
Xy
1
2
3

Zx
4
Initial State
  a 0 b1
 0    00
1
a 0  00  11   b 1  00  11 

2
C-Not gate
Teleportation
Circuit

H
 00
{
0

M1


M2
Xy
1
2
3

Zx
4
After Applying the C-Not gate to Alice’s bits:
1
a 0  00  11   b 1  10  01 
1 
2
Teleportation
Circuit
Hadamard gate

 00
H
{

0

M1

M2
Xy
1
2
3

Zx
4
After applying the Hadamard gate to the first qubit:
2
1

2
 00 a 0  b 1   01 a 1  b 0   


 10 a 0  b 1   11 a 1  b 0  
Teleportation
Circuit
Measurement
devices

 00
H
{
0

M1


M2
Xy
1
2
3

Zx
4
After Alice observes/measures her 2 qubits, she
sends the resulting classical information to Bob:
00   3  a 0  b 1 , 01   3  a 1  b 0 ,
10   3  a 0  b 1 , 11   3  a 1  b 0 .
Teleportation
Circuit

 00
H
{
0

M1


M2
Xy
1
2
3

Zx
4
Bob applies the appropriate quantum gate to his
qubit based on the classical information from Alice:
00 : Z 0 X 0  3  I  3   4 ,
01 : Z 0 X 1  3  aX 1  bX 0  a 0  b 1   4 ,
10 : Z 1 X 0  3  aZ 0  bZ 1  a 0  b 1   4 ,
11 : Z 1 X 1  3  Z aX 1  bX 0   aZ 0  bZ 1  a 0  b 1   4 .
Teleportation
Bob finally recovers the initial qubit  4  
that Alice teleported to him.
Conclusion
• Brief Review of Quantum Mechanics
• Quantum Circuits/Gates
–
–
–
–
Classical Gates vs. Quantum Gates
Hadamard Gates
C-not Gates
Bell States
Conclusion, cont.
• No-Cloning
• Distinguishability of Quantum States
• Superdense Coding
- Pauli Matrices
- The Conditions
- How it Works
Conclusion, cont.
• Quantum Teleportation
- What is it used for?
- Teleportation Circuit
Bibliography
http://en.wikipedia.org/wiki/Inner_product_space
http://vergil.chemistry.gatech.edu/notes/quantrev/node14.html
http://en2.wikipedia.org/wiki/Linear_operator
http://vergil.chemistry.gatech.edu/notes/quantrev/node17.html
http://www.doc.ic.ad.uk/~nd/surprise_97/journal/vol4/spb3/
http://www-theory.chem.washington.edu/~trstedl/quantum/quantum.html
Gudder, S. (2003-March). Quantum Computation. American Mathmatical
Monthly. 110, no. 3,181-188.
Special Thanks to:
Dr. Steve Deckelman