Download Toffoli gate

1
The state 2 ( 00 + 11 )
 Is entangled, to prove it we assume the
contrary 1 ( 00 + 11 ) = (a0 0 + a1 1 )(b0 0 + b1 1 ) =
2

!
= a0b0 00 + a0b1 01 + a1b0 10 + a1b1 11 "
a0b0 =
1
2
a0b1 = 0
a1b0 = 0
1
a1b1 =
2
contradiction
!
Toffoli gate
T : F23 " F23 ,T(x1, x 2 , x 3 ) = (x1, x 2 , x1 # x 2 $ x 3 )
This gate is called Toffoli gate
 Toffoli gate does not change bits x1 and
x2
 It computes the not-operation on x3 only
if x1=1 and x2=1

!

Symbol for Toffoli gate
1

All Boolean circuits can be simulated by using only
reversible gates


Not gates are reversible
And gate are simulated by Toffoli gate with x3=0
T(x1, x 2 , x 3 ) = (x1, x 2 , x1 " x 2 # x 3 )
T(x1, x 2 ,0) = (x1, x 2 , x1 " x 2 )


x1∨x2=¬(¬x1∧¬x2)
Fanout (multiple wires leaving a gate) is simulated by the
controlled not-gate with x2=0
!
C(x1, x 2 ) = (x1, x1 " x 2 ) C(x1,0) = (x1, x1 )
!
Spectral Representation of
Unitary Operators

Functions eiT defined on a self-adjoint operator
defined by spectral representation
T = "1 x1 x1 + "2 x 2 x 2 + L + "n x n x n
e iT = e i"1 x1 x1 + e i"2 x 2 x 2 + L + e i"n x n x n
iT *
!
!
!
(e )


= e"i#1 x1 x1 + e"i#2 x 2 x 2 + L + e"i#n x n x n = (e iT )
eiT is unitary
"1
Each unitary mapping can be represented
as eiT, where T is a self-adjoint operator
2
Shor’s Algorithm




Shor’s quantum algorithm for factoring relies upon a
result from number theory
Relates the period of a particular periodic function to
the factor of an integer
Given an integer n (number to be factored) construct a
function
fn(a)=xa mod n


where x is an integer chosen at random that is a coprime to n
Coprime, means that the greatest common divisor of x and n is
1, gcd(x,n)=1
3

Why is this function interesting with respect to
the problem of factoring n





It turns out that fn(a) is periodic
For a=0,1,2,3,.. the values of the function
fn(0),fn(1),fn(2),fn(3),.. fall into repeating pattern
eventually
Different values of x give rise to different patterns
The number of values in between the repeating
pattern, for a particular value x is called period of x
modulo n indicated by r
xr=1 mod n
Shor’s Algorithm for factoring n
1) Pick a number q (with small prime factors)
such that
2n 2 " q " 3n 2
!
2) Pick a random integer x that is coprime to n
3) Repeat steps labeled (a) through (g) order
log(q) times, using the same random number x
each time
4
(a) Create a quantum memory register
and partition the qubits into two sets,
called Register1 and Register2
 If the qubits in Register1 are in the state
reg1 and those in Register2 are in the
state reg2, we represent the joint state of
both registers as (decimally)

reg1,reg2
!

(c) Apply exploiting quantum parallelism
the transformation xa mod n to each
number in Register1 and Place the
results in Register2
q#1
1
" =
a, x a mod n
$
q a= 0
!
5
(d) Measure the state of Register2
obtaining some result k
 This has the effect of projecting out the
state of Register1 to be a superposition
of just those values of a such that


xa mod n=k
" =

1
A
$ a',k
a' #A
Where A={a’: xa mod n=k} and ||A|| is the
number of elements in this set
!

How to find the period r of xa mod n=k?
We will compute the Fourier transform of |a’>
 Fourier transform can be represented by an
unitary operator (Quantum Fourier transform)

• Unitary because of the Parseval’s identity

States corresponding to integer multiplies of
the inverse period, and these close to them
have a higher value (greater amplitude)
6
(e) We compute the discrete Fourier
transform of the projected state
inRegister1
 |a’> is mapped into

q%1
1
a' "
e 2 #ia$ c / q c
&
q c= 0
' =
1
A
q%1
&
a' (A
1
e 2 #ia$ c / q c,k
&
q c= 0
!

(f) Measure the state of Register1
• The discrete Fourier transform is sampled

This returns some number c’
c' #
"
q r
!
7
(g) To determine the period r we need to
estimate λ
 Accomplished by computing the
convergent of the fraction expansion of
c’/q and retaining the closest such
fraction of λ/r

Grover’s Amplification

Operators which we will use:

We need a query operator which calls for
value fy uses n qubits for the source register
and one target bit
y " F2n
V f x = ("1) f (x ) x

!
!
# 1,if x = y
f y (x) = $
%0, otherwise
We need a quantum operator Rn defined on
n qubits and operating as
Rn 0 = " 0
!
and
Rn x = x ,x # 0
8
Amplitude Amplification

Finding y by the quantum operator
• Gn=-HnRnHnVf
• Working on n qubits representing elements x
• HnRnHn can be written as a 2nx2n matrix
#
2
%1" 2 n
% 2
%" n
% 2
H n Rn H n = % 2
"
% 2n
% M
%% " 2
$ 2n
2
2n
2
1" n
2
2
" n
2
M
2
" n
2
"
2
2n
2
" n
2
2
1" n
2
M
2
" n
2
"
L
L
L
O
L
2 &
2n (
2 (
" n (
2 (
2
" n (
2 (
M (
2
1" n ((
2 '
"
!

HnRnHn can be also expressed as


HnRnHn=I-2P
Where I is a 2nx2n identity matrix and P is
a 2nx2n projection matrix whose every
entry is 1/2n
9

In this example we consider function
y " F2n
# 1,if x = y
f 5 (x) = $
%0, otherwise
• The search begins with superposition
!
1
2n
#x
x "F2n
c 0 = c1 = c 2 = Lc 2 n $1 =
1
2n
!
Vf5 is applied o change the sign of x=y
 Those amplitudes that are coefficients of
a vector |x> satisfying f5(x)=1 become
negative, c5 becomes negative

10

The average of the amplitude is now
A=

!
1# n
1
1 &
1 #
2&
2 "1) n " n ( =
1"
%
n %(
n(
2 $
2
2 '
2n $ 2 '
Inversion about the average-operator
-HnRnHn will perform a transformation
1
2
"
n
a 2A "
1
2
n
a 2A +
1
2
1
#
n
1
2
2n
# 3$
n
1
2n
!
1
2
"

n
a 2A "
1
2
n
a 2A +
1
2
n
#
1
2
n
1
2n
# 3$
1
2n
The probability to find the answer is 9/2n by a single query, 4.5
times better than a classical randomized search can do
!
11

Iterative use of the mapping

Gn=-HnRnHnVf

Instead of a blackbox function that
assumes only one solution, we will study
a general function f having k solutions

By using a quantum circuit, any problem
in NP can be solved with a nonvanishing
correctness probability in time
(
O 2 n p(n)

!
)
Where p is polynomial depending on the
particular problem
12
No teleportation theorem

A classical information channel can not
transmit quantum information
Remember: no cloning theorem?
 Quantum states can not be copied!

A quantum state cannot be determined
via a single measurement
 Once converted to classical information,
quantum information cannot be recovered






The entangled bits or qubits of a state are called an ebit
An ebit is a shared resource
An ebit is allways disrtributed between two particles
(qubits)
1
An ebit provides a channel
11 )
( 00for+ communication
2
Once either particle comprising the ebit is measured,
the states of both particles become definite
!
13

Bell basis describes four orthogonal
states
1
+
( 00
2
1
"# =
( 00
2
1
$+ =
( 01
2
1
$# =
( 01
2
"
=
+ 11 )
# 11 )
+ 10
)
# 10
)
!


The three particle state shown above thus becomes the
following four-term superposition in the new basis:
%# (
%$ (
%+$ ( 0
1 - + %# (
+
+
+
"
+
"
+
,
+
,
/
2
'
*
'
*
'
*
' * part
We2 have& $done
a change
of basis
on Alice's
of the
)B
&+$ ) B
&# ) B
& # )B 1
.
system into the orthogonal basis (Bell state)
• No operation has been performed and the three particles are still
in the same state
!
14

Given the density matrix p, von Neumann defined the
entropy as
S( p) = "Tr( pln p)


!

It is a proper extension of the Gibbs entropy (and the
Shannon entropy) to the quantum case
We note that the entropy S(p) times the Boltzmann
constant equals the thermodynamical or physical
entropy
It can be shown that if two observables
are measured simultaneously, the
uncertainty in their joint values must
always obey the inequality (Heisenberg
Uncertainty) ˆ ˆ 1 ˆ ˆ
"A"B #
[ ]
A, B
2
1
"Aˆ "Bˆ #
x Aˆ , Bˆ x
2
1
Var$ (A)Var$ (B) #
x Aˆ , Bˆ x
4
[ ]
[ ]
!
2
15
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