Download Quantum Computing Devices Quantum Bits

10/10/13
Quantum Computing Devices!
Quantum Bits!
n  A
quantum bit, qubit, is a two-level
quantum system!
n  A two dimensional Hilbert space H2 is a
quantum qubit!
n  H2 has the basis !B = { 0! , 1 } !
!
!
!so-called computational basis!
n  States !0 , 1 !are called basis states!
€
€
1
10/10/13
n 
A general state of a single quantum bit is a
vector!
ω 0 0 + ω1 1
2
2
ω 0 + ω1 = 1
•  Having unit length!
€Observation of a quantum bit in such a state will
give 0 or 1 as an outcome with probabilities!
n 
2
ω 0 , ω1
2
€
n  An
operation on a qubit, called unary
quantum gate, is a unitary mapping !
!U: H2 → H2!
n  Unary
quantum gate defines a linear
operation, such that the corresponding matrix
is unitary !
•  a* stands for the complex conjugate number a!
0 → a 0 + b1
1 →c 0 + d1
# a b &# a*
and %
(% *
$ c d '$ b
c * & #1 0&
(=%
(
d * ' $0 1'
€
2
10/10/13
n  Let
use the coordinate representation!
T
T
0 = (1,0)
1 = (0,1)
n  The
unitary matrix defines an action!
"0 1%
M¬ = $
'
#1 0&
€
n  The
unitary quantum gate defined by M¬
is called a quantum-not gate!
€
M¬ 0 = 1 , M¬ 1 = 0
€
M¬ 0 = 1 , M¬ 1 = 0
n  Can
€
be written using the XOR operation!
M¬ x = x ⊕ 1
0 ⊕1 = 1
1⊕1 = 0
€
€
€
3
10/10/13
n 
Another quantum gate!
#1+ i
%
M¬ = % 2
1− i
%
$ 2
1+ i
1− i
0 +
1
2
2
1− i
1+ i
M¬ 1 =
0 +
1
2
2
1− i &
2 (
1+ i (
(
2 '
M¬ 0 =
2
2
1+ i
1− i
1
=
=
2
2
2
€
€
n 
0 and 1€ with a probability 1/2, because!
M¬ ⋅ M¬ = M¬
n 
Is called square root of the not-gate!
€
n  Gate
#
%
W2 = %
%
$
€
defined by W2!
1
2
1
2
1 &
(
2 (
1
(
−
2'
1
1
0 +
1
2
2
1
1
W2 1 =
0 −
1
2
2
W2 0 =
n  W2
is called Walsh matrix, Hadamard matrix
€
or Hamarad-Walsh
matrix!
•  Quantum gates a linear!
" 1
% 1
1
1 1
1 1
W 2$
( 0 + 1 )'& = 2 W 2 0 + 2 W 2 1 = 2 2 ( 0 + 1 ) + 2 2 ( 0 − 1 ) = 0
# 2
€
4
10/10/13
n  Let
use the coordinate representation!
0 =
1
T
(1,1)
2
1 =
1
T
(1,−1)
2
n  What
is the unitary matrix representation
€ of the quantum-not gate in this basis?!
M¬ 0 = 1 , M¬ 1 = 0
€
The closure Relation!
a basis set! { x1 , x 2 ,…, x n }
n  The identity operator can be written as!
n  Given
n
!
€
∑x
i
xi = I
€
i=1
n
n
#n
&
a = I a = % ∑ x i x i ( a = ∑ x i x i a = ∑ω i x i
$ i=1
'
i=1
i=1
€
5
10/10/13
One quantum bit!
I= 0 0 + 1 1
I=
1 1 0 0
+
0 0 1 1
"1
I =$
#0
"1
I =$
#0
0% "0 0%
'+$
'
0& #0 1&
0%
'
1&
€
Representation of Operators!
&
#
& #
A = IAI = %∑ x i x i ( A%%∑ x j x j ((
$ i
' $ j
'
A = ∑ xi A x j xi x j
i, j
" x1 A x1
$
$ x 2 A x1
$
$
# x n A x1
€
x1 A x 2
x2 A x2


x1 A x n %
'

'
'
'
xn A xn &
€
6
10/10/13
n  Matrix
representation of an operator with
respect to a certain basis!
" u1 A u1
$
$ u2 A u1
$
$
# un A u1
u1 A u2
u2 A u2


u1 A un %
'

'
'
'
un A un &
€
Operators in two dimensional space!
n  Convention
with respect to the
computational basis!
" 0 A0
A =$
# 1A0
0 A1 %
'
1 A1 &
€
7
10/10/13
Z operator!
Z 0 = 0,
#
Z =%
$
#
Z =%
$
Z 1 = −1
0 Z 1 & # 0 (Z 0 )
(=%
1 Z 1 ' $ 1 (Z 0 )
0Z0
1Z 0
0 (Z 1 )&
(
1 (Z 1 ) '
− 0 1 & #1 0 &
(=%
(
− 1 1 ' $ 0 −1'
00
10
€
Unitary Transformation!
n  Change
xi
of basis !
to
ui
" u1 x1
U =$
# u2 x1
u1 x 2 %
'
u2 x 2 &
a( = U a
A( = UAU *
€
8
10/10/13
n  Let
use the coordinate representation!
T
T
0 = (1,0)
1 = (0,1)
n  The
unitary matrix defines an action!
"0 1%
M¬ = $
'
#1 0&
€
n  The
new basis!
+ =
€
1
T
(1,1)
2
− =
1
T
(1,−1)
2
€
#+0
+1&
U =%
(
−1 '
$ −0
1 #1 1 &
U=
%
( = W2
2 $1 −1'
!
n  Hadarmad
€
n 
matrix (also called H)!
!+ , −
n  The
basis is called Hadamard basis !
€
9
10/10/13
U = U −1 = U *
1 #1 1 &#0 1&#1 1 &
UM¬U = %
(%
(%
(
2 $1 −1'$1 0'$1 −1'
#1 0 &
UM¬U = %
(
$0 −1'
€
Quantum Register!
n  A
system of two quantm bits is a fourdimensional Hilbert space! H 4 = H 2 ⊗ H 2
n  With the orthonormal basis!
{0
0 , 0 1,1 0 ,1 1}
€
0 0 = 00 , 0 1 = 01 , 1 0 = 10 , 1 1 = 11
n  We
€
€
write:!
n  A
state of a two-qubit system is a unitlength vector!
ω 0 00 + ω1 01 + ω 2 10 + ω 3 11
2
2
2
2
ω 0 + ω1 + ω 2 + ω 3 = 1
€
10
10/10/13
!
ω 0 00 + ω1 01 + ω 2 10 + ω 3 11
n  Observation
€
€
of two-qubit system gives,
00, 01, 10, 11 with probabilities!
2
2
2
2
ω 0 , ω1 , ω 2 , ω 3
n  Observation of one qubit (out of two) give
0 and one with probabilities for the first
and probabilities for the second!
2
2
2
ω 0 + ω1 , ω 2 + ω 3
2
2
2
2
2
( ω 0 + ω 2 , ω1 + ω 3 )
€
n 
The tensor product of vectors does not
commute!
0 1 ≠1 0
#0& #0&
% ( % (
%1( ≠ %0(
%0( %1(
% ( % (
$0' $0'
n 
We use linear ordering from left to right to
address the qubits individually!
€
11
10/10/13
n  A
state z ∈ H4 of a two-qubit system is
decomposable if z can be written as a
product of states in H2!
z=x⊗y
n  A
state that is not decomposable is
entangled!
€
n  The
1
state! 2 ( 00 + 01 + 10 + 11 )
n  Is decomposable because!
€
1
00 + 01 + 10 + 11 ) =
(
2
1
1
1
= (0 0 + 0 1 + 1 0 + 1 1)=
0 + 1 )⋅
(
(0 + 1)
2
2
2
€
"1%
$'
1 $1' 1 "1%
1 "1%
=
$ '⊗
$'
2 $1'
2 #1&
2 #1&
$'
#1&
(1)
Ψ
( 2)
⊗ Ψ
(2) (
%ω (1)
%ω 00 (
0 ω0
' (1) (2) *
' *
(2) (
%ω (1)
(
%
ω 01
ω
ω
ω
1,2
= ' 0(1) * ⊗ ' 0(2) * = ' 0(1) 1(2) * = Ψ( ) = ' *
'
*
'ω10 *
&ω1 ) &ω1 ) ω1 ω 0
' (1) (2) *
' *
& ω11 )
&ω1 ω1 )
€
€
12
10/10/13
1
state! 2 ( 00 + 11 )
n  Is entangled, to prove it we assume the
contrary! 1 ( 00 + 11 ) = (a0 0 + a1 1 )(b0 0 + b1 1 ) =
2
n  The
€
= a0b0 00 + a0b1 01 + a1b0 10 + a1b1 11 →
a0b0 =
1
2
a0b1 = 0
a1b0 = 0
1
a1b1 =
2
contradiction
€
1
If two qubits are entangled state! 2 ( 00 + 11 )
then observing one of them will give 0 or 1, both
with probability 1/2!
n  It is not possible to observe
€ different values on
the qubits!
n 
Experiments have shown that this correlation can
remain even if the qubits are separated more than 10
km!
n  Opportunities for quantum communication
(teleportation) !
n 
•  Such as state is also called EPR-pair (Einstein, Podlsky,
Rosen, regarded distant correlation as a source of paradox
for quantum physics) !
13
10/10/13
n  A
1
( 00 + 11 )
2
•  is called EPR-pair (Einstein, Podolsky and Rosen)!
pair of bits in the state!
n  If
both qubits are run through a Hadamard
gate, the resulting €state is again! 1 ( 00 + 11 )
2
n  GHZ (Greenberger-Horne-Zeilinger) state
state is as well entangled!
GHZ =
1
0
2
(
⊗M
! M = 3, 1 ( 000
2
+1
⊗M
), M > 2
€
+ 111 )
€
n  A
binary quantum gate is a unitary
mapping H4 → H4!
n  We
use the coordinate representation!
T
T
T
00 = (1,0,0,0) , 01 = (0,1,0,0) , 10 = (0,0,1,0) , 11 = (0,0,0,1)
T
€
14
10/10/13
n 
Gate Mcnot!
M cnot
"1
$
0
=$
$0
$
#0
0
1
0
0
0
0
0
1
0%
'
0'
1'
'
0&
M cnot 00 = 00 , M cnot 01 = 01 , M cnot 10 = 11 , M cnot 11 = 10
n 
€
Is called controlled not, since the second qubit
(target qubit) is flipped if and only if the first
(control qubit) is 1!
Tensor product, Kornecker product of matrices!
" b11 b12
 a1s %
'
$
 a2s '
b
b
,B = $ 21 22
$
  '

'
$
 ars &
# bt1 bt 2
" a11B a12 B  a1sB %
$
'
a21B a22 B  a2sB'
$
A⊗B=
$ 


 '
$
'
# ar1B ar 2 B  arsB &
" a11 a12
$
a
a
A = $ 21 22
$ 

$
# ar1 ar 2
 b1u %
'
 b2u '
  '
'
 btu &
€
15
10/10/13
n 
n 
n 
If M1 and M2 are 2 x 2 matrices that describe
unitary quantum gates, then it is easy to verify
that the joint actions of M1 of the first qubis and
M2 on the second are described by M1 ⊗ M2!
This generalize to quantum systems of any size!
If matrices M1 and M2 define unitary mappings
on Hilbert soace Hn and Hm, then nm x nm
matrix M1 ⊗ M2 defines a unitary mapping on
the space Hn ⊗ Hm!
n  Let
M1=M2=W2 be the Hadamard matrix!
$1 1 1 1 '
&
)
1 &1 −1 1 −1)
W4 = W2 ⊗ W2 =
2 &1 1 −1 −1)
&
)
%1 −1 −1 1 (
W 4 x 0 x1 =
€
1
( 0 + (−1)x0 1 )( 0 + (−1)x1 1 )
2
1
= ( 00 + (−1) x1 01 + (−1) x 0 10 + (−1) x 0 +x1 11 )
2
x 0 , x1 ∈ {0,1}
€
16
10/10/13
n  The
result of action of W4 in our example
is decomposable. Why?! H 4 = H 2 ⊗ H 2
n  Input state is decomposable!
x 0 x1 = x 0 , x1 = x 0 x1 = x 0 ⊗ x1
€
x 0 , x1 ∈ {0,1}
n  W4 is decomposable!
€
W4 = W2 ⊗ W2
€
€
n  Mcnot
cannot be expressed as a tensor
product of 2 x 2 matrices!
n  Consider 00 (…….operator is linear) !
U t (ω1 x1 + ω 2 x 2 + …ω n x n ) = ω1U t x1 + ω 2U t x 2 + …ω nU t x n
1
1
(0 + 1)0 =
( 00 + 10
€
2
2
1
1
M cnot
( 00 + 10 ) =
(M cnot 00 + M cnot 10 )
2
2
1
1
M cnot
( 00 + 10 ) =
( 00 + 11
2
2
W2 0 0 =
€
€
17
10/10/13
n  The
action of Mcnot turns a decomposable
state into an entangled state!
n  ..it
cannot be a tensor product of two
unary operators!
M cnot
M cnot
1
( 00 + 11 ) =
2
1
( 00 + 11 ) =
2
1
(M cnot 00 + M cnot 11 )
2
1
1
( 00 + 10 =
(0 + 1)0
2
2
1
W 2 ( 0 + 1 ) 0 = 0 0 = 00
2
€
18
10/10/13
n 
n 
n 
By a quantum register of length m we
understand an ordered system of m qubits!
State is the m-fold tensor product!
H2m = H2 ⊗  ⊗ H2
The basis states are!
{ x

m
x ∈ {0,1}
}

x = x1 … x m
€n  We can also say that the basis states of m-qubit
register are!
{a
€
a ∈ {0,1,…,2 m −1}}
€
n 
A peculiar feature feature of exponential
packing density associated with quantum
system of m bits corresponds to!
c 0 0 + c1 1 +  + c 2 m −1 2 m −1
2
2
2
c 0 + c1 +  + c 2 m −1 = 1
n 
€
n 
A general description of an m two-state
quantum system requires 2m complex numbers!
Time evolution is described by unitary matrix of
2 m x 2 m!
19
10/10/13
n  Time
evolution is described by unitary
matrix of 2m x 2m!
n  Exponential in the physical size of the
system!
n  Effective deterministic simulation
difficult!
n  Difficult
to simulate on a probabilistic
computer (Interference) !
n 
n 
n 
n 
A set of n elements can be identified with
vectors of an ortonormal basis of n-dimensional
complex vector space Hn!
A state is a unit-length vector in the state space
2
with probability !ω i
The state space of a compound system,
consisting of two subsystems is the tensor
product €
of the subsystem state space !
State transformations are length-preserving
mappings (Unitary mappings)!
20