vorlesung quantum computing ss ‘08 1 quantum bits conventional bit on 3.2 - 5.5 v 1 off -0.5 - 0.8...

Vorlesung Quantum Computing SS ‘08

1

quantum bits

conventional bit

on <=> 3.2 - 5.5 V <=> 1

off <=> -0.5 - 0.8 V <=> 0

quantum mechanical bit (qubit)

| 0 <=> <=>

| 1 <=> <=>

10(

(

01(

(

a1| 0 + a2| 1 = a1

a2( )

superposition:


2

quantum computing

H H-1

calculation

U

preparation

read-out

|A|

time

time

quantum-bit (qubit)

0 1

a10 + a21 =a1a2


3

boolean algebra and logic gates

classical (irreversible) computing

gateinout

1-bit logic gates: identity

x NOT x

0 11 0

x Id

0 01 1

NOT

x NOT x


4

quantum gates

1-bit logic gate: NOT a1|1 + a2| 0 (a1| 0 + a2| 1 ) =

manipulation in quantum mechanics is done by linear operators operators have a matrix representationX ≡

0

01

1matrix representation for the NOT gate:

X =0

01

1a1

a2

a1

a2

=a2

a1

X X-1 =1

10

0


5

quantum parallelism

a1 F |00>+

a2 F |01>+

a3 F |10>+

a4 F |11>

}{a1 |00>

+a2 |01>

+a3 |10>

+a4 |11>

}{input

b1 |00>+

b2 |01>+

b3 |10>+

b4 |11>

}{=

output

F


6

how to create superposition

start in ground state ≡| 0 ≡1

0

manipulation with a unitary transformation

H = 1√2

H =1

-11

11√2

Hadamard Gate

cmeyer

tafel rechnen zur Darstellung der Basis


7

quantum computing

H H-1

calculation

U

preparation

read-out

|A|

time

time

classical bit

1 ON 3.2 – 5.5 V

0 OFF -0.5 – 0.8 V

quantum-bit (qubit)

0 1

a10 + a21 =a1a2


8

NOT:|1 |0

Bloch Sphere

the 2 dimensional Hilbertspace of a single qubit can be represented by the Bloch-Sphere

H:|0 |0 |1|1 |0 |1

|0

|1source: http://www.c3.lanl.gov/~knill/qip/nmrprhtml/node5.html

operations on a single qubit are represented by rotations on this sphere

| = cos( ) + eisin( )


9

Bloch Sphere

| = a1 + a2

| |a1|2 + |a2|2 = 1

| = r1ei + r2ei polar coordinates:

multiply with global phase e-i | = r1 + r2ei

(

| = r1 + r2eir1 + (x + iy)


10

Bloch Sphere

| = r1 + r2eir1 + (x + iy) with normalization constraint:

|r1|2 + | x + iy |2 = r12 + (x – iy) (x + iy)

= r12 + x2 + y2 = 1

3 dim unit sphere

x = r sin cos y = r sin sin z = r cos

| = cos( ) + eisin( )

= z + (x + iy) = cos + sin (cos + i sin ) = cos + ei sin

0 ≤ ≤ 0 ≤ ≤ 2

cmeyer

beweis für halbe winkel


11

infinitesimal unitary transformation

finite transformations can be decomposed in successive infinitesimal transformations

U() = I + iF^ ^ ^

(I + iF)^ ^ (I iF*) = I^ ^ ^ F hermitian, infinitesimal small and real

^with

F can be determined by the change L of an observable L ^ ^ ^

L’ = L + L = U()LU()* = L + i[F,L]^ ^ ^ ^ ^ ^ ^ ^ ^

U = eiF̂^


12

how to rotate a qubit

rotation about z-axis

U(x) = (R-1x) ≈ (x+y, y-x, z)

≈ (x,y,z) +(y /x – x /y)

= (1 – i/ħ [xpy – ypx]) = (1 – i/ħ J3)

and finite angle

^

^ ^

U() = (U())n = (1 – J3)n → ei ħ n

iħ

J3^ ^

cmeyer

J = r x pJ3, dritte Komponente des Drehimpulses

Ihr Benutzername

rotationsmatrix cos a -sin aR-1 = sin a cos a Näherung für kleine a: cos =1, sin = aNäherung für lineare Änderung: Taylorreihe


13

spin as basis

Pauli spin matrices form a complete base using spin as basis is very convenient for all implementations

Sz = = Z

1

-10

0ħ2

ħ2

Sx = = X

0

01

1ħ2

ħ2

Sy = = Y

0

0i

-iħ2

ħ2

|0

|1

2

NOT : e = = e iħ

Sx 0

0-i

-i i 0

01

1


14

Superposition of one and more qubits

H =1

-11

11√2

e =

iħ

(Sx+ Sz)

√2 1

-11

11i√2

H2=H2H1 = √2

√2

= 12

cmeyer

check global phase


15

entanglement

QC

→1√2

1√2

states that can be factorized:

live in subspaces H1 and H2

1√2

→states that cannot be factorized:

live in product space HQC only

BellBellstatesstates


16

Bell states

The Bell State has the property that upon measuring the

1st qubit one obtains two possible results.

- 0 with probability ½ leaving the post measurement state

- 1 with probability ½ leaving the post measurement state

- The measurement of the 2nd qubit always gives the result depending on the measurement of the 1st qubit.

- ie: The measurements are CORRELATED

‘

‘


17

Bell basis

– superpositions connected by the outline

– Bell states connected by diagonals

= 1√2

= 1√2


18

Bell’s inequalities

restrictions due to assumption of hidden classical variablesinequalities are violated by quantum mechanicsz

x

a,a’ = 1

z’x’

g,g’ = 1no influence between measurements,they are done at different spacetime points

f := (a+a’)g – (a-a’)g’

(a and a’ are either equal or opposite)

f := p(a,a’,g,g’) f ≤ 2

aa’ g’gga’aa’

g’g

1

1-1-1-1

-1

1

111

ag + a’g – ag’ + a’g’ ≤ 2

cmeyer

diese bekannt als CHSH (Clauser, Horne, Shimony, Holt)


19

Bell’s inequalities

z

x

a,a’ = 1

z’x’

g,g’ = 1

1

-10

0

0

01

1

a =

a’ =

1

-11

1g = – 1√2

1

-1-1

-1g’ =

1√2

AGAG

1√2 AAG G

ag = | a g | = 1√2

a’g = a’g’ = 1√2

ag’ = 1√2

ag+a’g – ag’+ a’g’ = 2 √2

cmeyer

diese bekannt als CHSH (Clauser, Horne, Shimony, Holt)


20

experiment

source of entangled photons

(use spontaneous parametric down conversion of a non-linear, birefringent crystal)


21

experiment

G. Weihs et al, Phys. Rev. Lett. 81, 5039 (1998)

spacetime separation measurement apparatus

measurement time: 100 ns

physical random number generator


22

uncorrelated measurements

measurementapparatus

measurementapparatus

random number generator


23

boolean algebra and logic gates

2-bit logic gates:

y x OR y

0

1

0

1

0

0

1

1

x

0

1

1

1

y x AND y

0

1

0

1

0

0

1

1

x

0

0

0

1

x

yx OR y

x

yx AND y

all other operations can be constructed from NOT, OR, and AND

x XOR y = (x OR y) AND NOT (x AND y)

cmeyer

aufgabe: truth table für XOR


24

classical binary addition

two one-bit digit in, one two-bit digit out: a0 + b0 = c0 + c1

X

&

a0b0

c0 (add mod2)

c1 (carry bit)

fanout0 1

0 00 011 01 10

truth table

+

++

X

++

X

++

X

1 2 3 4

a3

b1

a2

b2

a1

b3

a0

b0 c1

c2

c0

c3

c4

more thanone bit...


25

2 qubit gates

base vectors of a two–qubit register:

a, a ba, b

00 0001 0110 1111 10

CNOT:


26

2 qubit gates

– switch on the interaction Hamiltonian– use free evolution of the system

|1

|0

00 0001 0110 1111 10

CNOT:

source: http://www.c3.lanl.gov/~knill/qip/nmrprhtml/node7.html


27

the CNOT gate

control

target

iħ2

eSy

iħ2

e- Sy

iħ2

eSz


28

create entanglement

Ca= 1

Cb= 1

H1

√2

ab

ab

ab→ no factorization into product states possible1√2

UCNOT = =

1√2

1√2

1√2

1√2


29

no-cloning theorem

is a “no-copying theorem”

classical: copy with XOR

y x XOR y

0

1

0

1

0

0

1

1

x

0

1

1

0(x,0) → (x,x)

quantum mechanical: copy with CNOT ?


30

no-cloning theorem

“control” qubit is used as source“target” qubit is initialized to

try to copy a0a1

a0 a1

=

a1

a0

a1

a0

= a0 a1

≠

Bell stateBell state


31

toward n qubits

a0a1a2a3

two–qubit state:

n–qubit state:

Hilbertspace: 2n 2n ai ii=0

e.g., n = 5:

2 -1n


32

universal computing

all possible operations can be done by using 1-qubit-rotations, phase-shifts and the CNOT gate

→ this set of gates is therefore called “universal”

(in a classical computer NOT and NAND are a universal set)

single universal gate: Toffoli gate (3 qubits)

a

b

c

a

b

c(a b)

Toffoligate


33

Toffoli gate

a, b, c (ab)a, b, c

000 000001 001010 010011 011100

110111

101100101111110

Table of Truth

Matrix

UTF =

a

b

c c(a b)

Toffoligate

b

a

vorlesung quantum computing ss ‘08 1 quantum bits conventional bit on 3.2 - 5.5 v 1 off -0.5 - 0.8...

Documents