Download presentation

1
**( ( ) quantum bits conventional bit**

off <=> V <=> 0 quantum mechanical bit (qubit) | 0 <=> <=> | 1 <=> <=> 1 ( superposition: a1 a2 ( ) a1| 0 + a2| 1 =

2
**quantum computing |Y0 U H H-1 Y|A|Y quantum-bit (qubit) 0 **

0 1 a1 0 + a2 1 = a1 a2 preparation |Y0 calculation read-out time U H H-1 Y|A|Y time

3
**boolean algebra and logic gates**

classical (irreversible) computing gate in out 1-bit logic gates: identity NOT x Id 1 x NOT x 1 x NOT x

4
**quantum gates X ≡ X = X X-1 = 1-bit logic gate: NOT**

(a1| 0 + a2| 1 ) = a1|1 + a2| 0 manipulation in quantum mechanics is done by linear operators operators have a matrix representation X ≡ 1 matrix representation for the NOT gate: X = 1 a1 a2 X X-1 = 1

5
**quantum parallelism a1 |00> + a2 |01> a3 |10> a4 |11>**

input b1 |00> + b2 |01> b3 |10> b4 |11> = output a1 F |00> + a2 F |01> a3 F |10> a4 F |11> F

6
**how to create superposition**

≡ | 0 ≡ 1 start in ground state manipulation with a unitary transformation Hadamard Gate H = 1 -1 √2 H |0 = 1 √2 (|0 + |1)

7
**quantum computing |Y0 U H H-1 Y|A|Y classical bit**

1 ON – 5.5 V 0 OFF – 0.8 V quantum-bit (qubit) 0 1 a1 0 + a2 1 = a1 a2 preparation |Y0 calculation read-out time U H H-1 Y|A|Y time

8
**Bloch Sphere |y = cos( ) |0 + eij sin( )|1 |0 NOT: |1 |0 H:**

the 2 dimensional Hilbertspace of a single qubit can be represented by the Bloch-Sphere operations on a single qubit are represented by rotations on this sphere |y = cos( ) |0 + eij sin( )|1 q 2 |0 |1 source: NOT: |1 |0 H: |0 |0 + |1 |1 |0 - |1

9
**Bloch Sphere y |y = 1 |y = a1|0 + a2|1 |a1|2 + |a2|2 = 1**

|y = r1eij|0 + r2eiJ|1 polar coordinates: multiply with global phase e-ij: |y = r1|0 + r2eif|1 f = (J - j) |y = r1|0 + r2eif|1 = r1|0 + (x + iy)|1

10
**Bloch Sphere 3 dim unit sphere |y = z |0 + (x + iy)|1**

|y = r1|0 + r2eif|1 = r1|0 + (x + iy)|1 with normalization constraint: |r1|2 + | x + iy |2 = r12 + (x – iy) (x + iy) = r12 + x2 + y2 = 1 3 dim unit sphere |y = z |0 + (x + iy)|1 = cos q |0 + sin q (cos f + i sin f)|1 = cos q |0 + eif sin q |1 x = r sin q cos f y = r sin q sin f z = r cos q |y = cos( ) |0 + eif sin( )|1 q 2 0 ≤ q ≤ p, 0 ≤ f ≤ 2p

11
**infinitesimal unitary transformation**

finite transformations can be decomposed in successive infinitesimal transformations U(e) = I + ieF ^ (I + ieF) ^ (I - ieF*) = I ^ ^ F hermitian, e infinitesimal small and real ^ with F can be determined by the change dL of an observable L ^ L’ = L + dL = U(e)LU(e)* = L + ie[F,L] ^ U = eieF ^

12
**how to rotate a qubit ^ UY (x) = Y (R-1x) ≈ Y (x+ey, y-ex, z)**

rotation about z-axis ^ UY (x) = Y (R-1x) ≈ Y (x+ey, y-ex, z) ≈ Y (x,y,z) + e (y Y /x – x Y /y) = Y (1 – ie/ħ [xpy – ypx]) = Y (1 – ie/ħ J3) ^ ^ and finite angle q i ħ - qJ3 ^ ^ i q ħ n U(q) = (U(e))n = (1 – J3)n → e

13
**spin as basis Sz = = Z Sx = = X Sy = = Y 1 ħ 2 -1 |0 ħ 2 1 -i ħ |1 2**

Pauli spin matrices form a complete base using spin as basis is very convenient for all implementations Sz = = Z 1 -1 ħ 2 |0 |1 Sx = = X 1 ħ 2 Sy = = Y i -i ħ 2 p 2 NOT : e = = e i ħ - pSx -i - i 1

14
**Superposition of one and more qubits**

H = 1 -1 √2 e = i ħ (Sx+ Sz) p √2 1 -1 i√2 H2=H2H1 |00 = √2 (|0 + |1) = (|00 + |01 + |10 + |11) 1 2

15
**entanglement Bell states |cQC = |Y1 |Y2**

√2 states that can be factorized: live in subspaces H1 and H2 Bell states 1 √2 | c → (|01 - |10) states that cannot be factorized: live in product space HQC only

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 |Y‘ = |01 |Y‘ = |10

17
**Bell basis |01 |10 |11 |00 |Y = (|01 |10) |Y = (|00 |11)**

superpositions connected by the outline Bell states connected by diagonals |Y = (|01 |10) 1 √2 |Y = (|00 |11) 1 √2

18
**Bell’s inequalities -1 1 1 1 1 1 -1 -1 -1 1 a’**

restrictions due to assumption of hidden classical variables inequalities are violated by quantum mechanics z’ x’ z x no influence between measurements, they are done at different spacetime points g,g’ = 1 a,a’ = 1 f := (a+a’)g – (a-a’)g’ -1 1 1 1 1 (a and a’ are either equal or opposite) f := S p(a,a’,g,g’) f ≤ 2 1 -1 -1 -1 1 a’ ag + a’g – ag’ + a’g’ ≤ 2 a’ a’ a g’ g g’ g g a

19
**Bell’s inequalities |Y = | A | A | G | G | A | A | G | G 1 √2**

z’ x’ z x g,g’ = 1 a,a’ = 1 ag = Y | a g |Y = 1 √2 a’g = a’g’ = 1 √2 ag’ = 1 √2 ag+a’g – ag’+ a’g’ = 2 √2 1 -1 a = a’ = 1 -1 g = – √2 g’ =

20
**experiment source of entangled photons**

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

21
**experiment spacetime separation measurement apparatus**

measurement time: 100 ns physical random number generator G. Weihs et al, Phys. Rev. Lett. 81, 5039 (1998)

22
**uncorrelated measurements**

random number generator measurement apparatus measurement apparatus

23
**boolean algebra and logic gates**

2-bit logic gates: x y x OR y x y x AND y y x OR y 1 x y x AND y 1 x all other operations can be constructed from NOT, OR, and AND x XOR y = (x OR y) AND NOT (x AND y)

24
**classical binary addition**

two one-bit digit in, one two-bit digit out: a0 + b0 = c0 + c1 X & a0 b0 c0 (add mod2) c1 (carry bit) fanout 0 1 truth table + X 1 2 3 4 a3 b1 a2 b2 a1 b3 a0 b0 c1 c2 c0 c3 c4 more than one bit...

25
**2 qubit gates |00 = |01 = |10 = |11 = 1 CNOT: 1 00 01 10**

base vectors of a two–qubit register: |00 = 1 |01 = |10 = |11 = a, a b a, b 00 01 10 11 1 CNOT:

26
**2 qubit gates switch on the interaction Hamiltonian**

use free evolution of the system |1 |0 00 00 01 01 10 11 11 10 CNOT: source:

27
the CNOT gate control target i ħ p 2 e Sy Sy i ħ p 2 e Sz

28
**create entanglement 1 |00 = |01 = |10 = |11 = Ca= |1 |a |b**

Cb= |1 H 1 √2 (|0 - |1) |a |b 1 UCNOT = = - √2 |00 = 1 |01 = |10 = |11 = |a |b → (|01 - |10) no factorization into product states possible 1 √2

29
**no-cloning theorem 1 1 1 1 is a “no-copying theorem”**

classical: copy with XOR y x XOR y 1 x (x,0) → (x,x) 1 1 1 1 quantum mechanical: copy with CNOT ?

30
**no-cloning theorem |Y |0 = a0 |0 |0 + a1 |1 |0 1 =**

“control” qubit is used as source “target” qubit is initialized to |0 try to copy |Y = a0 |0 + a1|1 |Y |0 = a0 |0 |0 + a1 |1 |0 Bell state 1 = a1 a0 a1 a0 = a0 |0 |0 + a1 |1 |1 ≠ |Y |Y

31
**toward n qubits |Y = S ai |i |Y = a0|00 + a1|01 + a2|10 + a3|11**

two–qubit state: |Y = a0|00 + a1|01 + a2|10 + a3|11 n–qubit state: 2 -1 n |Y = S ai |i Hilbertspace: 2n 2n i=0 e.g., n = 5: |00101 = |5 |00110 = |6

32
**universal computing |a |b |c |c(a b) Toffoli gate**

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 |c(a b) Toffoli gate

33
**Toffoli gate 1 UTF = Table of Truth 000 Matrix 001 010 011**

|c |c(a b) Toffoli gate a, b, c (ab) a, b, c 000 001 010 011 100 110 111 101 Table of Truth Matrix 1 UTF =

Similar presentations

OK

Factor P 16 8(8-5ab) 4(d² + 4) 3rs(2r – s) 15cd(1 + 2cd) 8(4a² + 3b²)

Factor P 16 8(8-5ab) 4(d² + 4) 3rs(2r – s) 15cd(1 + 2cd) 8(4a² + 3b²)

© 2018 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Plant life cycle for kids ppt on batteries Ppt on learning styles Ppt on limits and derivatives calculus Free download ppt on probability for class 9 Ppt on acid-base indicators image Ppt on generation transmission and distribution of electricity in india Seminar ppt on blue brain Ppt on underwater wireless communication Ppt on classical economics time Ppt on marie curie pictures