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Some remarkable differences between quantum and classical information Mika Hirvensalo Department of Mathematics and Statistics University of Turku mikhirve@utu.fi Rio de Janeiro, June 2015 Mika Hirvensalo Some remarkable ... 1 of 43

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SLIDE 1

Some remarkable differences between quantum and classical information

Mika Hirvensalo

Department of Mathematics and Statistics University of Turku mikhirve@utu.fi

Rio de Janeiro, June 2015

Mika Hirvensalo Some remarkable ... 1 of 43

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SLIDE 2

Quantum Computing

Mika Hirvensalo Some remarkable ... 2 of 43

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SLIDE 3

Quantum Computing

Ingredients of the term

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SLIDE 4

Quantum Computing

Ingredients of the term Quantum mechanics

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SLIDE 5

Quantum Computing

Ingredients of the term Quantum mechanics Computing

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SLIDE 6

Bullets

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SLIDE 7

Waves

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SLIDE 8

Neutrons

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SLIDE 9

Quantum Mechanics

Max Planck (1858–1947)

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SLIDE 10

Quantum Mechanics

Max Planck (1858–1947) Black body radiation (1900)

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SLIDE 11

Quantum Mechanics

Niels Bohr (1885–1962)

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SLIDE 12

Quantum Mechanics

Niels Bohr (1885–1962) Hydrogen atom model (1913)

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SLIDE 13

Quantum Mechanics

Erwin Schr¨

  • dinger (1887–1961)

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SLIDE 14

Quantum Mechanics

Erwin Schr¨

  • dinger (1887–1961)

d dt ψ = iHψ (1926)

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SLIDE 15

Quantum Mechanics

Louis de Broglie (1892–1987)

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SLIDE 16

Quantum Mechanics

Louis de Broglie (1892–1987) Wawe-particle duality

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SLIDE 17

Mechanics

Newtonian equation of motion F = ma

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SLIDE 18

Mechanics

Newtonian equation of motion F = ma = m d

dt v

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SLIDE 19

Mechanics

Newtonian equation of motion F = ma = m d

dt v = d dt mv

Mika Hirvensalo Some remarkable ... 10 of 43

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SLIDE 20

Mechanics

Newtonian equation of motion F = ma = m d

dt v = d dt mv = d dt p

Mika Hirvensalo Some remarkable ... 10 of 43

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SLIDE 21

Mechanics

Newtonian equation of motion F = ma = m d

dt v = d dt mv = d dt p

Total energy H = 1

2mv2 + V (x)

Mika Hirvensalo Some remarkable ... 10 of 43

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SLIDE 22

Mechanics

Newtonian equation of motion F = ma = m d

dt v = d dt mv = d dt p

Total energy H = 1

2mv2 + V (x) = p2 2m −

x

x0

F(s) ds

Mika Hirvensalo Some remarkable ... 10 of 43

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SLIDE 23

Mechanics

Newtonian equation of motion F = ma = m d

dt v = d dt mv = d dt p

Total energy H = 1

2mv2 + V (x) = p2 2m −

x

x0

F(s) ds Hamiltonian reformulation

d dt x = ∂ ∂pH, d dt p = − ∂ ∂x H

Mika Hirvensalo Some remarkable ... 10 of 43

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SLIDE 24

Mechanics

Classical

d dt x = ∂ ∂pH, d dt p = − ∂ ∂x H

Mika Hirvensalo Some remarkable ... 11 of 43

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SLIDE 25

Mechanics

Classical

d dt x = ∂ ∂pH, d dt p = − ∂ ∂x H

Quantum

∂ ∂t ψ = −iHψ, where ψ is the wave function

Mika Hirvensalo Some remarkable ... 11 of 43

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SLIDE 26

Wave Function

Max Born’s interpretation |ψ(x, t)|2 is the probability density of the particle position at time t

Mika Hirvensalo Some remarkable ... 12 of 43

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SLIDE 27

Wave Function

Max Born’s interpretation |ψ(x, t)|2 is the probability density of the particle position at time t So: P(a ≤ x ≤ b) = b

a

|ψ(x, t)|2 dx

Mika Hirvensalo Some remarkable ... 12 of 43

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SLIDE 28

Wave Function

Max Born’s interpretation |ψ(x, t)|2 is the probability density of the particle position at time t So: P(a ≤ x ≤ b) = b

a

|ψ(x, t)|2 dx At the same time (omitting t):

  • ψ(p) = F[ψ(x)](p) =

∞

−∞

ψ(x)e−2πixp dx is the probability density of the particle momentum.

Mika Hirvensalo Some remarkable ... 12 of 43

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SLIDE 29

Wave Function

At the same time (omitting t):

  • ψ(p) = F[ψ(x)](p) =

∞

−∞

ψ(x)e−2πixp dx is the probability density of the particle momentum. So: P(a ≤ p ≤ b) = b

a

  • ψ(p)
  • 2

dp

Mika Hirvensalo Some remarkable ... 13 of 43

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SLIDE 30

Wave Function

At the same time (omitting t):

  • ψ(p) = F[ψ(x)](p) =

∞

−∞

ψ(x)e−2πixp dx is the probability density of the particle momentum. So: P(a ≤ p ≤ b) = b

a

  • ψ(p)
  • 2

dp Wavefunction ψ gives the full characterization of the system at a fixed time

Mika Hirvensalo Some remarkable ... 13 of 43

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Finite Quantum Systems

Nuclear spin Photon polarization Wavefunction ψ defined on a finite set.

Mika Hirvensalo Some remarkable ... 14 of 43

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SLIDE 32

Finite Quantum Systems

Nuclear spin Photon polarization Wavefunction ψ defined on a finite set. Formally a (pure) state ψ = α1ψ1 + α2ψ2 + . . . + αnψn, where {ψ1, . . . , ψn} is an

  • rthonormal basis of n-dimensional complex vector space Hn.

Mika Hirvensalo Some remarkable ... 14 of 43

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Finite Quantum Systems

Nuclear spin Photon polarization Wavefunction ψ defined on a finite set. Formally a (pure) state ψ = α1ψ1 + α2ψ2 + . . . + αnψn, where {ψ1, . . . , ψn} is an

  • rthonormal basis of n-dimensional complex vector space Hn.

For mixed states, representation must be generalized.

Mika Hirvensalo Some remarkable ... 14 of 43

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SLIDE 34

Computability

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SLIDE 35

Computability

Alan Turing (1912–1954)

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Turing Machine

Tape → I N P U T

✻

← Read-write head

✒ ✏

p, q, r, . . . State set → (∼ Program)

Mika Hirvensalo Some remarkable ... 16 of 43

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SLIDE 37

Turing Machine

Tape → I N P U T

✻

← Read-write head

✒ ✏

p, q, r, . . . State set → (∼ Program) In state p: Read a, write b (b depends only on p and a) Move read-write head (direction depends only on p and a) Go to state q (q depends only on p and a)

Mika Hirvensalo Some remarkable ... 16 of 43

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SLIDE 38

Computability

Churc-Turing thesis Algorithmic computability = Turing Machine computability

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SLIDE 39

Computability

Churc-Turing thesis Algorithmic computability = Turing Machine computability Not provable

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SLIDE 40

Computability

Richard Feynman (1918–1988):

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SLIDE 41

Computability

Richard Feynman (1918–1988): Simulating Physics with Computers (1982)

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Computability

David Deutsch (1954–): Quantum Turing Machine (1985)

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Computability

David Deutsch (1954–): Quantum Turing Machine (1985), a “proof” of Church-Turing thesis

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SLIDE 44

Quantum Computing

Peter Shor (1959–): Fast Quantum Factoring Algorithm (1994)

Mika Hirvensalo Some remarkable ... 20 of 43

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Quantum Computing

Lov Grover (1961–): O( √ N) Quantum Search (1996)

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SLIDE 46

Quantum bit (Qubit)

✲ ✻

|0 |1

✚✚✚✚✚✚✚✚✚✚✚✚ ✚ ❃

a |0 + b |1 |a|2 + |b|2 = 1

Superposition of states |0 ja |1

Mika Hirvensalo Some remarkable ... 22 of 43

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SLIDE 47

Quantum bit (Qubit)

✲ ✻

|0 |1

✚✚✚✚✚✚✚✚✚✚✚✚ ✚ ❃

a |0 + b |1 |a|2 + |b|2 = 1

Superposition of states |0 ja |1 ❅ ❘

  • ✠

Amplitudes

Mika Hirvensalo Some remarkable ... 22 of 43

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SLIDE 48

Quantum bit (Qubit)

✲ ✻

|0 |1

✚✚✚✚✚✚✚✚✚✚✚✚ ✚ ❃

a |0 + b |1 |a|2 + |b|2 = 1

Superposition of states |0 ja |1 ❅ ❘

  • ✠

Amplitudes Measurement in basis {|0 , |1}: p(0) = |a|2, p(1) = |b|2

Mika Hirvensalo Some remarkable ... 22 of 43

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SLIDE 49

Quantum bit (Qubit)

✲ ✻

|0 |1

✚✚✚✚✚✚✚✚✚✚✚✚ ✚ ❃

a |0 + b |1 |a|2 + |b|2 = 1

Superposition of states |0 ja |1 ❅ ❘

  • ✠

Amplitudes Measurement in basis {|0 , |1}: p(0) = |a|2, p(1) = |b|2 Minimal interpretation

Mika Hirvensalo Some remarkable ... 22 of 43

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SLIDE 50

Quantum bit (Qubit)

✲ ✻

  • ✒

|0 |1

1 √ 2 |0 + 1 √ 2 |1

Basis 1: {|0 , |1} p(0) = 1

2

Mika Hirvensalo Some remarkable ... 23 of 43

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SLIDE 51

Quantum bit (Qubit)

✲ ✻

  • ✒

|0 |1

1 √ 2 |0 + 1 √ 2 |1

Basis 1: {|0 , |1} p(0) = 1

2

Basis 2: { 1

√ 2 |0 + 1 √ 2 |1 = |0′ , 1 √ 2 |0 − 1 √ 2 |1 = |1′}

p(0′) = 1 = |0′ |1′

  • ✒

❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❘

Mika Hirvensalo Some remarkable ... 23 of 43

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SLIDE 52

Quantum bit (Qubit)

✲ ✻

  • ✒

|0 |1

1 √ 2 |0 + 1 √ 2 |1

Basis 1: {|0 , |1} p(0) = 1

2

Basis 2: { 1

√ 2 |0 + 1 √ 2 |1 = |0′ , 1 √ 2 |0 − 1 √ 2 |1 = |1′}

p(0′) = 1 = |0′ |1′

  • ✒

❅ ❅ ❅ ❅ ❅ ❅ ❅ ❅ ❘

Pure state = (generalized) probability distribution

Mika Hirvensalo Some remarkable ... 23 of 43

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Quantum gate

Example: W : C2 → C2 W |0 = 1 √ 2 |0 + 1 √ 2 |1 W |1 = 1 √ 2 |0 − 1 √ 2 |1 is unitary (Hadamard-Walsh transform)

Mika Hirvensalo Some remarkable ... 24 of 43

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SLIDE 54

Interference / Walsh transform once

Mika Hirvensalo Some remarkable ... 25 of 43

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SLIDE 55

Interference / Walsh transform once

|0 |0 |1

✓ ✓ ✓ ✓ ✓ ✴ ❙ ❙ ❙ ❙ ❙ ✇

|0

1 √ 2 |0 + 1 √ 2 |1 1 √ 2 1 √ 2

Mika Hirvensalo Some remarkable ... 25 of 43

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SLIDE 56

Interference / Walsh transform twice

|0 |0 |0 |1 |1 |0 |1

✓ ✓ ✓ ✓ ✓ ✴ ❙ ❙ ❙ ❙ ❙ ✇ ✁ ✁ ✁ ✁ ✁ ☛ ❆ ❆ ❆ ❆ ❆ ❯ ✁ ✁ ✁ ✁ ✁ ☛ ❆ ❆ ❆ ❆ ❆ ❯

|0

1 √ 2 |0 + 1 √ 2 |1 1 2 |0 + 1 2 |1

+ 1

2 |0 − 1 2 |1 = |0 1 √ 2 1 √ 2 1 √ 2 1 √ 2 1 √ 2

− 1

√ 2

Mika Hirvensalo Some remarkable ... 26 of 43

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SLIDE 57

Interference / Walsh transform twice

|0 |0 |0 |1 |1 |0 |1

✓ ✓ ✓ ✓ ✓ ✴ ❙ ❙ ❙ ❙ ❙ ✇ ✁ ✁ ✁ ✁ ✁ ☛ ❆ ❆ ❆ ❆ ❆ ❯ ✁ ✁ ✁ ✁ ✁ ☛ ❆ ❆ ❆ ❆ ❆ ❯

|0

1 √ 2 |0 + 1 √ 2 |1 1 2 |0 + 1 2 |1

+ 1

2 |0 − 1 2 |1 = |0 1 √ 2 1 √ 2 1 √ 2 1 √ 2 1 √ 2

− 1

√ 2

❅ ❅ ❅ ■

  • ✒

Constructive interference Mika Hirvensalo Some remarkable ... 26 of 43

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SLIDE 58

Interference / Walsh transform twice

|0 |0 |0 |1 |1 |0 |1

✓ ✓ ✓ ✓ ✓ ✴ ❙ ❙ ❙ ❙ ❙ ✇ ✁ ✁ ✁ ✁ ✁ ☛ ❆ ❆ ❆ ❆ ❆ ❯ ✁ ✁ ✁ ✁ ✁ ☛ ❆ ❆ ❆ ❆ ❆ ❯

|0

1 √ 2 |0 + 1 √ 2 |1 1 2 |0 + 1 2 |1

+ 1

2 |0 − 1 2 |1 = |0 1 √ 2 1 √ 2 1 √ 2 1 √ 2 1 √ 2

− 1

√ 2

❅ ❅ ❅ ■

  • ✒

Constructive interference

❅ ❅ ❅ ■

  • ✒

Destructive interference Mika Hirvensalo Some remarkable ... 26 of 43

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SLIDE 59

n quantum bits

Tensor product notation: |❛ ⊗ |❜ = |❛ |❜ = |❛ ❜. ① ① ① ① ① ① ① ① ① ① ① ① ①

Mika Hirvensalo Some remarkable ... 27 of 43

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SLIDE 60

n quantum bits

Tensor product notation: |❛ ⊗ |❜ = |❛ |❜ = |❛ ❜. For example, |0 |0 = |00, |0 |1 = |01, etc. ① ① ① ① ① ① ① ① ① ① ① ① ①

Mika Hirvensalo Some remarkable ... 27 of 43

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SLIDE 61

n quantum bits

Tensor product notation: |❛ ⊗ |❜ = |❛ |❜ = |❛ ❜. For example, |0 |0 = |00, |0 |1 = |01, etc. General state

  • ①∈{0,1}n

c① |① (2n-dimensional Hilbert space), where

  • ①∈{0,1}n

|c①|2 = 1 ① ① ① ① ① ① ① ①

Mika Hirvensalo Some remarkable ... 27 of 43

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SLIDE 62

n quantum bits

Tensor product notation: |❛ ⊗ |❜ = |❛ |❜ = |❛ ❜. For example, |0 |0 = |00, |0 |1 = |01, etc. General state

  • ①∈{0,1}n

c① |① (2n-dimensional Hilbert space), where

  • ①∈{0,1}n

|c①|2 = 1 If Uf |① |0 = |① |f (①) can be realized, then Uf 1 √ 2n

  • ①∈{0,1}n

|① |0 = 1 √ 2n

  • ①∈{0,1}n

|① |f (①) (Quantum parallelism)

Mika Hirvensalo Some remarkable ... 27 of 43

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SLIDE 63

n quantum bits

If Uf |① |0 = |① |f (①) can be realized, then Uf 1 √ 2n

  • ①∈{0,1}n

|① |0 = 1 √ 2n

  • ①∈{0,1}n

|① |f (①) (Quantum parallelism) ① ① ① ①

Mika Hirvensalo Some remarkable ... 28 of 43

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SLIDE 64

n quantum bits

If Uf |① |0 = |① |f (①) can be realized, then Uf 1 √ 2n

  • ①∈{0,1}n

|① |0 = 1 √ 2n

  • ①∈{0,1}n

|① |f (①) (Quantum parallelism) P(|① |f (①)) =

  • 1

√ 2n

  • 2

= 1

2n

① ①

Mika Hirvensalo Some remarkable ... 28 of 43

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SLIDE 65

n quantum bits

If Uf |① |0 = |① |f (①) can be realized, then Uf 1 √ 2n

  • ①∈{0,1}n

|① |0 = 1 √ 2n

  • ①∈{0,1}n

|① |f (①) (Quantum parallelism) P(|① |f (①)) =

  • 1

√ 2n

  • 2

= 1

2n

Observation “collapses” the system into |① |f (①) (Projection postulate)

Mika Hirvensalo Some remarkable ... 28 of 43

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SLIDE 66

n quantum bits

If Uf |① |0 = |① |f (①) can be realized, then Uf 1 √ 2n

  • ①∈{0,1}n

|① |0 = 1 √ 2n

  • ①∈{0,1}n

|① |f (①) (Quantum parallelism) P(|① |f (①)) =

  • 1

√ 2n

  • 2

= 1

2n

Observation “collapses” the system into |① |f (①) (Projection postulate) Direct method offers no advantage over probabilistic guessing!

Mika Hirvensalo Some remarkable ... 28 of 43

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SLIDE 67

Nondeterministic Computing

Mika Hirvensalo Some remarkable ... 29 of 43

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SLIDE 68

Nondeterministic Computing

Good computational paths should be supported

Mika Hirvensalo Some remarkable ... 29 of 43

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SLIDE 69

Nondeterministic Computing

Good computational paths should be supported Seems impossible in classical computing

Mika Hirvensalo Some remarkable ... 29 of 43

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SLIDE 70

Nondeterministic Computing

Good computational paths should be supported Seems impossible in classical computing Sometimes possible in quantum computing

Mika Hirvensalo Some remarkable ... 29 of 43

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SLIDE 71

Nondeterministic Computing

Good computational paths should be supported Seems impossible in classical computing Sometimes possible in quantum computing The “efficiency” of quantum computing is based on interference

Mika Hirvensalo Some remarkable ... 29 of 43

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SLIDE 72

Quantum algorithms

Interference should favour the good computation paths Difficult to control in algorithm design

Mika Hirvensalo Some remarkable ... 30 of 43

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SLIDE 73

Quantum algorithms

Interference should favour the good computation paths Difficult to control in algorithm design Main methods Quantum Fourier transform Grover iteration Adiabatic quantum computing Quantum random walks

Mika Hirvensalo Some remarkable ... 30 of 43

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SLIDE 74

Quantum Fourier transform

Discrete Fourier transform on coefficients of c0 |00 . . . 0 + c1 |00 . . . 1 + . . . + c2n−1 |11 . . . 1 , Can be implemented in time Poly(n) (instead of 2n) Exponential advantage for problems with periodic structure Main ingredient in Shor’s factoring algorithm

Mika Hirvensalo Some remarkable ... 31 of 43

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SLIDE 75

Grover iteration

Basic idea: Using k calls of function f , the superposition 1 √ 2n

  • |00 . . . 0 + |00 . . . 1 + . . . + |11 . . . 1
  • ,

coefficients of such vectors |① for which f (①) = 1 can be increased to ≈ C ·

k √ 2n , hence the probability of seeing such

an element becomes ≈ |C|2 k2

2n .

Provides a quadratic advantage over classical algorithms Works on all search problems

Mika Hirvensalo Some remarkable ... 32 of 43

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SLIDE 76

EPR Paradox

EPR pair 1 √ 2 (|00 + |11) (entangled state), perfect correlation

Mika Hirvensalo Some remarkable ... 33 of 43

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SLIDE 77

EPR Paradox

EPR pair 1 √ 2 (|00 + |11) (entangled state), perfect correlation; cf. with 1 2(|00 + |01 + |10 + |11)

Mika Hirvensalo Some remarkable ... 33 of 43

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SLIDE 78

EPR Paradox

EPR pair 1 √ 2 (|00 + |11) (entangled state), perfect correlation; cf. with 1 2(|00 + |01 + |10 + |11) = 1 √ 2 (|0 + |1) 1 √ 2 (|0 + |1) (decomposable state).

Mika Hirvensalo Some remarkable ... 33 of 43

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SLIDE 79

Compound Systems

Correlation over distance also possible in classical mechanics: Probability distribution

1 2[00] + 1 2[11]

Mika Hirvensalo Some remarkable ... 34 of 43

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SLIDE 80

Compound Systems

Correlation over distance also possible in classical mechanics: Probability distribution

1 2[00] + 1 2[11]

But 1 √ 2 |00 + 1 √ 2 |11 violates a Bell inequality.

Mika Hirvensalo Some remarkable ... 34 of 43

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SLIDE 81

John Bell

John Steward Bell (1928–1990)

Mika Hirvensalo Some remarkable ... 35 of 43

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SLIDE 82

EPR Paradox

Einstein, Podolsky, Rosen: Can Quantum-Mechanical Description of Physical Reality Be Considered Com- plete? Physical Review 47, 777–780 (1935)

Niels Bohr (1885–1962) & Albert Einstein (1879–1955)

Mika Hirvensalo Some remarkable ... 36 of 43

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SLIDE 83

EPR Paradox (Bohm formulation)

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 84

EPR Paradox (Bohm formulation)

Einstein: The physical world is local and realistic

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 85

EPR Paradox (Bohm formulation)

Einstein: The physical world is local and realistic Assume distant qubits in state

1 √ 2 |00 + 1 √ 2 |11

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 86

EPR Paradox (Bohm formulation)

Einstein: The physical world is local and realistic Assume distant qubits in state

1 √ 2 |00 + 1 √ 2 |11

Quantum mechanics: neither qubit has definite pre-observation value

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 87

EPR Paradox (Bohm formulation)

Einstein: The physical world is local and realistic Assume distant qubits in state

1 √ 2 |00 + 1 √ 2 |11

Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 88

EPR Paradox (Bohm formulation)

Einstein: The physical world is local and realistic Assume distant qubits in state

1 √ 2 |00 + 1 √ 2 |11

Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit ⇒ The value of the second qubit is known certainly

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 89

EPR Paradox (Bohm formulation)

Einstein: The physical world is local and realistic Assume distant qubits in state

1 √ 2 |00 + 1 √ 2 |11

Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit ⇒ The value of the second qubit is known certainly (without “touching” or “disturbing” it)

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 90

EPR Paradox (Bohm formulation)

Einstein: The physical world is local and realistic Assume distant qubits in state

1 √ 2 |00 + 1 √ 2 |11

Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit ⇒ The value of the second qubit is known certainly (without “touching” or “disturbing” it) ⇒ The value if the second qubit is “an element of reality”

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 91

EPR Paradox (Bohm formulation)

Einstein: The physical world is local and realistic Assume distant qubits in state

1 √ 2 |00 + 1 √ 2 |11

Quantum mechanics: neither qubit has definite pre-observation value Observe the first qubit ⇒ The value of the second qubit is known certainly (without “touching” or “disturbing” it) ⇒ The value if the second qubit is “an element of reality” ⇒ Quantum mechanics is an incomplete theory

Mika Hirvensalo Some remarkable ... 37 of 43

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SLIDE 92

Bell Inequalities

Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989)

Mika Hirvensalo Some remarkable ... 38 of 43

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SLIDE 93

Bell Inequalities

Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls

Mika Hirvensalo Some remarkable ... 38 of 43

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SLIDE 94

Bell Inequalities

Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic

Mika Hirvensalo Some remarkable ... 38 of 43

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SLIDE 95

Bell Inequalities

Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden

Mika Hirvensalo Some remarkable ... 38 of 43

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SLIDE 96

Bell Inequalities

Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden 30 red and wooden?

Mika Hirvensalo Some remarkable ... 38 of 43

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SLIDE 97

Bell Inequalities

Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden 30 red and wooden? Then 80+60-30=110 are red or wooden. No way!

Mika Hirvensalo Some remarkable ... 38 of 43

slide-98
SLIDE 98

Bell Inequalities

Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden 30 red and wooden? Then 80+60-30=110 are red or wooden. No way! In other words: (0.8, 0.6, 0.3) does not express probabilities (p1, p2, p12) of two events and their intersection.

Mika Hirvensalo Some remarkable ... 38 of 43

slide-99
SLIDE 99

Bell Inequalities

Itamar Pitowsky: Quantum Probability – Quantum Logic, Springer (1989) Ballot box of 100 balls Each red or blue, wooden or plastic 80 red, 60 wooden 30 red and wooden? Then 80+60-30=110 are red or wooden. No way! In other words: (0.8, 0.6, 0.3) does not express probabilities (p1, p2, p12) of two events and their intersection. Reason: P(1 ∨ 2) = p1 + p2 − p12 is a probability, too.

Mika Hirvensalo Some remarkable ... 38 of 43

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SLIDE 100

Bell Inequalities

Lemma (p1, p2, p12) is an “eligible” probability vector if and only if 0 ≤ p12 ≤ p1, p2 ≤ 1 and 0 ≤ p1 + p2 − p12 ≤ 1 These are Bell inequalities!

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slide-101
SLIDE 101

Bell Inequalities

In fact, (p1, p2, p12) = (1 − p2 − p2 + p12)(0, 0, 0) + (p2 − p12)(0, 1, 0) + (p1 − p12)(1, 0, 0) + p12(1, 1, 1). However, the representation is not generally unique.

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SLIDE 102

CHSH Inequality

Probabilities → Expected values Let A1, A2, B1, B2 be ±1-valued observables.

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SLIDE 103

CHSH Inequality

Probabilities → Expected values Let A1, A2, B1, B2 be ±1-valued observables. Then A1B1 + A1B2 + A2B1 − A2B2 = A1(B1 + B2) + A2(B1 − B2) ∈ {−2, 2}

Mika Hirvensalo Some remarkable ... 41 of 43

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SLIDE 104

CHSH Inequality

Probabilities → Expected values Let A1, A2, B1, B2 be ±1-valued observables. Then A1B1 + A1B2 + A2B1 − A2B2 = A1(B1 + B2) + A2(B1 − B2) ∈ {−2, 2} Hence |E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2)| ≤ 2

Mika Hirvensalo Some remarkable ... 41 of 43

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SLIDE 105

EPR Paradox Resolved

①

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SLIDE 106

EPR Paradox Resolved

Assume Alice and Bob share state ① =

1 √ 2 |00 + 1 √ 2 |11.

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SLIDE 107

EPR Paradox Resolved

Assume Alice and Bob share state ① =

1 √ 2 |00 + 1 √ 2 |11.

Define observables A1 = 1 1

  • , A2 =

1 −1

  • ,

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slide-108
SLIDE 108

EPR Paradox Resolved

Assume Alice and Bob share state ① =

1 √ 2 |00 + 1 √ 2 |11.

Define observables A1 = 1 1

  • , A2 =

1 −1

  • ,

B1 = 1 √ 2 (A1 + A2), B2 = 1 √ 2 (A1 − A2) (eigenvalues = potential values =±1)

Mika Hirvensalo Some remarkable ... 42 of 43

slide-109
SLIDE 109

EPR Paradox Resolved

Assume Alice and Bob share state ① =

1 √ 2 |00 + 1 √ 2 |11.

Define observables A1 = 1 1

  • , A2 =

1 −1

  • ,

B1 = 1 √ 2 (A1 + A2), B2 = 1 √ 2 (A1 − A2) (eigenvalues = potential values =±1)

Mika Hirvensalo Some remarkable ... 42 of 43

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SLIDE 110

EPR Paradox Resolved

For these observables, E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2) = 2 √ 2, which contradicts the CHSH inequality |E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2)| ≤ 2.

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slide-111
SLIDE 111

EPR Paradox Resolved

For these observables, E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2) = 2 √ 2, which contradicts the CHSH inequality |E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2)| ≤ 2. Conclusion:

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slide-112
SLIDE 112

EPR Paradox Resolved

For these observables, E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2) = 2 √ 2, which contradicts the CHSH inequality |E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2)| ≤ 2. Conclusion: Locality, realism, and quantum mechanics form a contradictory set

  • f assumptions.

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slide-113
SLIDE 113

EPR Paradox Resolved

For these observables, E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2) = 2 √ 2, which contradicts the CHSH inequality |E(A1B1) + E(A1B2) + E(A2B1) − E(A2B2)| ≤ 2. Conclusion: Locality, realism, and quantum mechanics form a contradictory set

  • f assumptions.

From them, you can derive anything.

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