Elements of quantum theory from limited information and - - PowerPoint PPT Presentation

▶
elements of quantum theory from limited information and
SMART_READER_LITE
LIVE PREVIEW

Elements of quantum theory from limited information and - - PowerPoint PPT Presentation

Aug 24, 2023 663 likes •1.09k views

Elements of quantum theory from limited information and complementarity Philipp H ohn Perimeter Institute FFP 14 Marseille July 16 th , 2014 July 16 th , 2014 P. H ohn (PI) 1 / 14 What makes quantum theory special? which physical

slide-1
SLIDE 1

Elements of quantum theory from limited information and complementarity

Philipp H¨

  • hn

Perimeter Institute

FFP ’14 Marseille

July 16th, 2014

  • P. H¨
  • hn (PI)

July 16th, 2014 1 / 14

slide-2
SLIDE 2

What makes quantum theory special?

which physical statements characterize QT within landscape of probabilistic theories? ⇒ wave of QT reconstructions within ‘generalized probabilistic theories’ framework

[Hardy, Masanes, M¨ uller, Brukner, Dakic, D’Ariano, Chiribella, Perinotti......]

complementary to this: understand quantum theory as framework for information inference [Rovelli, Zeilinger, Brukner, Fuchs, Spekkens,......] ⇒ Can one characterize/derive QT with primacy on information inference? Why useful?

1 gain novel perspective on QT within inference theory space 2 why or why not QT in its present form a fundamental theory

  • P. H¨
  • hn (PI)

July 16th, 2014 2 / 14

slide-3
SLIDE 3

Operational setup

Observer O interrogating system S with binary questions Qi, i = 1, . . . each Qi non-trivial 1-bit question (info measure later) O has tested identical S sufficiently

  • ften to have gained info about set of

possible “states” Bayesian viewpoint: for specific S, O assigns probabilities pi to Qi accord. to his info about

particular S, and about possible set of “states”

pi encode all O can say about S ⇒ state of S (rel. to O): collection of pi

  • P. H¨
  • hn (PI)

July 16th, 2014 3 / 14

slide-4
SLIDE 4

Question types

require symmetry under relabeling ‘yes’ ↔ ‘no’ for any Qi ∃ special state of ‘no information’ pi = 1

2 ∀ i ⇒ call totally mixed

state Qi, Qj are: independent if, relative to totally mixed state of S, answer to only Qi gives O no information about answer to Qj (and vice versa) ⇒ p(Qi, Qj) = pi · pj factorizes compatible if O may know answers to both simultaneously ⇒ pi, pj can be simultaneously 0, 1 complementary if knowledge of Qi disallows O to know Qj at the same time (and vice versa) ⇒ pi = 0, 1, then pj = 1/2

  • P. H¨
  • hn (PI)

July 16th, 2014 4 / 14

slide-5
SLIDE 5

Basic postulates (for N qubit systems)

for now, forget about QT, postulate: LI (limited information): “O can acquire maximally N ∈ N independent bits of information about S at the same time.” ∃ Qi, i = 1, . . . , N (mutually) independent compatible C (complementarity): “O can always get up to N new (independent) bits

  • f information about S. Whenever O asks a new question he

experiences no net loss of information.” ∃ Q′

i, i = 1, . . . , N independent compatible but Qi, Q′ j=i

complementary

(Postulates motivated by Rovelli’s Relational Quantum Mechanics and work by Brukner/Zeilinger)

⇒ postulates imply elements of qubit QT

  • P. H¨
  • hn (PI)

July 16th, 2014 5 / 14

slide-6
SLIDE 6

Number of degrees of freedom

How many independent Qi for N qubits? N = 1: only individual Qi, i = 1, . . . , D1 ⇒ D1 =? (know D1 ≥ 2)

  • P. H¨
  • hn (PI)

July 16th, 2014 6 / 14

slide-7
SLIDE 7

Number of degrees of freedom

How many independent Qi for N qubits? N = 1: only individual Qi, i = 1, . . . , D1 ⇒ D1 =? (know D1 ≥ 2) N = 2: 2D1 individual Qi vertex: individual question Qi, Q′

j

qubit 1 qubit 2 Q1 Q2 Q3 QD1 Q′

1

Q′

2

Q′

3

Q′

D1

. . . . . . . . .

  • P. H¨
  • hn (PI)

July 16th, 2014 6 / 14

slide-8
SLIDE 8

Number of degrees of freedom

How many independent Qi for N qubits? N = 1: only individual Qi, i = 1, . . . , D1 ⇒ D1 =? (know D1 ≥ 2) N = 2: 2D1 individual Qi + D2

1 composite questions:

Qij := Qi ↔ Q′

j “Are answers to Qi and Q′ j the same?”

+ ??? vertex: individual question Qi, Q′

j

edge: composite question Qij qubit 1 qubit 2 Q1 Q2 Q3 QD1 Q′

1

Q′

2

Q′

3

Q′

D1

. . . . . . . . . Q11

  • P. H¨
  • hn (PI)

July 16th, 2014 6 / 14

slide-9
SLIDE 9

Number of degrees of freedom

How many independent Qi for N qubits? N = 1: only individual Qi, i = 1, . . . , D1 ⇒ D1 =? (know D1 ≥ 2) N = 2: 2D1 individual Qi + D2

1 composite questions:

Qij := Qi ↔ Q′

j “Are answers to Qi and Q′ j the same?”

+ ??? vertex: individual question Qi, Q′

j

edge: composite question Qij qubit 1 qubit 2 Q1 Q2 Q3 QD1 Q′

1

Q′

2

Q′

3

Q′

D1

. . . . . . . . . Q11 Q22 QD1D1 Q31 Q23

  • P. H¨
  • hn (PI)

July 16th, 2014 6 / 14

slide-10
SLIDE 10

Number of degrees of freedom

How many independent Qi for N qubits? N = 1: only individual Qi, i = 1, . . . , D1 ⇒ D1 =? (know D1 ≥ 2) N = 2: 2D1 individual Qi + D2

1 composite questions:

Qij := Qi ↔ Q′

j “Are answers to Qi and Q′ j the same?”

+ ??? vertex: individual question Qi, Q′

j

edge: composite question Qij qubit 1 qubit 2 Q1 Q2 Q3 QD1 Q′

1

Q′

2

Q′

3

Q′

D1

. . . . . . . . .

  • P. H¨
  • hn (PI)

July 16th, 2014 6 / 14

slide-11
SLIDE 11

Compatibility and independence of composite questions

assume Specker’s principle: if n Qi are pairwise compatible then they are also mutually compatible Specker’s principle + Complementarity ⇒ {Qij}D1

i,j=1 pairwise

independent

  • P. H¨
  • hn (PI)

July 16th, 2014 7 / 14

slide-12
SLIDE 12

Compatibility and independence of composite questions

assume Specker’s principle: if n Qi are pairwise compatible then they are also mutually compatible Specker’s principle + Complementarity ⇒ {Qij}D1

i,j=1 pairwise

independent Qij complementary if corresponding edges intersect . . .

  • P. H¨
  • hn (PI)

July 16th, 2014 7 / 14

slide-13
SLIDE 13

Compatibility and independence of composite questions

assume Specker’s principle: if n Qi are pairwise compatible then they are also mutually compatible Specker’s principle + Complementarity ⇒ {Qij}D1

i,j=1 pairwise

independent Qij complementary if corresponding edges intersect Specker + C ⇒ Qij compatible if corresponding edges non-intersecting (otherwise net loss of information possible) ⇒ entanglement: > 1 bit in Qij [Brukner, Zeilinger] . . .

  • P. H¨
  • hn (PI)

July 16th, 2014 7 / 14

slide-14
SLIDE 14

Compatibility and independence of composite questions

assume Specker’s principle: if n Qi are pairwise compatible then they are also mutually compatible Specker’s principle + Complementarity ⇒ {Qij}D1

i,j=1 pairwise

independent Qij complementary if corresponding edges intersect Specker + C ⇒ Qij compatible if corresponding edges non-intersecting (otherwise net loss of information possible) ⇒ entanglement: > 1 bit in Qij [Brukner, Zeilinger] . . .

  • P. H¨
  • hn (PI)

July 16th, 2014 7 / 14

slide-15
SLIDE 15

Compatibility and independence of composite questions

assume Specker’s principle: if n Qi are pairwise compatible then they are also mutually compatible Specker’s principle + Complementarity ⇒ {Qij}D1

i,j=1 pairwise

independent Qij complementary if corresponding edges intersect Specker + C ⇒ Qij compatible if corresponding edges non-intersecting (otherwise net loss of information possible) ⇒ entanglement: > 1 bit in Qij [Brukner, Zeilinger] how many independent Qs for N = 2? ⇒ depends on D1 . . .

  • P. H¨
  • hn (PI)

July 16th, 2014 7 / 14

slide-16
SLIDE 16

Why is the Bloch sphere 3-dim.?

[see also M¨ uller, Masanes, v. Weizsaecker,....]

Logical argument from N = 2 case: Qii, i =, . . . , D1 pairwise independent, compatible O can acquire answers to all D1 composites Qii simultaneously (Specker) . . . Q11 Q22 QD1D1 Q33

  • P. H¨
  • hn (PI)

July 16th, 2014 8 / 14

slide-17
SLIDE 17

Why is the Bloch sphere 3-dim.?

[see also M¨ uller, Masanes, v. Weizsaecker,....]

Logical argument from N = 2 case: Qii, i =, . . . , D1 pairwise independent, compatible O can acquire answers to all D1 composites Qii simultaneously (Specker) LI: O cannot know more than N = 2 independent bits about S . . . Q11 Q22 QD1D1 Q33

  • P. H¨
  • hn (PI)

July 16th, 2014 8 / 14

slide-18
SLIDE 18

Why is the Bloch sphere 3-dim.?

[see also M¨ uller, Masanes, v. Weizsaecker,....]

Logical argument from N = 2 case: Qii, i =, . . . , D1 pairwise independent, compatible O can acquire answers to all D1 composites Qii simultaneously (Specker) LI: O cannot know more than N = 2 independent bits about S ⇒ answers to any two Qii determine answers to all other Qjj . . . Q11 Q22 QD1D1 Q33

  • P. H¨
  • hn (PI)

July 16th, 2014 8 / 14

slide-19
SLIDE 19

Why is the Bloch sphere 3-dim.?

[see also M¨ uller, Masanes, v. Weizsaecker,....]

Logical argument from N = 2 case: Qii, i =, . . . , D1 pairwise independent, compatible O can acquire answers to all D1 composites Qii simultaneously (Specker) LI: O cannot know more than N = 2 independent bits about S ⇒ answers to any two Qii determine answers to all other Qjj . . . Q11 Q22 QD1D1 Q33 e.g., truth table for any three Qii (a = b): ⇒ Q33 = Q11 ↔ Q22 or ¬(Q11 ↔ Q22) Q11 Q22 Q33 1 a 1 a 1 1 b b

  • P. H¨
  • hn (PI)

July 16th, 2014 8 / 14

slide-20
SLIDE 20

Why is the Bloch sphere 3-dim.?

[see also M¨ uller, Masanes, v. Weizsaecker,....]

Logical argument from N = 2 case: Qii, i =, . . . , D1 pairwise independent, compatible O can acquire answers to all D1 composites Qii simultaneously (Specker) LI: O cannot know more than N = 2 independent bits about S ⇒ answers to any two Qii determine answers to all other Qjj . . . Q11 Q22 QD1D1 Q33 e.g., truth table for any three Qii (a = b): ⇒ Q33 = Q11 ↔ Q22 or ¬(Q11 ↔ Q22) ⇒ holds for all compatible sets of Qij: 2 ≤ D1 ≤ 3 Q11 Q22 Q33 1 a 1 a 1 1 b b

  • P. H¨
  • hn (PI)

July 16th, 2014 8 / 14

slide-21
SLIDE 21

Why is the Bloch sphere 3-dim.?

[see also M¨ uller, Masanes, v. Weizsaecker,....]

Logical argument from N = 2 case: Qii, i =, . . . , D1 pairwise independent, compatible O can acquire answers to all D1 composites Qii simultaneously (Specker) LI: O cannot know more than N = 2 independent bits about S ⇒ answers to any two Qii determine answers to all other Qjj . . . Q11 Q22 QD1D1 Q33 e.g., truth table for any three Qii (a = b): ⇒ Q33 = Q11 ↔ Q22 or ¬(Q11 ↔ Q22) ⇒ holds for all compatible sets of Qij: 2 ≤ D1 ≤ 3 ⇒ # DoFs: 15 if D1 = 3; 9 if D1 = 2 Q11 Q22 Q33 1 a 1 a 1 1 b b

  • P. H¨
  • hn (PI)

July 16th, 2014 8 / 14

slide-22
SLIDE 22

Odd and even correlations

know: Q33 =

?

¬(Q11 ↔ Q22) =

?

¬(Q12 ↔ Q21)

Q1 Q2 Q3 Q′

1

Q′

2

Q′

3

Q11 Q22 Q33 Q1 Q2 Q3 Q′

1

Q′

2

Q′

3

Q12 Q21 Q33

Hence, (a) Q11 ↔ Q22 = Q12 ↔ Q21, or (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21)

  • P. H¨
  • hn (PI)

July 16th, 2014 9 / 14

slide-23
SLIDE 23

Odd and even correlations

know: Q33 =

?

¬(Q11 ↔ Q22) =

?

¬(Q12 ↔ Q21) Hence, (a) Q11 ↔ Q22 = Q12 ↔ Q21, or (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21) case (a): e.g., suppose Q11 = Q22 = 1 ⇒ Q12 ↔ Q21 = 1

Q11 Q22 Q12 Q21

  • P. H¨
  • hn (PI)

July 16th, 2014 9 / 14

slide-24
SLIDE 24

Odd and even correlations

know: Q33 =

?

¬(Q11 ↔ Q22) =

?

¬(Q12 ↔ Q21) Hence, (a) Q11 ↔ Q22 = Q12 ↔ Q21, or (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21) case (a): e.g., suppose Q11 = Q22 = 1 ⇒ Q12 ↔ Q21 = 1 ⇒ diagrams can be consistently joined

  • P. H¨
  • hn (PI)

July 16th, 2014 9 / 14

slide-25
SLIDE 25

Odd and even correlations

know: Q33 =

?

¬(Q11 ↔ Q22) =

?

¬(Q12 ↔ Q21) Hence, (a) Q11 ↔ Q22 = Q12 ↔ Q21, or (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21) case (a): e.g., suppose Q11 = Q22 = 1 ⇒ Q12 ↔ Q21 = 1 ⇒ diagrams can be consistently joined ⇒ Q1 ↔ Q2 = Q′

1 ↔ Q′ 2

  • P. H¨
  • hn (PI)

July 16th, 2014 9 / 14

slide-26
SLIDE 26

Odd and even correlations

know: Q33 =

?

¬(Q11 ↔ Q22) =

?

¬(Q12 ↔ Q21) Hence, (a) Q11 ↔ Q22 = Q12 ↔ Q21, or (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21) case (a): e.g., suppose Q11 = Q22 = 1 ⇒ Q12 ↔ Q21 = 1 ⇒ diagrams can be consistently joined ⇒ Q1 ↔ Q2 = Q′

1 ↔ Q′ 2

⇒ illegal complementary info ⇒ would obtain same diagram in ‘hidden variable model’

  • P. H¨
  • hn (PI)

July 16th, 2014 9 / 14

slide-27
SLIDE 27

Odd and even correlations

know: Q33 =

?

¬(Q11 ↔ Q22) =

?

¬(Q12 ↔ Q21) Hence, (a) Q11 ↔ Q22 = Q12 ↔ Q21, or (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21) case (b): e.g., suppose Q11 = Q22 = 1 ⇒ Q12 ↔ Q21 = 0

Q11 Q22 Q12 Q21

  • P. H¨
  • hn (PI)

July 16th, 2014 9 / 14

slide-28
SLIDE 28

Odd and even correlations

know: Q33 =

?

¬(Q11 ↔ Q22) =

?

¬(Q12 ↔ Q21) Hence, (a) Q11 ↔ Q22 = Q12 ↔ Q21, or (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21) case (b): e.g., suppose Q11 = Q22 = 1 ⇒ Q12 ↔ Q21 = 0 ⇒ diagrams cannot be joined consistently (while assigning values simultaneously to compl. Qs)

Q11 Q22 Q12 Q21

⇒ no illegal info can be extracted

  • P. H¨
  • hn (PI)

July 16th, 2014 9 / 14

slide-29
SLIDE 29

Odd and even correlations

know: Q33 =

?

¬(Q11 ↔ Q22) =

?

¬(Q12 ↔ Q21) Hence, (a) Q11 ↔ Q22 = Q12 ↔ Q21, or (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21) case (b): e.g., suppose Q11 = Q22 = 1 ⇒ Q12 ↔ Q21 = 0 ⇒ diagrams cannot be joined consistently (while assigning values simultaneously to compl. Qs)

Q11 Q22 Q12 Q21

⇒ no illegal info can be extracted ⇒ Complementarity implies (b) Q11 ↔ Q22 = ¬(Q12 ↔ Q21)

  • P. H¨
  • hn (PI)

July 16th, 2014 9 / 14

slide-30
SLIDE 30

Correlation structure for qubits (N = 2 and D1 = 3)

Compatibility structure of Qs ⇒ correlation structure for 2 qubits in QT

Q, Q′ compatible if connected by edge, otherwise complementary

+ + + − − − Q11 Q22 Q12 Q33 Q13 Q13 Q21 Q23 Q23 Q31 Q32

identify identify

− A B C

⇔ odd correlation A = ¬(B ↔ C), etc...

+ A B C

⇔ even correlation A = B ↔ C, etc...

  • P. H¨
  • hn (PI)

July 16th, 2014 10 / 14

slide-31
SLIDE 31

Correlation structure for rebits (N = 2 and D1 = 2)

similarly for 2 rebits + − Q11 Q22 Q12 Q33 Q21 key difference rebits vs. qubits: Q33 = ¬(Q11 ↔ Q22) non-local (∄ Q3, Q′

3)

Q11 Q22

complementary to all indiv. Qs ⇒ determines entanglement

− A B C

⇔ odd correlation A = ¬(B ↔ C), etc...

+ A B C

⇔ even correlation A = B ↔ C, etc...

  • P. H¨
  • hn (PI)

July 16th, 2014 11 / 14

slide-32
SLIDE 32

N = 3 qubits and monogamy

(3 + 1)3 − 1 = 63 independent questions (for D1 = 3): individual: QxA, QxB, QxC , . . . (3 × 3 = 9) bipartite: QxAxB, QxAxC , QxBxC, . . . ( 3

2

  • 32 = 27)

tripartite: QxAxBxC := QxA ↔ QxBxC = QxAxB ↔ QxC = QxAxC ↔ QxB, . . . ( 3

3

  • 33 = 27)

implies monogamy of entanglement: e.g., suppose O knows QxAxB, QyAyB ⇒ which composite Qs involving C compatible? A B C QxAxB QyAyB

  • P. H¨
  • hn (PI)

July 16th, 2014 12 / 14

slide-33
SLIDE 33

N = 3 qubits and monogamy

(3 + 1)3 − 1 = 63 independent questions (for D1 = 3): individual: QxA, QxB, QxC , . . . (3 × 3 = 9) bipartite: QxAxB, QxAxC , QxBxC, . . . ( 3

2

  • 32 = 27)

tripartite: QxAxBxC := QxA ↔ QxBxC = QxAxB ↔ QxC = QxAxC ↔ QxB, . . . ( 3

3

  • 33 = 27)

implies monogamy of entanglement: e.g., suppose O knows QxAxB, QyAyB ⇒ which composite Qs involving C compatible? A B C QxAxB QyAyB QyByC

  • P. H¨
  • hn (PI)

July 16th, 2014 12 / 14

slide-34
SLIDE 34

N = 3 qubits and monogamy

(3 + 1)3 − 1 = 63 independent questions (for D1 = 3): individual: QxA, QxB, QxC , . . . (3 × 3 = 9) bipartite: QxAxB, QxAxC , QxBxC, . . . ( 3

2

  • 32 = 27)

tripartite: QxAxBxC := QxA ↔ QxBxC = QxAxB ↔ QxC = QxAxC ↔ QxB, . . . ( 3

3

  • 33 = 27)

implies monogamy of entanglement: e.g., suppose O knows QxAxB, QyAyB ⇒ which composite Qs involving C compatible? Specker + C: only correlations

  • f QxC, QyC , QzC with

QxAxB, QyAyB A B C QxAxB QyAyB QyByC

  • P. H¨
  • hn (PI)

July 16th, 2014 12 / 14

slide-35
SLIDE 35

N = 3 qubits and monogamy

(3 + 1)3 − 1 = 63 independent questions (for D1 = 3): individual: QxA, QxB, QxC , . . . (3 × 3 = 9) bipartite: QxAxB, QxAxC , QxBxC, . . . ( 3

2

  • 32 = 27)

tripartite: QxAxBxC := QxA ↔ QxBxC = QxAxB ↔ QxC = QxAxC ↔ QxB, . . . ( 3

3

  • 33 = 27)

implies monogamy of entanglement: e.g., suppose O knows QxAxB, QyAyB ⇒ which composite Qs involving C compatible? Specker + C: only correlations

  • f QxC, QyC , QzC with

QxAxB, QyAyB

A B C QxAxB QyAyB QxAxB xC QxA QyA QzA QxB QyB QzB QxC QyC QzC

  • P. H¨
  • hn (PI)

July 16th, 2014 12 / 14

slide-36
SLIDE 36

N = 3 qubits and monogamy

(3 + 1)3 − 1 = 63 independent questions (for D1 = 3): individual: QxA, QxB, QxC , . . . (3 × 3 = 9) bipartite: QxAxB, QxAxC , QxBxC, . . . ( 3

2

  • 32 = 27)

tripartite: QxAxBxC := QxA ↔ QxBxC = QxAxB ↔ QxC = QxAxC ↔ QxB, . . . ( 3

3

  • 33 = 27)

implies monogamy of entanglement: e.g., suppose O knows QxAxB, QyAyB ⇒ which composite Qs involving C compatible? Specker + C: only correlations

  • f QxC, QyC , QzC with

QxAxB, QyAyB ⇒ equivalent to asking QxC , QyC or QzC ⇒ only indiv. info about C

A B C QxAxB QyAyB QxAxB xC QxA QyA QzA QxB QyB QzB QxC QyC QzC

  • P. H¨
  • hn (PI)

July 16th, 2014 12 / 14

slide-37
SLIDE 37

Reconstruction attempt of QT

State of S relative to O:

  • PO→S =

   p1 . . . pDN    , pi prob. that Qi = 1, Qi indep.

  • P. H¨
  • hn (PI)

July 16th, 2014 13 / 14

slide-38
SLIDE 38

Reconstruction attempt of QT

State of S relative to O:

  • PO→S =

   p1 . . . pDN    , pi prob. that Qi = 1, Qi indep.

further postulates P3: Any PO→S permissible, s.t. info in PO→S compatible with LI and C P4: O’s total info about S preserved between interrogations P5: Time evolution of PO→S continuous

  • P. H¨
  • hn (PI)

July 16th, 2014 13 / 14

slide-39
SLIDE 39

Reconstruction attempt of QT

State of S relative to O:

  • PO→S =

   p1 . . . pDN    , pi prob. that Qi = 1, Qi indep.

further postulates P3: Any PO→S permissible, s.t. info in PO→S compatible with LI and C P4: O’s total info about S preserved between interrogations P5: Time evolution of PO→S continuous which info measure? ⇒ P4 + P5 (+ operational conditions on measure): O’s info about Qi αi = (2pi − 1)2, O’s total info about S IO→S = ||2 PO→S − 1||2

  • P. H¨
  • hn (PI)

July 16th, 2014 13 / 14

slide-40
SLIDE 40

Reconstruction attempt of QT

State of S relative to O:

  • PO→S =

   p1 . . . pDN    , pi prob. that Qi = 1, Qi indep.

further postulates P3: Any PO→S permissible, s.t. info in PO→S compatible with LI and C P4: O’s total info about S preserved between interrogations P5: Time evolution of PO→S continuous which info measure? ⇒ P4 + P5 (+ operational conditions on measure): O’s info about Qi αi = (2pi − 1)2, O’s total info about S IO→S = ||2 PO→S − 1||2 further results: reversible time evolution, Bloch sphere for N = 1, results concerning information distribution,....

  • P. H¨
  • hn (PI)

July 16th, 2014 13 / 14

slide-41
SLIDE 41

Conclusion and outlook

quantum theory as operational framework for information inference quantum state as state of information of O about S what are physical statements that single out QT from set of inference theories? ⇒ limited information and complementarity imply many structural features of QT

Outlook:

complete approach to full reconstruction statements characterizing QT ⇒ justify why or why not to modify QT for “fundamental theory”? in quantum gravity/cosmology ‘wave fct. of the universe’ as global state of information?

  • r rather: no global/absolute state and universe as ‘information

exchange network’ of subsystems

  • P. H¨
  • hn (PI)

July 16th, 2014 14 / 14