Quantum Probability 1982-2017

Hans Maassen, Universities of Nijmegen and Amsterdam

Nijmegen, June 23, 2017. 1582: Pope Gregory XIII audaciously replaces the old-fashioned Julian calender by the new Gregorian calender, which still prevails.

1982: A bunch of mathematicians and physicists replaces the old-fashioned Kolmogorovian probability axioms by new ones: quantum probability, which still prevail.

Two historic events 1982: A bunch of mathematicians and physicists replaces the old-fashioned Kolmogorovian probability axioms by new ones: quantum probability, which still prevail.

Two historic events

1582: Pope Gregory XIII audaciously replaces the old-fashioned Julian calender by the new Gregorian calender, which still prevails. Two historic events

1582: Pope Gregory XIII audaciously replaces the old-fashioned Julian calender by the new Gregorian calender, which still prevails.

1982: A bunch of mathematicians and physicists replaces the old-fashioned Kolmogorovian probability axioms by new ones: quantum probability, which still prevail. Two historic events

1582: Pope Gregory XIII audaciously replaces the old-fashioned Julian calender by the new Gregorian calender, which still prevails.

1982: A bunch of mathematicians and physicists replaces the old-fashioned Kolmogorovian probability axioms by new ones: quantum probability, which still prevail. ∗ 400 years time difference;

∗ They happened in the same hall, in Villa Mondragone, 30 km south-west of Rome, in the Frascati hills.

Villa Mondragone

First conference in quantum probability, 1982

How are these two events related? ∗ They happened in the same hall, in Villa Mondragone, 30 km south-west of Rome, in the Frascati hills.

Villa Mondragone

First conference in quantum probability, 1982

How are these two events related?

∗ 400 years time difference; Villa Mondragone

First conference in quantum probability, 1982

How are these two events related?

∗ 400 years time difference;

∗ They happened in the same hall, in Villa Mondragone, 30 km south-west of Rome, in the Frascati hills. Villa Mondragone

First conference in quantum probability, 1982

How are these two events related?

∗ 400 years time difference;

∗ They happened in the same hall, in Villa Mondragone, 30 km south-west of Rome, in the Frascati hills. How are these two events related?

∗ 400 years time difference;

∗ They happened in the same hall, in Villa Mondragone, 30 km south-west of Rome, in the Frascati hills.

Villa Mondragone

First conference in quantum probability, 1982 (A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds .

The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible); The new scheme

(A, ϕ) to replace(Ω , Σ, µ) Quantum probability space A: C*-algebra containing 1l (of operators on a Hilbert space). ϕ: state (positive complex functional on A with ϕ(1l) = 1). Interpretation: p ∈ A projection ←→ question, proposition; 1l − p ←→ not-p ; pq = qp ←→ p and q are compatible propositions; pq ←→ p and q (if compatible); p + q − pq ←→ p or q (if compatible);

p1 + ... + pn = 1l ←→ exactly one of p1,... pn holds . From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ Where is the innovation?’

Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. Reactions to the new framework

From probability: ∗ ‘We understand that random variables form an algebra, ∗ but what is this funny non-commutativity? ∗ Is not Kolmogorov good enough?’ From quantum physics: ∗ ‘We understand that observables do not commute, ∗ but what is the role of this funny algebra? ∗ Is not Hilbert space good enough?’ In general: ∗ ‘This is nothing new, just operator algebras in quantum mechanics. ∗ Where is the innovation?’ To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics.

Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. Short answers To the probabilists: ∗ Non-commutativity is at least very convenient. Ask the physicists! ∗ Kolmogorov is indeed not good enough. This approach would require non-local models, which we don’t like in physics. To the physicists: ∗ Algebras often give better insight in the structure of the problem. Commutants come into view, and the distiction between classical and quantum information. Ask the probabilists about their σ-algebras! ∗ Hilbert space is indeed not good enough: For big (infinite) systems A the Hilbert space changes with the state ϕ. For instance different temperatures require inequivalent Hilbert spaces representations in statistical mechanics. It makes a connection. Over that bridge ideas could be carried to the other side.

About innovation:

Quantum probability was built from known ingredients. Over that bridge ideas could be carried to the other side.

About innovation:

Quantum probability was built from known ingredients. It makes a connection. About innovation:

Quantum probability was built from known ingredients. It makes a connection. Over that bridge ideas could be carried to the other side. Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum groups (Woronovic)

Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic) Developments since 1982 Internal: • Quantum stochastic calculus (Hudson, Parthasarathy) • Markov dilations of dynamical semigroups(Frigerio, K¨ummerer) • Quantum statistics and estimation (Holevo, Gill, Guta) • Quantum filtering and control (Belavkin, van Handel, Bouten,Gough) • Ergodic theory of quantum trajectories (K¨ummerer) • Bialgebras, Hilbert modules, quantum groups (Sch¨urmann, Skeide, Franz) External: • Bell inequalities and Aspect’s experiment • Free probability (Voiculescu, Speicher) • Quantum information theory (Bennett, Shor, Werner) • Quantum computing (Shor) • Quantum trajectories (Gisin, Carmichael,Mølmer) • Quantum groups (Woronovic) 2. Quantum information: the impossibility of copying. 3. Inequivalent representations. 4. Ergodic decomposition of quantum trajectories.

Topics for today

1. The need for a non-commutative theory, locality. 3. Inequivalent representations. 4. Ergodic decomposition of quantum trajectories.

Topics for today

1. The need for a non-commutative theory, locality. 2. Quantum information: the impossibility of copying. 4. Ergodic decomposition of quantum trajectories.

Topics for today

1. The need for a non-commutative theory, locality. 2. Quantum information: the impossibility of copying. 3. Inequivalent representations. Topics for today

1. The need for a non-commutative theory, locality. 2. Quantum information: the impossibility of copying. 3. Inequivalent representations. 4. Ergodic decomposition of quantum trajectories. A = M3, the C*-algebra of all complex 3 × 3 matrices , ϕ = hψ, aψi, where ψ ∈ C3, kψk = 1 .

This is the only case in quantum prhysics where the (real part of the) Hilbert space C3 coincides with our own Euclidean space. We can play a game with the ρ-meson: We ask: ‘Is your spin zero in this direction?’ It answers by saying ‘yes’ or ‘no’. A coordinate frame now becomes a set of three mutually exclusive propositions!

Basic didactic example: particle of spin 1 (ρ-meson) ϕ = hψ, aψi, where ψ ∈ C3, kψk = 1 .

This is the only case in quantum prhysics where the (real part of the) Hilbert space C3 coincides with our own Euclidean space. We can play a game with the ρ-meson: We ask: ‘Is your spin zero in this direction?’ It answers by saying ‘yes’ or ‘no’. A coordinate frame now becomes a set of three mutually exclusive propositions!

Basic didactic example: particle of spin 1 (ρ-meson)

A = M3, the C*-algebra of all complex 3 × 3 matrices , This is the only case in quantum prhysics where the (real part of the) Hilbert space C3 coincides with our own Euclidean space. We can play a game with the ρ-meson: We ask: ‘Is your spin zero in this direction?’ It answers by saying ‘yes’ or ‘no’. A coordinate frame now becomes a set of three mutually exclusive propositions!

Basic didactic example: particle of spin 1 (ρ-meson)

A = M3, the C*-algebra of all complex 3 × 3 matrices , ϕ = hψ, aψi, where ψ ∈ C3, kψk = 1 . We can play a game with the ρ-meson: We ask: ‘Is your spin zero in this direction?’ It answers by saying ‘yes’ or ‘no’. A coordinate frame now becomes a set of three mutually exclusive propositions!

Basic didactic example: particle of spin 1 (ρ-meson)

A = M3, the C*-algebra of all complex 3 × 3 matrices , ϕ = hψ, aψi, where ψ ∈ C3, kψk = 1 .

This is the only case in quantum prhysics where the (real part of the) Hilbert space C3 coincides with our own Euclidean space. We ask: ‘Is your spin zero in this direction?’ It answers by saying ‘yes’ or ‘no’. A coordinate frame now becomes a set of three mutually exclusive propositions!

Basic didactic example: particle of spin 1 (ρ-meson)

A = M3, the C*-algebra of all complex 3 × 3 matrices , ϕ = hψ, aψi, where ψ ∈ C3, kψk = 1 .

This is the only case in quantum prhysics where the (real part of the) Hilbert space C3 coincides with our own Euclidean space. We can play a game with the ρ-meson: It answers by saying ‘yes’ or ‘no’. A coordinate frame now becomes a set of three mutually exclusive propositions!

Basic didactic example: particle of spin 1 (ρ-meson)

A = M3, the C*-algebra of all complex 3 × 3 matrices , ϕ = hψ, aψi, where ψ ∈ C3, kψk = 1 .

This is the only case in quantum prhysics where the (real part of the) Hilbert space C3 coincides with our own Euclidean space. We can play a game with the ρ-meson: We ask: ‘Is your spin zero in this direction?’ A coordinate frame now becomes a set of three mutually exclusive propositions!

Basic didactic example: particle of spin 1 (ρ-meson)

A = M3, the C*-algebra of all complex 3 × 3 matrices , ϕ = hψ, aψi, where ψ ∈ C3, kψk = 1 .

This is the only case in quantum prhysics where the (real part of the) Hilbert space C3 coincides with our own Euclidean space. We can play a game with the ρ-meson: We ask: ‘Is your spin zero in this direction?’ It answers by saying ‘yes’ or ‘no’. Basic didactic example: particle of spin 1 (ρ-meson)

A = M3, the C*-algebra of all complex 3 × 3 matrices , ϕ = hψ, aψi, where ψ ∈ C3, kψk = 1 .

This is the only case in quantum prhysics where the (real part of the) Hilbert space C3 coincides with our own Euclidean space. We can play a game with the ρ-meson: We ask: ‘Is your spin zero in this direction?’ It answers by saying ‘yes’ or ‘no’. A coordinate frame now becomes a set of three mutually exclusive propositions! Two players: A(lice) and B(ob). Each asks the question: ‘Is the rabbit behind this door?’

1. A silly game with a rabbit

Three doors on stage. Behind one of them sits a rabbit. ‘Is the rabbit behind this door?’

1. A silly game with a rabbit

Three doors on stage. Behind one of them sits a rabbit.

Two players: A(lice) and B(ob). Each asks the question: 1. A silly game with a rabbit

Three doors on stage. Behind one of them sits a rabbit.

Two players: A(lice) and B(ob). Each asks the question: ‘Is the rabbit behind this door?’ 1. A silly game with a rabbit

Three doors on stage. Behind one of them sits a rabbit.

Two players: A(lice) and B(ob). Each asks the question: ‘Is the rabbit behind this door?’ If Alice and Bob point at the same door:

1 (yes, yes) with probability 3 2 (no, no) with probability 3

If Alice and Bob point at different doors:

1 (yes, no) with probability 3 1 (no, yes) with probability 3 1 (no, no) with probability 3

These phenomena can be reproduced even without a rabbit.

phenomena: 1 (yes, yes) with probability 3 2 (no, no) with probability 3

If Alice and Bob point at different doors:

1 (yes, no) with probability 3 1 (no, yes) with probability 3 1 (no, no) with probability 3

These phenomena can be reproduced even without a rabbit.

phenomena:

If Alice and Bob point at the same door: 2 (no, no) with probability 3

If Alice and Bob point at different doors:

1 (yes, no) with probability 3 1 (no, yes) with probability 3 1 (no, no) with probability 3

These phenomena can be reproduced even without a rabbit.

phenomena:

If Alice and Bob point at the same door:

1 (yes, yes) with probability 3 If Alice and Bob point at different doors:

1 (yes, no) with probability 3 1 (no, yes) with probability 3 1 (no, no) with probability 3

These phenomena can be reproduced even without a rabbit.

phenomena:

If Alice and Bob point at the same door:

1 (yes, yes) with probability 3 2 (no, no) with probability 3 1 (yes, no) with probability 3 1 (no, yes) with probability 3 1 (no, no) with probability 3

These phenomena can be reproduced even without a rabbit.

phenomena:

If Alice and Bob point at the same door:

1 (yes, yes) with probability 3 2 (no, no) with probability 3

If Alice and Bob point at different doors: 1 (no, yes) with probability 3 1 (no, no) with probability 3

These phenomena can be reproduced even without a rabbit.

phenomena:

If Alice and Bob point at the same door:

1 (yes, yes) with probability 3 2 (no, no) with probability 3

If Alice and Bob point at different doors:

1 (yes, no) with probability 3 1 (no, no) with probability 3

These phenomena can be reproduced even without a rabbit.

phenomena:

If Alice and Bob point at the same door:

1 (yes, yes) with probability 3 2 (no, no) with probability 3

If Alice and Bob point at different doors:

1 (yes, no) with probability 3 1 (no, yes) with probability 3 These phenomena can be reproduced even without a rabbit.

phenomena:

If Alice and Bob point at the same door:

1 (yes, yes) with probability 3 2 (no, no) with probability 3

If Alice and Bob point at different doors:

1 (yes, no) with probability 3 1 (no, yes) with probability 3 1 (no, no) with probability 3 phenomena:

If Alice and Bob point at the same door:

1 (yes, yes) with probability 3 2 (no, no) with probability 3

If Alice and Bob point at different doors:

1 (yes, no) with probability 3 1 (no, yes) with probability 3 1 (no, no) with probability 3

These phenomena can be reproduced even without a rabbit. Between the rooms no communication is allowed. The same phenomena are obtained.

This time, we do need the rabbits! They must be placed behind the doors before the questioning begins.

Another silly game with two correlated rabbits

Two rooms, one for A and one for B. Each room has three doors next to each other. Behind corresponding doors rabbits are sitting. The same phenomena are obtained.

This time, we do need the rabbits! They must be placed behind the doors before the questioning begins.

Another silly game with two correlated rabbits

Two rooms, one for A and one for B. Each room has three doors next to each other. Behind corresponding doors rabbits are sitting.Between the rooms no communication is allowed. This time, we do need the rabbits! They must be placed behind the doors before the questioning begins.

Another silly game with two correlated rabbits

Two rooms, one for A and one for B. Each room has three doors next to each other. Behind corresponding doors rabbits are sitting.Between the rooms no communication is allowed. The same phenomena are obtained. Another silly game with two correlated rabbits

Two rooms, one for A and one for B. Each room has three doors next to each other. Behind corresponding doors rabbits are sitting.Between the rooms no communication is allowed. The same phenomena are obtained.

This time, we do need the rabbits! They must be placed behind the doors before the questioning begins. A game with two entangled ρ-mesons

Two rooms, one for A, one for B. Each room contains a ρ-meson; the pair of mesons is described by

A = M3 ⊗ M3 , 3 1 X ϕ(x)= hψ, xψi, where ψ = √ ei ⊗ ei , 3 i=1 3 Let pi ∈ M3 denote the projection onto ei ∈ R .Then the answers to the questions p1 ⊗ 1l and 1l ⊗ p1 have the probabilities 1 ϕ([yes, yes]) = ϕ(p ⊗ p ) = , 1 1 3 2 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 1 3

The answers to p1 ⊗ 1l and 1l ⊗ p2 have probabilities 1 ϕ([yes, no]) = ϕ(p ⊗ (1l − p )) = , 1 2 3 1 ϕ([no, yes]) = ϕ((1l − p ) ⊗ p ) = , 1 2 3 1 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 2 3 These results are obtained in all frames.

Quantum mechanics predicts:.... Then the answers to the questions p1 ⊗ 1l and 1l ⊗ p1 have the probabilities 1 ϕ([yes, yes]) = ϕ(p ⊗ p ) = , 1 1 3 2 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 1 3

The answers to p1 ⊗ 1l and 1l ⊗ p2 have probabilities 1 ϕ([yes, no]) = ϕ(p ⊗ (1l − p )) = , 1 2 3 1 ϕ([no, yes]) = ϕ((1l − p ) ⊗ p ) = , 1 2 3 1 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 2 3 These results are obtained in all frames.

Quantum mechanics predicts:.... 3 Let pi ∈ M3 denote the projection onto ei ∈ R . 1 ϕ([yes, yes]) = ϕ(p ⊗ p ) = , 1 1 3 2 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 1 3

The answers to p1 ⊗ 1l and 1l ⊗ p2 have probabilities 1 ϕ([yes, no]) = ϕ(p ⊗ (1l − p )) = , 1 2 3 1 ϕ([no, yes]) = ϕ((1l − p ) ⊗ p ) = , 1 2 3 1 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 2 3 These results are obtained in all frames.

Quantum mechanics predicts:.... 3 Let pi ∈ M3 denote the projection onto ei ∈ R .Then the answers to the questions p1 ⊗ 1l and 1l ⊗ p1 have the probabilities The answers to p1 ⊗ 1l and 1l ⊗ p2 have probabilities 1 ϕ([yes, no]) = ϕ(p ⊗ (1l − p )) = , 1 2 3 1 ϕ([no, yes]) = ϕ((1l − p ) ⊗ p ) = , 1 2 3 1 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 2 3 These results are obtained in all frames.

Quantum mechanics predicts:.... 3 Let pi ∈ M3 denote the projection onto ei ∈ R .Then the answers to the questions p1 ⊗ 1l and 1l ⊗ p1 have the probabilities 1 ϕ([yes, yes]) = ϕ(p ⊗ p ) = , 1 1 3 2 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 1 3 1 ϕ([yes, no]) = ϕ(p ⊗ (1l − p )) = , 1 2 3 1 ϕ([no, yes]) = ϕ((1l − p ) ⊗ p ) = , 1 2 3 1 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 2 3 These results are obtained in all frames.

Quantum mechanics predicts:.... 3 Let pi ∈ M3 denote the projection onto ei ∈ R .Then the answers to the questions p1 ⊗ 1l and 1l ⊗ p1 have the probabilities 1 ϕ([yes, yes]) = ϕ(p ⊗ p ) = , 1 1 3 2 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 1 3

The answers to p1 ⊗ 1l and 1l ⊗ p2 have probabilities These results are obtained in all frames.

Quantum mechanics predicts:.... 3 Let pi ∈ M3 denote the projection onto ei ∈ R .Then the answers to the questions p1 ⊗ 1l and 1l ⊗ p1 have the probabilities 1 ϕ([yes, yes]) = ϕ(p ⊗ p ) = , 1 1 3 2 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 1 3

The answers to p1 ⊗ 1l and 1l ⊗ p2 have probabilities 1 ϕ([yes, no]) = ϕ(p ⊗ (1l − p )) = , 1 2 3 1 ϕ([no, yes]) = ϕ((1l − p ) ⊗ p ) = , 1 2 3 1 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 2 3 Quantum mechanics predicts:.... 3 Let pi ∈ M3 denote the projection onto ei ∈ R .Then the answers to the questions p1 ⊗ 1l and 1l ⊗ p1 have the probabilities 1 ϕ([yes, yes]) = ϕ(p ⊗ p ) = , 1 1 3 2 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 1 3

The answers to p1 ⊗ 1l and 1l ⊗ p2 have probabilities 1 ϕ([yes, no]) = ϕ(p ⊗ (1l − p )) = , 1 2 3 1 ϕ([no, yes]) = ϕ((1l − p ) ⊗ p ) = , 1 2 3 1 ϕ([no, no]) = ϕ((1l − p ) ⊗ (1l − p )) = , 1 2 3 These results are obtained in all frames. According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ .

Calculation:

3 3 3 ! 3 ! 1 X 1 X X X (r ⊗ r)ψ = √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames (r ⊗ r)ψ = ψ .

Calculation:

3 3 3 ! 3 ! 1 X 1 X X X (r ⊗ r)ψ = √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: Calculation:

3 3 3 ! 3 ! 1 X 1 X X X (r ⊗ r)ψ = √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ . 3 3 3 ! 3 ! 1 X 1 X X X (r ⊗ r)ψ = √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ .

Calculation: 3 3 3 ! 3 ! 1 X 1 X X X √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ .

Calculation:

(r ⊗ r)ψ = 3 3 ! 3 ! 1 X X X = √ rji ej ⊗ rki ek 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ .

Calculation:

3 1 X (r ⊗ r)ψ = √ rei ⊗ rei 3 i=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ .

Calculation:

3 3 3 ! 3 ! 1 X 1 X X X (r ⊗ r)ψ = √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ .

Calculation:

3 3 3 ! 3 ! 1 X 1 X X X (r ⊗ r)ψ = √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1

The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ .

Calculation:

3 3 3 ! 3 ! 1 X 1 X X X (r ⊗ r)ψ = √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since The same answers in all frames According to quantum mechanics the model
M3 ⊗ M3, hψ, ·ψi gives the same answers in all frames, since it is invariant for joint rotations r: (r ⊗ r)ψ = ψ .

Calculation:

3 3 3 ! 3 ! 1 X 1 X X X (r ⊗ r)ψ = √ rei ⊗ rei = √ rji ej ⊗ rki ek 3 i=1 3 i=1 j=1 k=1 3 3 ! 1 X X = √ rji rki ej ⊗ ek = ψ 3 j,k=1 i=1 since 3 3 X X ∗ ∗ rji rki = rji (rik ) = (rr )jk = δjk . i=1 i=1 There exists no local classical model for the entangled ρ-meson game. I.e.: there do not exist a(Ω , Σ, P) and random variables 2 Xα, Yβ :Ω → {0, 1} , (α, β ∈ S ) , such that for all α, β ∈ S 2 and almost all ω ∈ Ω

 1 P([Xα = Yβ = 1]) = 3 α = β =⇒ 2 , (1) P([Xα = Yβ = 0]) = 3  1 P([Xα = 1, Yβ = 0]) = 3  1 α ⊥ β =⇒ P([Xα = 0, Yβ = 1]) = 3 (2)  1 P([Xα = 0, Yβ = 0]) = 3

Theorem I.e.: there do not exist a(Ω , Σ, P) and random variables 2 Xα, Yβ :Ω → {0, 1} , (α, β ∈ S ) , such that for all α, β ∈ S 2 and almost all ω ∈ Ω

 1 P([Xα = Yβ = 1]) = 3 α = β =⇒ 2 , (1) P([Xα = Yβ = 0]) = 3  1 P([Xα = 1, Yβ = 0]) = 3  1 α ⊥ β =⇒ P([Xα = 0, Yβ = 1]) = 3 (2)  1 P([Xα = 0, Yβ = 0]) = 3

Theorem

There exists no local classical model for the entangled ρ-meson game. 2 Xα, Yβ :Ω → {0, 1} , (α, β ∈ S ) , such that for all α, β ∈ S 2 and almost all ω ∈ Ω

 1 P([Xα = Yβ = 1]) = 3 α = β =⇒ 2 , (1) P([Xα = Yβ = 0]) = 3  1 P([Xα = 1, Yβ = 0]) = 3  1 α ⊥ β =⇒ P([Xα = 0, Yβ = 1]) = 3 (2)  1 P([Xα = 0, Yβ = 0]) = 3

Theorem

There exists no local classical model for the entangled ρ-meson game. I.e.: there do not exist a(Ω , Σ, P) and random variables such that for all α, β ∈ S 2 and almost all ω ∈ Ω

 1 P([Xα = Yβ = 1]) = 3 α = β =⇒ 2 , (1) P([Xα = Yβ = 0]) = 3  1 P([Xα = 1, Yβ = 0]) = 3  1 α ⊥ β =⇒ P([Xα = 0, Yβ = 1]) = 3 (2)  1 P([Xα = 0, Yβ = 0]) = 3

Theorem

There exists no local classical model for the entangled ρ-meson game. I.e.: there do not exist a(Ω , Σ, P) and random variables 2 Xα, Yβ :Ω → {0, 1} , (α, β ∈ S ) ,  1 P([Xα = 1, Yβ = 0]) = 3  1 α ⊥ β =⇒ P([Xα = 0, Yβ = 1]) = 3 (2)  1 P([Xα = 0, Yβ = 0]) = 3

Theorem

There exists no local classical model for the entangled ρ-meson game. I.e.: there do not exist a(Ω , Σ, P) and random variables 2 Xα, Yβ :Ω → {0, 1} , (α, β ∈ S ) , such that for all α, β ∈ S 2 and almost all ω ∈ Ω

 1 P([Xα = Yβ = 1]) = 3 α = β =⇒ 2 , (1) P([Xα = Yβ = 0]) = 3 Theorem

There exists no local classical model for the entangled ρ-meson game. I.e.: there do not exist a(Ω , Σ, P) and random variables 2 Xα, Yβ :Ω → {0, 1} , (α, β ∈ S ) , such that for all α, β ∈ S 2 and almost all ω ∈ Ω

 1 P([Xα = Yβ = 1]) = 3 α = β =⇒ 2 , (1) P([Xα = Yβ = 0]) = 3  1 P([Xα = 1, Yβ = 0]) = 3  1 α ⊥ β =⇒ P([Xα = 0, Yβ = 1]) = 3 (2)  1 P([Xα = 0, Yβ = 0]) = 3 From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED.

Proof From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED.

Proof

From (1) we must have Xα = Yα almost surely. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED.

Proof

From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED.

Proof

From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED.

Proof

From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED.

Proof

From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED.

Proof

From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED.

Proof

From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Contradiction! QED.

Proof

From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Proof

From (1) we must have Xα = Yα almost surely. From (2) we find for all α ⊥ β that P[Xα = Xβ = 1] = 0. Hence for all orthogonal frames α, β, γ we have P[Xα + Xβ + Xγ > 1] = 0. But also

P[Xα + Xβ + Xγ > 0] = P[Xα = 1 or Xβ = 1 or Xγ = 1] = P[Xα = 1] + P[Xβ = 1] + P[Xγ = 1] 1 1 1 = + + = 1 . 3 3 3

We conclude that Xα(ω) + Xβ(ω) + Xγ(ω) = 1 for almost all ω ∈ Ω. Now choose ω ∈ Ω sucht that this holds for all 16 frames (α, β, γ) ocurring in the ”Sudoku of Asher Peres”. Contradiction! QED. 33 directions; 16 frames involved. This ‘sudoku’ has no solution.

Asher Peres 1991 33 directions; 16 frames involved. This ‘sudoku’ has no solution.

Asher Peres 1991 33 directions; 16 frames involved. This ‘sudoku’ has no solution.

Asher Peres 1991 16 frames involved. This ‘sudoku’ has no solution.

Asher Peres 1991

33 directions; This ‘sudoku’ has no solution.

Asher Peres 1991

33 directions; 16 frames involved. Asher Peres 1991

33 directions; 16 frames involved. This ‘sudoku’ has no solution. Asher Peres 1991

33 directions; 16 frames involved. This ‘sudoku’ has no solution. We have seen that there exists no local classical model for the behaviour of the entangled ρ-mesons.

However, the quantum model is local in the sense that the answer to the question α of A remains the same proposition pα ∈ M3, whatever question β ⊥ α his opponent B asks.:

Xα = pα ⊗ 1l ;

Yβ = 1l ⊗ pβ .

Locality However, the quantum model is local in the sense that the answer to the question α of A remains the same proposition pα ∈ M3, whatever question β ⊥ α his opponent B asks.:

Xα = pα ⊗ 1l ;

Yβ = 1l ⊗ pβ .

Locality

We have seen that there exists no local classical model for the behaviour of the entangled ρ-mesons. However, the quantum model is local in the sense that the answer to the question α of A remains the same proposition pα ∈ M3, whatever question β ⊥ α his opponent B asks.:

Xα = pα ⊗ 1l ;

Yβ = 1l ⊗ pβ .

Locality

We have seen that there exists no local classical model for the behaviour of the entangled ρ-mesons. Locality

We have seen that there exists no local classical model for the behaviour of the entangled ρ-mesons.

However, the quantum model is local in the sense that the answer to the question α of A remains the same proposition pα ∈ M3, whatever question β ⊥ α his opponent B asks.:

Xα = pα ⊗ 1l ;

Yβ = 1l ⊗ pβ . Related result: Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 .

Proof. By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 .

Proof. By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Related result: There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 .

Proof. By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Related result: Lemma R(α) + R(β) + R(γ) = 1 .

Proof. By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Related result: Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 Proof. By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Related result: Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 . By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Related result: Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 .

Proof. R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Related result: Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 .

Proof. By Gleason’s theorem, such a functioon would have to be of the form for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Related result: Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 .

Proof. By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) But 0 and 1 do not satisfy the rquirement.

There is no rabbit function on the sphere Related result: Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 .

Proof. By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant. There is no rabbit function on the sphere Related result: Lemma There exists no function R : S 2 → {0, 1} with R(−x) = R(x) such that for every frame (α, β, γ) in R3 R(α) + R(β) + R(γ) = 1 .

Proof. By Gleason’s theorem, such a functioon would have to be of the form R(α) = tr(ρpα) for some positive 3 × 3-matrix ρ with trace 1. Hence R has to be continuous, hence constant.But 0 and 1 do not satisfy the rquirement. However, there is no answer to all questions jointly.

Such knowledge we might call quantum information.

ANY, BUT NOT ALL!

A kind of quantum double think. (Orwell, ‘1984’.) (Einstein hated this.)

Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked. Such knowledge we might call quantum information.

ANY, BUT NOT ALL!

A kind of quantum double think. (Orwell, ‘1984’.) (Einstein hated this.)

Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked.

However, there is no answer to all questions jointly. ANY, BUT NOT ALL!

A kind of quantum double think. (Orwell, ‘1984’.) (Einstein hated this.)

Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked.

However, there is no answer to all questions jointly.

Such knowledge we might call quantum information. ANY, BUT NOT ALL!

A kind of quantum double think. (Orwell, ‘1984’.) (Einstein hated this.)

Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked.

However, there is no answer to all questions jointly.

Such knowledge we might call quantum information. A kind of quantum double think. (Orwell, ‘1984’.) (Einstein hated this.)

Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked.

However, there is no answer to all questions jointly.

Such knowledge we might call quantum information.

ANY, BUT NOT ALL! A kind of quantum double think. (Orwell, ‘1984’.) (Einstein hated this.)

Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked.

However, there is no answer to all questions jointly.

Such knowledge we might call quantum information.

ANY, BUT NOT ALL! (Einstein hated this.)

Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked.

However, there is no answer to all questions jointly.

Such knowledge we might call quantum information.

ANY, BUT NOT ALL!

A kind of quantum double think. (Orwell, ‘1984’.) (Einstein hated this.)

Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked.

However, there is no answer to all questions jointly.

Such knowledge we might call quantum information.

ANY, BUT NOT ALL!

A kind of quantum double think. (Orwell, ‘1984’.) Quantum Information (informal)

The ρ-meson knows the answer to any question it is asked.

However, there is no answer to all questions jointly.

Such knowledge we might call quantum information.

ANY, BUT NOT ALL!

A kind of quantum double think. (Orwell, ‘1984’.) (Einstein hated this.) A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information

A classical system can be copied as follows. This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information

A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information

A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information

A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information

A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information

A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information

A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’).

2. Copying quantum information

A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 2. Copying quantum information

A classical system can be copied as follows.

c :Ω → Ω × Ω: ω 7→ (ω, ω) . This operation lifts in a contravariant way to an action on the algebra of observables A = C(Ω).: C : A ⊗ A → A :( Ch)(ω) = h(c(ω)) = h(ω, ω) . Properties: 1. C ∗(ϕ) = ϕ ⊗ ϕ for ϕ pure (‘C ∗ is a copier’) 2. C(f ⊗ g) = f · g = g · f (‘C is abelian multiplication’); 3. C(f ⊗ 1l) = C(1l ⊗ f ) = f (‘C ∗ is a broadcaster’); 3’.( τ ⊗ id)C ∗ = (id ⊗ τ)C ∗ = id (‘C ∗ is a broadcaster’). For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) : C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f .

No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: No-cloning theorem For any C*-algebra A and any a completely positive unital operation C : A ⊗ A → A the properties 1, 2, 3, and 3’ are equivalent. In particular, an operation with these properties exists if and only if A is commutative. Proof (2) =⇒ (1) =⇒ (3) is easy. (3) =⇒ (2) :Since C is completely positive, we have for all x ∈ A ⊗ A: C(x ∗x) ≥ C(x)∗C(x) . It follows that for all states ϕ on A the quadratic form on A ⊗ A given by ∗ ∗
Qϕ(x, y) := ϕ C(x y) − C(x) C(x) is positive definite. Now, for all f ∈ A: f ∗f = C(1l⊗f ∗f ) = C(1l⊗f )∗(1l⊗f )
≥ C(1l⊗f )∗C(1l⊗f ) = f ∗f . Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A, By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 . Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A: In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf . C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again.

Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. Therefore, for all f ∈ A,

Qϕ(1l ⊗ f , 1l ⊗ f ) = 0 .

By Cuachy-Schwarz for the quadratic form Qϕ it follows that for all f , g ∈ A:

Qϕ(g ⊗ 1l, 1l ⊗ f ) = 0 . In other words: C(g ⊗ 1l)(1l ⊗ f )
= C(g ⊗ 1l)C(1l ⊗ f ) = gf .

But the left hand side can also be written as C(g ⊗ f ) = C(1l ⊗ f )(g ⊗ 1l)
= fg . So A must be commutative. The rest is easy again. The Monty Hall problem

door quizmaster wrong/right position prize

ω M

notepad Q position prize door player P wrong/right µ door quizmaster

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