Computational Power of Observed Quantum Turing Machines Simon - PowerPoint PPT Presentation

Computational Power of Observed Quantum Turing Machines Simon Perdrix PPS, Universit´ e Paris Diderot & LFCS, University of Edinburgh New Worlds of Computation, January 2009

Quantum Computing Basics State space in a classical world of computation: countable A . Hilbert space C A in a quantum world: ket map | . � : A → C A s.t. {| x � , x ∈ A } is an orthonormal basis of C A Arbitrary states � Φ = α x | x � x ∈ A x ∈ A | α x | 2 = 1 s.t. �

Quantum Computing Basics bra map L ( C A , C ) � . | : A → s.t. ∀ x, y ∈ A , � 1 if x = y � y | | x � = “ Kronecker ” 0 otherwise ∀ v, t ∈ A, | v �� t | : C A → C A : � | v � if t = x ( | v �� t | ) | x � = | v � ( � t | | x � ) = “ | v �� t | ≈ t �→ v ” otherwise 0 Evolution of isolated systems: Linear map U ∈ L ( C A , C A ) � U = u x,y | y �� x | x,y ∈ A which is an isometry ( U † U = I ).

Observation � Let Φ = α x | x � x ∈ A (Full) measurement in standard basis: The probability to observe a ∈ A is | α a | 2 . If a ∈ A is observed, the state becomes Φ a = | a � . Partial measurement in standard basis: Let K = { K λ , λ ∈ Λ } be a partition of A . a ∈ K λ | α a | 2 The probability to observe λ ∈ Λ is p λ = � If λ ∈ Λ is observed, the state becomes 1 1 � Φ λ = P λ Φ = α a | a � √ p λ √ p λ a ∈ K λ where P λ = � a ∈ K λ | a �� a | .

Observation � Let Φ = α x | x � x ∈ A (Full) measurement in standard basis: The probability to observe a ∈ A is | α a | 2 . If a ∈ A is observed, the state becomes Φ a = | a � . Partial measurement in standard basis: Let K = { K λ , λ ∈ Λ } be a partition of A . a ∈ K λ | α a | 2 The probability to observe λ ∈ Λ is p λ = � If λ ∈ Λ is observed, the state becomes 1 1 � Φ λ = P λ Φ = α a | a � √ p λ √ p λ a ∈ K λ where P λ = � a ∈ K λ | a �� a | .

Deterministic Turing Machine (DTM) Classical Turing machine ( Q, Σ , δ ) : δ : Q × Σ → Q × Σ × {− 1 , 0 , 1 } 0 1 1 0 0 1 0 1 1 0 0 0 1 0 ( q, T, x ) ∈ Q × Σ ∗ × Z is a classical configuration.

Deterministic Turing Machine (DTM) Classical Turing machine ( Q, Σ , δ ) : δ : Q × Σ → Q × Σ × {− 1 , 0 , 1 } 0 1 1 0 1 1 0 1 1 0 0 0 1 0 ( q, T, x ) ∈ Q × Σ ∗ × Z is a classical configuration.

Deterministic Turing Machine (DTM) Classical Turing machine ( Q, Σ , δ ) : δ : Q × Σ → Q × Σ × {− 1 , 0 , 1 } 0 1 1 0 1 1 0 1 1 0 0 0 1 0 ( q, T, x ) ∈ Q × Σ ∗ × Z is a classical configuration.

Deterministic Turing Machine (DTM) Classical Turing machine ( Q, Σ , δ ) : δ : Q × Σ → Q × Σ × {− 1 , 0 , 1 } 0 1 1 0 1 1 1 1 1 0 0 0 1 0 ( q, T, x ) ∈ Q × Σ ∗ × Z is a classical configuration.

Quantum Turing Machine (QTM) Quantum Turing machine M = ( Q, Σ , δ ) : δ : Q × Σ × Q × Σ × {− 1 , 0 , 1 } → C 0 1 1 0 0 1 0 0 1 1 0 1 0 1 1 0 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 A quantum configuration is a superposition of classical configurations � α q,T,x | q, T, x � ∈ C Q × Σ ∗ × Z q ∈ Q,T ∈ Σ ∗ ,x ∈ Z

Evolution operator � δ ( p, T x , q, σ, d ) | q, T σ U M = x , x + d � � p, T, x | p,q ∈ Q,σ ∈ Σ ,d ∈{− 1 , 0 , 1 } ,T ∈ Σ ∗ ,x ∈ Z A QTM ( Q, Σ , δ ) has to satisfy some well-formedness conditions...

Well-formedness conditions Definition: A QTM M is well-formed iff U M is an isometry, i.e. U † M U M = I • The evolution of the machine does not violate the postulates of quantum mechanics. • During the computation, the machine is isolated from the rest of the universe.

Well-formedness conditions Definition: A QTM M is well-formed iff U M is an isometry, i.e. U † M U M = I • The evolution of the machine does not violate the postulates of quantum mechanics. • During the computation, the machine is isolated from the rest of the universe. Environment 1 0 0 1 0 1 0 1 0 1 1 0 1 0 0 1 0 1 1 0

Halting of QTM Environment 0 0 1 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 At the end of the computation, the QTM is ‘observed’.

Halting of QTM Environment 0 0 1 1 0 1 1 0 0 At the end of the computation, the QTM is ‘observed’.

Halting of QTM Environment 0 0 1 1 0 1 0 1 1 0 1 0 1 0 1 0 0 1 0 1 At the end of the computation, the QTM is ‘observed’.

Halting of QTM Environment 0 1 1 0 1 0 0 1 If the halting state is not reached, the computation is useless.

Halting of QTM 0 Environment 1 0 0 0 1 0 1 1 0 1 1 0 1 0 0 1 0 1 0 1 Halting qubit (Ad hoc)

Halting of QTM 1 Environment 1 0 0 0 1 0 1 1 0 1 1 0 1 0 0 1 0 1 0 1 Halting qubit (Ad hoc)

Halting of QTM 1 Environment 0 0 1 1 0 1 1 0 0 Halting qubit (Ad hoc)

‘Un-isolated’ QTM Isolation assumption is probably too strong • technical issues like the halting of QTM, • models of QC (one-way model, measurement-only model) based on measurements. • PTM and DTM are not well-formed QTM (reversible DTM does) • quest of a universal QTM: a classical control is required.

‘Un-isolated’ QTM Isolation assumption is probably too strong • technical issues like the halting of QTM, • models of QC (one-way model, measurement-only model) based on measurements. • PTM and DTM are not well-formed QTM (reversible DTM does) • quest of a universal QTM: a classical control is required. Environment 1 0 0 1 0 1 0 1 0 1 1 0 1 0 0 1 0 1 1 0

Modelling Environment: Observed QTM Environment is modelled as a partial measurement of the configuration, characterised by a partition K = { K λ } λ ∈ Λ of Q × Σ ∗ × Z . Definition: For a given QTM M = ( Q, Σ , δ ) and a given partition K = { K λ } λ ∈ Λ of Q × Σ ∗ × Z , [ M ] K is an Observed Quantum Turing Machine (OQTM).

Evolution of OQTM One transition of [ M ] K is composed of: 1. partial measurement K of the quantum configuration; 2. transition of M ; 3. partial measurement K of the quantum configuration. Definition: An OQTM [ M ] K is well-observed iff � P λ U † M U M P λ = I λ ∈ Λ where P λ = � ( p,T,x ) ∈ K λ | p, T, x � � p, T, x | .

a weaker condition Lemma: If a QTM M is well-formed then [ M ] K is a well-observed OQTM for any K . Proof: λ ∈ Λ P λ U † � M U M P λ = � λ ∈ Λ P λ P λ = � λ ∈ Λ P λ = � � p,T,x ∈ K λ | p, T, x � � p, T, x | λ ∈ Λ = � p,T,x ∈ Q × Σ ∗ × Z | p, T, x � � p, T, x | = I

a weaker condition Lemma: If a QTM M is well-formed then [ M ] K is a well-observed OQTM for any K . Proof: λ ∈ Λ P λ U † � M U M P λ = � λ ∈ Λ P λ P λ = � λ ∈ Λ P λ = � � p,T,x ∈ K λ | p, T, x � � p, T, x | λ ∈ Λ = � p,T,x ∈ Q × Σ ∗ × Z | p, T, x � � p, T, x | = I

Example: halting of QTM For a given QTM M = ( Q, Σ , δ ) s.t. q h ∈ Q is the unique halting state. { q h } × Σ ∗ × Z K h = K ¯ = K \ K h h [ M ] { K h ,K ¯ h } evolves as follows: 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 0 1 1 0

Example: halting of QTM For a given QTM M = ( Q, Σ , δ ) s.t. q h ∈ Q is the unique halting state. { q h } × Σ ∗ × Z K h = K ¯ = K \ K h h [ M ] { K h ,K ¯ h } evolves as follows: 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 0 1 1 0

Example: halting of QTM For a given QTM M = ( Q, Σ , δ ) s.t. q h ∈ Q is the unique halting state. { q h } × Σ ∗ × Z K h = K ¯ = K \ K h h [ M ] { K h ,K ¯ h } evolves as follows: 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 0 1 1 0

Example: halting of QTM For a given QTM M = ( Q, Σ , δ ) s.t. q h ∈ Q is the unique halting state. { q h } × Σ ∗ × Z K h = K ¯ = K \ K h h [ M ] { K h ,K ¯ h } evolves as follows: 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 0 1 1 0

Example: halting of QTM For a given QTM M = ( Q, Σ , δ ) s.t. q h ∈ Q is the unique halting state. { q h } × Σ ∗ × Z K h = K ¯ = K \ K h h [ M ] { K h ,K ¯ h } evolves as follows: 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 0 1 1 0

Example: halting of QTM For a given QTM M = ( Q, Σ , δ ) s.t. q h ∈ Q is the unique halting state. { q h } × Σ ∗ × Z K h = K ¯ = K \ K h h [ M ] { K h ,K ¯ h } evolves as follows: 0 1 1 0 1 0 1 0 0 1 0 1 0 1 0 1 1 0 1 0 1 0 0 1 1 0 1 0 0 1 1 0

OQTM more expressive than QTM Lemma: For any DTM M = ( Q, Σ , δ ) , [ M ] {{ c } ,c ∈ Q × Σ ∗ × Z } is a well-observed OQTM.

OQTM, a too powerful model ?!? Theorem: There is a well-observed OQTM [ M h ] K h for deciding (with high probability), for any DTM M and any input u , whether M halts on input u .