Computational Power of Observed Quantum Turing Machines Simon Perdrix PPS, Universit´eParis Diderot & LFCS, University of Edinburgh New Worlds of Computation, January 2009 Quantum Computing Basics State space in a classical world of computation: countable A. inaquantumworld: Hilbertspace CA ket map . : A CA | i → s.t. x , x A is an orthonormal basis of CA {| i ∈ } Arbitrary states Φ = αx x | i x A X∈ 2 s.t. x A αx =1 ∈ | | P Quantum Computing Basics bra map . : A L(CA, C) h | → s.t. x, y A, ∀ ∈ 1 if x = y y x = “ Kronecker ” h | | i (0 otherwise v,t A, v t : CA CA: ∀ ∈ | ih | → v if t = x ( v t ) x = v ( t x )= | i “ v t t v ” | ih | | i | i h | | i (0 otherwise | ih |≈ 7→ Evolution of isolated systems: Linear map U L(CA, CA) ∈ U = ux,y y x | ih | x,y A X∈ which is an isometry (U †U = I). Observation Let Φ = αx x | i x A X∈ (Full) measurement in standard basis: 2 The probability to observe a A is αa . ∈ | | If a A is observed, the state becomes Φa = a . ∈ | i Partial measurement in standard basis: Let K = Kλ, λ Λ be a partition of A. { ∈ } 2 The probability to observe λ Λ is pλ = a Kλ αa ∈ ∈ | | If λ Λ is observed, the state becomes P ∈ 1 1 Φλ = PλΦ = αa a √pλ √pλ | i a Kλ X∈ where Pλ = a Kλ a a . ∈ | ih | P Observation Let Φ = αx x | i x A X∈ (Full) measurement in standard basis: 2 The probability to observe a A is αa . ∈ | | If a A is observed, the state becomes Φa = a . ∈ | i Partial measurement in standard basis: Let K = Kλ, λ Λ be a partition of A. { ∈ } 2 The probability to observe λ Λ is pλ = a Kλ αa ∈ ∈ | | If λ Λ is observed, the state becomes P ∈ 1 1 Φλ = PλΦ = αa a √pλ √pλ | i a Kλ X∈ where Pλ = a Kλ a a . ∈ | ih | P Deterministic Turing Machine (DTM) Classical Turing machine (Q, Σ,δ): δ : Q Σ Q Σ 1, 0, 1 × → × × {− } 01100 1 0 1 1 0 00 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 × → × × {− } 01101 1 0 1 1 0 00 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 × → × × {− } 01101 1 0 1 1 0 00 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 × → × × {− } 011011 1 1 1 0 00 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 × × × × {− } → 0110 10010000101 1 0 1 1 10 01 0 1 01 10 01 A quantum configuration is a superposition of classical configurations ∗ Q Σ Z αq,T,x q,T,x C × × | i ∈ q Q,T Σ∗,x Z ∈ X∈ ∈ Evolution operator σ U = δ(p,Tx, q, σ, d) q,T , x + d p,T,x M | x i h | p,q Q,σ Σ,d 1,0,1 ,T Σ∗,x Z ∈X∈ ∈{− } ∈ ∈ A QTM (Q, Σ,δ) has to satisfy some well-formedness conditions... Well-formedness conditions Definition: A QTM M is well-formed iff UM is an isometry, i.e. UM† UM = 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 UM is an isometry, i.e. UM† UM = 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 0100101 1 0011 001 1 0 01 Halting of QTM Environment 0100101 1 0011 001 1 0 01 At the end of the computation, the QTM is ‘observed’. Halting of QTM Environment 00 10 1 1 0 0 At the end of the computation, the QTM is ‘observed’. Halting of QTM Environment 0100101 1 0011 001 1 0 01 At the end of the computation, the QTM is ‘observed’. Halting of QTM Environment 011 0 1 0 0 1 If the halting state is not reached, the computation is useless. Halting of QTM 0 Environment 0100101 1 0011 001 1 0 01 Halting qubit (Ad hoc) Halting of QTM 1 Environment 0100101 1 0011 001 1 0 01 Halting qubit (Ad hoc) Halting of QTM 1 Environment 00 10 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 0100101 1 0011 001 1 0 01 Modelling Environment: Observed QTM Environment is modelled as a partial measurement of the configuration, characterised by a partition K = Kλ λ Λ of Q Σ∗ Z. { } ∈ × × Definition: For a given QTM M = (Q, Σ,δ) and a given partition K = Kλ 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 PλUM† UM Pλ = I λ Λ X∈ where Pλ = (p,T,x) Kλ p,T,x p,T,x . ∈ | i h | P 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 PλUM† UM Pλ = λ Λ PλPλ ∈ ∈ = λ Λ Pλ P P ∈ = λ Λ p,T,x Kλ p,T,x p,T,x P ∈ ∈ | i h | = ∗ Z p,T,x p,T,x Pp,T,xPQ Σ | i h | = I ∈ × × P 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 PλUM† UM Pλ = λ Λ PλPλ ∈ ∈ = λ Λ Pλ P P ∈ = λ Λ p,T,x Kλ p,T,x p,T,x P ∈ ∈ | i h | = ∗ Z p,T,x p,T,x Pp,T,xPQ Σ | i h | = I ∈ × × P Example: halting of QTM For a given QTM M = (Q, Σ,δ) s.t. qh Q is the unique halting state. ∈ Kh = qh Σ∗ Z { }× × K¯ = K Kh h \ [M] Kh,Kh¯ evolves as follows: { } 0110 10010000101 1 0 1 1 10 01 0 1 01 10 01 Example: halting of QTM For a given QTM M = (Q, Σ,δ) s.t. qh Q is the unique halting state. ∈ Kh = qh Σ∗ Z { }× × K¯ = K Kh h \ [M] Kh,Kh¯ evolves as follows: { } 0110 10010000101 1 0 1 1 10 01 0 1 01 10 01 Example: halting of QTM For a given QTM M = (Q, Σ,δ) s.t. qh Q is the unique halting state. ∈ Kh = qh Σ∗ Z { }× × K¯ = K Kh h \ [M] Kh,Kh¯ evolves as follows: { } 0110 10010000101 1 0 1 1 10 01 0 1 01 10 01 Example: halting of QTM For a given QTM M = (Q, Σ,δ) s.t. qh Q is the unique halting state. ∈ Kh = qh Σ∗ Z { }× × K¯ = K Kh h \ [M] Kh,Kh¯ evolves as follows: { } 0110 10010000101 1 0 1 1 10 01 0 1 01 10 01 Example: halting of QTM For a given QTM M = (Q, Σ,δ) s.t. qh Q is the unique halting state. ∈ Kh = qh Σ∗ Z { }× × K¯ = K Kh h \ [M] Kh,Kh¯ evolves as follows: { } 0110 10010000101 1 0 1 1 10 01 0 1 01 10 01 Example: halting of QTM For a given QTM M = (Q, Σ,δ) s.t. qh Q is the unique halting state. ∈ Kh = qh Σ∗ Z { }× × K¯ = K Kh h \ [M] Kh,Kh¯ evolves as follows: { } 0110 10010000101 1 0 1 1 10 01 0 1 01 10 01 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 [Mh]Kh for deciding (with high probability), for any DTM M and any input u, whether M halts on input u. (Proof) Hadamard QTM Let Mh = ( q , q , q , qh, q¯ , Σ,δh) be a well-formed QTM, s.t. qh and { 0 1 2 h} q¯ are the halting states and for σ Σ h ∈ δh(q0, σ, q1, σ, 0) = 1/√2 δh(q0, σ, q2, σ, 0) = 1/√2 δh(q1, σ, qh, σ, 0) = 1/√2 δh(q1, σ, qh¯ , σ, 0) = 1/√2 δh(q2, σ, qh, σ, 0) = 1/√2 δh(q , σ, q¯ , σ, 0) = 1/√2 2 h − w Σ∗, ∀ ∈ U 2 q , w = U ( 1 ( q , w + q , w )) Mh 0 Mh √2 1 2 | i 1 | i | i = ( qh, w + q¯ , w + qh, w q¯ , w ) 2 | i | h i | i − | h i = qh, w | i For any DTM M and any input u, let wM,u Σ∗ be an ‘encoding’ of M and u.