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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

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Computational Power of Observed Quantum Turing Machines

Simon Perdrix

PPS, Universit´ e Paris Diderot & LFCS, University of Edinburgh

New Worlds of Computation, January 2009

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Quantum Computing Basics

State space in a classical world of computation: countable A. in a quantum world: Hilbert space CA ket map |. : A → CA s.t. {|x , x ∈ A} is an orthonormal basis of CA Arbitrary states Φ =

  • x∈A

αx |x s.t.

x∈A |αx|2 = 1

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Quantum Computing Basics

bra map .| : A → L(CA, C) s.t. ∀x, y ∈ A, y| |x =

  • 1

if x = y

  • therwise

“ Kronecker ” ∀v, t ∈ A, |vt| : CA → CA: (|vt|) |x = |v (t| |x) =

  • |v

if t = x

  • therwise

“ |vt| ≈ t → v ” Evolution of isolated systems: Linear map U ∈ L(CA, CA) U =

  • x,y∈A

ux,y |yx| which is an isometry (U †U = I).

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Observation

Let Φ =

  • x∈A

αx |x (Full) measurement in standard basis: The probability to observe a ∈ A is |αa|2. If a ∈ A is observed, the state becomes Φa = |a. Partial measurement in standard basis: Let K = {Kλ, λ ∈ Λ} be a partition of A. The probability to observe λ ∈ Λ is pλ =

a∈Kλ |αa|2

If λ ∈ Λ is observed, the state becomes Φλ = 1 √pλ PλΦ = 1 √pλ

  • a∈Kλ

αa |a where Pλ =

a∈Kλ |aa|.

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Observation

Let Φ =

  • x∈A

αx |x (Full) measurement in standard basis: The probability to observe a ∈ A is |αa|2. If a ∈ A is observed, the state becomes Φa = |a. Partial measurement in standard basis: Let K = {Kλ, λ ∈ Λ} be a partition of A. The probability to observe λ ∈ Λ is pλ =

a∈Kλ |αa|2

If λ ∈ Λ is observed, the state becomes Φλ = 1 √pλ PλΦ = 1 √pλ

  • a∈Kλ

αa |a where Pλ =

a∈Kλ |aa|.

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Deterministic Turing Machine (DTM)

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

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Deterministic Turing Machine (DTM)

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

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Deterministic Turing Machine (DTM)

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

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Deterministic Turing Machine (DTM)

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

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Quantum Turing Machine (QTM)

Quantum Turing machine M = (Q, Σ, δ): δ : Q × Σ × Q × Σ × {−1, 0, 1} → C 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 A quantum configuration is a superposition of classical configurations

  • q∈Q,T ∈Σ∗,x∈Z

αq,T,x |q, T, x ∈ CQ×Σ∗×Z

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Evolution operator

UM =

  • p,q∈Q,σ∈Σ,d∈{−1,0,1},T ∈Σ∗,x∈Z

δ(p, Tx, q, σ, d) |q, T σ

x , x + d p, T, x|

A QTM (Q, Σ, δ) has to satisfy some well-formedness conditions...

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Well-formedness conditions

Definition: A QTM M is well-formed iff UM is an isometry, i.e. U †

MUM = 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.

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Well-formedness conditions

Definition: A QTM M is well-formed iff UM is an isometry, i.e. U †

MUM = 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. 1 1 1 1 1 1 1 1 1 1 Environment

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Halting of QTM

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

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Halting of QTM

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

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Halting of QTM

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

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Halting of QTM

1 1 1 1 Environment If the halting state is not reached, the computation is useless.

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Halting of QTM

1 1 1 1 1 1 1 1 1 1 Environment Halting qubit (Ad hoc)

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Halting of QTM

1 1 1 1 1 1 1 1 1 1 Environment 1 Halting qubit (Ad hoc)

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Halting of QTM

1 1 1 1 Environment 1 Halting qubit (Ad hoc)

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‘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.
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‘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.

1 1 1 1 1 1 1 1 1 1 Environment

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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).

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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

  • λ∈Λ

PλU †

MUMPλ = I

where Pλ =

(p,T,x)∈Kλ |p, T, x p, T, x|.

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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 PλU †

MUMPλ

=

  • λ∈Λ PλPλ

=

  • λ∈Λ Pλ

=

  • λ∈Λ
  • p,T,x∈Kλ |p, T, x p, T, x|

=

  • p,T,x∈Q×Σ∗×Z |p, T, x p, T, x|

= I

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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 PλU †

MUMPλ

=

  • λ∈Λ PλPλ

=

  • λ∈Λ Pλ

=

  • λ∈Λ
  • p,T,x∈Kλ |p, T, x p, T, x|

=

  • p,T,x∈Q×Σ∗×Z |p, T, x p, T, x|

= I

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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¯

h

= K \ Kh [M]{Kh,K¯

h} evolves as follows:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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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¯

h

= K \ Kh [M]{Kh,K¯

h} evolves as follows:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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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¯

h

= K \ Kh [M]{Kh,K¯

h} evolves as follows:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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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¯

h

= K \ Kh [M]{Kh,K¯

h} evolves as follows:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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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¯

h

= K \ Kh [M]{Kh,K¯

h} evolves as follows:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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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¯

h

= K \ Kh [M]{Kh,K¯

h} evolves as follows:

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

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OQTM more expressive than QTM

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

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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.

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(Proof) Hadamard QTM

Let Mh = ({q0, q1, q2, qh, q¯

h}, Σ, δh) be a well-formed QTM, s.t. qh and

q¯

h are the halting states and for σ ∈ Σ

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, q¯

h, σ, 0)

= 1/ √ 2 δh(q2, σ, qh, σ, 0) = 1/ √ 2 δh(q2, σ, q¯

h, σ, 0)

= −1/ √ 2 ∀w ∈ Σ∗, U 2

Mh |q0, w

= UMh( 1

√ 2(|q1, w + |q2, w))

=

1 2(|qh, w + |q¯ h, w + |qh, w − |q¯ h, w)

= |qh, w

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For any DTM M and any input u, let wM,u ∈ Σ∗ be an ‘encoding’ of M and u. K0 = {(q1, wM,u) s.t. M(u) does not halt} ∪ {(q¯

h, w)}

K1 = {(q, w) s.t. (q, w) / ∈ K1} What is the evolution of [Mh]{K0,K1} if the initial configuration is (q0, wM,u)? Evolution of Mh: |q0, wM,u →

1 √ 2(|q1, wM,u + |q2, wM,u) → |qh, wM,u

  • If M(u) halts, then (q1, wM,u), (q2, wM,u) ∈ K1, thus the evolution
  • f [Mh]{K0,K1} is

|q0, wM,u →∗ |qh, wM,u

  • If M(u) does not halt, then (q1, wM,u) ∈ K0, and (q2, wM,u) ∈ K1

moreover (q¯

h, wM,u) ∈ K0 and (qh, wM,u) ∈ K1, thus:

|q0, wM,u →∗

  • |qh, wM,u

with probability 1/2 |q¯

h, wM,u

with probability 1/2

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For any DTM M and any input u, let wM,u ∈ Σ∗ be an ‘encoding’ of M and u. K0 = {(q1, wM,u) s.t. M(u) does not halt} ∪ {(q¯

h, w)}

K1 = {(q, w) s.t. (q, w) / ∈ K1} What is the evolution of [Mh]{K0,K1} if the initial configuration is (q0, wM,u)? Evolution of Mh: |q0, wM,u →

1 √ 2(|q1, wM,u + |q2, wM,u) → |qh, wM,u

  • If M(u) halts, then (q1, wM,u), (q2, wM,u) ∈ K1, thus the evolution
  • f [Mh]{K0,K1} is

|q0, wM,u →∗ |qh, wM,u

  • If M(u) does not halt, then (q1, wM,u) ∈ K0, and (q2, wM,u) ∈ K1

moreover (q¯

h, wM,u) ∈ K0 and (qh, wM,u) ∈ K1, thus:

|q0, wM,u →∗

  • |qh, wM,u

with probability 1/2 |q¯

h, wM,u

with probability 1/2

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For any DTM M and any input u, let wM,u ∈ Σ∗ be an ‘encoding’ of M and u. K0 = {(q1, wM,u) s.t. M(u) does not halt} ∪ {(q¯

h, w)}

K1 = {(q, w) s.t. (q, w) / ∈ K1} What is the evolution of [Mh]{K0,K1} if the initial configuration is (q0, wM,u)? Evolution of Mh: |q0, wM,u →

1 √ 2(|q1, wM,u + |q2, wM,u) → |qh, wM,u

  • If M(u) halts, then (q1, wM,u), (q2, wM,u) ∈ K1, thus the evolution
  • f [Mh]{K0,K1} is

|q0, wM,u →∗ |qh, wM,u

  • If M(u) does not halt, then (q1, wM,u) ∈ K0, and (q2, wM,u) ∈ K1

moreover (q¯

h, wM,u) ∈ K0 and (qh, wM,u) ∈ K1, thus:

|q0, wM,u →∗

  • |qh, wM,u

with probability 1/2 |q¯

h, wM,u

with probability 1/2

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For any DTM M and any input u, let wM,u ∈ Σ∗ be an ‘encoding’ of M and u. K0 = {(q1, wM,u) s.t. M(u) does not halt} ∪ {(q¯

h, w)}

K1 = {(q, w) s.t. (q, w) / ∈ K1} What is the evolution of [Mh]{K0,K1} if the initial configuration is (q0, wM,u)? Evolution of Mh: |q0, wM,u →

1 √ 2(|q1, wM,u + |q2, wM,u) → |qh, wM,u

  • If M(u) halts, then (q1, wM,u), (q2, wM,u) ∈ K1, thus the evolution
  • f [Mh]{K0,K1} is

|q0, wM,u →∗ |qh, wM,u

  • If M(u) does not halt, then (q1, wM,u) ∈ K0, and (q2, wM,u) ∈ K1

moreover (q¯

h, wM,u) ∈ K0 and (qh, wM,u) ∈ K1, thus:

|q0, wM,u →∗

  • |qh, wM,u

with probability 1/2 |q¯

h, wM,u

with probability 1/2

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Towards a new definition of OQTM

  • Initial proposition: K is a partition of Q × Σ∗ × Z.
  • Focus on the (classical) control: K is a partition of Q × Z.

Theorem: There is a QTM M ′

h and a partition K of Q × Z, s.t.

[Mh]K is well observed and decides (with high probability), for any DTM M and any input u, whether M halts on input u.

  • Finite partition: K is a partition of Q × Σ. (Q: internal states; Σ:

symbol pointed out by the head.)

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Towards a new definition of OQTM

  • Initial proposition: K is a partition of Q × Σ∗ × Z.
  • Focus on the (classical) control: K is a partition of Q × Z.

Theorem: There is a QTM M ′

h and a partition K of Q × Z, s.t.

[Mh]K is well observed and decides (with high probability), for any DTM M and any input u, whether M halts on input u.

  • Finite partition: K is a partition of Q × Σ. (Q: internal states; Σ:

symbol pointed out by the head.)

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Towards a new definition of OQTM

  • Initial proposition: K is a partition of Q × Σ∗ × Z.
  • Focus on the (classical) control: K is a partition of Q × Z.

Theorem: There is a QTM M ′

h and a partition K of Q × Z, s.t.

[Mh]K is well observed and decides (with high probability), for any DTM M and any input u, whether M halts on input u.

  • Finite partition: K is a partition of Q × Σ. (Q: internal states; Σ:

symbol pointed out by the head.)

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Simulation

Theorem: For any well-observed OQTM [M]K there exists a well-formed QTM M ′ which simulates [M]K within a quadratic slowdown.

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Step one

If M = (Q, Σ, δ) and K = {Kλ}λ∈Λ, let ˜ M = (Q, Σ, Λ, ˜ δ) be a 2-tape QTM s.t. ˜ δ(p, τ, , q, σ, λ, d, +1) =

  • δ(p, τ, q, σ, d)

if (p, τ) ∈ Kλ

  • therwise

1 1 1 1 1 1 1 1 1 1 1 1 µ λ µ λ µ λ Lemma: ˜ M is well formed.

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Step two

Lemma: [ ˜ M] ˜

K simulates [M]K, where ˜

K = {Q × Σ × {λ}}λ∈Λ 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 µ λ µ λ µ λ

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Step three

Since they act on distinct systems (the second head always moves to the right), the measurements can be postponed to the end of the computation: Lemma: ˜ M simulates [ ˜ M] ˜

K.

Lemma: There exists a well-formed 1-tape QTM M ′ which simulates ˜ M within a quadratic slowdown.

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Conclusion

  • OQTM: extension of QTM with measurements;
  • a more expressive (but not overpowerfull) machine: QTM, DTM,

halting QTM. Perspectives:

  • Universal quantum Turing machine;
  • what is the minimal k for which any OQTM [M]K can be efficiently

simulated by an OQTM [M ′]K′ where all regions of K′ have a size less than k ?