Quantum control: Introduction and some mathematical results. Rui Vilela Mendes UTL and GFM 12/16/2004

Transcription

1 Quantum control: Introduction and some mathematical results Rui Vilela Mendes UTL and GFM 12/16/2004

2 Introduction to quantum control

3 Quantum control : A chemists dream Bond-selective chemistry with high-intensity precise frequency laser pulses Control based on pulse shape or quantum interference

4 Applications of quantum control To produce new molecules or exotic states, molecular devices, programmable optics and nanomachines To control electrons in solid-state quantum wells To regulate the flow of electrons in semiconductors by interference. Fast optical switches To select sharply defined states as products of chemical reactions To use interference to devise lasers for ultrashort pulses of light To use molecules for quantum computation To manipulate cold ions on linear traps Gamma ray lasers to control nuclear reactions

5 The main quantum control techniques Coherent control with two continuous-wave laser beams Pulse timing control (interference of two short pulses) Control by theoretically shaped pulses (frequencychirped laser pulses) Adaptive control. Teaching lasers to control molecules

6 Coherent control Controlled chemical dissociation

7 Coherent control (Lu, Park, Xie, Gordon, J. Chem. Phys. 94 (1992) 6613

8 Pulse timing control Two-pulse laser control

9 Pulse timing control (Scherer et al.; J. Chem. Phys. 93 (1990) 856)

10 Control by theoretically shaped pulses Optimal control theory used to design frequency-chirped laser pulses (Kohler et al.; Phys. Rev. Lett. 74 (1995) 3360)

11 Adaptive control. Teaching lasers to control molecules (A. Assion et al., Science 282 (1988) 919)

12 Controlability results Quantum dynamics r l= 1 d ih ψ = H ψ + u () t Hψ 0 l l dt ψ f ψ 0

13 Controlability results C={adH 0j H k k=1,...,r; j=0,1,...} LA A={H 0,H 1,...,H r } LA B={H 1,..., H r } LA Thm. (Huang, Tarn, Clark) If [C,B] B and dim C(φ) < the system is strongly controllable (C(φ) = tangent space at φ) Systems with a finite no. of states : Thm. System is controllable if A contains U(N)

14 Some mathematical results

15 Motivations 1) Nonlinear techniques in quantum control? Remark: Stable (dissipative) techniques in classical control (Ex.: sliding mode)

16 Motivations 2) Control in infinite dimensions (HTC) - No control on the the -dim. Hilbert sphere if A is a finite-dimensional Lie algebra Remarks: - An incommensurable one-dim. orbit approaches any point in T n as closely as desired. - Baire category theorem X (metric complete) i Ν Μ i Ν (nowhere dense) However = X closure ( i Ν Μ i Ν )?

17 The Strocchi map Ψ>= Ψ k > k k 1 Ψ = q + ip k 2 ( ) k k 1 < Ψ' Ψ>= ( qq ' + pp ' ) + iqp ( ' pq ' ) k k k k k k k k 2 k 1 = { G( Ψ ', Ψ) + iω( Ψ ', Ψ)} 2 k

18 The Strocchi map Kähler structure ' G( Ψ, Ψ) = Ω( Ψ, JΨ) Hamiltonian 1 H = {( q q + p p ) Re H + i( p q q p )Im H 2 k, j ' J = complex structure in real Hilbert space k j k j kj k j k j kj d dt q d dt p k k = p H k = q H k Hamiltonian system equivalent to the Schrödinger equation Role of the symplectic form Ω

19 The Strocchi map Role of the Riemannian metric Given ψ S H and V a the subspace associated to the value a of an observable A. If the measurement of A yields the value a, the state is projected on the state in V a closest to ψ in the G metric

20 Unitary (Hamiltonian) evolution and measurements (jumps) in the Strocchi phase space

21 Decoherence = Splitting of densities in Strocchi s phase space Phase damping channel 0 0 E 1 p 0 0 E + p 0 1 E 1 0 E 1 p 1 0 E + p 1 2 E with p = Γdt using Lindblad s equation d dt ρ = L µ ρl µ Γρ µ>0 L 1 = 10 Γ ; M 00 2 = 00 Γ 01 For an initial condition ρ 0 =(ψ ψ 1 1)(ψ ψ 1 1 ) ψ0 ψ ρ (t) = 0 e Γt ψ 0 ψ 1 e Γt ψ 1 ψ 0 ψ 1 ψ 1 In the SM phase space, defining the densities ρ a = δ (µ 0 q 0 ) δ (µ 1 q 1 ) δ (ν 0 p 0 ) δ (ν 1 p 1 ) ρ b = δ (µ 0 q 0 ) δ (ν 0 p 0 ) δ (µ 1 ) δ (ν 1 ) +δ (µ 0 ) δ (ν 0 ) δ (µ 1 q 1 ) δ (ν 1 p 1 ) thesolutionis ρ (µ, ν) = 1 e Γt ρ b + e Γt ρ a t = 0 t > 0

22 Consequences Control (by measurement) of non-controllable systems Thm 1. (RVM, V. I. Man ko) Given a goal state Ψ, there is a family of observables such that measurement of one of these plus unitary evolution leads to Ψ if A is either O(N) or Sp(N/2) Control by measurement plus unitary evolution

23 Consequences Formulation of quantum optimal control analog to Pontryagin s maximum principle Sliding mode techniques in quantum control

24 Quantum control in infinite dimensions Control in 2 (Z) 2 (Z) is isomorphic to any - dimensional separable Hilbert space a={...,a -2,a -1,a 0,a 1,...} ; Basis e k ={...,0,0,1 k,0,...} a=σa k e k Control operators : U + e k = e k+1 U + -1 e k = e k+1 (Magnetic pulse) Π e 0 = e 1, Π e k = e k k 0,1 Π e 1 = e 0 Π n =U +n Π U + -n U(2) in {e 0, e 1 } identity in 2 (Z) \ {e 0, e 1 } (Rabi rotation)

25 Quantum control in infinite dimensions Results : (W. Karwowski, RVM) Thm 2. For any a 2 (Z) the linear span of G(U +, Π) a is dense in 2 (Z) Thm 3. For any a 2 (Z) the set G(U +, U(2)) a is dense in the 2 (Z) Hilbert sphere

26 References Huang, Tarn and Clark ; J. Math. Phys. 24 (1983) 2608 Rabitz, Vivie-Riedle, Motzkus, Kompa; Science 288 (2000) 824 F. Strocchi; Rev. Mod. Phys. 38 (1966) 36 RVM, V. I. Man ko; Phys. Rev. A67 (2003) W. Karwowski, RVM; Phys. Lett A332 (2004) 282

27 The End (for the general public)

28 Theorem 1: Given any goal state ψ f,thereisa family of observables M ψ f such that measurement of one of these observables on any ψ 0 plus unitary evolution leads to ψ f if G(A) is either O (N) or Sp 1 2 N. Proof: If G(A) =O (N) or Sp 1 2 N we may choose an orthonormal basis {φ i } for S N 1 in the orbit G(A)ψ f. Construct an observable M = i a ip φi, P φi being the projector on φ i. Measuring this observable on any state ψ 0 and recording the measured value a k the state becomes φ k Then, by unitary evolution, ψ f maybereached. Remarks: (i) There is a large family of observables appropriate for this type of control. (ii) If both ψ 0 and ψ f arefixedamuchsimpler set of controlling interactions H j may be sufficient.

29 Lemma 1. Given a 2 (Z),k Z,l Z, the linear operator Π k,k+l (l N) defined by Π k,k+1 = Π k and Π k,k+l a = Π k Π k+1 Π k+l 2 Π k+l 1 Π k+1 Π k a for l 2, exchanges the coefficients of e k and e k+l in a, thatisπ k,k+l a = a k+l e k + a k e k+l + r=k,k+l a r e r Theorem 2. LetG (U +, Π) stand for the group generated by U +,U+ 1 and Π. Then for any 0 = a 2 (Z) the linear span of G (U +, Π) a is dense in 2 (Z). Proof :Itissufficienttoshowthatb G (U +, Π) a implies b =0. Suppose b = e k for some k Z. Sincea = 0there is l N\{0} such that at least one of the numbers a k+l or a k l is different from zero. Then (b, Π k,k+l a)=a k+l or (b, Π k l,k a)=a k l, a contradiction.

30 Similarly if both a and b are terminating sequences. Suppose now that b is terminating but a is not. Let b k = 0for k > N. Then N N such that (b, a) = N N b k a k =0and either b N a N = 0or b N a N = 0. Then there is l such that a N+l = a N or a N l = a N. Hence (b, Π N,N+l a)= N 1 N b k a k + b N a N+l = 0or (b, Π N, N l a)= N 1 N b k a k +b N a N l = 0, a contradiction. Similarly for a terminating and b nonterminating. If neither a nor b terminates, then there are pairs a k = a l and b m = b n. With appropriate g, g G (U +, Π) we obtain (b, ga) =b ma k + b na l + b ka m + b l a n + b ra r =0 r=k,l,m,n b, g a = b na k + b ma l + b ka m + b l a n + r=k,l,m,n b ra r =0 Hence b ma k +b na l = b na k +b ma l, which is possible only if either b m = b n or a k = a l, a contradiction.

31 Theorem 3. For any a 2 (Z), a =1,theset G (U +,U(2)) a is dense in the 2 (Z) Hilbert sphere. Lemma 2. Suppose a is a terminating normalized sequence. Then, there is g G (U +,U(2)) such that ge 0 = a. Proof :Let a = a N e N + + a o e a N e N By U (2) transformations in the {e 0,e 1 } subspace and use of the Π k,k+l operators one constructs with operators g i G (U +,U(2)) the following sequence g 1 e 0 = x 1 e 0 + a N e N = α 1 g 2 α 1 = x 2 e 0 + a N+1 e N+1 + a N e N = α 2 g 2N α 2N 1 = x 2N e 0 + N N a ke k = α 2N g 2N+1 α 2N = a Finally g 2N+1 g 2N g 2 g 1 e 0 = a

32 Proof of theorem 3 : Consider a, b 2 (Z) with a = b =1.Chooseε and N such that N α = a k e k > 1 ε N N β = b k e k > 1 ε N By the lemma 2 there are g,g G (U +,U(2)) such that N g a k e k = αe 0 N g (αe 0 )=(α/β) Hence b g ga 2ε + N b k e k N 1 α β 3ε

33 The end for everybody