Adiabatic evolution and dephasing

Transcription

1 Adiabatic evolution and dephasing Gian Michele Graf ETH Zurich January 26, 2011 Fast and slow, adiabatic and geometric effects in nonequilibrium dynamics Institut Henri Poincaré

2 Outline The Landau-Zener model An adiabatic theorem Optimal parametrization Collaborators: Y. Avron, M. Fraas, P. Grech, O.Kenneth

3 Outline The Landau-Zener model An adiabatic theorem Optimal parametrization

4 A motivating example: Landau-Zener tunnelling The Hamiltonian case H(s) = 1 2 x σ on C2 ( x(s) = (s, 0, )) = x 2 P + x 2 P σ(h(s)) e + (s) s eigenvalues e ± (s) = ± x(s) /2 eigenprojections P ± (s) (1±σ x )/2, (s ± ) e (s) H(s) is general form of single-parameter avoided crossing

5 Landau-Zener: Hamiltonian case (cont.) scaled time s = εt: or i dψ dt iε ψ = H(s)ψ initial state: spin down = H(εt)ψ ( = d/ds) (ψ(s), P + (s)ψ(s)) 0 (s ) tunnelling probability (ψ(s), P + (s)ψ(s)) T (s + ) Landau, Zener (1932) T = e π 2 /2ε (exponentially small in ε 0).

6 Adiabatic tunnelling: Hamiltonian case More generally, let H(s) smooth H(s) (or P ± (s)) constant near s = ±, e.g. at s = s 0, s 1. s 0 σ(h(s)) s 1 e +(s) e (s) s

7 Adiabatic tunnelling: Hamiltonian case More generally, let H(s) smooth H(s) (or P ± (s)) constant near s = ±, e.g. at s = s 0, s 1. s 0 σ(h(s)) s 1 e +(s) e (s) s Then: T(s, s 0 ) = O(ε 2 ) (s generic) At intermediate times s, down state contains a coherent admixture O(ε) of the up state. T(s 1, s 0 ) = O(ε n ) (n = 1, 2,...)

8 Adiabatic tunnelling: Hamiltonian case More generally, let H(s) smooth H(s) (or P ± (s)) constant near s = ±, e.g. at s = s 0, s 1. s 0 σ(h(s)) s 1 e +(s) e (s) s Then: T(s, s 0 ) = O(ε 2 ) (s generic) At intermediate times s, down state contains a coherent admixture O(ε) of the up state. T(s 1, s 0 ) = O(ε n ) (n = 1, 2,...) Essentially no memory is retained at the end: tunnelling is reversible.

9 Lindblad evolution System coupled to Bath: Evolution of a mixed state ρ = ρ S ( ρ φ t (ρ) = tr B Ut (ρ ρ B )Ut ) with joint unitary evolution U t (U t+s = U t U s ) Properties: trφ t (ρ) = trρ φ t completely positive φ t+s = φ t φ s approximately, if time scales of Bath time scales of System exactly, if bath is white noise Generator: L := dφ t dt t=0 Theorem (Lindblad, Sudarshan-Kossakowski-Gorini 1976) The general form of the generator is L(ρ) = i[h, ρ] + 1 (2Γ α ργ α Γ 2 αγ α ρ ργ αγ α ) α

10 Dephasing Lindbladians with L(ρ) = i[h, ρ] + 1 (2Γ α ργ α Γ 2 αγ α ρ ργ αγ α ) α [Γ α, P i ] = 0 for H = e i P i i Then L(P i ) = 0, resp. φ t (P i ) = P i : Like in the Hamiltonian case, eigenstates P i are invariant. Example: 2-level system L(ρ) = i[h, ρ] γ(p ρp + + P + ρp ) (γ 0) Evolution turns coherent into incoherent superpositions within a time γ 1. Is a model for measurement of H. Application: Nuclear magnetic resonance

11 Dephasing 2-level Lindbladian with L(s)(ρ) = i[h(s), ρ] γ(s)(p (s)ρp + (s) + P + (s)ρp (s)) H(s) = x(s) σ/2 γ(s) 0 x(s) with x(s) x(± ), (s ± ). Lindblad equation for ρ = ρ(s) Result ε ρ = L(s)(ρ) γ(s) T = ε x(s) 2 + γ(s) 2 tr( Ṗ (s) 2) ds + O(ε 2 ) Tunnelling has memory and is irreversible.

12 Dephasing Landau-Zener Lindbladian x(s) = (s, 0, ): For γ(s) constant: tr ( Ṗ (s) 2) = 2 2 x 4 T = πε 4 2 Q(γ/ ) + O(ε2 ) Q(x) = π 2 x( x 2 ) 1 + x 2 ( 1 + x 2 + 1) Q(x) Figure: The function Q(x). It has a maximum at x =

13 Dephasing Landau-Zener Lindbladian x(s) = (s, 0, ): For γ(s) constant: tr ( Ṗ (s) 2) = 2 2 x 4 T = πε 4 2 Q(γ/ ) + O(ε2 ) Q(x) = π 2 x( x 2 ) 1 + x 2 ( 1 + x 2 + 1) Q(x) 0.6 linear at small γ Figure: The function Q(x). It has a maximum at x =

14 Dephasing Landau-Zener Lindbladian x(s) = (s, 0, ): For γ(s) constant: tr ( Ṗ (s) 2) = 2 2 x 4 T = πε 4 2 Q(γ/ ) + O(ε2 ) Q(x) = π 2 x( x 2 ) 1 + x 2 ( 1 + x 2 + 1) Q(x) 0.6 linear at small γ Zeno effect at large γ Figure: The function Q(x). It has a maximum at x =

15 Outline The Landau-Zener model An adiabatic theorem Optimal parametrization

16 A question Recall: Hamiltonian case reversible tunnelling, oblivion Deph. Lindbladian case irreversible tunnelling, memory

17 A question Recall: Hamiltonian case reversible tunnelling, oblivion Deph. Lindbladian case irreversible tunnelling, memory Question: Is there a common point of view making this evident?

18 The scheme Setup: V linear space, finite-dimensional. L(s) : V V, x L(s)x linear in x V, smooth in 0 s 1. Assumptions: 0 is an eigenvalue of L(s), isolated uniformly in s. V = ker L ran L. In particular: L is invertible on ran L: L 1 1 = P + Q (projections), x = a + b (decomposition) Evolution equation for x = x(s): εẋ = L(s)x Parallel transport T(s, s ) : V V with s T(s, s ) = [Ṗ(s), P(s)]T(s, s ), T(s, s ) = 1 implying P(s)T(s, s ) = T(s, s )P(s )

19 The theorem i) εẋ = L(s)x admits solutions of the form with x(s) = N ε n (a n (s) + b n (s)) + ε N+1 r N (ε, s) n=0 a n (s) ker L(s), b n (s) ran L(s) a n (0) ker L(0), r N (ε, 0) V arbitrary ii) Coefficients (n = 0, 1,...): b 0 (s) = 0 a n (s) = T(s, 0)a n (0) + s 0 T(s, s )Ṗ(s )b n (s )ds b n+1 (s) = L(s) 1 (Ṗ(s)a n(s) + Q(s)ḃn(s)) iii) If L(s) generates a contraction semigroup, then r N (ε, s) is uniformly bounded in ε and in s, if so at s = 0

20 A corollary Recall: b 0 (s) = 0 a n (s) = T(s, 0)a n (0) + s 0 T(s, s )Ṗ(s )b n (s )ds b n+1 (s) = L(s) 1 (Ṗ(s)a n(s) + Q(s)ḃn(s)) (Note: b 0 a 0 b 1 a 1...) Corollary If P(s) is constant near s = s 0, then b n (s 0 ) = 0, (n = 0, 1, 2,...)

21 A corollary Recall: b 0 (s) = 0 a n (s) = T(s, 0)a n (0) + s 0 T(s, s )Ṗ(s )b n (s )ds b n+1 (s) = L(s) 1 (Ṗ(s)a n(s) + Q(s)ḃn(s)) (Note: b 0 a 0 b 1 a 1...) Corollary If P(s) is constant near s = s 0, then b n (s 0 ) = 0, (n = 0, 1, 2,...) Answer: the a n s carry the memory, the b n s don t.

22 A corollary Recall: b 0 (s) = 0 a n (s) = T(s, 0)a n (0) + s 0 T(s, s )Ṗ(s )b n (s )ds b n+1 (s) = L(s) 1 (Ṗ(s)a n(s) + Q(s)ḃn(s)) (Note: b 0 a 0 b 1 a 1...) Corollary If P(s) is constant near s = s 0, then b n (s 0 ) = 0, (n = 0, 1, 2,...) Answer: the a n s carry the memory, the b n s don t. Next: One result, different applications.

23 Appl. to Quantum Mechanics: Hamiltonian case V = H, x = ψ iε ψ = H(s)ψ e(s) : isolated, simple eigenvalue of H(s) Set ψ(s) = ψ(s) exp(iε 1 s e(s )ds ) and rewrite ε ψ = i(h(s) e(s)) ψ L(s) ψ with 0 isolated, simple eigenvalue of L(s).

24 Appl. to Quantum Mechanics: Hamiltonian case V = H, x = ψ iε ψ = H(s)ψ e(s) : isolated, simple eigenvalue of H(s) Set ψ(s) = ψ(s) exp(iε 1 s e(s )ds ) and rewrite ε ψ = i(h(s) e(s)) ψ L(s) ψ with 0 isolated, simple eigenvalue of L(s). Tunnelling out of e(s) is motion out of ker L(s). Hence reversible.

25 Appl. to QM: Dephasing Lindbladian case V = {operators on H}, x = ρ, L(s) = L(s) For simplicity dim H = 2, hence dim V = 4. L(ρ) = i[h, ρ] γ(p ρp + + P + ρp ) with γ 0 and H ψ i = e i ψ i, (i = ±) Basis of V : E ij = ψ i ψ j In particular, P i = E ii.

26 Appl. to QM: Dephasing Lindbladian case V = {operators on H}, x = ρ, L(s) = L(s) For simplicity dim H = 2, hence dim V = 4. L(ρ) = i[h, ρ] γ(p ρp + + P + ρp ) with γ 0 and H ψ i = e i ψ i, (i = ±) Basis of V : E ij = ψ i ψ j In particular, P i = E ii. L(P i ) = 0 L(E + ) = ( i(e + e ) γ)e + λ + E +, λ + = λ + ker L = span(p +, P ), ran L = span(e +, E + ) Tunnelling T(s, 0) = tr(p + (s)ρ(s)) for ρ(0) = P (0) is motion within ker L. Hence irreversible.

27 Dephasing Lindbladian case: Quantitative result Recall: In the general scheme, solution of εẋ = L(s)x of the form x(s) = a 0 (s) + ε(a 1 (s) + b 1 (s)) + O(ε 2 ) a n (s) = T(s, 0)a n (0) + s 0 T(s, s )Ṗ(s )b n (s )ds b n+1 (s) = L(s) 1 (Ṗ(s)a n(s) + Q(s)ḃn(s)) In the application x = ρ, a 0 (0) = P (0), a 1 (0) = 0 one obtains a 0 (s) = P (s) s a 1 (s) = ( P (s) + P + (s)) α(s )ds (loss & gain) 0 α(s) = (λ + (s) 1 + λ + (s) 1 ) tr ( P + (s)ṗ (s) 2 P + (s) ) as claimed. (λ λ 1 + ) = 2γ (e + e ) 2 + γ 2

28 Outline The Landau-Zener model An adiabatic theorem Optimal parametrization

29 Optimal parametrization: Hamiltonian case Family of Hamiltonians H(q), (0 q 1) Allotted time 1/ε to get from q = 0 to q = 1: q = q(s) = q(εt) with q : [0, 1] [0, 1], s q; q(0) = 0, q(1) = 1. 2-level system, iε ψ = H(s)ψ (ψ(0), P + (0)ψ(0)) = 0, (ψ(1), P + (1)ψ(1)) =: T[q]

30 Optimal parametrization: Hamiltonian case Family of Hamiltonians H(q), (0 q 1) Allotted time 1/ε to get from q = 0 to q = 1: q = q(s) = q(εt) with q : [0, 1] [0, 1], s q; q(0) = 0, q(1) = 1. 2-level system, iε ψ = H(s)ψ (ψ(0), P + (0)ψ(0)) = 0, (ψ(1), P + (1)ψ(1)) =: T[q] Question: Given ε > 0. Which are the parametrizations q minimizing T[q]?

31 Optimal parametrization: Hamiltonian case Family of Hamiltonians H(q), (0 q 1) Allotted time 1/ε to get from q = 0 to q = 1: q = q(s) = q(εt) with q : [0, 1] [0, 1], s q; q(0) = 0, q(1) = 1. 2-level system, iε ψ = H(s)ψ (ψ(0), P + (0)ψ(0)) = 0, (ψ(1), P + (1)ψ(1)) =: T[q] Question: Given ε > 0. Which are the parametrizations q minimizing T[q]? Answer: For small ε, minimizers (with T[q] = 0) are ubiquitous.

32 A theorem with inf q x(q) g 0 > 0. H(q) = 1 x(q) σ 2

33 A theorem with inf q x(q) g 0 > 0. H(q) = 1 x(q) σ 2 Theorem Let ε/g 0 be small enough. Then, for any parametrization q = q(s), q C 1 [0, 1] there is a parametrization q ε such that q ε q 2π q ε/g 0 T( q ε ) = 0

34 An illustration H(q) = 1 x(q) σ, 2 x(q) = (cos πq, 0, sinπq) q s s Figure: Left: Three parametrizations q(s). Right: log T[q] as function of ε

35 Proof Evolution is by precession about ĝ(q) = x(q)/ x(q) Partition [0, 1] into intervals [q, q + ] with q + q = 2πε/g 0. Part of it is enough to precess ĝ(q ) to ĝ(q + )

36 Optimal parametrization: Deph. Lindbladian case Family of 2-level Lindbladians parametrized by 0 q 1 L(q)(ρ) = i[h(q), ρ] γ(q)(p (q)ρp + (q) + P + (q)ρp (q)) Allotted time 1/ε to get from q = 0 to q = 1: q = q(s) = q(εt) with q : [0, 1] [0, 1], s q; q(0) = 0, q(1) = 1. Tunnelling T[q] = tr(p + (s)ρ(s)) s=1 for ρ(0) = P (0)

37 Optimal parametrization: Deph. Lindbladian case Family of 2-level Lindbladians parametrized by 0 q 1 L(q)(ρ) = i[h(q), ρ] γ(q)(p (q)ρp + (q) + P + (q)ρp (q)) Allotted time 1/ε to get from q = 0 to q = 1: q = q(s) = q(εt) with q : [0, 1] [0, 1], s q; q(0) = 0, q(1) = 1. Tunnelling T[q] = tr(p + (s)ρ(s)) s=1 for ρ(0) = P (0) Question: Given ε > 0. Which are the parametrizations q minimizing T[q]?

38 Optimal parametrization: Deph. Lindbladian case Family of 2-level Lindbladians parametrized by 0 q 1 L(q)(ρ) = i[h(q), ρ] γ(q)(p (q)ρp + (q) + P + (q)ρp (q)) Allotted time 1/ε to get from q = 0 to q = 1: q = q(s) = q(εt) with q : [0, 1] [0, 1], s q; q(0) = 0, q(1) = 1. Tunnelling T[q] = tr(p + (s)ρ(s)) s=1 for ρ(0) = P (0) Question: Given ε > 0. Which are the parametrizations q minimizing T[q]? Aside: In the Hamiltonian case, for small ε, minimizers (with T[q] = 0) are ubiquitous.

39 Optimal parametrization: The functional L(q)(ρ) = i[h(q), ρ] γ(q)(p (q)ρp + (q) + P + (q)ρp (q)) To leading order in ε T[q] = ε 1 with f(s) := f(q(s)) for f = x, γ and 0 γ(s) x(s) 2 + γ(s) 2 tr( Ṗ (s) 2) ds Ṗ (s) = P (q(s)) q(s), ( = d ds, = d dq ) Functional of Lagrangian type T[q] = 1 0 L(q(s), q(s), s)ds L(q, q, s / γ(q) ) = ε x(q) 2 + γ(q) 2 tr( P (q) 2) q 2 (weighted Fubini-Study metric)

40 Optimal parametrization: Results T[q] = 1 0 L(q(s), q(s), s)ds L(q, q, s / γ(q) ) = ε x(q) 2 + γ(q) 2 tr( P (q) 2) q 2 Minimizing parametrization has conserved energy L q q L = L Theorem The parametrization minimizes tunnelling iff it has constant tunnelling rate. In particular: velocity q is large, where gap x(q) is large small, where projection P (q) changes rapidly

41 Summary Dephasing Lindbladians describe open systems with several invariant states. Tunnelling between them if Lindbladian is changed adiabatically. Tunnelling has memory, unlike for Hamiltonian dynamics Analog of Landau-Zener formula for Hamiltonian 2-level systems General adiabatic theorem encompassing Lindbladian and Hamiltonian dynamics Optimal parametrization: No unique minimizers in Hamiltonian case. Unique minimizers in Lindbladian case, characterized by constant tunnelling rate.