Generators of quantum Markov Semigroups Matt Ziemke University of - - PowerPoint PPT Presentation

Generators of quantum Markov Semigroups

Matt Ziemke

University of South Carolina

Virginia Operator Theory and Complex Analysis Meeting (VOTCAM) November 7th, 2015

Matt Ziemke Generators of QMS

Paper

(with G. Androulakis), ”Generators of quantum Markov semigroups”, J. Math. Phys. (2015).

Matt Ziemke Generators of QMS

Outline

• 1. Definitions
• 2. Background
• 3. Known Results
• 4. Some Results
• 5. Examples

Matt Ziemke Generators of QMS

Definitions

The σ-weak topology Let A be a von Neumann algebra. Then A has a predual A∗ and the σ-weak topology on A is the σ(A, A∗) topology, that is, the weak∗ topology when A is viewed as the dual of A∗. Note 1: Every von Neumann algebra (when viewed as a Banach space) has a predual (Sakai). Note 2: We are mostly interested in the case when A = B(H), where H is a separable Hilbert space. In this case B(H)∗ = S1(H).

Matt Ziemke Generators of QMS

Definitions cont.

Completely positive operators Let H be a Hilbert space and let T : B(H) → B(H) be a bounded linear operator. Let B(H) ⊗ Mn be the ∗-algebra of n × n matrices with coefficients in B(H). We say the operator T is completely positive if for any n ∈ N, any positive element [Aij]1≤i,j≤n ∈ B(H) ⊗ Mn, and any h1, h2, . . . hn ∈ H we have

n

• i,j=1

hi, T(Aij)hj ≥ 0.

Matt Ziemke Generators of QMS

Definitions cont.

Quantum dynamical semigroup Let A be a von Neumann algebra. A quantum dynamical semigroup (QDS) is a one-parameter family (Tt)t≥0 of σ-weakly continuous, completely positive, linear operators on A such that (i) T0 = 1 (ii) Tt+s = TtTs (iii) for a fixed A ∈ A, the map t → Tt(A) is σ-weakly continuous. Further, if Tt(1) = 1 for all t ≥ 0 then we say the quantum dynamical semigroup is Markovian or we simply refer to it as a quantum Markov semigroup (QMS). If the map t → Tt is norm continuous then we say the semigroup is uniformly continuous.

Matt Ziemke Generators of QMS

Generator of a QMS Given a QDS (Tt)t≥0, we say that an element A ∈ A belongs to the domain of the infinitesimal generator L of (Tt))t≥0, denoted by D(L), if lim

t→0

1 t (TtA − A) converges in the σ-weak topology and, in this case, define the infinitesimal generator to be the generally unbounded operator L such that L(A) = σ-weak- lim

t→0

1 t (TtA − A) , A ∈ D(L). If (Tt)t≥0 is uniformly continuous then the generator L is bounded and given by L = lim

t→0

1 t (Tt − 1) where the limit is taken in the norm topology. In this case Tt = etL.

Matt Ziemke Generators of QMS

Background

Lindblad (‘76) If (Tt)t≥0 is a uniformly continuous QMS on B(H) then there exists G ∈ B(H) and a completely positive map φ : B(H) → B(H) such that the infinitesimal generator L of (Tt)t≥0 is given by L(A) = φ(A) + GA + AG ∗ for all A ∈ B(H). Note 1: Lindblad proved this for a uniformly continuous QMS on a hyperfinite factor A of B(H) (which includes the case A = B(H) by Topping (‘71)). Note 2: Christensen and Evans proved this for uniformly continuous QMS on arbitrary von Neumann algebras in ‘79.

Matt Ziemke Generators of QMS

Background cont.

Stinespring (‘55) Let B be a C ∗-subalgebra of the algebra of all bounded operators

• n a Hilbert space H and let A be a C ∗-algebra with unit. A linear

map T : A → B is completely positive if and only if it has the form T(A) = V ∗π(A)V where (π, K) is a unital ∗-representation of A on some Hilbert space K, and V is a bounded operator from H to K.

Matt Ziemke Generators of QMS

Background cont.

Lindblad + Stinespring Let L be the generator of a uniformly continuous QMS on B(H). Then there exists an operator G ∈ B(H), a unital ∗-representation

• f B(H) on some Hilbert space K, and a V ∈ B(H, K) such that

L(A) = V ∗π(A)V + GA + AG ∗ for all A ∈ B(H). Note: Due to a result of Kraus (‘70), there exists a sequence (Vj)j≥1 ⊆ B(K, H) such that V ∗π(A)V =

∞

• j=1

V ∗

j AVj

where the series ∞

j=1 V ∗ j AVj converge strongly.

Matt Ziemke Generators of QMS

Background cont.

Question Does the generator of a general QMS (that is, one which is not uniformly continuous) have a similar form? Note 1: Many important examples of QMS are not uniformly continuous (for example, the QMS associated to the noncommutative heat equation). Note 2: The QMS (Tt)t≥0 is uniformly continuous if and only if L is bounded.

Matt Ziemke Generators of QMS

Known results

Davies (‘79) Let Tt : S1(H) → S1(H) be a semigroup which satisfies: T ∗

t (C(H)) ⊆ C(H) for all t ≥ 0,

There exists e ∈ H\{0} such that Tt(|ee|) = |ee|,and the map [0, ∞) ∋ t → Tt(A) ∈ B(H) is SOT-continuous for all A ∈ B(H). Then there exists a dense linear subspace D of H and linear

• perators G : D → H and Ln : D → H such that the infinitesimal

generator L of (Tt)t≥0 is given by L(A) =

∞

• n=1

LnAL∗

n + GA + AG ∗

for all A ∈ (G − 1)−1S1(H)(G ∗ − 1)−1.

Matt Ziemke Generators of QMS

Known results cont.

Holevo (‘95) Let (Tt)t≥0 be a QMS on B(H). Assume that there exists a dense linear subspace D of H such that lim

t→0

• x, TtA − A

t y

• exists for all A ∈ B(H) and all x, y ∈ D. Then there exists a linear
• perator G : D → H, a separable Hilbert space H0, and a linear
• perator L : D → H ⊗ H0 such that

x, L(A)y = Lx, (A ⊗ 10)(Ly)H⊗H0 + Gx, Ay + x, AGy for all A ∈ B(H) and all x, y ∈ D.

Matt Ziemke Generators of QMS

Results

Notation If H is a Hilbert space and D is a linear subspace of H, let S(D) denote the set of sesquilinear forms on D × D. Definition Let D be a linear subspace of H and A be a linear subspace of B(H). A linear map φ : A → S(D) is called D-completely positive if for any k ∈ N, and any positive operator A = (Ai,j)1≤i,j≤k ∈ A ⊗ Mk(C) and for all x1, . . . , xk ∈ D,

k

• i,j=1

φ(Ai,j)(xi, xj) ≥ 0.

Matt Ziemke Generators of QMS

Definition Let (Tt)t≥0 be a QDS, L be its generator and Dom(L) its domain. Then A = {A ∈ Dom(L) : A∗A, AA∗ ∈ Dom(L)} is the domain algebra and is equal to the largest ∗-subalgebra of Dom(L) by Arveson (‘02).

Matt Ziemke Generators of QMS

Results cont.

Androulakis, Z. (‘15) Let L be the infinitesimal generator of a QMS on B(H) and let A be its domain algebra. Assume that there exists e ∈ H such that |ee| :∈ Dom(L). Let De = {x ∈ H : |xe| ∈ A}. Then there exists a linear map G : De → H and a De-completely positive map φ : A → S(De) such that x, L(A) = φ(A)(x, y) + x, GAy + GA∗x, y for all A ∈ A and x, y ∈ De.

Matt Ziemke Generators of QMS

Results cont.

Androulakis, Z. (‘15) Let A be a unital ∗-subalgebra of B(H), D be a linear subspace of H, and φ : A → S(D) be a D-completely positive map. Then there exists a Hilbert space K, a ∗-representation π : A → B(K) and a linear map V : D → K such that φ(A)(x, y) = Vx, π(A)VyK for all x, y ∈ D.

Matt Ziemke Generators of QMS

Results cont.

Corollary Let L be the infinitesimal generator of a QMS on B(H) and let A be its domain algebra. Assume that there exists e ∈ H such that |ee| ∈ Dom(L). Let De = {x ∈ H : |xe| ∈ A}. Then there exists a Hilbert space K, a ∗-representation π : A → B(K), and a linear map V : D → K such that x, L(A)y = Vx, π(A)VyK + x, GAy + GA∗x, y for all A ∈ A and x, y ∈ De. Note: Can take G to be G(x) = L(|xe|)e − 1 2e, L(|ee|)ex.

Matt Ziemke Generators of QMS

Example 1

(Parthasarathy (‘92)). Let (Bt)t≥0 be standard Brownian motion, V be a selfadjoint operator on H and define Tt : B(H) → B(H) by Tt(A) = E[eiBtV Ae−iBtV ]. Then (Tt)t≥0 is a QMS. If H = L2(R), V = i d

dx , e(t) = e−t2/2 then De is dense in L2(R)

and x, L(A)y = Vx, AVy + x, −1 2V 2Ay + −1 2V 2A∗x.y for all x, y ∈ Ue and A ∈ A .

Matt Ziemke Generators of QMS

Example 2

(Fagnola (‘00), Arveson, (‘02)). Let H = L2[0, ∞), and let Ut : H → H be defined by Ut(g)(s) = g(s − t) if s ≥ t

• therwise

Define Et : L2[0, ∞) → L2[0, t) the natural projection. Define ω : B(H) → C by ω(A) = f , Af where f ∈ H is defined by f (s) = e−s (s ∈ [0, ∞)). Define Tt : B(H) → B(H) by Tt(A) = ω(A)Et + UtAU∗

t .

Then (Tt)t≥0 is a QMS.

Matt Ziemke Generators of QMS

Example 2 cont.

Fix e ∈ L2[0, ∞) such that De ∈ L2[0, ∞) (where D is the differentiation operator), and e, f = 0. Then De ⊆ {x ∈ L2[0, ∞) : x, f = 0} hence De is not dense in H. Also A is not SOT dense in B(H). We have x, L(A)y = ω(A)x(0)y(0) + x, DAy + DA∗x, y for all x, y ∈ De and A ∈ A.

Matt Ziemke Generators of QMS

Thank you!

Matt Ziemke Generators of QMS