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Specified finite GNS detailed balance

Definition

Let d≥1 be finite, let H=Cd with its standard inner product, and write Md(C) for its matrices. Let Hs=Hs∗∈Md(C) be a Hamiltonian in joules, let ℏ>0 have units joule-seconds, and let m≥0 be an integer with L1,…,Lm∈Md(C) a finite family of jump matrices in seconds−1/2. Matrices on which the generators act are dimensionless. The Schrödinger generator L is the finite GKSL generator of Finite GKSL generators and units.

Its trace-dual observable generator is

G(A)=iℏ[Hs,A]+D(A),D(A)=∑j=1m(Lj∗ALj−12{Lj∗Lj,A}).

Here [U,V]=UV−VU, {U,V}=UV+VU, and A∗ is the conjugate transpose. The units of G, D, and L are seconds−1. The generator preserves the identity, D(I)=0, so etD(I)=I.

Let σ=σ∗>0 be a dimensionless density matrix on H, so Tr⁡σ=1. Define

⟨A,B⟩σ:=Tr⁡(σB∗A),∥A∥σ:=⟨A,A⟩σ1/2.

This inner product is linear in its first variable. It is positive definite because σ>0. The data satisfy the specified finite GNS detailed-balance convention when [Hs,σ]=0 and D is self-adjoint for this inner product, meaning ⟨D(A),B⟩σ=⟨A,D(B)⟩σ for all A,B∈Md(C). The Hamiltonian part of G is kept separate and is anti-self-adjoint. Quantum detailed-balance conventions vary; in particular, this definition does not identify GNS symmetry with KMS or Kubo–Mori symmetry.

Facts & Assumptions

Given: The finite GKSL data d,Hs,ℏ,(Lj)j=1m, a faithful density matrix σ, and the finite trace.

[F1]

Finite GKSL generators and units: L(X)=−iℏ[Hs,X]+∑j(LjXLj∗−12{Lj∗Lj,X}).

[F2]

Finite matrix spectral and state calculus: finite Hermitian matrices have basis-independent spectral calculus; a finite density is positive semidefinite with trace one.

[F3]

Finite matrix trace cyclicity and nondegeneracy: the finite trace cycles rectangular and square matrix products.

Proof

1.1givenF2F3algebra

The square root σ1/2 exists by [F2] and is invertible because σ>0. Thus ⟨A,A⟩σ=Tr⁡((Aσ1/2)∗(Aσ1/2))>0 for every nonzero A. Linearity in the first variable is immediate, and cyclicity [F3] gives Hermitian symmetry; hence the displayed form is a positive-definite inner product.

2.1givenF1F3step 1.1algebra

For the Hamiltonian term, cyclicity in [F3] gives Tr⁡(−i[Hs,X]A/ℏ)=Tr⁡(X i[Hs,A]/ℏ). Indeed, Tr⁡(HsXA)=Tr⁡(XAHs) and Tr⁡(XHsA) is already in the desired input-first order.

3.1givenF1F3step 2.1algebra∎

For each jump matrix, Tr⁡(LjXLj∗A)=Tr⁡(XLj∗ALj). The first anticommutator term satisfies Tr⁡(Lj∗LjXA)=Tr⁡(XALj∗Lj) by cyclicity, and the second is already Tr⁡(XLj∗LjA). Thus the trace-dual dissipator is exactly the displayed D. Substituting A=I cancels Lj∗Lj−12(Lj∗Lj+Lj∗Lj)=0 term by term, so D(I)=0.

Remarks

The cited Alicki–Lendi convention uses the same dissipative self-adjointness condition with the Hamiltonian part separated. Its inner product is written conjugate-linear in the first variable; swapping the two slots and conjugating gives the first-linear convention used here, without changing self-adjointness. Carlen–Maas explicitly distinguish the GNS inner product from the KMS inner product. No equivalence with another detailed-balance convention is assumed.

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