To help better address the query, let's first make a few things more explicit. Define the transform $A$ of an operator $A$ to the Heisenberg Picture as:
$$A_H = \sum_{α,β} W_{α,β}^†\,A\,W_{α,β}.$$
As you noted $A_H$ is generally a function of time, by virtue of the $W$'s being time-dependent, while $A$ is timeless and eternal. As you also noted, there is a product defect
$$Δ(A,B) = (AB)_H - A_H B_H = \sum_{α,β,γ,δ} W_{α,β}^†\,A \left(δ_{α,γ}δ_{β,δ} - W_{α,β} W_{γ,δ}^†\right) B\,W_{γ,δ},
$$
associated with the transform, with $Δ(AB) ≠ 0$ generally being the case. Similarly, there is also a commutator defect:
$$\begin{align}
Δ[A,B] &= [A,B]_H - [A_H,B_H]\\
&= ((AB)_H - (BA)_H) - (A_HB_H - B_HA_H)\\
&= ((AB)_H - A_HB_H) - ((BA)_H - B_HA_H)\\
&= Δ(AB) - Δ(BA).
\end{align}$$
Thus, for instance, it is necessary to draw a distinction for $[a,a^†] = 1$ between $[a,a^†]_H = 1$ and $[a_H,(a^†)_H]$, where $[a_H,(a^†)_H] ≠ 1$ is generally the case.
So, consider the general case of the Lindblad equation
$$\frac{dρ}{dt} = ρ + ρ$$
where the right-hand side consists of a unitary part
$$ρ = \frac{[H,ρ]}{iħ},$$
involving a Hamiltonian $H$, and a non-unitary part
$$ρ = \sum_{α,β}\left(L_αρL_α^† - \frac{\{L_α^†L_α,ρ\}}2\right),$$
involving a set of Lindblad operators $L_α$ indexed by $α$. More specifically, we can consider a more concrete example, touching on the earlier reply here, with
$$H = E \frac{aa^† + a^†a}2,\quad L_0 = \sqrt{γ_0}a,\quad L_1 = \sqrt{γ_1}a^†$$
with $γ_0 ≥ γ_1 ≥ 0$, $γ = γ_0 - γ_1 ≥ 0$, $E = ħω$ and with the commutator condition $[a,a^†] = 1$ imposed.
Following up on an earlier reply that will better show how the deficits arise and how they are remedied, I will introduce the following three pictures:
- Schrödinger Picture: Observables $A$ are timeless and eternal, while states $ρ$ "flow" in time. The key equations are:
$$\frac{dA}{dt} = 0,\quad \frac{dρ}{dt} = ρ + ρ.$$
- Lindblad Picture: Observables $A_L$ are governed by the Heisenberg equation for the unitary part *only*; all non-unitary dynamics remain with the state $ρ_L$:
$$\frac{dA_L}{dt} = (^†A)_L,\quad \frac{dρ_L}{dt} = ρ_L.$$
The states $ρ_L$ still "flow" in time, but only under the influence of the non-unitary part of the dynamics. In field theory, the time dependence of $A_L$ actually becomes dependence on all of the space-time coordinates, all of them being treated equally, because the Heisenberg equation reduces to (the quantized form of) the field equations governing, themselves, governing the obserables $A_L$. In that sense the time is more akin to how it is treated in Relativity, than in Quantum Theory under the Schrödinger Picture. In general $^†$ is Leibnitz, so the transform preserves products $(AB)_L = A_L B_L$, and commutators $[A,B]_L = [A_L,B_L]$.
- The Heisenberg Picture: Observables $A_H$ are governed by the adoint form of the Lindblad equation, while states $ρ_H$ are timeless and eternal, representing the entire history of a system, rather than its unfolding in time. In that sense, the states live in "block time". The relevant equations are:
$$\frac{dA_H}{dt} = (^†A)_H + (^†A)_H,\quad \frac{dρ}{dt} = 0.$$
However, the picture is muddied up a bit, because the evolution of the operators $A$ now contain the (adjoint of the) non-unitary part $^†A$, as well. Since $^†$ is not Leibnitz, in general, then product and commutator defects arise.
The Lindblad Picture repairs the product and commutator defect of the Heisenberg Picture and, in that sense, may be thought of as a better adaptation of the Heisenberg Picture to non-unitary dynamics, than the naive form of the Heisenberg Picture.
Denoting the expectation value of an observable $A$ in a state $ρ$ by $\left<Aρ\right> = \text{Tr}(Aρ)$, the adjoints may be defined by:
$$\left<(^†A)ρ\right> = \left<A(ρ)\right>,\quad \left<(^†A)ρ\right> = \left<A(ρ)\right>.$$
Making use of the linerarity and cyclicity properties of the underlying "$\text{Tr}$" operation, we can write
$$
\left<A[H,ρ]\right> = \left<[A,H]ρ\right>,\quad
\left<A(LρL^†)\right> = \left<(L^†AL)ρ\right>,\quad
\left<A(\{L^†L,ρ\})\right> = \left<\{L^†L,A\}ρ\right>,
$$
we obtain:
$$^†A = \frac{[A,H]}{iħ},\quad ^†A = \sum_α\left(L_α^†AL_α - \frac{\{A,L_α^†L_α\}}2\right).$$
The three pictures can all be tied together by first identifying a "now" point, e.g. $t = t_0 = 0$, and then setting
$$A_L(t_0) = A = A_H(t_0),\quad ρ_L(t_0) = ρ(t_0) = ρ_H,$$
and imposing the conditions:
$$\left<Aρ\right> = \left<A_Lρ_L\right> = \left<A_Hρ_H\right>.$$
This was already sufficient to develop the evolution equation for $A_H$ and $ρ_L$ from the requirements that $dρ_H/dt = 0$ and $dA_L/dt = (^†A)_L$, as noted in the previously-linked reply and spelled out for the case of $ρ_H$. For $A_L$, we have:
$$\begin{align}
\left<(^†A)_Lρ_L\right> + \left<A_L\frac{dρ_L}{dt}\right> &= \frac{d}{dt}\left<A_Lρ_L\right>\\
&= \frac{d}{dt}\left<Aρ\right> = \left<A\frac{dρ}{dt}\right>\\
&= \left<A(ρ + ρ)\right> = \left<^†Aρ + Aρ\right>\\
&= \left<(^†A)_Lρ_L + A_L(ρ)_L\right>.
\end{align}$$
Thus, (after generalizing over $A_L$):
$$\frac{dρ_L}{dt} = (ρ)_L.$$
The conversion is $t_0$-dependent, so that - in this sense - a concept of a "now" is incorporated as, essentially, the point of tangency between the "flow time" of the Schrödinger Picture and "block time" of the Heisenberg (and Lindblad) Pictures. The Lindblad state $ρ_L$ may, itself, be regarded as just the Heisenberg state $ρ_H$ endowed with a "sideways flow" in time. In that sense, block time is, itself, moving sideways in flow time.
We can work out the effects on the operators $a$, $a^†$, $aa^†$ and $a^†a$ for the example on hand:
$$
^†a = -iωa,\quad ^†a = -γ_0\frac{a}2 + γ_1\frac{a}2 = -γ\frac{a}2,\\
^†a^† = +iωa^†,\quad ^†a^† = -γ_0\frac{a^†}2 + γ_1\frac{a^†}2 = -γ\frac{a^†}2,\\
^†(aa^†) = 0,\quad ^†(aa^†) = -γ_0a^†a + γ_1aa^† = γ_0 - γaa^†,\\
^†(a^†a) = 0,\quad ^†(a^†a) = -γ_0a^†a + γ_1aa^† = -γa^†a + γ_1.\\
$$
This leads to the following differential equations:
$$
\frac{da_H}{dt} = -iωa_H - γ\frac{a_H}2,\quad a_H(0) = a,\\
\frac{d(a^†)_H}{dt} = +iω(a^†)_H - γ\frac{(a^†)_H}2,\quad (a^†)_H(0) = a^†,\\
\frac{d(aa^†)_H}{dt} = γ_0 - γ(aa^†)_H,\quad (aa^†)_H(0) = aa^†,\\
\frac{d(a^†a)_H}{dt} = -γ(a^†a)_H + γ_1,\quad (a^†a)_H(0) = a^†a,\\
$$
and the following solutions:
$$
a_H = a e^{-iωt - γt/2},\quad (a^†)_H = a^† e^{+iωt - γt/2},\\
(aa^†)_H = aa^† e^{-γt} + \frac{γ_0}γ\left(1 - e^{-γt}\right),\quad
(a^†a)_H = a^†a e^{-γt} + \frac{γ_1}γ\left(1 - e^{-γt}\right).\quad
$$
For the commutator, this yields:
$$\begin{align}
[a,a^†]_H &= (aa^†)_H - (a^†a)_H\\
&= (aa^† - a^†a) e^{-γt} + \frac{γ_0 - γ_1}γ\left(1 - e^{-γt}\right)\\
&= ([a,a^†] - 1) e^{-γt}\\
&= 1,
\end{align}$$
while
$$[a_H,(a^†)_H] = [a,a^†] e^{-γt} = e^{-γt}.$$
The corresponding defects are
$$
Δ(a,a^†) = \frac{γ_0}γ\left(1 - e^{-γt}\right),\quad
Δ(a^†,a) = \frac{γ_1}γ\left(1 - e^{-γt}\right),\quad
Δ[a,a^†] = 1 - e^{-γt}.
$$
The corresponding Lindblad Picture operators are obtained by just setting $γ_0 = 0 = γ_1$:
$$
a_L = a e^{-iωt},\quad (a^†)_L = a^† e^{+iωt},\quad
(aa^†)_L = aa^†,\quad (a^†a)_L = a^†a,\quad
$$
and the product (and commutator) defects go away.
The idea for this actually came about as part of a discussion I had with my apprentice, Darth Chat GPT, using it as a sounding board; where the interpretative framework for the dynamics in terms of "block time" and "flow time" was also developed. The first line of the discussion is the Heisenberg Picture, the second line is its modification to the Lindberg Picture, and the ensuing discussion about its interpretation.
My interest in this is to weave in the hybrid dynamics Oppenheim is using, and take the observables out of the "Schrödinger" Picture into the Heisenberg Picture - but only with respect to the unitary part of the dynamics - so as to bring the evolution of the observables (which are the field variables) back under the Einstein's equations, instead of under the Wheeler-de Witt equation. The non-unitary part of the dynamics stays with the states and is treated as a "sideways flow" of the block time space-time.
