# Markovian quantum master equation with Poincaré symmetry

Kaito Kashiwagi<EMAIL_ADDRESS>Akira Matsumura
<EMAIL_ADDRESS>Department of Physics, Kyushu University,
Fukuoka, 819-0395, Japan

###### Abstract

We investigate what kind of Markovian quantum master equation (QME) in the
Gorini-Kossakowski-Sudarshan-Lindblad (GKSL) form is realized under Poincaré
symmetry. The solution of the Markovian QME is given by a quantum dynamical
semigroup, for which we introduce invariance under Poincaré transformations.
Using the invariance of the dynamical semigroup and applying the unitary
representation of Poincaré group, we derive the Markovian QME for a
relativistic massive spin-0 particle. Introducing the field operator of the
massive particle and examining its evolution, we find that the field follows a
dissipative Klein-Gordon equation. In addition, we show that any two local
operators for spacelike separated regions commute with each other. This means
that the microcausality condition is satisfied for the dissipative model of
the massive particle.

###### Contents

  1. I Introduction
  2. II Markovian quantum master equation
  3. III Dynamical semigroup with Poincaré symmetry
  4. IV Model of a massive spin-0 particle and its field dynamics
  5. V Derivation of the Markovian quantum master equation
  6. VI Discussion
  7. VII Conclusion and outlook
  8. A Proof that the GKSL generator in Eq.(25) is Poincaré invariant
  9. B Proof of the commutation relations Eqs.(40) and and (44)
  10. C Time evolution of the product of operators
  11. D Derivation of the Markovian quantum master equation Eq.(78)

##  I Introduction

Markovian quantum master equation (QME) is of extensive importance in modern
physics. A typical study using Markovian QME is on the dynamics of a quantum
system coupled to its surrounding environment Breuer2002 ; Davies1976 . Such a
quantum system is called an open quantum system, and its Markovian process is
governed by the Markovian QME. In the theory of open quantum systems, the
celebrated Gorini-Kossakowski-Sudarshan-Lindblad (GKSL) form of the Markovian
QME Gorini1976 ; Lindblad1976  is crucial for describing the decoherence and
dissipative phenomena of an open quantum system. The description based on the
GKSL form has been applied in quantum information science Breuer2002 ;
Gardinar2000 ; Nielsen2002 ; Kraus2008 ; Wiseman2010 .

Another study applying Markovian QME is on the interplay of quantum and
gravity physics. In modern physics, it has not been elucidated whether gravity
follows quantum mechanics. This stimulates two theoretical standpoints: one is
that “gravity is quantum” and the other is that “gravity is classical”. In the
former standpoint, perturbative quantum gravity Donoghue1994  was established
as an effective theory of quantum gravity. In this effective theory, we can
describe the perturbative dynamics of canonically quantized matters and
gravitational fields. In the latter standpoint, the Kafri-Taylor-Milburn model
Kafri2014  and the Diósi-Penrose or the Diósi-Tilloy model Diosi1987 ;
Diosi1989 ; Penrose1996 ; Tilloy2016  were proposed. In these model, gravity
does not obey the principle of quantum superposition and is assumed to be in a
classically definite state. Such gravity causes the collapse of the wave
function of matter, which is described by a non-relativistic model based on a
Markovian QME.

As mentioned so far, Markovian QME has been broadly used for the description
of dissipative and collapse dynamics. In these contexts, it is intriguing to
discuss the Markovian QME for the dissipative and collapse dynamics of
relativistic quantum systems. For example, the decay process of an unstable
particle, such as pion decay, is considered to be a relativistic dissipative
phenomenon. This phenomenon would be governed by a Markovian QME for a long
time scale at which the Markovian approximation is valid. Such an unstable
particle might be described by relativistic theories of open quantum fields
Baidya2017 ; Meng2021 ; Wang2022 . Within the gravity-induced collapse of wave
function, a relativisitic model using a Markovian (or possibly non-Markovian)
QME should be preferred if the collapse occurs at a fundamental level and for
a relativistic quantum matter. In Ref.Oppenheim2022 , for formulating such a
model, the covariant theory of genuinely classical gravity coupled to quantum
matters was discussed. Also, regarding collapse dynamics not directly related
to gravity, relativistic collapse models have been discussed in the literature
Bedingham2014 ; Bedingham2019 ; Jones2021a ; Jones2021b ; Pearle2015 ;
Kurkov2011 .

Towards understanding the dissipative and collapse dynamics of relativistic
quantum systems, in this paper, we discuss a relativistic Markovian QME in the
GKSL form. This equation is developed by adopting Poincaré symmetry as a
guiding principle, which is commonly employed in relativistic theories, such
as quantum field theories and perturbative quantum gravity in the Minkowski
background. To respect the symmetry, we define the Poincaré invariance of the
quantum dynamical semigroup that gives the solution of Markovian QME. In the
previous work Matsumura2023 , the Poincaré invariant reduced dynamics of a
massive spin-0 particle was discussed. However, its time-local evolution
equation based on Markovian QME was not fully understood. In the present
paper, leveraging the Poincaré invariance of the dynamical semigroup and the
unitary representation of the Poincaré group, we construct the Markovian QME
of a relativistic massive spin-0 particle. We further investigate the
evolution of the particle field and assess whether the evolution is consistent
with microcausality. We find that the field obeys a Klein-Gordon (KG) equation
with a dissipative term and the microcausality is satisfied. This suggests
that our approach has the potential to provide dissipative quantum field
theories.

The structure of this paper is as follows. In Sec.II, we explain the basics of
Markovian QME in the GKSL form. In Sec.III, we introduce the Poincaré group,
its unitary representation, and the Poincaré algebra. We also define the
condition that a Markovian QME has Poincaré symmetry, which is crucial for
formulating the relativistic theory of Markovian QME. In Sec. IV we present
the concrete model of a massive particle of spin 0 and investigate the field
dynamics of the particle. We then find the evolution equation of the field and
that the microcausality is satisfied. In Sec.V, we explain the derivation of
the Markovian QME of the massive particle. In Sec.VI, the conclusion and the
future prospects of this paper are devoted. Throughout this paper, we use the
natural unit $\hbar=c=1$ and adopt the convention of the Minkowski metric as
$\eta_{\mu\nu}=\text{diag}[-1,1,1,1]$ used for lowering and raising indices.
Also, the commutator and the anticommutator are defined as
$[\hat{A},\hat{B}]=\hat{A}\hat{B}-\hat{B}\hat{A}$ and
$\\{\hat{A},\hat{B}\\}=\hat{A}\hat{B}+\hat{B}\hat{A}$, respectively.

##  II Markovian quantum master equation

Our main purpose is to formulate Markovian QMEs applicable to relativistic
quantum systems. To this end, in this section, we introduce the Markovian QME
in a general GKSL form Breuer2002 ; Davies1976 ; Gorini1976 ; Lindblad1976 ,
which can describe the dissipative dynamics of open quantum system and the
collapse of the wave function of quantum system. The Markovian QME for the
density operator $\rho(t)$ of a quantum system at time $t$ is

$\frac{d}{dt}\rho(t)=\mathcal{L}[\rho(t)],$ (1)

where the GKSL generator $\mathcal{L}$ is defined as

$\mathcal{L}[\rho(t)]=-i\left[\hat{M},\rho(t)\right]+\sum_{\lambda}\left[\hat{L}_{\lambda}\rho(t)\hat{L}^{\dagger}_{\lambda}-\frac{1}{2}\left\\{\hat{L}^{\dagger}_{\lambda}\hat{L}_{\lambda},\rho(t)\right\\}\right]$
(2)

with a Hermitian operator $\hat{M}$ and noise operators $\hat{L}_{\lambda}$
called Lindblad operators. In Eq.(2), the first term describes the unitary
evolution of the system, and the second term leads to the non-unitary
evolution of the system, such as dissipation and wave function collapse. The
GKSL generator $\mathcal{L}$ does not change under the following
transformations Breuer2002 ; Holevo1998 ,

$\displaystyle\hat{L}_{\lambda}\longrightarrow\hat{L}^{\prime}_{\lambda}=\sum_{\lambda^{\prime}}\mathcal{U}_{\lambda\lambda^{\prime}}\hat{L}_{\lambda^{\prime}}+\alpha_{\lambda}\hat{\mathbb{I}},$
(3)
$\displaystyle\hat{M}\longrightarrow\hat{M}^{\prime}=\hat{M}+\frac{1}{2i}\sum_{\lambda,\lambda^{\prime}}\left[\alpha^{\ast}_{\lambda}\mathcal{U}_{\lambda\lambda^{\prime}}\hat{L}_{\lambda^{\prime}}-\alpha_{\lambda}\mathcal{U}^{\ast}_{\lambda\lambda^{\prime}}\hat{L}_{\lambda^{\prime}}^{\dagger}\right]+\beta\hat{\mathbb{I}},$
(4)

where $\alpha_{\lambda}$ are complex number, $\beta$ is real and
$\mathcal{U}_{\lambda\lambda^{\prime}}$ is unitary matrix satisfying
$\sum_{\lambda}\mathcal{U}^{*}_{\lambda_{1}\lambda}\mathcal{U}_{\lambda_{2}\lambda}=\sum_{\lambda}\mathcal{U}^{*}_{\lambda\lambda_{1}}\mathcal{U}_{\lambda\lambda_{2}}=\delta_{\lambda_{1},\lambda{2}}$.
This invariance is important in discussing Poincaré symmetry in the later
sections. The formal solution of the QME (1) is

$\rho(t)=e^{\mathcal{L}t}\left[\rho(0)\right],$ (5)

where $e^{\mathcal{L}t}$ maps the initial state $\rho(0)$ onto the evolved
state $\rho(t)$. The map $e^{\mathcal{L}t}$ has two properties, complete
positivity and trace-preserving, and the one parameter family
$\\{e^{\mathcal{L}t}:t\geq 0\\}$ is called a quantum dynamical
semigroupBreuer2002 . In the next section, we will discuss Poincaré symmetry
and introduce the Poincaré invariance for the quantum dynamical semigroup
considered here. The invariance condition will be used to identify the
Markovian QME of a massive spin-0 particle in Sec.IV.

##  III Dynamical semigroup with Poincaré symmetry

To make our formulation clear, in this section, we first explain the Poincaré
symmetry in quantum theory Weinberg1995  and then define the Poincaré
invariance of quantum dynamical semigroup. The Poincaré symmetry is respected
for relativistic theories and requires invariance under the coordinate
transformation $x^{\mu}\rightarrow x^{\prime\mu}=\Lambda^{\mu}_{\
\nu}x^{\nu}+a^{\mu}$, where $\Lambda^{\mu}{}_{\nu}$ give the Lorentz
transformation matrix satisfying
$\eta_{\mu\nu}=\eta_{\rho\sigma}\Lambda^{\rho}{}_{\mu}\Lambda^{\sigma}{}_{\nu}$
and $a^{\mu}$ are the parameters of spacetime translation. This transformation
is called a Poincaré transformation and preserves the line element of the
Minkowski spacetime, $ds^{2}=\eta_{\mu\nu}dx^{\mu}dx^{\nu}$. In the following,
we only consider the continuous transformation, that is, the proper
orthochronous Lorentz transformation with $\Lambda^{0}{}_{0}\geq 1$ and
$\det\Lambda=1$.

Denoting the Poincaré transformation as the map
$g(\Lambda,a):x^{\mu}\rightarrow x^{\prime\mu}=\Lambda^{\mu}_{\
\nu}x^{\nu}+a^{\mu}$, this map satisfies the multiplication rule

$g(\Lambda^{\prime},a^{\prime})g(\Lambda,a)=g(\Lambda^{\prime}\Lambda,\Lambda^{\prime}a+a^{\prime}).$
(6)

From Eq.(6), the inverse transformation and the identical transformation can
be given as $g(I,0)$ and $g(\Lambda^{-1},-\Lambda^{-1}a)$, respectively, where
$I$ is the identity matrix and $\Lambda^{-1}$ is the inverse of the Lorentz
transformation matrix $\Lambda$. Therefore, the whole of Poincaré
transformation $g(\Lambda,a)$ forms a group, which is called Poincaré group.

In quantum theory, the Poincaré transformation $g(\Lambda,a)$ is represented
on a Hilbert space as a unitary operator $\hat{U}(\Lambda,a)$ satisfying

$\hat{U}(\Lambda^{\prime},a^{\prime})\hat{U}(\Lambda,a)=\hat{U}(\Lambda^{\prime}\Lambda,\Lambda^{\prime}a+a^{\prime}).$
(7)

This unitary operator $\hat{U}(\Lambda,a)$ is called the unitary
representation of Poincaré group. Considering the infinitesimal Poincaré
transformation $g(I+\omega,\epsilon)$ with

$\Lambda^{\mu}_{\ \nu}=\delta^{\mu}_{\ \nu}+\omega^{\mu}_{\ \nu},\quad
a^{\mu}=\epsilon^{\mu},$ (8)

where $\omega^{\mu}_{\ \nu}$ with $\omega_{\mu\nu}=-\omega_{\nu\mu}$ and
$\epsilon^{\mu}$ are infinitesimal parameters, and $\delta^{\mu}_{\ \nu}$ is
Kronecker’s symbol, we can expand the unitary operator
$\hat{U}(I+\omega,\epsilon)$ as

$\hat{U}(I+\omega,\epsilon)=\hat{I}+\frac{i}{2}\omega_{\mu\nu}\hat{J}^{\mu\nu}-i\epsilon_{\mu}\hat{P}^{\mu}+(\mathrm{high\
order\ term\ with\ respect\ to\ }\omega\ \mathrm{and}\ \epsilon),$ (9)

where the Hermitian operators $\hat{J}^{\mu\nu}$ with
$\hat{J}^{\mu\nu}=-\hat{J}^{\nu\mu}$ and $\hat{P}^{\mu}$ in the Schrödinger
picture are the generators of Lorentz transformation and spacetime
translation, respectivelyWeinberg1995 . These generators satisfy the following
commutation relations,

$\displaystyle\left[\hat{J}_{k},\hat{J}_{\ell}\right]$
$\displaystyle=i\epsilon_{k\ell m}\hat{J}^{m},$ (10)
$\displaystyle\left[\hat{J}_{k},\hat{K}_{\ell}\right]$
$\displaystyle=i\epsilon_{k\ell m}\hat{K}^{m},$ (11)
$\displaystyle\left[\hat{K}_{i},\hat{K}_{j}\right]$
$\displaystyle=-i\epsilon_{ijk}\hat{J}^{k},,$ (12)
$\displaystyle\left[\hat{J}_{k},\hat{P}_{\ell}\right]$
$\displaystyle=i\epsilon_{k\ell m}\hat{P}^{m},$ (13)
$\displaystyle\left[\hat{K}_{j},\hat{H}\right]$ $\displaystyle=i\hat{P}_{j}$
(14) $\displaystyle\left[\hat{K}_{i},\hat{P}_{j}\right]$
$\displaystyle=i\delta_{ij}\hat{H}$ (15)
$\displaystyle\left[\hat{P}_{i},\hat{P}_{j}\right]$ $\displaystyle=0$ (16)
$\displaystyle\left[\hat{P}_{i},\hat{H}\right]$
$\displaystyle=0=\left[\hat{J}_{i},\hat{H}\right]$ (17)

where we used the notations
$\hat{P}^{\mu}=(\hat{H},\hat{P}^{1},\hat{P}^{2},\hat{P}^{3})$,
$\hat{J}^{k}=\frac{1}{2}\epsilon^{ijk}\hat{J}_{ij}=(\hat{J}^{23},\hat{J}^{31},\hat{J}^{12})$,
and $\hat{K}^{k}=\hat{J}^{k0}=(\hat{J}^{10},\hat{J}^{20},\hat{J}^{30})$. The
above commutation relations are called the Poincaré algebra. From this
Poincaré algebra, it turns out that the boost operators $\hat{K}^{k}$ do not
commute with the Hamiltonian $\hat{H}$. In relativistic unitary theories such
as quantum field theories, the boost operator in the Heisenberg picture,
$\hat{K}^{k}_{\text{H}}=e^{i\hat{H}t}\hat{K}^{i}e^{-i\hat{H}t}$, is conserved
by the Noether theorem. This leads to that $\hat{K}^{k}$ in the Schrödinger
picture is $\hat{K}^{i}_{0}-t\hat{P}^{i}$, where $\hat{K}^{i}_{0}$ is
$\hat{K}^{i}$ at $t=0$. Hence the unitary $\hat{U}(\Lambda,a)$ has the
explicit time dependence in the Schrödinger picture. In the following, we
denote $\hat{K}^{i}$ and $\hat{U}(\Lambda,a)$ as $\hat{K}^{i}_{t}$ and
$\hat{U}_{t}(\Lambda,a)$, respectively, to specify the time dependence in the
Schrödinger picture.

We are in a position to define the Poincaré invariance of quantum dynamical
semigroup. If the dynamical semigroup $\\{e^{\mathcal{L}t}:t\geq 0\\}$ given
in (5) satisfies the following condition

$\hat{U}_{t}(\Lambda,a)e^{\mathcal{L}t}\left[\rho(0)\right]\hat{U}^{\dagger}_{t}(\Lambda,a)=e^{\mathcal{L}t}\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right],$
(18)

we regard that $\\{e^{\mathcal{L}t}:t\geq 0\\}$ has the Poincaré invariance
(see also Matsumura2023 ). The similar invariance condition for symmetry group
has been discussed in Refs.Cirstoiu2020 ; Marvian2014 ; Marko2017 . The
unitary operator $\hat{U}_{t}(\Lambda,a)$ is generated by $\hat{P}^{\mu}$,
$\hat{J}^{k}$, and $\hat{K}^{i}_{t}=\hat{K}^{i}_{0}-t\hat{P}^{i}$, which
satisfies the group multiplication rule
$\hat{U}_{t}(\Lambda^{\prime},a^{\prime})\hat{U}_{t}(\Lambda,a)=\hat{U}_{t}(\Lambda^{\prime}\Lambda,\Lambda^{\prime}a+a^{\prime})$.
In unitary relativistic theories, the time evolution of a density operator
$\rho(t)$ is given as

$\rho(t)=\mathcal{U}_{t}[\rho(0)]=e^{-i\hat{H}t}\rho(0)e^{i\hat{H}t},$ (19)

and this map $\mathcal{U}_{t}$ satisfies the condition Eq.(18) as

$\hat{U}_{t}(\Lambda,a)\mathcal{U}_{t}\left[\rho(0)\right]\hat{U}^{\dagger}_{t}(\Lambda,a)=\mathcal{U}_{t}\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right].$
(20)

So, the invariance condition (18) for $\\{e^{\mathcal{L}t}:t\geq 0\\}$ is a
generalization of that for $\\{\mathcal{U}_{t}:t\geq 0\\}$. We can get the
derivative form of the invariance condition (18). Differentiating the both
sides of the condition (18) with respect to time $t$, we obtain

$\frac{d}{dt}\rho^{\prime}(t)=\mathcal{L}\rho^{\prime}(t),$ (21)

where
$\rho^{\prime}(t)=\hat{U}_{t}(\Lambda,a)\rho(t)\hat{U}^{\dagger}_{t}(\Lambda,a)$.
Eq.(21) means that the generator of the time evolution, that is, the GKSL
generator $\mathcal{L}$ is invariant under the Poincaré transformation
$\rho(t)\rightarrow\rho^{\prime}(t)$.

Here, we explain our motivation why the Poincaré invariance is adopted. In the
context of dissipative dynamics, the decay of an unstable particle is a
relativistic dissipative phenomenon. In the context of collapse dynamics,
possibly induced by gravity, if the collapse of wave function is fundamental,
its relativistic description should be prefered. These dissipative and
collapse dynamics may be relativistically symmetric. As a step to exploring
this possibility, in this paper, we are considering a framework with the
Poincaré symmetry as a relativistic symmetry.

To get a simple invariance condition, we reduce Eq. (18) for
$\\{e^{\mathcal{L}t}:t\geq 0\\}$ to the condition for the GKSL generator
$\mathcal{L}$. Evaluating Eq.(18) for a small $t$ with the help of

$\hat{K}^{i}_{t}=e^{-i\hat{H}t}\hat{K}^{i}_{0}e^{i\hat{H}t}$ (22)

and

$\hat{U}_{t}(\Lambda,a)=e^{-i\hat{H}t}\hat{U}_{0}(\Lambda,a)e^{i\hat{H}t},$
(23)

we obtain

$\hat{U}_{0}(\Lambda,a)\left(\mathcal{L}-\mathcal{H}\right)\left[\rho(0)\right]\hat{U}^{\dagger}_{0}(\Lambda,a)=\left(\mathcal{L}-\mathcal{H}\right)\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right],$
(24)

where $\mathcal{H}[\rho(t)]$ means $\mathcal{H}[\rho(t)]=-i[\hat{H},\rho(t)]$.
Our task is to get a Markovian QME for relativistic quantum systems from the
GKSL generator $\mathcal{L}$ which satisfies Eq.(24). In the next section, as
an example of the Markovian QME derived from (24), we will present the
Markovian QME of a massive relativistic particle of spin 0. After the next
section, we will derive the QME.

##  IV Model of a massive spin-0 particle and its field dynamics

In this section, we exemplify the Markovian QME of a massive spin-0 particle
with a mass, whose the GKSL generator $\mathcal{L}$ satisfies(24). The
Markovian QME is

$\frac{d}{dt}\rho(t)=\mathcal{H}[\rho(t)]+\mathcal{D}[\rho(t)],$ (25)

where the first term $\mathcal{H}[\rho(t)]=-i\left[\hat{H},\rho(t)\right]$ is
given by the free Hamiltonian,

$\hat{H}=\int d^{3}p\ E_{\bm{p}}\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),$
(26)

with the energy of the particle, $E_{\bm{p}}=\sqrt{\bm{p}^{2}+m^{2}}$, and the
second dissipation term is

$\mathcal{D}[\rho(t)]=\gamma\int
d^{3}p\left[\hat{a}(\bm{p})\rho(t)\hat{a}^{\dagger}(\bm{p})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\rho(t)\right\\}\right]$
(27)

with a non-negative dissipation rate $\gamma$. The superoperator $\mathcal{D}$
in (27) is called a dissipator. The annihilation and creation operators of the
massive particle, $\hat{a}(\bm{p})$ and $\hat{a}^{\dagger}(\bm{p})$, satisfy
the following commutation relations

$\left[\hat{a}(\bm{p}),\hat{a}^{\dagger}(\bm{p}^{\prime})\right]=\delta^{3}(\bm{p}-\bm{p}^{\prime}),\quad\left[\hat{a}(\bm{p}),\hat{a}(\bm{p}^{\prime})\right]=0.$
(28)

It is easy to check the GKSL generator $\mathcal{L}=\mathcal{H}+\mathcal{D}$
of the Markovian QME (25) satisfies the invariance condition (24), whose
detail is provided in Appendix A. The equation (25) will be explicitly derived
in the next section.

In this section, we investigate the properties of the model by introducing the
field operator of the massive particle,

$\hat{\Phi}(\bm{x})=\frac{1}{(2\pi)^{\frac{3}{2}}}\int d^{3}p\
\frac{1}{\sqrt{2E_{\bm{p}}}}\left(e^{i\bm{p}\cdot\bm{x}}\hat{a}(\bm{p})+e^{-i\bm{p}\cdot\bm{x}}\hat{a}^{\dagger}(\bm{p})\right),$
(29)

where $\bm{x}$ is a spatial coordinate. Let us first seek the time evolution
equation of the massive spin-0 field in the Heisenberg picture. The time
evolved field $\hat{\Phi}_{\text{H}}(t,\bm{x})$ is formally given as

$\hat{\Phi}_{\text{H}}(t,\bm{x})=e^{\mathcal{L}^{\dagger}t}\left[\hat{\Phi}(\bm{x})\right],$
(30)

where the adjoint of the GKSL generator $\mathcal{L}=\mathcal{H}+\mathcal{D}$,
denoted by $\mathcal{L}^{\dagger}$, is defined as

$\mathcal{L}^{\dagger}\left[\hat{\Phi}_{\text{H}}\right]=\mathcal{H}^{\dagger}\left[\hat{\Phi}_{\text{H}}\right]+\mathcal{D}^{\dagger}\left[\hat{\Phi}_{\text{H}}\right],$
(31)

with

$\mathcal{H}^{\dagger}[\hat{\Phi}_{\text{H}}]=i\left[\hat{H},\hat{\Phi}_{\text{H}}\right],\quad\mathcal{D}^{\dagger}\left[\hat{\Phi}_{\text{H}}\right]=\gamma\int
d^{3}p\left(\hat{a}^{\dagger}(\bm{p})\hat{\Phi}_{\text{H}}\hat{a}(\bm{p})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\hat{\Phi}_{\text{H}}\right\\}\right).$
(32)

For convenience, we consider the following operator
$\hat{\Phi}_{\text{I}}(t,\bm{x})$ given as

$\hat{\Phi}_{\text{I}}(t,\bm{x})=e^{i\hat{H}t}\hat{\Phi}(\bm{x})e^{-i\hat{H}t}.$
(33)

The relation between $\hat{\Phi}_{\text{H}}(t,\bm{x})$ and
$\hat{\Phi}_{\text{I}}(t,\bm{x})$ is

$\hat{\Phi}_{\text{H}}(t,\bm{x})=e^{\mathcal{D}^{\dagger}t}\left[\hat{\Phi}_{\text{I}}(t,\bm{x})\right].$
(34)

Calculating the operation $\mathcal{D}^{\dagger}$ for
$\hat{\Phi}_{\text{I}}(t,\bm{x})$, we get the time evolution
$\hat{\Phi}_{\text{H}}(t,\bm{x})$ from Eq.(34). Substituting (29) into (33),
we obtain the operator $\hat{\Phi}_{\text{I}}(t,\bm{x})$ as

$\hat{\Phi}_{\text{I}}(t,\bm{x})=\frac{1}{(2\pi)^{\frac{3}{2}}}\int\frac{d^{3}p}{\sqrt{2E_{\bm{p}}}}\left(e^{i(\bm{p}\cdot\bm{x}-E_{\bm{p}}t)}\hat{a}(\bm{p})+e^{-i(\bm{p}\cdot\bm{x}-E_{\bm{p}}t)}\hat{a}^{\dagger}(\bm{p})\right),$
(35)

where we used the Baker-Campbell-Hausdorff formula and the commutation
relations
$\left[\hat{H},\hat{a}^{\dagger}(\bm{p})\right]=E_{\bm{p}}\hat{a}^{\dagger}(\bm{p})$
and $\left[\hat{H},\hat{a}(\bm{p})\right]=-E_{\bm{p}}\hat{a}(\bm{p})$. We
compute $\mathcal{D}^{\dagger}[\hat{\Phi}_{\text{I}}(t,\bm{x})]$ as

$\displaystyle\mathcal{D}^{\dagger}\left[\hat{\Phi}_{\text{I}}(t,\bm{x})\right]$
$\displaystyle=\gamma\int
d^{3}p^{\prime}\left(\hat{a}^{\dagger}(\bm{p}^{\prime})\hat{\Phi}_{\text{I}}(t,\bm{x})\hat{a}(\bm{p}^{\prime})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p}^{\prime})\hat{a}(\bm{p}^{\prime}),\hat{\Phi}_{\text{I}}(t,\bm{x})\right\\}\right)$
$\displaystyle=\frac{\gamma}{2}\frac{1}{(2\pi)^{\frac{3}{2}}}\int\frac{d^{3}p}{\sqrt{2E_{\bm{p}}}}\left(e^{i(\bm{p}\cdot\bm{x}-E_{\bm{p}}t)}\left[\hat{N},\hat{a}(\bm{p})\right]+e^{-i(\bm{p}\cdot\bm{x}-E_{\bm{p}}t)}\left[\hat{a}^{\dagger}(\bm{p}),\hat{N}\right]\right)$
$\displaystyle=-\frac{\gamma}{2}\hat{\Phi}_{\text{I}}(t,\bm{x}),$ (36)

where $\hat{N}$ is the number operator defined as

$\hat{N}=\int d^{3}p\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}).$ (37)

Repeating the operation of $\mathcal{D}^{\dagger}$ for
$\hat{\Phi}_{\text{I}}(t,\bm{x})$, we get
$\left(\mathcal{D}^{\dagger}\right)^{n}\left[\hat{\Phi}_{\text{I}}(t,\bm{x})\right]=\left(-\frac{\gamma}{2}\right)^{n}\hat{\Phi}_{\text{I}}(t,\bm{x})$
and hence

$\hat{\Phi}_{\text{H}}(t,\bm{x})=e^{\mathcal{D}^{\dagger}t}\left[\hat{\Phi}_{\text{I}}(t,\bm{x})\right]=\sum_{n=0}^{\infty}\frac{\left(\mathcal{D}^{\dagger}t\right)^{n}}{n!}\left[\hat{\Phi}_{\text{I}}(t,\bm{x})\right]=\sum_{n=0}^{\infty}\frac{1}{n!}\left(-\frac{\gamma}{2}t\right)^{n}\hat{\Phi}_{\text{I}}(t,\bm{x})=e^{-\frac{\gamma}{2}t}\hat{\Phi}_{\text{I}}(t,\bm{x}).$
(38)

Noticing that the field operator $\hat{\Phi}_{\text{I}}(t,\bm{x})$ follows the
usual KG equation,
$\partial^{2}_{t}\hat{\Phi}_{\text{I}}(t,\bm{x})=(\nabla^{2}-m^{2})\hat{\Phi}_{\text{I}}(t,\bm{x})$,
we find that $\hat{\Phi}_{\text{H}}(t,\bm{x})$ obeys a dissipative KG
equation,

$\left[\frac{\partial^{2}}{\partial t^{2}}+\gamma\frac{\partial}{\partial
t}+\frac{\gamma^{2}}{4}\right]\hat{\Phi}_{\text{H}}(t,\bm{x})=\left(\nabla^{2}-m^{2}\right)\hat{\Phi}_{\text{H}}(t,\bm{x}).$
(39)

If the dissiapation rate vanishes, $\gamma=0$, (39) becomes the usual KG
equation. In the case where $\gamma\neq 0$, the term
$\gamma\partial_{t}\hat{\Phi}_{\text{H}}$ represents the friction effect,
which represents the dissipative dynamics of the field.

We examine whether microcausality holds in the dissipative dynamics. The
microcausality condition suggests that any two local operators at spacelike
distance commute with each other. In special relativity, there is no
transmission beyond the speed of light, and this fact is reflected in the
notion of microcausality. Using the solution of the field
$\hat{\Phi}_{\text{H}}(t,\bm{x})$, we can check that the field satisfies the
microcausality as

$\left[\hat{\Phi}_{\text{H}}(x^{0},\bm{x}),\hat{\Phi}_{\text{H}}(y^{0},\bm{y})\right]=0,$
(40)

for any spacelike distance $(x^{0}-y^{0})^{2}-(\bm{x}-\bm{y})^{2}<0$. Here, we
introduce a local operator

$\hat{\Pi}(\bm{x})=-\frac{i}{(2\pi)^{\frac{3}{2}}}\int
d^{3}p\sqrt{\frac{E_{\bm{p}}}{2}}\left[e^{i\bm{p}\cdot\bm{x}}\hat{a}(\bm{p})-e^{-i\bm{p}\cdot\bm{x}}\hat{a}^{\dagger}(\bm{p})\right],$
(41)

and the evolved operator
$\hat{\Pi}_{\text{H}}(t,\bm{x})=e^{\mathcal{L}^{\dagger}t}[\hat{\Pi}(\bm{x})]$
is given as

$\hat{\Pi}_{\text{H}}(t,\bm{x})=e^{-\frac{\gamma}{2}t}\hat{\Pi}_{\text{I}}(t,\bm{x})$
(42)

with

$\displaystyle\hat{\Pi}_{\text{I}}(t,\bm{x})$
$\displaystyle=\frac{\partial}{\partial t}\hat{\Phi}_{\text{I}}(t,\bm{x})$
$\displaystyle=-\frac{i}{(2\pi)^{\frac{3}{2}}}\int
d^{3}p\sqrt{\frac{E_{\bm{p}}}{2}}\left[e^{i\left(\bm{p}\cdot\bm{x}-E_{\bm{p}}t\right)}\hat{a}(\bm{p})-e^{-i\left(\bm{p}\cdot\bm{x}-E_{\bm{p}}t\right)}\hat{a}^{\dagger}(\bm{p})\right].$
(43)

For the field $\hat{\Phi}_{\text{H}}$ and the operator $\hat{\Pi}_{\text{H}}$,
the simultaneous time commutation relation
$[\hat{\Phi}_{\text{H}}(t,\bm{x}),\hat{\Pi}_{\text{H}}(t,\bm{y})]=ie^{-\gamma
t}\delta^{3}(\bm{x}-\bm{y})$ holds. In this paper, the local operator
$\hat{\Pi}_{\text{H}}$ is just called the conjugate momentum of
$\hat{\Phi}_{\text{H}}$ even though these operators do not follow the usual
canonical commutation relations. We then find that the field
$\hat{\Phi}_{\text{H}}$ and its conjugate momentum $\hat{\Pi}_{\text{H}}$
satisfy

$\left[\hat{\Pi}_{\text{H}}(x^{0},\bm{x}),\hat{\Pi}_{\text{H}}(y^{0},\bm{y})\right]=0=\left[\hat{\Phi}_{\text{H}}(x^{0},\bm{x}),\hat{\Pi}_{\text{H}}(y^{0},\bm{y})\right]$
(44)

for $(x^{0}-y^{0})^{2}-(\bm{x}-\bm{y})^{2}<0$. The detailed proof of the above
three commutation relations are provided in Appendix B. Furthermore, we can
show that the results (40) and (44) holds for more general local operators. As
shown in Ref.Weinberg1995 , arbitrary operator of particle is expanded by the
creation and annihilation operators of the particle. This leads to that
arbitrary local operator $\hat{O}$ at time $t=0$ for a finite spatial region O
is expanded by $\hat{\Phi}(\bm{x})$ and $\hat{\Pi}(\bm{x})$, and the time
evolution of $\hat{O}$ is given as

$\displaystyle\hat{O}_{\text{H}}(t)$ $\displaystyle=\sum_{N,M}\int
d^{3}x^{\prime}_{1}\cdots d^{3}x^{\prime}_{N}\cdots d^{3}x_{1}\cdots
d^{3}x_{M}\sum_{i^{\prime}_{1},\cdots,i^{\prime}_{N}=1,2}\sum_{i_{1},\cdots,i_{M}=1,2}O^{i^{\prime}_{1}\cdots
i^{\prime}_{N}i_{1}\cdots
i_{M}}_{NM}(\bm{x}^{\prime}_{1},\cdots,\bm{x}^{\prime}_{N},\bm{x}_{1},\cdots,\bm{x}_{M})$
$\displaystyle\times
e^{\mathcal{L}^{\dagger}t}[\hat{\xi}^{i^{\prime}_{1}}(\bm{x}^{\prime}_{1})\cdots\hat{\xi}^{i^{\prime}_{N}}(\bm{x}^{\prime}_{N})\hat{\xi}^{i_{1}}(\bm{x}_{1})\cdots\hat{\xi}^{i_{M}}(\bm{x}_{M})],$
(45)

where the expansion coefficients $O^{i^{\prime}_{1}\cdots
i^{\prime}_{N}i_{1}\cdots
i_{M}}_{NM}(\bm{x}^{\prime}_{1},\cdots,\bm{x}^{\prime}_{N},\bm{x}_{1},\cdots,\bm{x}_{M})$
only have nonzero values for the region O, and
$\hat{\xi}^{j}(\bm{x})=\left(\hat{\Phi}(\bm{x}),\hat{\Pi}(\bm{x})\right)$.
Note that in the non-unitary theory considered here,
$e^{\mathcal{L}t}[\hat{\xi}^{i^{\prime}_{1}}(\bm{x}^{\prime}_{1})\cdots\hat{\xi}^{i^{\prime}_{N}}(\bm{x}^{\prime}_{N})\hat{\xi}^{i_{1}}(\bm{x}_{1})\cdots\hat{\xi}^{i_{M}}(\bm{x}_{M})]\neq\hat{\xi}_{\text{H}}^{i^{\prime}_{1}}(t,\bm{x}^{\prime}_{1})\cdots\hat{\xi}_{\text{H}}^{i^{\prime}_{N}}(t,\bm{x}^{\prime}_{N})\hat{\xi}_{\text{H}}^{i_{1}}(t,\bm{x}_{1})\cdots\hat{\xi}_{\text{H}}^{i_{M}}(t,\bm{x}_{M})$.
In Appendix C, we show that the time evolution of the product
$\hat{\xi}^{i^{\prime}_{1}}(\bm{x}^{\prime}_{1})\cdots\hat{\xi}^{i^{\prime}_{N}}(\bm{x}^{\prime}_{N})\hat{\xi}^{i_{1}}(\bm{x}_{1})\cdots\hat{\xi}^{i_{M}}(\bm{x}_{M})$
is given as the product of operators chosen from
$\hat{\xi}_{\text{H}}^{i^{\prime}_{1}}(t,\bm{x}^{\prime}_{1}),\dots,\hat{\xi}^{i^{\prime}_{N}}_{\text{H}}(t,\bm{x}^{\prime}_{N}),\hat{\xi}_{\text{H}}^{i_{1}}(t,\bm{x}_{1}),\dots,\hat{\xi}_{\text{H}}^{i_{M}}(t,\bm{x}_{M})$.
This implies that the commutation relation of arbitrary local operators is
computed from the commutation relations
$[\hat{\xi}^{i}_{\text{H}}(x^{0},\bm{x}),\hat{\xi}^{j}_{\text{H}}(y^{0},\bm{y})]$,
which vanish for $(x^{0}-y^{0})^{2}-(\bm{x}-\bm{y})^{2}<0$ as shown in (40)
and $\eqref{eq:PiPhi}$. Hence, the time-evolved operators
$\hat{A}_{\text{H}}(t)$ and $\hat{B}_{\text{H}}(t^{\prime})$ of any two local
operators $\hat{A}$ at time $t=0$ for a spatial region A and $\hat{B}$ at time
$t^{\prime}=0$ for a spatial region B commutes if $\hat{A}_{\text{H}}(t)$ and
$\hat{B}_{\text{H}}(t^{\prime})$ are spacelike separated. Therefore, the
microcausality holds for the present model of the massive particle.

In the following section, we will derive the Markovian QME of the massive
particle considered here.

##  V Derivation of the Markovian quantum master equation

In this section, we derive the Markovian QME of the massive spin-0 particle,
which was presented in the previous section. For the derivation, let us first
return to the condition Eq.(24) of the Poincaré invariance,
$\hat{U}_{0}(\Lambda,a)\left(\mathcal{L}-\mathcal{H}\right)\left[\rho(0)\right]\hat{U}^{\dagger}_{0}(\Lambda,a)=\left(\mathcal{L}-\mathcal{H}\right)\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right]$.
Because the superoperator $\mathcal{L}-\mathcal{H}$ is given as

$(\mathcal{L}-\mathcal{H})\left[\rho\right]=-i\left[\hat{M}-\hat{H},\rho\right]+\sum_{\lambda}\left[\hat{L}_{\lambda}\rho\hat{L}^{\dagger}_{\lambda}-\frac{1}{2}\left\\{\hat{L}^{\dagger}_{\lambda}\hat{L}_{\lambda},\rho\right\\}\right],$
(46)

the condition Eq.(24) is rewritten as

$\displaystyle(\mathcal{L-H})[\rho]$
$\displaystyle=\hat{U}^{\dagger}_{0}(\Lambda,a)\left(\mathcal{L}-\mathcal{H}\right)\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right]\hat{U}_{0}(\Lambda,a)$
$\displaystyle=-i\left[\hat{U}^{\dagger}_{0}(\Lambda,a)(\hat{M}-\hat{H})\hat{U}_{0}(\Lambda,a),\rho(0)\right]+\sum_{\lambda}\Big{[}\left(\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{L}_{\lambda}\hat{U}_{0}(\Lambda,a)\right)\rho(0)\left(\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{L}^{\dagger}_{\lambda}\hat{U}_{0}(\Lambda,a)\right)$
$\displaystyle-\frac{1}{2}\left\\{\left(\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{L}_{\lambda}\hat{U}_{0}(\Lambda,a)\right)\left(\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{L}^{\dagger}_{\lambda}\hat{U}_{0}(\Lambda,a)\right),\rho(0)\right\\}\Big{]},$
(47)

Eq.(47) means that the Markovian QME has invariance for the transformations
$\hat{M}-\hat{H}\longrightarrow\hat{U}^{\dagger}_{0}(\Lambda,a)(\hat{M}-\hat{H})\hat{U}_{0}(\Lambda,a)$
and
$\hat{L}_{\lambda}\longrightarrow\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{L}_{\lambda}\hat{U}_{0}(\Lambda,a)$.
Recalling that the form of Markovian QME does not change under the
transformations (3) and (4), we get the following rules,

$\displaystyle\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{L}_{\lambda}(\Lambda,a)\hat{U}_{0}(\Lambda,a)=\sum_{\lambda^{\prime}}\mathcal{U}_{\lambda\lambda^{\prime}}(\Lambda,a)\hat{L}_{\lambda^{\prime}}+\alpha_{\lambda}(\Lambda,a)\hat{\mathbb{I}},$
(48)
$\displaystyle\hat{U}^{\dagger}_{0}(\Lambda,a)(\hat{M}-\hat{H})\hat{U}_{0}(\Lambda,a)=\hat{M}-\hat{H}+\frac{1}{2i}\sum_{\lambda,\lambda^{\prime}}\left[\alpha^{*}_{\lambda}\mathcal{U}_{\lambda\lambda^{\prime}}(\Lambda,a)\hat{L}_{\lambda^{\prime}}-\alpha_{\lambda}(\Lambda,a)\mathcal{U}^{\ast}_{\lambda\lambda^{\prime}}(\Lambda,a)\hat{L}^{\dagger}_{\lambda^{\prime}}\right]+\beta(\Lambda,a)\hat{\mathbb{I}},$
(49)

For convenience, we introduce the vectors $\hat{\bm{L}}$ and $\bm{\alpha}$
with the components $\hat{L}_{\lambda}$ and $\alpha_{\lambda}$, respectively.
Then, the transformation rules (48) and (49) are rewritten as

$\displaystyle\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{\bm{L}}\hat{U}_{0}(\Lambda,a)=\mathcal{U}(\Lambda,a)\hat{\bm{L}}+\bm{\alpha}(\Lambda,a)\mathbb{I},$
(50)
$\displaystyle\hat{U}^{\dagger}_{0}(\Lambda,a)(\hat{M}-\hat{H})\hat{U}_{0}(\Lambda,a)=\hat{M}-\hat{H}+\frac{1}{2i}\left[\bm{\alpha}^{\dagger}(\Lambda,a)\mathcal{U}(\Lambda,a)\hat{\bm{L}}-\hat{\bm{L}}^{\dagger}\mathcal{U}^{\dagger}(\Lambda,a)\bm{\alpha}(\Lambda,a)\right]+\beta(\Lambda,a)\mathbb{I}$
(51)

Using that the operator $\hat{U}_{0}(\Lambda,a)$ is the unitary representation
satisfying
$\hat{U}_{0}(\Lambda^{\prime}\Lambda,a^{\prime}+\Lambda^{\prime}a)=\hat{U}_{0}(\Lambda^{\prime},a^{\prime})\hat{U}_{0}(\Lambda,a)$,
we can give the conditions of the unitary matrix $\mathcal{U}(\Lambda,a)$, the
vector $\bm{\alpha}(\Lambda,a)$, and the real parameter $\beta(\Lambda,a)$:

$\displaystyle\mathcal{U}(\Lambda^{\prime}\Lambda,a^{\prime}+\Lambda^{\prime}a)=\mathcal{U}(\Lambda^{\prime},a^{\prime})\mathcal{U}(\Lambda,a)$
(52)
$\displaystyle\bm{\alpha}(\Lambda^{\prime}\Lambda,a^{\prime}+\Lambda^{\prime}a)=\mathcal{U}(\Lambda^{\prime},a^{\prime})\bm{\alpha}(\Lambda,a)+\bm{\alpha}(\Lambda^{\prime},a^{\prime})$
(53)
$\displaystyle\beta(\Lambda^{\prime}\Lambda,a^{\prime}+\Lambda^{\prime}a)=\beta(\Lambda,a)+\beta(\Lambda^{\prime},a^{\prime})+\frac{1}{2i}\left[\bm{\alpha}^{\dagger}(\Lambda^{\prime},a^{\prime})\mathcal{U}(\Lambda^{\prime},a^{\prime})\bm{\alpha}(\Lambda,a)-\bm{\alpha}^{\dagger}(\Lambda,a)\mathcal{U}^{\dagger}(\Lambda^{\prime},a)\bm{\alpha}(\Lambda^{\prime},a^{\prime})\right]$
(54)

It is easy to derive these conditions. Eqs.(52) and (53) are given by
comparing the both side of
$\hat{U}^{\dagger}_{0}(\Lambda^{\prime}\Lambda,a^{\prime}+\Lambda^{\prime}a)\hat{\bm{L}}\hat{U}_{0}(\Lambda^{\prime}\Lambda,a^{\prime}+\Lambda^{\prime}a)=\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{U}^{\dagger}_{0}(\Lambda^{\prime},a^{\prime})\hat{\bm{L}}\hat{U}_{0}(\Lambda^{\prime},a^{\prime})\hat{U}_{0}(\Lambda,a)$.
Eq.(54) is obtained by comparing the both side of
$\hat{U}^{\dagger}_{0}(\Lambda^{\prime}\Lambda,a^{\prime}+\Lambda^{\prime}a)(\hat{M}-\hat{H})\hat{U}_{0}(\Lambda^{\prime}\Lambda,a^{\prime}+\Lambda^{\prime}a)=\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{U}^{\dagger}_{0}(\Lambda^{\prime},a^{\prime})(\hat{M}-\hat{H})\hat{U}_{0}(\Lambda^{\prime},a^{\prime})\hat{U}_{0}(\Lambda,a)$.

Our task is to derive the expression of the self-adjoint operator
$\hat{M}-\hat{H}$ and the Lindblad operator $\hat{\bm{L}}$ satisfying Eq.(50)
and (51). In particular, Eq.(50) is decomposed into irreducible representation
subspaces for ease of conputation. Hence, the irreducible unitary
representations of the Poincaré group are useful for our analysis. We
introduce the standard momentum $\ell^{\mu}$ and the Lorentz transformation
$\left(S_{q}\right)^{\mu}_{\ \nu}$ with

$q^{\mu}=\left(S_{q}\right)^{\mu}_{\ \nu}\ell^{\nu}.$ (55)

From (52), the unitary matrix $\mathcal{U}(\Lambda,a)$ is written as

$\mathcal{U}(\Lambda,a)=\mathcal{U}(I,a)\mathcal{U}(\Lambda,0)=\mathcal{T}(a)\mathcal{V}(\Lambda),,$
(56)

where $I$ is the identity matrix, $\mathcal{U}(I,a)=\mathcal{T}(a)$,
$\mathcal{U}(\Lambda,0)=\mathcal{V}(\Lambda)$. The unitary matrix
$\mathcal{U}(\Lambda,a)$ is the unitary representation on a vector space, and
then we define the following vector on the vector space:

$R^{\mu}\bm{v}_{\ell,\xi}=\ell^{\mu}\bm{v}_{\ell,\xi},$ (57)

where $R^{\mu}$ is the generator of spacetime translation, the label $\xi$
describes the degrees of freedom which can not be specified by momentum
$\ell^{\mu}$. Also, $\mathcal{T}(a)$ can be written as $e^{-iR_{\mu}a^{\mu}}$
by using the generator of spacetime translation $R^{\mu}$. Now, we define the
eigenvector $\bm{v}_{q,\xi}$ which belongs to the eigenvalue $q^{\mu}$ of
$R^{\mu}$ as

$\bm{v}_{q,\xi}=N_{q}\mathcal{V}(S_{q})\bm{v}_{\ell,\xi},$ (58)

where $N_{q}$ is the normalization. Then, we can obtain the following rules
$\bm{v}_{q,\xi}$:

$\displaystyle\mathcal{T}(a)\bm{v}_{q,\xi}=e^{-iq^{\mu}a_{\mu}}\bm{v}_{q,\xi},$
(59) $\displaystyle\mathcal{V}(\Lambda)\bm{v}_{q,\xi}=\frac{N_{q}}{N_{\Lambda
q}}\sum_{\xi^{\prime}}\mathcal{D}_{\xi\xi^{\prime}}\left(W(\Lambda,q)\right)\bm{v}_{\Lambda
q,\xi^{\prime}},$ (60)

where $W(\Lambda,q)=S^{-1}_{\Lambda q}\Lambda S_{q}$ is an element of the
little group and satisfies $W^{\mu}_{\ \nu}\ell^{\nu}=\ell^{\mu}$. For the
derivations of Eqs.(59) and (60), see Ref.Weinberg1995 . The matrix
$\mathcal{D}(W)$ with the components $\mathcal{D}_{\xi\xi^{\prime}}(W)$ forms
the unitary representation of the little group. For the various four-momentum,
we give the standard momenta and the little groups corresponded to each
standard momentum in Table 1. For simplicity, $\xi$ is assumed to be the label
of basis vectors of the irreducible representation subspaces of the little
group.

Standard Momentum $\ell^{\mu}$ | Little Group
---|---
$\ell^{\mu}=[M,0,0,0],\ M>0$ | SO(3)
$\ell^{\mu}=[-M,0,0,0],\ M>0$ | SO(3)
$\ell^{\mu}=[\kappa,0,0,\kappa],\ \kappa>0$ | ISO(2)
$\ell^{\mu}=[-\kappa,0,0,\kappa],\ \kappa>0$ | ISO(2)
$\ell^{\mu}=[0,0,0,N],\ N^{2}>0$ | SO(2,1)
$\ell^{\mu}=[0,0,0,0]$ | SO(3,1)
Table 1: Classification of the standard momentum $\ell^{\mu}$ and the little
group associated with $\ell^{\mu}$, $M$ is mass and $\kappa$ is arbitrary
positive energy.

Next, we consider the unitary operator $\hat{U}_{0}(\Lambda,a)$. As in the
unitary matrix $\mathcal{U}(\Lambda,a)$, the operator $\hat{U}_{0}(\Lambda,a)$
is decomposed in the same way. Therefore, it is written as

$\hat{U}_{0}(\Lambda,a)=\hat{U}_{0}(I,a)\hat{U}_{0}(\Lambda,0)=\hat{T}(a)\hat{V}(\Lambda),$
(61)

where $\hat{U}_{0}(I,a)=\hat{T}(a)=e^{-i\hat{P}_{\mu}a^{\mu}}$ with
$\hat{P}^{\mu}=[\hat{H},\hat{P}^{1},\hat{P}^{2},\hat{P}^{3}]$ and
$\hat{U}_{0}(\Lambda,0)=\hat{V}(\Lambda)$ with the generators $\hat{J}^{i}$
and $\hat{K}^{i}_{0}$. Let us focus on the transformation rule Eq.(50). From
the rule for $\Lambda=I$, we obtain

$\hat{T}^{\dagger}(a)\hat{\bm{L}}\hat{T}(a)=\mathcal{T}(a)\hat{\bm{L}}+\bm{\alpha}(I,a)\hat{\mathbb{I}}.$
(62)

In Eq. (50) for $a^{\mu}=0$, we obtain

$\hat{V}^{\dagger}(\Lambda)\hat{\bm{L}}\hat{V}(\Lambda)=\mathcal{V}(\Lambda)\hat{\bm{L}}+\bm{\alpha}(\Lambda,0)\hat{\mathbb{I}}.$
(63)

Introducing $\hat{L}_{q,\xi}=\bm{v}_{q,\xi}^{\dagger}\hat{\bm{L}}$ and
$\alpha_{q,\xi}=\bm{v}_{q,\xi}^{\dagger}\bm{\alpha}$, w e can rewrite (62) and
(63) as follows:

$\displaystyle\hat{T}^{\dagger}(a)\hat{L}_{q,\xi}\hat{T}(a)$
$\displaystyle=e^{-iq_{\mu}a^{\mu}}\hat{L}_{q,\xi}+\alpha_{q,\xi}(I,a)\hat{\mathbb{I}},$
(64) $\displaystyle\hat{V}^{\dagger}(\Lambda)\hat{L}_{q,\xi}\hat{V}(\Lambda)$
$\displaystyle=\frac{N^{\ast}_{q}}{N^{\ast}_{\Lambda^{-1}q}}\sum_{\xi^{\prime}}\mathcal{D}^{\ast}_{\xi^{\prime}\xi}\left(W(\Lambda^{-1},q)\right)\hat{L}_{\Lambda^{-1}q,\xi^{\prime}}+\alpha_{q,\xi}(\Lambda,0)\hat{\mathbb{I}}.$
(65)

Because of these transformation rules, we get the expression of the complex
number $\alpha_{q,\xi}(\Lambda,a)$. Since (50) can be rewritten as

$\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{L}_{q,\xi}\hat{U}_{0}(\Lambda,a)=\bm{v}^{\dagger}_{q,\xi}\mathcal{U}(\Lambda,a)\hat{\bm{L}}+\alpha_{q,\xi}(\Lambda,a)\hat{\mathbb{I}},$
(66)

we get $\alpha_{q,\xi}(\Lambda,a)$ as

$\alpha_{q,\xi}(\Lambda,a)=e^{-iq_{\mu}a^{\mu}}\alpha_{q,\xi}(\Lambda,0)+\alpha_{q,\xi}(I,a)$
(67)

by using (64) and (65) and comparing the both side of Eq.(66). In Eq.(65) for
$\Lambda=S_{q}$, we obtain

$\hat{V}^{\dagger}(S_{q})\hat{L}_{q,\xi}\hat{V}(S_{q})=N^{\ast}_{q}\hat{L}_{\ell,\xi}+\alpha_{q,\xi}(S_{q},0)\hat{\mathbb{I}},$
(68)

where we used the facts that $N_{\ell}=1$ and
$W\left(S^{-1}_{q},q\right)=S^{-1}_{S^{-1}_{q}q}S^{-1}_{q}S_{q}=S^{-1}_{\ell}=I$
hold. These facts can be checked by using Eq.(58) and the definition of
$W(\Lambda,q)$. Eq.(68) means that the Lindblad operator $\hat{L}_{q,\xi}$ is
determined from the Lindblad operator $\hat{L}_{\ell,\xi}$ with the standard
momentum $\ell^{\mu}$. To discuss the form of the Lindblad operator
$\hat{L}_{\ell,\xi}$, we focus on the following equations obtained from
Eqs.(64) and (65) for $q^{\mu}=\ell^{\mu}$ and $\Lambda=Q$ with $Q^{\mu}_{\
\nu}\ell^{\nu}=\ell^{\mu}$, respectively:

$\displaystyle\hat{T}^{\dagger}(a)\hat{L}_{\ell,\xi}\hat{T}(a)=e^{-il_{\mu}a^{\mu}}\hat{L}_{\ell,\xi}+\alpha_{\ell,\xi}(I,a)\hat{\mathbb{I}},$
(69)
$\displaystyle\hat{V}^{\dagger}(Q)\hat{L}_{\ell,\xi}\hat{V}(Q)=\sum_{\xi^{\prime}}\mathcal{D}^{\ast}_{\xi^{\prime}\xi}(Q^{-1})\hat{L}_{\ell,\xi^{\prime}}+\alpha_{\ell,\xi}(Q,0)\hat{\mathbb{I}},$
(70)

where note that $N_{\ell}=N_{Q^{-1}\ell}=1$.

Now, we can get a model of the Markovian QME for a massive spin-0 particle by
assuming that the Lindblad operator $\hat{L}_{q,\xi}$ is given as

$\hat{L}_{q,\xi}=\int d^{3}p\ f_{q,\xi}(\bm{p})\hat{a}(\bm{p}),$ (71)

and that the self-adjoint operator $\hat{M}-\hat{H}$ is given as

$\hat{M}-\hat{H}=\int d^{3}p\ d^{3}p^{\prime}\
g(\bm{p},\bm{p}^{\prime})\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}^{\prime}),$
(72)

where $g(\bm{p},\bm{p}^{\prime})=g^{*}(\bm{p}^{\prime},\bm{p})$ is satisfied
because the $\hat{M}-\hat{H}$ is self-adjoint. The Markovian QME with the
above operators gives an evolution from a Gausssian state to another Gausssian
state. Furthermore, in the dynamics, there are no particle creations because
the GKSL generator $\mathcal{L}$ has no creation processes. The Poincaré
transformation rules of $\hat{a}^{\dagger}(\bm{p})$ shown in Ref.Peres2004 ;
Weinberg1995

$\displaystyle\hat{T}(a)\hat{a}^{\dagger}(\bm{p})\hat{T}^{\dagger}(a)=e^{-ip^{\mu}a_{\mu}}\hat{a}^{\dagger}(\bm{p}),$
(73)
$\displaystyle\hat{V}(\Lambda)\hat{a}^{\dagger}(\bm{p})\hat{V}^{\dagger}(\Lambda)=\sqrt{\frac{E_{\bm{p}_{\Lambda}}}{E_{\bm{p}}}}\
\hat{a}^{\dagger}(\bm{p}_{\Lambda}),$ (74)

are useful to obtain the model of the massive particle, where
$E_{\bm{p}}=p^{0}$, $E_{\bm{p}_{\Lambda}}=(\Lambda p)^{0}$ and
$(\bm{p}_{\Lambda})^{i}$ is the vector with the elements written as
$(\bm{p}_{\Lambda})^{i}=(\Lambda p)^{i}$ Weinberg1995 . Also,
$W(\Lambda,p)=S^{-1}_{\Lambda p}\Lambda S_{p}$ with $(S_{p})^{\mu}_{\
\nu}k^{\nu}=p^{\mu}$ is an element of the little group and satisfies
$W^{\mu}_{\ \nu}k^{\nu}=k^{\mu}$, where $k^{\mu}$ is the standard momentum for
a massive particle ( $k^{\mu}=[m,0,0,0],m>0$ ). Substituting the ansatz of
$\hat{L}_{q,\xi}$ (71) into Eqs.(69) and (70) and using the transformation
rules (73) and (74), we obtain the equations

$\displaystyle f_{\ell,\xi}(\bm{p})e^{-ip^{\mu}a_{\mu}}$
$\displaystyle=f_{\ell,\xi}(\bm{p})e^{-i\ell^{\mu}a_{\mu}},$ (75)
$\displaystyle\sqrt{\frac{E_{\bm{p}_{Q}}}{E_{\bm{p}}}}f_{\ell,\xi}(\bm{p}_{Q})$
$\displaystyle=\sum_{\xi^{\prime}}\mathcal{D}^{*}_{\xi^{\prime}\xi}(Q^{-1})f_{\ell,\xi^{\prime}}(\bm{p}).$
(76)

In addition, we get $\alpha_{q,\xi}(\Lambda,0)=\alpha_{q,\xi}(I,a)=0$, and
then Eq.(67) leads to $\alpha_{q,\xi}(\Lambda,a)=0$. Also, we can obtain the
condition of the coefficient $g(\bm{p},\bm{p}^{\prime})$ by substituting the
ansatz of $\hat{M}-\hat{H}$ (72) into Eq.(49) as

$\sqrt{\frac{E_{\bm{p}_{\Lambda}}E_{\bm{p}^{\prime}_{\Lambda}}}{E_{\bm{p}}E_{\bm{p}^{\prime}}}}\
g(\bm{p}_{\Lambda},\bm{p}^{\prime}_{\Lambda})e^{i\left((\Lambda
p)^{\mu}-(\Lambda p^{\prime})^{\mu}\right)a_{\mu}}=g(\bm{p},\bm{p}^{\prime}).$
(77)

From the fact that there is no term proportional to the identity operator
$\hat{\mathbb{I}}$ in $\hat{M}-\hat{H}$ and the condition
$\alpha_{q,\xi}(\Lambda,a)=0$, the real number $\beta(\Lambda,a)$ vanishes.

Solving Eqs.(75), (76), and (77), we can derive the Markovian QME for a
massive spin-0 particle as follows:

$\frac{d}{dt}\rho(t)=-i\left[\hat{H}+g\hat{N},\rho(t)\right]+\gamma\int
d^{3}p\left[\hat{a}(\bm{p})\rho(t)\hat{a}^{\dagger}(\bm{p})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\rho(t)\right\\}\right],$
(78)

where a non-negative parameter $\gamma$, a real parameter $g$, and $\hat{N}$
is the number operator (37). The Markovian QME, Eq.(25) presented in Sec.IV,
is Eq.(78) for $g=0$. The detailed derivation of Eq.(78) is presented in
Appendix D.

##  VI Discussion

In this section, we discuss the previous works on relativistic Markovian QME
Poulin2017 ; Alicki1986 ; Diosi2022  in comparing it with the presented
Markovian QME. In Poulin2017 ; Alicki1986 , the following master equation of a
massive spin-0 particle is proposed:

$\frac{d}{dt}\rho(t)=\mathcal{H}[\rho(t)]+\tilde{\mathcal{D}}[\rho(t)],$ (79)

where $\mathcal{H}[\rho(t)]=-i[\hat{H},\rho(t)]$ with the Hamiltonian
$\hat{H}$ (26), and

$\tilde{\mathcal{D}}[\rho(t)]=\kappa\int
d^{3}pE_{\bm{p}}\left[\hat{a}(\bm{p})\rho(t)\hat{a}^{\dagger}(\bm{p})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\rho(t)\right\\}\right].$
(80)

The dissipation rate associated with the dissipator $\tilde{\mathcal{D}}$ is
read out as $\kappa E_{\bm{p}}$, which depends on the energy of the particle.
The master equation (79) was derived by respecting the Poincaré (or Lorentz)
covariance Poulin2017 ; Alicki1986 . From this master equation (79), we have
the field equation of evolution,

$\frac{\partial^{2}}{\partial
t^{2}}\hat{\varphi}_{\text{H}}(t,\bm{x})+\kappa\sqrt{-\nabla^{2}+m^{2}}\frac{\partial}{\partial
t}\hat{\varphi}_{\text{H}}(t,\bm{x})=\left(1+\frac{\kappa^{2}}{4}\right)(\nabla^{2}-m^{2})\hat{\varphi}_{\text{H}}(t,\bm{x}),$
(81)

where
$\hat{\varphi}_{\text{H}}(t,\bm{x})=e^{\tilde{\mathcal{L}}^{\dagger}t}[\Phi(\bm{x})]$
with the adjoint $\tilde{\mathcal{L}}^{\dagger}$ of
$\tilde{\mathcal{L}}=\mathcal{H}+\tilde{\mathcal{D}}$ and the initial
condition $\hat{\Phi}(\bm{x})$ given in (29). We also find that the field
$\hat{\varphi}_{\text{H}}(t,\bm{x})$ does not satisfy the microcausality, that
is
$[\hat{\varphi}_{\text{H}}(x^{0},\bm{x}),\hat{\varphi}_{\text{H}}(y^{0},\bm{y})]\neq
0$ even for $(x^{0}-y^{0})^{2}-(\bm{x}-\bm{y})^{2}<0$, and the violation of
microcausality typically appears in the range $1/m$ Poulin2017 ; Diosi2022 .
In contrast, our QME is based on not the Poincaré covariance but the
invariance defined by (18). This results in the dissipator $\mathcal{D}$ in
our QME, which gives the dissipation rate independent of the energy of the
particle. Furthermore, as we checked, the microcausality holds in our model.

It is interesting to highlight the difference between the two dissipators
$\mathcal{D}$ and $\tilde{\mathcal{D}}$ by following the discussion in
Ref.Diosi2022 . We pick up two critiques given in Diosi2022  on the above
master equation (79). One is that the mater equation (79) does not fully
respect the Lorentz symmetry. Concretely, this means that the dissipator
$\tilde{\mathcal{D}}$ is not invariant under the Lorentz boost. On the
contrary, our dissipator $\mathcal{D}$ in (25) is invariant under the Lorentz
boost since $\mathcal{D}=\mathcal{L}-\mathcal{H}$ satisfying (24) is invariant
under any Poincaré transformations including the Lorentz boost.

Another critique is that the field of the massive particle does not follow the
covariant equation of evolution. Indeed, it seems that not only the above
field equation (81) but also our field equation (39) are not Lorentz
covariant. However, we guess that our field is not scalar and this fact makes
its equation covariant. To clarify this argument, it may be important to note
that the field transformation in quantum theory is identified through unitary
transformations. The scalar field of a massive spin-0 particle, $\phi(x)$,
changes as $\phi(x)\rightarrow\phi^{\prime}(x^{\prime})=\phi(x)$ under Lorentz
transformations $x^{\prime\mu}=\Lambda^{\mu}{}_{\nu}x^{\nu}$. In unitary
relativistic quantum theories, such as quantum field theories, the
corresponding transformation is

$\hat{\phi}_{\text{H}}(x)\rightarrow\hat{V}^{\dagger}(\Lambda)\hat{\phi}_{\text{H}}(x^{\prime})\hat{V}(\Lambda)=\hat{\phi}_{\text{H}}(x),$
(82)

where the unitary operator $\hat{V}(\Lambda)=\hat{U}_{0}(\Lambda,0)$ defined
around (61) represents the Lorentz transformation, and
$\hat{\phi}_{\text{H}}(t,\bm{x})=e^{i\hat{H}t}\hat{\Phi}(\bm{x})e^{-i\hat{H}t}$.
On the other hand, the field operator $\hat{\Phi}_{\text{H}}$ obeying the
dissipative KG equation (39) in our model is not transformed like (82). The
field $\hat{\Phi}_{\text{H}}$ satisfies
$\hat{V}^{\dagger}(\Lambda)e^{\frac{\gamma}{2}x^{\prime
0}}\hat{\Phi}_{\text{H}}(x^{\prime})\hat{V}(\Lambda)=e^{\frac{\gamma}{2}x^{0}}\hat{\Phi}_{\text{H}}(x)$
because
$\hat{\Phi}_{\text{I}}(x)=e^{\frac{\gamma}{2}x^{0}}\hat{\Phi}_{\text{H}}(x)$
is the solution of usual free KG equation, which plays a role of Lorentz
scalar field. Hence the field transformation of $\hat{\Phi}_{\text{H}}$ is
given as

$\hat{\Phi}_{\text{H}}(x)\rightarrow\hat{V}^{\dagger}(\Lambda)\hat{\Phi}_{\text{H}}(x^{\prime})\hat{V}(\Lambda)=e^{-\gamma(x^{\prime
0}-x^{0})/2}\hat{\Phi}_{\text{H}}(x).$ (83)

This means that $\hat{\Phi}_{\text{H}}(x)$ is not Lorentz scalar field, which
is crucial to discuss the Lorentz covariance of (39). If the field follows the
transformation (83), the dissipative KG equation (39) is certainly Lorentz
covariant, which is easily observed from the fact that
$\hat{\Phi}_{\text{I}}(x)$ is just a Lorentz scalar field satisfying the free
KG equation. We think that the present discussion on the covariance of the
field equation is very speculative. We would need a clear understanding on
this concern as a future issue.

##  VII Conclusion and outlook

In this study, we discussed the Markovian QME in the GKSL form, whose solution
is given by a quantum dynamical semigroup with the Poincaré invariance. In
particular, we derived the Markovian QME of a relativistic massive spin-0
particle and investigated its field dynamics. First, it turned out that the
field of the massive particle follows the KG equation with a dissipation term.
Besides, we also showed that not only field operators but also any local
operators commute with each other when these operators are space-like
separated. This means that the present theory specified on the Markovian QME
has the microcausality property. Our formulation for the Markovian QME
respecting the Poincaré symmetry may give dissipative quantum field theories.

In this paper, we just provided a theoretical framework for yielding the
Markovian QME of relativistic particles. We expect that this framework may be
valid for relativistic dissipative phenomena, such as the decay of unstable
relativistic particles. Confirming this expectation is a future subject of
this present paper, which would deepen the understanding of our framework.

Another future subject is to build a theoretical framework including the two
standpoints mentioned in the Introducition: one is that “gravity is quantum”,
and the other is that “gravity is classical”. For this purpose, we will expand
the present theory to that can take into account gravitational interactions.
If such a framework is established, we can provide candidate theories that
describe the interplay regime of quantum and gravity phenomena. This leads to
an interesting theme of what candidates will be accepted from future quantum
gravity experiments. In particular, we believe that recent experimental
technologiesCripe2019 ; Matsumoto2020 ; Westphal2021  for testing quantum
mechanics and gravity theory would explore the consistent theory unifying
quantum and gravity physics. We hope that the present work will be a help for
the research of quantum gravity.

###### Acknowledgements.

We thank D. Carney, K. Gallock-Yoshimura, B. L. Hu, S. Kanno, Y. Kuramochi, Y.
Nambu, J. Oppenheim, and K. Yamamoto for variable discussions and comments
related to this paper. A.M. was supported by JSPS KAKENHI (Grants No.
JP23K13103 and No. JP23H01175).

##  Appendix A Proof that the GKSL generator in Eq.(25) is Poincaré invariant

We prove the Poincaré invariance of the GKSL generator $\mathcal{L}$ in
Eq.(25). Since the GKSL generator $\mathcal{L}$ has the form
$\mathcal{L}=\mathcal{H}+\mathcal{D}$, the condition of Poincaré invariance
(24) is nothing but the invariance condition for the dissipator $\mathcal{D}$,

$\hat{U}_{0}(\Lambda,a)\mathcal{D}\left[\rho(0)\right]\hat{U}^{\dagger}_{0}(\Lambda,a)=\mathcal{D}\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right],$
(84)

where $\mathcal{D}$ is yielded as

$\mathcal{D}[\rho(0)]=\gamma\int
d^{3}p\left[\hat{a}(\bm{p})\rho(0)\hat{a}^{\dagger}(\bm{p})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\rho(0)\right\\}\right].$
(85)

Substituting $\mathcal{D}[\rho(0)]$ into the left hand side of the condition
(84), the left hand side can be rewritten as follow:

$\displaystyle\hat{U}_{0}(\Lambda,a)\mathcal{D}\left[\rho(0)\right]\hat{U}^{\dagger}_{0}(\Lambda,a)$
$\displaystyle\quad=\gamma\int d^{3}p\
\hat{U}_{0}(\Lambda,a)\left[\hat{a}(\bm{p})\rho(0)\hat{a}^{\dagger}(\bm{p})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\rho(0)\right\\}\right]\hat{U}^{\dagger}_{0}(\Lambda,a)$
$\displaystyle\quad=\gamma\int
d^{3}p\left[\hat{U}_{0}(\Lambda,a)\hat{a}(\bm{p})\rho(0)\hat{a}^{\dagger}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a)-\frac{1}{2}\hat{U}_{0}(\Lambda,a)\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\rho(0)\right\\}\hat{U}^{\dagger}_{0}(\Lambda,a)\right]$
$\displaystyle\quad=\gamma\int d^{3}p\
\hat{U}_{0}(\Lambda,a)\hat{a}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a)\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right]\hat{U}_{0}(\Lambda,a)\hat{a}^{\dagger}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a)$
$\displaystyle\quad\quad-\frac{\gamma}{2}\int
d^{3}p\left\\{\hat{U}_{0}(\Lambda,a)\hat{a}^{\dagger}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{U}_{0}(\Lambda,a)\hat{a}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a),\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right\\}.$
(86)

Using the transformation rules (73) and (74), which lead to

$\hat{U}_{0}(\Lambda,a)\hat{a}^{\dagger}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a)=\sqrt{\frac{E_{\bm{p}_{\Lambda}}}{E_{\bm{p}}}}e^{-i(\Lambda
p)^{\mu}a_{\mu}}\hat{a}^{\dagger}(\bm{p}_{\Lambda}),$ (87)

we have

$\displaystyle\hat{U}_{0}(\Lambda,a)\mathcal{D}\left[\rho(0)\right]\hat{U}^{\dagger}_{0}(\Lambda,a)$
$\displaystyle\quad=\gamma\int d^{3}p\
\hat{U}_{0}(\Lambda,a)\hat{a}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a)\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right]\hat{U}_{0}(\Lambda,a)\hat{a}^{\dagger}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a)$
$\displaystyle\quad\quad-\frac{\gamma}{2}\int
d^{3}p\left\\{\hat{U}_{0}(\Lambda,a)\hat{a}^{\dagger}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{U}_{0}(\Lambda,a)\hat{a}(\bm{p})\hat{U}^{\dagger}_{0}(\Lambda,a),\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right\\}$
$\displaystyle\quad=\gamma\int d^{3}p\
\frac{E_{\bm{p}_{\Lambda}}}{E_{\bm{p}}}\left[\hat{a}(\bm{p}_{\Lambda})\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{a}^{\dagger}(\bm{p}_{\Lambda})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p}_{\Lambda})\hat{a}(\bm{p}_{\Lambda}),\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right\\}\right]$
$\displaystyle\quad=\gamma\int
d^{3}p_{\Lambda}\left[\hat{a}(\bm{p}_{\Lambda})\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\hat{a}^{\dagger}(\bm{p}_{\Lambda})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p}_{\Lambda})\hat{a}(\bm{p}_{\Lambda}),\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right\\}\right]$
$\displaystyle\quad=\mathcal{D}\left[\hat{U}_{0}(\Lambda,a)\rho(0)\hat{U}^{\dagger}_{0}(\Lambda,a)\right],$
(88)

where we used the fact that $d^{3}p/E_{\text{p}}$ is the Lorentz ionvariant
measure in the third equality. Certainly, it is confirmed that the GKSL
generator $\mathcal{L}$ of the Markovian quantum master equation (25)
satisfies the condition Eq.(24).

##  Appendix B Proof of the commutation relations Eqs.(40) and and (44)

Before the proofs of them, we mention the Lorentz invariance of the measure
$d^{3}p/2E_{\bm{p}}$. This measure can be rewirtten as follow:

$\frac{d^{3}p}{2E_{\bm{p}}}=d^{4}p\
\delta(p_{\mu}p^{\mu}+m^{2})\theta(p^{0}),$ (89)

where $\theta(p^{0})$ is step function defined as

$\theta(p^{0})=\left\\{\,\begin{aligned} 1\ (p^{0}\geq 0)\\\ 0\
(p^{0}<0)\end{aligned}\right.,$ (90)

and $\eta_{\mu\nu}$ is the Minkowski metric tensor. Under the Lorentz
transformation $p^{\mu}\rightarrow
p^{\prime\mu}=\Lambda^{\mu}{}_{\nu}p^{\nu}$, the right hand side of (89) is

$\displaystyle
d^{4}p^{\prime}\,\delta(p^{\prime}_{\mu}p^{\prime\mu}+m^{2})\theta(p^{\prime
0})$ $\displaystyle=d^{4}p\ \delta(p_{\mu}p^{\mu}+m^{2})\theta(p^{\prime 0}),$
(91)

where the matrix $\Lambda^{\mu}_{\ \nu}$ is a proper orthochronous Lorentz
transformation matrix. If $p^{\prime 0}$ is positive, the step function is
Lorentz invariant too. Looking the delta function, the integration have a
meaningful value when $p^{\rho}p_{\rho}+m^{2}=0$. The condition
$p^{\rho}p_{\rho}+m^{2}=0$ means that $p^{\rho}$ is time-like vector. Since we
consider a proper orthochronous Lorentz transformation,
$\theta(p^{0})=\theta(p^{\prime 0})=1$ holds. Thus, we obtain the fact that
the measure $d^{3}p/2E_{\bm{p}}$ is Lorentz invariant.

We first prove Eq.(40). The commutator
$\left[\hat{\Phi}_{\text{H}}(x^{0},\bm{x}),\hat{\Phi}_{\text{H}}(y^{0},\bm{y})\right]$
is computed as

$\displaystyle\left[\hat{\Phi}_{\text{H}}(x^{0},\bm{x}),\hat{\Phi}_{\text{H}}(y^{0},\bm{y})\right]$
$\displaystyle=e^{-\frac{\gamma}{2}(x^{0}+y^{0})}\left[\hat{\Phi}_{\text{I}}(x^{0},\bm{x}),\hat{\Phi}_{\text{I}}(y^{0},\bm{y})\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int\frac{d^{3}p}{\sqrt{2E_{\bm{p}}}}\int\frac{d^{3}p^{\prime}}{\sqrt{2E_{\bm{p}^{\prime}}}}$
$\displaystyle\quad\times\left[e^{i(p^{\mu}x_{\mu}-p^{\prime\mu}y_{\mu})}\left[\hat{a}(\bm{p}),\hat{a}^{\dagger}(\bm{p}^{\prime})\right]-e^{-i(p^{\mu}x_{\mu}-p^{\prime\mu}y_{\mu})}\left[\hat{a}(\bm{p}),\hat{a}^{\dagger}(\bm{p}^{\prime})\right]\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int\frac{d^{3}p}{2E_{\bm{p}}}\left[e^{ip^{\mu}(x_{\mu}-y_{\mu})}-e^{-ip^{\mu}(x_{\mu}-y_{\mu})}\right],$
(92)

where we used the notations $p^{\mu}=[E_{\bm{p}},\bm{p}]$,
$x_{\mu}=[-x^{0},\bm{x}]$, $y_{\mu}=[-y^{0},\bm{y}]$, and Eqs.(28). For any
spacelike distance $(x^{0}-y^{0})^{2}-(\bm{x}-\bm{y})^{2}<0$, we can introduce
$x^{\prime}_{\rho}-y^{\prime}_{\rho}=\Lambda^{\mu}_{\ \rho}(x_{\mu}-y_{\mu})$
and choose a proper orthochronous Lorentz transformation $\Lambda$ such that
$x^{\prime}_{0}-y^{\prime}_{0}=0$. Thus, we obtain the following result

$\displaystyle\left[\hat{\Phi}_{\text{H}}(x^{0},\bm{x}),\hat{\Phi}_{\text{H}}(y^{0},\bm{y})\right]$
$\displaystyle=\frac{1}{(2\pi)^{3}}e^{-\frac{\gamma}{2}(x^{0}+y^{0})}\int\frac{d^{3}p}{2E_{\bm{p}}}\left[e^{i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}-e^{-i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}\right]$
$\displaystyle=\frac{1}{(2\pi)^{3}}e^{-\frac{\gamma}{2}(x^{0}+y^{0})}\int\frac{d^{3}p}{2E_{\bm{p}}}e^{i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}-\frac{1}{(2\pi)^{3}}e^{-\frac{\gamma}{2}(x^{0}+y^{0})}\int\frac{d^{3}\tilde{p}}{2E_{\bm{\tilde{p}}}}e^{i\bm{\tilde{p}}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}$
$\displaystyle=0,$ (93)

where $\bm{\tilde{p}}\equiv-\bm{p}$ in the second line.

Next, we will prove the Eqs.(44), respectively. In the same manner performed
to prove Eq.(40), we just use the Lorentz invariance of the measure
$d^{3}p/2E_{\bm{p}}$ and choose the Lorentz transformation $\Lambda$ such that
$x^{\prime}_{0}-y^{\prime}_{0}=0$ for
$(x^{0}-y^{0})^{2}-(\bm{x}-\bm{y})^{2}<0$. Therefore, the following results
will be obtained

$\displaystyle\left[\hat{\Pi}_{\text{H}}(x^{0},\bm{x}),\hat{\Pi}_{\text{H}}(y^{0},\bm{y})\right]$
$\displaystyle=e^{-\frac{\gamma}{2}(x^{0}+y^{0})}\left[\hat{\Pi}_{\text{I}}(x^{0},\bm{x}),\hat{\Pi}_{\text{I}}(y^{0},\bm{y})\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int\frac{d^{3}p}{2E_{\bm{p}}}\cdot
E^{2}_{\bm{p}}\left[e^{ip^{\mu}(x_{\mu}-y_{\mu})}-e^{-ip^{\mu}(x_{\mu}-y_{\mu})}\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int
d^{4}p\ \delta(p_{\mu}p^{\mu}+m^{2})\theta(p^{0})\
(p^{0})^{2}\left[e^{ip^{\mu}(x_{\mu}-y_{\mu})}-e^{-ip^{\mu}(x_{\mu}-y_{\mu})}\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int
d^{4}p^{\prime}\ \delta(p^{\prime}_{\mu}p^{\prime\mu}+m^{2})\theta(p^{\prime
0})(p^{\prime
0})^{2}\left[e^{ip^{\prime\mu}(x_{\mu}-y_{\mu})}-e^{-ip^{\prime\mu}(x_{\mu}-y_{\mu})}\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int
d^{4}p\ \delta(p_{\mu}p^{\mu}+m^{2})\theta(p^{0})\ (\Lambda^{0}_{\
\nu}p^{\nu})^{2}\left[e^{ip^{\mu}(x^{\prime}_{\mu}-y^{\prime}_{\mu})}-e^{-ip^{\mu}(x^{\prime}_{\mu}-y^{\prime}_{\mu})}\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int\frac{d^{3}p}{2E_{\bm{p}}}(\Lambda^{0}{}_{0}E_{\bm{p}}+\Lambda^{0}{}_{j}p^{j})^{2}\left[e^{i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}-e^{-i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int\frac{d^{3}p}{2E_{\bm{p}}}\cdot
2\Lambda^{0}_{\ 0}\Lambda^{0}_{\
j}E_{\bm{p}}p^{j}\left[e^{i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}-e^{-i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}\right]$
$\displaystyle=\frac{e^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\Lambda^{0}_{\
0}\Lambda^{0}_{\ j}\int d^{3}p\ (-i)\frac{\partial}{\partial
x^{\prime}_{j}}\left[e^{i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}+e^{-i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}\right]$
$\displaystyle=-2ie^{-\frac{\gamma}{2}(x^{0}+y^{0})}\ \Lambda^{0}_{\
0}\Lambda^{0}_{\ j}\frac{\partial}{\partial
x^{\prime}_{j}}\delta^{3}(\bm{x}^{\prime}-\bm{y}^{\prime})=0,$ (94)

where we used the fact that the vectors $\bm{x}^{\prime}$ and
$\bm{y}^{\prime}$ are space-like separated since the vectors $x^{\mu}$ and
$y^{\mu}$ are space-like separated. We further show that

$\displaystyle\left[\hat{\Phi}_{\text{H}}(x^{0},\bm{x}),\hat{\Pi}_{\text{H}}(y^{0},\bm{y})\right]$
$\displaystyle=e^{-\frac{\gamma}{2}(x^{0}+y^{0})}\left[\hat{\Phi}_{\text{I}}(x^{0},\bm{x}),\hat{\Pi}_{\text{I}}(y^{0},\bm{y})\right]$
$\displaystyle=\frac{ie^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int\frac{d^{3}p}{2E_{\bm{p}}}\,E_{\bm{p}}\left[e^{ip^{\mu}(x_{\mu}-y_{\mu})}+e^{-ip^{\mu}(x_{\mu}-y_{\mu})}\right]$
$\displaystyle=\frac{ie^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int
d^{4}p\ \delta(p_{\mu}p^{\mu}+m^{2})\theta(p^{0})\
p^{0}\left[e^{ip^{\mu}(x_{\mu}-y_{\mu})}+e^{-ip^{\mu}(x_{\mu}-y_{\mu})}\right]$
$\displaystyle=\frac{ie^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int
d^{4}p\ \delta(p_{\mu}p^{\mu}+m^{2})\theta(p^{0})\ \Lambda^{0}_{\
\nu}p^{\nu}\left[e^{ip^{\rho}(x^{\prime}_{\rho}-y^{\prime}_{\rho})}+e^{-ip^{\rho}(x^{\prime}_{\rho}-y^{\prime}_{\rho})}\right]$
$\displaystyle=\frac{ie^{-\frac{\gamma}{2}(x^{0}+y^{0})}}{(2\pi)^{3}}\int\frac{d^{3}p}{2E_{\bm{p}}}\Lambda^{0}_{\
0}E_{\bm{p}}\left[e^{i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}+e^{-i\bm{p}\cdot(\bm{x}^{\prime}-\bm{y}^{\prime})}\right]$
$\displaystyle=i\Lambda^{0}_{\ 0}\
e^{-\frac{\gamma}{2}(x^{0}+y^{0})}\delta^{3}(\bm{x}^{\prime}-\bm{y}^{\prime})=0,$
(95)

where we used the fact that the vectors $x^{\mu}$ and $y^{\mu}$ are space-like
separated. We conclude the proofs of the three commutation relations.

##  Appendix C Time evolution of the product of operators

In Sec.IV, we stated that any operators commutate if they are space-like
separated under the model of the master equation derived in this study. We
prove this statement in this section. To this end, we have to find the time
evolution of the product
$\hat{\xi}^{i^{\prime}_{1}}(\bm{x}^{\prime}_{1})\cdots\hat{\xi}^{i^{\prime}_{N}}(\bm{x}^{\prime}_{N})\hat{\xi}^{i_{1}}(\bm{x}_{1})\cdots\hat{\xi}^{i_{M}}(\bm{x}_{M})$.
With reference Eqs.(33) and (34), the following relation is satisfied:

$e^{\mathcal{L}^{\dagger}t}\left[\hat{\xi}^{i^{\prime}_{1}}(\bm{x}^{\prime}_{1})\cdots\hat{\xi}^{i^{\prime}_{N}}(\bm{x}^{\prime}_{N})\hat{\xi}^{i_{1}}(\bm{x}_{1})\cdots\hat{\xi}^{i_{M}}(\bm{x}_{M})\right]=e^{\mathcal{D}^{\dagger}t}\left[\hat{\xi}_{\mathrm{I}}^{i^{\prime}_{1}}(t,\bm{x}^{\prime}_{1})\cdots\hat{\xi}_{\mathrm{I}}^{i^{\prime}_{N}}(t,\bm{x}^{\prime}_{N})\hat{\xi}_{\mathrm{I}}^{i_{1}}(t,\bm{x}_{1})\cdots\hat{\xi}_{\mathrm{I}}^{i_{M}}(t,\bm{x}_{M})\right]$
(96)

If the action $(\mathcal{D}^{\dagger})^{k}$ can be computed, the right hand
side of (96) is obtained. We introduce the normal ordered product and use the
Wick’s theorem to calculate the right hand side of (96).

The operator $\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})$ is decomposed into the
term which contains annihilation operators $\hat{\zeta}^{i}_{+}(t,\bm{x})$ and
the term which contains creation operators $\hat{\zeta}^{i}_{-}(t,\bm{x})$:

$\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})=\hat{\zeta}^{i}_{+}(t,\bm{x})+\hat{\zeta}^{i}_{-}(t,\bm{x}),$
(97)

where $\hat{\zeta}^{i}_{+}(t,\bm{x}_{i})$ and
$\hat{\zeta}^{i}_{-}(t,\bm{x}_{i})$ are called the positive frequency term and
the negative frequency term, respectively. For two field operators
$\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})$ and
$\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})$, the following product
$\mathcal{N}[\hat{\xi}^{i}_{\text{I}}(t,\bm{x})\hat{\xi}^{j}_{\text{I}}(t,\bm{y})]$
is called the normal ordered product,

$\mathcal{N}\left[\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})\right]=\hat{\zeta}^{i}_{+}(t,\bm{x})\hat{\zeta}^{j}_{+}(t,\bm{y})+\hat{\zeta}^{j}_{-}(t,\bm{y})\hat{\zeta}^{i}_{+}(t,\bm{x})+\hat{\zeta}^{i}_{-}(t,\bm{x})\hat{\zeta}^{j}_{+}(t,\bm{y})+\hat{\zeta}^{i}_{-}(t,\bm{x})\hat{\zeta}^{j}_{-}(t,\bm{y}).$
(98)

The convenient point of the normal ordered product is that the calculation of
$\mathcal{D}^{\dagger}$ is simple. The normal ordered product
$\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$
of the $n$ fields is

$\displaystyle\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$
$\displaystyle\quad=\hat{\zeta}^{i_{1}}_{+}(t,\bm{x}_{1})\hat{\zeta}^{i_{2}}_{+}(t,\bm{x}_{2})\cdots\hat{\zeta}^{i_{n}}_{+}(t,\bm{x}_{n})$
$\displaystyle\quad\quad+\hat{\zeta}^{i_{1}}_{-}(t,\bm{x}_{1})\hat{\zeta}^{i_{2}}_{+}(t,\bm{x}_{2})\hat{\zeta}^{i_{3}}_{+}(t,\bm{x}_{3})\cdots\hat{\zeta}^{i_{n}}_{+}(t,\bm{x}_{n})+\hat{\zeta}^{i_{2}}_{-}(t,\bm{x}_{2})\hat{\zeta}^{i_{1}}_{+}(t,\bm{x}_{1})\hat{\zeta}^{i_{3}}_{+}(t,\bm{x}_{3})\cdots\hat{\zeta}^{i_{n}}_{+}(t,\bm{x}_{n})+\cdots$
$\displaystyle\quad\quad+\hat{\zeta}^{i_{1}}_{-}(t,\bm{x}_{1})\hat{\zeta}^{i_{2}}_{-}(t,\bm{x}_{2})\hat{\zeta}^{i_{3}}_{+}(t,\bm{x}_{3})\cdots\hat{\zeta}^{i_{n}}_{+}(t,\bm{x}_{n})+\hat{\zeta}^{i_{1}}_{-}(t,\bm{x}_{1})\hat{\zeta}^{i_{3}}_{-}(t,\bm{x}_{3})\hat{\zeta}^{i_{2}}_{+}(t,\bm{x}_{2})\cdots\hat{\zeta}^{i_{n}}_{+}(t,\bm{x}_{n})+\cdots$
$\displaystyle\hskip 170.71652pt\vdots$
$\displaystyle\quad\quad+\hat{\zeta}^{i_{1}}_{-}(t,\bm{x}_{1})\cdots\hat{\zeta}^{i_{n-1}}_{-}(t,\bm{x}_{n-1})\hat{\zeta}^{i_{n}}_{+}(t,\bm{x}_{n})+\hat{\zeta}^{i_{1}}_{-}(t,\bm{x}_{1})\cdots\hat{\zeta}^{i_{n}}_{-}(t,\bm{x}_{n})\hat{\zeta}^{i_{n-1}}_{+}(t,\bm{x}_{n-1})+\cdots$
$\displaystyle\quad\quad+\hat{\zeta}^{i_{1}}_{-}(t,\bm{x}_{1})\hat{\zeta}^{i_{2}}_{-}(t,\bm{x}_{2})\cdots\hat{\zeta}^{i_{n}}_{-}(t,\bm{x}_{n}).$
(99)

To make $\mathcal{D}^{\dagger}$ work for
$\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$,
we can use the following relation:If the product with $k$ pieces of
$\hat{\zeta}_{-}$ lined up on the left, the action of $\mathcal{D}^{\dagger}$
on that is

$\mathcal{D}^{\dagger}[\underbrace{\hat{\zeta}_{-}\cdots\hat{\zeta}_{-}}_{k\
\mathrm{pieces}}\underbrace{\hat{\zeta}_{+}\cdots\hat{\zeta}_{+}}_{n-k\
\mathrm{pieces}}]=-\frac{n}{2}\gamma\underbrace{\hat{\zeta}_{-}\cdots\hat{\zeta}_{-}}_{k\
\mathrm{pieces}}\underbrace{\hat{\zeta}_{+}\cdots\hat{\zeta}_{+}}_{n-k\
\mathrm{pieces}}.$ (100)

Eq.(100) holds for any $k$, so
$\mathcal{D}^{\dagger}[\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]]$
is computed as

$\mathcal{D}^{\dagger}\left[\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]\right]=-\frac{n}{2}\gamma\
\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right].$
(101)

Since what we originally want to know is
$\mathcal{D}^{\dagger}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$,
the relation between
$\mathcal{D}^{\dagger}[\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]]$
and
$\mathcal{D}^{\dagger}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$
is needed. The relation is given by Wick’s theorem as follows: the product
$\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})$
is represented by normal ordered product,

$\displaystyle\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})$
$\displaystyle=\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$
$\displaystyle\quad+\sum_{1-\mathrm{pair}}\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction[1ex]{}{\cdots}{}{\cdots}\cdots\cdots\cdots\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$
$\displaystyle\quad+\sum_{2-\mathrm{paris}}\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction{}{\cdots}{\cdots}{\cdots}\contraction[2ex]{\cdots}{\cdots}{\cdots}{\cdots}\cdots\cdots\cdots\cdots\cdots\
\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$
$\displaystyle\quad+(3-\mathrm{pairs\ or\ more\ terms})\cdots,$ (102)

where $\contraction{}{\cdots}{}{\cdots}\cdots\cdots$ is called the Wick
contraction and is given as the expectation value taken in a vacuum for the
product of two operators connected by the line. Considering, for example, the
case of the product
$\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})$,
the Wick’s theorem gives

$\displaystyle\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})$
$\displaystyle=\mathcal{N}\left[\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})\right]+\contraction{}{\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})}{}{\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})}\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})$
$\displaystyle=\mathcal{N}\left[\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})\right]+\langle
0|\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})|0\rangle.$
(103)

Since $\mathcal{D}^{\dagger}$ is a linear superoperator, using (101) and
(102),
$\mathcal{D}^{\dagger}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]$
is computed as

$\displaystyle\mathcal{D}^{\dagger}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle=\mathcal{D}^{\dagger}\left[\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]\right]$
$\displaystyle+\sum_{1-\mathrm{pair}}\mathcal{D}^{\dagger}\left[\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction[1ex]{}{\cdots}{}{\cdots}\cdots\cdots\cdots\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]\right]$
$\displaystyle+\sum_{2-\mathrm{pairs}}\mathcal{D}^{\dagger}\left[\mathcal{N}[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction{}{\cdots}{\cdots}{\cdots}\contraction[2ex]{\cdots}{\cdots}{\cdots}{\cdots}\cdots\cdots\cdots\cdots\cdots\
\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})]\right]$
$\displaystyle+(3-\mathrm{pairs\ or\ more\ terms})$
$\displaystyle=-\frac{n}{2}\gamma\
\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle-\sum_{1-\mathrm{pair}}\frac{n-2}{2}\gamma\
\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction[1ex]{}{\cdots}{}{\cdots}\cdots\cdots\cdots\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle-\sum_{2-\mathrm{pairs}}\frac{n-4}{2}\gamma\
\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction{}{\cdots}{\cdots}{\cdots}\contraction[2ex]{\cdots}{\cdots}{\cdots}{\cdots}\cdots\cdots\cdots\cdots\cdots\
\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle+(3-\mathrm{pairs\ or\ more\ terms}).$ (104)

Furthermore, $(\mathcal{D}^{\dagger})^{k}$ operates as,

$\displaystyle(\mathcal{D}^{\dagger})^{k}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad=\left(-\frac{n}{2}\gamma\right)^{k}\
\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+\sum_{1-\mathrm{pair}}\left(-\frac{n-2}{2}\gamma\right)^{k}\
\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction[1ex]{}{\cdots}{}{\cdots}\cdots\cdots\cdots\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+\sum_{2-\mathrm{pairs}}\left(-\frac{n-4}{2}\gamma\right)^{k}\
\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction{}{\cdots}{\cdots}{\cdots}\contraction[2ex]{\cdots}{\cdots}{\cdots}{\cdots}\cdots\cdots\cdots\cdots\cdots\
\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+(3-\mathrm{pairs\ or\ more\ terms}).$ (105)

Therefore, we can calculate the Heisenberg operator
$e^{\mathcal{L}^{\dagger}t}[\hat{\xi}^{i_{1}}(\bm{x}_{1})\hat{\xi}^{i_{2}}(\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}(\bm{x}_{n})]$
because of Eqs.(96) and (105):

$\displaystyle
e^{\mathcal{L}^{\dagger}t}\left[\hat{\xi}^{i_{1}}(\bm{x}_{1})\hat{\xi}^{i_{2}}(\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}(\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad=\exp(-\frac{n}{2}\gamma t)\
\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+\exp(-\frac{n-2}{2}\gamma t)\
\sum_{1-\mathrm{pair}}\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction[1ex]{}{\cdots}{}{\cdots}\cdots\cdots\cdots\cdots\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+\exp(-\frac{n-4}{2}\gamma t)\
\sum_{2-\mathrm{pair}}\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{I}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{I}}(t,\bm{x}_{2})\contraction{}{\cdots}{\cdots}{\cdots}\contraction[2ex]{\cdots}{\cdots}{\cdots}{\cdots}\cdots\cdots\cdots\cdots\cdots\
\hat{\xi}^{i_{n}}_{\mathrm{I}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+(3-\mathrm{pairs\ or\ more\ terms})$
$\displaystyle\qquad\qquad=\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{H}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{H}}(t,\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}_{\mathrm{H}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+\exp(\gamma
t)\sum_{1-\mathrm{pair}}\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{H}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{H}}(t,\bm{x}_{2})\contraction[1ex]{}{\cdots}{}{\cdots}\cdots\cdots\cdots\cdots\hat{\xi}^{i_{n}}_{\mathrm{H}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+\exp(2\gamma
t)\sum_{2-\mathrm{pair}}\mathcal{N}\left[\hat{\xi}^{i_{1}}_{\mathrm{H}}(t,\bm{x}_{1})\hat{\xi}^{i_{2}}_{\mathrm{H}}(t,\bm{x}_{2})\contraction{}{\cdots}{\cdots}{\cdots}\contraction[2ex]{\cdots}{\cdots}{\cdots}{\cdots}\cdots\cdots\cdots\cdots\cdots\hat{\xi}^{i_{n}}_{\mathrm{H}}(t,\bm{x}_{n})\right]$
$\displaystyle\qquad\qquad+(3-\mathrm{pairs\ or\ more\ terms}),$ (106)

where we used the relation Eq.(38) in the last equality. The factors, such as
$\exp(\gamma t)$ and $\exp(2\gamma t)$, appear since the Wick contraction of
$\hat{\xi}^{i}_{\mathrm{I}}$ connects with that of
$\hat{\xi}^{i}_{\mathrm{H}}$ as

$\contraction{}{\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})}{}{\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})}\hat{\xi}^{i}_{\mathrm{I}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})=\exp(\gamma
t)\contraction{}{\hat{\xi}^{i}_{\mathrm{H}}(t,\bm{x})}{}{\hat{\xi}^{j}_{\mathrm{I}}(t,\bm{y})}\hat{\xi}^{i}_{\mathrm{H}}(t,\bm{x})\hat{\xi}^{j}_{\mathrm{H}}(t,\bm{y}).$
(107)

The result (106) means that the time evolution
$e^{\mathcal{L}^{\dagger}t}[\hat{\xi}^{i_{1}}(\bm{x}_{1})\hat{\xi}^{i_{2}}(\bm{x}_{2})\cdots\hat{\xi}^{i_{n}}(\bm{x}_{n})]$
is written with the combinations of the products of the Heisenberg operators
chosen from
$\hat{\xi}^{i_{1}}_{\mathrm{H}}(t,\bm{x}_{1}),....,\hat{\xi}^{i_{n}}_{\mathrm{H}}(t,\bm{x}_{n})$
by repeatedly using the Wick’s theorem.

The above result implies that the commutation relation of the arbitrary two
operators can be calculated from the commutation relations
$[\hat{\xi}^{i_{k}}_{\mathrm{H}},\hat{\xi}^{j_{l}}_{\mathrm{H}}]$. Therefore,
supposing that $\hat{A}$ is a operator defined for a spatial region A at time
$t=0$ and $\hat{B}$ is a operator defined for a spatial region B at time
$t^{\prime}=0$, it turns out that the time-evolved operators
$\hat{A}_{\mathrm{H}}(t)$ and $\hat{B}_{\mathrm{H}}(t^{\prime})$ commutes if
they are spacelike separated.

##  Appendix D Derivation of the Markovian quantum master equation Eq.(78)

We assume that the massive spectrum of $\hat{P}^{\mu}$ satisfies

$\hat{P}^{\mu}\hat{P}_{\mu}=-m^{2},\ \hat{P}^{0}>0.$ (108)

The equation (108) leads to the Hamiltonian $\hat{H}=\hat{P}^{0}$ given as
$\hat{H}=\sqrt{\hat{P}^{k}\hat{P}_{k}+m^{2}}$. In this appendix, thinking
about the cases in Table 1, we derive the Lindblad operator $\hat{L}_{q,\xi}$
and the self-adjoint operator $\hat{M}$, which give the Markovian QME Eq.(25).

Lindblad operator for the case $\,\ell^{\mu}=[\pm M,0,0,0],\,M>0$:
Substituting $a^{\mu}=(0,\bm{a})$ into Eq.(75), this equation leads to

$f_{\ell,\xi}(\bm{p})e^{-i\bm{p}\cdot\bm{a}}=f_{\ell,\xi}(\bm{p})\,\therefore\,f_{\ell,\xi}(\bm{p})=f_{\ell,\xi}\delta^{3}(\bm{p}).$
(109)

Eq.(109) can be checked by integrating both sides. Next, we think (75) in the
case of $a^{\mu}=(a,\bm{0})$, which leads to

$f_{\ell,\xi}(\bm{p})e^{iE_{\bm{p}}a}=f_{\ell,\xi}(\bm{p})e^{\pm iMa}.$ (110)

This equation has a meaning only when $\bm{p}=\bm{0}$ because of (109). So, we
eventually get

$f_{\ell,\xi}e^{ima}=f_{\ell,\xi}e^{\pm iMa}.$ (111)

Since the mass $m$ is positive, to get a nontrivial solution as
$f_{\ell,\xi}\neq 0$, we should choose $+M$ with $M=m$. The above analysis
implies that the Lindblad operator $\hat{L}_{\ell,\xi}$ with
$\ell^{\mu}=[m,0,0,0]$ has

$\hat{L}_{\ell,\xi}=\int d^{3}p\
f_{\ell,\xi}\hat{a}(\bm{p})\delta^{3}(\bm{p})=f_{\ell,\xi}\hat{a}(\bm{0})$
(112)

Eq.(68) tells us that

$\hat{L}_{q,\xi}=N^{*}_{q}\hat{V}(S_{q})\hat{L}_{\ell,\xi}\hat{V}^{\dagger}(S_{q})=N^{\ast}_{q}f_{\ell,\xi}\sqrt{\frac{E_{\bm{q}}}{m}}\hat{a}(\bm{q})=f_{\ell,\xi}\hat{a}(\bm{q}),$
(113)

where $E_{\bm{q}}=(S_{q}\ell)^{0}$, $q^{i}=(S_{q}\ell)^{i}$ and
$N_{q}=\sqrt{m/E_{\bm{q}}}$. The normalization $N_{q}$ is given by setting the
inner product $\bm{v}^{\dagger}_{q^{\prime},\xi^{\prime}}\bm{v}_{q,\xi}$ as

$\bm{v}^{\dagger}_{q^{\prime},\xi^{\prime}}\bm{v}_{q,\xi}=\delta^{3}(\bm{q}^{\prime}-\bm{q})\delta_{\xi^{\prime}\xi}.$
(114)

Under this inner product, we can get the following completeness condition,

$\int d^{3}q\sum_{\xi}\bm{v}_{q,\xi}\bm{v}^{\dagger}_{q,\xi}=I.$ (115)

We then derive a part of $\mathcal{D}[\rho]$ as

$\mathcal{D}[\rho]\supset\sum_{\xi}|f_{\ell,\xi}|^{2}\int
d^{3}p\left[\hat{a}(\bm{p})\rho(t)\hat{a}^{\dagger}(\bm{p})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\rho(t)\right\\}\right].$
(116)

Lindblad operator for the case
$\,\ell^{\mu}=[\pm\kappa,0,0,\kappa],\,\kappa>0$: Substituting
$a^{\mu}=[0,\bm{a}]$ into Eq.(75), we have

$f_{\ell,\xi}(\bm{p})e^{-i\bm{p}\cdot\bm{a}}=f_{\ell,\xi}(\bm{p})e^{-i\bm{\ell}\cdot\bm{a}}\
\therefore f_{\ell,\xi}(\bm{p})=f_{\ell,\xi}\delta^{3}(\bm{p}-\bm{\ell}),$
(117)

where $\bm{\ell}=[0,0,\kappa]^{\mathrm{T}}$. Also, substituting
$a^{\mu}=[a,0,0,0]$ into Eq.(75), we obtain the following result:

$f_{\ell,\xi}(\bm{p})e^{iE_{\bm{p}}a}=f_{\ell,\xi}(\bm{p})e^{\pm i\kappa a}.$
(118)

Because of $f_{\ell,\xi}(\bm{p})=f_{\ell,\xi}\delta^{3}(\bm{p}-\bm{\ell})$ ,
we get

$f_{\ell,\xi}e^{i\sqrt{\kappa^{2}+m^{2}}a}=f_{\ell,\xi}e^{\pm i\kappa a}\
\therefore f_{\ell,\xi}=0,$ (119)

where $E_{\bm{\ell}}=\sqrt{\bm{\ell}^{2}+m^{2}}=\sqrt{\kappa^{2}+m^{2}}$, and
hence $f_{\ell,\xi}(\bm{p})=0$. Combined with the above analysis, the Lindblad
operator $\hat{L}_{\ell,\xi}$ vanishies, and then

$\hat{L}_{q,\xi}=N^{\ast}_{q}\hat{V}(S_{q})\hat{L}_{\ell,\xi}\hat{V}^{\dagger}(S_{q})=0.$
(120)

Lindblad operator for the case $\ell^{\mu}=[0,0,0,N]$, $N^{2}>0$: Eq.(75) for
all $a^{\mu}=[a,0,0,0]$ leads to

$f_{\ell,\xi}(\bm{p})e^{iE_{\bm{p}}a}=f_{\ell,\xi}(\bm{p})\ \therefore
f_{\ell,\xi}(\bm{p})=0,$ (121)

where note that $E_{\bm{p}}=\sqrt{\bm{p}^{2}+m^{2}}\neq 0$. For this case, the
Lindblad operator $\hat{L}_{\ell,\xi}$ vanishes, and we have

$\hat{L}_{q,\xi}=N^{\ast}_{q}\hat{V}(S_{q})\hat{L}_{\ell,\xi}\hat{V}^{\dagger}(S_{q})=0.$
(122)

Lindblad operator for the case $\ell^{\mu}=[0,0,0,0]$: For this case, Eq.(75)
for all $a^{\mu}=[a,0,0,0]$ is written as

$f_{\xi}(\bm{p})e^{iE_{\bm{p}}a}=f_{\xi}(\bm{p})\ \therefore
f_{\xi}(\bm{p})=0,$ (123)

where we dropped the label $\ell$ and note that
$E_{\bm{p}}=\sqrt{\bm{p}^{2}+m^{2}}\neq 0$. Hence, the Lindblad operator
vanishes:

$\hat{L}_{\xi}=0.$ (124)

Self-adjoint operator $\hat{M}$: Now that we can get the form of
$\mathcal{D}[\rho]$, let us move to the analysis the self-adjoint operator
$\hat{M}-\hat{H}$. Adopting $\Lambda=I$ and $a^{\mu}=[0,\bm{a}]$ in Eq.(77),
we can obtain

$g(\bm{p},\bm{p}^{\prime})e^{i(\bm{p}-\bm{p}^{\prime})\cdot\bm{a}}=g(\bm{p},\bm{p}^{\prime})\
\therefore\
g(\bm{p},\bm{p}^{\prime})=g(\bm{p})\delta^{3}(\bm{p}-\bm{p}^{\prime}).$ (125)

This result can be checked by integrating the both sides. Also, adopting
$a^{\mu}=0$ and substituting into Eq.(77), we have

$g(\bm{p}_{\Lambda})=g(\bm{p}),$ (126)

where we used
$E_{\bm{p}}\delta^{3}(\bm{p}-\bm{p}^{\prime})=E_{\bm{p}_{\Lambda}}\delta^{3}(\bm{p}_{\Lambda}-\bm{p}^{\prime}_{\Lambda})$.
For $\bm{p}=\bm{0}$ and $\Lambda=S_{q}$ with $(S_{q})^{\mu}_{\
\nu}k^{\nu}=q^{\mu}$ for $k^{\mu}=[m,0,0,0]$ in Eq.(126), we obtain

$g(\bm{q})=g(\bm{0}).$ (127)

By defining $g(\bm{0})$ as $g$, the self-adjoint operator $\hat{M}$ is given
as

$\hat{M}=\hat{H}+g\hat{N},$ (128)

where $\hat{N}$ is the number operator (37).

Summary: Combining the above results (116), (120), (122), (124) and (128), we
get the following form of the Markovian QME,

$\frac{d}{dt}\rho(t)=-i\left[\hat{H}+g\hat{N},\rho(t)\right]+\gamma\int
d^{3}p\left[\hat{a}(\bm{p})\rho(t)\hat{a}^{\dagger}(\bm{p})-\frac{1}{2}\left\\{\hat{a}^{\dagger}(\bm{p})\hat{a}(\bm{p}),\rho(t)\right\\}\right],$
(129)

where we defined $\gamma$ as $\gamma=\sum_{\xi}|f_{\ell,\xi}|^{2}$.

## References

  * (1) H.-P. Breuer and F. Petruccione, “The Theory of Open Quantum Systems” (Oxford University Press, New York, 2002).
  * (2) E. B. Davies, “Quantum Theory of Open Systems” (Academic, New York, 1976).
  * (3) V. Gorini, A. Kossakowski, and E. C. G. Sudarshan, “Completely Positive Dynamical Semigroup of n-Level System”, J. Math. Phys. (N.Y.) 17, 821 (1976).
  * (4) G. Lindblad, “On the Generators of Quantum Dynamical Semigroups”, Commun. Math. Phys. 48, 119 (1976).
  * (5) C.W. Gardiner and P. Zoller, “Quantum Noise”, (Springer, Berlin, 2000).
  * (6) M. A. Nielsen and I. Chuang, “Quantum Computation and Quantum Information” (Cambridge University Press, Cambridge, England, 2002).
  * (7) B. Kraus, H. P. Büchler, S. Diehl, A. Kantian, A. Micheli, and P. Zoller, “Preparation of entangled states by quantum markov processes”, Phys. Rev. A 78, 042307(2008).
  * (8) H. M. Wiseman and G. J. Milburn, “Quantum Measurement and Control” (Cambridge University Press, Cambridge, England, 2010)
  * (9) J. F. Donoghue, “General relativity as an effective field theory: The leading quantum corrections,” Phys. Rev. D 50 (1994) 3874-3888.
  * (10) D. Kafri, J. M. Taylor, and G. J. Milburn, “A classical channel model for gravitational decoherence”, New J. Phys. 16, 065020 (2014).
  * (11) L. Diósi, “A universal master equation for the gravitational violation of quantum mechanics”, Phys. Lett. A 120 377–381 (1987)
  * (12) L. Diósi, “Models for universal reduction of macroscopic quantum fluctuations”, Phys. Rev. A 40, 1165 (1989).
  * (13) R. Penrose, “On Gravity’s role in Quantum State Reduction”, Gen. Relativ. and Gravt. 28, 581–600 (1996).
  * (14) A. Tilloy and L. Diósi, “Sourcing semiclassical gravity from spontaneously localized quantum matter”, Phys. Rev. D 93, 024026 (2016).
  * (15) A. Baidya, C. Jana, R. Loganayagam, and A. Rudra, “Renormalization in open quantum field theory. Part I. Scalar field theory”, J. High Energ. Phys. 11 (2017) 204.
  * (16) X. Meng, “Double-trace deformation in open quantum field theory”, Phys. Rev. D 104, 016016 (2021).
  * (17) P. Wang, “Relativistic quantum field theory of stochastic dynamics in the Hilbert space”, Phys. Rev. D 105, 115037 (2022).
  * (18) J. Oppenheim and Z. Weller-Davies, “The constraints of post-quantum classical gravity”, J. High Energ. Phys. 02 (2022) 080.
  * (19) C. Jones, T. Guaita, and A. Bassi, “Impossibility of extending the Ghirardi-Rimini-Weber model to relativistic particles”, Phys. Rev. A 103, 042216 (2021).
  * (20) C. Jones, G. Gasbarri, and A. Bassi, “Mass-coupled relativistic spontaneous collapse models”, J. Phys. A 54, 295306 (2021).
  * (21) D. Bedingham, D. Dürr, G. Ghirardi, S. Goldstein, R. Tumulka, and N. Zanghì, “Matter Density and Relativistic Models of Wave Function Collapse”, J. Stat. Phys. 154, 623 (2014).
  * (22) D. Bedingham and P. Pearle “Continuous-spontaneous-localization scalar-field relativistic collapse model”, Phys. Rev. Research 1, 033040 (2019).
  * (23) P. Pearle, “Relativistic dynamical collapse model”, Phys. Rev. D 91, 105012 (2015).
  * (24) M. A. Kurkov and V. A. Franke, “Local Fields Without Restrictions on the Spectrum of 4-Momentum Operator and Relativistic Lindblad Equation”, Found. Phys. 41, 820 (2011).
  * (25) A. Matsumura, “Reduced dynamics with Poincaré symmetry in an open quantum system”, Phys. Rev. A 108, 042217 (2023).
  * (26) A. S. Holevo, “Covariant quantum dynamical semigroups: Unbounded generators”, in Irreversibility and Causality Semigroups and Rigged Hilbert Spaces, edited by A. Bohm, H.-D. Doebner, and P. Kielanowski (Springer, Berlin, 1998) pp. 67– 81\.
  * (27) S. Weinberg, “The Quantum Theory of Fields, Vol. I” (Cambridge University Press, Cambridge, England, 1995).
  * (28) C. Cîrstoiu, K. Korzekwa, and D. Jennings, “Robustness of Noether’s Principle: Maximal Disconnects between Conservation Laws and Symmetries in Quantum Theory”, Phys. Rev. X 10, 041035 (2020).
  * (29) I. Marvian and R. W. Spekkens, “Extending Noether’s theorem by quantifying the asymmetry of quantum states”, Nat. Commun. 5, 3821 (2014).
  * (30) M. Toro$\check{\text{s}}$, “Constraints on the spontaneous collapse mechanism:theory and experiments”, Ph.D thesis.
  * (31) A. Peres and D. R. Terno, “Quantum information and relativity theory”, Rev. Mod. Phys. 76, 93 (2004).
  * (32) D. Poulin, Information loss in quantum field theories.
https://online.kitp.ucsb.edu//online/qinfo_c17/poulin/

  * (33) E. Alicki, M. Fannes, and A. Verbeure, “Unstable particles and the Poincare semigroup in quantum field theory”, J. Phys. A 19, 919 (1986)
  * (34) L. Diósi, “Is there a relativistic Gorini-Kossakowski-Lindblad-Sudarshan master equation?”, Phys. Rev. D 106, L051901 (2022).
  * (35) J. Cripe, N. Aggarwal, R. Lanza, A. Libson, R. Singh, P. Heu, D. Follman, G. D. Cole, N. Mavalvala, and T. Corbitt, “Measurement of quantum back action in the audio band at room temperature”, Nature (London) 568, 364 (2019).
  * (36) S. B. Cataño-Lopez, J. G. Santiago-Condori, K. Edamatsu, and N. Matsumoto, “High-Q Milligram-Scale Monolithic Pendulum for Quantum-Limited Gravity Measurements”, Phys. Rev. Lett. 124, 221102, (2020).
  * (37) T. Westphal, H. Hepach, J. Pfaff, and M. Aspelmeyer, “Measurement of gravitational coupling between millimetre-sized masses”, Nature (London) 591, 225 (2021).


* [125] K. Rasul, C. Seward, I. Schuster, and R. Vollgraf, “Autoregressive denoising diffusion models for multivariate probabilistic time series forecasting,” in _International Conference on Machine Learning_. PMLR, 2021, pp. 8857–8868.