Eliminating unphysical photon components from Dirac-Maxwell Hamiltonian quantized in the Lorenz gauge
Shinichiro Futakuchi\footnote{futakuchi@math.sci.hokudai.ac.jp
Lorenz_DM_USUI_20150325.tex
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\title{Eliminating unphysical photon components from Dirac-Maxwell Hamiltonian
quantized in the Lorenz gauge}
\author{Shinichiro Futakuchi\footnote{futakuchi@math.sci.hokudai.ac.jp} and Kouta Usui\footnote{kouta@math.sci.hokudai.ac.jp}\\
\small{Department of Mathematics, Hokkaido University, Sapporo, 060-0810, Japan.}}
\date{}

\begin{document}
\maketitle
\abstract{We study the Dirac-Maxwell model quantized in the Lorenz gauge. 
In this gauge, the space of quantum mechanical state vectors inevitably be an indefinite 
metric vector space so that the canonical commutation relation (CCR) is realized
in a Lorentz covariant manner. 
In order to obtain a physical subspace in which no negative norm state exists, the method first proposed by Gupta and Bleuler
is applied with mathematical rigor.
It is proved that a suitably defined
 physical subspace has a positive semi-definit metric, and naturally induces a physical Hilbert space with a positive definite metric.
 The original Dirac-Maxwell Hamiltonian naturally defines an induced Hamiltonian on the physical Hilbert space which is essentially self-adjoint.
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Introduction}
We consider a quantum system of $N$ Dirac particles under an external potential $V$ interacting with a quantized gauge field (so called Dirac-Maxwell model). %in the Lorenz gauge. 
If we apply the informal perturbation theory to this model, quantitative predictions are obtained
such as the Klein-Nishina formula for the cross section of the Compton scattering of an electron and a photon \cite{Nishijima1973a}, which
agrees with the experimental results very well. 
Hence, the Dirac-Maxwell model is expected to describe a certain class of realistic quantum phenomena 
and thus is worth the analysis with mathematical rigor, even though it may suffer from 
so called ``negative energy problem". The mathematically rigorous study of this model was initiated by Arai in Ref. \cite{MR1765584},
and several mathematical aspects of the model was analyzed so far (see, e.g.,
Refs. \cite{MR1981623}, \cite{MR2260374}, \cite{MR2810826}, \cite{MR2178588}, and \cite{MR2377946}). %All of these studies are discussed in the Coulomb gauge.

 The motivation of the present study is to treat this model in the Lorentz gauge in which the Lorenz covariance is manifest.
 In analyzing gauge theories such as quantum electrodynamics (QED) in a Lorentz covariant gauge, a difficulty always arises, since one  inevitably 
adopt ``an indefinite metric Hilbert space" as a space of all the state vectors (for instance, see Ref. \cite{MR509200}).
In such cases, we have to identify a positive definite subspace as a physical 
state vector space by eliminating unphysical photon modes with negative norms. 
The most general and elegant method to identify the physical subspace for non-abelian gauge theories quantized in a covariant gauge
was given by the celebrated work by Kugo and Ojima \cite{Kugo:1977zq,MR557151},
which is based on the BRST symmetry, the remnant gauge symmetry of the Lagrangian density after imposing some gauge fixing condition. 
The Kugo-Ojima formulation reduces to 
Nakanishi and Lautrup's $B$-field theory \cite{Lautrup:1967,Nakanishi:1973fu,Nakanishi:1974pr,Nakishi:1966di} in the case where the gauge field is abelinan and after integrating out the auxiliary Nakanishi-Lautrup's $B$-field, it is reduced to the condition first proposed by Gupta and Bleuler \cite{MR0038883,MR0036166}.

The Gupta and Bleuler condition says that a state vactor belongs to the phyisical subspace if and only if it has the   
vanishing expectation value of the operator $\partial_\mu A^{\mu}$, the four component divergence of the gauge field:
\begin{align}\label{GB-cond}
\inprod{\Psi}{\partial_\mu A^\mu (t,\xx) \Psi} = 0,
\end{align}
at every spacetime point $(t,\xx)\in\R^4$.
From the equations of motion $\Box A_\mu =- j_\mu$ (with ``mostly plus metric") and the current conservation equation $\partial_\mu j^\mu=0$, 
we heuristically find that $\partial_\mu A^\mu (t,\xx)$ satisfies the Klein-Gordon equation so that \eqref{GB-cond}
is written as
\begin{align}
[\partial_\mu A^\mu]^{+} (t,\xx) \Psi=0,
\end{align}
where $[\partial_\mu A^\mu]^{+} (t,\xx)$ denotes the positive frequency part of the free field $\partial_\mu A^\mu$.
However, in order to rigorously perform this procedure, one has to answer the following questions.
Firstly, how to identify $A(t,\xx)$ at a time $t\in \R$? Since the present state vector space is not an ordinary Hilbert space with a
positive definite metric, the Hamiltonian can not be defined as a self-adjoint operator in the ordinary sense.
Thus, it is far from trivial if there is a solution of quantum Heisenberg equations of motion.
Secondly, is it possible to identify the positive frequency part of the operator satisfying Klein-Gordon 
equation even in an indefinite metric space? The first problem is solved by the general construction method of
time evolution operator generated by a non-self-adjoint operator given by the authors \cite{FutakuchiUsui2013}, and as to the second 
one, the general definition of ``positive frequency part" of a quantum field satisfying Klein-Gordon equation is given in 
Ref. \cite{MR2533876}. 

Mathematically rigorous study of concrete models of QED in the Lorentz covariant gauge (see, for instance, Refs \cite{MR574710,MR2412280,MR2533876,IR-GB})
was only given for a solvable models as far as we know. However, the model treated here, the Dirac-Maxwell model, is \textit{not} solvable
in the sense that an explicit expression of the time-dependent gauge field is not easily found.
Thus, the problem has to be abstractly considered, not relying on the explicit expression of the time dependent gauge field
but only on the abstract existence theorem. In this paper, we establish the existence of a time evolution operator and give a definition of the 
``positive frequency part" of a free field safisfying Klein-Gordon equation in an abstract setup. 
Our definition of ``positive frequency part" given here is different from that given in Ref \cite{MR2533876}, but results in the same
consequence when applied to the concrete models. 
We then apply to the abstract theory to the concrete Dirac-Maxwell model and identify the physical Hilbert space.
We also prove that the original Dirac-Maxwell Hamiltonian naturally defines a self-adjoint ``physical Hamiltoninan" on the Hilbert space,
which is essentially equivalent to the Dirac-Maxwell Hamiltonian in the Coulomb gauge discussed in Ref. \cite{MR1765584}. 

The mathematical tools developed here would have some interests in its own right. The time evolution operator 
generated by \textit{unbounded, non-self-adjoint} Hamiltonians has been constructed in Ref. \cite{FutakuchiUsui2013}. 
In this paper, we further develop the theory in several aspects.
Firstly, we define a general class of operators, which we will call $\mathcal{C}_n$-class operators, and prove that
a $\mathcal{C}_n$-class operator $B$ has a time evolution $B(t)$ for $t\in\R$ which solves the Heisenberg equation,
and $B(t)\xi$ is $n$-times strongly differentiable in $t$ for $\xi$ belonging to a dense subset $D'$.  
Moreover, the $n$-th derivative of $B(t)$ enjoys the natural expression in terms of a weak commutator defined in a suitable sense.
Secondly, we define a more restricted class of operators, which will be called $\mathcal{C}_\omega$-class operators,
and prove that $\mathcal{C}_\omega$-class operators has a time evolution which is analytic in $t\in\R$.
Furthermore, it would be interesting to see that the following Taylor expansion formula \eqref{taylor} remains valid 
even for an unbounded, non-unitary time evolution.
Thirdly, the Klein-Gordon equation is analyzed in this abstract setup.
We generally define a ``free field"  as a solution of the generalized Klein-Gordon equation,
and then we prove that generalized Klein-Gordon equation is explicitly solved under some suitable conditions, even for an unbounded, non-unitary time evolution,
even in a vector space with an indefinite metric. 
From this explicit solution, positive
and negative frequency parts are defined unambiguously. 
Fourthly, abstract study with mathematical rigor makes it clear how and why the Gupta-Bleuler method works well and 
what is essentially responsible for the possibility of the method
is the electro-magnetic current conservation. 

%In fact, although we can not find the explicit expression of the time dependent gauge field,
%we can explicitly write down the positive frequency part of $\partial_\mu A^\mu$,
%because it satisfies the Klein-Gordon equation due to the current conservation law.
%We also emphasize that the mathematical generality enables us to define the physical Hamiltonian of the Dirac-Maxwell model somewhat abstractly 
%in terms of the time evolution operator.
%The present study also differs from previous ones in that the electrons are treated as
%relativistic particles for which a covariant formulation would be more desirable when doing some concrete
%computations. 
%

The paper is organized as follows. In Section \ref{abstract}, after the brief 
review of the results obtained in \cite{FutakuchiUsui2013}, we discuss the abstract theory
of time evolution generated by non-self-adjoint Hamiltonians and give 
a definition of general free field to which the positive frequency part is defined.
In Section \ref{model}, we intoduce the Difinition of the model and how to mathematically deal with an 
indefinite metric space. 
In Section \ref{time-evolution}, we show that the Dirac-Maxwell Hamiltonian is essentially self-adjoint
with respect to the indefinite metric. We also show that the general theory developed so far is applicable to the Dirac-Maxwell Hamiltonian
and to study the time evolution of quantum fields. We derive the current conservation equation 
and the equation of motion.
In Section \ref{Gupta-Bleuler}, we identify the positive frequency part of the operator $\partial_\mu A^\mu(t,\xx)$
by using the abstract theory of generalized free field developed in Section \ref{time-evolution}.
Then, we define the physical subspace and the physical Hilbert space.
We also show that the original Hamiltonian naturally defines the physical Hamiltonian
which is essentially self-adjoint on the physical Hilbert space. The relation to the Coulomb gauge Hamiltonian 
is discussed there. Finally, we show that if the ultraviolet cutoff function of the gauge field is infrared singular, 
then the physical subspace has to become trivial.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section{Time-evolution Generated by Non-self-adjoint Hamiltoninans}\label{abstract}
One of the main obstacles to mathematical analysis in the Lorenz gauge is
that the existence of the time-evolution operator is
not trivial at all because we can not use the
usual functional calculus to define the time-evolution operator $e^{-itH}$ if the generator $H$ is not self-adjoint.
The existence of a time-evolution is established by the general theory which the authors developed 
in Ref. \cite{FutakuchiUsui2013}. We summarize and further extend the 
results obtained in Ref. \cite{FutakuchiUsui2013} in an abstract setup. 

\subsection{Existence of time-evolution}
Let $\mathcal{H}$ be a Complex Hilbert space and $\ip {\cdot} {\cdot}$ its inner product, and
$\norm{\cdot}$ its norm. The inner product is linear in the second variable. 
For a linear operator $ T $ in $ \mathcal{H} $, we denote its domain (resp. range) by $ D(T) $ (resp. $ R(T) $). We also denote the adjoint of $T$ by $ T^* $ and the closure by $ \overline{T} $ if these exist. 
For a self-adjoint operator $ T $,  $ E_T (\cdot ) $ denotes the spectral measure of $ T $.
The symbol $T\upharpoonright D$ denotes the restriction of a linear operator $T$ 
to the subspace $D$.

Let $H_0$ be a self-adjoint operator on $\mathcal{H}$. Suppose that there is a nonnegative self-adjoint operator $A$ 
which is strongly commuting with $H_0$. We use the notations
\begin{align}
V_L&:=E_A([0,L]),\quad L\ge 0, \\
D&:=\bigcup_{L\ge 0} V_L,\\
D'&=D\cap D(H_0).
\end{align}
Let us define a family of linear operators $\mathcal{C}_0$ as follows:
\begin{Def}
We say that a linear operator $B$ is in $\mathcal{C}_0$-class if $B$ satisfies
\begin{enumerate}[(i)]
\item $B$ is densely defined and closed.
\item $B$ and $B^*$ are $A^{1/2}$- bounded.
\item There is a constant $b>0$ such that $\xi\in V_L$ implies $B\xi$ and $B^*\xi$ belong to $V_{L+b}$.
\end{enumerate}
\end{Def}
The set of all $\mathcal{C}_0$-class operators is also denoted by the same symbol $\mathcal{C}_0$.
We remark that if $B$ is in $\mathcal{C}_0$, then so $B^*$ is.
We consider an operator 
\begin{align}
H=H_0+H_1
\end{align}
with $H_1\in\mathcal{C}_0$. The following Propositions summarize the results obtained in Ref. \cite{FutakuchiUsui2013}.
\begin{Prop}\label{main-thm1} For each $ t,t' \in \R , \, \xi \in D $, the series:
\begin{align}
U(t,t') \xi := \xi + (-i) \int _{t'} ^t d\tau _1 \, H_1(\tau _1) \xi + (-i)^2 \int _{t'} ^t d\tau _1 \int _{t'} ^{\tau _1} d\tau _2 \, H_1 (\tau _1) H_1 (\tau _2) \xi + \cdots 
\end{align}
converges absolutely, where each of integrals are strong integrals. 
Furthermore, $D\subset D(U(t,t') ^*)$.
%%% 
\if0
 and the following (i) and (ii) hold.
\begin{enumerate}[(i)]
\item  For fixed $ t' \in \Real  $ and $ \xi \in D $, the $\mathcal{H}$-valued functions $ \R\ni t\mapsto U(t,t') \xi $ 
and $\R\ni t\mapsto U(t,t')^*\xi $ is strongly continuously differentiable, 
and $ U(t,t') \xi \in D(H_1 (t)) $. Moreover, $ U(t,t') \xi $ satisfies
\begin{align}\label{DE1}
\frac{\partial }{\partial t} U(t,t') \xi &= -i H_1 (t) U(t,t') \xi ,\\
 \frac{\partial }{\partial t} U(t,t') ^* \xi &= i U(t,t')^* H_1 (t) ^* \xi . 
\end{align}
\item  For fixed $ t \in \Real  $ and $ \xi \in D $, the $\mathcal{H}$-valued function $\R \ni t'\mapsto  U(t,t') \xi $
and $\R \ni t'\mapsto  U(t,t')^* \xi $  is strongly continuously differentiable,
 and satisfies
\begin{align}\label{DE2}
\frac{\partial }{\partial t'} U(t,t') \xi &= i U(t,t') H_1 (t') \xi ,\\
\frac{\partial }{\partial t'} U(t,t') ^* \xi &= -i H_1 (t') ^* U(t,t') ^* \xi .
\end{align}
\end{enumerate}
\fi
%%%
 \end{Prop}
 %%%
\if0
The time evolution operator $U(t,t')$ has the following properties.
\begin{Prop}\label{main-thm2} The following (i) and (ii) hold.
\begin{enumerate}[(i)]
\item For all $ \xi \in D , \, t,t',t'' \in \R $, $ U(t,t) \xi = \xi $ and the operator equality
\begin{align}\label{associative}
\overline{U(t,t')} U(t' , t'') = U(t,t'').
\end{align}
holds.
\item For any $ s, t,t' \in \Real $, the operator equality
\begin{align}\label{translation}
e^{isH_0} \overline{U(t,t')} e^{-isH_0} = \overline{U(t+s, t'+s)}
\end{align}
holds.
\end{enumerate}
\end{Prop}
\fi
%%%
\if0 %%%
If we assume in addition that $ H_1 $ is symmetric, the stronger results follow:
\begin{Prop}\label{uniqueness} 
Let $H_1$ be a closed symmetric operator. Then, $ \overline{U(t,t')} $ is unitary and the following properties hold.
\begin{enumerate}[(i)] 
\item The operator $U(t,t')$ satisfies the following operator equalities:
\begin{align} 
\overline{U(t,t)} =I , \q \overline{U(t,t')} \; \overline{U(t',t'')} = \overline{U(t,t'')} ,
\end{align}
where $ I $ denotes the identity operator.
\item $U(t,t')$ is unique in the following sense. 
If there exist a dense subspace $\widetilde{D}$ in $\mathcal{H}$ and an operator valued function $ V (t,t') \, (t,t' \in \Real ) $ such that 
$\widetilde{D}\subset D(V(t,t'))$ for all $t,t'\in\R$ and for $ \xi \in \widetilde{D} $, $ V(t,t')\xi $ is strongly differentiable with respect to $ t $, and $ V(t,t') \xi \in D(H_1 (t)) $, which satisfies
\begin{align}
V (t,t)\xi =\xi , \q \frac{\partial }{\partial t} V(t,t') \xi = -i H_1 (t) V(t,t') \xi , \q \xi \in \widetilde{D} , \q t,t' \in \R ,
\end{align}
then $V(t,t')\upharpoonright \widetilde{D}$ is closable and $ \overline{V (t,t')\upharpoonright \widetilde{D}} = \overline{U(t,t')} $. In particular, if $ D(V(t,t')) =\H $ and $ V(t,t') $ is bounded for all $ t,t' \in \R $, then $ V(t,t') = \overline{U(t,t')} $. 
\end{enumerate}
\end{Prop}
\fi %%%

We discuss the existence of the dynamics generated by $H$.
Let 
\begin{align}
W(t) := e^{-itH_0} \overline{U(t,0)} , \q t\in \R .
\end{align}
Then, we have
\begin{Prop}\label{sch-existence} For each $ \xi \in  D' $, the vector valued functions $ t\mapsto \xi (t) := W(t )\xi $ and $ t\mapsto \xi^*(t) := W(t )^* \xi $ 
 are strongly differentiable in $t\in \R$. Moreover, the followings hold:
\begin{align}\label{sch30}
\frac{d}{dt} \xi (t) &= -iH \xi (t) = -iHW(t)\xi = -i W(t)H\xi , \\
\frac{d}{dt} \xi^* (t) & = -iH^* \xi^* (t)= -iH^*W(t)^*\xi = -i W(t)^*H^*\xi .
\end{align}
Further, $W(t)\xi$ and $W(t)^*\xi$ belong to $D(A^{k/2})\cap D(H_0)$ for all $k=0,1,2,\dots$.
\end{Prop}



The existence of a solution of the Heisenberg equation 
is ensured if $B\in \mathcal{C}_0$ in a weak sense.
\begin{Prop}\label{w-Hei} 
Let $B\in\mathcal{C}_0$. Then,
\begin{enumerate}[(i)]
\item $ D \subset D(W(-t) B W(t)) $.
\if0
\item $ D \subset D(W(-t)^* B^* W(t)^*) $ and 
\begin{align}
B(t)^*\xi = W(-t)^* B^* W(t)^*\xi
\end{align}
for all $\xi \in D$.
\item For all $\xi\in D'$, the function $t\mapsto BW(t)\xi$ is strongly differentiable in $t\in\Real$ with
\begin{align}
\frac{d}{dt}BW(t)\xi = -iBW(t)H\xi.
\end{align}
\fi
\item The operator-valued function $ B(t) $ defined as 
\begin{align}
D(B(t)) := D , \q B(t) \xi & := W(-t) B W(t) \xi , \q \xi \in D, \q t \in \R ,
\end{align}
is a solution of the weak Heisenberg equation:
\begin{align}\label{wHei2}
\frac{d}{dt} \left\langle \eta , B(t) \xi \right\rangle = \left\langle (iH)^* \eta , B(t) \xi \right\rangle - \left\langle B(t) ^* \eta , iH \xi \right\rangle , \q  \xi , \eta \in  D' .
\end{align}
\end{enumerate}
\end{Prop}

In Ref. \cite{FutakuchiUsuiSelf-ad}, a new criterion to prove the essential self-adjointness is proposed
as an application of these results. We refer two results obtained there for later use.

\begin{Prop}\label{a-or-b}
 Let $ T $ be a symmetric operator in $ \H $. If there exists a dense subspace $ V $ such that for any $ \xi \in V $ the initial value problem 
\begin{align}
\frac{d}{dt} \xi (t) = -i T \xi (t) , \q \xi (0 ) = \xi ,
\end{align}
has a strong solution $ \R \ni t \mapsto \xi (t ) \in D(T) $, then, 
exactly one of the following (a) or (b) holds.
\begin{enumerate}[(a)] 
\item $ T $ has no self-adjoint extension.

\item $ T $ is essentially self-adjoint.
\end{enumerate}
\end{Prop}
By using this Proposition, one readily finds 
\begin{Prop}\label{s.a.}
Let $H_1$ be in $\mathcal{C}_0$ and symmetric. Then, $H$ is also symmetric and for the symmetric operator $H$, exactly one of the following (a) and (b) holds:
\begin{enumerate}[(a)]
\item $H$ has no self-adjoint extension.
\item $H$ is essentially self-adjoint.
\end{enumerate} 
\end{Prop}

\subsection{$N$-th derivatives and Taylor expansion}
In this section, we develop a general theory concerning $n$-times differentiability of the operator $B(t)=W(-t)BW(-t)$
on a suitable subspace. Here, we take a slightly different formulation from that of Ref. \cite{FutakuchiUsui2013}
so that the generalization to $n$-th differentiability is easier.

To begin with, we prove a simple property of $B(t)$, which is not explicitly stated in Ref. \cite{FutakuchiUsui2013}.  
\begin{Lem}\label{s-conti-c_0}
Let $B\in\mathcal{C}_0$. Then the mapping
\begin{align}
t\mapsto W(-t)BW(t)\xi 
\end{align}
is strongly continuous in $t\in\Real$ for all $\xi \in D$.
\end{Lem}
\begin{proof}
Define an operator $U_n(t,t')$ on $D$, following Ref. \cite{FutakuchiUsui2013}, as
\begin{align}
U_n(t,t')\xi := (-i)^n\int_{t'}^t ds_1 \dots \int_{t'}^{s_{n-1}}ds_n \,H_1(s_1)\dots H_n(s_n)\xi.
\end{align}
Then the estimate 
\begin{align}\label{basic-estimate}
\norm{U_n(t,t')\xi} \le \frac{C^n}{n!}  |t-t'|^n (L+(n-1)b+1)^{1/2}\dots (L+1)^{1/2}\norm{\xi},
\end{align}
holds for some $C>0$ and $b>0$ (\cite{FutakuchiUsui2013} Lemma 3.4). From this estimate and the assumption that $B$ is in $\mathcal{C}_0$,
it is straightforward to check the mapping
\begin{align}
t\mapsto e^{itH_0}U_n(-t,0)Be^{-itH_0}U_m(t,0)\xi,\q \xi\in D
\end{align}
is strongly continuous. Since $W(-t)BW(t)\xi$ can be expanded in
a series converging absolutely and locally uniformly in $t\in\R$:
\begin{align}
BW(t)\xi = \sum_{n,m=0}^{\infty} e^{itH_0}U_n(-t,0)Be^{-itH_0}U_m(t,0)\xi,
\end{align}
the limit function
\begin{align}
t\mapsto W(-t)BW(t)\xi,\q \xi\in D
\end{align}
is also strongly continuous.
\end{proof}

\begin{Def}\label{C_1}
We say an operator $B$ is in $\mathcal{C}_1$-class if it satisfies 
\begin{enumerate}[(i)]
\item $B$ is in $\mathcal{C}_0$-class.
\item There is an operator $C\in\mathcal{C}_0$ such that
\begin{align}\label{w-comm}
\ip{(iH)^*\xi}{B\eta} - \ip{B^*\xi}{iH\eta}=\ip{\xi}{C\eta}
\end{align}
for all $\xi,\eta\in D'$.
\end{enumerate}
\end{Def}
We remark that the above operator $C\in\mathcal{C}_0$ is not unique in general. But one finds
\begin{Lem}
Let $B\in\mathcal{C}_1$ and $C$ be an operator mentioned in Definition \ref{C_1}. Then the operator
\begin{align}
\ad(B):=\overline{C\upharpoonright D(A^{1/2})}
\end{align}
does not depend on the choice of $C$ and determined only by $B$.
\end{Lem} 
\begin{proof}
Suppose that two operators $C_1$ and $C_2$ fulfills the condition. Then for all $\xi\in D'$,
we have
\begin{align}
C_1\xi=C_2\xi.
\end{align}
Take arbitrary $\eta\in D(A^{1/2})$ and put $\eta_n=E_A([0,n])E_{H_0}([-n,n])\eta$.
Then clearly $\eta_n\in D'$ and 
\begin{align}
\eta_n\to\eta,\q (A+1)^{1/2}\eta_n\to(A+1)^{1/2}\eta,
\end{align}
as $n$ tends to infinity. Therefore, we have
\begin{align}
C_1\eta &= \lim_{n\to\infty} C_1(A+1)^{-1/2}(A+1)^{1/2}\eta_n \no\\
	&= \lim_{n\to\infty} C_2(A+1)^{-1/2}(A+1)^{1/2}\eta_n \no\\
	&=C_2 \eta,
\end{align}
since both $C_1(A+1)^{-1/2}$ and $C_2(A+1)^{-1/2}$ are bounded. Thus one obtains
\begin{align}
C_1\upharpoonright D(A^{1/2})=C_2\upharpoonright D(A^{1/2}).
\end{align}
Taking the closure of both sides proves the assertion.
\end{proof}
\begin{Lem}\label{extended-w-comm}
Let $B\in\mathcal{C}_1$. Then the relation 
\eqref{w-comm} remains valid for all $\xi,\eta\in D(H_0)\cap D(A^{1/2})$. 
\end{Lem}
\begin{proof}
Let $\xi,\eta\in D(H_0)\cap D(A^{1/2})$. Put 
\begin{align}
\xi_n:=E_{A}([0,n])\xi, \q \eta_n:=E_{A}([0,n])\eta.
\end{align}
Then, clearly $\xi_n,\eta_n\in D'$ and 
\begin{align}
\xi_n\to\xi, \q H_0\xi_n\to H_0\xi,\q B\xi_n\to B\xi
\end{align}
as $n$ tends to infinity for all $B\in\mathcal{C}_0$, and the same is true for $\eta$.
The relation \eqref{w-comm} holds for $\xi_n $ and $\eta_n$. Thus, by the limiting argument,
the assertion follows.
\end{proof}
The strong Heisenberg equation is satisfied if $B$ is in $\mathcal{C}_1$-class.
\begin{Thm}\label{C_1-equation}
For $B\in \mathcal{C}_1$ and $\xi \in D'$ the mapping
\begin{align}
\R\ni t\mapsto B(t)\xi=W(-t)BW(t)\xi\in\mathcal{H}
\end{align}
is strongly continuously differentiable in $t\in\R$ and satisfies the Heisenberg equation of motion
\begin{align}
\frac{d}{dt} B(t)\xi = W(-t)\ad (B) W(t) \xi.
\end{align}
\end{Thm}
\begin{proof}
By Propositions \ref{sch-existence} and
\ref{w-Hei}, and Lemma \ref{extended-w-comm}, we have 
\begin{align}
\frac{d}{dt} \left\langle \eta , B(t) \xi \right\rangle &= \left\langle (iH)^* \eta , B(t) \xi \right\rangle - \left\langle B(t) ^* \eta , iH \xi \right\rangle , \no\\
&=\left\langle (iH)^* W(-t)^*\eta , BW(t) \xi \right\rangle - \left\langle B ^*W(-t)^* \eta , iH W(t) \xi \right\rangle \no\\
&=\ip{W(-t)^*\eta}{\ad(B)W(t)\xi}.
\end{align}
Since $\ad(B)$ is in $\mathcal{C}_0$, $\ad(B)W(t)\xi\in D(W(-t))$ and one gets
\begin{align}
\frac{d}{dt} \ip{\eta} {B(t)\xi} = \ip{\eta}{W(-t)\ad(B)W(t)\xi}.
\end{align}
Hence we have 
\begin{align}
\ip{\eta} {B(t)\xi} &= \ip{\eta}{\ad(B)\xi} + \int_0^t ds\, \ip{\eta}{W(-s)\ad(B)W(s)\xi} \no\\
	&=\ip{\eta}{\ad(B)\xi} +  \ip{\eta}{\int_0^t ds\,W(-s)\ad(B)W(s)\xi},
\end{align}
by Lemma \ref{s-conti-c_0}. Thus we obtain
\begin{align}
B(t)\xi=\ad(B)\xi + \int_0^t ds\,W(-s)\ad(B)W(s)\xi.
\end{align}
This equation shows that $B(t)\xi$ is continuously strongly differentiable and
\begin{align}
\frac{d}{dt}B(t)\xi = W(-t)\ad(B)W(t)\xi, \q \xi\in D'.
\end{align}
This completes the proof.
\end{proof}
One of the merits of the present formulation of the strong Heisenberg equation is that 
it is easy to extend for $n$-th differentiability. 
\begin{Def}
We define $\mathcal{C}_n$-class and $\ad^n(B)$ for $n=0,1,\dots$ inductively. That is, we say that an operator $B$ is in $\mathcal{C}_n$-class if $B$ is in $\mathcal{C}_{n-1}$-class and
$\ad(B)$ is in $\mathcal{C}_{n-1}$-class. For $B\in\mathcal{C}_n$, we write
\begin{align}
\ad^n(B):=\ad(\ad^{n-1}(B)),\q n=1,2,\dots.
\end{align}
It is clear that $\mathcal{C}_n\subset\mathcal{C}_{n-1}$ for $n=1,2,\dots$ and that if $B\in\mathcal{C}_n$, then $\ad(B)\in\mathcal{C}_{n-1}$, $\ad^2(B)\in\mathcal{C}_{n-2}$, \dots
$\ad^n(B)\in\mathcal{C}_0$. We define $\ad^0(B):=B$. An operator $B$ is said to be in $\mathcal{C}_\infty$-class if $B$ is in $\mathcal{C}_n$ for all $n\in\Natural$. Namely,
\begin{align}
\mathcal{C}_\infty := \bigcap_{n=0}^\infty \mathcal{C}_n.
\end{align}
\end{Def}

The following Theorem is important and useful but the proof is almost trivial by induction.
\begin{Thm}\label{n-th-diff}
Let $B$ is in $\mathcal{C}_n$-class. Then, for all $\xi\in D'$, $B(t)\xi$ is $n$-times strongly continuously differentiable 
in $t\in\R$ and 
\begin{align}
\frac{d^k}{dt^k}B(t)\xi = W(-t)\ad^k(B)W(t)\xi,\q k=0,1,2,\dots,n.
\end{align} 
In particular, if $B\in\mathcal{C}_\infty$, then
\begin{align}
\frac{d^k}{dt^k}B(t)\xi = W(-t)\ad^k(B)W(t)\xi,\q k=0,1,2,\dots.
\end{align} 
\end{Thm}
From Theorem \ref{n-th-diff}, we immediately have
\begin{Thm}
Let $B\in\mathcal{C}_n$ and $\xi\in D'$. Then, there is a $\theta\in (0,1)$ such that 
\begin{align}\label{finite-Taylor}
B(t)\xi = \sum_{k=0}^{n-1} \frac{t^k}{k!} \ad^k(B) \xi + \frac{t^n}{n!}W(-\theta t)\ad^n(B)W(\theta t)\xi. 
\end{align}
\end{Thm}
To obtain a Taylor series expansion for $B(t)\xi$ for $\xi\in D'$, we need one more concept.
\begin{Def}\label{c-omega}
We say that an operator $B$ is in $\mathcal{C}_\omega$-class if 
\begin{enumerate}[(i)]
\item $B\in\mathcal{C}_\infty$,
\item The operator norm 
\begin{align}
a_n:=\norm{\ad^n(B)(A+1)^{-1/2}}
\end{align} 
satisfies 
\begin{align}
\lim_{n\to\infty}\frac{t^na_n}{n!}=0, \q t>0.
\end{align}
\item There exists some constant $b>0$ such that
for all $n\ge 0$, $\xi\in V_L$ implies that $\ad^n(B)\xi$ belongs to $V_{L+b}$.
\end{enumerate}
\end{Def}
We then arrive at the following simple result.

\begin{Thm}\label{Taylor}
Suppose that $B\in\mathcal{C}_\omega$. Then, for each $\xi\in D'$, $B(t)\xi$ has the
norm-converging power series expansion formula
\begin{align}\label{taylor}
B(t)\xi = \sum_{n=0}^\infty \frac{t^n}{n!}\ad^n(B) \xi, \q t\in\R.
\end{align}
\end{Thm}
\begin{proof}
By Theorem \ref{n-th-diff}, all we have to show is that the norm of the reminder term in \eqref{finite-Taylor} vanishes:
\begin{align}
\lim_{n\to\infty} \norm{\frac{t^n}{n!}W(-\theta t)\ad^n(B)W(\theta t)\xi} = 0.
\end{align}
Since $B$ is in $\mathcal{C}_\omega$-class, there is a constant $b'>0$, which is independent of $n$, such that
$\xi\in V_{L}$ implies $\ad^n(B)\xi \in V_{B+b'}$. Choose $b>0$ such that 
$\xi\in V_{L}$ implies $H_1\xi \in V_{B+b}$ and $H_1^*\xi \in V_{B+b}$.
Put 
\begin{align}
C:=\norm{H_1(A+1)^{-1/2}}
\end{align}
which is finite since $B\in\mathcal{C}_\omega$, and let $\xi\in V_L$ for some $L\ge 0$.
Note that $W(-\theta t)\ad^n(B)W(\theta t)\xi$ is expanded in the norm-converging series and is estimated as
\if0
\begin{align}
\sup_{n}\norm{W(-\theta t) \ad^n(B) W(\theta t)\xi} &\le \sup_{n}\sum_{k,l=0}^\infty \norm{U_k(0,\theta t) e^{itH_0}\ad^n(B)e^{-itH_0}U_l(\theta t, 0)\xi} \no\\
&\le \sum_{k,l=0}^\infty \frac{C^{k+l}a_n\,|\theta t|^{k+l}}{k!\,l!} (L+(k+l-1)b+b'+1)^{1/2}\dots \times \no\\
&\q\q \times(L+(l-1)b+b'+1)^{1/2}(L+(l-1)b+1)^{1/2}\dots (L+1)^{1/2}\norm\xi \no\\
&<\infty.
\end{align}
Therefore we conclude for sufficiently large $n\in\N$ that
\fi 
\begin{align}
\norm{\frac{t^n}{n!}W(-\theta t)\ad^n(B)W(\theta t)\xi} 
&\le \frac{|t|^n}{n!}\sum_{k,l=0}^\infty \norm{U_k(0,\theta t) e^{i\theta tH_0}\ad^n(B)e^{-i\theta tH_0}U_l(\theta t, 0)\xi}\no\\
&\le \frac{|t|^na_n}{n!}\sum_{k,l=0}^\infty \frac{C^{k+l}\,|\theta t|^{k+l}}{k!\,l!} (L+(k+l-2)b+b'+1)^{1/2}\dots\times\no\\
&\q\q \times(L+(l-1)b+b'+1)^{1/2}(L+(l-1)b+1)^{1/2}\dots (L+1)^{1/2}\norm\xi\no\\
&\to 0,
\end{align}
as $n$ tends to infinity.
This completes the proof.
\end{proof}

\subsection{Generalized Klein-Gordon equation and free fields}\label{GFF}
In this section, we investigate the solution of Klein-Gordon equation in mathematically
general formulation which is suitable to the present context.
Let $\mathfrak{h}$ be a complex Hilbert space and $T$ be a nonnegative self-adjoint operator
on $\mathfrak{h}$. 
\if0
Define for nonnegative integers $k,n$,
\begin{align}
\mathfrak{d}_k^n := D(T^{-1/2+k+n}),\no\\
\mathfrak{d}_k^\infty:=\bigcap_{n=0}^\infty D(T^{-1/2+k+n}).
\end{align}
It is clear by definition that for $k\ge 1$
\begin{align}
\mathfrak{d}_k^0\supset\mathfrak{d}_k^1\supset\cdots \supset \mathfrak{d}_k^\infty,
\end{align}
for $k=0$,
\begin{align}
\mathfrak{d}_0^1\supset\mathfrak{d}_0^2\supset\cdots \supset \mathfrak{d}_0^\infty,
\end{align}
and $\mathfrak{d}_k^\infty$ is dense in $\mathfrak{h}$.
\fi 
\begin{Def}
A mapping $\phi:\mathfrak{h}\to\mathcal{C}_0$ is said to be $T$-free field if and only if
for $f\in D(T^2)$, $\phi(f)$ belongs to $\mathcal{C}_2$-class, and $\phi(t,f):=W(-t)\phi(f)W(t)$ satisfies the differential equation:
\begin{align}\label{KG}
\frac{d^2}{dt^2}\phi(t,f)\xi-\phi(t,-T^2 f)\xi=0,\quad \xi\in D',
\end{align}
where the differentiation is the strong one.
\end{Def}

We call this equation \textit{generalized Klein-Gordon equation}, since in the case where $\mathfrak{h}=L^2(\Real^3)$ and
$T=\sqrt{-\Delta}$, the equation \eqref{KG} gives the ordinary Klein-Gordon equation for a quantum field $\phi$.
As in the ordinary case, generalized Klein-Gordon equation can be explicitly solved,
in spite of the fact that, in the present case, the time evolution is not generated by a self-adjoint Hamiltonian.

We denote 
\begin{align}
C^\infty(T):=\bigcap_{n=1}^\infty D(T^n).
\end{align}

\begin{Lem}\label{even-odd}
Let $\phi$ be a $T$-free field. Then, for all $f\in C^\infty(T)$, $\phi(f)\in\mathcal{C}_\infty$ and
\begin{align}
\ad^{2n}[\phi(f)]\xi &= \phi((-1)^nT^{2n} f) \xi, \label{even}\\
\ad^{2n+1}[\phi(f)]\xi &= \ad[\phi((-1)^nT^{2n} f)] \xi, \label{odd}
\end{align}
for all $\xi\in D'$ and $n\ge 0$.
\end{Lem}
\begin{proof}
By Theorem \ref{n-th-diff}, the generalized Klein-Gordon equation \eqref{KG} is equivalent to 
\begin{align}
W(-t)\ad^2[\phi(f)]W(t)= W(-t)\phi(-T^2f)W(t)
\end{align}
on $D'$. At $t=0$ we have in particular
\begin{align}
\ad^2[\phi(f)]=\phi(-T^2f)
\end{align}
on $D'$. Since $f\in C^\infty(T)$, the right hand side belongs to $\mathcal{C}_2$,
which implies that $\phi(f)\in\mathcal{C}_4$ and 
\begin{align}
\ad^3[\phi(f)]&=\ad[\phi(-T^2f)], \\
\ad^4[\phi(f)]&=\ad^2[\phi(-T^2f)]=\phi((-T^2)^2f) \label{2nd-step}
\end{align}
on $D'$. But again the right hand side of \eqref{2nd-step} belongs to $\mathcal{C}_2$,
we obtain $\phi(f)\in\mathcal{C}_6$ and 
\begin{align}
\ad^5[\phi(f)]&=\ad[\phi((-T^2)^2f)], \\
\ad^6[\phi(f)]&=\ad^2[\phi((-T^2)^2f)]=\phi((-T^2)^3f). \label{3rd-step}
\end{align}
By repeating this argument, one finds that $\phi(f)\in\mathcal{C}_\infty$ and
\eqref{even}, \eqref{odd} hold.
\end{proof}

To solve the generalized Klein-Gordon equation, we introduce a well-behaved $T$-free field:
\begin{Def}\label{analytic-free}
A $T$-free field $\phi(\cdot)$ is said to be analytic if 
\begin{enumerate}[(i)]
\item For all $f$ which belongs to the subspace
\begin{align}
\bigcup_{N\in\N} E_T\left(\left[\frac{1}{N},N\right]\right),
\end{align}
$\phi(f)$ is in $\mathcal{C}_\omega$ class.
\item For $f\in D(T)$, $\phi(f)\in\mathcal{C}_1$.
\item $f_n\to f$ implies  
\begin{align}
\phi(f_n)\xi\to\phi(f)\xi ,\q \xi\in D',
\end{align}
and $Tf_n\to Tf$ implies 
 \begin{align}
 \ad[\phi(f_n)]\xi \to \ad[\phi(f)]\xi,\q \xi\in D'.
\end{align}
\end{enumerate}
\end{Def}
\begin{Thm}\label{KG-sol}
Let $\phi$ be an analytic $T$-free field. Then, for all $f\in D(T)$, we find
\begin{align}\label{sol-of-KG}
\phi(t,f)=\phi((\cos tT)f)+\ad\left[\phi\left(\left(\frac{\sin tT}{T}\right)f\right)\right]
\end{align} 
on $D'$.
\end{Thm}
\begin{proof}
Take $f\in R(E_T([1/N,N]))$ for some $N\in\N$. Then, $\phi(f)\in\mathcal{C}_\omega$. 
Thus we have
\begin{align}\label{exp-1}
\phi(t,f)=\sum_{n=0}^\infty \frac{t^n}{n!}\ad^n (\phi(f))
\end{align}
on $D'$ by Theorem \ref{Taylor}.
By Lemma \ref{even-odd}, one has for all $n\ge 0$,
\begin{align}
\ad^{2n}[\phi(f)]&=\phi((-1)^n T^{2n} f), \\
\ad^{2n+1}[\phi(f)]&=\ad[\phi((-1)^n T^{2n} f)],
\end{align}
on $D'$.
Then the Taylor expansion \eqref{exp-1} becomes
\begin{align}
\phi(t,f)=\sum_{n=0}^{\infty} \frac{t^{2n}}{(2n)!}\phi((-1)^n T^{2n} f) +\sum_{n=0}^{\infty} \frac{t^{2n+1}}{(2n+1)!}\ad(\phi((-1)^n T^{2n+1}T^{-1} f)) 
\end{align}
on $D'$. Note that as $N$ tends to infinity
\begin{align}
\sum_{n=0}^{N} \frac{(-1)^n t^{2n}}{(2n)!}T^{2n} f  \to (\cos tT) f,\q T\left(\sum_{n=0}^{N} \frac{(-1)^nt^{2n+1}}{(2n+1)!}T^{2n+1} T^{-1} f \right) \to T\left(\frac{\sin tT}{T}f\right).
\end{align}
Thus, by the assumption (iii) in Definition \ref{analytic-free}, 
we conclude that 
\begin{align}
\phi(t,f)=\phi((\cos tT)f )+\ad\left[\phi\left(\left(\frac{\sin tT}{T}\right)f\right)\right]
\end{align}
on $D'$ for all $f\in\cup_{N\in\N}R(E_T([1/N,N]))$. Here, the operators
$\cos tT$ and $\sin tT /T$ are defined through functional calculus.

Take arbitrary $f\in D(T)$ and put 
\begin{align}
f_N=E_T\left(\left[\frac{1}{N},N\right]\right)f.
\end{align}
One sees that the operators $\cos tT$ and $T^{-1}\sin t T $ are both bounded and
that $f_N$ converges to $f$ as $N$ tends to infinity. 
Hence, by the limiting argument, we find that $\eqref{sol-of-KG}$ remains valid for all $f\in D(T)$.
\end{proof}

This Theorem \ref{KG-sol} enables us to define positive and negative frequency parts of $\phi$:
\begin{Def}
Let $\phi$ be an analytic $T$-free field. We define for $f\in D(T^{-1})$, on $D'$
\begin{align}
\phi^{+}(t,f)&:=\phi\left(\frac{e^{-itT}}{2}f\right)-\ad\left[\phi\left(\frac{e^{-itT}}{2iT}f\right)\right],\label{positive-fp}\\
\phi^{-}(t,f)&:=\phi\left(\frac{e^{itT}}{2}f\right)+\ad\left[\phi\left(\frac{e^{itT}}{2iT}f\right)\right]
\end{align} 
and call $\phi^{+}$ (resp. $\phi^{-}$) positive (resp. negative) frequency part of $\phi$.
\end{Def}
The operators in the parentheses are defined though the 
functional calculus of self-adjoint operator $T$. 
It is clear by definition that an analytic $T$-free field $\phi$ can be written as a sum of
its positive and negative frequency parts,
\begin{align}
\phi(t,f)=\phi^{+}(t,f)+\phi^-(t,f),\q t\in\R, \q f\in D(T^{-1}),
\end{align}
on the subspace $D'$.


\section{Definition of the Dirac-Maxwell Hamiltonian in the Lorenz Gauge}\label{model}
In this section, we introduce the Dirac-Maxwell Hamiltonian $ H_\rm{DM}(V,N) $ quantized in the Lorenz gauge.
This Hamiltonian describes 
a quantum system consisting of a Dirac particle under a potential $V$ and a gauge field minimally interacting
with each other.
We use the unit system in which the speed of light and $ \hbar $, the Planck constant
devided by $2\pi$, are set to be unity.

\if0 In the present paper, we employ the following notations. 
For a linear operator $ T $ in $ \H $, we denote, in general, its domain or range by $ D(T) $ or $ R(T) $, respectively. We also denote the adjoint of $T$ by $ T^* $, the closure by $ \overline{T} $, if these exist. For a self-adjoint operator $ T $,  $ E_T (\cdot ) $ denotes the one dimensional spectral measure of $ T $.
\fi

\subsection{Dirac particle sector}
Let us denote the mass and the charge of the Dirac particle by $ M>0 $ and $ q \in \Real $, respectively.   The Hilbert space of state vectors for the Dirac particle is taken to be 
\begin{align}
\H _\mathrm{D} := L^2 (\Real_{\xx} ^3; \C ^4),
\end{align}
the square integrable functions on $\Real_{\xx}^3 = \{ \xx = (x^1 , x^2 , x^3) \, | \, x^j \in \R , \, j=1,2,3 \} $
 into $\C^4$. The vector space $\R^3_\xx$ here represents
the position space of the Dirac particle. We sometimes omit the subscript $\xx$ and just write $\R^3$ instead of $\R^3_\xx$
when no confusion may occur.
The target space $\C^4$ realizes a representation of the four dimensional 
Clifford algebra accompanied by the four dimensional Minkowski vector space. The generators 
$\{\gamma^\mu\}_{\mu=0,1,2,3}$ 
satisfy the anti-commutation relations
\begin{align}
\{\gamma^\mu, \gamma^\nu\}:=\gamma^\mu\gamma^\nu+\gamma^\nu\gamma^\mu=-2\eta^{\mu\nu},\q \mu,\nu=0,1,2,3,
\end{align}
where the Minkowski metric tensor $\eta = (\eta_{\mu\nu} )$ is given by
\begin{align}
\eta= \left(\begin{matrix}
-1 & 0& 0& 0 \\
0 & 1 & 0& 0 \\
0& 0& 1 &0 \\
0&0&0& 1
\end{matrix}\right) .
\end{align}
We set $ \eta ^{-1} = (\eta ^{\mu \nu}) $, the inverse matrix of $ \eta $, Then we have $ \eta ^{\mu \nu} = \eta _{\mu \nu} , \; \mu , \nu =0,1,2,3 $.
We assume $\gamma^0$ to be Hermitian and $\gamma^j$'s ($j=1,2,3$) be anti-Hermitian.
We use the notations following Dirac:
\begin{align}
\beta:=\gamma^0 , \q \alpha^j := \gamma^0 \gamma^j,\q j=1,2,3.
\end{align}
Then, $\alpha^j$s and $\beta$ satisfy the anti-commutation relations
\begin{align}
& \{ \alpha ^i , \alpha ^j \} = 2 \delta ^{ij} , \q i,j =1,2,3, \\
& \{ \alpha ^j , \beta \} =0 , \q \beta ^2 =1, \q j=1,2,3,
\end{align}
where $ \delta _{ij} $ is the Kronecker delta.
The momentum operator of the Dirac particle is given by
\begin{align}
\mathbf{p} := (p_1 , p_2 , p_3) := (-i D_1 , -iD_2 , -iD_3)
\end{align}
with $ D_j $ being the generalized partial differential operator on $ L^2 (\Real ^3; \C ^4) $ 
with respect to the variable $ x^j $, the $ j $-th component of $ \bvec{x} = (x^1 , x^2 , x^3) \in \R ^3 $. 
We write in short
\[  { \boldsymbol \alpha } \cdot \bvec{p} := \sum _{j=1} ^3 \alpha ^j p_j .\]
The potential is represented by a $ 4\times 4 $ Hermitian matrix-valued function $ V $ on $ \R ^3_\xx$ with each matrix components 
being Borel measurable. Note that the function $V$ naturally defines a linear operator acting in $\H_\rm{D}$ and
we denote it by the same symbol $V$.
The Hamiltonian of the Dirac particle under the influence of this external potential $V$ is then given by the Dirac operator
\begin{align}
H_\mathrm{D} (V) := { \boldsymbol \alpha } \cdot \bvec{p} +M \beta +V 
\end{align}
acting in $\H_{\mathrm{D}}$,
with the domain $ D(H_\mathrm{D} (V)) := H^1 (\Real ^3 ; \C ^4) \cap D(V) $, where $ H^1 (\Real ^3; \C ^4 ) $
denotes the $ \C ^4 $-valued Sobolev space of order one. 
%We will assume that the potential satisfies the following assumption \ref{ass-V}.
Let $C$ be the conjugation operator in $\mathcal{H}_\rm{D}$ defined by
\[ (Cf)(\xx)=f(\xx)^* ,\q f\in \H_\rm{D},\q x\in \R^3,\]
where $*$ means the usual complex conjugation.
By Pauli's lemma \cite{MR1219537}, there is a $4\times 4$ unitary matrix $U$ satisfying
\begin{align}
U^2&=1, \q UC=CU, \label{Pauli-matrix1}\\
U^{-1}\alpha^j U &= \overline{\alpha^j}, \q j=1,2,3, \q U^{-1}\beta U = -\beta, \label{Pauli-matrix2}
\end{align}
where for a matrix $A$, $\overline{A}$ denotes its complex-conjugated matrix
and $1$ the identity matrix.
We assume that the potential $V$ satisfies the following conditions :
\begin{Ass}\label{ass-V}
\begin{enumerate}[(I)]
\item Each matrix component of $V$ belongs to
\[ L^2_\rm{loc}(\R^3):=\left\{f:\R^3\to \C\, \Bigg|\, \text{Borel measurable and } \int_{|\xx|\le R}|f(\xx)|^2<\infty \,\text{ for all $R>0$.} \right\}.\]
\item $V$ is Charge-Parity (CP) invariant in the following sense:
\begin{align}
U^{-1}V(\xx)U = V (-\xx)^*, \q \rm{a.e.} \,x\in\R^3.
\end{align}
\item $H_\rm{D}(V)$ is essentially self-adjoint.
\end{enumerate}
\end{Ass}
\noindent Hereafter, we denote the closure of $H_\rm{D}(V)$, which is self-adjoint by Assumption
\ref{ass-V}, by the same symbol.
The important remark is that the Coulomb type potential 
\begin{align}\label{Cou}
V(\xx)=-\frac{Zq^2}{|\xx|} 
\end{align}
satisfies Assumption \ref{ass-V} provided that $Zq^2 <1/2$, or more concretely,
$Z\le 68$ if we put $q=e$, the elementary charge \cite{MR1219537}.

Suppose that there are $N$ Dirac particles in the external potential $V$. In this case, the Hilbert space should be
\begin{align}
\H_\rm{D}(N):=\wedge^N \H_\rm{D} := \ot_\rm{as}^N L^2(\Real^3;\C^4)= L^2_{\rm{as}}((\R^3\times\{1,2,3,4\})^{N}),
\end{align}
where $\ot_\rm{as}^N$ denotes the $N$-fold anti-symmetric tensor product.
The $a$-th component of 
\[ \XX = (\xx^1,l^1;\dots;\xx^N,l^N) \in (\R^{3}\times\{1,2,3,4\})^N \]
represents the position and the spinor of the $a$-th Dirac particle.
For notational simplicity, we denote the position-spinor space of one electron by
$\mathcal{X}=\R^3\times\{1,2,3,4\}$ in what follows. We regard $\mathcal{X}$ as
a topological space with the product topology of the ordinary one on $\R^3$ and 
the discrete one on $\{1,2,3,4\}$.
The $N$ particle Hamiltonian is then given by
\begin{align}
H_D(V,N)=\sum_{a=1}^N \big(1\otimes  \dots \otimes \stackrel{a\text{-th}}{H_D(V)}\otimes\dots\otimes 1\big),
\end{align}
which is written as 
\begin{align}
H_D(V,N)=\sum_{a=1}^N\big(\alpha^a\cdot \bvec p^a + \beta^a M^a + V^a \big).
\end{align}
with $\pp^a = (-iD^a_{1},-iD^a_{2},-iD^a_{3}) $ denoting the generalized differential operator with respect to the $a$-th coordinate, $\alpha^a$ and $\beta^a$ denoting the operators $ 1\otimes \dots \otimes \stackrel{a\text{-th}}{\alpha}\otimes\dots\otimes 1 $ and $ 1\otimes \dots \otimes \stackrel{a\text{-th}}{\beta}\otimes\dots\otimes 1 $ acting in $\ot^N \C^4$, respectively, 
$M^a$ being the mass of the $a$-th Dirac particle, 
and $V^a$ the matrix-valued multiplication operator in $\wedge^N \H_\rm{D}$
 by the matrix-valued function
$\mathcal{X}^N \ni (\xx^1,l^1;\dots;\xx^N,l^N)\mapsto V(\xx^a) $, acting as
\begin{align}
(V^a\Psi)(\xx^1,l^1;\dots,\xx^N,l^N)=\sum_{k^a}V_{l^ak^a}(\xx^a)\Psi(\xx^1,l^1;\dots; \xx^a,k^a;\dots ;\xx^N,l^N).
\end{align}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Free Hamiltonian of gauge field} 
Next, we introduce the free gauge field Hamiltonian in the Lorenz gauge.
We adopt as the one-photon Hilbert space
\begin{align}
\H _\mathrm{ph} := L^2 (\R ^3 _\kk ; \C ^4) .
\end{align}
The above $ \R ^3 _\kk := \{ \mathbf{k} = (k^1,k^2,k^3) \, | \, k^j \in \R , \, j=1,2,3 \} $ represents the momentum space of photons, and the the target space $\C^4$ represents the
degrees of freedom coming from the polarization of photons. As is well known, there are four
polarizations, two of which are physical (or transverse) ones while the other two 
are unphysical, scalar and longitudinal ones. 
We often omit the subscript $ \kk $ in $ \R ^3 _\kk $, and just denote it by $\R^3$, when there is no danger of confusion.
The Hilbert space for the quantized gauge field in the Lorentz gauge is given by 
\begin{align}
\F _\mathrm{ph} :=  \op _{n=0} ^\infty \ot _\rm{s} ^n \H _\rm{ph} = \Big\{ \Psi = \{ \Psi ^{(n)} \} _{n=0} ^\infty \, \Big| \, \Psi ^{(n)} \in \ot _\rm{s} ^n \H _\rm{ph} , \, \, ||\Psi ||^2:=\sum _{n=0} ^\infty \| \Psi ^{(n)} \| ^2 <\infty \Big\} ,
\end{align}
the Boson Fock space over $ \H _\mathrm{ph} $, where $ \ot _\rm{s} ^n $ denotes the $ n $-fold symmetric tensor product with the convention $ \ot _\rm{s} ^0 \H _\rm{ph} := \C $.
Let $ \omega (\kk ) := |\kk | , \, \kk \in \R ^3 $, the energy of a photon with momentum $\kk\in
\Real^3$. The multiplication operator by the $4\times 4$ matrix-valued function 
\begin{align}
\kk\mapsto
\begin{pmatrix}
 \omega (\kk) & 0 & 0& 0\\
 0 &\omega(\kk) &0&0 \\
 0&0&\omega(\kk) &0\\
 0&0&0&\omega(\kk)
\end{pmatrix}
 \end{align}
acting in $\H_\rm{ph}$ is self-adjoint, and we also denote it by the same symbol $ \omega $. This operator 
$\omega$ is a one-photon Hamiltonian
in $\mathcal{H}_{\rm{ph}}$, and 
the free Hamiltonian (kinetic term) of the quantum gauge field is given by its second quantization
\begin{align}
H_\mathrm{ph} &:= \mathrm{d}\Gamma _\bb (\omega ) :=\op_{n=0}^\infty \overline{ \Big( \sum _{j=1} ^n 1 \otimes \dots \otimes  1 \otimes \stackrel{j\text{-th}}{\omega} \otimes 1 \otimes \dots \otimes 1 \Big) \upharpoonright \hot ^n D(\omega ) } ,
\end{align}
where $\hot$ denotes the algebraic tensor product.
The operator $ H_\mathrm{ph} $ is self-adjoint.

\subsection{Canonical commutation relations and indefinite metric}
The main obstacle for the mathematical treatment of the Lorentz covariant gauge is 
that in order to realize the canonical commutation relations in a Lorentz covariant manner,
we have to employ an indefinite metric vector space as a space of state vectors.
In the above Hilbert space $\H _\mathrm{ph}$ for photon fields, the ordinary positive definite inner
product 
\begin{align}
\langle F, G\rangle:=\sum_{\mu=0}^3 \int_{\Real^3} F^\mu(\kk)^*G^\mu(\kk) \,d\xx
\end{align}
for $F(\kk)=(F^0(\kk),\dots,F^3(\kk))$ and $G(\kk)=(G^0(\kk),\dots,G^3(\kk))$ in $\F _\mathrm{ph}$, \textit{plays
no physical role} and just define the Hilbert space topology on the vector space $\F _\mathrm{ph}$. 
 For instance, the probability amplitude that a state $\Psi\in\F _\mathrm{ph}$
is observed in a state $\Phi\in\F _\mathrm{ph}$ is not given by $\langle\Psi,\Phi\rangle$. 
The ``physical" inner product is given by the mapping
\begin{align}
\langle F| G\rangle:= \int_{\Real^3} \eta_{\mu\nu} F^\mu(\kk)^*G^\nu(\kk) \,d\xx,
\end{align} 
where the summation over $\mu,\nu= 0,1,2,3$ is understood (In what follows, we omit the summation
symbol whenever the summation is taken with respect to one upper and one lower Lorentz indicies).
The indefinite metric naturally induced on $\F_\rm{ph}$ from this indefinite metric on $\H_\rm{ph}$ is a physical one in the sense that $\langle\Psi | \Phi\rangle$ gives the probability amplitude that a state $\Psi\in\F _\mathrm{ph}$
is observed in a state $\Phi\in\F _\mathrm{ph}$.
Note that $\langle\Psi | \Psi\rangle$ may become negative and thus $\F_\rm{ph}$ contains a lot of unphysical state vectors with
``negative probability", which should be eliminated.
Note also that the self-adjointness which is to be required for the Hamiltonian has to be that with respect to the indefinite metric $\langle\cdot|\cdot\rangle$,
in stead of the ordinary one with respect to the unphysical inner product $\Expect{\cdot,\cdot}$.  

We make precise the statement that a linear operator 
is self-adjoint with respect to the indefinite metric $\langle\cdot|\cdot\rangle$.
The metric tensor $\eta=(\eta_{\mu\nu})$ is considered to be a linear operator 
on $\H_\rm{ph}$ by
\[ (\eta F)^\nu (\kk)=\eta_{\mu\nu} F^\mu(\kk),\q F=(F^\mu)_{\mu=0}^3=(F^0,\dots,F^3) \]
and we also denote by $\eta$ the second quantization of $\eta$,
\[ \eta :=\mathrm{\Gamma}_\bb(\eta):= \op_{n=0}^\infty  \Big(\eta \otimes \dots \otimes  \eta  \Big), \]
whenever no confusion may occur.
Then $ \eta $ is unitary and satisfies $ \eta ^* = \eta , \, \eta ^2 =I $.
We remark that the physical inner product can be written as
\[ \langle\Psi|\Phi\rangle = \langle\Psi,\eta\Phi\rangle. \]
\begin{Def}
For a densely defined linear operator $T$, we denote 
\begin{align}
T^\dagger = \eta T^* \eta.
\end{align}
and call it physical adjoint or $\eta$-adjoint of $T$, where $T^*$ denotes the ordinary adjoint with respect to the positive 
definite inner product $\langle\cdot,\cdot\rangle$. 
\end{Def}
Clearly, it follows that
\begin{align}
\inprod \Psi  {T\Phi } = \inprod{T^\dagger \Psi} \Phi  , \q \Psi \in D(T^\dagger ) , \q \Phi \in D(T) .
\end{align}
The notions on self-adjointness with respect to $\inprod\cdot\cdot$ are defined as follows :
\begin{Def}\label{eta-sa}
\begin{enumerate}[(i)] 
\item A densely defined linear operator $ T $ is symmetric with respect to
$\inprod\cdot\cdot$ or $ \eta $-symmetric if $ T \subset T^\dagger $.

\item A densely defined linear operator $ T $ is self-adjoint with respect to
$\inprod\cdot\cdot$ or $ \eta $-self-adjoint if $ T^\dagger =T $.

\item A densely defined linear operator $ T $ is essentially self-adjoint with respect to
$\inprod\cdot\cdot$ or essentially $ \eta $-self-adjoint if $ \overline{T} $ is $ \eta $-self-adjoint.

%\item A densely defined linear operator $ T $ is unitary with respect to
%$\inprod\cdot\cdot$ or $ \eta $-unitary if $ T $ is injective and $ T ^\dagger = T^{-1} $.
\end{enumerate}
\end{Def}

\begin{Lem}\label{eta-lem} 
\begin{enumerate}[(i)] 
\item $ T $ is $ \eta $-symmetric if and only if $ \eta T $ is symmetric.

\item $ T $ is $ \eta $-self-adjoint if and only if $ \eta T $ is self-adjoint. 

\item $ T $ is essentially $ \eta $-self-adjoint if and only if $ \eta T $ is essentially self-adjoint.

\item If $ T $ is $ \eta $-symmetric then $ T $ is closable.

\item Let $ T $ be $ \eta $-self-adjoint and $ \eta T $ is essentially self-adjoint on a subspace $ D $, Then $ D $ is a core of $ T $.
\end{enumerate}
\end{Lem}

\begin{proof} See \cite{MR2533876}.
\end{proof}
Note that the free Hamiltonian $ H_\rm{ph} $ is self-adjoint and $ \eta $-self-adjoint.

The creation operator $ c^\dagger(F) $ with $ F \in \H _\mathrm{ph} $ is a densely defined closed linear operator on $ \F _\mathrm{ph} $ given by
\begin{align}
(c^\dagger(F)\Psi ) ^{(0)} = 0 , \q (c^\dagger (F) \Psi ) ^{(n)} = \sqrt{n} S_n (F \otimes \Psi ^{(n-1)}) , \q n \ge 1 , \q \Psi \in D(c^\dagger(F)), 
\end{align}
where $ S_n $ denotes the symmetrization operator on $ \otimes ^n \H _\rm{ph} $, i.e. $ S_n (\otimes ^n \H _\rm{ph}) = \otimes _\rm{s} ^n \H _\rm{ph} $. We note that $ c^\dagger (F) $ linear in $ F $.
For $f\in L^2(\R)$, we introduce the components of $c^\dagger(\cdot)$ with \textit{lower} indices by
\begin{align}
& c^\dagger_{0} (f) := c^\dagger(f,0,0,0) , \q c^\dagger_{1} (f) := c^\dagger (0,f,0,0) , \\
& c^\dagger_{2} (f) := c^\dagger(0,0,f,0) , \q c^\dagger_{3} (f) := c^\dagger(0,0,0,f).
\end{align}
Then, for $F=(F^\mu)\in\mathcal{H}_\mathrm{ph}$, we have
\begin{align}\label{cov-1}
c^\dagger (F) = c^\dagger_\mu(F^\mu).
\end{align}

Next, we introduce the annihilation operators. 
The annihilation operator is labeled by the element of $\mathcal{H}_\mathrm{ph}^*$, the dual space of
$\mathcal{H}_\mathrm{ph}$. In the ordinary case, the dual space of 
a Hilbert space is naturally identified with the original Hilbert space
via the anti-linear mapping
\begin{align}
F\mapsto \Expect{F,\cdot}.
\end{align}
But, in the present case, the identification mapping we should employ is
\begin{align}
F\mapsto \inprod{F}{\cdot}.
\end{align}
That is, $\phi\in\mathcal{H}_\mathrm{ph}^*$ is identified with the vector 
 $\overline{F}\in\mathcal{H}_\mathrm{ph}$ by the relation
\begin{align}
	\phi(G)=\inprod{\overline{F}}{G}.
\end{align}
Note that there is another vector $F\in\mathcal{H}_\mathrm{ph}$ which is also identified with $\phi\in\mathcal{H}_\mathrm{ph}^*$
via
\begin{align}
	\phi(G)=\Expect{F,G},
\end{align}	
and they are related by
\begin{align}\label{eta-cord}
 \eta F = \overline{F}. 
\end{align}
We employ the notations in which $\mu$-th component of $\overline{F}$ in the direct sum decomposition
\begin{align}
\mathcal{H}_\mathrm{ph} = \op_{\mu=0}^3 L^2(\R^3)
\end{align}
is written as $\overline{F}=(\overline{F}_\mu)$ by \textit{lower} indices and \eqref{eta-cord} can be read as
\begin{align}
\eta_{\mu\nu}F^\nu = \overline{F}_\mu.
\end{align}
Now, the annihilation operator which annihilates one photon with the state $F\in\mathcal{H}_\mathrm{ph}$
is given by $c(\overline{F})=c(\eta F)$, where $c(\cdot)$ is the usual annihilation operator 
of the photon Fock space $\mathcal{F}_\mathrm{ph}$ (for $F\in\mathcal{H}_\mathrm{ph}$, 
$c(F)=c^\dagger(F)^*$). One sees that
\begin{align}
(c^\dagger (F))^\dagger = c (\overline{F}).
\end{align}
It is natural to introduce the components of $c(\cdot)$ with \textit{upper} indices by
\begin{align}
& c^{0} (f) := c(f,0,0,0) , \q c^{1} (f) := c (0,f,0,0) , \\
& c^{2} (f) := c(0,0,f,0) , \q c^{3} (f) := c(0,0,0,f),
\end{align}
for $f\in L^2(\R^3)$. Then, it follows that 
\begin{align}\label{cov-2}
c(\overline{F})=c^\mu(\overline{F}_\mu)=c^\mu(\eta_{\mu\nu}F^\nu).
\end{align}
Form \eqref{cov-1} and \eqref{cov-2}, we have the natural identity
\begin{align}
c^\dagger_\mu(f)=(\eta_{\mu\nu} c^\nu(f))^\dagger,
\end{align}
for all $f\in L^2(\R^3)$.
 
 As is well known, the creation and annihilation operators leave the finite particle subspace 
\begin{align}
\F _{\bb , 0} (\H _\rm{ph}) := \Big\{ \{ \Psi ^{(n)} \} _{n=0} ^\infty \in \F _\rm{ph} \, \Big| \,  \Psi ^{(n)} =0 \, \text{for all $ n\ge n_0 $ with some $n_0$} \Big\} 
\end{align}
invariant and satisfy the canonical commutation relations, which leads in the present case
\begin{align}
[c(\overline{F}) , c^\dagger(F')  ] = \inprod{F}{F'}, \q [c(\overline{F}) , c(\overline{F'} ) ] = [c^\dagger(F)  , c^\dagger(F' ) ]=0,
\end{align}
on $ \F _{\bb, 0} (\H _\rm{ph}) $. 

Let us introduce photon polarization vectors $\{e_\lambda \}_{\lambda=0,1,2,3}$.
Photon polarization vectors are 
$ \R_\kk ^4 $-valued measurable functions defined on $\Real^3$, $ e_\lambda (\cdot)$ ($\lambda=0,1,2,3$),  
such that, for all $ \kk \in M_0  := \R ^3 \backslash \{ (0,0, k^3) \, | \, k^3 \in \R \} $ , 
\begin{align}\label{pol-1}
e _{\lambda} (\kk ) \cdot e _{\sigma} (\kk ) = \eta_{\lambda \sigma} , \q e _{\lambda} (\kk )\cdot k=0 , \q \lambda =1,2,
\end{align}
where the above $\cdot$ means the Minkowski inner product defined by
\[ e_\lambda (\kk)\cdot e _{\sigma} (\kk ) = \eta_{\mu\nu}e^\mu_{\;\;{\lambda}}(\kk ) e^\nu_{\;\;\sigma} (\kk),\q e _{\lambda} (\kk ) \cdot k = \eta_{\mu\nu}e^\mu_{\;\;\lambda} (\kk )  k^\nu,
\q k = (|\kk|, \kk)\in\Real^4,\]
with $e^\mu_{\;\;\lambda}$ being the $\mu$-th component of $e_\lambda$ with respect to the standard basis in $\R^4$.
 Note that such vector valued functions can be chosen so that they are continuous
 on $M_0$, for instance, we may choose
\begin{align}\label{pol-2}
e_0(\kk)=(1,\bvec{0}),\q e_1(\kk)=(0,\ee_1(\kk)),\q e_2(\kk)=(0,\ee_2(\kk)),\q e_3(\kk)=(0,\kk/|\kk|) 
 \end{align}
by using $\{\ee_r (\kk)\}_{r=1,2}$ satisfying the relations
\[\mathbf{e}_r(\kk ) \cdot \mathbf{e}_{r'} (\kk ) = \delta _{rr'} , \q \mathbf{e}_r (\kk ) \cdot \kk =0, \q r,r' =1,2. \]
In this paper, we assume the photon polarization vectors are chosen in this way.

For each $ f\in L^2 (\R _\kk ^3) $ and $ \mu =0,1,2,3 $, we define
\begin{align}
 a^\mu (f) &:= c\big( f e^\mu_{\;\;0} ,\,  f e^\mu_{\;\;1}  ,\, f e^\mu_{\;\;2}  ,\, f e^\mu_{\;\;3}  \big) \no\\
	&= c^\nu(e^\mu_{\;\;\nu} f),
\end{align}
and 
\begin{align}
a_\mu(f):=\eta_{\mu\nu} a^\nu(f).
\end{align}
Then the adjoints are given by
\begin{align}
& a^{\dagger\mu}(f):=(a^\mu(f))^\dagger=\eta^{\sigma\lambda}c^\dagger_\sigma (e^\mu_{\;\:\lambda}f), \\
& a^\dagger_\mu(f):=(a_\mu(f))^\dagger=\eta^{\sigma\lambda}c^\dagger_\sigma (e_{\mu\lambda}f), 
\end{align}
where we have defined 
\begin{align}
e_{\mu\lambda}:=\eta_{\mu\nu}e^{\nu}_{\;\;\lambda}.
\end{align}
The operators $ a_\mu (f) $ and $ a_\mu ^\dagger (f) $ are closed, and satisfy the
Lorentz covariant canonical commutation relations:
\begin{align*}
[a_\mu (f) , a_\nu ^\dagger (g) ] & = \eta_{\mu\nu } \left\langle f,g \right\rangle _{L^2 (\R ^3)} , \\
[a_\mu (f) , a_\nu (g)] & = [a_\mu ^\dagger (f) , a_\nu ^\dagger (g) ] =0 ,
\end{align*}
on $ \F _{\bb , 0} (\H _\rm{ph}) $.

\subsection{Gauge field operator}
For all $ f \in L^2 (\R ^3 _\xx ) $ satisfying $ \hat{f}/\sqrt{\omega} \in L^2 (\R ^3 _\kk ) $, we set 
\begin{align}
A_\mu (f) := 
\frac{1}{\sqrt{2}}\left(a_\mu \left( \frac{\hat{f^*}}{\sqrt{\omega }} \right) + a_\mu ^\dagger \left( \frac{\hat{f}}{\sqrt{\omega}} \right)\right) ,
\end{align}
where $ \hat{f} $ denotes the Fourier transform of $ f $, and $ f^* $ denotes the complex conjugate of $ f $. The functional $ \mathscr{S} (\R ^3 _\xx ) \ni f \mapsto A_\mu (f) $ gives an operator-valued distribution (Cf. \cite{MR2412280} Definition 2.5)
 acting on $ (\F _\rm{ph} , \F _{\bb , 0} (\H _\rm{ph})) $ and it is called the quantized gauge field at time $ t=0 $.
Now, fix $ \chi _\rm{ph} \in L^2 (\R ^3_\xx ) $ which is real and satisfies $\chi_\rm{ph}(\bvec x)=\chi_\rm{ph}(-\xx)$ and $ \hat{\chi _\rm{ph}} / \sqrt{\omega} \in L^2 (\R ^3 _\kk ) $. We set
\begin{align}
& A_\mu (\xx ) := A_\mu (\chi _\rm{ph} ^\xx ) , \\
& \chi _\rm{ph} ^\xx (\yy ) := \chi _\rm{ph} (\yy - \xx ) , \q \yy \in \R ^3 .
\end{align}
$ A_\mu (\xx ) $ is called the point-like quantized gauge field with a momentum cutoff $ \hat{\chi _\mathrm{ph}} $.
Note that $\hat{\chi _\mathrm{ph}}$ is real-valued since $\chi_\rm{ph}(\bvec x)=\chi_\rm{ph}(-\xx)$.
As will be seen later, for real-valued $ f $, the closures of $ A_\mu (f) , \, \mu =0,1,2,3, $ are $ \eta $-self-adjoint but not even normal in the positive definite inner product $\Expect{\cdot,\cdot}$. 
%%%%%%%%%%%
\subsection{Interaction between electrons and gauge field and total Hamiltonian}
Next, we introduce the interaction Hamiltonian and total Hamiltonian 
in the Hilbert space of state vectors 
for the coupled system, which is
taken to be 
\begin{align}
\F _\mathrm{DM}(N) := \H_\rm{D}(N)\ot \F _\mathrm{ph} .
\end{align}
We remark that this Hilbert space can be naturally identified with
\begin{align}
\F _\mathrm{DM} (N)= L^2_\rm{as} (\mathcal{X}^{N} ; \F _\mathrm{ph}) = A_N\int ^{\oplus} _{\mathcal{X}^{N}} d\XX \, \F _\mathrm{ph},
\end{align}
the Hilbert space of $ \F _\mathrm{ph}$-valued functions on $ \mathcal{X}^{N}=\mathcal{X}\times\dots\times\mathcal{X}$
which are square integrable
with respect to the Borel measure (the product measure of the Lebesgue measure on $\R^3$ and the counting measure on $\{1,2,3,4\}$)
 and which are anti-symmetric in the arguments,
that is, the exchange of the $a$-th electron with the $b$-th electron
\begin{align}
(\xx^1,l^1;\dots;\xx^a,l^a;\dots;\xx^b,l^b;\dots;\xx^N,l^N)\mapsto (\xx^1,l^1;\dots;\xx^b,l^b;\dots;\xx^a,l^a\dots;\xx^N,l^N)
\end{align}
gives a minus sign.
The last expression is the constant fibre direct integral with the base space $ (\mathcal{X}^{N} , d\XX ) $ and fibre $\F _\mathrm{ph}$. We freely use this identification. 
 
The mappings $ \XX \mapsto \chi_\rm{ph}^{\xx^a} $ ($a=1,2,\dots,N$) from $ \mathcal{X}^{N} $ to $ \H _\mathrm{ph} $ is strongly continuous, 
and thus we can define a decomposable self-adjoint operator $A_\mu$ by
\begin{align}
A_\mu^a := \int ^\oplus _{\mathcal{X}^{N}} d\XX \, A_\mu(\xx^a ) ,\q \mu=0,1,2,3,\q a=1,2,\dots,N,
\end{align}
acting in $ \int ^\oplus _{\mathcal{X}^{N}} d\XX \, \F _\mathrm{ph}$.
Now we are ready to define the minimal interaction
Hamiltonian $H_1$, between the Dirac particle and the quantized gauge field with the UV cutoff $ \chi_\rm{ph} $. It is given by
\begin{align}
H_1 := q \sum_{a=1}^N \alpha^a \cdot A^a=q\sum_{a=1}^N \alpha^{a\mu} A_\mu^a, \q \alpha^{a\mu}=(1,\bvec{\alpha})=(1,\alpha^{a1},\alpha^{a2},\alpha^{a3}).
\end{align}
The total Hamiltonian of the coupled system is then given by
\begin{align}
H_\mathrm{DM} (V,N) &:= H_0  + H_1, \\
 H_0& :=\overline{H_\mathrm{D} (V,N) + H_\mathrm{ph}}.
\end{align} 
This is the $N$-particle \textit{Dirac-Maxwell Hamiltonian} in the Lorenz gauge. 
We remark that there seems to be no term for the Coulomb interaction between Dirac particles in the Hamiltonian
at a first glance. 
In the Lorenz gauge, this contribution is included in the photon kinetic term $\rm{d\Gamma}_\rm{b}(\omega)$. In fact,
as will be seen later, the Coulomb interaction of Dirac particles emerges from the photon kinetic term via
gauge transformation.

\section{Self-adjointness, Current Conservation and Equations of Motion}\label{time-evolution}
In the present section, we consider the self-adjointness of the Dirac-Maxwell Hamiltonian and the 
time evolution generated by it. The abstract theory developed in Section \ref{abstract}
is applied with $A=N_\bb:=1\otimes\rm{d}\Gamma_\bb(1)$ (See Lemma \ref{C_0-lemma}).
The subspace $V_L$ then becomes
\begin{align}
V_L= R(E_{N_\bb}([0,L]))=\H_\rm{D}(N)\ot \left( \op_{n=0}^L \ot_{s}^n \H_\rm{ph} \right), \q L\in\N,
\end{align}
and $D=\cup_{L}V_L$ is
\begin{align}
D= \hot_{n=0}^\infty \left(\H_\rm{D}(N)\ot \left(\ot_{s}^n \H_\rm{ph} \right)\right).
\end{align}
In what follows, by the word commutator of the linear operator $A$ and $B$, we \textit{always} mean the weak commuatator 
which is denoted by $[A,B]$.
For instance, if we say $[A,B]=C$ on a subspace $\mathcal{D}$, it means that
$\mathcal{D}\subset D(A)\cap D(A^*)\cap D(B)\cap D(B^*)\cap D(C)$ and
\begin{align}
\ip{A^*\psi}{B\phi} - \ip{B^*\psi}{A\phi}= \ip{\psi}{C\phi}, \q \psi,\phi\in \mathcal{D}.
\end{align} 
\subsection{self-adjointness}
\begin{Thm}\label{ess-s.a.}
Under Assumption \ref{ass-V}, $H_{\rm{DM}}(V,N)$ is essentially $\eta$-self-adjoint.
\end{Thm}
To prove Theorem \ref{ess-s.a.}, we prepare some Lemmas.

\begin{Lem}\label{field-lemma}
For all real-valued $f\in L^2(\R)$ with $\hat{f}/\sqrt{\omega}\in L^2(\R)$, the estimate
\begin{align}\label{A_mu-bound}
\norm{A_\mu(f) \Psi} \le 4\sqrt{2} \norm{\frac{\hat{f}}{\sqrt{\omega}}}\norm{(\rm{d}\Gamma_\bb(1)+1)^{1/2}\Psi}
\end{align}
holds for $\Psi\in D(\rm{d}\Gamma(1)^{1/2})$.
\end{Lem}
\begin{proof}
Let $\Psi\in D(\rm{d}\Gamma(1)^{1/2})$. By the well-know inequalities, we have for $\mu=0,1,2,3$,
\begin{align}
\norm{c^\mu(f)\Psi}&\le \norm{f}\norm{\rm{d}\Gamma_\bb(1)^{1/2}\Psi}, \\
\norm{c_\mu^\dagger(f) \Psi} &\le \norm{f}\norm{(\rm{d}\Gamma_\bb(1)+1)^{1/2}\Psi}.
\end{align}
Form these inequalities, we find for $\mu=0,1,2,3$,
\begin{align}\label{annihilation-bound}
\norm{a_\mu\left(\frac{\hat{f}}{\sqrt{\omega}}\right)\Psi}&\le \sum_{\lambda=0}^3\norm{e_{\mu\lambda}\frac{\hat{f}}{\sqrt{\omega}}}\norm{\rm{d}\Gamma_\bb(1)^{1/2}\Psi}\no\\
&\le 4 \norm{\frac{\hat{f}}{\sqrt{\omega}}} \norm{\rm{d}\Gamma_\bb(1)^{1/2}\Psi},
\end{align}
since $|e^\mu_{\;\lambda}(\bvec k)| \le 1$ for almost every $\bvec k\in\R^3$. Similarly, we find
\begin{align}\label{creation-bound}
\norm{a^\dagger_\mu\left(\frac{\hat{f}}{\sqrt{\omega}}\right)\Psi}\le 4 \norm{\frac{\hat{f}}{\sqrt{\omega}}}\norm{(\rm{d}\Gamma_\bb(1)+1)^{1/2}\Psi}
\end{align}
for $\mu=0,1,2,3$. Inequalities \eqref{annihilation-bound} and \eqref{creation-bound} imply \eqref{A_mu-bound}.
\end{proof}

\begin{Lem}\label{C_0-lemma}
The interaction term of the Dirac-Maxwell Hamiltonian $H_1$ satisfies
\begin{enumerate}[(i)]
\item $H_1$ is densely defined, closed.
\item $H_1$ and $H_1^*$ are $N_\bb^{1/2}$-bounded.
\item $\Psi\in V_L:=R(E_{N_\bb}([0,L]))$ implies both $H_1\Psi$ and $H_1^*\Psi$ belong to the subspace $V_{L+1}$.
\end{enumerate}
Therefore, $H_1$ belongs to $\mathcal{C}_0$ class.
\end{Lem}
\begin{proof}
The statement (i) is obvious.

We prove (ii). 
Let $\Psi\in D(N_\bb^{1/2})$. Then, for all $\bvec X\in\mathcal{X}^{N}$,
$\Psi(\bvec X)\in D(\rm{d}\Gamma_\bb(1)^{1/2})$ and
\begin{align}
\int_{\mathcal{X}^{N}}\norm{\rm{d}\Gamma_\bb(1)^{1/2}\Psi(\bvec X)}^2 d\bvec X < \infty.
\end{align}
Thus, one finds $\Psi(\bvec X)\in D(A_\mu(\bvec x^a))$ ($a=1,2,\dots ,N$) and by Lemma \ref{field-lemma} 
\begin{align}
\int_{\mathcal{X}^{N}}\norm{A_\mu(\bvec x^a)\Psi(\bvec X)}^2 d\bvec X&\le 32\norm{\frac{\hat{\chi}_\rm{ph}}{\sqrt{\omega}}}\int_{\mathcal{X}^{N}}
\norm{(\rm{d}\Gamma_\bb(1)+1)^{1/2}\Psi(\bvec X)}^2 d\bvec X \no\\
&= 32\norm{\frac{\hat{\chi}_\rm{ph}}{\sqrt{\omega}}}
\norm{(N_\bb+1)^{1/2}\Psi}^2  \no\\
&<\infty.
\end{align}
This means $\Psi\in D (A^a_\mu)$ ($\mu=0,1,2,3$, $a=1,2,\dots, N$) and 
\begin{align}
\norm{A^a_\mu\Psi} \le 32  \norm{\frac{\hat{\chi}_\rm{ph}}{\sqrt{\omega}}} \norm{(N_\bb+1)^{1/2}\Psi},
\end{align}
and therefore $\Psi\in D(H_1)$ with
\begin{align}
\norm{H_1\Psi}&\le 32|q|\sum_{a=1}^N\norm{\alpha^a{}^\mu} \norm{\frac{\hat{\chi}_\rm{ph}}{\sqrt{\omega}}}\norm{(N_\bb+1)^{1/2}\Psi}\no\\
&\le 32N|q|\norm{\frac{\hat{\chi}_\rm{ph}}{\sqrt{\omega}}}\norm{(N_\bb+1)^{1/2}\Psi}.
\end{align}
Since $H_1^*\Psi=\eta H_1 \eta \Psi$ and $\eta$ commutes with $N_\bb$, $H_1^*\Psi$ satisfies the same estimate.
This proves (ii).

We prove (iii). Let $\Psi\in V_L$ for some $L\ge 0$. It is clear by definition that both $H_1$ and $H_1^*$ create at most one photon.
Thus, $H_1\Psi$ and $H_1^*\Psi$ belong to $V_{L+1}$. 
\end{proof}
\begin{proof}[Proof of Theorem \ref{ess-s.a.}]
One easily sees that $H_0$ is $\eta$-self-adjoint and $H_{\rm{DM}}(V,M)$ is $\eta$-symmetric. 
By utilizing Proposition \ref{s.a.}, Lemma \ref{eta-lem}, and Lemma \ref{C_0-lemma},
we find that it suffices to prove $\eta H_\rm{DM}(V,M)$ has at least one self-adjoint extension.
Following Arai's proof given in Ref. \cite{MR1765584}, we will prove this by applying von Neumann's Theorem (\cite{MR0493420}, Theorem
X.3), which asserts that if there is a conjugation $J$
with $J\eta H_\rm{DM}=\eta H_{\rm{DM}}J$ then $\eta H_\rm{DM}$ has a self-adjoint extension.   

Let $J_\rm{cp}$ be an anti-linear operator in $\mathcal{F}_\rm{DM}(N)$ defined by
\begin{align}
(J_\rm{cp}\Psi)(\bvec x^1,l^1;\dots;\xx^N,l^N):= \sum_{m^1,\dots,m^N} U_{l^1 m^1}\dots U_{l^N m^N}\Psi(-\xx^1,m^1;\dots;-\xx^N,m^N)^*,\label{CP}
\end{align}
where $U$ is a unitary $4\times 4$ matrix satisfying
\eqref{Pauli-matrix1} and \eqref{Pauli-matrix2}. Note that $J_\rm{cp}$ can also be written as
\begin{align}
J_\rm{cp}=\left(\ot_{a=1}^N PUC\right)\otimes \left(\op_{n=0}^\infty \ot_n C \right),
\end{align}
where $P$ is the parity transformation on $L^2(\R^3;\C^4)$ defined by
\[ (Pf)(\bvec x) = f(-\xx) ,\q f\in L^2(\R^3;\C^4) \]
 and $C$ is the complex conjugation.
 Cleary, $J_\rm{cp}$ is anti-linear, norm preserving and satisfies $J_\rm{cp}^2=1$. Thus $J_\rm{cp}$ is a conjugation on $\mathcal{F}_\rm{DM}(N)$. 
 
 We claim 
 \begin{align}\label{claim-1}
 J_\rm{cp} \eta (H_\rm{DM}(V,N)-\sum_{a}\beta^aM^a) \subset \eta( H_\rm{DM}(V,N)-\sum_a\beta^aM^a) J_\rm{cp}.
 \end{align}
 Note that \eqref{claim-1} implies the assertion. In fact, since $J_\rm{cp}$ is a conjugation, \eqref{claim-1} implies
 the operator equality (Ref. \cite{MR1765584}, Lemma 3.1):
 \begin{align}
  J_\rm{cp} \eta (H_\rm{DM}(V,N)-\sum_{a}\beta^aM^a) = \eta( H_\rm{DM}(V,N)-\sum_a\beta^aM^a) J_\rm{cp}.
 \end{align}
 Thus, by von Neumann's Theorem, $\eta(H_\rm{DM}(V,N)-\sum_{a}\beta^aM^a)$ has a self-adjoint extension, say $\eta \tilde{H}$.
 But, since $ \sum_{a}\beta^aM^a $ is a bounded operator, it is obvious that $\eta (\tilde{H}+\sum_{a}\beta^aM^a)$ is a
 self-adjoint extension of $\eta H_\rm{DM}(V,N)$. 
 Note also that $\eta$ and $J_\rm{cp}$ are commuting with each other, and thus \eqref{claim-1} is equivalent to
  \begin{align}\label{claim-2}
 J_\rm{cp} (H_\rm{DM}(V,N)-\sum_{a}\beta^aM^a) \subset ( H_\rm{DM}(V,N)-\sum_a\beta^aM^a) J_\rm{cp}.
 \end{align}
We prove \eqref{claim-2}.
 Take $\Psi\in D(H_0)$. Since the algebraic tensor product
 \begin{align}
 D_0:=\hot_{\rm{as}}^N C^\infty_0(\R^3;\C^4)\hot D(H_\rm{ph})
 \end{align}
 is a core of $H_0$ (Ref. \cite{AraiFock}, Corollary 2-27), there is $\Phi_n\in D_0$
 such that
 \begin{align}
 \Phi_n\to \Psi,\q (H_0-\sum_{a}\beta^aM^a)\Phi_n \to (H_0-\sum_{a}\beta^aM^a)\Psi
 \end{align}
 as $n$ tends to infinity. It is straightforward to verify that $J_\rm{cp}\Phi_n\in D(H_0)$ and 
 \begin{align}\label{app-1}
 J_\rm{cp} (H_0-\sum_{a}\beta^aM^a)\Phi_n  = ( H_0-\sum_a\beta^aM^a) J_\rm{cp}\Phi_n,
 \end{align} 
 if the potential $V$ satisfies Assumption \ref{ass-V} (II). Since $J_\rm{cp}$ is continuous,
 By taking the limit $n\to \infty$ on the both sides of \eqref{app-1}, we find
 $J_\rm{cp}\Psi$ belongs to the domain of $H_0$ and
 \begin{align}
J_\rm{cp} (H_0-\sum_a \beta^a M^a)\Psi =  (H_0-\sum_a \beta^a M^a)J_\rm{cp}\Psi,
 \end{align}
 That is,
\begin{align}\label{H_0-conj}
J_\rm{cp} (H_0-\sum_a \beta^a M^a) \subset  (H_0-\sum_a \beta^a M^a)J_\rm{cp}.
 \end{align}
 Next, take $\Psi \in D(H_1)$. Then, by \eqref{CP}, $J_\rm{cp}\Psi\in D(H_1)$ and
 \begin{align}
 &(J_\rm{cp} H_1 \Psi)(\xx^1,l^1;\dots;\xx^N,l^N)\no\\
 &=\sum_{m^1,\dots,m^N}U_{l^1m^1}\dots U_{l^Nm^N}\sum_{a=1}^N(q\alpha^{a\mu} A_\mu^a\Psi)^*(-\xx^1,m^1;\dots;-\xx^N,m^N) \no\\
 &=q\sum_{a=1}^N\sum_{m^1,\dots,m^N}U_{l^1m^1}\dots (U\alpha^*)_{l^am^a}U_{l^Nm^N}\left\{A_\mu(-\xx^a)\Psi(-\xx^1,m^1;\dots;-\xx^N,m^N)\right\}^* \no\\
 &=q\sum_{a=1}^N\sum_{m^1,\dots,m^N}U_{l^1m^1}\dots (\alpha U)_{l^am^a}U_{l^Nm^N} A_\mu([\chi_\rm{ph}^{-\xx^a}]^*)\Psi(-\xx^1,m^1;\dots;-\xx^N,m^N)^* \no\\
  &=q\sum_{a=1}^N\sum_{m^1,\dots,m^N}U_{l^1m^1}\dots (\alpha U)_{l^am^a}U_{l^Nm^N} A_\mu(\chi_\rm{ph}^{\xx^a})\Psi(-\xx^1,m^1;\dots;-\xx^N,m^N)^* \no\\
    &=q\sum_{a=1}^N\alpha^{a\mu} A_\mu(\xx^a)(J_\rm{cp}\Psi)(\xx^1,m^1;\dots;\xx^N,m^N) \no\\
&=( H_1J_\rm{cp}\Psi)(\xx^1,l^1;\dots;\xx^N,l^N).
 \end{align}
 Hence,
 \begin{align}\label{H_1-conj}
J_\rm{cp} (H_1-\sum_a \beta^a M^a) \subset  (H_1-\sum_a \beta^a M^a)J_\rm{cp}.
 \end{align}
Combining \eqref{H_0-conj} and \eqref{H_1-conj}, we obtain \eqref{claim-2}, which completes the proof.
\end{proof}

\subsection{Time evolution operator  and current conservation}
By Lemma \ref{C_0-lemma}, we can construct the time evolution operator $\{W(t)\}_{t\in\R}$ from the
non-self-adjoint Hamiltonian $H_\rm{DM}(V,N)$. Here we consider the time evolution of the electro-magnetic current density operator 
 $j^\mu(t,f)$ which is an operator-valued distribution defined by
\begin{align}\label{current}
j^\mu(f) := q\sum_{a=1}^N \alpha^{a\mu}\int_{\mathcal{X}^N}^\oplus f(\xx^a) \,d\bvec X, \q f\in\S(\R^3).
\end{align}
As is easily checked, \eqref{current} certainly defines an operator valued tempered distribution (Ref. \cite{MR2412280} Definition 2.5) .
\if0
 \begin{Def}
 A $d$-dimensional operator-valued tempered distribution $\phi$, acting on a dense subspace $\mathcal{D}\subset\F_\rm{DM}$, 
 is a mapping which maps a rapidly decreasing function $f\in\S(\R^d)$
 onto a densely defined linear operator $\phi(f)$ on $\F_\rm{DM}$, satisfying
 \begin{enumerate}[(i)]
 \item For all $f\in\S(\R^d)$, $\mathcal{D}\subset D(\phi(f))\cap D(\phi(f)^*)$.
 \item For all $\Psi\in\mathcal{D}$, the mapping $f\mapsto \phi(f)\Psi$ is complex linear.
 \item For all $\Psi,\Phi\in\mathcal{D}$, the $\C$-valued function $f\mapsto \inprod{\Psi}{\phi(f)\Phi}$ is 
 continuous.
 \end{enumerate}
 \end{Def}
As is easily checked, the current density $j^\mu(\cdot)$ defines an operator-valued tempered distribution.
\begin{Prop}
The mapping 
\begin{align}
\S(\R^3)\ni f\mapsto j^\mu(f)\in \mathcal{B}(\F_\rm{DM})
\end{align}
is complex linear and continuous as a mapping from $\S(\R^3)$ into $\mathcal{B}(\F_\rm{DM})$,
the set of all bounded operators in $\F_\rm{DM}$, with the operator norm topology.
In particular, this defines an operator valued tempered distribution which acts on $\F_\rm{DM}$. 
\end{Prop}
\fi
\if0
 \begin{proof}
 The linearity is obvious. The other assertions follows immediately from the inequality
 \begin{align}
 \norm{j^\mu(f)} \le |q|N \norm{f}_\infty.
 \end{align}
 \end{proof}
 \fi
 We remark that $\alpha^{a\mu}$ is the $\mu$-component of the four-component-velocity of the $a$-th electron (See Appendix), and thus, the informal definition of the current density operator
 should be 
 \begin{align}
 j^\mu (\xx)=q\sum_{a=1}^N \alpha^{a\mu}\delta(\xx -\xx^a).
 \end{align}
 If the informal point-like current density $j^{\mu}(\xx)$ is smeared with the smooth function $f$, it gives $j^\mu(f)$ in \eqref{current}.
 
% Let $\mathfrak{h}$ be a vector space consisting of functions on $\R^3$ and let $\phi$ be a mapping which maps
% $f\in\mathfrak{h}$ into a (not necessarily bounded) linear operator $\phi(f)$ acting on $\F_\rm{DM}(V,N)$.
Let $\phi$ be an operator-valued tempered distribution on $\R^3$. We define its derivative in $x^k$ ($k=1,2,3$) by
\begin{align}
\partial_k\phi(f):= \pdfrac{\phi(f)}{x^k} := - \phi\left(\pdfrac{f}{x^k}\right) , \q k=1,2,3.
\end{align} 
If $\phi$ also has a strongly differentiable time-dependence, namely, $\phi$ maps 
$(t,f)\in \R\times \mathscr{S}(\R^3)$ into $\phi(t,f)$, a linear operator in $\F_\rm{DM}$, and   
the mapping $\R\ni t\mapsto \phi(t,f)\Psi\in\F_\rm{DM}(V,N)$, is strongly differentiable in $t\in\R$ for fixed
$f\in\mathscr{S}(\R^3)$ and for $\Psi\in D(\phi(f))$, the strong derivative in $t$ is denoted by
\begin{align}
\partial_0\phi(t,f)\Psi:=\pdfrac{\phi(t,f)}{t}\Psi:=\frac{d}{dt}\phi(t,f)\Psi.
\end{align}   
 
 The most important property of the current density is that it fulfills the 
 conservation equation:
 \begin{Thm}\label{current-conservation-thm}
 The current density $j^\mu(f)$ is in $\mathcal{C}_0$-class for all $f\in \mathscr{S}(\R^3)$ and thus
 the time-dependent current density $j^\mu(t,f):=W(-t)j^\mu(f)W(t)$ exists. The zero-th component 
 $j^0(f)$ is in $\mathcal{C}_1$ and thus satisfies the strong Heisenberg equation of motion. 
 Furthermore, the current density 
 satisfies the conservation equation
 \begin{align}\label{current-conservation}
 \partial_\mu j^{\mu}(t,f):=\pdfrac{j^0(t,f)}{t} + \sum_{k=1,2,3}\pdfrac{j^k(t,f)}{x^k} =0,
 \end{align}
 on $D'$.
 \end{Thm}
 \begin{proof}
 First, we have $j^\mu(f)\in\mathcal{C}_0$, since it is bounded and leave the subspace $V_L$
 invariant.
 To prove $j^0(f)\in\mathcal{C}_1$, we compute the commutator with $iH$ on $D'$:
 \begin{align}\label{commutator-1}
 [iH,j^0(f)]&=\sum_{a}i[\bvec \alpha^a\cdot \pp^a + \beta^aM^a+V^a, j^0(f)] + [i\rm{d}\Gamma_\bb(\omega), j^0(f)]+[iH_1,j^0(f)]\no\\
 &=iq\sum_a\sum_b \alpha^{ak} \left[p^{a}_k, \int_{\mathcal{X}^N}^\oplus f(\xx^b)\,d\XX \right]\no\\
 &=q\sum_a \alpha^{ak} \int_{\mathcal{X}^N}^\oplus  \partial_k f(\xx^a)\,d\XX \no\\
 &=j^{k}(\partial_k f).
 \end{align}
 Since $j^{k}(\partial_k f)\in\mathcal{C}_0$ by the above observation, it follows that $j^0(f)\in\mathcal{C}_1$ with
 $\ad(j^0(f))=j^{k}(\partial_k f)$.
 By Theorem \ref{C_1-equation}, \eqref{commutator-1} also implies that 
 \begin{align}
 \partial_0 j^0(t,f)+\sum_{k=1,2,3}\partial_kj^{k}(t,f)=0
 \end{align}
 on $D'$.
 \end{proof}
 
 \subsection{Time evolution of the gauge field and equation of motion}
The time dependent gauge field $A_\mu(t,f)$ 
 generated by $W(t)$ fulfills the following equation of motion:
 \begin{Thm}
 The gauge field $A_\mu(f)$ is in $\mathcal{C}_2$ class for all $f\in \mathscr{S}(\R^3)$ and the 
 time-dependent field $A_\mu(t,f):=W(-t)A_\mu(f)W(t)$ satisfies the equation of motion
 \begin{align}\label{eq-of-mot}
 \Box A_\mu(t,f):=\partial_\nu\partial^\nu A_\mu(t,f)=j_\mu(t,\chi_\rm{ph}*f), \q t\in\R,\q f\in\mathscr{S}(\R^3).
 \end{align}
 \end{Thm}
 \begin{proof}
 We compute the commutator $\Pi_\mu(f):=[iH,A_\mu(f)]$ on $D'$:
 \begin{align}\label{pi-0}
 \Pi_\mu(f)&:=[iH,A_\mu(f)]=[i\rm{d}\Gamma_\bb(\omega),A_\mu(f)] \no\\
 &=\frac{1}{\sqrt{2}}\left(a_\mu\left((i\omega)\frac{\hat{f^*}}{\sqrt{\omega}}\right)+a^\dagger_\mu\left((i\omega)\frac{\hat{f}}{\sqrt{\omega}}\right)\right).
 \end{align}
 This shows that $\Pi_\mu(f)$ belongs to $\mathcal{C}_0$, and thus $A_\mu(f)\in\mathcal{C}_1$ with $\ad (A_\mu(f))=\Pi_\mu(f)$.
 Again, we have
 \begin{align}
 [iH,\Pi_\mu(f)]&=[i\rm{d}\Gamma_\bb(\omega),\Pi_\mu(f)]+[iH_1,\Pi_\mu(f)]\no\\
 &=\frac{1}{\sqrt{2}}\left(a_\mu\left((i\omega)^2\frac{\hat{f^*}}{\sqrt{\omega}}\right)+a^\dagger_\mu\left((i\omega)^2\frac{\hat{f}}{\sqrt{\omega}}\right)\right)
 +iq\sum_{a}\alpha^{a\nu}\int^\oplus_{\mathcal{X}^N}[A_\nu(\xx^a),\Pi_\mu(f)]\,d\XX \no\\
 &=A_\mu(\Delta f)+\frac{iq}{2}\sum_a\alpha_\mu^a\int_{\mathcal{X}^N}^\oplus \left\{ \ip{\hat{\chi_\rm{ph}^{\xx^a}}}{i\hat{f}}-\ip{i\hat{f^*}}{\hat{\chi_\rm{ph}^{\xx^a}}}\right\} \,d\XX \no \\
 &=A_\mu(\Delta f)-q\sum_a\alpha_\mu^a\int_{\mathcal{X}^N}^\oplus \chi_\rm{ph}*f(\xx^a)\,d\XX \no\\
 &=A_\mu(\Delta f)-j_\mu(\chi_\rm{ph}*f),
\end{align}
which shows that $\Pi_\mu(f)\in\mathcal{C}_1$ with 
\begin{align}\label{pi-1}
 \ad(\Pi_\mu(f))=\ad^2(A_\mu(f))=A_\mu(\Delta f)-j_\mu(\chi_\rm{ph}*f).
 \end{align}
 Therefore, by Theorem \ref{n-th-diff}, we conclude that $A_\mu(t,f)=W(-t)A_\mu(f)W(t)$ is twice differentiable in $t$ on $D'$ and
 \begin{align}
 \frac{d^2}{dt^2}A_\mu(t,f)=A_\mu(t,\Delta f)-j_\mu(t,\chi_\rm{ph}*f),
 \end{align}
 which is equivalent to \eqref{eq-of-mot}.
 \end{proof}
 
 It is clear by Theorem \ref{current-conservation-thm}, \eqref{pi-0} and \eqref{pi-1} that $A_0(t,f)$ ($f\in \mathscr{S}(\R^3)$) is three times differentiable in $t$ and
 \begin{align}
 \partial^\mu\Box A_\mu(t,f)\Psi = 0, \q \Psi\in D'.
 \end{align}
 This implies that the vector-valued function
 \begin{align}
 \partial^\mu A_\mu(t,f)\Psi,\q \Psi\in D'
 \end{align}
 satisfies the Klein-Gordon equation
 \begin{align}
 \Box  \partial^\mu A_\mu(t,f)\Psi = 0,\q \Psi\in D'.
 \end{align}
 It is straightforward to obtain for $f\in\mathscr{S}(\R^3)$
 \begin{align}\label{dA}
 \partial^\mu A_\mu(t,f)\Big|_{t=0}\Psi = -\frac{1}{\sqrt{2}}\left[a_\mu\left(ik^\mu\frac{\hat{f^*}}{\sqrt{\omega}}\right)+
 a^\dagger_\mu\left(ik^\mu\frac{\hat{f}}{\sqrt{\omega}}\right)\right]\Psi,\q \Psi\in D',
 \end{align}
 where $k^0:=\omega(\kk)$.
 
 \section{Gupta-Bleuler's condition and physical subspace}\label{Gupta-Bleuler}
 In the present section, we prove that the operator $\partial_\mu A^\mu(t,f)$ is 
 a generalized free field in the sense of Section \ref{GFF}, and by utilizing this fact, we
 identify the physical subspace following the Gupta-Bleuler's method. Furthermore, 
 the physical subspace naturally induces the Hilbert space with a positive definite metric,
 which we regard as the Hilbert space consisting of all the physical state vectors for 
 the quantum system under consideration.

Let $\mathfrak{h}=D((-\Delta)^{1/4})$ regarded as a Hilbert space by the inner product given by
\begin{align}
\ip{f}{g}_\mathfrak{h}:=\ip{f}{g}_{L^2(\R^3)}+\ip{(-\Delta)^{1/4}f}{(-\Delta)^{1/4}g}_{L^2(\R^3)}.
\end{align} 
A linear operator $T$ in $L^2(\R^3)$ can be 
considered to be an operator $T_\mathfrak{h}$ in $\mathfrak{h}$  with
\begin{align}
D(T_\mathfrak{h})&:=\{ f\in \mathfrak{h}\cap D(T) \,|\, Tf\in\mathfrak{h} \}, \\
T_\mathfrak{h}f&:=Tf,\quad f\in D(T_\mathfrak{h}).
\end{align}
For instance, if we put $T:=(-\Delta)^{1/2}$, then $D(T_\mathfrak{h})=D((-\Delta)^{3/4})$.

In view of \eqref{dA}, it is natural to define a mapping $\Omega:\mathfrak{h}\to\mathcal{C}_0$
 \begin{align}\label{def-omega}
\Omega(f):= -\frac{1}{\sqrt{2}}\left[a_\mu\left(ik^\mu\frac{\hat{f^*}}{\sqrt{\omega}}\right)+
 a^\dagger_\mu\left(ik^\mu\frac{\hat{f}}{\sqrt{\omega}}\right)\right]
 \end{align}
for each $f\in \mathfrak{h}$. 

\subsection{Physical Subspace, Hilbert space and Hamiltonian}
In this subsection, we see that the physical subspace determined by the Gupta-Bleuler method has a positive semi-definite metric and 
naturally induces the positive definite Hilbert space as the completion of the quotient space by
the subspace spanned by the null vectors.
The resulting Hilbert space should be regarded as the physical Hilbert space of the quantum system under consideration.
We see that the original Hamiltonian $H_\rm{DM}(V,N)$ induces the Hamiltonian on the Hilbert space which is essentially self-adjoint. 
\begin{Thm}\label{KG-omega}
Let $T=(-\Delta)^{1/2}$. The mapping $f\mapsto \Omega(f)$ defines an analytic $T_\mathfrak{h}$-free field with
\begin{align}\label{ad-omega}
\ad[\Omega(f)]&= -\frac{1}{\sqrt{2}}\left[a_\mu\left(ik^\mu(i\omega)\frac{\hat{f^*}}{\sqrt{\omega}}\right)+
 a^\dagger_\mu\left(ik^\mu(i\omega)\frac{\hat{f}}{\sqrt{\omega}}\right)\right] \no \\
&\hskip 4mm -\frac{iq}{2}\sum_a \alpha^{a\mu}\int^\oplus_{\mathcal{X}^N}d\bvec X \left[\ip{\frac{\hat{\chi_{\rm{ph}}^{{\bvec x}^a}}}{\sqrt{\omega}}}{\frac{ik_\mu\hat{f}}{\sqrt{\omega}}}
 -\ip{\frac{ik_\mu\hat{f^*}}{\sqrt{\omega}}}{\frac{\hat{\chi_{\rm{ph}}^{{\bvec x}^a}}}{\sqrt{\omega}}}\right]
\end{align}
for $f\in D((-\Delta)^{3/4})$. In particular, 
\begin{align}
\Omega(t,f)=\Omega((\cos t\sqrt{-\Delta})f)+\ad\left[\Omega\left(\frac{\sin t \sqrt{-\Delta}}{\sqrt{-\Delta}}f\right)\right]
\end{align}
on $D'$.
\end{Thm}
\begin{proof}
First, note that $D(T_\mathfrak{h})=D((-\Delta)^{3/4})$ and $D(T_\mathfrak{h}^2)=D((-\Delta)^{5/4})$.
By a direct computation, we obtain
\begin{align}
[iH,[iH,\Omega(f)]]=\Omega(\Delta f)
\end{align}
on $D'$, and thus $\Omega(f)$ belongs to $\mathcal{C}_2$ for $f\in D(T_\mathfrak{h})=D((-\Delta)^{3/4})$ with $\ad^2(\Omega(f))=\Omega(\Delta f)$.
Hence, we see that $\Omega(\cdot)$ is a $T_\mathfrak{h}$-free field by Theorem \ref{n-th-diff}. We prove it is analytic in the sense of Definition \ref{analytic-free}.

For $f\in D(T_\mathfrak{h})$, \eqref{ad-omega} follows from direct computation, which proves (ii) in Definition \ref{analytic-free}.
It is clear by \eqref{def-omega} and \eqref{ad-omega} that the statement (iii) in Definition \ref{analytic-free} is valid.

To prove (i) in Definition \ref{analytic-free}, let $f$ belong to the range of $E_T([1/N,N])$ for some $N\in\Natural$. By Lemma \ref{even-odd},
we have for $n\in\N$
\begin{align}
\ad^{2n}[\Omega(f)]=\Omega((-1)^nT^{2n}f),\label{even-o} \\
\ad^{2n+1}[\Omega(f)]=\ad[\Omega(-1)^n T^{2n}f],\label{odd-o}
\end{align}
on $D'$. By these equations, \eqref{annihilation-bound} and \eqref{creation-bound}, we find that
there is a constant $C>0$ such that
\begin{align}
\norm{\ad^n[\Omega(f)](N_b+1)^{-1/2}}\le C\norm{\frac{ik_\mu(i\omega)^n \hat{f}}{\sqrt{\omega}}}.
\end{align}
But since $\hat{f}$ belongs to the range of $E_\omega([1/N,N])$, one obtains
\begin{align}\label{esti-omega}
\norm{\ad^n[\Omega(f)](N_b+1)^{-1/2}}\le CR^n
\end{align}
for some $R>0$. The equations \eqref{even-o}, \eqref{odd-o}, and \eqref{esti-omega} implies that
$\Omega(f)$ is in $\mathcal{C}_\omega$ class. 

The last assertion immediately follows from Theorem \ref{KG-sol}.
\end{proof}


We have now arrived at the stage to discuss Gupta-Bleuler's definition of physical subspace.
From Theorem \ref{KG-omega} and \eqref{positive-fp}, we can define the positive frequency part of $\Omega(t,f)$ by
\begin{align}\label{omega+}
\Omega^+(t,f)&:=\Omega\left(\frac{e^{-i\sqrt{-\Delta}t }}{2}f\right)-\ad\left[\Omega\left(\frac{e^{-i\sqrt{-\Delta}t }}{2i\sqrt{-\Delta}}f\right)\right]\no\\
&=-\frac{1}{\sqrt{2}}a_\mu\left(\frac{ik^\mu e^{i\omega t}\hat{f^*}}{\sqrt{\omega}}\right)+\frac{iq}{2}\sum_a \int^\oplus_{\mathcal{X}^N} d\bvec X
\ip{\frac{\hat{\chi_\rm{ph}^{{\bvec x}^a}}}{\sqrt{\omega}}}{\frac{e^{-i\omega t}\hat{f}}{\sqrt{\omega}}},
\end{align} 
for $f\in D(T_\mathfrak{h}^{-1})$ on $D'$. (Note that the domain of the operator $T_\mathfrak{h}^{-1}$ is equal to the range of
$T_\mathfrak{h}$ which is $\{(-\Delta)^{1/2}f\,|\, f\in D((-\Delta)^{3/4})\}$). 
Once the positive frequency part of $\Omega(t,f)$ is known, the physical is defined following
Gupta and Bleuler by the relation
\begin{align}
V_\rm{phys}:=\overline{\{\Psi\in D' \,|\, \Omega^+(t,f)\Psi = 0, \;\text{for all}\; t\in\R ,\, f\in D(T_\mathfrak{h}^{-1}) \}}.
\end{align}

To find an explicit expression of the physical subspace $V_\rm{phys}$, we define two unitary operators following Refs. \cite{MR2533876,MR2412280}.
The first one is:
\begin{align}
W:=\op_{n=0}^\infty\ot^n \overline{w}, 
\end{align}
with
\begin{align}
\overline{w}=(w_\mu^{\;\;\nu})=\begin{pmatrix}
1/\sqrt{2} & 0 & 0& -1/\sqrt{2} \\
 0&1&0&0 \\
 0&0&1&0\\
 1/\sqrt{2}&0&0&1/\sqrt{2}
\end{pmatrix} : \mathcal{H}_\rm{ph}^*\to \mathcal{H}_\rm{ph}^*.
\end{align}
To define the other unitary operator we assume that $\hat{\chi_\rm{ph}}\in D(\omega^{-3/2})$. Under this assumption,
we put 
\begin{align}
g^{{\bvec x}^a}=(g^{{\bvec x}^a\mu})=\begin{pmatrix}
 0\\0\\0\\ i\hat{\chi_\rm{ph}^{{\bvec x}^a}}/\omega^{3/2}
\end{pmatrix}\in \mathcal{H}_\rm{ph}
\end{align} and consider the unitary operator $e^{iG}$ with the self-adjoint operator $G$ defined by
\begin{align}\label{G}
G&:=-q\sum_a \int^\oplus_{\mathcal{X}^N} d\bvec X \frac{1}{\sqrt{2}}\overline{\left[c^\mu \left(g^{{\bvec x}^a}_\mu\right)+c^{\dagger}_\mu\left(
g^{{\bvec x}^a\mu}\right) \right]} \no\\
&=-q\sum_a \int^\oplus_{\mathcal{X}^N} d\bvec X \frac{1}{\sqrt{2}}\overline{\left[c^3\left(\frac{i\hat{\chi_\rm{ph}^{{\bvec x}^a}}}{\omega^{3/2}}\right)+c^{\dagger}_3\left(
\frac{i\hat{\chi_\rm{ph}^{{\bvec x}^a}}}{\omega^{3/2}}\right) \right]}.
\end{align}
By these two unitary transformations $W$ and $e^{iG}$, $\Omega^+(t,f)$ is fairly simplified:

\begin{Lem}\label{g-trans}
Let $\hat{\chi_\rm{ph}}\in D(\omega^{3/2})$, $f\in D(T_\mathfrak{h}^{-1})$ and $h\in \mathcal{H}_\rm{ph}$ be
\begin{align}
h:=(h^\mu)=\begin{pmatrix}
i\sqrt{\omega} e^{i\omega t}\hat{f^*} \\ 0 \\ 0 \\ 0
\end{pmatrix}.
\end{align}
Then, we have
\begin{align}
We^{iG}\Omega^+(t,f)e^{-iG}W=c^\mu\left(h_\mu \right)
\end{align}
on $We^{iG}D'$.
\end{Lem}
\begin{proof}
On the subspace $D'$, we have
\begin{align}
\left[iG, \frac{1}{\sqrt{2}}a_\mu\left(\frac{ik^\mu e^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)\right]&= \frac{iq}{2}\sum_a \int^\oplus_{\mathcal{X}^N} d\bvec X \left[a_\mu\left(\frac{ik^\mu }{\sqrt{\omega}}\hat{f^*}\right), c^\dagger_\nu(g^{{\bvec x}^a\nu})\right]\no\\
&=\frac{iq}{2}\sum_a \int^\oplus_{\mathcal{X}^N} d\bvec X\ip{\frac{\hat{\chi_\rm{ph}^{{\bvec x}^a}}}{\sqrt{\omega}}}{e^{-it\omega}\hat{f}}.
\end{align}
Thus, it follows that 
\begin{align}
-\frac{1}{\sqrt{2}}e^{-iG}a_\mu\left(\frac{ik^\mu e^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)e^{iG}&=-\frac{1}{\sqrt{2}}a_\mu\left(\frac{ik^\mu e^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)
+\frac{1}{\sqrt{2}}\left[iG,a_\mu\left(\frac{ik^\mu e^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)\right] \no\\
&=-\frac{1}{\sqrt{2}}a_\mu\left(\frac{ik^\mu e^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)+\frac{iq}{2}\sum_a \int^\oplus_{\mathcal{X}^N} d\bvec X\ip{\frac{\hat{\chi_\rm{ph}^{{\bvec x}^a}}}{\sqrt{\omega}}}{e^{-it\omega}\hat{f}}\no\\
&=\Omega^+(t,f)
\end{align}
on the subspace $D'$, which implies for $\Psi \in e^{iG}D'$,
\begin{align}\label{part-1}
e^{iG}\Omega^+(t,f)e^{-iG}\Psi=\frac{1}{\sqrt{2}}a_\mu\left(\frac{ik^\mu e^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)\Psi.
\end{align}

Introduce four-component vectors $\overline{e}_{\nu}(\bvec k)$ ($\nu=0,1,2,3$ and $\kk\in\Real^3$) by
\begin{align}
\overline{e}^{\mu}_{\;\;\nu}(\kk):= e^\mu_{\;\;\rho}(\kk) w_{\nu}^{\;\;\rho}. 
\end{align}
Then, we have the relations from \eqref{pol-1}, \eqref{pol-2},
\begin{align}
\overline{e}^\mu_{\;\;\nu}(\kk)\overline{e}_{\mu\lambda}(\kk)&=\overline{\eta}_{\nu\lambda}, \no\\
\overline{e}^\mu_{\;\;\nu}(\kk)k_\mu&= \overline{\kappa}_\nu(\kk),
\end{align}
with
\begin{align}
(\overline{\eta}_{\nu\lambda}):=\begin{pmatrix}
0 &0&0& -1 \\
0&1&0&0\\
0&0&1&0\\
-1&0&0&0
\end{pmatrix},\q
(\overline{\kappa}^\mu(\kk))=\begin{pmatrix}
-\sqrt{2}|\kk| \\
0\\
0\\
0
\end{pmatrix}.
\end{align}
Then, one finds
\begin{align}
Wa_\mu\left(\frac{ik^\mu e^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)W&=a_\mu\left(w_\rho^{\;\;\mu}\frac{ik^\rho e^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)\no\\
&=c^\nu\left(\overline{\kappa}_\nu \frac{ie^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right),
\end{align}
which, combined with \eqref{part-1} implies
\begin{align}
We^{iG}\Omega^+(t,f)e^{-iG}W\Psi = \frac{1}{\sqrt{2}}c^\mu\left(\overline{\kappa}_\mu \frac{ie^{i\omega t}}{\sqrt{\omega}}\hat{f^*}\right)\Psi=c^\mu(h_\mu)\Psi
\end{align}
for all $\Psi\in We^{iG}D'$.
\end{proof}

Let $\mathcal{F}_{\rm{TL}}$ be the closed subspace of $\F_\rm{DM}(N)$
\begin{align}
\mathcal{F}_{\rm{TL}}:=\H_\rm{D}(N)\otimes\left(\C\Omega\otimes\mathcal{F}_\bb(L^2(\R^3))\otimes\mathcal{F}_\bb(L^2(\R^3))\otimes\mathcal{F}_\bb(L^2(\R^3))\right).
\end{align}
As is well known, for a dense subspace $V\subset L^2(\R^3)$, $\F_\rm{TL}$ is characterized as
\begin{align}\label{vac-character}
\mathcal{F}_{\rm{TL}}=\left.\left\{\Psi\in \bigcap_{f\in V}D(c^0(f)) \,\right|\, c^0(f)\Psi=0, f\in V \right\} = \bigcap_{f\in V}\ker c^0(f).
\end{align}
From Lemma \ref{g-trans}, we can identify the physical subspace $V_\rm{phys}$:
\begin{Thm}
\begin{align}
V_\rm{phys}=e^{-iG}W\mathcal{F}_\rm{TL}.
\end{align}
\end{Thm}
\begin{proof}
It is obvious that the right hand side is included by the left.
To prove the converse, take $\Psi\in D'$ for which $\Omega^+(t,f)\Psi$
vanishes for all $t\in\R$ and $f\in D(T_\mathfrak{h}^{-1})$.
Put $\Phi=We^{iG}\Psi$. Then, 
\begin{align}
c^\mu(h_\mu)\Phi=We^{iG}\Omega^+(t,f)\Psi=0
\end{align}
for all $t\in\R$ and $f\in D(T_\mathfrak{h}^{-1})$. By \eqref{vac-character}, such a $\Phi$ must be an element of $\mathcal{F}_\rm{TL}$. Hence,
$\Psi \in e^{-iG}W\mathcal{F}_\rm{TL}$.
\end{proof}

We investigate the properties of $\Vp$ in detail in order to reveal how the unphysical photon modes are eliminated by
the Gupta-Bleuler formulation. Here, we remark the basic fact 
\begin{align}\label{ip-wick}
&\inprod{c^\dagger_{\mu_1}(F_1^{\mu_1})\dots c^\dagger_{\mu_n}(F_n^{\mu_n})\Omega }{c^\dagger_{\nu_1}(G_1^{\nu_1})\dots c^\dagger_{\nu_m}(G_n^{\nu_m})\Omega}\no\\
&\hskip 4mm =\delta_{nm}\sum_{\sigma\in \mathfrak{S}_n} \inprod{F_1}{G_{\sigma(1)}}\dots\inprod{F_n}{G_{\sigma(n)}} \no\\
&\hskip 4mm=\delta_{nm}\sum_{\sigma\in \mathfrak{S}_n}\eta_{\mu_1\nu_{\sigma(1)}}\ip{F_1^{\mu_1}}{G_{\sigma(1)}^{\nu_{\sigma(1)}}}\dots \eta_{\mu_n\nu_{\sigma(n)}}\ip{F_n^{\mu_n}}{G_{\sigma(n)}^{\nu_{\sigma(n)}}}.
\end{align}
For a closed subspace $\V\subset \H_\rm{ph}$, we define $\mathcal{F}_\bb(\V)$ by
\begin{align}
\mathcal{F}_\bb(\V)&:=\mathcal{\overline{L}}(\{c^\dagger(F_1)\dots c^\dagger(F_n)\Omega, \Omega\,|\, F_1,\dots,F_n\in \V, n\in\N\}), \\
\mathcal{F}^+_\bb(\V)&:=\mathcal{\overline{L}}(\{c^\dagger(F_1)\dots c^\dagger(F_n)\Omega |\, F_1,\dots,F_n\in \V, n\in\N\}),
\end{align}
where $\mathcal{\overline{L}}(\{S\})$ denotes a closed subspace spanned by the vectors in a subset $S\subset\H_\rm{ph}$.
We also define for a direct decomposed subspace $\V=\V_1\op\V_2\subset\H_\rm{ph}$
\begin{align}
\mathcal{F}^+_\bb(\V;\V_1)&:=\mathcal{\overline{L}}(\{c^\dagger(F_1)\dots c^\dagger(F_n)\Omega |\, F_1,\dots,F_n\in \V, n\in\N,\text{ at least one $F_j$ belongs to } \V_1\}) \\
&=\mathcal{F}_\bb(\V)\cap \mathcal{F}_\bb(\V_2)^\perp.
\end{align}
Note that $\F_\bb^+(\V;\V)$ is naturally identified with the tensor product space
\begin{align}
\mathcal{F}^+_\bb(\V_1)\otimes \mathcal{F}_\bb(\V_2).
\end{align}
Let $w$ be a matrix $\eta \overline{w} \eta$ regarded as a bounded operator on $\H_\rm{ph}$:
\begin{align}
w&=(w^\mu_{\;\;\nu})\no\\
&=(\eta^{\mu\nu}w_\nu^{\;\;\rho}\eta_{\rho\nu})=\begin{pmatrix}
-1/\sqrt{2} & 0 & 0& 1/\sqrt{2} \\
 0&1&0&0 \\
 0&0&1&0\\
 1/\sqrt{2}&0&0&1/\sqrt{2}
\end{pmatrix} : \mathcal{H}_\rm{ph}\to \mathcal{H}_\rm{ph}.
\end{align}
Then, $W\F_\rm{TL}$ can be written as
\begin{align}\label{TL}
W\F_\rm{TL}=\H_\rm{D}(N)\ot\F_\bb(w\H_\rm{TL}),
\end{align}
where 
\begin{align}
\H_\rm{TL}:=\{F\in\H_\rm{ph}\,|\, F^0=0\}, 
\end{align}
and
\begin{align}
w\H_\rm{TL}:=\{wF\,|\, F\in\H_{\rm{TL}}\}=\{F\in\H_\rm{ph}\,|\,F^0=F^3\}.
\end{align}
We also define
\begin{align}
\H_\rm{L}&:=\{F\in\H_\rm{TL}\,|\, F^1=F^2=0\}, \\
\H_\rm{T}&:=\{F\in\H_\rm{TL}\,|\, F^3=0\}.
\end{align}
Then, $w\H_\rm{LT}$ is furnished with the direct sum decomposition
\begin{align}\label{direct-sum}
w\H_\rm{TL}=w\H_\rm{L}\op w\H_\rm{T},
\end{align}
with
\begin{align}
w\H_\rm{L}&=\{F\in\H_\rm{ph}\,|\,F^0=F^3, F^1=F^2=0\}, \\
w\H_\rm{T}&=\{F\in\H_\rm{ph}\,|\,F^0=F^3=0\}.
\end{align}
We remark here that if we write the direct sum decomposition of $F,G\in w\H_\rm{TL}$ with respect to \eqref{direct-sum}
by $F=F_T+ F_L$ and $G=G_T+ G_L$, then we find
\begin{align}\label{oth-1}
\inprod{F}{G}=\inprod{F_T}{G_T}=\ip{F_T}{G_T},
\end{align}
since for arbitrary $F_T\in w\H_\rm{T}$ and $G_L\in w\H_\rm{L}$,
\begin{align}\label{oth-2}
\inprod{F_T}{G_L}=0.
\end{align}
\begin{Lem}\label{positive-lemma} The following statements hold:
\begin{enumerate}[(i)]
\item The physical subspace $\Vp$ is non-negative. That is, for all $\Psi\in\Vp$, $\inprod{\Psi}{\Psi}\ge 0$.
\item For $\Psi\in\Vp$, $\inprod{\Psi}{\Psi}=0$ if and only if $\Psi$ belongs to
the closed subspace 
\begin{align}
\mathcal{N} := e^{-iG}(\H_\rm{D}(N)\ot\mathcal{F}^+_\bb(w\H_\rm{TL};w\H_\rm{L})).
\end{align}
\end{enumerate}
\end{Lem}
\begin{proof}
First of all, we remark that $e^{iG}$ is $\eta$-unitary so that 
\begin{align}\label{inprod-inv}
\inprod{e^{-iG}\Psi}{e^{-iG}\Psi}=\inprod{\Psi}{\Psi},
\end{align} 
since it commutes with $\eta$.
Let $\{e_n\}_{n\in\N}$ and $\{f_m\}_{m\in\N}$ be complete orthonormal systems on $w\H_\rm{L}$ and 
$w\H_\rm{T}$, respectively. 
Take arbitrary $\Psi\in \F_\bb(w\H_\rm{TL})$. Then, $\Psi$ can be written as
\begin{align}
\Psi = \sum_{j=0}^\infty\sum_{k=0}^\infty\sum_{n_1,\dots,n_j\in\N}\sum_{m_1,\dots,m_k\in\N}\alpha_{j,k}(n_1,\dots,n_j ; m_1,\dots, m_k) c^\dagger(e_{n_1})\dots c^\dagger(e_{n_j})
c^\dagger(f_{m_1})\dots c^\dagger(f_{m_k})\Omega,
\end{align}
with coefficient mappings $\alpha_{j,k}$ 
\begin{align}
\alpha_{j,k}: \N^j\times \N^k \to \C,\q j,k \ge 0
\end{align}
satisfying 
\begin{align}
\ip{\Psi}{\Psi}=\sum_{j,k=0}^\infty\sum_{n_1,\dots,n_j}\sum_{m_1,\dots,m_k}C_{j,k}(n_1,\dots,n_j;m_1,\dots,m_k)|\alpha_{j,k}(n_1,\dots,n_j ; m_1,\dots, m_k)|^2 < \infty.
\end{align}
Here, 
\begin{align}
C_{j,k}:\N^j\times \N^k \to \C,\q j,k \ge 0
\end{align}
 is defined as follows. Let $\{n_1,\dots,n_j\}=\{\nu_1,\dots,\nu_p\}$ and $\{m_1,\dots,m_k\}=\{\mu_1,\dots,\mu_q\}$
with $\nu$'s and $\mu$'s are mutually different. We denote by $N_l$ ($l=1,2,\dots,p$) the number of $\nu_l$'s in $\{n_1,\dots, n_p\}$
and denote by $M_{l'}$ ($l'=1,2,\dots,q$) the number of $\mu_{l'}$'s in $\{m_1,\dots, m_p\}$, that is,
\begin{align}
N_l&:=\sharp\{ i\,|\, \nu_l=n_i \} \\
M_{l'}&:=\sharp\{ i\,|\, \mu_{l'}=m_i \}.
\end{align}
Then, 
\begin{align}
C_{j,k}(n_1,\dots,n_j;m_1,\dots,m_k):=N_1!\dots N_p! M_1!\dots M_q!.
\end{align}
By \eqref{ip-wick} and \eqref{oth-2} one obtains
\begin{align}\label{inprod}
\inprod{\Psi}{\Psi}=\sum_{k=0}^\infty\sum_{m_1,\dots,m_k}C_{0,k}(m_1,\dots,m_k)|\alpha_{0,k}( m_1,\dots, m_k)|^2.
\end{align}

From \eqref{inprod-inv} and \eqref{inprod}, the assertion (i) is obvious. 
The assertion (ii) also immediately follows from the fact that $\Psi\in \F_\bb^+(w \H_\rm{TL}; w\H_\rm{L})$ if and only if
$\alpha_{0,k}=0$ for all $k\ge 0$.
 \if0
Next, for the direct sum decomposition 
\begin{align}
w\H_\rm{TL}=w \H_\rm{L}\op w\H_\rm{T},
\end{align}
we denote $F=F_T\op F_L$. Then, $\inprod{F}{G}=\inprod{F_T}{G_T}$ and thus
it is clear that, for $F\in w\H_\rm{TL}$ and $G\in w\H_\rm{L}$, $\inprod{F}{G}=0$.

Therefore, the statement (i) is obvious from \eqref{ip-wick} and \eqref{TL} and the fact that 
$w\H_\rm{TL}$ is non-negative, that is, for all $F\in w\H_\rm{TL}$, $\inprod{F}{F}=\eta_{\mu\nu}\ip{F^\mu}{F^\nu}\ge0$.

We prove (ii). For the direct sum decomposition 
\begin{align}
w\H_\rm{TL}=w \H_\rm{L}\op w\H_\rm{T},
\end{align}
we denote $F=F_T\op F_L$. Then, $\inprod{F}{G}=\inprod{F_T}{G_T}$ and thus
it is clear that, for $F\in w\H_\rm{TL}$ and $G\in w\H_\rm{L}$, $\inprod{F}{G}=0$.
From this observation and \eqref{ip-wick}, the ``if" part is obvious.

 In order to prove the ``only if" part, take arbitrary $\Psi \in \Vp$. There is $\Phi\in\F_\bb(w\H_\rm{TL})$ with $\Psi=e^{-iG}\Phi$.
 By the remark given at the beginning of the proof, it suffices to show $\inprod{\Phi}{\Phi}=0$ implies $\Phi\in \mathcal{F}^+_\bb(w\H_\rm{TL};w\H_\rm{L})$.
 The inner product $\inprod{\Phi}{\Phi}$ is given by an infinite summation of non-negative terms of the form
 \begin{align}\label{non-zero}
 \inprod{c^\dagger(F_1)\dots c^\dagger(F_n)\Omega}{c^\dagger(F_1)\dots c^\dagger(F_n)\Omega}, \q F_1,\dots F_n\in w\H_\rm{TL}.	
 \end{align}
 Therefore, from the formula \eqref{ip-wick}, one finds that for each $n\ge1$,
 \begin{align}\label{vanish}
\inprod{F_1}{F_{\sigma(1)}}\dots\inprod{F_n}{F_{\sigma(n)}}=0,
 \end{align}
 for all $\sigma\in\mathfrak{S}_n$ and $F_1,\dots F_n\in w\H_\rm{TL}$.
  Consider the case where $\sigma$ is identity. The condition \eqref{vanish} means
 $\inprod{F_j}{F_j}=\inprod{F_{jT}}{F_{jT}}=0$ for some $j$. But $\inprod{\cdot}{\cdot}$ is
 certainly positive definite if restricted to the subspace $w\H_\rm{T}$, because for all $G\in w\H_\rm{T}$,
 $\eta_{\mu\nu}\inprod{G^\mu}{G^\nu} = (G^1)^2+(G^2)^2$. Hence, $F_{jT}=0$, and $F_j\in w\H_\rm{L}$ for some $j$.
This implies $\Phi\in \mathcal{F}^+_\bb(w\H_\rm{TL};w\H_\rm{L})$ completing the proof.
\fi
 \end{proof}
 
By Lemma \ref{positive-lemma}, the quotient vector space $\Vp/\mathcal{N}$ becomes a pre-Hilbert space with 
respect to the naturally induced metric from $\eta$, and its completion
 \begin{align}
 \H_\rm{phys}:=\overline{\Vp/\mathcal{N}}
 \end{align}
is a Hilbert space with the induced metric, which we also denote by $\inprod{\cdot}{\cdot}$. This Hilbert space $\H_\rm{phys}$,
which we call physical Hilbert space, consists of all 
the physical state vectors $\Psi$ satisfying $\inprod{\Psi}{\Psi}\ge 0$.  
Let $\V$ be an orthogonal complement
to $\mathcal{N}$ in $\V_\rm{phys}$:
\begin{align}
\V_\rm{phys}=\V\oplus \mathcal{N},
\end{align}
and $P$ be an orthgonal projection onto the subspace $\V$.
Then, via unitary transformation $U:\V\to \H_\rm{phys}$,
\begin{align}\label{U}
U:\Psi \mapsto [\Psi],\q U^{-1}:[\Psi]\mapsto P\Psi, 
\end{align}
$(\V,\ip{\cdot}{\cdot})$ is identified with $(\H_\rm{phys},\inprod{\cdot}{\cdot})$ ($[\Psi]$ denotes the equivalent class to which $\Psi$ belongs.). 
We remark that $\V$ is equal to $e^{-iG}\H_\rm{D}(N)\ot\F_\bb(wH_\rm{T})$ by Lemma \ref{positive-lemma} and
that $\eta$ is identity on $\V$, namely, $\inprod{\Psi}{\Phi}=\ip{\Psi}{\Phi}$ for all $\Psi,\Phi\in\V$.

For a bounded operator $A$ in $\F_\rm{DM}$ satisfying $AV_\rm{phys}\subset V_\rm{phys}$
and $A\mathcal{N}\subset\mathcal{N}$, the bounded operator $\tilde{A}$ in $\H_\rm{phys}$ is 
defined by the relation
\begin{align}
\tilde{A}[\Psi]:=[A\Psi],\q\Psi \in V_\rm{phys}.
\end{align}
In this case, the operator $\tilde{A}$ is identified with $PAP$ by the unitary transformation $U$ given by \eqref{U}.
On the other hand, if $A$ is unbounded, the situation may become complicated since we have to be careful to deal with the operator domain. 
It may happen, in general, that even if $D(A)$ is dense in $\F_\rm{DM}$, $PD(A)=\{0\}$. 
Thus, it is not a trivial problem to define the physical Hamiltonian in a suitable manner.
We define the physical Hamiltonian by the generator of the time evolution which is naturally induced by $W(t)$.

\begin{Lem}\label{group-prop}
\begin{enumerate}[(i)]
Let $\psi\in D$ and $s,t,\lambda\in\R$, and $B\in\mathcal{C}_0$. Then,
\item $e^{i\lambda G}\psi\in D(W(s)W(t))$ and
\begin{align}
W(s)W(t)e^{i\lambda G}\psi=W(s+t)e^{i\lambda G}\psi.
\end{align}
\item $e^{i\lambda G}\psi\in D(W(-t)B(s)W(t))$ and
\begin{align}
W(-t)B(s)W(t)e^{i\lambda G}\psi = B(s+t)e^{i\lambda G}\psi.
\end{align}
\end{enumerate}
\end{Lem}
\begin{proof}
\begin{enumerate}[(i)]
\item First, the estimate for some $C>0,b>0$ and $L_\psi > 0$
\begin{align}
&\norm{\sum_{l,m,n}e^{-isH_0}U_l(s,0)e^{-itH_0}U_m(t,0)\frac{(i\lambda)^n}{n!}G^n\psi }\no\\
\le &\sum_{l,n,m}\frac{|\lambda|^n }{n!} \norm{U_l(s+t,t)U_m(t,0)G^n\psi} \no\\
\le &\sum_{l,m,n}\frac{|\lambda|^n }{n!} \frac{|s|^m}{m!}\frac{|t|^n}{n!}C^{m+n} (L_\psi+b(l+n+m-3)+1)^{1/2}\cdots(L_\psi+1)^{1/2}\norm{\psi}\no\\
\le &\sum_{N=0}^\infty \frac{1}{N!}(|\lambda|+C|s|+C|t|)^N(L_\psi+b(N-3)+1)^{1/2}\cdots(L_\psi+1)^{1/2}\norm{\psi},
\end{align} 
shows that the series 
\begin{align}
\sum_{l,m,n}e^{-isH_0}U_l(s,0)e^{-itH_0}U_m(t,0)\frac{(i\lambda)^n}{n!}G^n\psi
\end{align}
is absolutely convergent and thus defines a vector in $\mathcal{F}_\rm{DM}(V,N)$.
On the other hand, one similarly sees that the series
\begin{align}
&\sum_{n}\frac{(i\lambda)^n }{n!} G^n\psi, \\
&\sum_{m,n}e^{-itH_0}U_m(t,0)\frac{(i\lambda)^n}{n!}G^n\psi 
\end{align}
are also absolutely convergent. Since the operators $e^{i\lambda G}$, $W(s)$ and $W(t)$
are closed operators, this implies that $\psi\in D(W(s)W(t)e^{i\lambda G})$ and 
\begin{align}\label{expansion-of-LHS}
W(s)W(t)e^{i\lambda G}=\sum_{l,m,n}e^{-isH_0}U_l(s,0)e^{-itH_0}U_m(t,0)\frac{(i\lambda)^n}{n!}G^n\psi.
\end{align}
But the right hand side of \eqref{expansion-of-LHS} is equal to
\begin{align}
\sum_{N,n}e^{-i(s+t)H_0}U_N(s+t,0)\frac{(i\lambda)^n}{n!}G^n\psi = W(s+t)e^{i\lambda G\psi},
\end{align}
since the equality
\begin{align}\label{sum-rule}
\sum_{l+m=N}U_l(s+t,t)U_m(t,0)=U_N(s+t,0)
\end{align}
hols on $D$.
\item The proof is so similar to that of (i) that we omit it.
\end{enumerate}
\end{proof}

\begin{Lem}
For all $\psi \in e^{i\lambda G}D$ ($\lambda\in\R$) and $t\in\R$, 
\begin{align}\label{unitarity}
\inprod{W(t)\psi}{W(t)\psi}=\inprod{\psi}{\psi}.
\end{align}
\end{Lem}
\begin{proof}
By the easily checked equality
\begin{align}
U_n(s,t)^\dagger=U_n(t,s),\q n=0,1,2,\dots,
\end{align}
on $D$, and \eqref{sum-rule}, we have for $\psi\in e^{i\lambda G} D$ with $\phi\in D$,
\begin{align}
\inprod{W(t)\psi}{W(t)\psi}&=\inprod{W(t)e^{i\lambda G}\phi}{W(t)e^{i\lambda G}\phi}\no\\
&=\sum_{k,l,n,m}\frac{(i\lambda )^k}{k!}\frac{(i\lambda )^l}{l!}\inprod{G^k\phi}{U_n(0,t)U_m(t,0)G^l\phi}\no\\
&=\sum_{k,l}\sum_{N=0}^\infty\frac{(i\lambda )^k}{k!}\frac{(i\lambda )^l}{l!}\inprod{G^k\phi}{U_N(0,0)G^l\phi}\no\\
&=\inprod{e^{i\lambda G}\phi}{U(0,0)e^{i\lambda G}\phi}\no\\
&=\inprod{\psi}{\psi}.
\end{align}
This completes the proof.
\end{proof}
%%%
\if0
Let $S$ be a set
\begin{align}
S:=\{B, e^{i\lambda B'},e^{isH_0}, \overline{U(t,t')}\,|\, \lambda,s,t,t'\in\Real , B,B' \in\mathcal{C}_0, B' \text{ is self-adjoint}\}.
\end{align} 
Then, by using the estimate \eqref{basic-estimate} and the fact that operators in $S$ are 
closed, we find, for all $A_1,A_2,\dots,A_k,G\in S$, $D\subset D(A_1A_2\dots A_k)$ and
\begin{align}
A_1A_2\dots A_k D\subset D(G).
\end{align}
This means if we denote the set of all polynomials of the elements in $S$ by $\mathfrak{A}_S$,
then 
\begin{align}
\mathfrak{A}_S D\subset D(C),\q C\in \mathfrak{A}_S,
\end{align} 
In particular, $\mathfrak{A}_SD\subset D(W(t))$.

One can extend basic properties, such as group property of $W(t)$ 
\begin{align}
W(s)W(t)=W(s+t),
\end{align}
which have been established on $D$, to the subspace $\mathfrak{A}_SD$
by the limiting argument. In particular,
the operator valued function of $t\in\R$, $B(t)=W(-t)BW(t)$ with $B\in\mathcal{C}_0$,
fulfills
\begin{align}
W(-s)B(t)W(s)\Psi=B(s+t)\Psi,
\end{align}
and
\begin{align}
\inprod{W(t)\Psi}{W(t)\Psi}=\inprod{\Psi}{\Psi}.
\end{align}
for all $\Psi\in  e^{-iG}D$.
\fi
%%%
We employ the notations
\begin{align}
V_{\rm{phys},0}&:=e^{-iG}\left(W\F_\rm{TL}\cap D\right)=e^{-iG}\left(\H_\rm{D}(N)\ot \F_\bb(w\H_\rm{TL})\cap D\right), \\
\mathcal{N}_0&:=e^{-iG}\left(\H_\rm{D}(N)\ot\F_{\bb}(wH_\rm{T})\cap D\right).
\end{align}
Clearly, $V_\rm{phys,0}$ and $\mathcal{N}_0$ are dense subspaces of
$V_\rm{phys}$ and $\mathcal{N}$ respectively.
Then, we have

\if0
 let
\begin{align}
D_1:=\{f=(f^\mu)\in wH_\rm{TL}\,|\,  f^\mu\in C_0^\infty(\R^3)\}.
\end{align}
Then, the subspace 
\begin{align}
\mathcal{D}_1:=C_0^\infty(\R^3;\C^4)\hat{\otimes}\F_{\bb,\rm{fin}}(D_1)
\end{align}
satisfies
\begin{align}
\mathcal{D}_1\subset D, \q e^{-iG}\mathcal{D}_1\subset D(H_0)
\end{align}
so that 
\begin{align}
e^{-iG}\mathcal{D}_1\subset \mathfrak{A}_S D\cap D(H_0).
\end{align}
We ramark that $e^{-iG}\mathcal{D}_1$ is dense in $V_\rm{phys}$.
\fi
%%%%%%%%%%%% cut $$$$$$$$$$$$$$$$$$$
\begin{Lem}\label{well-def}
\begin{enumerate}[(i)]
\item $W(t)V_{\rm{phys},0}\subset V_\rm{phys}$.
\item $W(t)\mathcal{N}_0\subset \mathcal{N}$.
\end{enumerate}
\end{Lem}
\begin{proof}
Let $\Psi\in\F_{\bb,0}(w\H_\rm{TL})$. Then, the assertion (i) follows from the computation using Lemma \ref{group-prop}
\begin{align}
\Omega^+(s,f)W(t)e^{-iG}\Psi=W(t)\Omega^+(s+t,f)e^{-iG}\Psi = 0,
\end{align}
for $e^{-iG}\Psi\in V_\rm{phys}$. This means $W(t)e^{-iG}\Psi\in V_\rm{phys}$.
The assertion (ii) is obvious from \eqref{unitarity}.
\end{proof}

From Lemma \ref{well-def}, we can define the densely defined operator $\{\tilde{W}(t)\}_{t\in\R}$ on $\H_\rm{phys}$ as follows.
Let $[V_{\rm{phys},1}]$ be a dense subspace of $\H_\rm{phys}$ spanned by the vectors equivalent to a vector belonging to the subspace
\begin{align}
V_{\rm{phys},1}:=\{W(t)\Psi\,|\, t\in\R, \, \Psi\in V_{\rm{phys},0} \}.
\end{align}
On the subspace $[V_{\rm{phys},1}]$, we define a linear operator $\tilde{W}(s)$ ($s\in\R$) by  
\begin{align}
D(\tilde{W}(s))&:=[V_{\rm{phys},1}]=\{[\Psi]\,|\,\Psi=W(t)\Phi \text{ for some $t\in\R$ and $\Psi\in V_{\rm{phys},0}$}\} ,\\
\tilde{W}(s)[W(t)\Phi]&:=[W(s+t)\Phi].
\end{align}

The next lemma is crucial:
\begin{Lem}\label{unitary-group}
$\tilde{W}(t)$ defines a strongly continuous one-parameter unitary group $\{U(t)\}_t$ on $\H_\rm{phys}$.
\end{Lem}
\begin{proof}
First, we see $\tilde{W}(t)$ has the unitary extension. In fact, for all $\Psi\in V_{\rm{phys},1}$, it follows that
\begin{align}
\inprod{\tilde{W}(t)[\Psi]}{\tilde{W}(t)[\Psi]}=\inprod{W(t)\Psi}{W(t)\Psi}=\inprod{\Psi}{\Psi}=\inprod{[\Psi]}{[\Psi]}
\end{align}
which shows $\tilde{W}(t)$ is isometric. Furthermore, since $R(\tilde{W}(t))\supset [V_{\rm{phys},1}]$, $\tilde{W}(t)$ has
the unitary extension which we denote by $U(t)$.

To prove that $\{U(t)\}_t$ has the stated property, let $s,t\in\R$. Then,
\begin{align}
\tilde{W}(s)\tilde{W}(t)[\Psi]=\tilde{W}(s+t)[\Psi],\q \Psi\in V_{\rm{phys},1}.
\end{align}
Hence, we have the group property
\begin{align}
U(s)U(t)=U(s+t).
\end{align}
The continuity follows from the fact that $\tilde{W}(\cdot)[\Psi]$ is strongly continuous for $\Psi\in V_{\rm{phys},1}$.
\end{proof}

Lemma \ref{unitary-group} implies that the physical Hamiltonian $H_\rm{phys}$ should be defined as:
\begin{Def}
The physical Hamiltonian $H_\rm{phys}$ is the generator of $\{U(t)\}_t$.
\end{Def}



The following Lemma follows from lengthy but straightforward computation:
\begin{Lem}\label{transf}
On the subspace 
\begin{align}
\mathcal{D}_1:=\left(\hot_{\rm{as}}^N C^\infty_0(\R^3;\C^4)\right)\hat{\otimes}\F_{\bb,\rm{fin}}(C^\infty_0(\R^3;\C^4))
\end{align}
 one finds
\begin{align}\label{H-transf}
e^{iG}H_\rm{DM}(V,N)e^{-iG}&=H_0+q\sum_{a}\alpha^{a\mu}\int^\oplus_{\mathcal{X}^N}d\bvec X\frac{1}{\sqrt{2}}
\left[ c^\nu\left(Q_{\mu\nu}(\xx^a)\right)+c^\dagger_\nu\left(Q_\mu^{\;\;\nu}(\xx^a)\right)\right] \no\\
&\hskip 4mm -\frac{1}{2}q^2 \sum_{a,b}\alpha^{a\mu}\int^\oplus_{\mathcal{X}^N}d\bvec X \ip{k_\mu g^{\xx^a}_\nu}{g^{\xx^b \nu}},
\end{align}
where
\begin{align}
Q_\mu^{\;\;\nu}(\xx):=ik_\mu g^{\xx \nu}+e_{\mu}^{\;\;\nu}\frac{\chi_\rm{ph}^{\xx}}{\sqrt{\omega}}\in\mathcal{H}_\rm{ph},\q \xx\in\Real^3.
\end{align}
\end{Lem}
\begin{proof}
Let us compute the commutators on $\mathcal{D}_1$:
\[[iG,H_\rm{D}(V,N)],\q [iG,\rm{d}\Gamma_\bb (\omega)],\q [iG, H_1].\] 
The first one is 
\begin{align}
[iG,H_\rm{D}(V,N)]&=\left[iG,\sum_a(\bvec \alpha^a\cdot\bvec p^a+\beta^a M^a + V^a)\right] \no\\
&=\left[iG,\sum_a \bvec \alpha^a\cdot\bvec p^a\right] \no\\
&=-\frac{iq}{\sqrt{2}}\sum_{a,b}\left[\int_{\mathcal{X}^N}^\oplus d\bvec X\,\left\{c^\mu(g^{{\bvec x}^a}_\mu)+c^\dagger_\mu(g^{{\bvec x}^a\mu})\right\},\bvec \alpha^b\cdot \bvec p^b\right] \no\\
&=\frac{q}{\sqrt{2}}\sum_{a}\alpha^{aj}\int_{\mathcal{X}^N}^\oplus\left\{c^\mu(ik^j g^{{\bvec x}^a}_\mu)+c^\dagger_\mu(ik^j g^{{\bvec x}^a\mu})\right\}.
\end{align}
The next one is
\begin{align}
[iG,\rm{d}\Gamma_\bb (\omega)]&=-\frac{iq}{\sqrt{2}}\sum_a \int^\oplus_{\mathcal{X}^N} d\bvec X\,\left[c^\mu(g^{{\bvec x}^a}_\mu)+c^\dagger_\mu(g^{{\bvec x}^a\mu}),\rm{d}\Gamma_\bb (\omega)\right]\no\\
&=-\frac{q}{\sqrt{2}}\sum_a \int^\oplus_{\mathcal{X}^N} d\bvec X\,\left\{c^\mu(i\omega g^{{\bvec x}^a}_\mu)+c^\dagger_\mu(i\omega g^{{\bvec x}^a\mu})\right\}.
\end{align}
The last one is
\begin{align}
[iG,H_1]&=-\frac{iq}{\sqrt{2}}\sum_{a} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\left[c^\mu(g^{{\bvec x}^a}_\mu)+c^\dagger_\mu(g^{{\bvec x}^a\mu}),q\sum_b\alpha^{b\mu}A_\mu({\bvec x}^b)\right]\no\\
&=-\frac{iq^2}{2}\sum_{a,b}\alpha^{b\nu} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\left[c^\mu(g^{{\bvec x}^a}_\mu)+c^\dagger_\mu(g^{{\bvec x}^a\mu}),c^\lambda(e_{\nu\lambda}\chi_\rm{ph}^{{\bvec x}^b}\omega^{-1/2})+c^\dagger_\lambda(e_{\nu}^{\;\;\lambda}\chi_\rm{ph}^{{\bvec x}^b}\omega^{-1/2})\right]\no\\
&=-\frac{iq^2}{2}\sum_{a,b}\alpha^{b\nu} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\left\{ \eta^{\mu}_{\;\;\lambda}\ip{g_\mu^{{\bvec x}^a}}{e_{\nu}^{\;\;\lambda}\chi_\rm{ph}^{{\bvec x}^b}\omega^{-1/2}}-\eta^\lambda_{\;\;\mu} \ip{e_{\nu\lambda}\chi_\rm{ph}^{{\bvec x}^b}\omega^{-1/2}}{g^{{\bvec x}^a\mu}} \right\}\no\\
&=q^2\sum_{a,b}\alpha^{b\nu} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\rm{Im}\, \ip{g_\mu^{{\bvec x}^a}}{e_{\nu}^{\;\;\mu}\chi_\rm{ph}^{{\bvec x}^b}\omega^{-1/2}}\no\\
&=0.
\end{align}

Taking the commutator with $iG$ once again, we find
\begin{align}\label{second-comm-hd}
[iG,[iG,H_\rm{D}(V,N)]]&=-\frac{iq^2}{2}\sum_{a,b}\alpha^{bj} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\left[c^\nu(g^{{\bvec x}^a}_\nu)+c^\dagger_\nu(g^{{\bvec x}^a\nu}),c^\mu(ik^j g^{{\bvec x}^b}_\mu)+c^\dagger_\mu(ik^j g^{{\bvec x}^b\mu})\right]\no\\
&=-\frac{iq^2}{2}\sum_{a,b}\alpha^{bj} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\left\{\ip{g^{{\bvec x}^a}_\mu}{ik^j g^{{\bvec x}^b\mu}}-\ip{ik^j g^{{\bvec x}^b}_\mu}{g^{{\bvec x}^a\mu}}\right\}\no\\
&=q^2\sum_{a,b}\alpha^{bj} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\rm{Im}\,\ip{g^{{\bvec x}^a}_\mu}{ik^j g^{{\bvec x}^b\mu}},
\end{align}
and
\begin{align}\label{second-comm-gamma}
[iG,[iG,\rm{d}\Gamma_\bb (\omega)]]&=\frac{iq^2}{2}\sum_{a,b} \int^\oplus_{\mathcal{X}^N} d\bvec X\, \left[c^\nu(g^{{\bvec x}^a}_\nu)+c^\dagger_\nu(g^{{\bvec x}^a\nu}),c^\mu(i\omega g^{{\bvec x}^b}_\mu)+c^\dagger_\mu(i\omega g^{{\bvec x}^b\mu})\right]\no\\
&=\frac{iq^2}{2}\sum_{a,b} \int^\oplus_{\mathcal{X}^N} d\bvec X\, \left\{\ip{g^{{\bvec x}^a}_\nu}{i\omega g^{{\bvec x}^b\mu}}-\ip{i\omega g^{{\bvec x}^b}_\mu}{g^{{\bvec x}^a\nu}}\right\} \no\\
&=-q^2\sum_{a,b} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\rm{Im}\,\ip{g^{{\bvec x}^a}_\mu}{i\omega g^{{\bvec x}^b\mu}}.
\end{align}
Note that the final expressions \eqref{second-comm-hd} and \eqref{second-comm-gamma} are commuting with $iG$.

Summarizing these results in a covariant manner, one has on the subspace $\mathcal{D}_1$
\begin{align}
[iG,H_0]&=\frac{q}{\sqrt{2}}\sum_a \alpha^{a\nu}\int^\oplus_{\mathcal{X}^N} d\bvec X\, \left\{c^\mu(ik_\nu g^{{\bvec x}^a}_\mu)+c^\dagger_\mu(ik_\nu g^{{\bvec x}^a\mu})\right\}, \no\\
[iG,H_1]&=0, \no\\
[iG,[iG,H_0]]&=q^2\sum_{a,b}\alpha^{b\nu} \int^\oplus_{\mathcal{X}^N} d\bvec X\,\rm{Im}\,\ip{g^{{\bvec x}^a}_\mu}{ik_\nu g^{{\bvec x}^b\mu}},
\end{align}
and higher commutators with $iG$ vanish.

From the formula
\begin{align}
e^{i\lambda G} H_\rm{DM}(V,N)e^{-i\lambda}G = \sum_{n=0}^\infty \frac{\lambda^n}{n!}[iG ,[ iG,[\dots [iG, H_\rm{DM}(V,N)]\dots] ],
\end{align}
we have \eqref{transf} at least heuristically. To make this computation a mathematical proof, we have to be careful on the operator domains.
But this can be done by the argument of closedness, which is letft to the reader.
\end{proof}

Regard the above $Q_\mu^{\;\;\nu}(\xx)$ as a $\nu$-th component of four-component vector $Q_\mu(\xx)=(Q_\mu^{\;\;\nu}(\xx))\in\H_\rm{ph}$ ($\mu=0,1,2,3$).
Then, it immediately follows that
\begin{Lem}\label{inv}
The four component vectors $Q_\mu(\xx)$ ($\mu=0,1,2,3$) belong to $w\H_\rm{TL}$. In particular, $H_\rm{DM}(V,N)|_{e^{-iG}\mathcal{D}_1}$ leaves $V_\rm{phys}$ and $\mathcal{N}$ invariant.
\end{Lem} 
\begin{proof}
By a direct computation, we find 
\begin{align}\label{Q-concrete}
Q_0(\xx) = (Q_0^{\;\;\nu}(\xx))_\nu=\begin{pmatrix}
\dfrac{\hat{\chi^\xx}}{\sqrt{\omega}} \\ 0\\0\\ \dfrac{\hat{\chi^\xx}}{\sqrt{\omega}}
\end{pmatrix}\in wH_\rm{L}, \q
Q_k(\xx)=(Q_k^{\;\;\nu}(\xx))_\nu=\begin{pmatrix}
0 \\ e_k^{\;\;1} \dfrac{\hat{\chi^\xx}}{\sqrt{\omega}}\\ e_k^{\;\;2} \dfrac{\hat{\chi^\xx}}{\sqrt{\omega}}\\ 0
\end{pmatrix}\in w H_\rm{T} ,\q k=1,2,3.
\end{align}
Note that the second term in \eqref{H-transf} can be written as
\begin{align}
q\sum_{a}\int^\oplus_{\mathcal{X}^N}d\bvec X\frac{1}{\sqrt{2}}
\left[ c\left(\overline{Q}_{0}(\xx^a)\right)+c^\dagger\left(Q_0(\xx^a)\right)\right] 
+q\sum_k\sum_{a}\alpha^{ak}\int^\oplus_{\mathcal{X}^N}d\bvec X\frac{1}{\sqrt{2}}
\left[ c\left(\overline{Q}_k(\xx^a)\right)+c^\dagger\left(Q_k(\xx^a)\right)\right] 
\end{align}
where
\begin{align}
\overline{Q}_\mu(\xx) = (Q_{\mu\nu}(\xx))_\nu = (\eta_{\nu\lambda} Q_\mu^{\;\;\lambda}(\xx) )_\nu,
\end{align}
and $F\in wH_\rm{L}$ implies $\overline{F}\in wH_\rm{S}$ while $wH_\rm{T}$ is $\eta$-invariant.
 Then, the assertions obviously follow.
\end{proof}

\begin{Thm}\label{ess-sa}
The operator $\tilde{H}$ defined by the relations
\begin{align}
D(\tilde{H})&:=[\left(e^{-iG}\mathcal{D}_1\right)\cap V_\rm{phys}], \\
\tilde{H}[\Psi] &:= [H_\rm{DM}(V,N)\Psi] ,
\end{align}
is essentially self-adjoint and the unique self-adjoint extension is 
equal to $H_\rm{phys}$.
\end{Thm}
\begin{proof}
 For $\Psi\in (e^{-iG}\mathcal{D}_1)\cap V_\rm{phys}$, 
 \begin{align}
 \frac{d}{dt}U(t)[\Psi]\Bigg|_{t=0}=[-iH_\rm{DM}(V,N)\Psi]=-i\tilde{H}[\Psi].
 \end{align}
 Thus the assertion follows from Proposition \ref{a-or-b}.
\end{proof}
\subsection{Relation to the Coulomb gauge Hamiltonian}
In the Coulomb gauge, the Hilbert space $\mathcal{H}_\rm{Coulomb}$ of state vectors of the quantum system considered is naturally taken to be 
\begin{align}
\H_\rm{Coulomb}:=\wedge_{a=1}^NL^2(\R^3;C^4)\ot \F_\bb(L^2(\R^3;\C^2)).
\end{align}
In the Coulomb gauge, only two of the the photon polarization, that is, the two transverse ones, are
in the model, so that the photon Hilbert space is chosen to be the boson Fock space over $L^2(\R^3;\C^2))$,
the target space $\C^2$ representing the two photon polarization degrees of freedom.
In our present formulation in the Lorentz gauge, the closed subspace 
\begin{align}
\H_\rm{D}(N)\ot \F_\bb(\H_\rm{T})
\end{align} 
is naturally identified with $\H_\rm{Coulomb}$. But $w$ is just an identity matrix when restricted on $\H_\rm{T}$,
it is equal to the closed subspace of $\F_\rm{DM}(V,N)$
\begin{align}
\H_\rm{D}(N)\ot \F_\bb(w\H_\rm{T}).
\end{align}
As we have ever seen, the physical Hilbert space with positive definite metric is naturally identified
with the closed subspace
\begin{align}
\mathcal{V}=e^{-iG}\left(\H_\rm{D}(N)\ot \F_\bb(w\H_\rm{T})\right),
\end{align}
via unitary transformation $U$ given by \eqref{U}.
Hence, the self-adjoint operator 
\begin{align}
U_G^{-1} H_\rm{phys} U_G,\quad U_G:=Ue^{-iG}
\end{align}
is to be considered as a gauge transformed Hamiltonian from the Lorenz gauge into the Coulomb gauge. 
Let $\rm{d}\Gamma^\rm{C}_\bb(\omega)$ be a second quantization operator of $\omega$ when regarded as 
an operator in $\H_\rm{T}$, that is, the multiplication operator of the matrix valued function
\begin{align}
\kk\mapsto \begin{pmatrix}
\omega(\kk) & 0 \\
0 & \omega(\kk)
\end{pmatrix}. 
\end{align}
\begin{Thm}
We have the operator equality 
\begin{align}
U_G^{-1}H_\rm{phys}U_G=\overline{H_D(V,N)+\rm{d}\Gamma^\rm{C}_\bb(\omega)+q\sum_{k=1,2,3}\int^\oplus_{\mathcal{X}^N} d\XX\, \alpha^{ak} A^{\rm{C}}_k(\xx^a)+E^\rm{C}},
\end{align}
where 
\begin{align}\label{equivalent-Coulomb}
A_k^\rm{C}(\xx)&:= \frac{1}{\sqrt{2}}\sum_{r=1,2}\left\{c^{r}\left(e_{rk}\frac{\hat{\chi^{\xx^a}_\rm{ph}}}{\sqrt{\omega}}\right)+
c^\dagger_{r}\left(e^r_{\;\;k}\frac{\hat{\chi^{\xx^a}_\rm{ph}}}{\sqrt{\omega}}\right)\right\},\quad k=1,2,3,\\
E^\rm{C}&:=-\frac{1}{2}q^2 \sum_{a,b}\alpha^{a\mu}\int^\oplus_{\mathcal{X}^N}d\bvec X \ip{k_\mu g^{\xx^a}_\nu}{g^{\xx^b \nu}}.
\end{align}
\end{Thm}
 \begin{proof}
 Let $\Phi\in \mathcal{D}_1\cap \left(\H_\rm{D}(N)\ot\F_\bb(w\H_\rm{T})\right)$. Then, 
 \begin{align}
 H_\rm{phys} U_G\Phi &= \tilde{H}[e^{-iG}\Phi] \no\\
 &=[H_\rm{DM}(V,N)e^{-iG}\Phi].
 \end{align}
 By Lemma \ref{transf}, this is the equivalence class to which the vector 
 \begin{align}
 e^{-iG}&\Bigg(H_0+q\sum_{a}\alpha^{a\mu}\int^\oplus_{\mathcal{X}^N}d\bvec X\frac{1}{\sqrt{2}}
\left[ c^\nu\left(Q_{\mu\nu}(\xx^a)\right)+c^\dagger_\nu\left(Q_\mu^{\;\;\nu}(\xx^a)\right)\right] \no\\
&\hskip 14mm -\frac{1}{2}q^2 \sum_{a,b}\alpha^{a\mu}\int^\oplus_{\mathcal{X}^N}d\bvec X \ip{k_\mu g^{\xx^a}_\nu}{g^{\xx^b \nu}}\Bigg)\Phi
 \end{align}
 belongs. If we apply $U^{-1}$ given in \eqref{U} and then $e^{-iG}$ to this equivalent class, the result is
  \begin{align}
 &\Bigg(H_0+q\sum_{a}\sum_{k=1,2,3}\alpha^{ak}\sum_{r=1,2}\int^\oplus_{\mathcal{X}^N}d\bvec X\frac{1}{\sqrt{2}}
\left[ c^r\left(Q_{kr}(\xx^a)\right)+c^\dagger_r\left(Q_k^{\;\;r}(\xx^a)\right)\right] \no\\
&\hskip 14mm -\frac{1}{2}q^2 \sum_{a,b}\alpha^{a\mu}\int^\oplus_{\mathcal{X}^N}d\bvec X \ip{k_\mu g^{\xx^a}_\nu}{g^{\xx^b \nu}}\Bigg)\Phi,
\end{align}
since by \eqref{Q-concrete} one sees that in the second term the summation over $\mu$ survives only for $\mu=1,2,3$ and 
summation over $\nu$ only for $\nu=1,2$.
But since the operator inside the bracket coincides with the one on the right hand side of \eqref{equivalent-Coulomb},
we obtain
\begin{align}\label{inc}
U_G^{-1}H_\rm{phys}U_G|_{\mathcal{D}_1\cap\left( \H_\rm{D}(N)\ot\F_\bb(w\H_\rm{T})\right)}\subset 
H_D(V,N)+\rm{d}\Gamma^\rm{C}_\bb(\omega)+q\sum_{k=1,2,3}\int^\oplus_{\mathcal{X}^N} d\XX\, \alpha^{ak} A^{\rm{C}}_k(\xx^a)+E^\rm{C}.
\end{align}
The operator on the right hand side of \eqref{inc} is clearly symmetric and the subspace $\mathcal{D}_1\cap \H_\rm{D}(N)\ot\F_\bb(w\H_\rm{T})$
 is a core of the self-adjoint operator $U_G^{-1}H_\rm{phys}U_G$ by Theorem \ref{ess-sa}. Hence, the assertion follows.
 \end{proof}
 At the last of the present subsection, we see that the ``residual" term $E^\rm{C}$ in \eqref{equivalent-Coulomb} certainly gives the Coulomb interaction energy
 between electrons in the limit where the ultraviolet cutoff is removed. We have to be careful to take the limit because the term $E^\rm{C}$ includes
 not only the Coulomb interaction between two different electrons but also electron's self-interaction energy through gauge field, which will diverge
 in the limit. This diverging self-interaction part should be adequately subtracted.
 
 Choose a special photon cutoff function whose Fourier transform is propotional to 
 $\chi_{\epsilon, \Lambda}$, by which we denote the characteristic function of the set 
 \begin{align}
 \{\kk\in\R^3\,|\, \epsilon\le |\kk| \le \Lambda \}.
 \end{align} 
 The parameter $\epsilon$ and $\Lambda$ represent the infrared and ultraviolet cutoffs, respectively.
 Then,
  \begin{Thm}
 Let $E^\rm{C}(\epsilon,\Lambda)$ be $E^\rm{C}$ when a photon cutoff function is taken so that $\hat{\chi_\rm{ph}}=\chi_{\epsilon,\Lambda}/(2\pi)^{3/2}$. 
 Then,
 \begin{align}\label{Coulomb}
 \lim_{\Lambda\to\infty}\lim_{\epsilon\to 0}\left(E^\rm{C}(\epsilon,\Lambda)+\frac{q^2}{8\pi^2}\Lambda\right)=-\frac{q^2}{4\pi}\sum_{a<b}\int^\oplus_{\mathcal{X}^N}d\XX\,\frac{1}{|\xx^a-\xx^b|},
 \end{align}
 where the limit is understood as the strong resolvent sense with operators on both sides being considered to be
 a self-adjoint operators.
 \begin{proof}
 By Lemma \ref{transf}, $E^\rm{C}(\epsilon,\Lambda)$ is a multiplication operator by a function
 \begin{align}
 \XX\mapsto -\frac{q^2}{2}\sum_{a,b}\alpha^{a\mu}\ip{k_\mu g^{\xx^a}_\nu}{g^{\xx^b\nu}}.
 \end{align}
 A direct computation shows that this function can be written as 
 \begin{align}
 -\frac{q^2}{2}\sum_{a,b}\alpha^{a\mu}\ip{k_\mu g^{\xx^a}_\nu}{g^{\xx^b\nu}}
 &= -\frac{q^2}{2}\cdot2 \sum_{a<b}\int\frac{d^3\kk}{(2\pi)^3}\frac{|\chi_{\epsilon,\Lambda}(\kk)|^2}{|\kk|^2}e^{i(\xx^a-\xx^b)\kk} -\frac{q^2}{2}\int\frac{d^3\kk}{(2\pi)^3}\frac{|\chi_{\epsilon,\Lambda}(\kk)|^2}{|\kk|^2}\no\\
 &=-\frac{q^2}{4\pi} \frac{2}{\pi}\sum_{a<b}\int_{\epsilon}^\Lambda dk\,\frac{\sin(|\xx^a-\xx^b | k)}{|\xx^a-\xx^b|k}-\frac{q^2}{8\pi^2}(\Lambda -\epsilon),
 \end{align}
 which converges uniformly in $\XX\in\mathcal{X}^N$ to 
 \begin{align}
 -\frac{q^2}{4\pi}\frac{2}{\pi} \sum_{a<b}\int_{0}^\Lambda dk\,\frac{\sin(|\xx^a-\xx^b | k)}{|\xx^a-\xx^b|k}-\frac{q^2}{8\pi^2}\Lambda,
 \end{align}
 as $\epsilon$ tends to zero. Hence, what we should show is that the multiplication operator 
 \begin{align}
E^\rm{C}_\rm{ren}(\Lambda):=-\frac{q^2}{4\pi} \frac{2}{\pi}\sum_{a<b} \int_{\mathcal{X}^N}^\oplus d\XX\, \int_0^\Lambda dk\,\frac{\sin(|\xx^a-\xx^b | k)}{|\xx^a-\xx^b|k} 
 \end{align}
 converges to the operator on the right hand side of \eqref{Coulomb}, which we call $E_\rm{ten}^\rm{C}$ hereafter, in the strong resolvent sense.
 
 To this end, take arbitrary $\Psi\in D(E_\rm{ren}^\rm{C})$. Since $D(E_\rm{ren}^\rm{C}) \subset D(E_\rm{ren}^{C}(\Lambda))$
 for all $\Lambda\ge 0$ as can be easily checked, it suffices to prove that 
 \begin{align}
 \norm{E_\rm{ren}^\rm{C}(\Lambda)\Psi - E_\rm{ren}^\rm{C}\Psi } \to 0
 \end{align}
 as $\Lambda\to\infty$. But this follows from the well known integral formula
 \begin{align}
 \lim_{L\to\infty}\int_0^L \frac{\sin y}{y}\,dy =\frac{\pi}{2},
 \end{align}
 as well as the estimate
 \begin{align}
 \norm{E_\rm{ren}^\rm{C}(\Lambda)\Psi - E_\rm{ren}^\rm{C}\Psi }&=\int_{\mathcal{X}^N} d\XX\,\norm{E_\rm{ren}^\rm{C}(\Lambda)(\XX)\Psi - E_\rm{ren}^\rm{C}(\XX)\Psi(\XX) }^2 \no\\
 &\le \frac{q^2}{4\pi}\sum_{a<b} \int_{\mathcal{X}^N} d\XX\,\norm{\left(\frac{2}{\pi}\int_0^{|\xx^a-\xx^b|\Lambda} \frac{\sin p}{p}\,dp - 1\right)\frac{1}{|\xx^a - \xx^b |}\Psi(\XX)}^2 \no\\
 &\to 0,
 \end{align}
as $\Lambda$ tends to infinity, justified by the Lebesgue dominant convergence theorem. 
 \end{proof}
 \end{Thm}
\subsection{Triviality of the physical subspace}
If $\hat{\chi_\rm{ph}}$ does not belong to the domain of $\omega^{-3/2}$, the definition of $G$ in \eqref{G}
makes no sense. In this case, the physical subspace is trivial: 
\begin{Thm}\label{trivial}
Suppose that $\hat{\chi_\rm{ph}}$ does not belong to $D(\omega^{-3/2})$. Then,
\begin{align}
V_\rm{phys}=\{0\}.
\end{align}
\end{Thm}
\if0
To prove Theorem \ref{trivial}, we note the following general fact:
\begin{Lem}
Let $(X,\mu)$ is a measure space and $\mathcal{K}$ be a separable Hilbert space. Let
\[ F:X\to \mathcal{K} \]
a mesurable $\mathcal{K}$-valued function with
\begin{align}
\int_{X}\norm{F(x)} d\mu(x) <\infty,
\end{align}
for which the strong integral
\begin{align}
\int_X F(x)d\mu(x) \in\mathcal{K}
\end{align}
is defined.
\end{Lem}
\fi
\begin{proof}
Let $\Psi\in D'$ and 
\begin{align}
\Omega^+(t,f)\Psi=0
\end{align}
for all $t\in\R$ and $f\in\mathfrak{h}\cap D((-\Delta)^{-1/4})$. Then, we have by \eqref{omega+} with $t=0$
\begin{align}
a_\mu\left(\frac{ik^\mu \hat{f^*}}{\sqrt{\omega}}\right)\Psi=\frac{iq}{\sqrt{2}}\sum_a\int_{\mathcal{X}^N}^{\oplus}d\bvec{X} \ip{\frac{\hat{\chi_\rm{ph}^{\xx^a}}}{\omega}}{f}\Psi(\XX).
\end{align}
Define $l:\mathfrak{h}\cap D((-\Delta)^{-1/4})\to \C$ by
\begin{align}
l(f)&:=\ip{\Psi}{a_\mu\left(\frac{ik^\mu \hat{f^*}}{\sqrt{\omega}}\right)\Psi} \no\\
&=\frac{iq}{\sqrt{2}}\sum_a\int_{\mathcal{X}^N}d\XX \ip{\Psi(\XX)}{ \ip{\frac{\hat{\chi_\rm{ph}^{\xx^a}}}{\omega}}{\hat{f}}\Psi(\XX)}.
\end{align}
We remark that, since 
\begin{align}
\int d\XX \norm{\Psi(X)}^2 \norm{ \frac{\hat{\chi_\rm{ph}^{\xx^a}}}{\omega} } <\infty,
\end{align}
the strong Bochner integral
\begin{align}
\int d\XX \norm{\Psi(X)}^2  \frac{\hat{\chi_\rm{ph}^{\xx^a}}}{\omega} \in L^2(\R^3)
\end{align}
is defined. Therefore, $l(f)$ can be rewritten as
\begin{align}\label{also}
l(f)= \ip{\frac{iq}{\sqrt{2}}\sum_a\int d\XX \norm{\Psi(\XX)}^2 \frac{\hat{\chi_\rm{ph}^{\xx^a}}}{\omega}}{\hat{f}}.
\end{align}\label{l-bound}
The definition of $l(f)$ and the estimate \eqref{annihilation-bound} implies there is a constant $C>0$ with
\begin{align}
| l(f) |\le C \norm{\Psi}\norm{N_\bb^{1/2}\Psi} \norm{\sqrt{\omega}\hat{f}}.
\end{align}
Take arbitrary $g\in D((-\Delta)^{-1/2})$. Then, $(-\Delta)^{-1/4}g \in  \mathfrak{h}\cap D((-\Delta)^{-1/4})$.
If we define $\tilde{l}: D((-\Delta)^{-1/2}))\to \C$ by
\begin{align}
\tilde{l}(g):=l((-\Delta)^{-1/4}g),
\end{align}
it follows by \eqref{l-bound} that 
\begin{align}
|\tilde{l}(g)| \le C_\Psi \norm {g} 
\end{align}
for some $\Psi$-dependent constant $C_\Psi$. From the Riesz Lemma, there is a $\Theta\in L^2(\R^3)$ such that
\begin{align}
\tilde{l}(g)&=\ip{\Theta}{\hat{g}} \no\\
&= \ip{\frac{iq}{\sqrt{2}}\sum_a\int d\XX \norm{\Psi(\XX)}^2 \frac{\hat{\chi_\rm{ph}^{\xx^a}}}{\omega}}{\frac{\hat{g}}{\sqrt{\omega}}},
\end{align}
where the second equality is obtained form \eqref{also}. This means that 
\begin{align}
\frac{iq}{\sqrt{2}}\sum_a\int d\XX \norm{\Psi(\XX)}^2 \frac{\hat{\chi_\rm{ph}^{\xx^a}}}{\omega}\in D(\omega^{-1/2}),
\end{align}
that is, the function
\begin{align}
\kk\mapsto \frac{1}{{\omega(\kk)}}\cdot \frac{iq}{\sqrt{2}}\sum_a\int d\XX \norm{\Psi(\XX)}^2 \frac{\hat{\chi_\rm{ph}^{\xx^a}}}{\omega}(\kk)
\end{align}
is square integrable. Hence, we find
\begin{align}\label{bounded-condition}
\frac{q^2}{2}\sum_{a,b}\int d^3\kk \frac{|\hat{\chi_\rm{ph}}(\kk)|^2}{\omega(\kk)^3} \phi^{a,b}(\kk) <\infty 
\end{align}
where
\begin{align}
\phi^{a,b}(\kk):=\int d\XX \int d\bvec Y \norm{\Psi(\XX)}^2\norm{\Psi(\bvec Y)}^2
e^{i\kk(\xx^a - \xx^b)}.
\end{align}

Suppose $\Psi\not=0$. Then 
\begin{align}
K:=\phi^{a,b}(\bvec 0) = \int d\XX \int d\bvec Y \norm{\Psi(\XX)}^2\norm{\Psi(\bvec Y)}^2 > 0.
\end{align}
By the Riemann-Lebesgue Theorem, $\phi^{a,b}$ is continuous in $\kk$, and thus there is an open ball $U\subset \R^3$
centered at the origin with
\begin{align}
\phi^{a,b}(\kk)\ge \frac{K}{2}, \q \kk \in U.
\end{align}
Since $\chi_\rm{ph}\in L^2(\R^3)$, one sees that 
\begin{align}
\left|\int_{\R^3\setminus U} d^3\kk \frac{|\hat{\chi_\rm{ph}}(\kk)|^2}{\omega(\kk)^3} \phi^{a,b}(\kk)\right| 
&\le \sup_{\kk\in\R^3\setminus U}  \left|\frac{\phi^{a,b}(\kk)}{\omega(\kk)^3}\right| \int_{\R^3\setminus U} d^3\kk |\hat{\chi_\rm{ph}}(\kk)|^2 <\infty.
\end{align}
Thus, \eqref{bounded-condition} says
\begin{align}\label{diverge}
\frac{q^2}{2}\sum_{a,b}\int_U d^3\kk \frac{|\hat{\chi_\rm{ph}}(\kk)|^2}{\omega(\kk)^3} \phi^{a,b}(\kk) <\infty .
\end{align}
But this is impossible when $\hat{\chi_\rm{ph}}\not\in D(\omega^{-3/2})$. To see this, 
note that
\begin{align}
\int_{\R^3\setminus U} \frac{|\hat{\chi_\rm{ph}}(\kk)|^2}{\omega(\kk)^3} \le 
\dfrac{1}{\inf_{\kk\in \R^3\setminus U}| \omega(\kk)|^3}\int_{\R^3\setminus U} |\hat{\chi_\rm{ph}}(\kk)|^2 <\infty,
\end{align}
since $U$ is centered at the origin and $\hat{\chi_\rm{ph}}$ belongs to $L^2(\R^3)$.
Thus, in the case where $\hat{\chi_\rm{ph}}\not\in D(\omega^{-3/2})$,
\begin{align}
\int_U d^3\kk \frac{|\hat{\chi_\rm{ph}}(\kk)|^2}{\omega(\kk)^3} =\infty.
\end{align}
Hence, the integration on the left hand side of \eqref{diverge} has to diverge in this case:
\begin{align}
\frac{q^2}{2}\sum_{a,b}\int_U d^3\kk \frac{|\hat{\chi_\rm{ph}}(\kk)|^2}{\omega(\kk)^3} \phi^{a,b}(\kk) \ge \frac{q^2}{2}\sum_{a,b}\int_U d^3\kk \frac{|\hat{\chi_\rm{ph}}(\kk)|^2}{\omega(\kk)^3}  \frac{K}{2}= \infty.
\end{align}
Therefore, $\hat{\chi_\rm{ph}}$ has to belong to $D(\omega^{-3/2})$ if $\Psi\not=0$. This comples the proof.
\end{proof}
\section*{Acknowlegdements}
The authors are grateful to Professor Asao Arai for valuable comments and discussions. 
This study is somewhat inspired by the lectures given by 
Assistant Professor Akito Suzuki at Hokkaido University. It is our pleasure to thank him.
\appendix
\if0
\section{Differential operator on a direct fiber integral Hilbert space}
Let $\mathcal{F}$ be a Hilbert space 
\begin{align}
 \mathcal{F}:=L^2(\Real^d;\mathcal{H})=\int_{\Real^d}^\oplus dx\,\mathcal{H}
\end{align}
with $\mathcal{H}$ being a separable complex Hilbert space. The Hilbert space $\mathcal{F}$ is unitary equivalent
to 
\begin{align}
L^2(\Real^d)\otimes\mathcal{H}
\end{align}
via the unitary transformation
\begin{align}
U: \mathcal{F}\to L^2(\Real^d)\otimes\mathcal{H}
\end{align}
which maps $f(\cdot)\Psi$ into $f\otimes\Psi$ for $f\in L^2(\Real^d)$ and $\Psi\in\mathcal{H}$.
We write $f\times \Psi$ to mean the function $x\mapsto f(x)\Psi$:
\[ f(\cdot)\Psi=f\times\Psi=U^{-1}f\otimes\Psi .\]
The generalized differential operator on $L^2(\Real^d)$ with respect to the $j$-th component of $x\in\Real^d$ is 
denoted by $D_j$ for $j=1,2,\dots,d$, and
the $j$-th component of the momentum operator by $p_j:=-iD_j$.
\begin{Prop}
Let 
\[A_j:=U(p_j\otimes 1)U^{-1}.\]
Then the followings hold:
\begin{align}
D(A_j)&=\{F \in\mathcal{F}\,|\, F(\cdot) \text{ is strongly differentiable in $x_j$ and the strong derivative } 
 F'(\cdot) \text{ belongs to } \mathcal{F}\},\label{want-domain-of-A_j}\\
 A_j F&=-iF' ,\quad F\in D(A_j).\label{want-action-of-A_j}
\end{align}
\end{Prop}
\begin{proof}
Define 
\[ V(t) = e^{itA_j} \]
for $t\in\Real$. Then, $\{V(t)\}_t$ forms a one parameter unitary group in $\mathcal{F}$ and 
$A_j$ is its unique generator. By Stone's theorem, we can identify the domain of definition of the generator $A_j$ as
\begin{align}\label{domain-of-A_j}
D(A_j)=\Bigg\{F\in\mathcal{F}\,\Bigg|\, \lim_{h\to 0}\frac{e^{ihA_j}-1}{h}F \text{ exists}.\Bigg\}
\end{align}
and its action
\begin{align}\label{action-of-A_j}
A_jF=-i\lim_{h\to 0}\frac{e^{ihA_j}-1}{h}F.
\end{align}

For $a\in\Real^d$, define $T_a$ as the translation in $\mathcal{F}$ by $a$. That is, for $F\in\Real^d$,
\begin{align}
(T_a F)(x):=F(x-a), \quad \text{a.e. }x\in\Real^d.
\end{align}
The operator $T_a$ is a unitary operator in $\mathcal{F}$. Similarly, we define $\tau_a$ as the translation 
by $a\in\Real^d$ in $L^2(\Real^d)$:
\begin{align}
(\tau_a f)(x):=f(x-a), \quad \text{a.e. }x\in\Real^d, \quad f\in L^2(\Real^d).
\end{align}
It is well known that the generator of the one parameter unitary group $\{\tau_{a_j}\}_{a_j}$ is $-p_j$\footnote{
We use a little bit lazy notation in which $a_j$ denotes both a real parameter and the element 
$(0,\dots,a_j,\dots,0)$ in $\Real^d$.}.

Now, take arbitrary $f\in L^2(\Real^d)$ and $\Psi\in\mathcal{H}$. Then, we obtain for all $h\in\Real$
\begin{align}
V(h)(f\times\Psi )&= Ue^{ih(p_j\otimes 1)} f\otimes\Psi\no\\
&=U(e^{ihp_j}f)\otimes \Psi\no\\
&=(\tau_{-h}f)\times \Psi \no\\
&=T_{-h}(f\times\Psi).
\end{align}
This implies that $V(h)=T_{-h}$ on the dense subspace 
spanned by the vectors of the form $f\times \Psi$. Since the subspace is dense and
both $V(h)$ and $T_{-h}$ are bounded, it follows that the operator equality 
\begin{align}
V(h)=T_{-h},\quad h\in\Real.
\end{align}
holds. This means that for $F\in\mathcal{F}$,
\begin{align}
[V(h)F](x)=[T_{-h}F](x)=F(x+h),\quad \text{a.e. }x\in\Real^d.
\end{align}
Hence, \eqref{domain-of-A_j} exactly coincides with \eqref{want-domain-of-A_j} and \eqref{action-of-A_j}
with \eqref{want-action-of-A_j}.
\end{proof}
\section{Leibniz rule}
Let 
\begin{align}
\mathcal{F}=L^2(\Real^d;\mathcal{F}_\mathrm{b}(\mathcal{H})).
\end{align}
The sigal field operator on $\mathcal{F}_\mathrm{b}(\mathcal{H})$ is denoted by $\phi(\cdot)$:
\begin{align}
\phi(f):=\frac{1}{\sqrt{2}}\left(a(f)+a(f)^*\right),\quad f\in \mathcal{H}.
\end{align}
Let $g$ be a mapping
\begin{align}
g:\Real^d\to \mathcal{H}
\end{align}
which is strongly differentiable and essentially bounded. We introduce a decomposable operator
\begin{align}
G=\int_{\Real^d}^\oplus dx\, \phi(g(x))
\end{align}
and consider the differentiability of the mapping
\[ x\mapsto \phi(g(x))\Psi(x) ,\quad \Psi\in D(G),\]
namely, of $G\Psi$.

\begin{Prop}
Let $N$ be the number operator on $\mathcal{F}_\mathrm{b}(\mathcal{H})$. Suppose that $\Psi\in D(1\otimes
(N+1)^{1/2})$ and 
$1\otimes(N+1)^{1/2}\Psi$ is strongly differentiable. Then, $\Psi\in D(G)$ and $G\Psi$ is also 
strongly differentiable with the derivative
\begin{align}
(G\Psi)'(x)=\phi(g'(x))\Psi(x)+\phi(g(x))\Psi'(x),
\end{align}   
where the prime symbol $'$ denotes the strong derivative.
\end{Prop}
\begin{proof}
First note that $G$ is $1\otimes(N+1)^{1/2}$ bounded because $\Psi\in D(1\otimes(N+1)^{1/2})$ implies 
$\Psi(x)\in D((N+1)^{1/2})$ for almost every $x\in\Real^d$, and then $\Psi(x)\in D(\phi(g(x))$ for
almost every $x\in\Real^d$ and 
\begin{align}\label{diff-esti}
\norm{G\Psi}^2&=\int_{\Real^d}dx\, \norm{\phi(g(x))\Psi(x)}^2 \no\\
&\le\sqrt{2} \int_{\Real^d}dx\,\norm{g(x)}^2\norm{(N+1)^{1/2}\Psi(x)}^2\no\\
&\le \text{Const.}\int_{\Real^d}dx\, \norm{(N+1)^{1/2}\Psi(x)}^2\no\\
&=\text{Const.}\norm{1\otimes(N+1)^{1/2}\Psi}^2.
\end{align}

Second, note also that under the present assumptions $\Psi(\cdot)$ is strongly differentiable
and $\Psi'(x) \in D((N+1)^{1/2})$. Furthermore, the equality
\begin{align}\label{consistency-N}
(N+1)^{1/2}\Psi'(x)=[1\otimes (N+1)^{1/2} \Psi]'(x)
\end{align}
holds.
In fact, for almost every $x\in\Real^d$, one has
\begin{align}
\frac{\Psi(x+h)-\Psi(x)}{h} &= (N+1)^{-1/2}(N+1)^{1/2}\frac{\Psi(x+h)-\Psi(x)}{h}\no\\
&=(N+1)^{-1/2}\frac{(1\otimes(N+1)^{1/2}\Psi)(x+h) - (1\otimes(N+1)^{1/2}\Psi)(x)}{h}\no\\
&\to (N+1)^{-1/2} (1\otimes(N+1)^{1/2}\Psi)'(x)
\end{align}
as $h$ goes to zero. This implies that the strong limit
\[ \Psi'(x) = \lim_{h\to 0}\frac{\Psi(x+h)-\Psi(x)}{h} \]
exists, belongs to $D((N+1)^{1/2})$ and satisfies
\[ (N+1)^{1/2}\Psi'(x) = (1\otimes(N+1)^{1/2}\Psi)'(x), \quad \text{a.e. }x\in\Real^d. \]

Now, what we should compute is
\begin{align}
&\norm{\frac{G\Psi(x+h)-G\Psi(x)}{h}-(\phi(g'(x))\Psi(x)+\phi(g(x))\Psi'(x))}\no\\
&\le\norm{\left\{\frac{\phi(g(x+h))-\phi(g(x))}{h}-\phi(g'(x))\right\}\Psi(x+h)}
+\norm{\phi(g(x))\left\{\frac{\Psi(x+h)-\Psi(x)}{h}-\Psi'(x)\right\}}\no\\
&\quad\quad+\norm{\phi(g'(x))(\Psi(x+h)-\Psi(x))}\no\\
\end{align}
The first term of \eqref{diff-esti} vanishes as $h$ tends to zero since one has
\begin{align}
\norm{\left\{\frac{\phi(g(x+h))-\phi(g(x))}{h}-\phi(g'(x))\right\}\Psi(x+h)}
&=\norm{\phi\left(\frac{g(x+h)-g(x)}{h}-g'(x)\right)\Psi(x+h)}\no\\
&\le\sqrt{2}\norm{\frac{g(x+h)-g(x)}{h}-g'(x)} \norm{(N+1)^{1/2}\Psi(x+h)}\no\\
&\to 0.
\end{align}
The second term of \eqref{diff-esti} also vanishes by the estimate
\begin{align}
\norm{\phi(g(x))\left\{\frac{\Psi(x+h)-\Psi(x)}{h}-\Psi'(x)\right\}}&\le\sqrt{2}\norm{g(x)}\norm{(N+1)^{1/2}\left\{\frac{\Psi(x+h)-\Psi(x)}{h}-\Psi'(x)\right\}}\no\\
&\to 0
\end{align}
as $h$ goes to zero. Here, we have used the relation \eqref{consistency-N}.
The third term of \eqref{diff-esti} can be estimated as follows.
\begin{align}
\norm{\phi(g'(x))(\Psi(x+h)-\Psi(x))}&\le \sqrt{2}\norm{g'(x)}\norm{(N+1)^{1/2}(\Psi(x+h)-\Psi(x))}
\end{align}
The right hand side vanishes in the limit $h\to0$ since $(N+1)^{1/2}\Psi(\cdot)$ is strongly differentiable
and so strongly continuous. This verifies the assertion.
\end{proof}
\fi
\section{Time-evolution of the position operator}\label{time-evolution-position}

In this section, we construct the time evolution of the position operator of the $a$-th Dirac particle, which
we denote by $\xx^a=(x^{a1},x^{a2},x^{a3})$. We will see that $\alpha^{aj}$ is the $j$-th component of the velocity operator 
(that is, the time derivative of the $j$-th component of the position operator) of the $a$-th Dirac particle.


For each $ r >0 $, we set
\begin{align}
\mathcal{M} _r := \{ f \in \H _\rm{D} \; \big| \; \supp f \subset \{ |\xx | \le r \} \} .
\end{align}

\begin{Lem}\label{L-Cone} For all $ r>0 $ and $ t \in \Real $,
\begin{align}
e^{-it H_\mathrm{D}} \mathcal{M} _r \subset \mathcal{M} _{r+ |t|} .
\end{align}
\end{Lem}

\begin{proof} See \cite[Section 1.5]{MR1219537}.
\end{proof}

For each $ r >0 $ and $ n \in \N $, we define the subspace $ \F _{r,n_0} \subset \F _\rm{DM} (N) $ as
\begin{align}
\F _{r,n_0} := ( \wg ^N \mathcal{M} _r ) \otimes \left(\op_{n=0}^{n_0} \ot _\rm{s} ^{n} \H _\rm{ph} \right) .
\end{align}


\begin{Thm} The followings hold:
\begin{enumerate}[(i)]
\item For all $ r >0 , \; n_0 \in \N $ and $ t\in \R $, $ \F _{r,n_0} \subset D(W(-t) x^{aj} W(t)) $. 

\item For all $ \Psi \in \cup _{r,n_0} \F _{r,n_0} \cap D(H_0) $, the mapping
\begin{align}
t \mapsto W(-t) x ^{aj} W(t) \Psi
\end{align}
is strongly continuously differentiable. Moreover, 
\begin{align}\label{WaW}
\frac{d}{dt} W(-t) x ^{aj} W(t) \Psi =  W(-t) \alpha ^{aj }W(t) \Psi 
\end{align}
holds.
\end{enumerate}
\end{Thm}

\begin{proof} 
\begin{enumerate}[(i)]
\item Using Lemma \ref{L-Cone} and the Trotter product formula \cite[Theorem VIII.31]{MR751959}, we see that $ e^{-itH_\rm{D} (V)} \mathcal{M} _r \subset \mathcal{M} _{r+ |t|} $, and thus $ e^{-itH_0} \mathcal{F} _{r,n_0} \subset \mathcal{F} _{r+ |t|,n_0}$. On the other hand, it is clear that $ H_1 \F _{r,n_0} \subset \F _{r,n_0+1} $. Hence we have
\begin{align}
& \sum _{m,n=0} ^\infty \norm{ e^{itH_0} U_m (-t, 0) x ^{a j} e^{-itH_0 } U_n (t, 0) \Psi } \no\\
& \le\sum _{m,n=0} ^\infty \frac{(C|t|)^{m+n}}{m!n!}(r+|t|+2n|t|)(n_0+m+n)^{1/2}\dots(n_0+1)^{1/2}\norm{\Psi} \no \\ 
&=\sum _{N=0}^\infty\sum_{n=0}^N \frac{(C|t|)^N}{N!} \begin{pmatrix} N \\ n
\end{pmatrix} (r+|t|+2n|t|) (n_0+N)^{1/2}\dots(n_0+1)^{1/2}\norm{\Psi} \no\\
& < \infty ,
\end{align}
for all $ \Psi \in \F _{r,n_0} $, which implies $ \Psi \in D(W(-t)x^{a j} W(t)) $.

\item By a direct calculation, we have
\begin{align}\label{w-derivative}
\Expect{(iH_\rm{DM}(V,N)) ^* \Psi , x ^{aj}  \Phi } - \Expect{x ^{aj} \Psi , iH_\rm{DM}(V,N) \Phi } = \ip{\Psi}{\alpha^{aj} \Phi } ,
\end{align}
for all $ \Psi , \Phi \in \cup _{r,n_0} \F _{r,n_0} \cap D(H_0) $. Using \eqref{w-derivative} and the fact that $\alpha^{aj}$ is bounded, we 
can mimic the proof of Theorem \ref{C_1-equation} to obtain \eqref{WaW}.
\end{enumerate}
\end{proof}

\if0
\section{Formulae for exponential transformation}
\begin{Def}
Let $\mathcal{D}$ be a dense subspace in $\mathcal{H}$, and $A,B$ are densely defined linear operator.
We say that $A$ and $B$ has a weak commutator on $\mathcal{D}$ if and only if
\begin{enumerate}[(i)]
\item $\mathcal{D}\subset D(A)\cap D(B)\cap D(A^*)\cap D(B^*)$.
\item There exists a linear operator $C$ such that 
$\mathcal{D}\subset D(C)$ and for all $\psi,\phi\in\mathcal{D}$,
\[ \inprod{A^*\psi}{B\phi} - \inprod{B^*\psi}{A\phi}=\inprod{\psi}{C\phi}.\]
\end{enumerate}
The linear operator $C|_\mathcal{D}$ is called the weak commutator of $A$ and $B$ on $\mathcal{D}$.
\end{Def}
The weak commutator of $A$ and $B$ on $\mathcal{D}$ is denoted by $[A,B]_\mathcal{D}$ or simply $[A,B]$
when there is no need to specify the subspace on which the weak commutator is defined. 
If $A$ and $[A,B]$ has a weak commutator on $\mathcal{D}$, we denote
\[ \ad(A)(B):=[A,[A,B]_\mathcal{D}]_\mathcal{D}. \]
Inductively, we define for $n=1,2,\dots$
\[ \ad(A)^n(B):=[A,\ad(A)^{n-1}(B)]=\underbrace{[A,[A,\dots,[A}_{n},B]\dots]], \]
if they exist. For $n=0$, we define
\[ \ad(A)^0(B):=B .\]
\begin{Def}
Let $\mathcal{D}$ be a dense subspace in $\mathcal{H}$. We say that the (not necessarily bounded)
operator valued function
\[  t\mapsto A(t) ,\quad t\in \Real \]
is weakly differentiable on $\mathcal{D}$ if and only if
for each $t\in \Real$ there is a linear operator $B(t)$ with the domain of definition being $\mathcal{D}$
such that
\begin{align}
\frac{d}{dt}\inprod\psi {A(t)\phi} = \inprod \psi {B(t)\phi}
\end{align} 
for all $\psi,\phi\in \mathcal{D}$. $B(\cdot)$ is called the weak derivative of $A(\cdot)$ on $\mathcal{D}$
and denoted by 
\[ B(t)=\mathrm{w}\frac{d}{dt}A(t), \quad t\in\Real. \]
\end{Def}

If $B(\cdot)$ is weakly differentiable on $\mathcal{D}$, its weak derivative is called 
the second weak derivative of $A(\cdot)$ and denoted by
\[ \mathrm{w}\frac{d^2}{dt^2}A(t).\]
The $n$-th weak derivatives are inductively defined for $n=1,2,\dots$.

\begin{Def}\label{W^n(T)-def}
Let $A$ is a densely defined linear operator in $\mathcal{H}$ and $T$ be a self-adjoint operator
in $\mathcal{H}$.
We assume that there exists a dense subspace $D_A$ satisfying
$D_A\subset D(T)$ and, for all $t\in\Real$, $e^{itT}D_A\subset D(A)$.
Put
\begin{align}
 A(t):=e^{itT}Ae^{-itT},\quad D_{A,T}:=\{e^{itT}\psi\,|\, t\in\Real,\,\psi\in D_A\}. 
 \end{align}
We say that $A$ is in $W^n(T)$-class for $n\in\Natural$ if and only if
\begin{enumerate}[(i)]
\item $A(t)$ is $n$ times weakly differentiable on $D_{A,T}$. 
\item The mapping 
\[ t\mapsto \mathrm{w}\dfrac{d^{n-1}}{dt^{n-1}}A(t)\psi \]
is strongly continuous for all $\psi\in D_A$.
\item $D_{A,T}$ is included by the domain of the adjoint operator of $\mathrm{w}\dfrac{d^{n-1}}{dt^{n-1}}A(0)$.
\end{enumerate}
\end{Def}

\begin{Thm}
Let $A$ be a densely defined linear operator and suppose that there exists 
a dense subspace $D_A\subset D(T)$ with 
\[ e^{itT}D_A\subset D(A), \quad t\in \Real. \]
The following (I) and (II) are equivalent:
\begin{enumerate}[(I)]
\item $A$ is in $W^n(T)$-class. 
\item $\ad(iT)^n(A)$ exists on $D_{A,T}$ and for each $\psi\in D_A$ the mapping
\[ t\mapsto e^{itT}\ad(iT)^{n-1} (A) e^{-itT}\psi ,\quad t\in\Real \]
is strongly continyous.
\end{enumerate}
Furthermore, if either (I) or (II) holds (and thus both (I) and (II) hold), then 
\begin{align}
\mathrm{w}\frac{d^n}{dt^n}A(t) = e^{itT}\ad(iT)^n(A)e^{-itT}=\ad(iT)^n(A(t)).
\end{align}
\end{Thm}

\begin{Lem}\label{prod-lemma}
Let $A(\cdot)$ be a mapping from $\Real$ into $\mathcal{B}(\mathcal{H})$ which is strongly continuous in $t\in\Real$ and
\[ \sup_{t\in\Real}\norm{A(t)}\le C,\]
for some $C>0$.
Let $\psi(\cdot)$ be a mapping from $\Real$ into $\mathcal{H}$ which is strongly continuous in $t\in\Real$.
Then, the mapping
\[ t\mapsto A(t)\psi(t) \]
 is strongly continuous in $t\in\Real$. 
\end{Lem}
\begin{proof}
\begin{align}\label{prod-esti}
\norm{A(t+h)\psi(t+h)-A(t)\psi(t)} &\le \norm{A(t+h)(\psi(t+h)-\psi(t))} + \norm{(A(t+h)-A(t))\psi(t)}\no\\
	&\le C\norm{\psi(t+h)-\psi(t)} + \norm{(A(t+h)-A(t))\psi(t)}
\end{align}
The first term of \eqref{prod-esti} tends to zero when $h$ goes to zero since $\psi(\cdot)$ is strongly continuous and
the second also vanishes since $A(\cdot)$ is strongly continuous in the same limit. This completes the proof.
\end{proof}
\begin{proof}
We prove by induction. Let $n=1$.

Suppose that (I) is true, that is, $A$ is in $W^1(T)$-class. By Lemma \ref{prod-lemma},
the mapping 
\[ t\mapsto e^{-itT}A(t)\psi = Ae^{-itT}\psi \]
 is strongly continuous. Take arbitrary $\psi,\phi\in D_{A,T}$.
 For $h>0$, one finds from the assumptions
\begin{align}\label{n=1}
\inprod{\psi}{\frac{A(t+h)-A(t)}{h}\phi} &= \inprod \psi{\frac{e^{i(t+h)T}A e^{-i(t+h)T} - e^{itT}Ae^{-itT}}{h}\phi}\no\\
	&=\inprod{\frac{e^{-i(t+h)T}-e^{-itT}}{h}\psi}{Ae^{-i(t+h)T}\phi} +  \inprod {e^{-itT}\psi}{A\frac{e^{-i(t+h)T}-e^{-itT}}{h}\phi}\no\\
	&=\inprod{\frac{e^{-i(t+h)T}-e^{-itT}}{h}\psi}{Ae^{-i(t+h)T}\phi} +  \inprod {A^*e^{-itT}\psi}{\frac{e^{-i(t+h)T}-e^{-itT}}{h}\phi}\no\\
	&\to \inprod{-iTe^{-itT}\psi}{Ae^{-itT}\phi}+ \inprod{A^*e^{-itT}\psi}{-iTe^{itT}\phi}\no\\
	&=\inprod{(iT)^*e^{-itT}\psi}{Ae^{-itT}\phi}- \inprod{A^*e^{-itT}\psi}{(iT)e^{itT}\phi}.
\end{align}
Then, from \eqref{n=1}, one obtains
\begin{align}
\inprod\psi{\mathrm{w}\frac{d}{dt}A(t)\phi} = \inprod{(iT)^*e^{-itT}\psi}{Ae^{-itT}\phi}- \inprod{A^*e^{-itT\psi}}{(iT)e^{itT}\phi}
\end{align}
for all $\psi,\phi\in D_{A,T}$. In particular, 
\begin{align}
\inprod\psi{\mathrm{w}\frac{d}{dt}A(0)\phi} = \inprod{(iT)^*\psi}{A\phi}- \inprod{A^*\psi}{(iT)\phi}
\end{align}
This means that $[iT,A]$ exists on $D_{A,T}$ and 
\begin{align}
[iT,A]=\mathrm{w}\frac{d}{dt}A(0).
\end{align}
Moreover, since $A$ is $W^1(T)$-class, the function
\[ t\mapsto A(t)\psi, \quad \psi\in D_{A} \]
 is strongly continuous. Thus, (II) holds.

Conversely, suppose that (II) is true. That is, $[iT,A]$ exists on $D_{A,T}$ and the mapping
\[ t\mapsto e^{itT}\ad(iT)^0(A) e^{-itT}\psi=A(t)\psi \]
 is strongly continuous for all $\psi\in D_A$. Then, from Lemma \ref{prod-lemma}, we have that
 \[t\mapsto Ae^{-itT}\psi \]
  is strongly continuous for all $\psi\in D_A$. Thus, the computation \eqref{n=1} is still valid and we have 
\begin{align}
\lim_{h\to 0}\inprod{\psi}{\frac{A(t+h)-A(t)}{h}\phi} &= \inprod{(iT)^*e^{-itT}\psi}{Ae^{-itT}\phi}- \inprod{A^*e^{-itT}\psi}{(iT)e^{itT}\phi}\no\\
	&=\inprod\psi{e^{itT}[iT,A]e^{-itT}\phi}.
\end{align}
This proves that $A(\cdot)$ is weakly differentiable on $D_{A,T}$ and
\begin{align}
\mathrm{w}\frac{d}{dt}A(t)=e^{itT}[iT,A]e^{-itT}.
\end{align}
The condition (ii) of Definition \ref{W^n(T)-def} is obvious.
The condition (iii) of Definition \ref{W^n(T)-def} follows since for $n=1$
\[ \left(\mathrm{w}\frac{d^{n-1}}{dt^{n-1}}A(0)\right)^*=A^* \]
and the assertion that $[iT,A]$ exists on $D_{A,T}$ includes the statement $D_{A,H}\subset D(A^*)$. 
Here, we remark that 
\[ A^*(t):=e^{itT}A^*e^{-itT}=A(t)^* \]
on $D_{A,T}$. In fact, for $\psi,\phi\in D_{A,T}$, we have from the assumptions
\begin{align}
\inprod{\psi}{e^{itT}Ae^{-itT}\phi} = \inprod{e^{itT}A^*e^{-itT}\psi}{\phi}.
\end{align}
This implies that 
$\psi\in D(A(t)^*)$ and 
\begin{align}
A(t)^*\psi = e^{itT}A^*e^{-itT}\psi=A^*(t)\psi,\quad t\in\Real.
\end{align}
Therefore, from \eqref{n=1}, it follows that 
\begin{align}
\lim_{h\to 0}\inprod{\psi}{\frac{A(t+h)-A(t)}{h}\phi} &= \inprod{(iT)^*\psi}{A(t)\phi}- \inprod{A(t)^*\psi}{(iT)\phi}.
\end{align}
Thus, we find
\begin{align}
\mathrm{w}\frac{d}{dt}A(t)=[iT,A(t)],
\end{align}
which completes the proof for $n=1$.

Suppose the assertion is valid for some $n\ge 1$. Then,
\begin{align}
\frac{d}{dt}\inprod\psi{\mathrm{w}\frac{d^n}{dt^n}A(t)\phi}&=\frac{d}{dt}\inprod\psi{e^{itT}\ad(iT)^n(A)e^{-itT}\phi}.
\end{align}

We prove (I) implies (II) for $n+1$. If we assume (I) for $n+1$, the mapping
\[ t\mapsto \mathrm{w}\frac{d^n}{dt^n} A(t)\psi = e^{itT}\ad(iT)^n (A)e^{-itT}\psi \]
is strongly continuous for all $\psi\in D_A$. By Lemma \ref{prod-lemma},
one concludes
\[ t\mapsto\ad(iT)^n (A)e^{-itT}\psi \]
 is strongly continuous. Thus, in the similar way to the computation \eqref{n=1}, we have
 \begin{align}\label{general-n}
\frac{d}{dt}\inprod\psi{\mathrm{w}\frac{d^n}{dt^n}A(t)\phi}&=\frac{d}{dt}\inprod\psi{e^{itT}\ad(iT)^n(A)e^{-itT}\phi}\no\\
	&=\inprod{(iT)^*e^{-itT}\psi}{\ad(iT)^n(A)e^{-itT}\phi}- \inprod{\ad(iT)^n(A)^*e^{-itT}\psi}{(iT)e^{itT}\phi}, 
 \end{align} 
 because 
 \[ \left(\mathrm{w}\frac{d^{n}}{dt^{n}}A(0)\right)^*=\ad(iT)(A)^* \]
 and combining the assumption (I) we have 
 \[ D_{A,T}\subset D(\ad(iT)(A)^*). \]
Since we now assume that $A(\cdot)$ is $n+1$ times weakly differentiable, \eqref{general-n} implies
\begin{align}
\inprod\psi{\mathrm{w}\frac{d^{n+1}}{dt^{n+1}}A(t)\phi}&=\inprod{(iT)^*e^{-itT}\psi}{\ad(iT)^n(A)e^{-itT}\phi}- \inprod{\ad(iT)^n(A)^*e^{-itT}\psi}{(iT)e^{itT}\phi}\no\\
&=\inprod{(iT)^*\psi}{e^{itT}\ad(iT)^n(A)e^{-itT}\phi}- \inprod{e^{itT}\ad(iT)^n(A)^*e^{-itT}\psi}{(iT)\phi}\no\\
&=\inprod{(iT)^*\psi}{\ad(iT)^n(A(t))\phi}- \inprod{\ad(iT)^n(A(t))^*\psi}{(iT)\phi}.
\end{align}
Thus, the weak commutator of $iT$ and $\ad(iT)^n(A(t))$ exists on $D_{A,T}$, and the relation
\begin{align}
\mathrm{w}\frac{d^{n+1}}{dt^{n+1}}A(t)&=[iT,\ad(iT)^n(A(t))]\no\\
&=\ad(iT)^{n+1}(A(t))
\end{align}
holds on $D_{A,T}$. Therefore, (II) for $n+1$ is derived.

Conversely, suppose (II) for $n+1$ is valid. By the assumption of the induction,
we conclude that the mapping
\[ t\mapsto \mathrm{w}\frac{d^n}{dt^n}A(t)\psi = e^{itT}\ad(iT)^n(A) e^{-itT}\psi \]
 is strongly continuous for all $\psi\in D_A$. Further, the assumption of the induction tells us that
 \[ \left(\mathrm{w}\frac{d^{n}}{dt^{n}}A(0)\right)^*=\ad(iT)(A)^* \]
and the assumption that $\ad(iT)^{n+1}(A)$ exists on $D_{A,T}$ implies
 \[ D_{A,T}\subset D(\ad(iT)(A)^*)= D\left(\left(\mathrm{w}\frac{d^{n}}{dt^{n}}A(0)\right)^*\right).\]
 Thus, the computation \eqref{general-n} is
 still valid. By (II) for $n+1$ the right hand side of \eqref{general-n}
 becomes
 \[ \inprod{e^{-itT}\psi}{[iT, \ad(iT)^{n}(A)]e^{-itT}\phi}=\inprod{\psi}{e^{itT}\ad(iT)^{n+1}(A)e^{-itT}\phi}. \]
 This means that $A(\cdot)$ is $n+1$ time weakly differentiable and 
 \begin{align}
\mathrm{w}\frac{d^{n+1}}{dt^{n+1}}A(t)=e^{itT}\ad(iT)^{n+1}(A)e^{-itT} 
 \end{align}
 on $D_{A,T}$. 
Hence, (I) for $n+1$ is derived, and the induction is completed.

\end{proof}
\fi 
\bibliographystyle{junsrt}
\bibliography{myref_master}

\end{document}
 %%%%%%%%%%%%%%%%%%%%% Updated %%%%%%%%%%%%
\section{Gauge transformation and physical subspace}
In this section, we study the gauge transformed Hamiltonian.
As is well known, the classical Lagrangian for electromagnetic fields has a local $U(1)$ gauge
symmetry, that is, the classical Lagrangian is invariant under the transformation
\begin{align}
A_\mu \mapsto A'_\mu = A_\mu-\partial_\mu \Lambda
\end{align}  
for arbitrary gauge function $\Lambda$.
If one desires that a new gauge potential $A'_\mu=(A'_0,-\bvec{A}')$ should satisfy the Coulomb 
gauge condition
\begin{align}
\nabla\cdot\bvec{A}'=0,
\end{align}  
at a classical level, then one should choose the function $\Lambda$ satisfying
\begin{align}
\Delta\Lambda = -\nabla\cdot\bvec{A}.
\end{align}




\section{Relation to the Coulomb gauge Hamiltonian}
\section{Gupta-Bleular condition}

Let
\begin{align}
& G := -q  \int ^{\oplus } d\xx _1 dots d\xx _N \; G(\xx _1 , ... , \xx _N ) , \\
& G(\xx _1 , ... , \xx _N )  := \sum _{j=1} ^N I \otimes \cdots \otimes \stackrel{j\text{-th}}{\alpha ^0} \otimes \cdots \otimes I \otimes \phi ^{(3)} \Big( \frac{\hat{\chi _\rm{ph} ^{\xx _j}}}{-i \omega ^{3/2}} \Big) \q \text{on} \q (\ot ^N \C ^4 )  \otimes \F _\rm{ph} .
\end{align}

We introduce a dense subspace $ D_\rm{fin} \subset \F _\rm{DM} (N) $ by 
\begin{align}
D_\rm{fin} := \hot _a ^N C_0 ^\infty (\R ^3 ; \C ^4) \hot \F _{\bb , \rm{fin}} (D(\omega )) .
\end{align}


\begin{Lem}
\begin{enumerate}
\item $ D_\fin $ forms an entire analytic vector space for $ G $.

\item If $ \hat{\chiph} / \omega ^{3/2}\in D(\omega ) $, then, it follows that $ D_\rm{fin} \subset D(e^{iG} H_\rm{DM} e^{-iG}) $ and 
\begin{align}
e^{iG} H_\rm{DM} e^{-iG} \Psi = \sum _{n=0} ^\infty \frac{(-ie)^n}{n!} \big( \rm{ad} ^n (G) H_\rm{DM} \big) \Psi , 
\end{align}
for every $ \Psi \in D_\fin $
\end{enumerate}
\end{Lem}

\begin{proof} Set
\begin{align}
\F _{\bb ,0} := \hop _{n=0} ^\infty \big( \wedge ^N \H _D \otimes (\ot _s ^n \H _\rm{ph}) \big)
\end{align}
Since $ \F _{\bb ,0} $ forms a entire analytic vector space for $ G $, we have
\begin{align}
e^{iG} \Psi ~ \slim _{N\to \infty } \sum _{n=0} ^N \frac{i^n}{n!} G^n \Psi .
\end{align}
We prove now that $  G^n \Psi \in D(H_\rm{DM}) $ for every $ \Psi \in D_\rm{fin} $. It is obvious that $ G^n \Psi \in D(H_\rm{int}) $ because $ G^n \Psi \in \F _{\bb , 0} $. Therefore, it suffices to prove that $ G^n \Psi \in D(H_0) $.

We first prove $ G^n \Psi \in D(H_\rm{ph}) $. For each $ \xx _1 , ..., \xx _N  \in \R ^3 $, it follows that $ G(\xx _1 , ... , \xx _N ) ^n \Psi \in D(H_\rm{ph}) $ from the present assumption. By lemma \ref{}, we have 
\begin{align}
[iH_\rm{ph} , \phi ^{(3)} \Big( \frac{\hat{\chi _\rm{ph} ^\xx }}{-i \omega ^{3/2}} \Big)] \Phi = \phi ^{(3)} \Big( \frac{\hat{\chi _\rm{ph} ^\xx }}{-\omega ^{1/2}} \Big) \Phi
\end{align}
for all $ \Phi \in \F _{\bb , \rm{fin}} (D(\omega )) $. It follows directly
\begin{align}
iH_\rm{ph} G(\xx _1 , ... , \xx _N ) ^n \Psi = \sum _{k=1} ^n G(\xx _1 , ... , \xx _N )  ^{k-1} \Big( \sum _{j=1} ^N I \otimes \cdots \otimes I \otimes \phi ^{(3)} \Big(  \frac{\hat{\chi _\rm{ph} ^{\xx _j}}}{-\omega ^{1/2}}  \Big)  G(\xx _1 , ... , \xx _N ) ^{n-k} \Psi + G(\xx _1 , ... , \xx _N ) ^n iH_\rm{ph} \Psi .
\end{align}
Hence we have 
\begin{align}
& \| iH_\rm{ph} G(\xx _1 , ... , \xx _N ) ^n \Psi \| \\
\le & n \Big( \frac{2}{\sqrt{2}}\Big) ^n \Big( \sum _{j=1} ^N \| \hat{\chi _\rm{ph} ^{\xx _j}} / \omega ^{3/2} \| \Big) ^{n-1} \Big( \sum _{j=1} ^N \| \hat{\chi _\rm{ph} ^{\xx _j}} / \omega ^{1/2} \| \Big) \| (N_\bb + n) ^{1/2} \cdots (N_\bb + 1) ^{1/2} \Psi \| \\
& + \Big( \frac{2}{\sqrt{2}}\Big) ^n \Big( \sum _{j=1} ^N \| \hat{\chi _\rm{ph} ^{\xx _j}} / \omega ^{3/2} \| \Big) ^{n} \| (N_\bb + n) ^{1/2} \cdots (N_\bb + 1) ^{1/2} \Psi \| .
\end{align}
Thus we see that
\begin{align}
\int d\xx _1 \dots d\xx _N \; \| H_\rm{ph} G(\xx _1 , \dots , \xx _N) ^n \Psi \| ^2 < \infty 
\end{align}
which implies that $ G^n \Psi \in D(H_\rm{ph}) $ and 
\begin{align}
\| H_\rm{ph} G^n \Psi \| ^2 & \le \int d\xx _1 \dots d\xx _N \; \| H_\rm{ph} G(\xx _1 , \dots \xx _N) ^n \Psi (\xx _1 , \dots , \xx _N) \| ^2 \\
& \le \int d\xx _1 \dots d\xx _N \Big( C n 2^{n/2}  (N_\Psi +n ) ^{1/2} \cdots (N_\Psi + 1) ^{1/2} \| \Psi (\xx _1 , \dots \xx _N) \| + C' 2^{n/2} (N_\Psi +n ) ^{1/2} \cdots (N_\Psi + 1) ^{1/2} \| H_\rm{ph} \Psi (\xx _1 , \dots \xx _N) \| \Big) ^2 .
\end{align}
Thus 
\begin{align}
& \sum _{n=0} ^\infty \frac{t^n }{n!} \| H_\rm{ph} G^n \Psi \| \\
\le &  \sum _{n=0} ^\infty \frac{t^n }{n!} (N_\Psi +n ) ^{1/2} \cdots (N_\Psi +1) ^{1/2} \Big( C n 2^{n/2} \| \Psi \| + C' 2^{n/2} \Big) \\
< & \infty .
\end{align}
Using the closedness of $ H_\rm{ph} $, we conclude that $ e^{iG} \Psi \in D(H_\rm{ph}) $.

Next, we prove $ e^{iG}\Psi \in D(H_\rm{D}) $. (以下未記入)
\end{proof}



\begin{Thm} It follows that
\begin{align}
e^{iG} H_\rm{DM} e^{-iG} = H^\CC + H_\rm{resid} + \rm{const.}
\end{align}
on $ D _\rm{fin} $.
\end{Thm}

\begin{proof} 
上のLemma を用いて，$ \rm{ad}^1 (G) H_\rm{DM} , \rm{ad}^2 (G) H_\rm{DM} $を計算して終わり．
\end{proof}



\begin{Thm} It follows that
\begin{align}
e^{iG} H_\rm{DM} e^{-iG} = H^\CC + H_\rm{resid} + \rm{const.}
\end{align}
on $ D _\rm{fin} $.
\end{Thm}




\begin{Def}[Operator-valued distribution] Let $ \K $ be a Klein space with metric $ \langle \cdot | \cdot \rangle $ and $ \L \subset \K $ a dense subspace and set $ d \in \N $. A map $ \varphi : \S (\R ^d) \ni f \mapsto \varphi (f) $ is called an operator-valued distribution on $ \R ^d  $ acting on $ (\K , \L ) $ if $ \varphi (f) \; (f \in \S (\R ^d )) $ are densely defined linear operators on $ \K $ and satisfy the following properties: (i) For all $ f \in \S (\R ^d) $, $ \L \subset D(\varphi (f)) \cap D(\varphi (f) ^\dagger ) $, (ii) for all $ f,g \in \S (\R ^d) $ and $ \alpha , \beta \in \C  $, $ \varphi (\alpha f + \beta g) = \alpha \varphi (f) + \beta \varphi (g) $ on $ \L $, and (iii) for all $ \psi , \phi \in \L $, the map $ f \mapsto \langle \psi | \varphi (f) \phi  \rangle $ is a tempered distribution on $ \R ^d $.
\end{Def}

\begin{Lem}\label{TO1}
\begin{enumerate}[(i)]
\item 
\begin{align}
\| A_\mu (0,f) \Psi \| & \le 2 \| \hat{f} / \sqrt{\omega } \| \, \| (N_\bb +1 ) ^{1/2} \Psi \| \\
& \le C_N \| \hat{f} \| _{N,0} (N_\Psi +1) ^{1/2} \| \Psi \|
\end{align}
where $ \| \cdot \| _{m , \alpha } $ is the norm of rapidly decreasing functions.

\item 
\begin{align}
\| U_n(t,t') \Psi  \| \le \frac{|t-t'| ^n }{n!} C^n (N_\Psi +n) ^{1/2} \cdots (N_\Psi +1) ^{1/2} \| \Psi \| . 
\end{align}
\end{enumerate}
\end{Lem}



\begin{Thm} For each $ t\in \R $, the map $ \S (\R ^d ) \ni f \mapsto _\mu (t,f)  $ is an operator-valued distribution on $ \R ^d $ acting on $ (\F  , \F _{\bb , 0}) $.
\end{Thm}

\begin{proof} (Time-orderのTheorem 2.7により) For all $ f\in \S (\R ^3) $,
\begin{align} 
\F _{\bb ,0} \subset D(A_\mu (t, f)) \cap D(A_\mu (t,f) ^\dagger ) .
\end{align}

For each $ \Psi \in \F _{\bb ,0} $, 
\begin{align}
A_\mu (t, \alpha f + \beta g) & = W(-t) A_\mu (0, \alpha f + \beta g) W(t) \Psi \\
& = \sum _{m,n=0} ^\infty e^{itH_0} U_m (-t,0) A_\mu (0, \alpha f+ \beta g) U_n (0 , -t) e^{-itH_0} \Psi \\
& = \alpha \sum _{m,n=0} ^\infty e^{itH_0} U_m (-t,0) A_\mu (0, f) U_n (0 , -t) e^{-itH_0} \Psi \\
& \q + \beta \sum _{m,n=0} ^\infty e^{itH_0} U_m (-t,0) A_\mu (0, g) U_n (0 , -t) e^{-itH_0} \Psi \\
& = \alpha A_\mu (0,f) + \beta A_\mu (0,g) .  
\end{align}
Thus, the linearity holds.

Using Lemma \ref{TO1}, we have
\begin{align}
\| A_\mu (t,f) \Psi \| \le \sum _{m,n =0} ^\infty \frac{|t| ^{m+n}}{m! n!} C^{m+n} C_N (N_\Psi + m+n+1) ^{1/2} \cdots (N_\Psi +1) ^{1/2} \| \Psi \| \times \| \hat{f} \| _{N,0} . 
\end{align}
Hence, the map $ f \mapsto \langle \Phi | A_\mu (t,f) \Psi \rangle $ is continuous.
\end{proof}


Let
\begin{align}
\pi _\mu (0,f) := \frac{i}{\sqrt{2}} \Big( a_\mu (\sqrt{\omega } \hat{f^*}) - a_\mu ^\dagger (\sqrt{\omega } \hat{f}) \Big) .
\end{align}

\begin{Lem}\label{pi1}
\begin{enumerate}[(I)]
\item $ \pi _\mu (0,f) $ and $ \pi _\mu (0, f) ^* $ are $ N_\bb ^{1/2} $- bounded and closed.

\item For all $ L \ge 0 $, $ \Psi \in R (E_{N_\bb } ([0, L])) $ implies that $ \pi_\mu (0,f) \Psi $ and $ \pi _\mu (0,f) ^* \Psi $ belong to $ R (E_{N_\bb } ([0, L+1]))  $.
\end{enumerate}
\end{Lem}



\begin{Thm} For each $ f\in \S (\R ^3) $ and $ \Psi \in \F _{\bb ,0} \cap D(H_0) $, the map $ \R \ni t\mapsto A_\mu (t,f) \Psi $ is two times strongly differentiable and 
\begin{align}\label{eq_of_motion}
\Box A_\mu (t,f) \Psi  = -q \sum _{j=1} ^N \alpha _{j,\mu } \int ^{\oplus } d\xx _j (f* \chi _\rm{ph} ) (\xx _j) 
\end{align}
holds.
\end{Thm}

\begin{proof} By using Theorem 2.8 of [Time order], 
\begin{align}
\frac{\pa }{\pa t} A_\mu (t,f) = W(-t) [iH , A_\mu (0,f)] W(t) . 
\end{align}
It follows directly
\begin{align}
[iH , A_\mu (0,f)] = - \pi _\mu (0,f)  
\end{align}
on $ \F _{\bb ,0} $. From Lemma \ref{pi1}, the map $ \R \ni t \mapsto W(-t) \pi _\mu (0,f) W(t) $ is strongly continuously differentiable in $ t\in \R $, and it follows that
\begin{align}
\frac{\pa }{\pa t} \Big( \frac{\pa }{\pa t} A_\mu (t,f) \Big) \Psi = - W(-t) [iH , \pi _\mu (0,f)] W(t) \Psi .
\end{align}
By direct calculation, we have
\begin{align}
[iH, \pi _\mu (0,f)] = \frac{1}{\sqrt{2}} \Big( a_\mu (\frac{\omega ^2 \hat{f^*}}{\sqrt{\omega }}) +  a_\mu ^\dagger (\frac{\omega ^2 \hat{f}}{\sqrt{\omega }}) \Big) + q \sum _{j=1} ^N \alpha _{j , \mu } \int ^{\oplus } d\xx _j (f* \chiph ) (\xx _j) .
\end{align}
In the operator-valued distribution sense, $ \Delta A_\mu (0,f) = A_\mu (0, \Delta f) $. Thus we obtain \eqref{eq_of_motion}.
\end{proof}

\begin{Thm} Let $ \varphi _\mu (t, \cdot ) \; (t\in \R ) $ be operator-valued distribution on $ \R ^3 $ acting on $ (\F , \F _{\bb ,0}) $ in the sense of Definition \ref{}. Assume that the following hold.

(A.1) For each $ f \in \S (\R ^3) $ and $ \Psi \in \F _{\bb ,0} $, the map $ \R \ni t \mapsto \varphi _\mu (t,f) \Psi \in \F  $ is two times strongly differentiable.

(A.2) For all $ f \in \varphi \in \S (\R ^3) $ and $ \Psi \in \F _{\bb , 0} $, 
\begin{align}
& \varphi _\mu (0,f) \Psi = A_\mu (0,f) \Psi , \\
& \frac{\pa }{\pa t} \varphi _\mu (t,f) \Psi \Big| _{t=0} = \frac{\pa }{\pa t} A_\mu (t,f) \Psi \Big| _{t=0} . 
\end{align}

(A.3) For all $ f\in \S (\R ^3) $ and $ \Psi \in \F _{\bb ,0} $,
\begin{align}
\Box \varphi _\mu (t,f) = -q \sum _j \alpha _{j , \mu} \int ^{\oplus } d\xx _j (f* \chiph ) (\xx _j) .
\end{align}

Then, for all $ f \in \S (\R ^3) $ and $ \Psi \in \F _{\bb ,0} $, 
\begin{align}
\varphi _\mu (t,f) \Psi = A_\mu (t,f) \Psi .
\end{align}
\end{Thm}


\begin{proof}(鈴木先生のとほぼ同じ証明方法) Let us fix arbitrary $ \Psi , \Phi \in \F _{\bb ,0} $ and set 
\begin{align}
C_\mu (t,f) := \langle \Psi | \varphi _\mu (t,f) \Phi  \rangle - \langle \Psi | A_\mu (t,f) \Phi \rangle , \q t\in \R , \q f\in \S (\R ^3) .
\end{align}
Then, $ \S (\R ^3) \ni f \mapsto C_\mu (t,f)  $ is a tempered distribution on $ \R ^3 $ and satisfies 
\begin{align}\label{uniqueness1}
\frac{\pa ^2 }{\pa t^2} C_\mu (t,f) = C_\mu (t, \Delta f)
\end{align}
with 
\begin{align}
C_\mu (t,f) =0 , \q \frac{\pa }{\pa t} C_\mu (t,f) \Big| _{t=0} =0 .
\end{align}
Hence we find
\begin{align}
C_\mu (t,f) =0 
\end{align}
for all $ t\in \R $ and $ f\in \S (\R ^3) $. (これはかなり省略しているので，以下の証明を載せるかどうか後で検討)

By using \eqref{uniqueness} repeatedly, we observe that $ C_\mu (t,f) $ is $ C^\infty $-class and satisfies
\begin{align}
\frac{d^n}{dt^n} C_\mu (t,f) \Big| _{t=0} =0 
\end{align}
for every $ n \in \Z _+ $. Therefore the function $ \R \ni t \mapsto C_\mu (t,f) $ is analytic, and it follows that
\begin{align}
C_\mu (t,f) = \sum _{n=0} ^\infty \frac{t^n }{n!} \Big( \frac{d^n }{dt^n} C_\mu (t,f) \Big| _{t=0}  \Big) =0 .
\end{align}
(ここまでが追加パート)
 
Since $ \Phi \in \F _{\bb ,0} $ is chosen arbitrary and $ \F _{\bb , 0} $ is dense in $ \F $, we get the desired result.
\end{proof}


Let 
\begin{align}
\Omega (t,f) := \pa _\mu A^\mu (t,f) .
\end{align}

\begin{Thm} It follows that
\begin{align}
\Omega (t,f) = & a_\mu \Big( \frac{ik^\mu e^{it\omega } \hat{f^*}}{\sqrt{2\omega }} \Big) + a_\mu ^\dagger  \Big( \frac{ik^\mu e^{it\omega } \hat{f}}{\sqrt{2\omega }} \Big) \\
& + \sum _{j=1} ^N iq \int ^{\oplus } d\xx _j \Big( \langle \frac{e^{it\omega } \hat{f^*}}{\sqrt{2\omega }} , \frac{\hat{\chiph ^{\xx _j}}}{\sqrt{2\omega }} \rangle - \langle \frac{\hat{\chiph ^{\xx _j}}}{\sqrt{2\omega }} , \frac{e^{it\omega } \hat{f}}{\sqrt{2\omega }} \rangle  \Big) .
\end{align}
\end{Thm}

\begin{proof} By \eqref{}, we have
\begin{align}
\Omega (t,f) = W(-t) \Big( a_\mu \Big( \frac{ik^\mu \hat{f^*}}{\sqrt{2\omega }}\Big) + a_\mu ^\dagger \Big( \frac{ik^\mu \hat{f}}{\sqrt{2\omega }}\Big) \Big) W(t) .
\end{align}
By using [Time order Theorem 2.8], 
\begin{align}
\frac{d}{dt} \Omega (t,f) = W(-t) [iH , \Omega (0, f)] W(t) .
\end{align}
By direct calculation,
\begin{align}
[iH , \Omega (0,f)] & = a_\mu \Big( \frac{i\omega i k^\mu \hat{f^*}}{\sqrt{2\omega }} \Big) + a_\mu ^\dagger \Big( \frac{i\omega i k^\mu \hat{f}}{\sqrt{2\omega }} \Big) \\
& \q - \sum _{j=1} ^N iq \alpha _{j ,\mu } \int ^{\oplus } d\xx _j \Big( \langle \frac{ ik^\mu \hat{f^*}}{\sqrt{2\omega }} , \frac{\hat{\chiph ^{\xx _j}}}{\sqrt{2\omega }} \rangle - \langle \frac{\hat{\chiph ^{\xx _j}}}{\sqrt{2\omega }} , \frac{ ik^\mu \hat{f}}{\sqrt{2\omega }} \rangle  \Big) .
\end{align}

以下，直接計算によって
\begin{align}
\Box \Omega (t,f) =0
\end{align}
がわかる．この場合，$ \pa _0 ^n \Omega (t,f) $を次々に計算していって，Taylor展開で$ \Omega (t,f) $を求める方針でいく．
\end{proof}

ロンスキアンを採用するかどうかは後で検討．とりあえず
\begin{align}
\Omega ^{(+)} (t,f) := a_\mu \Big( \frac{ik^\mu e^{it\omega } \hat{f^*}}{\sqrt{2\omega }} \Big) + \sum _{j=1} ^N iq \int ^{\oplus } d\xx _j \; \langle \frac{e^{it\omega } \hat{f^*}}{\sqrt{2\omega }} , \frac{\hat{\chiph ^{\xx _j}}}{\sqrt{2\omega }} \rangle ,
\end{align}
とおく．このとき，
\begin{align}
\V _\rm{phys} := \bigcap _{t \in \R , \, f\in \S (\R ^3)} \ker \Omega ^{(+)} (t,f) 
\end{align}
とおく．（$ \Omega ^{(+)} (F) \; (F \in \S (\R ^4)) $を基準にしてもＯＫだが，それもまた後で検討）


\begin{Lem} The subspace 
\begin{align}
\Big\{ \frac{i\omega e^{it\omega }\hat{f^*}}{\sqrt{2\omega }} \; \Big| \; f \in \S (\R ^3) , \; t\in \R  \Big\}
\end{align}
is dense in $ L^2 (\R ^3 ) $.
\end{Lem}

\begin{Lem}(これは不要っぽい) Let $ D \subset L^2 (\R ^3) $ be a dense subspace. Then 
\begin{align}
\F _3 = \bigcap _{f\in D} \ker c_0 (f) .
\end{align}
\end{Lem}






\begin{Thm}
\begin{enumerate}[(i)] 
\item $ \V _\rm{phys} $ is a closed subspace.  

\item $ \V _\rm{phys} $ is non-negative, that is, for all $ \Psi , \Phi \in \V _\rm{phys} $, $ \langle \Psi | \Phi \rangle \ge 0  $.

\item $ \V _\rm{phys} = U_G W \F _3 $.
\end{enumerate}
\end{Thm}

\begin{proof}
\begin{enumerate}[(i)]
\item By the closedness of $ \Omega ^{(+)} (t,f) $, $ \ker \Omega ^{(+)} (t,f) $ is closed subspace and so is $ \V _\rm{phys} $.

\item 未記入

\item これ以降，special Lorentz frameを使用．
\end{enumerate}
\end{proof}

\section{Scattering Problems}

\end{document}

Lorenz_DM_USUI_20150325.bbl
\begin{thebibliography}{10}

\bibitem{Nishijima1973a}
Kazuhiko Nishijima.
\newblock {\em Relativistic Quantum Mechanics (in Japanese)}.
\newblock Baihu-kan, 1973.

\bibitem{MR1765584}
Asao Arai.
\newblock A particle-field {H}amiltonian in relativistic quantum
  electrodynamics.
\newblock {\em J. Math. Phys.}, Vol.~41, No.~7, pp. 4271--4283, 2000.

\bibitem{MR1981623}
Asao Arai.
\newblock Non-relativistic limit of a {D}irac-{M}axwell operator in
  relativistic quantum electrodynamics.
\newblock {\em Rev. Math. Phys.}, Vol.~15, No.~3, pp. 245--270, 2003.

\bibitem{MR2260374}
Asao Arai.
\newblock Non-relativistic limit of a {D}irac polaron in relativistic quantum
  electrodynamics.
\newblock {\em Lett. Math. Phys.}, Vol.~77, No.~3, pp. 283--290, 2006.

\bibitem{MR2810826}
Asao Arai.
\newblock Heisenberg operators of a {D}irac particle interacting with the
  quantum radiation field.
\newblock {\em J. Math. Anal. Appl.}, Vol. 382, No.~2, pp. 714--730, 2011.

\bibitem{MR2178588}
Itaru Sasaki.
\newblock Ground state energy of the polaron in the relativistic quantum
  electrodynamnics.
\newblock {\em J. Math. Phys.}, Vol.~46, No.~10, pp. 102307, 6, 2005.

\bibitem{MR2377946}
Edgardo Stockmeyer and Heribert Zenk.
\newblock Dirac operators coupled to the quantized radiation field: essential
  self-adjointness \`a la {C}hernoff.
\newblock {\em Lett. Math. Phys.}, Vol.~83, No.~1, pp. 59--68, 2008.

\bibitem{MR509200}
Keiichi~R. Ito.
\newblock Canonical linear transformation on {F}ock space with an indefinite
  metric.
\newblock {\em Publ. Res. Inst. Math. Sci.}, Vol.~14, No.~2, pp. 503--556,
  1978/79.

\bibitem{Kugo:1977zq}
Taichiro Kugo and Izumi Ojima.
\newblock Manifestly covariant canonical formulation of yang-mills field
  theories: Physical state subsidiary conditions and physical s matrix
  unitarity.
\newblock {\em Phys. Lett.}, Vol. B73, p. 459, 1978.

\bibitem{MR557151}
Taichiro Kugo and Izumi Ojima.
\newblock Local covariant operator formalism of nonabelian gauge theories and
  quark confinement problem.
\newblock {\em Progr. Theoret. Phys. Suppl.}, No.~66, p. 130, 1979.

\bibitem{Lautrup:1967}
Benny Lautrup.
\newblock Canonical quantum electrodynamics in covariant gauges.
\newblock {\em Kgl. Danske Videnskab. Selskab. Mat.-fys. Medd}, Vol.~35,
  No.~11, p.~1, 1967.

\bibitem{Nakanishi:1973fu}
N.~Nakanishi.
\newblock {Indefinite-metric quantum theory of genuine and higgs-type massive
  vector fields}.
\newblock {\em Prog.Theor.Phys.}, Vol.~49, pp. 640--651, 1973.

\bibitem{Nakanishi:1974pr}
Noboru Nakanishi.
\newblock {Lehmann-Symanzik-Zimmermann Formalism for the Manifestly Covariant
  Quantum Electrodynamics}.
\newblock {\em Prog.Theor.Phys.}, Vol.~52, p. 1929, 1974.

\bibitem{Nakishi:1966di}
Noboru Nakishi.
\newblock {Covariant quantization of the electromagnetic field in the Landau
  gauge}.
\newblock {\em Prog.Theor.Phys.}, Vol.~35, p. 1111, 1966.

\bibitem{MR0038883}
K.~Bleuler.
\newblock Eine neue {M}ethode zur {B}ehandlung der longitudinalen und skalaren
  {P}hotonen.
\newblock {\em Helvetica Phys. Acta}, Vol.~23, pp. 567--586, 1950.

\bibitem{MR0036166}
Suraj~N. Gupta.
\newblock Theory of longitudinal photons in quantum electrodynamics.
\newblock {\em Proc. Phys. Soc. Sect. A.}, Vol.~63, pp. 681--691, 1950.

\bibitem{FutakuchiUsui2013}
Shinichiro Futakuchi and Kouta Usui.
\newblock {Construction of dynamics and time-ordered exponential for unbounded
  non-symmetric Hamiltonians}.
\newblock {\em Journal of Mathematical Physics}, Vol.~55, No. 062303, 2014.

\bibitem{MR2533876}
Fumio Hiroshima and Akito Suzuki.
\newblock Physical state for nonrelativistic quantum electrodynamics.
\newblock {\em Ann. Henri Poincar\'e}, Vol.~10, No.~5, pp. 913--953, 2009.

\bibitem{MR574710}
Keiichi~R. Ito.
\newblock Construction of two-dimensional quantum electrodynamics.
\newblock {\em J. Math. Phys.}, Vol.~21, No.~6, pp. 1473--1494, 1980.

\bibitem{MR2412280}
Akito Suzuki.
\newblock Physical subspace in a model of the quantized electromagnetic field
  coupled to an external field with an indefinite metric.
\newblock {\em J. Math. Phys.}, Vol.~49, No.~4, pp. 042301, 24, 2008.

\bibitem{IR-GB}
Simone Zerella.
\newblock Infrared gupta-bleuler qunantum electrodynamics: Solvable models and
  perturbative expansion.
\newblock {\em arXiv: 1411.2387}, 2014.

\bibitem{FutakuchiUsuiSelf-ad}
Shinichiro Futakuchi and Kouta Usui.
\newblock New criteria for self-adjointness and its application to
  dirac--maxwell hamiltonian.
\newblock {\em Letters in Mathematical Physics}, Vol. 104, No.~9, pp.
  1107--1119, September 2014.

\bibitem{MR1219537}
Bernd Thaller.
\newblock {\em The {D}irac Equation}.
\newblock Texts and Monographs in Physics. Springer-Verlag, Berlin, 1992.

\bibitem{MR0493420}
Michael Reed and Barry Simon.
\newblock {\em Methods of Modern Mathematical Physics. {II}. {F}ourier
  analysis, self-adjointness}.
\newblock Academic Press [Harcourt Brace Jovanovich Publishers], New York,
  1975.

\bibitem{AraiFock}
Asao Arai.
\newblock {\em Fock Spaces and Quantum Fields I, II (in Japanese)}.
\newblock Nippon-Hyoronsha, Tokyo, 2000.

\bibitem{MR751959}
Michael Reed and Barry Simon.
\newblock {\em Methods of Modern Mathematical Physics. {I}. Functional
  Analysis}.
\newblock Academic Press Inc. [Harcourt Brace Jovanovich Publishers], New York,
  second edition, 1980.

\end{thebibliography}
