text:
\section{Introduction}
The Gupta-Bleuler formulation of four-dimensional \linebreak quantization is well known from literature \cite{Gupta50} - \cite{WHK03}. \linebreak Mostly, it is assumed that the time-like photon states have negative norms. Unfortunately, states with such norms form a serious difficulty with the probability interpretation of quantum mechanics. There seems to be no experimental verification of such particles, therefore they are often called in literature unphysical or even ``ghosts" \cite{HPDER69}. In this paper the problem of negative norms of the time-like photons is profoundly investigated, introducing a different interpretation of the ladder operators.

For further analysis let us take a convention $c=\hbar=1$. Building a relativistic model for photons, we have to consider momentum and polarization of the photon. In this mathematical model these two quantities will be described in a tensor product structure, i.e.
\begin{eqnarray}
\textrm{photon momentum space}~
\otimes
~
\textrm{photon polarization space},
\nonumber
\end{eqnarray}
although it should be stressed that in relativistic context spin and momentum are not independent degrees of freedom.

First, in Section \ref{sec:4dconstruction} a construction coming from a covariant Hamiltonian for a four-dimensional polarization oscillator is shown.
Then we ask what is the consequence of creating particles on the energy of the whole system, i.e. does this raise the energy level or lower it? This is discussed in sections: \ref{sec:4dconstruction123} for the space-like polarizations and further in \ref{sec:4dconstruction0e+} and \ref{sec:4dconstruction0e-} two different interpretations of the time-like polarization are considered.

Next, some preliminary aspects on the Czachor's reducible representations of canonical commutation relations (CCR) are presented. Such model has strong arguments in its favor, mainly because it deals with most of the infrared and ultraviolet divergences. In Section \ref{sec:redmot} a basic concept of the model is introduced. Further, in Section \ref{sec:redladder} the ladder and the number operators are introduced for $N=1$ oscillator space.
In Section \ref{sec:redHam} the covariant Hamiltonian in the reducible representation algebra is presented.
How does the extension to $N$-oscillator space look like and what is the definition of the number operator in an arbitrary $N$ oscillator space are discussed in Section \ref{sec:redladderN}.
Next, the construction of vacuum is shown in \ref{sec:4dvacuum}.  Then we ask: how does such a theory, with four polarization degrees of freedom, correspond to Maxwell's electromagnetism theory. It turns out that vectors, denoted here by $\Psi_{EM}$, reproduce standard Maxwell electrodynamics. This is shown in Section \ref{sec:4dHem}. Finally, the electromagnetic field tensor for such representation is introduced in Section \ref{sec:4dF}.

In the last section a homogeneous Lorentz transformation for the four-dimensional oscillator is introduced. When taking into account the four-photon polarization degrees of freedom, the Lorentz transformation is accompanied by another transformation and this manifests itself also on the spin-frame level. In Section \ref{sec:LorentzBogoliubov} a corresponding transformation of the ladder operators is derived. Further, in Section \ref{sec:Lorentzrepresentation}, the generators for the irreducible representation are introduced. Next, in Section \ref{sec:Lorentzreducible}, we introduce Lorentz transformation in reducible representation algebra, whereas, in \ref{sec:Lorentzvacuum} we show vacuum transformation. In Section \ref{sec:gaugetransfor} we discuss gauge transformations of the four-vector potential and
in Section \ref{secmixing} we show that the ``ghost" operator and the covariant number operator are invariant in any gauge and any reference frame. Finally, in Section \ref{sec:P} the four-translations are introduced. Summary and final conclusions are drawn in Section \ref{sec:4dredconclusions}.

This paper closes with mathematical appendices, introducing  the Minkowski and null tetrad in Penrose and Rindler notation \cite{RPWR84} and proofs involving Lorentz transformation on electromagnetic potential electromagnetic field operator.
\section{Four-dimensional photon polarization space}
\subsection{Construction of four-dimensional polarization space}\label{sec:4dconstruction}
To construct a four-dimensional photon polarization space, let us first introduce a covariant Hamiltonian of the form
\begin{equation}
H\label{covH}
=
-
\frac{p_{{\bf a}}p^{{\bf a}}}{2}
-\frac{q_{{\bf a}}q^{{\bf a}}}{2},~~~ {\bf a}=0,1,2,3.
\end{equation}
Here $p_{{\bf a}}$ and  $q_{{\bf a}}$ are some canonical variables such that
\begin{equation}\label{canvar}
p_{{\bf a}}
=
i\partial_{{\bf a}}
=
i\frac{\partial}{\partial q^{{\bf a}}}
=
ig_{{\bf {ab}}}\frac{\partial}{\partial q_{{\bf {b}}}},
\end{equation}
with commutation relations
\begin{eqnarray}
[q_{{\bf a}},p_{{\bf {b}}}]
&=&
-ig_{{\bf{a}}}{_{{\bf{b}}}}.
\end{eqnarray}
The metric here is denoted as diag$(+,-,-,-)$. These canonical variables should not be mistaken with the position and momentum of the photon field. This construction is made strictly for the four degrees of photon polarization.  Now let us define non-hermitian operators
\begin{eqnarray}\label{acov}
a_{{\bf a}}
&=&
\frac{
	q_{{\bf a}}+ip_{{\bf a}}}{\sqrt{2}}
=
\frac{
	q_{{\bf a}}-\partial_{{\bf a}}}{\sqrt{2}},
\\
a_{{\bf a}}^{\dagger}\label{acovdag}
&=&
\frac{
	q_{{\bf a}}-ip_{{\bf a}}}{\sqrt{2}}
=
\frac{
	q_{{\bf a}}+\partial_{{\bf a}}}{\sqrt{2}},
\end{eqnarray}
which satisfy the following commutation relations
\begin{equation}\label{comaadag}
[a_{{\bf a}},a_{{\bf {b}}}^{\dagger}]
=
-g_{{\bf{a}}}{_{{\bf{b}}}},
\end{equation}
\begin{equation}
[a_{{\bf a}},a_{{\bf {b}}}]\label{comaa}
=
[a_{{\bf a}}^{\dag},a_{{\bf {b}}}^{\dag}]
=
0.
\end{equation}
As a consequence of such covariant formalism, there are four polarization degrees of freedom. They can be written in a four-dimensional tensor product space
\begin{eqnarray}\label{4a}
a_{1}=\textrm{a}_1\otimes 1\otimes 1\otimes 1,~~~~
a_{2}=1\otimes \textrm{a}_2\otimes 1\otimes 1,
\\
a_{3}=1\otimes 1\otimes \textrm{a}_3\otimes 1,~~~~
a_{0}=1\otimes 1\otimes 1\otimes \textrm{a}_0,
\end{eqnarray}
and such space is spanned by kets of the form
\begin{eqnarray}\label{ket4irr}
|n_1,n_2,n_3,n_0\rangle
&=&
|n_1\rangle \otimes |n_2\rangle \otimes |n_3\rangle \otimes |n_0\rangle.
\end{eqnarray}
Let us also take a closer look at
\begin{eqnarray}
a_{{\bf a}}^{\dagger}
a^{{\bf a}}
&=&
\frac{1}{2}q_{{\bf a}}q^{{\bf a}}
+2
+\frac{1}{2}
p_{{\bf a}}p^{{\bf a}}.
\end{eqnarray}
This implies that the Hamiltonian operator (\ref{covH}) can be written in terms of operators $a_{{\bf a}}^{\dagger}$ and $a^{{\bf a}}$ in the form
\begin{eqnarray}
H=
-a_{{\bf a}}^{\dagger}
a^{{\bf a}}
+2,
~~~~~ {\bf a}=0,1,2,3.
\end{eqnarray}
Furthermore, the following commutation relations hold
\begin{equation}
[H,a_{{\bf a}}]
=
[ -a_{{\bf {b}}}^{\dagger}
a^{{\bf {b}}}
,a_{{\bf a}}]
=
-a_{{\bf a}},
\end{equation}
\begin{equation}
[H,a_{{\bf a}}^{\dagger}]
=
[- a_{{\bf {b}}}^{\dagger}
a^{{\bf {b}}}
,a_{{\bf a}}^{\dagger}]
=
a_{{\bf a}}^{\dagger}.
\end{equation}
Let us also assume that the eigenvalue of the covariant Hamiltonian operator (\ref{covH}) acting on four-dimensional space is denoted by $E$.
\begin{eqnarray}
H|n_1,n_2,n_3,n_0\rangle
&=&
E|n_1,n_2,n_3,n_0\rangle.
\end{eqnarray}
At this point this is really a quantity corresponding to the number of particles, but in the upcoming Section \ref{sec:4dreducible} an extension of this model to reducible representation algebra is made and $E(1)$ will correspond to the total energy operator.
For now it can be shown that indeed $a_{{\bf a}}$ lowers and $a_{{\bf a}}^{\dag}$ raises $E$ by 1, i.e.
\begin{eqnarray}\label{defanihilation}
Ha_{{\bf a}}|n_1,n_2,n_3,n_0\rangle
=
(E-1)a_{{\bf a}}|n_1,n_2,n_3,n_0\rangle,
\\
\label{defcreation}
Ha_{{\bf a}}^{\dag}|n_1,n_2,n_3,n_0\rangle\
=
(E+1)a_{{\bf a}}^{\dag}|n_1,n_2,n_3,n_0\rangle.
\end{eqnarray}
\subsection{Construction for space-like photons }\label{sec:4dconstruction123}
From the previous section we learn that (\ref{defanihilation}) and (\ref{defcreation}) define lowering and raising energy operators respectively and the raising operators will be denoted with a dagger. These operators are also know in the literature as annihilation and creation operators, but at this point this terminology will not be used, since the definition of the ${\bf a}=0$ polarization degree (time-like) may be ambiguous.

It is quite evident that for the polarizations indexed by ${\bf a}=1,2,3$, the raising energy operators create new states. Let us then define vacuum states for these three dimensions as normalized states that are annihilated by lowering energy operators
\begin{eqnarray}\label{vac123}
\textrm{a}_{{\bf{j}}}|0\rangle
&=&
0,~~~~{{\bf{j}}}=1,2,3.
\end{eqnarray}
Now let us normalize these space-like states to 1. The state of $n_{{\bf{j}}}$ excitations must be proportional to $n$ raising energy operators acting on ground state
\begin{eqnarray}
|n_{{\bf{j}}}\rangle
~\sim~
\textrm{a}_{{\bf{j}}}^{\dag n}|0\rangle
,~~~~{{\bf{j}}}=1,2,3.
\end{eqnarray}
Then the scalar product
\begin{eqnarray}
\langle
n_{{\bf{i}}}|n_{{\bf{j}}}
\rangle
&~\sim~&
\nonumber\\
\langle 0|
\textrm{a}_{{\bf{i}}}^{n}
\textrm{a}_{{\bf{j}}}^{\dag n}|0\rangle
&=&
\delta_{{\bf{j}}}{_{{\bf{i}}}} n
\langle 0|
\textrm{a}_{{\bf{j}}}^{n-1}
\textrm{a}_{{\bf{i}}}^{\dag n-1}|0\rangle
\nonumber\\
&=&
\delta_{{\bf{j}}}{_{{\bf{i}}}} n(n-1)
\langle 0|
\textrm{a}_{{\bf{j}}}^{n-2}
\textrm{a}_{{\bf{i}}}^{\dag n-2}|0\rangle=...
\qquad
\end{eqnarray}
Going further with this recurrence, we get
\begin{eqnarray}
\langle
n_{{\bf{i}}}|n_{{\bf{j}}}
\rangle
~\sim~
\delta_{{\bf{i}}}{_{{\bf{j}}}} n!
\langle 0
|0\rangle.
\end{eqnarray}
Now we can give a normalized definition of bras and kets
\begin{eqnarray}
|n_{{\bf{j}}}\rangle
&=&
\frac{1}{\sqrt{n!}}
\textrm{a}_{{\bf{j}}}^{\dag n}|0\rangle,~~~~{\bf{j}}=1,2,3,
\nonumber\\
\langle n_{{\bf{j}}}|
&=&
\frac{1}{\sqrt{n!}}
\langle 0|
\textrm{a}_{{\bf{j}}}^n,~~~~{\bf{j}}=1,2,3.
\end{eqnarray}
This means that the action of raising and lowering operators is defined as follows
\begin{eqnarray}
\textrm{a}_{{\bf{j}}}^{\dagger}| n_{{\bf{j}}}\rangle  &=& \sqrt{n+1}| n_{{\bf{j}}}
+1\rangle,
\\
\textrm{a}_{{\bf{j}}}| n_{{\bf{j}}}\rangle &=& \sqrt{n}| n_{{\bf{j}}}-1\rangle.
\end{eqnarray}
We also define number operators for these three polarizations
\begin{eqnarray}~\label{n123}
\textrm{a}_{{\bf{j}}}^\dagger\textrm{a}_{{\bf{j}}}| n_{{\bf{j}}}\rangle
&=&
n_{{\bf{j}}}| n_{{\bf{j}}}\rangle,~~~~{\bf{j}}=1,2,3.
\end{eqnarray}
Furthermore, for the $q$ representation of vacuum (\ref{vac123}), we get a differential equation
\begin{eqnarray}\label{}
\frac{1}{\sqrt{2}}
\left(
q_{{\bf{j}}}+\frac{\partial}{\partial q_{{\bf{j}}}}
\right)
O(q_{{\bf{j}}})
=
0,~~~~{\bf{j}}=1,2,3,
\end{eqnarray}
with a simple solution
\begin{eqnarray}\label{}
O(q_{{\bf{j}}})
=
A\exp\left(
-\frac{q_{{\bf{j}}}^2}{2}
\right)
,~~~~{\bf{j}}=1,2,3.
\end{eqnarray}
Here $A$ is the normalization constant
\begin{eqnarray}\label{}
A
=
\left(
\frac{1}{\pi}
\right)^{\frac{1}{4}}.
\end{eqnarray}
The whole construction for these three polarization degrees of freedom is evident. Now without any confusion we can say that operators that raise the total energy, create particles, so can be called creation operators. Also those that lower the total energy should annihilate the vacuum. Not only the states have positive norms but also the vacuum in the canonical $q$ representation is proper, i.e. satisfies the requirement of going to zero at infinity, making it possible to normalize the function.
\subsection{Construction for time-like photons, where lowering energy operators annihilate vacuum}\label{sec:4dconstruction0e+}
The problem arises for the ${\bf a}=0$ degree. The question for the time-like polarization is: do the lowering energy operators annihilate particles and raising energy operators create ones or maybe do the lowering energy operators create particles and raising energy operators annihilate them.
First, let us assume that the lowering operator annihilates the vacuum, i.e.
\begin{eqnarray}\label{vac0}
\textrm{a}_{0}|0\rangle
&=&
0.
\end{eqnarray}
Then, the state of $n_{0}$ excitations is proportional to $n$ raising operators acting on the ground state, so that
\begin{eqnarray}
|n_{0}\rangle
~\sim~
\textrm{a}_{0}^{\dag n}|0\rangle
\end{eqnarray}
and the scalar product reads
\begin{eqnarray}
\langle
n_{0}|n_{0}
\rangle
&~\sim~&
\nonumber\\
\langle 0|
\textrm{a}_{0}^{n}
\textrm{a}_{0}^{\dag n}|0\rangle
&=&
n
\langle 0|(-)
\textrm{a}_{0}^{n-1}
\textrm{a}_{0}^{\dag n-1}|0\rangle
\nonumber\\
&=&
(-)^2 n(n-1)
\langle 0|
\textrm{a}_{0}^{n-2}
\textrm{a}_{0}^{\dag n-2}|0\rangle=...
\qquad
\end{eqnarray}
Going further with the recurrence we get
\begin{eqnarray}
\langle
n_{0}|n_{0}
\rangle
~\sim~
(-)^n n!
\langle 0
|0\rangle.
\end{eqnarray}
It looks like the states corresponding to odd values of $n$ give negative norms. Giving a normalized to 1 definition of bras and kets becomes a problem now. If we assume
\begin{eqnarray}
|n_{0}\rangle
&=&
\frac{1}{\sqrt{n!}}
(i \textrm{a}_{0}^{\dag })^n|0\rangle,
\nonumber\\
\langle n_{0}|
&=&
\frac{1}{\sqrt{n!}}
\langle 0|
\left(i \textrm{a}_{0}\right)^n,
\end{eqnarray}
then saving the positivity of the scalar product has an effect on the hermitian conjugate operation.
On the other hand we could follow Gupta \cite{Gupta50} and leave the metric indefinite.
But even then another problem arises, i.e. using the $q$ representation of vacuum, we get a differential equation
\begin{eqnarray}\label{}
\frac{1}{\sqrt{2}}
\left(
q_{0}-\frac{\partial}{\partial q_{0}}
\right)
O(q_{0})
=
0,
\end{eqnarray}
with a solution that is divergent at infinity, i.e.
\begin{eqnarray}\label{}
O(q_{0})
=
A\exp\left(
\frac{q_{0}^2}{2}
\right).
\end{eqnarray}
All these conclusions may suggest that another point of view on the time-like polarization is needed.
\subsection{Construction for time-like photons, where raising energy operators annihilate vacuum}\label{sec:4dconstruction0e-}
Now  let us assume that the raising operator annihilates the vacuum, which means that the energy spectrum is bounded from the top and to raise the total energy level we need to annihilate a particle, i.e.
\begin{eqnarray}\label{vac0}
\textrm{a}_{0}^{\dagger}|0\rangle
&=&
0.
\end{eqnarray}
Such a construction has lots of advantages which will be shown in this section.
Now the state of $n_{0}$ excitations is proportional to $n$ lowering energy operators acting on ground state, so that
\begin{eqnarray}
|n_{0}\rangle
~\sim~
\textrm{a}_{0}^{n}|0\rangle.
\end{eqnarray}
This means that creating new particles is lowering the total energy. Then the scalar product is
\begin{eqnarray}
\langle
n_{0}|n_{0}
\rangle
&~\sim~&
\nonumber\\
\langle 0|
\textrm{a}_{0}^{\dag n}\textrm{a}_{0}^{n}
|0\rangle
&=&
n
\langle 0|
\textrm{a}_{0}^{\dag n-1}
\textrm{a}_{0}^{n-1}|0\rangle
\nonumber\\
&=&
n(n-1)
\langle 0|
\textrm{a}_{0}^{\dag n-2}
\textrm{a}_{0}^{n-2}|0\rangle=...,
\end{eqnarray}
so that
\begin{eqnarray}
\langle
n_{0}|n_{0}
\rangle
~\sim~
n!
\langle 0
|0\rangle.
\end{eqnarray}
Giving a normalized definition of bras and kets is not a problem now
\begin{eqnarray}
|n_{0}\rangle
&=&
\frac{1}{\sqrt{n!}}
\textrm{a}_{0}^{n}|0\rangle,
\nonumber\\
\langle n_{0}|
&=&
\frac{1}{\sqrt{n!}}
\langle 0|
\textrm{a}_{0}^{\dag n}.
\end{eqnarray}
Furthermore, this means that for this representation the raising and lowering energy operators are defined as
\begin{eqnarray}
\textrm{a}_{0}| n_{0}\rangle  &=& \sqrt{n+1}| n_{0}+1\rangle,
\\
\textrm{a}_{0}^{\dagger}| n_{0}\rangle &=& \sqrt{n}| n_{0}-1\rangle.
\end{eqnarray}
It turns out that in this case we do not have to choose between the positivity of the scalar product and the hermitian conjugate operation. It should be stressed that the number operator for the time-like polarization should be defined carefully
\begin{eqnarray}~\label{n0}
\textrm{a}_{0}\textrm{a}_{0}^\dagger| n_{0}\rangle
&=&
n_{0}| n_{ 0}\rangle.
\end{eqnarray}
Now formulating the vacuum state is not problematic. Using the $q$ representation we get a differential equation
\begin{eqnarray}\label{}
\frac{1}{\sqrt{2}}
\left(
q_{0}+\frac{\partial}{\partial q_{0}}
\right)
O(q_{0})
=
0,
\end{eqnarray}
with a solution
\begin{eqnarray}\label{}
O(q_{0})
=
A\exp\left(
\frac{-q_{0}^2}{2}
\right).
\end{eqnarray}
Summarizing all the above, from now on we will use the $a_0^{\dag}$ for an operator that annihilates vacuum and $a_0$ for the creation operator of time-like polarization states.

Another interesting fact comes from such interpretation of the time-like photon. To see this we first split the Hamiltonian in two: for the transverse polarizations $H_{12}$ and the unmeasured ones $H_{03}$, such that
\begin{eqnarray}\label{H1203}
H|n_1,n_2,n_3,n_0\rangle
=
\left(
H_{12}+H_{03}
\right)
|n_1,n_2,n_3,n_0\rangle,
\end{eqnarray}
where
\begin{eqnarray}
\label{H12}
&&\hspace{-6pt}H_{12}|n_1,n_2,n_3,n_0\rangle
\nonumber\\
&&=
-\frac{p_1p^1+q_1q^1+p_2p^2+q_2q^2}{2}
|n_1,n_2,n_3,n_0\rangle
\nonumber\\
&&=
\left(
n_{1}
+
n_{2}
+1
\right)
|n_1,n_2,n_3,n_0\rangle,
\\
\label{H03}
&&\hspace{-6pt}H_{03}|n_1,n_2,n_3,n_0\rangle
\nonumber\\
&&=
-\frac{p_0p^0+q_0q^0+p_3p^3+q_3q^3}{2}
|n_1,n_2,n_3,n_0\rangle
\nonumber\\
&&=
\left(
-n_{0}
+
n_{3}
\right)
|n_1,n_2,n_3,n_0\rangle.
\end{eqnarray}
From this we see that for such a definition of $a_0$, the energy of the ground state comes only from transverse polarizations.
\section{Four-dimensional oscillator reducible representation  algebra}\label{sec:4dreducible}
\subsection{Motivation for reducible representation algebra proposed by Czachor}\label{sec:redmot}
In 1925 Heisenberg, Born and Jordan postulated that energies of classical free fields look in Fourier space like oscillator ensembles. It should be stressed that at that time Heisenberg, Born and Jordan did not know the notation of Fock space and may not fully understood the role of eigenvalues of operators. Having to consider oscillators with different frequencies they considered one oscillator for each frequency mode. In such case the  ensemble had to be infinite, since the number of modes was infinite.

It is a well known problem that the standard canonical procedures for field quantization result in various infinity problems.
It was shown by Czachor \cite{MC00} that the assumption of having one oscillator for each frequency mode may not be natural. This assumption continued in a series of papers on reducible representation of CCR \cite{MC03} - \cite{MWMC0904}. The main idea for reducible representation algebra is that each oscillator is a wave packet, a superposition of infinitely many different momentum states.
To describe this concept in more detail let us first introduce a spectral decomposition of the frequency operator $\Omega$
\begin{eqnarray}\label{omega}
\Omega
&=&
\int
d \Gamma ({\bf k})~
\omega({\bf k})~
|{\bf k}\rangle\langle{\bf k}|,
\end{eqnarray}
so that (\ref{omega}) fulfills the following eigenvalue problem
\begin{eqnarray}
\Omega|{\bf k}\rangle
&=&
\int
d \Gamma ({\bf k}')~
\omega({\bf k}')
|{\bf k}'\rangle\langle{\bf k}'|
{\bf k}\rangle
=
\omega({\bf k})
|{\bf k}\rangle.
\end{eqnarray}
Here $d \Gamma ({\bf k})$ is the Lorentz invariant measure
\begin{equation}\label{Lormeasure}
d\Gamma({\bf{k}})
=\frac{d^3k}{(2\pi)^{3} 2 |{\bf k}|}
.
\end{equation}
Furthermore, kets of momentum are normalized to
\begin{equation}
\langle{\bf k}|{\bf k}'\rangle
=
(2\pi)^{3}2|{\bf k}|\delta^{(3)}
({\bf k},{\bf k}')
=
\delta_{\Gamma}({\bf k},{\bf k}'),
\end{equation}
and the resolution of unity is
\begin{equation}
\int_{R^3}
d \Gamma ({\bf k})
|{\bf k}\rangle \langle {\bf k}|
=
I.
\end{equation}
The energy for photons, assuming the convention $\hbar=1$, is $E({\bf k})=\omega({\bf k})=|{\bf k}|$. Then the simplest Hamiltonian for one kind of polarization can be written in the form
\begin{eqnarray}
H
&=&
\Omega
\otimes
\left(
a^{\dag}a
+
\frac{1}{2}
\right)
\nonumber\\
&=&
\int
d \Gamma ({\bf k})~
\omega({\bf k})~
|{\bf k}\rangle\langle{\bf k}|
\otimes
\left(
a^{\dag}a
+
\frac{1}{2}
\right),
\end{eqnarray}
so that
\begin{eqnarray}
H
|{\bf k}, n
\rangle
&=&
\omega({\bf k})
\left(
n+\frac{1}{2}
\right)
|{\bf k}, n
\rangle.
\end{eqnarray}
Here ket $|n\rangle$ is the eigenvector of a ``standard theory" number operator
$a^{\dagger}a$, where $[a,a^{\dagger}]=1$.
Now let us introduce an operator that lives in both: the momentum and polarization spaces
\begin{eqnarray}\label{ared}
a({\bf k},1)
=
|{\bf k}\rangle\langle{\bf k}|
\otimes
a.
\end{eqnarray}
In bracket of $a({\bf k},1)$ in (\ref{ared}), ${\bf k}$ indicates that this representation is reducible and $1$ that it is the $N=1$ oscillator representation. Using the resolution of unity, we can also define an operator within the whole spectrum of frequencies
\begin{eqnarray}
a(1)
=
\int
d \Gamma ({\bf k})~
a({\bf k},1)
=
I
\otimes
a,
\end{eqnarray}
such that the commutator $[a(1),a^{\dagger}(1)]=I\otimes 1$. Here $1$ in the bracket of $a(1)$ denotes that this is the $N=1$ oscillator representation.
\subsection{Lie algerba}\label{sec:redladder}
The four-dimensional $N=1$ (or 1-oscillator) representation of CCR acts in the Hilbert space
${\cal H}(1)$ spanned by kets of the form
\begin{eqnarray}\label{ket41xy}
&&\hspace{-6pt}|{\bf k},n_1,n_2,n_3,n_0\rangle
\nonumber\\
&&\quad=
|{\bf k}\rangle\otimes
\frac{
	(a_1^{\dag})^{n_1}
	(a_2^{\dag})^{n_2}
	(a_3^{\dag})^{n_3}
	(a_0)^{n_0}}
{\sqrt{n_0!n_1!n_2!n_3!}}|0,0,0,0\rangle.
\quad
\end{eqnarray}
Operators $a_{1}, a_{2}, a_{3}, a_{0}$ satisfy the commutation relations typical for irreducible
representation of CCR (\ref{comaadag}). In (\ref{ket41xy}) $a_1^{\dag},~a_2^{\dag}$ stand for creation operators for linear polarized photons and $a_0$, $a_3^{\dag}$ are both creation operators for time-like and longitudinal photons respectively.
For the reducible representation we define the ladder operators
\begin{equation}
a_{{\bf a}}({\bf k},1)
=
|{\bf k}\rangle\langle{\bf k}|\otimes a_{{\bf a}},
~~~~~ {\bf a}=0,1,2,3.
\end{equation}
Then the following CCR algebra holds
\begin{eqnarray}\label{CCR41}
[a_{{\bf a}}({\bf k},1),a_{{\bf {b}}}
({\bf k}',1)^{\dagger}]
&=&
-g_{{\bf {ab}}}
\delta_{\Gamma}({\bf k},{\bf k}')
|{\bf k}\rangle\langle{\bf k}|\otimes 1_4
\nonumber\\
&=&
-
g_{{\bf {ab}}}
\delta_{\Gamma}({\bf k},{\bf k}')
I({\bf k} ,1)
.
\end{eqnarray}
This algebra representation is reducible, since the right-hand side of the commutator (\ref{CCR41}) is an operator-valued distribution $I({\bf k},1)=|{\bf k}\rangle\langle{\bf k}|\otimes 1_4$ belonging to the center of algebra, i.e.
\begin{equation}
[a_{{\bf a}}({\bf k},1),I({\bf k}',1)]
=
[a_{{\bf a}}({\bf k},1)^{\dagger},I({\bf k}',1)]
=
0,
\end{equation}
where operator $I({\bf k},1)$ forms the resolution of unity for
$\mathcal{H}(1)$ Hilbert space
\begin{equation}
\int d \Gamma({\bf k}) ~I({\bf k},1)
=I(1).
\end{equation}
The number operator for the reducible representation algebra in $\mathcal{H}(1)$ Hilbert space will be defined as
\begin{eqnarray}\label{n1}
n_{{\bf j}}({\bf k},1)
&=&
|{\bf k}\rangle\langle{\bf k}|\otimes a_{{\bf j}}^{\dagger}a_{{\bf j}},
~~~~~ {\bf j}=1,2,3,
\\
n_{0}({\bf k},1)
&=&
|{\bf k}\rangle\langle{\bf k}|\otimes a_{{0}}a_{{0}}^{\dagger}.
\end{eqnarray}
We can also define a number operator within the whole spectrum of frequencies as follows
\begin{eqnarray}\label{n(1)}
n_{{\bf a}}(1)
&=&
\int
d\Gamma({\bf k})
n_{\bf a}({\bf k},1),
~~~~~ {\bf a}=0,1,2,3.
\end{eqnarray}
Then the eigenvalue definition of the number operator, i.e. the number of photons of one kind of polarization within the whole frequency spectrum, would be
\begin{eqnarray}
n_{{\bf a}}(1)
|{\bf k},n_1,n_2,n_3,n_0\rangle
=
n_{{\bf a}}
|{\bf k},n_1,n_2,n_3,n_0\rangle.
\end{eqnarray}
Now the following Lie algebra for the reducible representation holds
\begin{equation}
[a_{{\bf a}}({\bf k},1),
a_{{\bf b}}({\bf k}',1)^{\dagger}]
=
-g_{{\bf a}{\bf b}}
\delta_{\Gamma}({\bf k},{\bf k}')
I({\bf k},1),
\end{equation}
\begin{equation}
[a_{{\bf a}}({\bf k},1),
n_{{\bf b}}({\bf k}',1)]
=
-g_{{\bf a}{\bf b}}
\delta_{\Gamma}({\bf k}, {\bf k}')
a_{{\bf b}}({\bf k},1),
\end{equation}
\begin{equation}
[a_{{\bf a}}({\bf k},1)^{\dagger},
n_{{\bf b}}({\bf k}',1)]
=
g_{{\bf a}{\bf b}}
\delta_{\Gamma}({\bf k}, {\bf k}')
a_{{\bf b}}({\bf k},1)^{\dagger} .
\end{equation}
Furthermore, for the representation within the whole frequency spectrum we have
\begin{equation}
[a_{{\bf a}}(1),
a_{{\bf b}}(1)^{\dagger}]
~=
-
g_{{\bf a}{\bf b}}
I(1),
\end{equation}
\begin{equation}
[a_{{\bf a}}(1),
n_{{\bf b}}(1)]
~=
-
g_{{\bf a}{\bf b}}
a_{\bf b}(1),
\end{equation}
\begin{equation}
[a_{{\bf a}}(1)^{\dagger},
n_{{\bf b}}(1)]
~=
g_{{\bf a}{\bf b}}
a_{{\bf b}}(1)^{\dagger} .
\end{equation}
As we can see, the Lie algebra for the whole frequency spectrum has the ``standard theory" structure.
\subsection{Hamiltonian}\label{sec:redHam}
Now, let us introduce a covariant Hamiltonian in the reducible representation algebra
\begin{eqnarray}
H(1)\label{covredH}
&=&
\Omega
\otimes
H.
\end{eqnarray}
Here $\Omega$ is the frequency operator (\ref{omega}) and $H$ is defined in (\ref{covH}).
Hamiltonian (\ref{covredH}) can be also written in terms of operators $a_{{\bf a}}^{\dagger}$ and  $a_{{\bf a}}$ in the form
\begin{eqnarray}
H(1)
&=&
\int
d \Gamma({\bf k})
~|{\bf k}|~
|{\bf k}\rangle \langle{\bf k}|
\otimes
\left(
-a_{{\bf a}}^{\dagger}
a^{{\bf a}}
+2
\right)
.
\end{eqnarray}
Now the following commutation relations hold for the reducible representation algebra
\begin{equation}
[H(1),a_{{\bf a}}({\bf k},1)]
=
-|{\bf k}|~a_{{\bf a}}({\bf k},1),
\end{equation}
\begin{equation}
[H(1),a_{{\bf a}}({\bf k},1)^{\dagger}]
=
|{\bf k}|~a_{{\bf a}}({\bf k},1)^{\dagger}.
\end{equation}
Moreover, within the whole frequency spectrum of the ladder operators, we get
\begin{equation}
[H(1),a_{{\bf a}}(1)]
=
-\int
d \Gamma({\bf k})~
|{\bf k}|~a_{{\bf a}}({\bf k},1)
=
-\Omega\otimes a_{{\bf a}},
\end{equation}
\begin{equation}
[H(1),a_{{\bf a}}(1)^{\dagger}]
=
\int
d \Gamma({\bf k})~
|{\bf k}|~a_{{\bf a}}({\bf k},1)^{\dagger}
=
\Omega\otimes a_{{\bf a}}^{\dag}.
\end{equation}
Let us also assume that the eigenvalue of the covariant Hamiltonian operator (\ref{covredH}) acting on the four-dimensional space of 1-oscillator reducible representation is the total energy denoted by $E(1)$
\begin{eqnarray}
&&\hspace{-6pt}H(1)|{\bf k}, n_1,n_2,n_3,n_0\rangle
\nonumber\\
&&=
E(1)|{\bf k}, n_1,n_2,n_3,n_0\rangle
\nonumber\\
&&=
|{\bf k}|~(n_1+n_2+n_3-n_0+1)|{\bf k}, n_1,n_2,n_3,n_0\rangle.\qquad
\end{eqnarray}
Now it can be shown that indeed $a_{{\bf a}}(1)$ lowers and $a_{{\bf a}}(1)^{\dag}$ raises the total energy by $|{\bf k}|$, i.e.
\begin{eqnarray}\label{defanihilationred}
&&\hspace{-6pt}H(1)a_{{\bf a}}(1)|{\bf k},n_1,n_2,n_3,n_0\rangle
\nonumber\\
&&=
(E(1)-|{\bf k}|)a_{{\bf a}}(1)|{\bf k},n_1,n_2,n_3,n_0\rangle,
\end{eqnarray}
\begin{eqnarray}
&&\hspace{-6pt}H(1)a_{{\bf a}}(1)^{\dag}|{\bf k},n_1,n_2,n_3,n_0\rangle\label{defcreationred}
\nonumber\\
&&=
(E(1)+|{\bf k}|)a_{{\bf a}}(1)^{\dag}|{\bf k},n_1,n_2,n_3,n_0\rangle.
\end{eqnarray}
\subsection{$N$-oscillator space}\label{sec:redladderN}
Now let us discuss an extension of this model to arbitrary $N$ oscillators. The parameter $N$ characterizes reducible representations but is not directly related to the number of photons. The Hilbert space for any $N$ oscillators reads
\begin{equation}
\mathcal{H}(N)
=
\underbrace{\mathcal{H}(1)\otimes\ldots\otimes\mathcal{H}(1)}_{N}
=
\mathcal{H}(1)^{\otimes N}.
\end{equation}
So that the $\mathcal{H}(N)$ Hilbert space is spanned by kets of the form
\begin{eqnarray}\label{ket41N}
|{\bf k}_1,n_1^{1},n_2^{1},n_3^{1},n_0^{1}
\rangle\otimes\dots\otimes |{\bf k}_N,n_1^{N},n_2^{N},n_3^{N},n_0^{N}\rangle.\qquad
\end{eqnarray}
Let us also define an operator
\begin{equation}\label{A^(n)}
A^{(n)}
=
\underbrace{I\otimes...\otimes I}_{n-1}
\otimes A\otimes\underbrace{I\otimes...\otimes I}_{N-n}.
\end{equation}
The top index $(n)$ shows the ``position" of the $A$ operator in $\mathcal H(N)$ space.
Then we construct the Hamiltonian for $N$ oscillators as follows
\begin{eqnarray}\label{covredHN}
H(N)
&=&
\sum_{n=1}^{N}
H(1)^{(n)}
.
\end{eqnarray}
A natural extension for the ladder operators to $N$-oscillator reducible representations would be
\begin{eqnarray}\label{aN}
a_{{\bf a}}({\bf k},N)
&=&
\frac{1}{\sqrt{N}}
\sum_{n=1}^{N}
a_{{\bf a}}({\bf k},1)^{(n)}.
\end{eqnarray}
Term $\frac{1}{\sqrt{N}}$ is the normalization factor for $N$-oscillator representations. The CCR algebras still hold
\begin{eqnarray}
[a_{{\bf b}}({\bf k},N),
a_{{\bf b}}({\bf k}',N)^{\dagger}]
=
-
g_{{\bf a}{\bf b}}
\delta_{\Gamma}({\bf k},{\bf k}')
I({\bf k},N),
\end{eqnarray}
where at the right-hand side there is an operator
\begin{equation}
I({\bf k},N)
=
\frac{1}{N}
\sum_{n=1}^N
I({\bf k},1)^{(n)}
,
\end{equation}
which for all $N$ is also in the center of algebra, since
\begin{equation}
[a_{{\bf a}}({\bf k},N),I({\bf k}',N)]
=
[a_{{\bf a}}({\bf k},N)^{\dagger},I({\bf k}',N)]
=0.
\end{equation}
Then $I({\bf k},N)$ forms the resolution of unity for
$\mathcal{H}(N)$ Hilbert space, i.e.
\begin{eqnarray}
\int d \Gamma({\bf k} )~I({\bf k},N)
=I(N)
=
\underbrace{I(1)\otimes...\otimes I(1)}_{N}
.\qquad
\end{eqnarray}
Again algebra representations within the whole frequency spectrum holds the ``standard theory" structure.
Using definition (\ref{n1}) of the number operator for $N=1$, we write
\begin{eqnarray}\label{nIIIN}
n_{{\bf a}}({\bf k},N)
&=&
\sum_{n=1}^N n_{{\bf a}}({\bf k}, 1)^{(n)}.
\end{eqnarray}
Furthermore, the following Lie algebras for reducible $N$-oscillator representations hold
\begin{equation}
[a_{{\bf a}}({\bf k},N),
n_{{\bf b}}({\bf k}',N)]
=
-g_{{\bf a}{\bf b}}
\delta_{\Gamma}({\bf k}, {\bf k}')
a_{{\bf b}}({\bf k},N),
\end{equation}
\begin{equation}
[a_{{\bf a}}({\bf k},N)^{\dagger},
n_{{\bf b}}({\bf k}',N)]
=
g_{{\bf a}{\bf b}}
\delta_{\Gamma}({\bf k}, {\bf k}')
a_{{\bf b}}({\bf k},N)^{\dagger} .
\end{equation}
\subsection{Vacuum and energy of vacuum}\label{sec:4dvacuum}
The subspace of vacuum states is spanned by vectors of the  form
\begin{eqnarray}
|{\bf k}_1,0,0,0,0\rangle\otimes\dots\otimes |{\bf k}_N,0,0,0,0\rangle.
\end{eqnarray}
Vacuum in this representation is any state annihilated by any annihilation operator.
\begin{equation}\label{defvac}
a_{{\bf j}}(1)|O(1)\rangle=0,~~~~{\bf{j}}=1,2,3,
\end{equation}
\begin{equation}
a_{0}(1)^{\dagger}|O(1)\rangle=0.
\end{equation}
Therefore, in $N=1$ oscillator representation, we may write
\begin{equation}\label{vacN1}
|O(1)\rangle
=
\int d\Gamma({\bf k}) O({\bf k})
|{\bf k},0,0,0,0\rangle.
\end{equation}
From the normalization condition
\begin{equation}
\langle O(1)
|O(1)\rangle
=1
\end{equation}
we get
\begin{equation}\label{sqrvac}
\int d\Gamma({\bf k}) |O({\bf k})|^{2}
=
\int d\Gamma({\bf k}) Z({\bf k})
=
1.
\end{equation}
Here the scalar field $Z({\bf k})=|O({\bf k})|^{2} $ represents vacuum probability density. Furthermore, square integrability of (\ref{sqrvac}) implies that $Z({\bf k})$ must decay at infinity. This point is of special importance for reducible representation algebra quantization.  It turns out that regularization can be a consequence of employing such special form of scalar field in the definition of vacuum.

Recalling (\ref{H03}), where it is shown that the energy of the vacuum comes only from the two transverse polarization degrees of freedom, we calculate
\begin{eqnarray}\label{vacN1E}
H(1)|O(1)\rangle
&=&
\int d \Gamma({\bf k})
~|{\bf k}|~
O({\bf k})
|{\bf k},0,0,0,0\rangle.
\end{eqnarray}
The extension to $N$-oscillator space is assumed to be a tensor product of $N=1$ vacuum states, i.e.
\begin{equation}
|O(N)\rangle
=
|O(1)\rangle^{\otimes N}
=
\underbrace{|O(1)\rangle\otimes...\otimes|O(1)\rangle}_{N}.
\end{equation}
Of course, the normalization condition for the $N$-oscillator representations still holds
\begin{equation}
\langle O(N)
|O(N)\rangle
=
\langle O(1)
|O(1)\rangle^{N}
=1.
\end{equation}
Let us also take definition (\ref{covredHN}) for the Hamiltonian in $N$-oscillator representations and write
\begin{eqnarray}\label{vacNE}
H(N)|O(N)\rangle
&=&
\sum_{n=1}^N
H(1)
^{(n)}
|O(N)\rangle
.
\end{eqnarray}
We see that the expectation value of the vacuum energy  does not depend on the $N$ parameter, i.e.
\begin{eqnarray}\label{vacNE}
&&\hspace{-6pt}\langle O(N)|
H(N)|O(N)\rangle
\nonumber\\
&&=
\langle O(1)
|O(1)\rangle^{N-1}
\langle O(1)|
H(1)|O(1)\rangle
\nonumber\\
&&=
\int d \Gamma({\bf k})
~|{\bf k}|~
Z({\bf k}).
\end{eqnarray}
This means that the energy of vacuum is not zero and depends only on the vacuum probability density $Z({\bf k})$. For the vacuum energy to be convergent we must demand
\begin{eqnarray}\label{vaccond}
\int d \Gamma({\bf k})
~|{\bf k}|~
Z({\bf k})
<\infty
.
\end{eqnarray}
Therefore, the convergence of vacuum is guaranteed by the proper choice of the vacuum probability density function and does not require the $N$ parameter at all. Furthermore, this analysis shows that the $N$ parameter may even be a finite number.
\subsection{Vector space corresponding to Maxwell's theory}\label{sec:4dHem}
Let us start by presenting the potential operator in reducible representation algebra for $N=1$ oscillator
\begin{eqnarray}\label{potN1}
A_{a}(x,1)
&=&
i\int d\Gamma({\bf k})
g_{a}{^{\bf a}}({\bf k})a_{{\bf a}}({\bf k},1)
e^{-ik\cdot x}+\rm{H.c.}\qquad
\end{eqnarray}
Here, not just two operators corresponding to the polarizations, but four types are introduced. The same four-dimensional quantization, where in the place of the annihilation operator of the Gupta-Bleuler type potential for the time-like degree stands a creation operator, was already formulated by Czachor and Naudts \cite{MCJN05} and further by Czachor and Wrzask \cite{MCKW09}. Let us point out that $a_0$ is indeed a lowering total energy operator which is a creation operator like in \cite{MCKW09}, only it is denoted here, on the contrary to mentioned papers, without a dagger. Notation in this paper comes straightforward from the analysis in Section \ref{sec:4dconstruction0e-} and is convenient for the possibility of writing collective formulas. In (\ref{potN1}) $g_{a}{^{\bf a}}(\bf k)$ is a field of Minkowski tetrads. More on the Minkowski and null tetrad in Penrose notation is presented in \ref{sec:AppA}. Furthermore, in \ref{sec:AppE} we show that the potential operator (\ref{potN1}) transforms as a four-vector.

The four-divergence of the potential operator in $N=1$ oscillator representation reads
\begin{eqnarray}
&&\hspace{-6pt}\partial^{a}A_a(x,1)\label{lorenzA}
\nonumber\\
&&=
\frac{1}{\sqrt{2}}
\int d\Gamma({\bf k})
\left(
a_0({\bf k},1)
-
a_3({\bf k},1)
\right)
e^{-ik\cdot x}+{\rm H.c.}\qquad\quad
\end{eqnarray}
As we can see this does not correspond to classical Maxwell theory, that is the Lorenz condition does not hold on the four-vector potential operator. Operator $a_0({\bf k},1)-a_3({\bf k},1)$ in D\"urr and Rudolph paper \cite{HPDER69} is called a ``bad ghost". This is a very adequate name because it spoils the classical electrodynamics correspondence.

Returning to our problem, it would be appreciable for the theory to eliminate ``bad ghosts" and this can be done by solving a weaker, i.e. in averages Lorenz condition
\begin{eqnarray}\label{weeklorenz}
\langle \Psi_{EM}(1)|
\partial^{a}A_a(x,1)
|\Psi_{EM}(1) \rangle
&=&0.
\end{eqnarray}
We will try to impose this condition on the Hilbert space instead of on the operators. Here $\Psi_{EM}(1)$ are vectors that satisfy (\ref{weeklorenz}). Later, it will be shown that such vectors give equivalence to standard free-field Maxwell electromagnetism theory. From (\ref{weeklorenz}) it follows that
\begin{eqnarray}\label{weeklorenz2}
\langle\Psi_{EM}(1)|
\int d\Gamma({\bf k})
\left(
a_0({\bf k},1)
-
a_3({\bf k},1)
\right)
|\Psi_{EM}(1) \rangle
=0.
\nonumber\\
\end{eqnarray}
Solving (\ref{weeklorenz2}) we find that $\Psi_{EM}(1)$  has a coherent-like structure.
Therefore, we define a displacement-like operator for time-like and longitudinal photons. We will start from the $N=1$ oscillator representation denoting
\begin{eqnarray}\label{DEMN1}
&&\hspace{-6pt}{\cal D}_{03}(\alpha,1)
\nonumber\\
&&=
\exp
\left(
\int d\Gamma({\bf k})
\Big(
\overline{\alpha({\bf k})}
(a_{0}({\bf k},1)
+
a_{3}({\bf k},1))
-
{\rm H.c.}
\Big)
\right).
\nonumber\\
\end{eqnarray}
Here $\alpha({\bf k})$ is a function corresponding to the ``amount of displacement" and can in general be dependent on ${\bf k}$.
Acting with operator (\ref{DEMN1}) on vacuum we get a vector state of the form
\begin{eqnarray}
&&\hspace{-6pt}
|\Psi_{EM}(1) \rangle=
{\cal D}_{03}(\alpha,1)|O(1)\rangle
\nonumber\\
&&=
\sum_{n_0,n_3=0}^{\infty}
\int d\Gamma({\bf k})
O({\bf k})
\exp
\left(-
|\alpha({\bf k})|^2
\right)
\nonumber\\
&&\hspace{8pt}\times
\frac{
	(\alpha({\bf k}))^{n_3} (\overline{\alpha({\bf k})})^{n_0}}
{\sqrt{n_3!n_0!}}
|{\bf k},n_1,n_2,n_3,n_0\rangle.
\end{eqnarray}
Operator (\ref{DEMN1}) shifts the ladder operators by $\alpha({\bf k})$
\begin{eqnarray}
&&\hspace{-6pt}
{\cal D}_{03}(\alpha,1)^{\dag}a_0({\bf k},1)
{\cal D}_{03}(\alpha,1)
=
a_0({\bf k},1)
+
\alpha({\bf k})I({\bf k},1),
\nonumber\\
\\
&&\hspace{-6pt}
{\cal D}_{03}(\alpha,1)^{\dag}a_3({\bf k},1)
{\cal D}_{03}(\alpha,1)
=
a_3({\bf k},1)
+
\alpha({\bf k})I({\bf k},1).
\nonumber\\
\end{eqnarray}
Therefore,
\begin{eqnarray}
&&\hspace{-6pt}
{\cal D}_{03}(\alpha,1)^{\dag}
\left(
a_0({\bf k},1)
-
a_3({\bf k},1)
\right)
{\cal D}_{03}(\alpha,1)
\nonumber\\
&&=
a_0({\bf k},1)
-
a_3({\bf k},1),
\end{eqnarray}
so that the week Lorenz condition (\ref{weeklorenz2}) holds. Furthermore, for $\Psi_{EM}(1)$ the number of time-like photons is equal to the number of longitudinal ones
\begin{eqnarray}\label{eqnon31}
\langle\Psi_{EM}(1)|
\left(
n_0({\bf k},1)
-
n_3({\bf k},1)
\right)
|\Psi_{EM}(1) \rangle
&=&0.\qquad
\end{eqnarray}
Therefore, their contribution cancels against each other.

The extension to arbitrary $N$-oscillator representations can be made by
\begin{eqnarray}
|\Psi_{EM}(N)\rangle
&=&
\underbrace{|\Psi_{EM}(1)\rangle\otimes...\otimes|\Psi_{EM}(1)\rangle}_{N}
=
|\Psi_{EM}(1)\rangle^{\otimes N}
,\nonumber\\
\end{eqnarray}
and the week Lorenz condition holds also for $N$-oscillator representations
\begin{eqnarray}\label{weeklorenzN}
&&\hspace{-6pt}
\langle\Psi_{EM}(N)|
\int d\Gamma({\bf k})
\left(
a_0({\bf k},N)
-
a_3({\bf k},N)
\right)
|\Psi_{EM}(N) \rangle
\nonumber\\
&&\hspace{4pt}=0.
\end{eqnarray}
Also for arbitrary $N$ oscillators, the number of longitudinal photons equals to the number of time-like ones, i.e.
\begin{eqnarray}\label{n_0n_3N}
\langle\Psi_{EM}(N)|
\left(
n_0({\bf k},N)
-
n_3({\bf k},N)
\right)
|\Psi_{EM}(N) \rangle
&=&0.\qquad
\end{eqnarray}

Let us note that this is not the usual Gupta-Bleuler condition such that
\begin{eqnarray}\label{guptacon2}
\left(
a_0
-
a_3
\right)
|\Psi \rangle
&=&0,
\end{eqnarray}
due to two aspects: a different definition of $a_0$ ladder operator and because it does not hold on the vector states but on the inner product.
\subsection{Electromagnetic field operator}\label{sec:4dF}
The electromagnetic field operator for $N=1$ oscillator representation is by definition
\begin{eqnarray}\label{e-mdef}
F_{ab}(x,1)
=
\partial_{a}A_{b}(x,1)-\partial_{b}A_{a}(x,1).
\end{eqnarray}
This can be written explicitly as
\begin{eqnarray}\label{e-m1}
&&\hspace{-6pt}
F_{ab}(x,1)
\nonumber\\
&&=
2\int d\Gamma({\bf k})
k_{[a}({\bf k})
g_{b]}{^{\bf{a}}}({\bf k})
a_{\bf{a}}({\bf k},1)
e^{-ik\cdot x}
+{\rm H.c.}
\nonumber\\
&&=\label{Fabtwofields}
\int d\Gamma({\bf k})
\left[
-k_a({\bf k})
x_{b}({\bf k})
+
k_{b}({\bf k})
x_{a}({\bf k})
\right]
a_{1}({\bf k},1)
e^{-ik\cdot x}
\nonumber\\
&&+
\int d\Gamma({\bf k})
\left[
-k_a({\bf k})
y_{b}({\bf k})
+
k_{b}({\bf k})
y_{a}({\bf k})
\right]
a_{2}({\bf k},1)
e^{-ik\cdot x}
\nonumber\\
&&+
\frac{1}{\sqrt{2}}
\int d\Gamma({\bf k})
\left[
t_a({\bf k})
z_{b}({\bf k})
-
t_{b}({\bf k})
z_{a}({\bf k})
\right]
\nonumber\\
&&\qquad\times
\left(
a_{0}({\bf k},1)
-
a_{3}({\bf k},1)
\right)
e^{-ik\cdot x}
+{\rm H.c.}
\end{eqnarray}
Here $k_a({\bf k})$ is the element of the null tetrad introduced in \ref{sec:AppA}.
The field tensor (\ref{Fabtwofields}) can be split into two parts: consisting the transverse photon operators and the ``bad ghost" operators corresponding to particles unmeasured in experiments. It can be shown that in $\Psi_{EM}(1)$ the electromagnetic field operator corresponds to standard electromagnetism theory.
Furthermore, let us also check the free Maxwell equations for the electromagnetic field operator
\begin{eqnarray}
&&\hspace{-6pt}
\partial_c F_{ab}(x,1)\label{maxwelleq}
+
\partial_a F_{bc}(x,1)
+
\partial_b F_{ca}(x,1)
=
0,
\\
&&\hspace{-6pt}
\partial_{a}\label{maxwelleq2}
F^{ab}(x,1)
\nonumber\\
&&
=
\frac{i}{\sqrt{2}}
\int d\Gamma({\bf k})
k^{b}({\bf k})
\left(
a_{0}({\bf k},1)
-
a_{3}({\bf k},1)
\right)
e^{-ik\cdot x}+{\rm H.c.}
\nonumber\\
\end{eqnarray}
The second equation (\ref{maxwelleq2}) consists of a ``bad ghost", but in $\Psi_{EM}(1)$ averages
\begin{eqnarray}
\langle \Psi_{EM}(1)|
\partial_{a}
F^{ab}(x,1)
|\Psi_{EM}(1) \rangle
&=&
0.
\end{eqnarray}
As in the potential operator, the extension of the electromagnetic field to $N$-oscillator representations is equivalent to the extension for $N$ representations of the creation and annihilation operators in (\ref{e-m1}).
~~\\
\section{Lorentz transformation}\label{sec:Lorentz}
\subsection{Bogoliubov type transformation}\label{sec:LorentzBogoliubov}
In \ref{sec:AppB} and \ref{sec:AppC} transformation properties of spin-frames and the field of Minkowski tetrads are introduced in detail. In this section we will concentrate on the corresponding transformation on the ladder operators.
At this point only the irreducible representation of CCR will be considered. Let us first define new ladder operators
\begin{equation}
b_{\bf{a}}
=
L_{\bf{a}}{^{\bf b}}(\Theta,\phi)a_{\bf{b}}.
\end{equation}
Transformation matrix $L_{\bf{a}}{^{\bf {b}}}(\Theta,\phi)$ can be written explicitly in terms of the Wigner phase $\Theta(\Lambda,{\bf k})$ and parameter  $\phi({\bf k})=|\phi({\bf k})|e^{i\xi({\bf k})}$ related to the gauge transformation. This is shown in (\ref{Labap}).
$L_{\bf{a}}{^{\bf b}}(\Theta,\phi)$ has the property of leaving the metric invariant, i.e.
\begin{equation}
g_{\bf{ab}}
=
L_{{\bf{a}}}{^{\bf{c}}}
(\Theta,\phi)
L_{\bf{b}}{^{\bf{d}}}
(\Theta,\phi)
g_{\bf{cd}},
\end{equation}
so that the new operators satisfy the same CCR
\begin{eqnarray}
[b_{\bf{a}},b_{\bf{b}}^{\dagger}]
&=&
[
L_{\bf{a}}{^{\bf {c}}}(\Theta,\phi)a_{\bf{c}}
~,
L_{\bf{b}}{^{\bf {d}}}(\Theta,\phi)a_{\bf{d}}^{\dagger}
]
\nonumber\\
&=&
-L_{\bf{a}}{^{\bf {c}}}(\Theta,\phi)
L_{\bf{b}}{^{\bf {d}}}(\Theta,\phi)
g_{\bf{cd}}
=
-g_{\bf{ab}}.\qquad
\end{eqnarray}
Therefore, there must exist an unitary Bogoliubov-type transformation $U(\Theta,\phi) $ satisfying
\begin{eqnarray}\label{bogulubovtrans}
b_{\bf{a}}
=
U(\Theta,\phi)^{\dag}
a_{\bf{a}}
U(\Theta,\phi)
=
L_{\bf{a}}{^{\bf b}}(\Theta,\phi)
a_{\bf{b}}.
\end{eqnarray}
In the next section an explicit representation of  $U(\Theta,\phi)$ will be given.
\subsection{Wigner rotations and gauge transformation in four-dimensional polarization space}\label{sec:Lorentzrepresentation}
In 1939 Wigner studied the subgroups of the Lorentz group, whose transformations leave the four-momentum of a given free particle invariant \cite{EPW39}. The maximal subgroup of the Lorentz group, which leaves the four momentum invariant is called the little group. This implies that the little group governs the internal space-time symmetries of relativistic particles. Wigner showed in his paper that invariant space-time symmetries are dictated by O(3)-like little groups in the case of massive particles and by E(2)-like little groups in the case of the massless ones. The application for photons has been discussed in many papers, among all \cite{JJ71}--\cite{HKS86}. It is also known that the Lorentz group is a very natural language for polarized light.
In order to explicitly construct transformation $U(\Theta,\phi)$ in (\ref{bogulubovtrans}), let us first introduce the representation of the Lie algebra generators of rotations $J_{\bf i}$ around an ${\bf i}$-th axis. This first will be done in canonical variables (\ref{canvar}) introduced earlier
\begin{equation}
J_{\bf i}=\varepsilon_{\bf {ijk}}q_{\bf j}p_{{\bf k}}.
\end{equation}
Using formulas (\ref{acov})-(\ref{acovdag}) we can write the generators of rotations in terms of the ladder operators
\begin{eqnarray}
&J_{\bf i}&=-i\varepsilon_{\bf {ijk}}a_{\bf j}^{\dagger}a_{{\bf k}}.
\end{eqnarray}
Also let us introduce boosts $K_{\bf i}$ along an ${\bf i}$-th axis, first in terms of canonical variables (\ref{canvar})
\begin{equation}
K_{\bf i}=p_{\bf i}q_{0}-p_{0}q_{\bf i}.
\end{equation}
Again using formulas (\ref{acov})-(\ref{acovdag}) we write the generators of boosts in terms of the ladder operators of general form
\begin{eqnarray}
&K_{\bf i}&=i (a_{\bf i}^{\dagger}a_{0}
-a_{0}^{\dagger}a_{\bf i}).
\end{eqnarray}
Interestingly, the same form of generators was introduced earlier in \cite{MCJN05}, although  without the knowledge of canonical variables (\ref{canvar}). These generators satisfy the following commutation relations
\begin{equation}
[J_{\bf i},J_{\bf j}]
=
i\varepsilon_{\bf {ijk}}J_{{\bf k}},
\end{equation}
\begin{equation}
[K_{\bf i},K_{\bf j}]
=
-i\varepsilon_{\bf {ijk}}J_{{\bf k}},
\end{equation}
\begin{equation}
[J_{\bf i},K_{\bf j}]
=
i\varepsilon_{\bf{ijk}}K_{{\bf k}}.
\end{equation}
Wigner in his paper \cite{EPW39} showed that the little group for massless particles moving along $z$ axis is generated by the rotation generators around $z$ axis $J_3$ and two other generators. These other two generators are combinations of $J_{\bf i}$ and $K_{\bf i}$ and form a representation of an Euclidean group E(2), i.e.
\begin{equation}\label{L1L2}
L_{1}=J_{1}+K_{2},
\qquad
L_{2}=J_{2}-K_{1},
\qquad
L_{3}=J_{3},
\end{equation}
with the following commutation relations
\begin{equation}
[L_{1},L_{3}]=-iL_{2}, \quad
[L_{2},L_{3}]=iL_{1}, \quad
[L_{1},L_{2}]=0.
\end{equation}
The physical variable associated with $J_3$ is the helicity degree of freedom of massless particles, but it was not clear what is the physical interpretation of generators $L_1$ and $L_2$. In 1971 Janner and Janssen \cite{JJ71} showed that those generators generate translations and are responsible for gauge transformations of the four potential. This will be discussed further in Section \ref{sec:gaugetransfor}.
Moreover, it should be stressed that only $L_3$ annihilates the vacuum states, since it is normally ordered, contrary to generators $L_1$ and $L_2$.

Now the Bogoliubov-type transformation can be written as
\begin{equation}\label{U(irr)}
U(\Theta,\phi)
=
\exp(i\alpha_{1}L_{1}+i\alpha_{2}L_{2}+i\alpha_{3}L_{3})
,
\end{equation}
with parameters
\begin{equation}
\alpha_{1}(\Theta,\phi)
=
-
\frac{\Theta(\Lambda, {\bf k})}{\sin\Theta(\Lambda, {\bf k})}
|\phi({\bf k})|\sin (\xi({\bf k})+\Theta(\Lambda, {\bf k})),
\end{equation}
\begin{equation}
\alpha_{2}(\Theta,\phi)
=
-
\frac{\Theta(\Lambda, {\bf k})}{\sin\Theta(\Lambda, {\bf k})}
|\phi({\bf k})|\cos(\xi({\bf k})+\Theta(\Lambda, {\bf k})),
\end{equation}
\begin{equation}
\alpha_{3}(\Theta)
=
-
2\Theta(\Lambda, {\bf k}).
\end{equation}
As we can see the parameter $\alpha_3$ depends only on the Wigner phase.
We can also write formula (\ref{U(irr)}) in following  forms
\begin{eqnarray}
&&\hspace{-6pt}U(\Theta,\phi)
\nonumber\\
&&=
\exp{\left(-i|\phi| \sin(\xi+2\Theta) L_1-i|\phi| \cos(\xi+2\Theta) L_2\right)}
\nonumber\\
&&\quad\times
\exp{\left(-2 i \Theta L_3\right)}\label{UGUR}
\\
&&=
\exp{\left(-2 i\Theta L_3\right)}\label{URUG}
\nonumber\\
&&\quad\times
\exp{\left(-i |\phi| \sin \xi L_1
	-i |\phi| \cos \xi L_2\right)}
\end{eqnarray}
and define transformations associated with the Lorentz transformation and gauge transformation respectively
\begin{equation}
U(\Lambda,{\bf k})
=
\exp{\left(-2 i\Theta(\Lambda, {\bf k}) L_3\right)},\label{UR}
\end{equation}
\begin{eqnarray}
U_G({\bf k})\label{UG}
=
\exp{\left(-i |\phi({\bf k})| \sin \xi({\bf k}) L_1
	-i |\phi({\bf k})| \cos \xi({\bf k}) L_2\right)},
\hspace{-1cm}
\nonumber\\
\end{eqnarray}
such that
\begin{equation}
U(\Lambda,{\bf k})^{\dag}
a_{\bf{a}}
U(\Lambda,{\bf k})
=
R_{{\bf{a}}}{^{\bf{b}}}(\Lambda,{\bf k})
a_{\bf{b}},
\end{equation}
\begin{equation}
U_G({\bf k})^{\dag}
a_{\bf{a}}
U_G({\bf k})
=
G_{{\bf{a}}}{^{\bf{b}}}({\bf k})
a_{\bf{b}},
\end{equation}
where $R_{{\bf{a}}}{^{\bf{b}}}(\Lambda,{\bf k})$ and
$G_{{\bf{a}}}{^{\bf{b}}}({\bf k})$ are transformation matrices (\ref{Rab}) and (\ref{Gab}) introduced in \ref{sec:AppC}.
\subsection{Lorentz transformation for the reducible representation algebra}\label{sec:Lorentzreducible}
To construct a Lorentz transformation for the reducible $N=1$ oscillator
representation the Bogoliubov-type transformation must be written as
\begin{equation}
U(\Lambda,0,N=1)\label{LorentzTransN1}\label{WU}
=
\int d\Gamma({\bf k})
|{\bf k}
\rangle\langle
{\bf{\Lambda^{-1}k}}|
\otimes
U(\Lambda,{\bf k}).
\end{equation}
Then, for the hermitian conjugate we can write
\begin{equation}\label{WU}
U(\Lambda,0,N=1)^{\dagger}
=
\int d\Gamma({\bf k})
|{\bf{\Lambda^{-1}k}}
\rangle\langle
{\bf k}|
\otimes U(\Lambda,{\bf k})^{\dag}.
\end{equation}
Let us denote
\begin{eqnarray}\label{Wop}
W(\Lambda)
=
\int d\Gamma({\bf k})
|{\bf k}
\rangle\langle
{\bf{\Lambda^{-1}k}}|.
\end{eqnarray}
This operator is not dependent on spin and acts only on momentum ${\bf k}$, i.e.
\begin{eqnarray}
W(\Lambda)|{\bf {k}}\rangle
=
|{\bf{\Lambda k}}\rangle.
\end{eqnarray}
Operator (\ref{Wop}) leaves the inner product invariant and therefore is an unitary operator. The hermitian conjugate of (\ref{Wop}) is
\begin{eqnarray}
&&\hspace{-6pt}W(\Lambda)^{\dag}\label{Wopdag}
=
W(\Lambda^{-1})
=
\int d\Gamma({\bf k})
|{\bf{\Lambda^{-1}k}}
\rangle\langle
{\bf k}|,
\\
&&\hspace{-6pt}	W(\Lambda)^{\dag}|{\bf {k}}\rangle
=
|{\bf{\Lambda^{-1}k}}\rangle.
\end{eqnarray}
For the $N$-oscillator extension we can write
\begin{eqnarray}
U(\Lambda,0,N)\label{LorentzTransN}
=
U(\Lambda,0,1)^{\otimes N}.
\end{eqnarray}
Then, the transformation rule for the ladder operators in $N$-oscillator reducible representations
\begin{eqnarray}\label{translora}
&&\hspace{-6pt}U(\Lambda,0,N)^{\dagger}
a_{\bf{a}}({\bf k},N)
U(\Lambda,0,N)
\nonumber\\
&&=
R_{{\bf{a}}}{^{\bf{b}}}(\Lambda,{\bf k})
a_{\bf{b}}({\bf{\Lambda^{-1}k}},N).
\end{eqnarray}
Furthermore, it is shown in \ref{sec:AppD} that
\begin{eqnarray}\label{Ulambdalambda'}
&&	U(\Lambda,0,1)U(\Lambda',0,1)
=
U(\Lambda\Lambda',0,1).
\end{eqnarray}
\subsection{Transformation properties of vacuum}\label{sec:Lorentzvacuum}
Let us remind ourselves that the definition of vacuum in this representation is not unique, i.e.
\begin{eqnarray}
|O(1)\rangle
=
\int d\Gamma({\bf k}) O({\bf k})
|{\bf k},0,0,0,0\rangle.
\nonumber
\end{eqnarray}
Then the Lorentz transformation acts on vacuum as follows
\begin{eqnarray}\label{vacuumtrans}
U(\Lambda,0,1)|O(1)\rangle
&=&
\int d\Gamma({\bf k}) O({\bf{\Lambda^{-1}k}})
|{\bf k},0,0,0,0\rangle
.\qquad
\end{eqnarray}
The transformed vacuum state is again a vacuum state, but the probability of finding ${\bf k}$ is modified by the \linebreak Doppler effect. As a byproduct we observe that the vacuum field transforms as a
scalar field
\begin{eqnarray}
O({\bf k})
~~\mapsto~~
O({\bf{\Lambda^{-1}k}})
.
\end{eqnarray}
This also implies the following transformation rule of the vacuum probability density
\begin{eqnarray}
Z({\bf k})
~~\mapsto~~
Z({\bf{\Lambda^{-1}k}})
.
\end{eqnarray}
Of course, the norm of such transformed vacuum, due to the Lorentz invariant measure (\ref{Lormeasure}), is invariant
\begin{eqnarray}
&&\hspace{-6pt}\langle O(1)| U(\Lambda,0,1)^{\dagger} U(\Lambda,0,1)
|O(1)\rangle
\nonumber\\
&&=
\int d\Gamma({\bf k}) |O({\bf{\Lambda^{-1}k}})|^{2}  =1.
\qquad
\end{eqnarray}
\subsection{Gauge transformation}\label{sec:gaugetransfor}
Quantum field theory is assumed to be gauge invariant. The change of gauge is a change in electromagnetic potential that does not produce any change in physical observables. We will show that for the covariant reducible representation algebra there exists a transformation that corresponds to a gauge transformation of the potential in $\Psi_{EM}(1)$ vector space.
Now let us start from the potential operator (\ref{potN1}) and see, how it transforms after transformation
\begin{equation}
U_G(\phi)=
\int
d \Gamma ({\bf k})
|{\bf k}\rangle \langle {\bf k}|
\otimes
U_G({\bf k})^{\dag} .
\end{equation}
Then
\begin{eqnarray}\label{potN1gauge3}
\tilde{A}_{a}(x,1)
&=&
U_G(\phi)^{\dag}
A_{a}(x,1)
U_G(\phi)
\nonumber\\
&=&
A_{a}(x,1)
+
\partial_a \varphi(x,1)
+
B_a (x,1).
\end{eqnarray}
Here
\begin{eqnarray}\label{potN1gauge4}
&&\hspace{-6pt}\varphi(x,1)
=
\sqrt{2}
\int d\Gamma({\bf k})
|\phi({\bf k})|
\left(
\cos\xi a_1({\bf k},1)\right.
\nonumber\\
&&\hspace{36pt}
\left.
+
\sin\xi a_2({\bf k},1)
\right)
e^{-ik\cdot x}
+\rm{H.c.},
\\
&&\hspace{-6pt}
B_a (x,1)\label{potN1gauge5}
=
i\int d\Gamma({\bf k})
\Big(
-x_{a}({\bf k})
|\phi|\cos \xi
+
y_{a}({\bf k})
|\phi|\sin \xi
\nonumber\\
&&\hspace{40pt}
+
\left(
z_{a}({\bf k})
-
t_{a}({\bf k})
\right)
\frac{|\phi|^2}{2}
\Big)
\nonumber\\
&&\hspace{40pt}
\times
\left(a_3({\bf k},1) -a_0({\bf k},1)
\right)
e^{-ik\cdot x} +\rm{H.c.}
\hspace*{-20pt}
\end{eqnarray}
The $B_a(x,1)$ term contains the ``bad ghost" operator, but in $\Psi_{EM}(1)$ vector space this contribution vanishes, i.e.
\begin{eqnarray}\label{}
&&\hspace{-6pt}\langle \Psi_{EM}(1)|
\tilde{A}_{a}(x,1)
|\Psi_{EM}(1) \rangle
\nonumber\\
&&=
\langle \Psi_{EM}(1)|
\left(
A_{a}(x,1)
+
\partial_a \varphi(x,1)
\right)
|\Psi_{EM}(1) \rangle.
\end{eqnarray}
Moreover, the potential operator (\ref{potN1gauge3}) under transformation (\ref{UG}) holds the weaker Lorenz condition, i.e.
\begin{eqnarray}\label{potN1Lorenzgauge}
\partial^a
\tilde{A}_{a}(x,1)
=
\partial^a
A_{a}(x,1),
\end{eqnarray}
so that
\begin{eqnarray}
\langle \Psi_{EM}(1)|
\partial^{a}\tilde{A}_{a}(x,1)
|\Psi_{EM}(1) \rangle
&=&0.
\end{eqnarray}
Earlier, in Section \ref{sec:4dHem}, it was shown that there exists a space denoted by $\Psi_{EM}(1)$ in which a weaker Lorenz condition (\ref{weeklorenz}) holds. This result should be compared with those of Janner and Janssen \cite{JJ71} followed by Han, Kim and Son \cite{HKS82}. They worked out a similar conclusion that $L_1$ and $L_2$ generators carry gauge transformations for the potential operator $(A_0, A_1, A_2, A_3)$, where $A_0=A_3$. Here the conclusion is the same for equal numbers of longitudinal and time-like photons, i.e. with $n_0=n_3$.
\subsection{Invariants in a combined homogeneous Lorentz and gauge transformation}\label{secmixing}
The combined homogeneous Lorentz and gauge transformation $U(\Theta,\phi)$ mixes transverse ladder operators $a_1,~a_2$ with excitations $a_3, ~a_0$.
Let us first take a closer look at the ``bad ghost" operator $a_{3}-a_0$, i.e.
\begin{eqnarray}
U(\Theta,\phi)^{\dag}  (a_{3}-a_0) U(\Theta,\phi)
&=&
a_{3} -a_{0}.
\end{eqnarray}
As we can see, this operator is invariant under transformation (\ref{U(irr)}).
It is also easy to show that the covariant total number of photons does not change due to the combined Lorentz and gauge transformation, i.e.
\begin{eqnarray}
&&\hspace{-6pt}U(\Theta,\phi)^{\dag}
\left(
n_1+n_2+n_3-n_0
\right)
U(\Theta,\phi)
\nonumber\\
&&=
n_1+n_2+n_3-n_0.
\end{eqnarray}
Furthermore for the Lorentz transformation we get
\begin{eqnarray}
&&\hspace{-6pt}U(\Lambda,{\bf k})^{\dag}
\left(
n_1+n_2
\right)
U(\Lambda,{\bf k})
=
n_1+n_2,
\\
&&\hspace{-6pt}U(\Lambda,{\bf k})^{\dag}
\left(
n_3-n_0
\right)
U(\Lambda,{\bf k})
=n_3-n_0.
\end{eqnarray}
This implies that the Lorentz transformation alone does not mix the two observable in experiments transverse polarizations with the two unobservable ones.
\subsection{Four-translations in four-dimensional oscillator representation}\label{sec:P}
Let us begin  with the $N=1$ oscillator representation and denote $U({\bf 1},y,1)= e^{iP(1)\cdot y}$.
The generator of four-translations, the four-momentum for the reducible representation reads
\begin{eqnarray}\label{P(1)}
P_{\bf{a}}(1)
&=&
{\textstyle\int} d\Gamma({\bf k})k_{\bf{a}}|{\bf k}\rangle
\langle {\bf k}|\otimes H,
\end{eqnarray}
where $P_0(1)$ is of course the covariant Hamiltonian (\ref{covredH}) introduced earlier in section \ref{sec:4dreducible}.
One can immediately verify that
\begin{eqnarray}
e^{iP(1)\cdot y}a_{\bf a}({\bf k},1)e^{-iP(1)\cdot y}
&=&
a_{\bf a}({\bf k},1) e^{-iy\cdot k},
\end{eqnarray}
implying the following transformation on the vector potential
\begin{eqnarray}
U({\bf 1},y,1)^{\dagger}A_{a}(x,1)U({\bf 1},y,1)
&=&
A_{a}(x-y,1).
\end{eqnarray}
Furthermore, the four-momentum for arbitrary \linebreak $N$-oscillator reads
\begin{eqnarray}\label{P(N)}
P_{\bf{a}}(N)
&=&
\sum_{n=1}^N P_{\bf{a}}(1)^{(n)}
.
\end{eqnarray}
Vectors (\ref{ket41N}) are simultaneously the eigenvectors \linebreak of $P_{\bf{a}}(N)$, i.e.
\begin{eqnarray}
&&\hspace{-6pt}P^{\bf{a}}(N) |{\bf k}_1,\dots,{\bf k}_N,
n_0^{1},\dots,n_3^{N}\rangle
\nonumber\\
&&=
\Big(k^{\bf{a}}_1\big(n_1^{1}+n_2^{1}+n_3^{1}-n_0^{1}+1\big)
+\dots
\nonumber\\
&&\quad
+~k^{\bf{a}}_N\big(n_1^{N}+n_2^{N}+n_3^{N}-n_0^{N}+1\big)\Big)
\nonumber\\
&&\quad\times~|{\bf k}_1,\dots,{\bf k}_N, n_0^{1},\dots,n_3^{N}\rangle.
\end{eqnarray}
Then the following transformation rule for the ladder operators in the $N$-oscillator reducible representation holds
\begin{eqnarray}
e^{iP(N)\cdot y}a_{\bf a}({\bf k},N)e^{-iP(N)\cdot y}
&=&
a_{\bf a}({\bf k},N) e^{-iy\cdot k},
\end{eqnarray}
implying
\begin{eqnarray}
U({\bf 1},y,N)^{\dagger}A_{a}(x,N)U({\bf 1},y,N) &=& A_{a}(x-y,N).\quad
\end{eqnarray}
Then the Poincar\'e group, i.e. the semi-direct product of homogeneous Lorentz transformation and space-time translation groups is
\begin{eqnarray}
U(\Lambda,y,1)
~=~
U({\bf 1},y,1) U(\Lambda,0,1),
\end{eqnarray}
and the composition law of two successive Poincar\'e transformations holds
\begin{eqnarray}
U(\Lambda_2,y_2,1) U(\Lambda_1,y_1,1)
~=~
U(\Lambda_2\Lambda_1,\Lambda_2y_1+y_2,1).\qquad
\end{eqnarray}
\section{Summary and conclusions}\label{sec:4dredconclusions}
In summary, we have proposed a construction for the four-dimensional polarization space coming from a definition of a covariant Hamiltonian (\ref{covH}).  Further analysis is presented for such formalism, especially regarding the interpretation of the ladder operators for the time-like polarization. Strong arguments are given in favor of an interpretation in which the operator annihilating vacuum is a raising energy operator. It turns out that assuming for the time-like photons the energy spectrum bounded from the top, we can preserve the positive norms as needed for the probability interpretation of quantum mechanics. Such an interpretation gives a non divergent vacuum representation and saves the hermiticity operation. Analysis presented in this paper is in agreement with the four-dimensional quantization of the potential operator in mentioned \cite{MCJN05}, \cite{MCKW09} papers. Further reducible representation algebras for the four-dimensional polarization oscillators are presented. Using the covariant Hamiltonian (\ref{covredHN}) for $N$-oscillator reducible representation algebras, we find that such formalism is free from vacuum energy divergences. We show that the convergence of vacuum is guaranteed by a proper choice of the vacuum probability density function $Z({\bf k})$.
Next, states $\Psi_{EM}$, which reproduce standard electromagnetic fields (i.e. photons with two transverse polarizations) from the four-dimensional covariant formalism (i.e. with two additional longitudinal and time-like photons) are shown explicitly and discussed. It is interesting that such states have a coherent-like structure.
Finally, a homogeneous Lorentz transformation for the four-dimensional oscillator is introduced.  Further, generators of these transformations coming from the canonical variables are shown.
We point out that there exists a transformation that corresponds to a gauge transformation of the potential for $\Psi_{EM}$ vectors. Next, invariants of introduced transformations are shown. Let us stress that the ``ghost operator" coming from the two extra degrees of photon polarization is an invariant.
\begin{acknowledgements}
	I am grateful to Marek Czachor and Gosia \'Smia\l{}ek-Telega for discussions and critical comments.
\end{acknowledgements}

redpajama_set_name: RedPajamaArXiv


Total length of males up to 112 mm, for females up to 120 mm. Adult males have a mental gland on the chin, which becomes more evident in mating season. Tail oval in cross section and slightly less than half the total length. Trunk on cross section square. Limbs well developed and hind legs slightly longer than front legs. Front feet with 4, hind feet with 5 flattened digits. Coloration as in other Hydromantes species highly variable, especially in the northern populations. Base color from light brown to black. Venter usually dark. Back often with spotted, blotched, striped or reticulated pattern. Pattern may be in red, yellow, gray or green and these colors can occur on the same animal, often with a metallic sheen. Different patterns can occur together in the same population. The populations in the provinces Bologna, Modena, Pistoia and Florence and southward usually have a typical "italicus"-like coloration, consisting of red or yellow/ochre spots or marbeled (Boehme et al 1999).
H. italicus is a northern and central Alpinnine endemic, ranging from the provinces of Reggio Emilia (Emillia-Romagna) and Lucca (Tuscany) southwards to the province of Pescara inclusive (Abruzzi). Specimens from Pian di Mugnone, just N of Florence, were introduced in 1983 naturalized in the cave "Buca del Nebbia". 11º14'11"E-43º20'09"N, 310m a.s.l., on the southern slope of Monte Maggio, near Monteriggioni, province of Siena (Tuscany). The altitudinal distribution ranges from 80 m a.s.l. in Garfagnana (near Anchiano, province of Lucca) up to 1594 m on Mount Corchia (Apuan Alps, province of Lucca) (Gasc 1997).
There is little data available on the reproduction of H. italicus in its natural habitat. The following account on the life history of H. italicus is derived from observations on H. ambrosii. A clutch contains 6-14 eggs of 5-6 mm in diameter. The females seem to keep in contact with their eggs. The eggs undergo direct development. After 5 months, the egg starts to swell due to increased water uptake. After 8 months, the egg has reached a diameter of 10mm. The egg then contracts until hatching after 10 months (all at 12º C). The young are 22-24 mm in length upon hatching. Development to sexual maturity takes 3 to 4 years. Direct observation in captivity has shown that this species may live up to six years. H. italicus seems to be an opportunistic hunter with a wide range of invertebrate prey (Boehme et al 1999).
The Mediterranean region is subject to increasing human habitation causing pollution, deforestation, fires, loss of surface waters and introduction of exotic species. Despite these factors, H. italicus is not endangered. This is mainly due to its relatively large distribution and its water-independent biology (Boehme et al 1999). H. italicusi is abundant in its range and should not be considered an endangered species. It is, however, less abundant or even relatively uncommon only in the province of Pescara, i.e. in the southernmost portion of its range (Gasc 1997).
Human activity has provided suitable habitats for this species in regions that are not so naturally rich in caves as the Karst area. Exploitation of "pietra serena" has yielded mounds of rocks near quarries and these and the rock walls used in traditional agriculture are now inhabited by the species.
Species Account Citation: AmphibiaWeb 2002 Hydromantes italicus: Italian Cave Salamander <http://amphibiaweb.org/species/4076> University of California, Berkeley, CA, USA. Accessed Apr 25, 2019.


A cool and fun audiotool track by one of my favorites, Grotzo!
I have learned that the company that has been supporting the Give & Get widget is discontinuing it in January. I will be taking down the page if I cannot find an alternative. This of course changes what I had been hoping to do with my music, but if nothing else, I will continue to distribute in hopes that people will enjoy the music. I hope people continue to find ways to give to charities and if possible look for ways to fight Mental Health stigma and help fund research for treatment.
Mental Illness isn't a choice or bandwagon to be jumped on.
Music made using Audiotool Uploaded via JustGo Music Grow your social audience today - http://bit.ly/186sjDQ.
Well I am giving up on making music for a while. I think the depression I fight with is winning right now. I've lost a lot of interest in things and am holding on to just get to work so my family can have money. Right now I feel it is foolish to ever have tried to make a difference or think anyone cares about my music. honestly it isn't to the standards that people want, but I tried. The fact that I tried is what I'm holding onto as a success so I don't completely beat myself up. I will keep posting things that I find interesting hoping that a few people will enjoy them with me. I will try to get out of this and maybe get back to making music. The music will probably be just forme, which makes me feel immensely lonely. Hopefully a few people will find it interesting enough to share the music with me. Thanks all for your support to date. Please give to charities that help fight mental illness. One place is my Give & Get page, but any other charity will do.


Agility Dog Training Agility Dog Training May be Enjoyable - Most of you've seen agility dog training on tv. This is when dogs are trained to follow directions like, jumping over bars, running via tunnels and weaving around poles. In the event you strategy on training your dog for agility competition, you'll require to discover a great training school or classes. At a school they ought to offer hurdles, jumps, weaving poles and much more. A course ought to consist of a tire hoops, a tunnel along with a see-saw.
A well trained dog has an benefit over an additional when beginning agility dog training. A dog that may work well off its leash is also much more desired. The handler ought to wear comfy clothing and shoes, creating it simple to run beside the dog becoming trained. The school will tell you which kind of collar and leash to bring for your dog and frequently will provide to sell you the right kind. A leather or webbed buckle collar and lengthy lead created of leather, nylon or rope will probably be essential. No chains will probably be allowed as these can turn out to be effortlessly entangled on jumps along with other equipment.
When beginning agility dog training you are able to anticipate the dog to be walked on a wide plank and taught to jump over extremely low hurdles or walls. A tunnel might be included at the beginning. As soon as the dog is comfy with these, a brand new set will probably be introduced at a taller range. Timing is essential to a dog on an agility run. Studying the commands and following a routine is essential to becoming a effective leader for your dog's agility course. Practice and praise for your dog can assist your dog succeed!
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Question about different dog leashes for agility dog training?
I am taking my year old Australian Shepherd female dog to agility training and I have a pull tab for her and i noticed someone has this leash that has a handle on it but it is very short about probably 6 inches to a foot and is black and adjustable and is just like you can drop it when doing stuff so the leash does not get tangled any idea what i am talking about anyone and where i could look for similar tings?
Sounds like a traffic lead, which is a very short leash, almost just a handle. Stick with your tab, it's much safer on the equipment for your dog. A tab should only hang to the middle of the dog's chest, no longer. Longer tabs or leashes can get caught on the equipment too easily. And either pull a jump over on a dog or hang them up on other equipment. Can seriously injure the dog or at the very least scare them as they are learning new things. Use the tab for off-leash work & then keep a leash handy to clip back on the dog in between turns or for any on-leash work.
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By Editor On Mar 27, 2019 Last updated Jul 17, 2019
BTV Highlights 5 Companies Spearheading Innovation
On this Business Television episode, find companies to invest and learn how innovative companies in the sectors of renewable energy, cannabis accessories, augmented reality, high performance Wi-Fi and bioscience are ready to takeoff.
High performance Wi-Fi
Edgewater Wireless Systems Inc. (YFI:TSX.V, KPIFF:OTCQB) has developed a new technology that promises to optimize Wi-Fi traffic. And the CEO Andrew Skafel describes the next steps to revolutionize this $33 billion dollars market.
The patented technology solves connection density issues. Now it is now an approved vendor of custom high-performance solutions for the residential market with the fifth largest cable operator in the US.
The universal problem with Wi-Fi is connection failure. Videos get pixelated, games are disconnected and connection slows down.
Density is what usually causes this problem. You can compare it to a single lane highway with cars trying to go the same direction, which makes all data traffic slow and not efficient.
Edgewater Wireless' patented solution is the only multi-lane highway for making Wi-Fi connections faster and more efficient.
Augmented Reality – a whole new solution for e-commerce
NexTech AR Solutions (OTC:NEXCF, CSE:NTAR) is turning the promise of innovation and imagination into tangible solutions for e-commerce, education, live streaming and video conferencing.
With augmented reality it is possible to see how a household appliance would fit in the space you have, or match with your decoration.
Nowadays, 90% of all shopping is still done in-store, which translates into a 30 trillion-dollar global marketplace.
Currently, e-commerce represents only about 3 trillion, but if people feel more comfortable buying from home, the conversion rate will increase from three percent to six to nine percent.
NexTech has also developed a way to integrate 3D advertising into Facebook, a platform used by over one and a half billion people.
This technology can be available to small, medium and large size e-commerce companies. The giant ones have been using it for some time now, without sharing it.
High Tide is surfing the wave of cannabis market
The cannabis industry is a growing sector and High Tide (CSE:HITI) is moving quickly into the markets in North America and the EU.
High Tide is a veteran in manufacturing, distributing, retail and wholesale of smoking accessories.
The founder of the company, Raj Grover, talks about his experience in distribution and retail prior to High Tide and how he could see the "Rising Tide of Cannabis" coming.
His instinct proved right and now High Tide has ten retail locations operating in Alberta, with another 26 locations in various stages of construction.
Also a leader in the e-Cig market, the company manufactures over 3,500 different cannabis accessories that are sold not only in Canada, but online and in over 3,000 stores worldwide.
And these numbers tend to increase for three reasons pointed by TheGreenEntrepeneur.com:
cannabis is becoming mainstream, which means more access to the end consumer
Latin America is expected to join in and dominate in short orders
Big Pharma will join the CBD party.
Renewable energy:
Renewable energy movement is in its beginning and has lots of potential to grow in the coming years.
Residential and small commercial power generation is the bet for parts of the world where electricity cost is considerably higher.
Eguana Technologies (TSX:EGT) has designed a power control system that enables different generation sources of energy to play nicely together. The founder, Brent Harris, talks about the patented inverter that takes DC power and connects it to the AC power on the grid.
The company made the necessary advances in power electronics, but received invaluable contribution from lithium-ion technology.
In partnership with LG Chem, a lithium battery supplier, Eguana Technologies has now over 6,000 systems installed around the world, 28 distributors in the US with focus on Puerto Rico and Hawaii.
In Australia, the company has 12 distributors and received a $ 200 million government subsidy to install 40,000 residential systems in South Australia, which translates into about half a billion dollar industry.
Solar System Company Hunwha Q Cells chose this technology to go along with their solar system in residential applications. By now there are over 17,000 rooftop systems installed.
If they reach the European average attachment rate of 50% using this technology, it would represent between 30 and 40 million in recognized revenue for this promising start up.
Bioscience applied to health:
Bioscience and a world-leading tobacco company have joined to develop unique oral nicotine products that promise to save 60,000 lives a year.
Lexaria Bioscience (CNSX:LXX) promises health improvements for any number of people who are ingesting drugs and molecules like pain relief, CBD and nicotine.
Everybody agrees that smoking is not healthy. Dehydra Tech is a technology that delivers CDB and nicotine through ingestion. The drug is carried across the intestine walls, getting more of it into the bloodstream more quickly. Studies in vitro in vivo and human clinical studies have proved them right.
With 10 patents approved and other 50 pending, Lexaria Bioscience sees the opportunity to change the population's smoking habits. And is partnering with companies that can help them do it in America and elsewhere in the world.
Analyst comments on what to look for when investing on small cap companies:
Brent Todd advises investors to be wary of small cap companies that change sectors overnight or commit to no complimentary businesses.
Secondly, always look at the market cap when buying penny stocks. If a company's share is low, it doesn't always mean it's a cheap stock.
For example, a company trading at a dollar a share may have 200 million shares, so it is a 200 million-dollar company.
BTV BROADCAST TIMES:
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Technology for Healthier Ingestion of Drugs
What to Watch For When Investing in Small Cap Companies
BTV Finds Opportunities in the Cannabis and Resource Sectors
BTV Visits Promising Companies in the Gold Sector
BTV Visits ESG and AI Companies
BTV Uncovers Companies in the Growing Cannabis Sector
Stock Market guide: How Lyft, Uber, Pinterest & AirBnb are…
American Manganese: Patented Battery Recycling Technology
Leagold Mining & Equinox Gold Join Forces
Stock Market Guide: When is the best time to sell a stock?
InMed Pharmaceuticals: Leading the Way in Rare Cannabinoids
Voyageur Pharmaceuticals: From the Earth to the Bottle
Cannabis: One Year Post Legalization
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Vietnamese government promises transparency in businessman's lawsuit
Thursday, August 31, 2017, 11:15 GMT+7
The government of Vietnam has promised to respond to a lawsuit filed by overseas businessman Trinh Vinh Binh with full transparency in order to maintain the complete confidence of current and potential investors in the country's various industries.
The statement was made by Vietnamese Minister and Chairman of the Government Office Mai Tien Dung on Wednesday as he took questions from reporters during a regular government press briefing.
According to Dung, the consistent stance of the Vietnamese government and its head has been to facilitate an equal business environment for foreign investors to thrive in Vietnam.
"If a certain local administration or authority violates their agreement [with foreign investors] or breaks the law, these investors reserve the right to sue the country's government," Dung said. "We can only wait, until the international court has finished investigating the case."
"The prime minister has given clear directions to approach this issue with transparency in order to maintain [investors'] confidence in our business environment," he added.
The minister and chairman promised to keep the press informed as more details become available.
Trinh Vinh Binh is a Vietnamese-Dutch businessman who was born in 1947 in the southern province of Soc Trang.
After spending time abroad, Binh returned to Vietnam in 1992 and invested heavily in real estate, purchasing over 2.5 million square meters of land and more than ten houses in the southern province of Ba Ria – Vung Tau.
Four years later, on December 5, 1996, Binh was arrested and prosecuted in Ba Ria – Vung Tau on charges of bribery and violating regulations on land management.
Binh was released on bail on June 25, 1998.
Between December 7 and 11 the same year, Binh and five others stood trial at the provincial court for their alleged crimes, with the businessman facing five years for his land management violation and eight years for bribery.
On May 4-5, 1999, the appellate court of the High People's Court of Ho Chi Minh City reduced Binh's total sentence to11 years.
However, Binh had fled abroad before his jail time was actually enforced.
The government of Vietnam later pardoned Binh and allowed him to return to his home country as a free man.
Binh recently filed a lawsuit against the Vietnamese government with the International Court of Arbitration in Paris, demanding US$1.2 billion in compensation for financial losses related to the ordeal and the return of his previously owned assets in Vietnam.
lawsuit vietnamese government trinh vinh binh
Vietnam collects $76.7mn in taxes from foreign service providers
Many foreign travelers come to Vietnam during Tet
330,000 visitors bring Hanoi $42.6mn on Tet
Vietnam's auto sales reached a record milestone last year
The tax revenue was collected on the first working day of this lunar year
Many international tourists chose to experience Vietnam's traditional Tet holiday


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Welcome to the Mike Dolbear Forums!
All users of the forums need to Register and be logged-in to comment.
General discussion forum for all things drumming. Use this group to exchange ideas and information with your fellow musicians.
Looking for a new kit or changing your existing setup? Whether it's shells, heads or hardware, you'll find your drum questions answered here.
From your ZXT's to you XS Pros and your 2002's to your Gen-X's. Anything you want to ask or share about your shiny, noisy metal discs, do it here!
All of your frequently asked (drumming) questions can be found here!
Are you a custom drum builder or do you fancy giving it a go? Discuss the issues with fellow drum builders here.
All the latest gigs and clinics coming up!
Whether you're a beginner or advanced drummer, get all your help and advice here.
Use this forum to discuss topics surrounding everything from triggers to mics and kits to modules.
Discussion group for lovers of classic kits, cymbals and everything old!
Want to share a cool video you've seen on YouTube / Vimeo etc? Then do it here!
Bargains, bull, or crazy auctions, if you've found it on eBay post it here.
Looking to sell your gear? Advertise here to other drummers and musicians!
Use this forum when you want to chat about non-drum related topics.


The Wayward Knitter: I've Got a Group!
Is that? Could it be? A sleeve! Yes indeed, my Easy A sample is moving along nicely/ Last night I finished the front, did the shoulder seams, did the applied i-cord to the neckline, added pockets, and picked up for the first sleeve. It's really starting to take shape and look like a sweater now.
I'm hoping to get the first sleeve done this week, and then hopefully have this thing finished over the weekend. That might be a bit ambitious, but I'm ready to go back to my Jean Pullover for a while. I'm also hoping that the universe is paying attention and that once I finish this sweater it'll get cold so I can wear it. I know that's probably a pipe dream, but I am ready for fall NOW, darn it.
In "news that I should have announced a few days ago but kept forgetting," I now have a Ravelry group! The link's in the right-side bar, with all my other social media links, so please come on over and join if you feel so inclined. There's a good group of people amassing so far, and I'd love to see you all over there too.
I enjoy reading your blog, and now I have joined your Ravelry group.


Maca with Goji Berries (a Chinese herb) and Korean Ginseng, as well as other beneficial herbs.
Native Peruvians have for centuries used a root that grows on the high Andes Plateau in Peru. This Maca root, they believe, helps with energy, fatigue, stamina and alertness. We have combined Maca with Goji Berries (a Chinese herb) and Korean Ginseng, as well as other beneficial herbs, which may increase the body's produc-tion of energy. Now you can enjoy chocolate and the health benefits it provides!


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There are three principle types of due diligence that can be considered in transactions, financial, commercial and legal. The success of an acquisition or investment in a business can be significantly influenced by the extent and scope of any due diligence undertaken.
There are a number of reasons why carrying out due diligence is critical, including challenging information provided by a vendor, assisting with negotiations and highlighting any risks and weaknesses.
This is a key aspect of investigating a business and although each assignment has a specific scope of work to be undertaken, there are a number of areas that typically will be part of the due diligence exercise including:
Historical trading performance and impact on cash flow and working capital requirements
Financial projections prepared by the company or the management team in the case of an MBO
Underlying financial systems and operations
Taxation matters and potential issues arising
We have provided our specialist services for a number of banks, ABLs and private equity providers in regard to acquisitions and MBOs. Regardless of the economic environment the value of due diligence is unquestionable. It is not possible to measure the benefits that due diligence can deliver, however the potential costs and issues that may arise if due diligence is not undertaken could turn out to be extremely expensive.
This looks at providing information on the operations and strategy elements to a business including customers, markets and competitors and how well the business is prepared for potential change both internally but also externally where they have little or no control.
Due Diligence undertaken should highlight the potential areas of concern, enabling the potential acquirers to be in a position that will allow them to make the necessary changes post deal to eradicate or at least dilute the impact of those areas discovered during Due Diligence.
Legal Due Diligence
As with the other forms of Due Diligence, legal Due Diligence is principally there to obtain sufficient information about a business to enable the acquirer or investor to decide on whether to proceed with the transaction. The information provided and the responses gained to questions raised as part of the exercise will also enable lawyers to tailor agreements and warranties in line with the comfort or not that is obtained during the Due Diligence process.
Having experienced corporate lawyers that are familiar with the process of Due Diligence and the information and time requirements that go hand in hand with it, is critical to ensure the transaction has the best opportunity of completing and within acceptable timeframes.
Director and Chairman
David Sharpe
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Pierce is a trading name of Pierce Group Ltd (09047081) and its associated companies. Pierce Group Ltd is the parent company of Pierce C.A. Ltd (04360541), Pierce Forensic Ltd (05969229) and Pierce Corporate Finance Ltd (05969217). All companies are registered in England & Wales and the registered office, list of all directors and VAT registration numbers are held at: Mentor House, Ainsworth Street, Blackburn, Lancashire BB1 6AY. Pierce C.A. Ltd is registered to carry on audit work in the UK and Ireland and regulated for a range of investment business activities by the Institute of Chartered Accountants in England and Wales - Chartered Accountants' Hall, PO Box 433, London EC2P 2BJ, Tel 020 7920 8100.


If you have Attention Deficit Disorder, I probably don't have to tell you about how easy it is for clutter to creep up on you. It's everywhere – the house, the car, our workspace, even our purses or briefcases.
I've often mentioned that one of the primary reasons that we ADDers are disorganized is because we tend to leave things out in the open so that we can find them when we need them. But what happens in a short while is that we've left so many things out that it all becomes a useless blur of clutter, with no clear way of seeing what's there.
We have a tendency to lose things, so we leave them out in the open.
We have a tendency to forget things, so we leave them out in the open.
We've learned that when we put things in a "safe place", we forget where that place is, so we leave them out in the open.
Another issue that contributes to clutter when you have ADHD is an inability to make a decision. Now indecisiveness by itself is not a symptom of Attention Deficit Disorder, but it can be a sort of side effect.
An inability to pay attention is the very heart of ADHD. So is lack of focus and impulsiveness. Take all of these things into consideration and you can see how it might be easy to make a decision that you later regret. Of course, everyone has made decisions that they've later regretted, but the chances of that happening with ADHD are greater.
So if you know that you sometimes don't make good decisions, what is the natural counter tactic? Spend more time thinking before acting. A good practice, most of the time. However, spending more time thinking often turns into no action at all, and that can result in clutter.
Should I accept this invitation? Should I go back to school? Should I give these things to charity? Will I need them?
While you're trying to decide, the invitation, the school catalog, and those old clothes are laying around your home, adding to the clutter.
Another cause of clutter is that we don't know what to do with it. Many people mistakenly think that neat piles of stuff about the house is cleaning and organizing. No, it's just neat clutter.
There are some great solutions that can help us conquer our clutter and lead more serene lives, but that's another post.
« Got An ADHD Kid?
Thank you for your post. I also just recently found out about this autism program, never knew that such was available. Did you ever hear about such programs before? Anyhows, keep up your blog posts, look forward to reading them.
Thank you for sharing that information with me and my readers.


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Important information for patients, families and visitors to read before coming to our sites.
First-of-its-kind study will examine beliefs about confidentiality in disclosing moral injury
Discovery Matters Here
Veterans currently living at Parkwood Institute will be participating in a first-of-its-kind study by researchers from Lawson Health Research Institute. The study will be the first to explore Canadian veterans' beliefs about confidentiality in mental health care and whether those beliefs act as a barrier to seeking treatment for a type of trauma known as moral injury.
Moral injury describes psychological distress following events where a person performs, witnesses or fails to prevent acts that conflict with deeply held moral standards. Evidence suggests that moral injuries are on the rise among deployed members of the Canadian Armed Forces (CAF), and that those exposed to such events are at a higher risk of developing post-traumatic stress disorder (PTSD) and depression.
Through the study, researchers aim to understand barriers that might prevent military personnel from disclosing details of moral injury.
Read the full story from Lawson Health Research Institute.
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