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\renewcommand{\@oddfoot}{--- ~March, 2002~ ---\hfil
\raisebox{0.3ex}{\tiny --- ~D.~A.~Arbatsky~ ``QED. VI.
Quantization.''~ ---}\hfil --- ~\thepage~ ---}
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\title{On quantization of electromagnetic field.\\
       VI. Quantization.}
\author{D.~A.~Arbatsky\footnote{ http://daarb.narod.ru/ , http://wave.front.ru/ }}
\date{March, 2002}
\maketitle

\begin{abstract}
Here we describe a general method of quantization of linear
fields. We introduce a conception of quantization invariant with
respect to action of some group. A space of quantum states of
relativistic fields is constructed in apparently
relativistic-invariant way. A connection with quantization of the
field oscillator is established. We substantiate the necessity of
using an indefinite scalar product for electromagnetic field. We
discuss additional condition for ``physically allowed'' states of
electromagnetic field. We discuss properties of the space of
states of electromagnetic field from the point of view of
functional analysis. We consider the question about origin of
anti-unitary transformations in quantum field theory.
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Other approaches to quantization.}
 Before we start description of construction of a quantized field,
let us give here a brief comparison of our construction with other
approaches to description of quantized fields.

\begin{itemize}

\item {\bf Method of formal manipulations.}
This method was used even in the earliest papers on quantum field
theory. Its essence is that we calculate Poisson brackets for
classical values and accept that for corresponding quantum values
commutators can be calculated by the formula:
\begin{equation}
 [\,\hat a\,,\,\hat b\,] = i\,\{\,a\,,\,b\,\}\widehat~  \ .
\label{CommRel}
\end{equation}
Of course, this formula can not be true for all values.
Nevertheless, if we use it cautiously enough, we can ascertain
many properties of fields, even without clarifying what they are.
In fact, this approach turned out to be the most effective from
the point of view of practical applications.

With regards to how Poisson brackets were calculated, it was
performed in such a way that it was very difficult to see
relativistic invariance even in the classical theory. Later
Peierls~\cite{Peierls1952} have suggested an alternative method of
obtaining commutation relations, which was apparently
relativistic-invariant. The corresponding classical bracket used
in the expression (\ref{CommRel}), was even named ``the Peierls
bracket''. A quite complicated ``physical'' grounding of the
relations (\ref{CommRel}), used for quantization in the Peierls
method, can be found in \cite{DeWitt1987}.

The approach described is most close to ours.

About classical description of relativistic fields we can say that
invariant Hamiltonian formalism allows to define the Poisson
bracket in apparently relativistic-invariant way. In this approach
the notion of the Poisson bracket changes in such a way that the
Peierls bracket turns out to be its very particular case. Methods
of practical calculations of Poisson brackets were described by me
in the paper~[I].

With regards to quantization of fields, in this paper in will be
performed {\it constructively}: on the base of the developed
in~[I] invariant Hamiltonian formalism, using usual algebraic
methods, we will construct in this paper quantum fields evidently.

\item {\bf Method of box.}
This method was also very popular in the field theory from its
birth. The essence of it is that we consider field not in the
infinite space, but in a large box with periodical boundary
conditions. So, the description of the field becomes similar to
the description of infinite number of oscillators.

The introduction of a box breaks the relativistic invariance in
essence, and this reason is sufficient to not consider this method
in detail.

Nevertheless, as a result of using this method a useful notion of
the field oscillator appeared. In this connection we should notice
that in our method we have an analogous notion (introduced in the
paper~[IV]). We introduce it in a somewhat more abstract way with
the use of the notion of an induced symplectic representation.
Only by this way it is possible to ascertain the correct
group-theory nature of the field oscillator and to connect
invariant quantizations of field with invariant quantizations of
oscillator.

\item {\bf Fock construction and Wigner-Mackey theory.}
In the paper~\cite{Fock1932} Fock suggested that the space of
states of a quantum field is a symmetrized tensor exponent of
one-particle subspace. Later Wigner and Mackey developed a method
of searching for all possible one-particle subspaces, based on the
theory of induced unitary representations of Poincare group (see,
for example, a review~\cite{Mackey1968}). A quite detailed account
of this approach is given in the book~\cite{Weinberg1996}.

The shortcomings of this approach are the following. First,
operators of creation and destruction of quanta are introduced
with quite complicated formulas from the very beginning. Second,
even after introducing of these operators quite complicated
reasoning is necessary for construction of local operators of {\it
field}. Third, in this approach we can not consider fields with
indefinite scalar product at all.

In our approach the initial object is not a one-particle quantum
space, but it is a classical field described by the language of
the invariant Hamiltonian formalism. So, the second shortcoming
disappears, because field operators are constructed during
quantization.

Operators of creation and destruction of quanta we can also
introduce. And formulas for them are just the consequence of our
algebraic construction of quantum field, and they are not
postulated from the beginning.

With regards to quantum fields with indefinite scalar product, in
our approach they appear as naturally as fields with
positive-definite scalar product. Our scheme is really more
general, and it can be seen even with the example of massive
scalar field. Due to this generality, the scheme naturally
includes electromagnetic field as a particular case.

Let us notice also that in our approach there is also an analog of
Wigner-Mackey theory. This is the theory of symplectic
representations of Poincare group~[IV]. The theory of induced
symplectic representations is richer than the theory of induced
unitary representations.

\item {\bf Approach connected with Stone-von Neumann theorem.}
In quantum mechanics when a quantum oscillator is studied the
Stone-von Neumann theorem turns out to be useful; it says that
canonical commutation relations for coordinates and momenta
uniquely define corresponding irreducible unitary representation.
Of course, people tried to bring this approach to quantum field
theory.

But the Stone-von Neumann theorem can not be directly changed for
the case of quantization of infinite dimensional systems (see, for
example, \cite{Schwarz1975}). Great researches were made to
overcome this difficulty. Despite of their undoubted importance, I
still think that they turned out to be too far from practical
applications.

It should be noticed here also that if we do not require
positive-definiteness of scalar product in the space of states,
the uniqueness of quantization is lost even in the case of
one-dimensional harmonic oscillator. Therefore, our approach
differs from this essentially: we refuse to require
positive-definiteness of scalar product, but instead we introduce
the conception of invariant quantization.

Let us mention here the following fact. When the Stone-von Neumann
theorem is formulated, it is necessary to write commutation
relations in the form of Weyl relations. But in our approach we
use commutation relations just for unbounded operators. And the
construction of quantization rigorously define in what sense they
must be understood.

\item {\bf Geometric quantization.}
The method of geometric quantization is that using geometry of the
classical phase space we construct the space of quantum states as
a some set of functions on classical phase space (see, for
example, \cite{Kirillov1985}).

From the point of view of the invariant Hamiltonian formalism it
is natural to apply this scheme to invariant phase space $Z$. But
let us notice that substantial difficulties are seen immediately
for this approach. First, it is not simple at all to make this
approach mathematically rigorous for systems with infinite number
of degrees of freedom. Second, the connection with Fock
construction in this case is given by quite non-evident formulas.
For these reasons this approach turns out to be not very
convenient for practical applications. Third, it is not clear how
we can describe  by this approach the case of quantization with
indefinite scalar product: even in the case of one-dimensional
harmonic oscillator great difficulties arise with definition of
indefinite scalar product.

\item {\bf Method of path integral.}
This method became very popular now because it is supposed to be
apparently relativistic-invariant. In fact, people work with path
integral as with a symbolic form of writing of some expressions.

On the other hand, it is possible to consider path integral as
rigorously defined mathematical object (see, for example,
\cite{Berez1980}). But such an approach meets so substantial
difficulties that we can not talk about any ``apparent''
relativistic invariance at all.

\end{itemize}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Quantization.} \label{quantization}
 Let us describe now, how having classical field we construct
quantum field.

\begin{enumerate}

\item
Consider the set of (complex) functions that are constant on the
whole space $Z$. Let us denote this set as $\mathcal C$. Consider
also the direct sum $\mathcal C\oplus Z^*_{\mathbb C}$. The
Poisson bracket turns this sum into a complex Lie algebra.

\item
Disengaging now from the algebraic structure of the set
 $\mathcal C\oplus Z^*_{\mathbb C}$ and considering all its elements
as completely independent let us call this set an {\it alphabet}
$\mathcal A$ and its elements {\it letters}. Let us make an
agreement that an element of $\mathcal C\oplus Z^*_{\mathbb C}$
considered as a letter of the alphabet $\mathcal A$ will be
written additionally with a ``hat'' on top or on side, for
example: $\hat a = a\widehat{~}$.

\item
Furthermore, we can introduce a formal product of elements of the
alphabet. Namely, let us suppose that if we multiply a (finite)
set of letters we get a {\it word} consisting of these letters
(the order of letters is important here):
\[
 \hat a_1 \cdot \hat a_2 \cdot \dots \cdot \hat a_k =
   \hat a_1 \hat a_2 \dots \hat a_k \ ,
 \qquad
 \hat a_1,\hat a_2, \dots,\hat a_k \in \mathcal A \ .
\]
Words we will also denote by symbols with the hat, for example:
$\hat w = w\widehat{~}$. Let us denote the set of all possible
words by the symbol $\mathcal W$; it is convenient to include in
this set the word of zero length that we can denote as
$()\widehat~$. It is natural to accept that the operation of
multiplication works on the set of words $\mathcal W$ also, and it
``joins'' words. So, $\mathcal A$ is a {\it system of free
generators}, and $\mathcal W$ is a {\it free semi-group with
neutral element} (the role of neutral element is played by the
special element $()\widehat~$).

\item
Similarly, let us define the formal sum of (finite) set of words
multiplied by arbitrary complex factors. Namely, let us call a
{\it phrase} a formal expression like:
\[
 \lambda_1 \hat w_1 + \lambda_2 \hat w_2 +
    \dots + \lambda_m \hat w_m \ ,
 \qquad
 \lambda_1,\lambda_2,\dots,\lambda_m\in\mathbb C \ , \ \
 \hat w_1,\hat w_2,\dots,\hat w_m\in\mathcal W \ .
\]
Let us also suppose two phrases to be equivalent, if for every
word they have the same sum of factors for this word. The set of
all such equivalence classes we will denote with the symbol
$\mathcal P$.

On the set of phrases $\mathcal P$ we can now naturally introduce
operations of addition and multiplication by a scalar. Namely, the
sum of phrases is the phrase that arise if we join initial phrases
by the symbol ``$+$''. Multiplication of a phrase by a number is
defined as multiplication of all factors by this number. So,
$\mathcal P$ gets natural structure of complex linear space
($\mathcal P$ is a {\it free $\mathbb C$-module generated by
$\mathcal W$}).

On the set of phrases $\mathcal P$ we can also naturally introduce
the operation of multiplication of two elements. Namely, the
product of two phrases, each of which contains one word, we define
by equality:
\[
 (\lambda_1 \hat w_1)\cdot(\lambda_2 \hat w_2) =
 (\lambda_1\lambda_2)\,(\hat w_1 \hat w_2)  \ ,
 \qquad
 \lambda_1,\lambda_2\in\mathbb C \ , \ \
 \hat w_1,\hat w_2\in\mathcal W \ .
\]
And we spread this definition to arbitrary phrases by the
requirement of distributivity:
\[
 \hat p_1\cdot(\hat p_2+\hat p_3) =
     \hat p_1 \hat p_2 + \hat p_1 \hat p_3 \ ,
 \qquad
 (\hat p_1 + \hat p_2)\cdot \hat p_3 =
     \hat p_1 \hat p_3 + \hat p_2 \hat p_3 \ ,
 \qquad
 \hat p_1,\hat p_2,\hat p_3\in\mathcal P \ .
\]

\item
So, the set of phrases $\mathcal P$ becomes an (associative)
algebra ($\mathcal P$ is a {\it semi-group algebra of the
semi-group $\mathcal W$} or a {\it free algebra over semi-group
$\mathcal W$}).

\item
Up to now when we constructed the algebra $\mathcal P$ the
alphabet $\mathcal A$ was considered as a completely arbitrary
set. Now, using structure of Lie algebra in
 $\mathcal C\oplus Z^*_{\mathbb C}$ and the correspondence
between the subspace $\mathcal C$ and the field of scalars
$\mathbb C$ we can additionally bring to the algebra $\mathcal P$
some {\it defining relations}. It is performed by factorization of
the algebra $\mathcal P$ with respect to the appropriate ideal.

Namely, consider all phrases of the following types:
\[
 (\lambda\, a)\widehat~ - \lambda\,\hat a \ ,
\]
\[
 (a+b)\widehat~ - (\hat a + \hat b) \ ,
\]
\[
 \{\,a\,,\,b\,\}\widehat~ + i\,(\hat a\,\hat b - \hat b\,\hat a) \ ,
\]
\[
 \hat1 - ()\widehat~ \ .
\]
 Here $\lambda\in\mathbb C$,
 $a,b\in\mathcal C\oplus Z^*_{\mathbb C}$,
 $\hat a,\hat b\in\mathcal A$.
 $1$~--- is a notation for the function that is equal to $1$
on the whole invariant phase space~$Z$, i.~e.
 $1\in\mathcal C\oplus Z^*_{\mathbb C}$; $\hat1$~---
the element of the alphabet that corresponds to it,
$\hat1\in\mathcal A$.

Let us span now on these phrases the two-sided ideal, i.~e. we
join to them the phrases that can be made from the given by using
finite number of operations of multiplication by a factor,
addition, and multiplication (from the left and the right) by
arbitrary phrases. Factorizing the algebra $\mathcal P$ with
respect to the obtained ideal we get an algebra that we will call
an {\it algebra of operators} and will denote as $\mathcal O$. The
elements of this algebra are called {\it operators} and are
denoted as phrases corresponding to them.

The performed factorization leads to the result that in the
algebra of operators we have the following defining relations:
\[
 (\lambda\, a)\widehat~ = \lambda\,\hat a \ ,
\]
\[
 (a+b)\widehat~ = \hat a + \hat b \ ,
\]
\[
 \{\,a\,,\,b\,\}\widehat~ = -\,i\,[\,\hat a\,,\,\hat b\,] \ ,
\]
\[
 \hat1 = ()\widehat~ \ .
\]
In the third relation we have used the standard notation for {\it
commutator} of two operators:
 $[\,\hat a\,,\,\hat b\,]=\hat a\,\hat b - \hat b\,\hat a$.

The fourth relation, in particular, allows us to escape using the
cumbersome notation $()\widehat~$, when we talk about operator and
to write always $\hat1$.

\item
Let us introduce now the operation of {\it conjugation}. This
operation will be always denoted by the symbol ${}^*$, without
distinction of an object that it is applied to.

Under conjugation of a complex number we will imply the usual
complex conjugation.

Elements of the space $\mathcal C\oplus Z^*_{\mathbb C}$ are
functions on $Z$. Under conjugation applied to such a function we
will imply a transition to the function with complex-conjugate
values. Obviously, in the result we also get an element of the
space $\mathcal C\oplus Z^*_{\mathbb C}$. Furthermore, so far as
the Poisson bracket in $\mathcal C\oplus Z^*_{\mathbb C}$ is a
complexified real Poisson bracket, we have the equality:
\begin{equation}
 \{\,a^*\,,\,b^*\,\}=\{\,a\,,\,b\,\}^* \ .
\label{poissconj}
\end{equation}

The definition of conjugation can be naturally spread to the
elements of the alphabet $\mathcal A$:
\[
 (\hat a)^*=(a^*)\widehat~ \ .
\]

To words and to phrases the operation of conjugation is spread by
the rules:
\[
 (\hat a\,\hat b)^* = \hat b^*\,\hat a^*    \ ,
 \qquad
 (\lambda\,\hat a)^* = \lambda^*\,\hat a^*  \ ,
 \qquad
 (\hat a + \hat b)^* = \hat a^* + \hat b^*  \ .
\]
Here $\hat a$ and $\hat b$ can be letters, words, and phrases;
$\lambda$ is a complex number.

Obviously, the operation of conjugation is correctly spread from
phrases to operators.

\item
Consider now the space $Z^*_{\mathbb C}$. Let us split it into a
direct sum of two subspaces:
\[
 Z^*_{\mathbb C} = Cr\oplus De \ .
\]
Let us require $Cr$ and $De$, first, to come to each other under
conjugation:
\begin{equation}
 Cr^*=De \ .
\label{cranconj}
\end{equation}
Second, let us require that the Poisson bracket on these subspaces
be zero\footnote{Taking into account~(\ref{poissconj}) and
(\ref{cranconj}), it is enough to require that it is zero on one
of these subspaces.}:
\[
 \{\,a\,,\,b\,\} = 0 \ , \qquad  a,b\in Cr \ ;
\]
\[
 \{\,a\,,\,b\,\} = 0 \ , \qquad  a,b\in De \ .
\]

The subspace $Cr$ we will call {\it creating}, and $De$~--- {\it
destructing}. In accordance with the construction given above, for
subspaces $Cr$ and $De$ we have some corresponding subspaces in
the algebra of operators; let us denote these subspaces
$Cr\widehat~$ and $De\widehat~$, correspondingly.

Let us introduce also a notation $\mathcal C\widehat~$ for the
subspace of the algebra of operators that corresponds to the
subspace $\mathcal C$ of the Lie algebra $\mathcal C\oplus
Z^*_{\mathbb C}$.

\item
Let us span now the left ideal on the subspace $De\widehat~$,
i.~e. let us join to operators there operators that are obtained
from them by finite number of operations of multiplication by
scalar, addition, and multiplication from the left with arbitrary
elements of the algebra $\mathcal O$.

Let us factorize now the algebra $\mathcal O$ with respect to the
constructed left ideal. In result we get some linear space. It is
not an algebra, but it is a left module, i.~e. there we have a
naturally defined action of operators from the left. Elements of
this factor-space are called {\it ket-vectors}. Ket-vectors are
denoted with the symbols like $|\,x\,\rangle$, where $x$ is any
symbol describing the given vector. The space of ket-vectors is
denoted $\mathcal H$ and is called {\it the space of states} of
the quantized field.

\item
We can perform the same construction with conjugate objects, i.~e.
instead of the subspace $De\widehat~$ we can consider the subspace
$Cr\widehat~$, span on it the right ideal instead of left, and,
factorizing, get the right module instead of left. The elements of
this module are called {\it bra-vectors}. And this module we will
call, if it does not make a misunderstanding, also {\it the space
of states} of the quantized field, and we will denote it also as
$\mathcal H$.

Bra-vectors are denoted by symbols like $\langle\,x\,|$. The
operation of conjugation is naturally spread to ket- and
bra-vectors and defines the one-to-one correspondence between
them. It is said that the ket- and bra-vectors, corresponding to
each other, belong to the same state of the quantized field.

\item
When we factorize the algebra of operators $\mathcal O$ and come
to the spaces of ket- and bra-vectors, the operator $\hat1$ turns
into some ket-vector $|\,0\,\rangle$ and bra-vector
$\langle\,0\,|$, correspondingly. These ket- and bra-vectors
describe the state which is called {\it vacuum}.

\item
Let us introduce now the so called {\it scalar product} of bra-
and ket- vectors. For an arbitrary bra-vector  $\langle\,x\,|$ and
a ket-vector $|\,y\,\rangle$ this product is written as
$\langle\,x\,|\cdot|\,y\,\rangle$, or, for shortness,
$\langle\,x\,|\,y\,\rangle$. As a result we get a complex number,
i.~e. $\langle\,x\,|\,y\,\rangle \in \mathbb C$.

Let us require that the operation of scalar product would satisfy
the following properties:
\[
 \langle\,0\,|\,0\,\rangle = 1 \ ,
\]
\[
 \big(\,\lambda\,\langle\,x\,|\,\big)\cdot|\,y\,\rangle =
               \lambda\cdot\langle\,x\,|\,y\,\rangle   \ ,
 \qquad
 \big(\,\langle\,x_1\,|+\langle\,x_2\,|\,\big)\cdot|\,y\,\rangle =
  \langle\,x_1\,|\,y\,\rangle+\langle\,x_2\,|\,y\,\rangle   \ ,
\]
\[
 \langle\,x\,|\cdot\big(\,\lambda\,|\,y\,\rangle\,\big) =
               \lambda\cdot\langle\,x\,|\,y\,\rangle   \ ,
 \qquad
 \langle\,x\,|\cdot\big(\,|\,y_1\,\rangle+|\,y_2\,\rangle\,\big) =
 \langle\,x\,|\,y_1\,\rangle+\langle\,x\,|\,y_2\,\rangle  \ ,
\]
\[
 \big(\,\langle\,x\,|\,\hat o\,\big)\cdot|\,y\,\rangle =
 \langle\,x\,|\,\cdot\big(\,\hat o\,|\,y\,\rangle\,\big)   \ .
\]
In the latter equality $\hat o\in\mathcal O$; this property allows
to write in such cases simply:
 $\langle\,x\,|\,\hat o\,|\,y\,\rangle$.

The given properties for the scalar product are defining, i.~e.
there is not more than one scalar product satisfying the given
properties.

\item
The scalar product is a Hermitian form in $\mathcal H$, i.~e.
always $\langle\,x\,|\,y\,\rangle=(\langle\,y\,|\,x\,\rangle)^*$.
If the scalar product turns out to be positive-definite, it
defines in $\mathcal H$ some topology. If we complete $\mathcal H$
with respect to this topology, it turns into a usual Hilbert
space. And operators appear to be defined on the corresponding
dense linear subspace in $\mathcal H$.

But the scalar product can be not positive-definite at all: it
depends on the Poisson bracket in $\mathcal C\oplus Z^*_{\mathbb
C}$ and the choice of the subspaces $Cr$ and $De$. The question
about definition of topology in this case for example of
electromagnetic field will be discussed in the
section~\ref{TFIPSEM}.

\end{enumerate}

So, in this section we have described how, if we have the Lie
algebra of observables of a classical field
 $\mathcal C\oplus Z^*_{\mathbb C}$, we construct the
algebra $\mathcal O$ of operators of the quantum field and the
space of states $\mathcal H$. This construction is called {\it
quantization} of a classical field.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Graduation of algebra $\mathcal O$ and of space
$\mathcal H$. Connection with Fock construction.}
 Consider now in the algebra of operators the subspaces
 $\mathcal C\widehat~$, $Cr\widehat~$ and $De\widehat~$.
Let us call them subspaces of the {\it grade} $0$, $+1$ and $-1$,
correspondingly. Making all possible products of these operators,
let us assign to each such product a grade that is equal to the
sum of grades of factors. If two products have the same grade, let
us assign the same grade to their sum. It is easy to see that the
algebra of operators, considered as a linear space, can be
represented as a sum of its linear subspaces belonging to
different grades:
\[
 \mathcal O = \dots \oplus
   \mathcal O_{-2} \oplus
   \mathcal O_{-1} \oplus
   \mathcal O_0    \oplus
   \mathcal O_{+1} \oplus
   \mathcal O_{+2} \oplus \dots
\]
And if $\hat a\in\mathcal O_i$ and $\hat b\in\mathcal O_j$, then
$\hat a\,\hat b\in\mathcal O_{i+j}$. Shortly speaking, the algebra
$\mathcal O$ is graded.

This graduation is naturally transferred to the space of states
$\mathcal H$. And negative grades in the space $\mathcal H$
correspond to trivial subspaces. In other words, the space of
states $\mathcal H$ is positive-graded:
\begin{equation}
 \mathcal H =
   \mathcal H_0 \oplus
   \mathcal H_1 \oplus
   \mathcal H_2 \oplus \dots
\label{gradH}
\end{equation}

If a ket-vector belongs to one of the subspaces $\mathcal H_n$, we
say that there are $n$ particles in this state.

It is obvious also, that all this construction is possible also
for conjugate objects, and decomposition~(\ref{gradH}) looks
exactly the same for bra-vectors.

It is not out of place to mention here that, according to the
introduced definition, the number of particles is always
non-negative, and it does not depend on whether or not the scalar
product in $\mathcal H$ is positive-definite.

It easy to see also that subspaces $\mathcal H_i$ and $\mathcal
H_j$ are orthogonal with respect to the scalar product. In other
words, if $\langle\,a\,|\in\mathcal H_i$ and
$|\,b\,\rangle\in\mathcal H_j$ and $i\neq j$, then
$\langle\,a\,|\,b\,\rangle=0$. Taking it into account, we can
write the decomposition~(\ref{gradH}) in the form:
\begin{equation}
 \mathcal H =
   \mathcal H_0 \dot+
   \mathcal H_1 \dot+
   \mathcal H_2 \dot+ \dots
\label{OrthSum}
\end{equation}
In the case when the scalar product is positive-definite, the
decomposition~(\ref{OrthSum}) allows to establish connection with
the usual Fock construction~\cite{Fock1932}. So far as this is
quite simple, we will not discuss this connection in more details.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Invariant quantization.} \label{invquant}
 In the section~\ref{quantization} quantization of a classical
field was described in ``internal'' notions of the invariant
Hamiltonian formalism. But for the choice of the subspaces $Cr$
and $De$ we have not suggested any concrete prescription: we have
just given some necessary conditions for the choice of these
subspaces. In practice it leads to the situation that for one
classical field we can get too many quantizations.

Let us suppose now that in the space $Z^*_{\mathbb C}$ there is a
linear symplectic representation of some group. Let us define then
{\it invariant quantization} by the requirement that subspaces
$Cr$ and $De$ are invariant with respect to the action of this
group. In other words, $Cr$ and $De$ are reducing subspaces of
this representation.

When we consider relativistic fields, as a main group of
invariance we have Poincare group~$\mathcal P$. A quantization,
invariant with respect to the Poincare group we will call
$\mathcal P$-invariant, for shortness. In the paper~[IV] we have
already seen, how the Poincare group acts in the space
$Z^*_{\mathbb C}$, and how its representations are reduced.

Quantization of relativistic fields we will consider a little
later, but now we will consider quantization of the harmonic
oscillator. Its symmetry group is just the one-parameter group of
time shifts.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Quantization of harmonic oscillator.} \label{harmosc}
 Consider harmonic oscillator. It is described by the Lagrangian:
\begin{equation}
 \textstyle
 L = \frac12 \dot\varphi^2 - \frac{m^2}2\varphi^2    \ .
\label{osclag}
\end{equation}
Here $\varphi(x)$ is a real function of one real argument (time).

The equation of the motion is:
\[
 (\partial^2 + m^2)\,\varphi = 0   \ .
\]

The invariant phase space $Z$ is a two-dimensional real space. The
symplectic structure on it is given  by the formula:
\[
 \omega = \dot\varphi(t) \wedge \varphi(t)    \ .
\]
Here forms $\dot\varphi$ and $\varphi$ are taken in the same
arbitrary moment of time $t$.

So far as the Lagrangian~(\ref{osclag}) is invariant with respect
to time shifts, in the space $Z$ we have the action of the
additive group $\mathbb R$. The field representation
 $Z^*_{\mathbb C}$ in this case, like in the case of
relativistic fields, decomposes into the direct sum of positive-
and negative-frequency subrepresentations:
\[
 Z^*_{\mathbb C} = Z^{*\,(+)}_{\mathbb C} \oplus
                   Z^{*\,(-)}_{\mathbb C} \ .
\]
As a reducing basis let us take the following two elements:
\[
 \textstyle
 a = \frac1{\sqrt{2m}}\,(m\varphi(0) + i\dot\varphi(0)) \ ,
 \qquad
 a^* = \frac1{\sqrt{2m}}\,(m\varphi(0) - i\dot\varphi(0)) \ .
\]
Really, so far as we have decomposition
 $\varphi(t) =
  \frac1{\sqrt{2m}}(a\,e^{-imt} + a^*e^{+imt})$,
we get that $a\in Z^{*\,(+)}_{\mathbb C}$,
 $a^*\in Z^{*\,(-)}_{\mathbb C}$.

Symplectic structure can be written by the forms $a$ and $a^*$ in
the following way:
\[
 \omega = i\,a^*\wedge a  \ .
\]
The Poisson bracket of the corresponding linear functions is:
\[
 \{\,a\,,\,a^*\,\} = -i \ .
\]

So, in accordance with what we have said in the
section~\ref{invquant}, we have exactly two invariant
quantizations: either $Cr=Z^{*\,(-)}_{\mathbb C}$ and
$De=Z^{*\,(+)}_{\mathbb C}$, or $Cr=Z^{*\,(+)}_{\mathbb C}$ and
$De=Z^{*\,(-)}_{\mathbb C}$. It is easy to see that the first
quantization leads to positive-definite scalar product. In the
case of second quantization scalar product turns out to be
indefinite. And in the decomposition~(\ref{OrthSum}) on the
subspaces with even number of particles the scalar product is
positive-definite, and on subspaces with odd~--- negative.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Connection of invariant quantizations of field and of
field oscillator.} \label{QuantFOsc}
 A field oscillator, as a finite-dimensional system, is more
convenient for investigation by algebraic means. On the other
hand, there is a close connection between quantization of a field
and quantization of its oscillator.

Really, it is easy to see that the choice of
 $\mathbb R\times\mathcal L_{k^{(0)}}$-invariant creation and
destruction subspaces for oscillator and the choice of
 $\mathcal P$-invariant creation and destruction subspaces
for field are connected by the operation of induction. So, there
is one-to-one correspondence between
 $\mathbb R\times\mathcal L_{k^{(0)}}$-invariant quantizations
of oscillator and $\mathcal P$-invariant quantizations of field.

Furthermore, corresponding quantizations of oscillator and field
have in the spaces of states scalar products either
positive-definite or indefinite simultaneously. So, we get the
convenient criterion of sign-definiteness of scalar product.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Quantization of scalar field.}
 In the paper [IV] we have shown that with respect to the action
of Poincare group the space of linear observables of scalar field
$Z^*_{\mathbb C}$ decomposes into the direct sum of the two
irreducible subspaces:
 $Z^*_{\mathbb C}=Z^{*\,(+)}_{\mathbb C}\oplus Z^{*\,(-)}_{\mathbb C}$.
So, there are exactly two possibilities: either we accept
$Cr=Z^{*\,(-)}_{\mathbb C}$ and $De=Z^{*\,(+)}_{\mathbb C}$, or we
accept $Cr=Z^{*\,(+)}_{\mathbb C}$ and
 $De=Z^{*\,(-)}_{\mathbb C}$. According to the formulas for
Poisson brackets of corresponding functions, obtained in the paper
[IV], both decompositions satisfy all requirements of the
section~\ref{quantization}.

So, the scalar field allows exactly two $\mathcal P$-invariant
quantizations. The field oscillator in this case is just the usual
one-dimensional real oscillator. Using results of the
section~\ref{harmosc} and the criterion from the
section~\ref{QuantFOsc} we get that for the first quantization the
scalar product in the space $\mathcal H$ turns out to be
positive-definite, and for the second quantization~- indefinite.

The first quantization is, in fact, conventional. Nevertheless, it
should not be thought that quantization with indefinite metric is
senseless, because it leads to ``negative probabilities''. In
principle, we can not reject a usefulness of such a theory where
scattering states of the field constitute a subspace where the
scalar product is positive-definite.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Quantization of electromagnetic field.}
 Consider now quantization of electromagnetic field. Under
electromagnetic field we will imply here ``non-physical''
electromagnetic field [I].

As we have explained in the paper [IV], with respect to the action
of the Poincare group $\mathcal P$ the space $Z^*_{\mathbb C}$ in
this case, like in the case of the scalar field, decomposes into
the direct sum of the two indecomposable subspaces. Therefore,
like in the case of the scalar field, we have exactly two
possibilities: either $Cr=Z^{*\,(-)}_{\mathbb C}$ and
 $De=Z^{*\,(+)}_{\mathbb C}$, or $Cr=Z^{*\,(+)}_{\mathbb C}$ and
$De=Z^{*\,(-)}_{\mathbb C}$.

Using the criterion from the section~\ref{QuantFOsc}, we easily
see that both quantizations lead to indefinite metric. Later on we
will consider only the first of the two quantizations, because
this quantization has useful physical interpretation.

The indefiniteness of the scalar product leads, of course, to
difficulties with probability interpretation.

In the paper [I] we have pointed out that scattering states of the
classical electromagnetic field satisfy additional condition~---
the Lorentz condition. It is natural to suppose that in the
quantum theory there is some analog of this condition. It is
important to emphasize, that this condition must appear in the
quantum theory not as a new postulate: it must be derivable from
the dynamical laws, like we have done it for the classical field.
Unfortunately, we do not have a satisfactory theory of interacting
fields yet. Because of this reason, we will not give here any
derivation\footnote{A formal derivation, of course, was present
even in~\cite{Bleuler1950}. On the other hand, it is possible to
appeal to Feynman rules, but this rules need grounding from the
point of view of this paper.}. The desired condition is:
\begin{equation}
 -\,i\,k_\mu\,\hat a_\mu^{(+)}(k)\,|\,\mathrm{rad}\,\rangle = 0 \ .
\label{Guptacond}
\end{equation}
So, a vector of state of radiated field satisfies this equality
for any~$k$.

The given condition looks the same like one of versions of this
condition in the classical theory. Nevertheless, it should be
noticed that it can not be written as
 $-\,i\,k_\mu\,\hat a_\mu^{(-)}(k)\,|\,\mathrm{rad}\,\rangle = 0$
or as $-\,i\,k_\mu\,\hat a_\mu(k)\,|\,\mathrm{rad}\,\rangle = 0$,
because even the vacuum does not satisfy these
conditions\footnote{In coordinate representation the condition
 $-\,i\,k_\mu\,\hat a_\mu(k)\,|\,\mathrm{rad}\,\rangle = 0$
looks as
 $\partial_\mu\hat A_\mu(x)\,|\,\mathrm{rad}\,\rangle = 0$.
In literature even up to now there is no unity of opinions on
whether or not it is possible to formulate the additional
condition in this way. In this way, on the grounding of analogy
with the classical field, it was formulated even by Fermi. It was
pointed out in the papers \cite{Ma1949,Belinf1949} that such an
additional condition leads to difficulties with normalizability of
states in the quantum case. In the paper \cite{Gupta1950} it was
substituted with
 $\partial_\mu\,\hat A_\mu^{(+)}(x)\,|\,\mathrm{rad}\,\rangle = 0$.
Later in literature some authors used condition
 $\partial_\mu\,\hat A_\mu(x)\,|\,\mathrm{rad}\,\rangle = 0$
\cite{Thirr1964,Umezawa1958,Dirac1979}. Other \cite{BogShir1984}
pointed that such a condition can not be applied, because there
appears some contradiction with commutation relations. The other
authors \cite{Prokh1988} claimed that such a condition can be
used, but special care is needed, because allowed vectors of
states turn out to be not-normalizable. These differences of
opinion have quite deep reasons:

\begin{enumerate}
\item No exact meaning was put into the term ``quantization''.
In fact, quantized electromagnetic field was studied by formal
manipulations with algebraic symbols, but it was not given any
constructive definition of this object. In such a situation the
additional condition in any form is not free from criticism.
\item There is no satisfactory formulation of the theory
of interacting fields (even in the frame of perturbation theory).
And the Feynman rules do not have any clear connection with the
operatoral formalism, and operatoral formalism turns out to be
separated from practical calculations (from scattering theory,
first of all).
\end{enumerate}

With regards to the first remark, I believe, the relation
 $\partial_\mu\,\hat A_\mu(x)\,|\,\mathrm{rad}\,\rangle = 0$
does not contradict directly to the commutation relations. It is
possible to say only that the construction of quantization
presented in this paper is badly co-ordinated with such a
condition.

The second remark is, of course, more important. When we
considered the dynamics of the classical field, it was pointed out
that the additional condition for scattering states automatically
follows from dynamics, and it is not an arbitrary postulate. We
have also clearly shown the role of the ``non-physical'' degrees
of freedom. And though in this paper we still do not formulate a
theory of quantum interacting fields, nevertheless, arguments
based on analogy with the classical theory give us reasons to
think that construction of theory with condition
 $\partial_\mu\,A_\mu(x)\,|\,\mathrm{rad}\,\rangle = 0$ is not a
realistic task.}.

Now we will show that the condition~(\ref{Guptacond}) leads to
positivity of scalar product.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Positivity of scalar product of electromagnetic
field.}\footnote{The content of this section does not have any
deep connection neither with constructivity of quantization, nor
with algebraic, nor with topological questions discussed in the
present papers. The only reason why we discuss here this old (and
very simple) question is that in all sources that I know an
erroneous proof is given (for some reason people think that
positivity of a quadratic form on vectors of some basis
necessarily leads to positivity of the form in general).}
 Consider the field oscillator of electromagnetic field. Let us
write here, for shortness, $a_\mu$ instead of $\hat a_\mu(+1)$ and
$a^*_\mu$ instead of $\hat a_\mu(-1)$. These operators satisfy
relations:
\[
 [\,a^*_\mu\,,\,a_\nu\,]\,= g_{\mu\nu}, \qquad
 [\,a^*_\mu\,,\,a^*_\nu\,]=0, \qquad
 [\,a_\mu\,,\,a_\nu\,]=0 \ .
\]
For the quantization under consideration we also have:
\[
 a_\mu\,|\,0\,\rangle=0 \ .
\]
The additional condition takes the form:
\begin{equation}
 k^{(0)}_\mu a_\mu\,|\,\mathrm{rad}\,\rangle = 0 \ ,
\label{OscAddCond}
\end{equation}
here $k^{(0)}_\mu$ is a fixed vector on the light cone. The
subspace of the vectors of states satisfying this condition we
will denote $\mathcal H^\mathrm{rad}$.

So far as the operator $k^{(0)}_\mu a_\mu$ has a definite grade
(its grade is equal to $-1$), the graduation of the space of
states $\mathcal H$ is transferred to the space
 $\mathcal H^\mathrm{rad}$:
\begin{equation}
 \mathcal H^\mathrm{rad}=
   \mathcal H^\mathrm{rad}_0 \dot+
   \mathcal H^\mathrm{rad}_1 \dot+
   \mathcal H^\mathrm{rad}_2 \dot+ \dots
\label{HradSum}
\end{equation}
In other words, the subspace, defined by the additional condition
(\ref{OscAddCond}), also decomposes into the orthogonal sum of
states with definite number of particles.

So far as the sum (\ref{HradSum}) is orthogonal, it is enough to
prove the non-negativity of the scalar product for each of the
subspaces $\mathcal H^\mathrm{rad}_n$.

An arbitrary state vector $|\,n\,\rangle$ from the subspace
 $\mathcal H^\mathrm{rad}_n$ can be represented in the form:
\[
 |\,n\,\rangle = T_{\mu\nu\dots\rho} \,
                 a^*_\mu\,a^*_\nu\dots a^*_\rho \,
                 |\,0\,\rangle    \ ,
\]
here $T_{\mu\nu\dots\rho}$ is an $n$-valent tensor. This tensor
can be supposed to be symmetrical, without loss of generality.

The condition (\ref{OscAddCond}) for the vector $|\,n\,\rangle$
turns out to be an additional condition for the tensor
$T_{\mu\nu\dots\rho}$:
\begin{equation}
 k^{(0)}_\mu T_{\mu\nu\dots\rho} = 0 \ .
\label{TOscAddCond}
\end{equation}

The scalar product for the vector $|\,n\,\rangle$ multiplied by
itself is:
\begin{equation}
 \langle\,n\,|\,n\,\rangle =
      (-1)^n\,n!\,\cdot\,
      T^*_{\mu\nu\dots\rho}\,T_{\mu\nu\dots\rho} \ .
\label{Tsum}
\end{equation}

The value
$(-1)^n\,n!\,\cdot\,T^*_{\mu\nu\dots\rho}\,T_{\mu\nu\dots\rho}$ is
a sum of positive numbers, some of them are included in the sum
with the sign ``plus'', and some of them with the sign ``minus''.
This sign is defined by the relativistic summation rule by
repeated indexes and by the factor $(-1)^n$. Let us show that this
sum is non-negative under condition (\ref{TOscAddCond}).

So far as the construction of the space of states is invariant
with respect to the choice of the vector $k^{(0)}_\mu$, this
vector can be supposed to be equal:
\begin{equation}
 k^{(0)}_\mu =
 \left(\begin{array}{c|ccc}
       k_0 & 0 & 0 & k_0
       \end{array}\right)_\mu    \ .
\label{kChoice}
\end{equation}

Consider now  in the sum (\ref{Tsum}) all addends of the type
$(-1)^n\,n!\,\cdot\,T^*_{0\nu\dots\rho}\,T_{0\nu\dots\rho}$.
Because of (\ref{TOscAddCond}) and (\ref{kChoice}), they are
completely neutralized by the addends of the type
$(-1)^n\,n!\,\cdot\,T^*_{3\nu\dots\rho}\,T_{3\nu\dots\rho}$.

Among remaining addends let us consider addends of the type
$(-1)^n\,n!\,\cdot\,T^*_{\mu0\dots\rho}\,T_{\mu0\dots\rho}$. They
are completely neutralized with the remaining addends of the type
$(-1)^n\,n!\,\cdot\,T^*_{\mu3\dots\rho}\,T_{\mu3\dots\rho}$.

Et cetera. Finally, in the sum (\ref{Tsum}) there remain only
addends that have indexes $1$ or $2$. All these addends are
included in the sum (\ref{Tsum}) with the sign ``plus''.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Topology and completeness of invariant phase space of
electromagnetic field.} \label{TFIPSEM}
 When we studied the invariant Hamiltonian formalism, at the beginning
we included into the invariant phase space $Z$ only solutions of
the equations of the motion that are smooth in coordinate
representation and finite in space direction. But the question
about what topology should be really introduced in the space $Z$
we have not discussed at all. In fact, we have not given even an
exact definition of the space $Z^*$; we have not clarified, how
exactly the functional spaces $Z$ and $Z^*$ turn out to be
isomorphic; an exact definition of the Poisson brackets on $Z^*$
was not given also. In this section we will show, that in the
space $Z$ we can introduce such a topology (and to complete it
with respect to this topology), that it becomes very similar to
the Hilbert space. And all necessary for invariant Hamiltonian
formalism properties of this topology will be satisfied.

A symplectic structure itself does not define a topology. But
between elements of the classical phase space $Z$ and one-particle
states of a quantized field there exist one-to-one correspondence:
\begin{equation}
 \underline c \ \longleftrightarrow\
 (I^{-1}\underline c)\widehat~ \ |\,0\,\rangle \ .
\label{ZHisom}
\end{equation}

Consider now the scalar field as an example. Among its invariant
quantizations there is a quantization with positive-definite
scalar product. The correspondence (\ref{ZHisom}), firstly,
introduces a complex structure in the space $Z$, and secondly,
transfers there the scalar product from the one-particle quantum
space $\mathcal H_1$. So, $Z$ has natural structure of complex
Hilbert space\footnote{The obtained topology is the most natural
for quantization. Bohr and Rosenfeld \cite{BohrRos1933} noticed
that for consideration of the quantized electromagnetic field it
is necessary to perform an averaging of the field in a small
space-time volume: the symbol $\hat A_\mu(x)$ itself is not a
physical value. Our consideration shows that this statement is in
the same way true for the classical field also, if its phase space
is provided with the ``proper'' topology.}.

The symplectic structure turns out to be continuous with respect
to the pair of its arguments with respect to the obtained
topology; the symplectic structure defines the isomorphism of the
spaces $Z$ and $Z^*$; the Poisson brackets are correctly defined
on the whole $Z^*$.

Let us represent now the scalar field in the Fourier
representation in the form:
\[
 \widetilde\varphi(k) = 2\pi\,\delta(k^2-m^2)\cdot a(k)  \ .
\]
The correspondence (\ref{ZHisom}) can be written more manifestly
as:
\[
 \underline c \ \longleftrightarrow\
 \int d\mu^+_m \cdot i\,a(k)^{\,\underline c} \cdot
               \hat a^*(k) \ |\,0\,\rangle \ ,
 \qquad \mbox{here} \ \
 d\mu^+_m = \frac{d^4\!k}{(2\pi)^4} \cdot
            2\pi\,\delta(k^2-m^2) \cdot
            \theta(k) \ .
\]
And the scalar product in the space $Z$ takes the form:
\begin{equation}
 \langle\,\underline c\,,\,\underline d\,\rangle =
 \int d\mu^+_m \cdot
 a^*(k)^{\,\underline c} \cdot
 a(k)^{\,\underline d} \ .
\label{ScalScalProd}
\end{equation}

The one-to-one transference of this scheme to the case of
electromagnetic field appears to be impossible, because, as we
have seen, among $\mathcal P$-invariant quantizations of
electromagnetic field there are  no quantizations with
positive-definite scalar product.

But here is another opportunity. Let us represent electromagnetic
field in the Fourier representation as
\[
\widetilde A_\mu(k) = 2\pi\,\delta(k^2) \cdot a_\mu(k)  \ .
\]
By analogy with the formula (\ref{ScalScalProd}) let us introduce
in the classical phase space $Z$ of electromagnetic field a scalar
product by the formula:
\begin{equation}
 \langle\,\underline c\,,\,\underline d\,\rangle =
 \int d\mu^+_m \cdot
 M_{\nu\rho} \cdot
 a^*_\nu(k)^{\,\underline c} \cdot
 a_\rho(k)^{\,\underline d} \ .
\label{EMScalProd}
\end{equation}
Here $M_{\nu\rho}$ is an arbitrary positive-definite Hermitian
matrix. In the capacity of such a matrix we can take, for example,
$M_{\nu\rho}=\mathrm{diag}\,(+1,+1,+1,+1)_{\nu\rho}$.

The scalar product (\ref{EMScalProd}) is positive-definite.
Therefore it defines some topology in the space $Z$. The important
is the fact that this topology does not depend on the concrete
choice of matrix $M_{\nu\rho}$.

From this it follows that the introduced topology is
relativistic-invariant.

With respect to the introduced topology the symplectic structure
turns out to be continuous with respect to the pair of its
arguments; the symplectic structure defines an isomorphism of the
spaces $Z$ and $Z^*$; the Poisson brackets turn out to be
correctly defined on the whole $Z^*$.

So, the invariant phase space of electromagnetic field has natural
structure of linear complex topological space, and the topology
there is defined by a set of equivalent (in the sense of topology)
scalar products. Such spaces we will call {\it spaces of Hilbert
type}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Tensor product of spaces of Hilbert type.}
 Now we will show that tensor product of spaces of Hilbert
type is a space of Hilbert type.

Consider two such spaces $X$ and $Y$. Their tensor product (more
exactly, algebraic tensor product) is defined as follows.

Consider all formal products of the type $x\diamond y$, where
$x\in X$ and $y\in Y$. We will call such products pairs. Let us
suppose that pairs can be formally multiplied by complex numbers
forming expressions of the type: $\lambda\cdot x\diamond y$. And
consider all formal sums of the type:
\[
 \lambda_1\cdot x_1\diamond y_1 +
 \lambda_2\cdot x_2\diamond y_2 +
 \dots +
 \lambda_n\cdot x_n\diamond y_n     \ .
\]
Let us consider two such sums to be equivalent, if they have the
same sums of factors for each pair. So, we come to the complex
vector space that we denote as  $X\diamond Y$ (this is a {\it free
$\mathbb C$-module generated by the Cartesian product
 $X\times Y$}).

In the constructed space consider elements of the type:
\[
 1\cdot (\lambda x)\diamond y - \lambda \cdot x\diamond y \ ,
\]
\[
 1\cdot x\diamond (\lambda y) - \lambda \cdot x\diamond y \ ,
\]
\[
 1\cdot (x_1+x_2)\diamond y -
   1\cdot x_1\diamond y - 1\cdot x_2\diamond y \ ,
\]
\[
 1\cdot x\diamond (y_1+y_2) -
   1\cdot x\diamond y_1 - 1\cdot x\diamond y_2  \ .
\]
The linear shell of these elements we denote as $X\circ Y$.

Factorizing $X\diamond Y$ with respect to $X\circ Y$ we come to
the tensor product:
\[
 X\otimes Y = (X\diamond Y)/(X\circ Y) \ .
\]

For shortness, let us introduce the notation $x\otimes y$. Namely,
let us suppose that under the defined factorization an element
$1\cdot x\diamond y$ of the space $X\diamond Y$ transfers into the
element $x\otimes y$ of the space $X\otimes Y$.

If $X$ and $Y$ are usual Hilbert spaces, then, as it is known,
their tensor product $X\otimes Y$ has natural structure of Hilbert
space. The scalar product in $X\otimes Y$ in this case at the
beginning is defined for pairs by the formula:
\begin{equation}
 \langle\, x_1\otimes y_1\,,
        \, x_2\otimes y_2\,\rangle_{X\otimes Y}=
 \langle\, x_1\,,\, x_2\,\rangle_X \cdot
 \langle\, y_1\,,\, y_2\,\rangle_Y            \ ,
\label{ScalProdInTensProd}
\end{equation}
and for other elements it is spread by the requirement of
linearity with respect to the second argument and anti-linearity
with respect to the first. Performing completion of $X\otimes Y$
with respect to the given scalar product we come to Hilbert space.

In the case of spaces of Hilbert type each of the spaces $X$ and
$Y$ has many equivalent scalar products. The space $X\otimes Y$,
in accordance with the given scheme, gets many scalar products.
Let us show that all these scalar products are equivalent.

So, suppose that in one of the two spaces, for example in $X$, we
change the scalar product $\langle\,\cdot\,,\,\cdot\,\rangle_X$ to
equivalent scalar product $(\,\cdot\,,\,\cdot\,)_X$. And the
scalar product $\langle\,\cdot\,,\,\cdot\,\rangle_{X\otimes Y}$ in
the space $X\otimes Y$ changes to
 $(\,\cdot\,,\,\cdot\,)_{X\otimes Y}$.

Consider some element of the space $X\otimes Y$:
\[
 z=x_1\otimes y_1 + x_2\otimes y_2 + \dots + x_n\otimes y_n \ .
\]
It is easy to see that this element can be represented in such a
form that in this sum all $y_1,y_2,\dots,y_n$ are orthogonal to
each other with respect to the scalar product in $Y$. Then the
scalar products of this element with itself take especially simple
form:
\begin{equation}
 \langle\,z\,,\,z\,\rangle_{X\otimes Y} =
 \langle\,x_1\,,\,x_1\,\rangle_X \cdot
         \langle\,y_1\,,\,y_1\,\rangle_Y +
 \langle\,x_2\,,\,x_2\,\rangle_X \cdot
         \langle\,y_2\,,\,y_2\,\rangle_Y + \dots +
 \langle\,x_n\,,\,x_n\,\rangle_X \cdot
         \langle\,y_n\,,\,y_n\,\rangle_Y  \ ,
\label{ZSqrAng}
\end{equation}
\begin{equation}
 (\,z\,,\,z\,)_{X\otimes Y} =
 (\,x_1\,,\,x_1\,)_X \cdot \langle\,y_1\,,\,y_1\,\rangle_Y +
 (\,x_2\,,\,x_2\,)_X \cdot \langle\,y_2\,,\,y_2\,\rangle_Y +\dots +
 (\,x_n\,,\,x_n\,)_X \cdot \langle\,y_n\,,\,y_n\,\rangle_Y  \ .
\label{ZSqrBra}
\end{equation}

A convenient necessary and sufficient criterion of equivalence of
scalar products $\langle\,\cdot\,,\,\cdot\,\rangle_{X\otimes Y}$
and $(\,\cdot\,,\,\cdot\,)_{X\otimes Y}$ is the following: there
exists such $\varepsilon\in\mathbb R$, $\varepsilon>0$, that for
any $z\in X\otimes Y$ we have the following conditions:
\[
 \varepsilon\cdot(\,z\,,\,z\,)_{X\otimes Y} <
 \langle\,z\,,\,z\,\rangle_{X\otimes Y} \ , \qquad
 \varepsilon\cdot\langle\,z\,,\,z\,\rangle_{X\otimes Y} <
 (\,z\,,\,z\,)_{X\otimes Y} \ .
\]

But if such $\varepsilon$ exists for the scalar products
$\langle\,\cdot\,,\,\cdot\,\rangle_X$ and
$(\,\cdot\,,\,\cdot\,)_X$, then in accordance with formulas
(\ref{ZSqrAng}) and (\ref{ZSqrBra}), it is appropriate also for
the scalar products
 $\langle\,\cdot\,,\,\cdot\,\rangle_{X\otimes Y}$ and
 $(\,\cdot\,,\,\cdot\,)_{X\otimes Y}$.

So, all scalar products in $X\otimes Y$ are equivalent. Performing
completion of $X\otimes Y$ with respect to the topology defined by
these scalar products we come to the space of Hilbert type.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{Topology of space of states of quantized
electromagnetic field.}
 As we have seen, the space of states of quantized electromagnetic
field $\mathcal H$ decomposes into orthogonal sum of subspaces
with definite number of particles:
\begin{equation}
 \mathcal H =
   \mathcal H_0 \dot+
   \mathcal H_1 \dot+
   \mathcal H_2 \dot+ \dots
\label{PartDecompEMF}
\end{equation}

The space $\mathcal H_0$ is one-dimensional. So, the question
about its topology does not arise.

The subspace $\mathcal H_1$, as it was shown in the
section~\ref{TFIPSEM}, can be in fact identified with the phase
space of the classical field $Z$. It is the space of Hilbert type.
So, the question about its topology is also solved.

Consider now the tensor product of the space $Z$ with itself:
$Z\otimes Z$. There we have naturally defined action of the group
of transpositions of two elements. The simplest elements are
transformed under action of this group as
 $\underline a\otimes\underline b\to
  \underline b\otimes\underline a$, and to other elements this
action is spread by linearity.

Let us choose among the scalar products in the space $Z\otimes Z$
only those invariant with respect to the action of the group under
consideration. In practice it is convenient just to restrict
ourself with the products for which in the right part of the
formula~(\ref{ScalProdInTensProd}) we have product of the same
scalar products.

Then the group of transpositions acts unitarily\footnote{Under
unitary transformation of a space of Hilbert type we imply an
automorphism of this space, i.~e. one-to-one transformation on
itself preserving linear structure and each of scalar products.}
in $Z\otimes Z$. And $Z\otimes Z$ decomposes into the orthogonal
sum of two invariant subspaces: symmetric and anti-symmetric.

Each of these two subspaces inherits topology from $Z\otimes Z$.

It is easy to see that the two-particle quantum space
 $\mathcal H_2$ can be naturally identified with the symmetric
subspace in $Z\otimes Z$ and therefore it gets corresponding
topology.

Et cetera. $n$-particle subspace $\mathcal H_n$ is identified with
fully symmetric subspace of $n$-th tensor power of $Z$. And
inherits topology from there.

So, each of the subspaces in the orthogonal sum
(\ref{PartDecompEMF}) is a space of Hilbert type. From practical
point of view it is very convenient, because for practical
applications it is much easier to define the described spaces
using bases (and not as factor-spaces of free modules).

Consider now the space $\mathcal H$ as a whole. Using topologies
introduced in its subspaces we can introduce different topologies
in the whole space. For example, convergence of a directed set to
zero can be understood as independent convergence to zero of all
projections of this set. Such a topology seems to be the most
natural in this case.

On the other hand, this requirement can be strengthened if we
require additionally, for example, that starting from some moment
only finite number of projections of the directed set differ from
zero.

So, there are many natural topologies in $\mathcal H$. And the
space $\mathcal H$ is not a space of Hilbert type.

In this connection I want to pay attention to the following. In
the paper of Gupta~\cite{Gupta1950} it was suggested to quantize
electromagnetic field in usual Hilbert space with
positive-definite metric. In fact, it is possible to do so
considering quantizations invariant with respect to a more narrow
group than the Poincare group (namely, with respect to the
subgroup of the Poincare group that leaves the direction of the
time axis unchanged). The subspaces with fixed number of particles
turn out to be the same, in fact, that we constructed above. But
the whole space $\mathcal H$ acquires such a topology that is not
relativistic-invariant. And it turns out that some states of the
field belong to the Hilbert space in one frame of reference and do
not belong in other. So, the old Gupta formalism is not
relativistic-invariant, even implicitly.

In his later papers Gupta tried to refuse from forced introduction
of a sign-definite metric. But his last paper on this
topic~\cite{Gupta1959} clearly showed that he still had no
apparently relativistic-invariant construction of quantized
electromagnetic field (to say nothing of the problems with
functional analysis).

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\myparagraph{About origin of anti-unitary transformations.}
 In accordance with the introduced procedure of quantization,
linear transformations of observables of a classical field that
preserve Poisson brackets and the subspaces $Cr$ and $De$ generate
unitary transformations of states of the quantized field. But it
is known that in quantum theory an important role is played also
by anti-unitary symmetry transformations. Consider their origin on
the example of the operation of time reversal~--- $T$.

Suppose that the Lagrangian of a classical field is $T$-invariant
(as examples we can use scalar and electromagnetic fields). Then
the action turns out to be $T$-invariant also. Therefore, field
functions satisfying the stationary action principle under the
operation of time reversal transfer into functions that also
satisfy the stationary action principle.

So, from the point of view of invariant Hamiltonian formalism, the
operation of time reversal is a one-to-one transformation of the
phase space $Z$ on itself. And how does the symplectic structure
change? It is obvious from the variational definition of
symplectic structure that it changes sign. It seems to be natural
to call such transformations, that change sign of symplectic
structure, {\it anti-symplectic}.

Consider now conjugate action of the operation of time reversal on
elements of the conjugate space $Z^*_{\mathbb C}$:
\[
 a \stackrel T{\longrightarrow} a_T \ , \qquad
 a,a_T \in Z^*_{\mathbb C} \ .
\]
The Poisson bracket changes sign under such a transformation:
\[
 \{\,a_T\,,\,b_T\,\} = - \,\{\,a\,,\,b\,\} \ .
\]
So, the operation of time reversal generates the automorphism of
the Lie algebra $\mathcal C \oplus Z^*_{\mathbb C}$.

In accordance with the procedure of quantization described in this
paper, it is natural to wish to construct corresponding
automorphism of the algebra of operators $\mathcal O$ for the
quantized field. But it is impossible to do it. The reason is that
for construction of the algebra of operators we used not only the
structure of the Lie algebra of the set
 $\mathcal C \oplus Z^*_{\mathbb C}$ but also the special
correspondence between the subspace $\mathcal C$ and the set of
scalars $\mathbb C$ (the relation $\hat1 = ()\widehat~$). It means
that for bringing of the operation of time reversal to the quantum
case we need some additional conventions, besides operation of
quantization that we have already introduced.

Let us agree that words (i.~e. elements of the semi-group
$\mathcal W$) under time reversal are transformed in accordance
with the following formula:
\[
 \hat a\,\hat b \dots \hat c \stackrel T{\longrightarrow}
 \hat c_T \dots \hat b_T\,\hat a_T \ , \qquad
 \hat a,\hat b,\dots,\hat c,\hat a_T,\hat b_T,\dots,\hat c_T
 \in\mathcal A \ ,
\]
i.~e. every letter is substituted by the corresponding, and after
that letters are written in the reverse order. It is obvious that
time reversal acts on a product of two words in the following way:
\[
 \hat p\,\hat q \stackrel T{\longrightarrow} \hat q_T\,\hat p_T \ ,
 \qquad \hat p,\hat q,\hat p_T,\hat q_T \in \mathcal W \ .
\]
A one-to-one transformation of a semi-group on itself that
satisfies such a property can be called an {\it
anti-auto\-mor\-phism}.

The constructed anti-automorphism of the semi-group of words
$\mathcal W$ by linearity spreads to anti-automorphism of the
algebra of phrases $\mathcal P$. Furthermore, it is easy to check
that this anti-automorphism of the algebra of phrases generates an
anti-automorphism of the algebra of operators $\mathcal O$.

Suppose now that under time reversal creation subspace $Cr$ and
destruction $De$ transfer to one another. This condition is
usually satisfied because these subspaces are negative- and
positive-frequency, correspondingly. The left ideal spanned on
$De\widehat~$ under time reversal transfers into the right ideal
spanned on $Cr\widehat~$. The constructed by factorization left
module transfers into the corresponding right module.

So, the operation of time reversal defines the one-to-one linear
correspondence of ket- and bra-vectors\footnote{Schwinger
suggested~\cite{Schwin1992} to interpret ket-vectors as symbols
denoting creation of system in the past; bra-vectors as symbols
denoting destruction of system in future; and operators as action
of instruments during experiment. Such an interpretation turns out
to be quite natural here.}.

As it can be easily seen, the scalar product is preserved under
time reversal:
\begin{equation}
 \langle\,x\,|\,y\,\rangle=\langle\,y_T\,|\,x_T\,\rangle \ .
\label{TScale}
\end{equation}

So far as there is a one-to-one anti-linear correspondence between
ket- and bra-vectors, we could make all our discussion inside one
space, for example, the space of ket-vectors. Then the operation
of time reversal could be considered as one-to-one anti-linear
transformation of this space. The relation~(\ref{TScale}) shows
that under this transformation the scalar product changes to
complex-conjugate. A transformation satisfying such properties is
called {\it anti-unitary}.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\bigskip
At the end I want to thank V.~M.~Shabaev, V.~S.~Buslaev,
L.~D.~Faddeev, T.~A.~Bolokhov, V.~A.~Franke, V.~D.~Lyakhovsky,
L.~V.~Prokhorov, V.~V.~Vereschagin and Yu.~M.~Pis'mak for fruitful
discussions.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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\end{document}
