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\renewcommand{\@oddfoot}{--- ~March, 2002~ ---\hfil
\raisebox{0.3ex}{\tiny --- ~D.~A.~Arbatsky~ ``QED. III. Formula
for generators\dots''~
---}\hfil --- ~\thepage~ ---}
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\title{On quantization of electromagnetic field.\\
       III. Formula for generators of infinitesimal\\
       linear canonical transformations.}
\author{D.~A.~Arbatsky\footnote{ http://daarb.narod.ru/ , http://wave.front.ru/ }}
\date{March, 2002}
\maketitle

\begin{abstract}
Here we suggest a formula for generators of infinitesimal linear
symplectic transformations of invariant phase space. We discuss
applications of this formula to classical and quantum field
theory. We show the existence of generators of the symmetry group
for quantum case.
\end{abstract}

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\myparagraph{Formula.}
 In the paper [I] we used a formula that gives generators of
infinitesimal linear transformations for invariant Hamiltonian
formalism. In view of importance and generality of this formula we
will give here its derivation and discussion.

So, consider some linear field which is described by the invariant
Hamiltonian formalism. Let us for definiteness look for the
generators of the Poincare group.

So far as the symplectic structure~$\omega$ of the invariant phase
space is supposed to be Poincare-invariant, the Poincare group
acts on~$Z$ as continuous linear symplectic group. For any
infinitesimal transformation from this group\footnote{I.~e. an
element of its Lie algebra.} we have a Hamiltonian flow on~$Z$.
The Hamiltonian of this flow is a function on~$Z$. Let us denote
it $G^{\,\underline .}$. Now we will find the explicit formula for
this function.

First, it is immediately seen that this function is defined up to
addition of a constant. We will fix this constant by the condition
$G^{\,\underline 0}=0$, i.~e. the function is equal to zero for
the zero vector from space~$Z$. Speaking in a more physical way,
we will suppose that the vacuum of a classical field has zero
energy, momentum etc.

Let us fix now in the space~$Z$ a point $\underline c$. The vector
of velocity of the flow in this point we denote
as~$\underline{\delta c}$. Let us draw the straight line segment
from the point~$\underline0$ to the point $\underline c$, i.~e.
let us consider the set of points like $\alpha\,\underline c$,
$\alpha\in[0;1]$. So far as transformations of the group under
consideration are linear, in the point $\alpha\,\underline c$ the
velocity of flow is equal to $\alpha\,\underline{\delta c}$. On
the other hand, the vector of the velocity of flow is connected
with the Hamiltonian by the relation:
\begin{equation}
 dG\big|_{\alpha\underline c}{}^{\underline b} =
 \omega^{\,\underline b\,;\,\alpha\underline{\delta c}} \ ,
\label{defg}
\end{equation}
which is true for any $\underline b\in Z$. Here the left side
denotes the differential of the function $G^{\,\underline .}$
calculated in the point ${\alpha\,\underline c}$ and taken on the
vector $\underline b$.

As a vector~$\underline b$ in the equation~(\ref{defg}) we can
take the vector $d\alpha\,\underline c$:
\[
 dG\big|_{\alpha\underline c}{}^{d\alpha\underline c} =
 \omega^{\,d\alpha\underline c\,;\,\alpha\underline{\delta c}} \ .
\]
Using bilinearity of the form $\omega$ we can write it as:
\[
 dG\big|_{\alpha\underline c}{}^{d\alpha\underline c} =
 \omega^{\,\underline c\,;\,\underline{\delta c}}\ \alpha\,d\alpha \ .
\]
Integrating this relation with respect to $\alpha$ in the limits
from $0$ to $1$ we get the formula for the generator:
\begin{equation}
 \begin{array}{|ccc|}
 \hline
 \phantom{1} & & \phantom{1}\\
  & G^{\,\underline c} =
              \frac12\,
              \omega^{\,\underline c\,;\,\underline{\delta c}} & \\
 \phantom{1} & & \phantom{1}\\
 \hline
 \end{array}
\label{basf}
\end{equation}

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\myparagraph{Application to classical field theory.}
 It should not be thought that the formula~(\ref{basf}) is just
other form of writing the Noether formula. First, it works only in
the linear case. Second, it is not supposed to be used for looking
for {\it integrals of motion}, because they are supposed to be
known for linear fields (the equations of the motion are solved).
Nevertheless, the formula~(\ref{basf}) turns out to be very
convenient in practical application for many reasons.

First, it is applied much easier. Second, in contrast to the
Noether formula, it is not connected with coordinate
representation. If we know a formula for the symplectic
structure~$\omega$, for example, in the Fourier-representation and
we know how in this representations the group under consideration
acts (for example, Poincare group) then we get the generators as
easily as in the coordinate representation.

Furthermore, as it was shown in the paper~[II], the structure of
the invariant Hamiltonian formalism does not change, if we add to
the Lagrangian a full divergence. So far as the
formula~(\ref{basf}) is written in the terms of this formalism
only, the independence of generators from this substitution
becomes apparent.

As a simplest example, consider the scalar field. In the paper~[I]
we have shown that in the Fourier representation the symplectic
structure of this field is given by the formula:
\begin{equation}
 \omega^{\,\underline\cdot\,;\,\underline\cdot} =
 \int d\mu_m \cdot i\,\varepsilon(k) \cdot
        a(-k)^{\,\underline\cdot} \, a(k)^{\,\underline\cdot} \ .
\label{sympl}
\end{equation}

Now consider a space-time shift of the field by infinitesimal
vector $\delta\varepsilon_\nu$. A state $\underline c$ changes to
 $\underline c + \underline{\delta c}$.
And:
\[
 a(k)^{\,\underline{\delta c}} =
     i\,\delta\varepsilon_\nu\,k_\nu\,a(k)^{\,\underline c}\ .
% \qquad
% a^*(k)^{\,\underline{\delta c}} =
%     {}-i\,\delta\varepsilon_\nu\,k_\nu\,a^*(k)^{\,\underline c} \ .
\]
Substituting this expression to the formula~(\ref{basf}), and
taking into account the formula~(\ref{sympl}), we get:
\[
 G^{\,\underline c} =
   -\,{\textstyle\frac12}\,\delta\varepsilon_\nu
   \int d\mu_m \cdot \varepsilon(k) \cdot
   k_\nu\,a(-k)^{\,\underline c}\,a(k)^{\,\underline c} =
  {}- \delta\varepsilon_\nu \displaystyle\int d\mu^+_m \cdot
   k_\nu\, a^*(k)^{\,\underline c}\,a(k)^{\,\underline c} \ .
\]

Let us denote the coefficient of $\delta\varepsilon_\nu$ as
$-P_\nu$. Omitting the argument $\underline c$ we get the formula
for the vector of energy and momentum:
\begin{equation}
 P_\nu =
   {\textstyle\frac12}
   \int d\mu_m \cdot \varepsilon(k) \cdot k_\nu\,a(-k)\,a(k) =
   \int d\mu^+_m \cdot k_\nu\, a^*(k)\,a(k) \ .
\label{MomScalF}
\end{equation}

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\myparagraph{About generators in quantum field theory.}
 In the paper~[VI] we will describe how construction of quantum
fields is performed. Let us discuss here how generators are found
in the quantum case.

As far as I know, there was not suggested any analog of Noether
theorem for the quantum field theory\footnote{Some authors allege
opposite. I allow myself just to ignore their opinion on this
question.}.

Note now, that in the case of invariant quantization of fields, in
accordance with the scheme given in the paper~[VI], the action of
group of invariance is naturally transferred to the quantum space
of states $\mathcal H$. And this action is unitary (in the case of
indefinite scalar product~--- pseudounitary).

If the scalar product in $\mathcal H$ is positive-definite, then
in accordance with the Stone theorem, we have the self-adjoint
quantum generator. If the scalar product in $\mathcal H$ is
indefinite, then everything is similar, but we need to introduce
some special mathematical terminology.

So, generators of the group of invariance can be rigorously
introduced in the quantum case also.

It seems to be of interest to get some useful formulas for quantum
generators also.

So far as the vacuum $|\,0\,\rangle$ of the quantum field remains
invariant under the action of the symmetry group, then, speaking
formally, we just need to write expressions like (\ref{MomScalF})
making normal ordering of operators in the integrand. In the case
of Poincare-invariant quantization of the scalar field that leads
to the positive-definite scalar product we get:
\begin{equation}
 \hat P_\nu = \int d\mu^+_m \cdot k_\nu\, \hat a^*(k)\,\hat a(k) \ .
\label{QuantMomScalF}
\end{equation}

But, nevertheless, the strict mathematical meaning of the obtained
expression is not fully clear\footnote{It will be seen from
discussion of quantization and topological questions in the paper
[VI] that symbols like $\hat a^*(k)$ have sense only after a
proper averaging. But in the given formula we need to integrate
{\it the product} of such symbols.}. It seems to be topical to
develop a theory of integrals like (\ref{QuantMomScalF}) and
prove that if the integrands are normally ordered, we get exactly
the generators introduced here.

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