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\renewcommand{\@oddfoot}{--- ~June, 2005~ ---\hfil
\raisebox{0.3ex}{\tiny --- ~D.~A.~Arbatsky~ ``The certainty
principle''~ ---}\hfil --- ~\thepage~ ---}
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\begin{document}

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

\title{The certainty principle}
\author{D.~A.~Arbatsky\footnote{ http://daarb.narod.ru/ , http://wave.front.ru/ }}
\date{June, 2005}
\maketitle

\begin{abstract}
The notion of the quantum angle is introduced. The quantum angle
turns out to be a metric on the set of physical states of a
quantum system. Its kinematics and dynamics are studied. The {\it
certainty}\/ principle for quantum systems is formulated and
proved. It turns out that the certainty principle is closely
connected with the Heisenberg uncertainty principle (it presents,
in some sense, an opposite point of view). But at the same time
the certainty principle allows to give rigorous formulations for
wider class of problems (it allows to rigorously interpret and
ground the analogous inequalities for the pairs of quantities like
time - energy, angle - angular momentum etc.).
\end{abstract}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{The quantum angle and the certainty principle}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\paragraph{The notion of the quantum angle.}
As it is known, the set of states of any quantum system forms a
complex Hilbert space. We will denote it $\mathcal H$.

Elements of the space $\mathcal H$ we will denote as
$a,b\dots\in\mathcal H$.

The scalar product in $\mathcal H$ we will write as
 $\langle a|b\rangle$. It is linear with respect to the second
argument and anti-linear with respect to the first one.

The norm of a vector $a$ we will denote as
 $ \|a\| = \langle a | a \rangle ^{1/2}$.

Consider two non-zero vectors $a,b\in\mathcal H$. Let us define
between them {\it the quantum angle} by the formula:
\[
 \angle(a,b) = \arccos
     \frac{\big|\langle a|b\rangle\big|}{\|a\|\,\|b\|}  \ .
\]

According to Cauchy-Bunyakovsky-Schwarz inequality, under the
function $\arccos$ we have the value that is not greater than
unity. Therefore, the quantum angle is a real number:
\[
 \angle(a,b) \in \mathbb R \ , \qquad
 0 \leqslant \angle(a,b) \leqslant \frac\pi2 \ .
\]

For simplification of formulas we will later on always work with
normalized vectors: $\|a\|=1$, $\|b\|=1$. In this case the formula
for the angle is written simply as:
\[
 \angle(a,b) = \arccos \big|\langle a|b\rangle\big|  \ .
\]

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

\paragraph{Geometry of quantum angle.}
Let us consider the two extreme cases: $\angle(a,b)=0$ and
$\angle(a,b)=\pi/2$.

According to the Parseval equality, the first case takes place
when the vectors differ only by phase factor:
\[
 \angle (a,b) = 0 \qquad \Longleftrightarrow \qquad
 a\parallel b \ .
\]
From physical point of view, we can say that the corresponding
quantum states are {\it identical}.

The second case takes place when vectors are orthogonal:
\[
 \angle(a,b)=\frac\pi2 \qquad \Longleftrightarrow \qquad
 a \perp b \ .
\]
In this case we can say that the corresponding quantum states are
{\it completely different}.

Implying this physical terminology, which is used in the
considered extreme cases, let us introduce also the following
definition. Let us say that the states described by the vectors
$a$ and $b$ {\it differ not-substantially}, if $\angle(a,b)<1$;
let us also say that the states {\it differ substantially}, if
$\angle(a,b)\geqslant 1$.

\mytheorem{}
 {\it For any three vectors $a,b,c\in\mathcal H$ the inequality
takes place (``the triangle inequality''):
\[
 \angle(a,c) \leqslant \angle(a,b) + \angle(b,c) \ .
\]}

In order to prove this theorem, let us first notice that, so far
as the inequality is proved for three vectors, we can bound
ourself with the case when the Hilbert space $\mathcal H$ is three
dimensional: $\mathcal H=\mathbb C^3$.

Multiplying the vectors $a$, $b$, and $c$ by appropriate factors
and choosing orthonormal basis in $\mathcal H$ appropriately, we
can achieve that components of these vectors take the form:
\[
 a = (\begin{array}{ccc} 1   & 0   & 0   \end{array}) \ , \qquad
 b = (\begin{array}{ccc} b_1 & z   & b_3 \end{array}) \ , \qquad
 c = (\begin{array}{ccc} c_1 & c_2 & 0   \end{array}) \ ,
\]
where $c_1,c_2,b_1,b_3\in[0;1]$ are real non-negative numbers, and
$z\in\mathbb C$ is complex.

Let us introduce also the auxiliary vector $b'$:
\[
 b'= (\begin{array}{ccc} b_1 & |z| & b_3 \end{array}) \ .
\]

We have:
\[
 \angle(b',c) =
   \arccos\big(b_1c_1+|z|\,c_2\big) \, \leqslant
   \arccos|\,b_1c_1+zc_2\,| =
   \angle(b,c) \ .
\]

So far as the three vectors $a$, $b'$, and $c$ have real
coordinates and unit lengths, the triangle inequality for them is
the well known triangle inequality on the sphere in the real
three-dimensional Euclidean space:
\[
 \angle(a,c) \leqslant \angle(a,b') + \angle(b',c) \ .
\]

Combining the obtained inequalities we get:
\[
 \angle(a,c) \leqslant
 \angle(a,b') + \angle(b',c) =
 \angle(a,b) + \angle(b',c) \leqslant
 \angle(a,b) + \angle(b,c) \qquad \blacksquare
\]

Summarizing what was said above, we can say that the quantum angle
$\angle$, considered as a function on the set of pairs of physical
quantum states (considered to phase factor), is a {\it metric}.

\mytheorem{}
 {\it The metric space of physical quantum states with
the metric $\angle$ is complete.}

The proof of this theorem is not difficult, but it requires
substantial technical work. So we omit it here.

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

\paragraph{Kinematics of quantum angle.}
Let now the vector $r$ depend on the real parameter $t$:
$t\in\mathbb R$, $r(t)\in\mathcal H$, $\|\,r(t)\,\|=1$.

Let us define {\it the quantum velocity} $v(t)$ by the formula:
\[
 v(t) = \dot r(t) =
    \lim_{\delta t\to0}
    \frac{r(t+\delta t)-r(t)}{\delta t} \ .
\]

Let us define also {\it the quantum angular speed} $\omega(t)$ by
the formula:
\[
 \omega(t) = \lim_{\delta t\to0}
     \frac{\angle\,\big(\,r(t+\delta t)\,,\,r(t)\,\big)}{|\,\delta t\,|} \ .
\]

In order to express $\omega(t)$ through $v(t)$ let us decompose
$v(t)$ into the two orthogonal components:
\[
 v_\parallel(t) = r(t)\,\langle r(t)|v(t)\rangle \ , \qquad
 v_\perp(t) = v(t) - v_\parallel(t) \ .
\]

\mytheorem{}
 {\it The quantum angular speed is equal to the norm of the orthogonal
component of the quantum velocity:
\[
 \omega(t)=\|\,v_\perp(t)\,\| \ .
\]}

In order to prove this theorem let us use the Parseval equality to
change $\arccos$ to $\arcsin$:
\[
 \angle\,\big(\,r(t+\delta t)\,,\,r(t)\,\big) \,=\,
 \arccos\,\big|\,\langle\,r(t+\delta t)\,|\,r(t)\,\rangle\,\big| \,=\,
\]
\[
 \,=\,
 \arcsin\,
   \big\|\,
   r(t+\delta t) -
   r(t) \,\langle\, r(t) \,|\, r(t+\delta t) \,\rangle
   \,\big\| \,=\,
\]
\[
 \,=\,
   \arcsin\,
    \big\|\,
    r(t+\delta t) \,-\, r(t) \,-\,
    r(t) \,
     \langle\,
        r(t) \,|\, r(t+\delta t)-r(t)
     \,\rangle
    \,\big\|
 \,=\,
\]
\[
 \,=\,
   \arcsin\,
    \big\|\,
    v(t)\,\delta t \,+\, o(\delta t) \,-\,
    r(t) \,
     \langle\,
        r(t) \,|\, v(t)\,\delta t + o(\delta t)
     \,\rangle
    \,\big\|
 \,=\,
\]
\[
 \,=\,
   \arcsin\,
    \Big\|\,
     \big(v(t)\,-\,
     r(t) \,
     \langle\,
        r(t) \,|\, v(t)
     \,\rangle\big)
     \,\delta t \,+\, o(\delta t)
    \,\Big\|
 \,=\,
\]
\[
 \,=\,
   \arcsin\,
    \big\|\,
    v_\perp(t)\,\delta t \,+\, o(\delta t)
    \,\big\|
 \,=\,
   \arcsin\,
    \big(\,
    \|v_\perp(t)\|\,|\delta t| \,+\, o(\delta t)
    \,\big)
 \,=\,
\]
\[
 \,=\,
 \|v_\perp(t)\|\,|\delta t| \,+\, o(\delta t)  \qquad \blacksquare
\]

\mytheorem{}
 {\it The quantum angle satisfy the estimate:
\begin{equation}
 \angle\,\big(\, r(t_2) \,,\, r(t_1) \,\big)
 \ \leqslant\
 \left|\
 \int_{t_1}^{t_2} \omega(t) \, dt
 \ \right|
 \ \ .
 \label{AngleIneq}
\end{equation}}

For the proof let us use the triangle inequality:
\[
 \Big|\,
 \angle \big(\, r(t+\delta t) \,,\, r(t_1) \,\big) -
 \angle \big(\, r(t) \,,\, r(t_1) \,\big)
 \,\Big|
 \ \leqslant\
 \angle \big(\, r(t+\delta t) \,,\, r(t) \,\big)
 \ .
\]
Dividing by $|\delta t|$ and taking the limit $\delta t\to 0$, we
get:
\[
 \left|\
 \frac{d}{dt}\,\angle \big(\, r(t) \,,\, r(t_1) \,\big)
 \ \right|
 \ \leqslant\
 \omega(t)
 \ .
\]
Performing integration from $t_1$ to $t_2$, we get the desired
inequality. $\blacksquare$

In fact, the estimate~(\ref{AngleIneq}) is the best. Namely, there
is the following

\mytheorem{}
 {\it The quantum angle between two vectors $r_1$ and $r_2$ can be
expressed by the formula:
\[
 \angle\,(\, r_2 \,,\, r_1 \,)
 \ =\
 \min\ \left|\
 \int_{t_1}^{t_2} \omega(t) \, dt
 \ \right|
 \ \ ,
\]
where the minimum is taken among all curves $r(t)$ with ends in
$r_1$ and $r_2$: $r(t_1)=r_1$, $r(t_2)=r_2$.}

For the proof of the theorem let us twist the phase of $r_2$,
 $r_2\to r_2'=e^{i\alpha}r_2$, so that
 $\langle r_2'|r_1\rangle$ become real and
 $\langle r_2'|r_1\rangle\in[0,1]$.

Let us consider the linear shell of $r_2'$ and $r_1$,
 $\mathcal L(r_2',r_1)$. There we can choose an orthonormal basis
so that
\[
 r_1 = (\begin{array}{ccc} 1 & 0 \end{array}) \ , \qquad
 r_2'= (\begin{array}{ccc} a & b \end{array}) \ ,
\]
where $a,b\in[0,1]$ are real non-negative numbers.

Considering then $r_2'$ and $r_1$ as real vectors on Euclidean
plane we see that it is possible to stretch the circular arc
between them where the estimation integral is exactly equal to the
quantum angle. $\blacksquare$

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

\paragraph{Dynamics of quantum angle.}
Let us have now a strongly continuous one-parameter unitary group
$U(\delta s)=e^{-\,i\,\delta s\,A/\hbar}$, where $A=A^*$ is a
self-adjoint operator in the space of states $\mathcal H$ (it is
called the infinitesimal generator of $U(\delta s)$); $\delta
s\in\mathbb R$ is the parameter of the group;
 $\hbar\in\mathbb R$ is the Planck's constant.

And suppose now that the dependence of state vector on the
parameter $\delta s$ is defined by the formula:
\[
 r(\delta s) \,=\, |\,\delta s\,\rangle \,=\,
 U(\delta s)\,\rangle \,=\,
 e^{-\,i\,\delta s\,A/\hbar} \,\rangle \ .
\]
Here $|\,\delta s\,\rangle\in\mathcal H$ is another notation for
the state vector connected with parameter equal to $\delta s$;
$\rangle\in\mathcal H$ is a fixed ket-vector of state.

Let us suppose that the function $r(\delta s)$ is differentiable.
Then the quantum velocity is expressed by the formula:
\[
 \textstyle
 v(\delta s) \,=\,
 \frac1{i\hbar}\, A \, e^{-\,i\,\delta s\,A/\hbar} \,\rangle \,=\,
 \frac1{i\hbar}\, A \, |\,\delta s\,\rangle \ .
\]

The mean of the operator $A$ does not depend on time:
\[
 \overline A \,=\,
 \langle\,\delta s\,|\, A \, |\,\delta s\,\rangle \,=\,
 \langle\,e^{+\,i\,\delta s\,A/\hbar} \, A \,
    e^{-\,i\,\delta s\,A/\hbar} \,\rangle \,=\,
 \langle\, A \, e^{+\,i\,\delta s\,A/\hbar} \,
    e^{-\,i\,\delta s\,A/\hbar} \,\rangle \,=\,
 \langle\, A \, \rangle \ .
\]

Therefore the components of the quantum velocity can be written
just as:
\[
 \textstyle
 v_\parallel(\delta s) \,=\,
 |\,\delta s\,\rangle \, \langle\,\delta s\,|\,
 \frac1{i\hbar}\, A \, |\,\delta s\,\rangle \,=\,
 \frac1{i\hbar}\,\overline A \, |\,\delta s\,\rangle
\]
\[
 \textstyle
 v_\perp(\delta s) \,=\,
 \frac1{i\hbar}\,\left(A-\overline A\right) \,
 |\,\delta s\,\rangle \ .
\]

The quantum angular speed turns out to be independent of the
parameter also:
\[
 \textstyle
 \omega(\delta s)
 \,=\,
 \|\,v_\perp(\delta s)\,\|
 \,=\,
 \frac1\hbar\,
 \langle\,\delta s\,|\,
    \left(A-\overline A\right)^2
 \, |\,\delta s\,\rangle^{1/2}
 \,=\,
 \frac1\hbar\,
 \langle\,
    e^{+\,i\,\delta s\,A/\hbar} \,
    \left(A-\overline A\right)^2 \,
    e^{-\,i\,\delta s\,A/\hbar}
 \,\rangle^{1/2}
 \,=\,
\]
\[
 \textstyle
 \,=\,
 \frac1\hbar\,
 \langle\,
    \left(A-\overline A\right)^2 \,
    e^{+\,i\,\delta s\,A/\hbar} \,
    e^{-\,i\,\delta s\,A/\hbar}
 \,\rangle^{1/2}
 \,=\,
 \frac1\hbar\,
 \langle\,
    \left(A-\overline A\right)^2
 \,\rangle^{1/2}
 \,=\,
 \frac1\hbar\, \Delta_\rangle A  \ .
\]
Here $\Delta_\rangle A$ is a short notation for the standard
deviation of $A$ in the state $\rangle$.

Consider now how the quantum angle $\angle(\, |\,\delta s\,\rangle
\,,\, \rangle \,)$ behaves in this case. Using for it the
estimate~(\ref{AngleIneq}) we have:
\[
 \angle(\, |\,\delta s\,\rangle \,,\, \rangle \,)
 \ \leqslant\
 \left|\
 \int_0^{\delta s} \omega(\sigma) \, d\sigma
 \ \right|
 \ =\
 \textstyle
 \frac1\hbar\,|\delta s|\,\Delta_\rangle A \ .
\]
From this inequality we obtain the

\mytheorem{}
 {\it So that under the action of strongly continuous one-parameter
unitary group $U(\delta s)=e^{-\,i\,\delta s\,A/\hbar}$ the
initial state vector $\rangle$ changes substantially, it is
necessary to satisfy the inequality:
\begin{equation}
 \begin{array}{|ccc|}
 \hline
 \phantom{1} & & \phantom{1}\\
  & |\delta s|\, \Delta_\rangle A \geqslant \hbar & \\
 \phantom{1} & & \phantom{1}\\
 \hline
 \end{array}
 \label{Certainty1}
\end{equation}}

By the example of the Schr\"odinger particle we will see that this
theorem turns out to be closely connected with the Heisenberg
uncertainty principle, but has other meaning. Taking into account
this connection, we can name this theorem {\it the certainty
principle}.

The inequality expressing the certainty principle can be written
also in the following way:
\begin{equation}
 \begin{array}{|ccc|}
 \hline
 \phantom{1} & & \phantom{1}\\
  & \Delta_\rangle (\delta s A) \geqslant \hbar & \\
 \phantom{1} & & \phantom{1}\\
 \hline
 \end{array}
 \label{Certainty2}
\end{equation}
In this form it can be naturally carried over to the case when
 $\delta s$ and $A$ are matrices.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\section*{Examples}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\paragraph{One-dimensional Schr\"odinger particle.}
 Consider the one-dimensional Schr\"odinger particle with the
coordinate defined by the variable $x$. Its state vector can be
written by the wave function $\psi(x)$. In its space of states the
group of shifts acts by the formula:
\[
 U(\delta x)\,\psi(x) = \psi(x-\delta x) \ .
\]
This group can be written in the form:
\[
 U(\delta x) = e^{-\,i\,\delta x\,P/\hbar} \ , \qquad
 P = -\,i\hbar\frac d{dx}   \ .
\]
Here $P$ is the operator of momentum.

Applying the certainty principle in the form~(\ref{Certainty1}),
we get:
\[
 |\delta x|\, \Delta_{\psi(x)} P \geqslant \hbar \ .
\]

If we take as $\psi(x)$ a well localized packet of de~Broglie
waves, that turns into zero outside of some interval $l$, then
from this inequality, in particular, we have that
 $l\geqslant\hbar/\Delta_{\psi(x)} P$: because when the packet
is moved to the distance $l$, the change of the quantum angle must
turn out to be greater than $1$ (namely, $\pi/2$).

So, the Heisenberg uncertainty principle, if it is understood in
qualitative sense, follows from the certainty principle.

But if we understand the Heisenberg uncertainty principle in
quantitative sense, according to the Pauli-Weyl inequality
\begin{equation}
 \Delta_{\psi(x)} X\,\Delta_{\psi(x)} P \,
 \geqslant\, \frac\hbar2 \ \ ,
 \label{PauliWeyl}
\end{equation}
then there is no direct connection between these two principles.

Furthermore, from physical point of view, the Heisenberg
uncertainty principle and the certainty principle are like two
points of view on the spread of the wave packet. The Heisenberg
uncertainty principle says, that the wave packet is spread,
because the {\it classical} state of the particle is {\it badly
defined}. On the other hand, the certainty principle states, that
the wave packet is spread, because the {\it quantum} state is {\it
well defined}.

To the three-dimensional case the certainty principle is easily
generalized in the form~(\ref{Certainty2}):
\[
 \Delta_{\psi(x)}\, (\,\delta x_i\, P_i\,) \geqslant \hbar \ ,
\]
where summation over $i$ is implied.

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

\paragraph{Schr\"odinger particle on circle.}
 Consider now a plane with Cartesian coordinates $x$ and $y$.
Let us have the circle $x^2+y^2=1$ defined on the plane. On this
circle as one-dimensional coordinate we can use the polar angle
$\varphi$, considered to $2\pi$.

The state vector of the Schr\"odinger particle on this circle can
be written by the wave function $\psi(\varphi)$. And
$\psi(\varphi+2\pi)=\psi(\varphi)$.

In the space of states the action of the rotation group is
naturally defined:
\[
 U(\delta \varphi)\,\psi(\varphi) = \psi(\varphi-\delta \varphi) \ .
\]
This group can be written in the form:
\[
 U(\delta\varphi) = e^{-\,i\,\delta\varphi\,J/\hbar} \ , \qquad
 J = -\,i\hbar\frac d{d\varphi}   \ .
\]
Here $J$ is the operator of angular momentum.

Using the certainty principle does not arise any difficulties:
\[
 |\delta\varphi|\, \Delta_{\psi(\varphi)} J \geqslant \hbar \ .
\]

As regards the uncertainty principle, a carrying over of the
inequality~(\ref{PauliWeyl}) to this case is
impossible\footnote{About some attempts that were made to suggest
an inequality like~(\ref{PauliWeyl}), see \cite{Dav1973}.}.

In the three-dimensional case the certainty principle also easily
gives:
\[
 \Delta_{\psi(x,y,z)}\, (\,\delta\varphi_i\, J_i\,)
 \geqslant \hbar \ .
\]

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

\paragraph{A system with Hamiltonian independent of time.}
 Let us have now a quantum system with a Hamiltonian $H$
independent of time. On the space of states we have the following
action of the group of time shifts:
\[
 U(\delta t)\,\rangle = e^{-\,i\,\delta t\,(-H)/\hbar}\rangle \ .
\]

The certainty principle in this case immediately gives:
\[
 |\delta t|\, \Delta_\rangle H \geqslant \hbar \ .
\]

If we apply this formula, for example, to estimation of the life
time of a quasi-stationary decaying state, then it states that its
typical life time is not less than Planck's constant divided by
the width of the corresponding energy level. And here all the
terms can be defined with exact mathematical sense.

As regards to attempts to formulate the Heisenberg uncertainty
principle for values time - energy, after Bohr has declared such a
principle (in qualitative sense) so many researches and
discussions were performed for clarification of its sense, that it
is possible to write about them a separate review. As far as I
know, a rigorous formulation of the uncertainty principle for this
case have not been formulated till now\footnote{See also the
discussion of this question by J.~Baez~\cite{Baez2000}.}.

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

\paragraph{Relativistic systems.}
 Consider now any relativistic quantum system.
On its space of states the Poincare group acts. As an example of
such a system any RCQ-quantized field can serve\footnote{All other
examples that I know either come from this or are mathematically
insolvent.} \cite{Arb2002,Arb2005wircq}. And let us restrict
ourself to the discussion of the case when the field turns out
quantized in the usual Hilbert space.

In this case the application of the certainty principle does not
meet any difficulties:
\[
 \textstyle
 \Delta_\rangle\,
   (\,-\,\delta x_\mu\, P_\mu\, +\,
      \frac12\,
      \delta\omega_{\mu\nu}\, J_{\mu\nu} \,) \geqslant \hbar \ ,
\]
where $P_\mu$ is the vector operator of energy-momentum,
$J_{\mu\nu}$ is the tensor operator of four-dimensional angular
momentum, $\delta x_\mu$ and $\delta\omega_{\mu\nu}$ are the
standard logarithmic coordinates of the Poincare group.

As regards to the application of the Heisenberg uncertainty
principle, it is unlikely to be possible. In the previous
paragraph we have seen that for the values time - energy it arose
great difficulties.

In this connection, even from the ideas of relativistic invariance it
is clear that even for coordinates and momenta the situation
cannot be simple. And it turns out to be so, because it is known
that all attempts to introduce the notion of coordinates (as
self-adjoint operators on the space of states) for relativistic
systems are quite artificial\footnote{And it turns out that for
some systems introduction of good operators of coordinates is not
possible at all~\cite{NewtWig1949}.}.

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

\begin{thebibliography}{99}

\bibitem{Dav1973}
A.~S.~Davydov {\it\rql Kvantovaya mekhanika\rqr}, 2-nd ed., M.:
Nauka (1973).

\bibitem{Baez2000}
J.~Baez ``The time-energy uncertainty relation'' (2000),
http://math.ucr.edu/home/baez/uncertainty.html .

\bibitem{Arb2002}
 D.~A.~Arbatsky ``On quantization of electromagnetic field''
(2002), \\
 http://daarb.narod.ru/qed-eng.html ,
 http://wave.front.ru/qed-eng.html ,
 arXiv:math-ph/0402003 . \\{}
 [Rus.: D.~A.~Arbatsky \rql O kvantovanii elektromagnitnogo
polya\rqr{} (2002), \\
 http://daarb.narod.ru/qed-rus.html ,
 http://wave.front.ru/qed-rus.html ,
 arXiv:math-ph/0402003 .]

\bibitem{Arb2005wircq}
 D.~A.~Arbatsky ``What is ``relativistic canonical
quantization''?'' (2005), \\
 http://daarb.narod.ru/wircq-eng.html ,
 http://wave.front.ru/wircq-eng.html ,
 arXiv:math-ph/0503029 . \\{}
 [Rus.: D.~A.~Arbatsky \rql Chto takoe \rql relyativistskoe
kanonicheskoe kvantovanie\rqr{}?\rqr{} (2005), \\
 http://daarb.narod.ru/wircq-rus.html ,
 http://wave.front.ru/wircq-rus.html ,
 arXiv:math-ph/0503029 .]

\bibitem{NewtWig1949}
T.~D.~Newton, E.~P.~Wigner ``Localized states for elementary
systems'', Rev. Mod. Phys. {\bf 21}(3), 400-406 (1949).
 [Rus.: \rql Lokalizovannye sostoyaniya elementarnykh sistem\rqr{},
in~\cite{Wigner1971}, p. 277.]

\bibitem{Wigner1971}
E.~Wigner {\it\rql Etyudy o simmetrii\rqr}, M.: Mir (1971).
 [Eng.: E.~P.~Wigner {\it ``Symmetries and reflections''},
Bloomington-London: Indiana univ. press (1970).]

\end{thebibliography}

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

\end{document}
