The state of nonrelativistic quantum system in a relativistic reference frame
I.V.~Sharph
We consider the problem of internal particle state transformation, which is a bound state of several particles, from the rest frame of a composite particle to the system in which it is relativistic. It is assumed that in the rest frame of the composite particle, its internal state could be considered in the nonrelativistic approximation. It has been shown, that this internal state is unchanged during the transition from one reference frame to another. Namely, the spherically symmetric particle in the rest frame stays spherically symmetric in any other reference frame. We discuss the possible application of these results for description of the hadrons scattering processes like bound states of quarks.
Pretvorenna_stanu_arXiv_RevTeX_v2.tex
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%\title[Sample title]
%\title[]{The state of nonrelativistic quantum system in a relativistic reference frame}% Force line breaks with \\PhysRevD version
%\title[]{Nonrelativistic state transformations under transition from one inertial reference frame to another}% arXiv
\title[]{Transformation of the nonrelativistic quantum system under transition from one inertial reference frame to another}% arXiv


%\thanks{Footnote to title of article.}
\author{I.V.~Sharph}
% \altaffiliation[Also at ]
% \altaffiliation{Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%Lines break automatically or can be forced with \\
\affiliation{Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%


% In alphabetic order

\author{M.A.~Deliyergiyev}%
\affiliation{ Department of Experimental Particle Physics, Jozef Stefan Institute, % \\
     Jamova 39, SI-1000 Ljubljana, Slovenia.}%

\author{A.G.~Kotanzhyan}%
\affiliation{ Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%

\author{K.K.~Merkotan}%
\affiliation{ Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%

\author{N.O.~Podolian}%
\affiliation{ Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%

\author{O.S.~Potiyenko}%
\affiliation{ Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%

\author{D.A.~Ptashynskyy}%
\affiliation{ Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%

\author{G.O.~Sokhrannyi}%
\affiliation{ Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%

\author{A.V.~Tykhonov}%
\affiliation{ Department of Experimental Particle Physics, Jozef Stefan Institute, % \\
     Jamova 39, SI-1000 Ljubljana, Slovenia.}%

\author{Yu.V.~Volkotrub}%
\affiliation{ Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%


% Rusov

\author{V.D.~Rusov}%
 \email{siiis@te.net.ua}
\affiliation{Odessa National Polytechnic University, Shevchenko av. 1, Odessa, 65044, Ukraine.}%
\affiliation{Department of Mathematics, Bielefeld University, %\\
      Universitatsstrasse 25, 33615 Bielefeld, Germany.}%

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\begin{abstract}
We consider the problem of internal particle state transformation, which is a bound state of several particles, from the rest frame of a composite particle to the system in which it is relativistic. It is assumed that in the rest frame of the composite particle, its internal state could be considered in the nonrelativistic approximation. It has been shown, that this internal state is unchanged during the transition from one reference frame to another. Namely, the spherically symmetric particle in the rest frame stays spherically symmetric in any other reference frame. We discuss the possible application of these results for description of the hadrons scattering processes like bound states of quarks.
\end{abstract}

%\pacs{Valid PACS appear here}% PACS, the Physics and Astronomy
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\keywords{nonrelativistic\*\  scattering processes \*\ longitudinal momenta \*\ quarks \*\ composite particle\*\ Laplace's method \*\ virtuality \*\ Regge theory\*\ reference frame}
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\section{Introduction}
%Recently, it has been shown in \cite{ujp}, that the hadron-hadron inelastic scattering processes could be described with Laplace method. However, calculations, in \cite{ujp}, have been done only for the model of scalar theory, so that allowed to replicate experimental results only on the qualitative level \cite{cej}. Obtained results force us to try to apply this method within the framework of more realistic theory $\textendash$ QCD. Herewith we run into the well-known problem, on the diagrams we have quark and gluon lines, while in the initial and finite states we have the bound quark states that are hadrons. 
Recently it has been shown that the hadron-hadron inelastic scattering processes can be successfully described with the Laplace's method~\cite{ujp}. However, calculations in~\cite{ujp} have been performed using a simplified scalar theory, rather than the quantum chromodynamics (QCD). This allowed to reproduce the experimental data for hadron-hadron scattering cross section on a qualitative level only~\cite{cej}. Obtained results force us to try to apply this method within the framework of more realistic theory, in QCD. Herewith we run into the well-known problem: there are quark and gluon lines in Feynman diagrams in QCD,  whereas there are no lines representing hadrons. 
In ``ordinary" scattering process the initial and finite state of the scattering particles are taken to be non-interacting, but for quarks inside interacting hadrons it is not true. %We surely can distribute the bound state of the particles over the eigenstates for momentum, notably over the states of free particles, however two problems appears. 
We surely can decompose the bound state of the particles over the momentum eigenstates, notably over the states of free particles, however two problems appears. 

The first one lays in the possibility to find coefficients of this series in the arbitrary reference frame. The second one $\textendash$ the Hamiltonian of that system in the initial and finite states does not approximate asymptotically to the free quarks Hamiltonian, hence the interaction process does not ``turn on" in the initial state and does not ``turn off" in the finite state, like in the ordinary scattering theory. %That results in that if we even distribute the bound state over free particles states, 
This leads to the fact that even we decompose the bound state over free particles states, the asymptotic time dependence of these states will not match with the dependence of free particles states. As the result, we cannot use the usual interaction representation, and we need to find another one, where the role of Hamiltonian of the free fields plays the Hamiltonian of the non-interaction hadrons, with accounting their quarks structure.

We can simplify solution of the mentioned problems with the following arguments. If we assume, that the free hadron in the initial and the finite state of the scattering process consists of certain numbers of specific quarks, then a new particles cannot be produced
as the result of these quarks interactions. Therefore, the internal state of the free hadron in the rest frame system can be described in the nonrelativistic approximation. Accordingly the Hamiltonian of the quarks inside this free hadron at the hadron rest frame system can be described in the nonrelativistic approximation. However, the initial and the finite state of the scattering process consists of a few hadrons, hence generally speaking we cannot find a reference frame that it would be the rest frame for all these hadrons.


In this way, it appears a problem of the transformation of the nonrelativistic internal state and the Hamiltonian in the transition from the rest frame of the particle into the reference frame where this particle is relativistic. 
%Therefore there is a problem of the transformation of the nonrelativistic internal state and the Hamiltonian in the transition from the rest frame of the particle into the reference frame where this particle is relativistic. 
%The essence of this problem could be explained by the following simple example. Suppose we have the most trivial quantum system $\textendash$ atom of hydrogen that stays, for example, in the simplest spherically symmetric ground state. We also may suppose that there is an inertial observer, which is moving with the relativistic speed relative to the hydrogen atom. What results will obtain this observer, if he/she will measure the coordinates or momentums of the particles that make up that quantum system? 
The essence of this problem can be illustrated with the following simple example. Consider the most trivial quantum system $\textendash$ the hydrogen atom in the simplest spherically symmetric ground state. Imagine an inertial observer moving with relativistic velocity with respect to it. The observer measures the coordinates (momenta) of particles that make up the system. What would be the result of this measurement? 
That is what probability amplitude can describe the results of these measurements and how this probability amplitude is connected with the probability amplitude in the rest frame of the hydrogen atom, namely with the center of mass frame of the particles which compose this atom. Further in the paper, for instance, we will consider a meson as a two-particle system of a quark and an antiquark and afterwards we will apply obtained results to more complex three-quark systems, i.e. baryons.

%Outlined problem is rather non-typical. Because usually we deal with the measurements of a specific quantities associated to the same event, which performed in the different reference frames. In the case with probability amplitude of the many-particle system the situation is rather different. If there are two inertial observers, which we call \textit{unprimed} and \textit{primed} then, for example, the probability amplitude of the two-particle system relative to the \textit{unprimed} observer (denoted as $\Psi \left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$) describes the result of the measurement particles coordinates that is simultaneous relative to the \textit{unprimed} system and taken at time $t$ of this system. 
Outlined problem is rather non-typical. Usually one deals with the measurements of a specific quantities associated with the same event, which are performed in the different reference frames. This is not the case if we deal with the probability amplitude for many-particle system. Consider two inertial observers, which we call \textit{unprimed} and \textit{primed} respectively. Then, the probability amplitude for two-particle system in the reference frame of \textit{unprimed} observer (denoted as $\Psi \left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$) describes the result of simultaneous measurement of particles' coordinates, performed at a certain moment of time $t$ in this frame. 
%In a similar way, the probability amplitude ${\Psi }'\left( {t}',{{{{\mathbf{r}}'}}_{1}},{{{{\mathbf{r}}'}}_{2}} \right)$ relative to the \textit{primed} observer describes the result of measurement, that is simultaneous relative to this observer at his clock time ${t}'$. However, measurements that are simultaneous relative to one observer will not be simultaneous to the another one, and vice versa. 
Likewise, the probability amplitude ${\Psi }'\left( {t}',{{{{\mathbf{r}}'}}_{1}},{{{{\mathbf{r}}'}}_{2}} \right)$ in the reference frame of the \textit{primed} observer describes the results of measurements, which are simultaneous with respect to this observer at his clock time ${t}'$. However, measurements that are simultaneous in the reference frame of the \textit{primed} observer will not be simultaneous in the reference frame of the \textit{unprimed} one, and vice versa. 
This is the essential distinction between the considered problem in comparison with the classical problem of the Lorentz contraction, where the coordinate measurement of the rod ends should be synchronized with the reference frame relative to which the rod moves, but it cannot be simultaneous in the rod rest frame. 
Due to this fact the rod length can be calculated by the coordinates of the same events but in different reference frames. 
%In our case the probability amplitudes $\Psi \left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$ and ${\Psi }'\left( {t}',{{{{\mathbf{r}}'}}_{1}},{{{{\mathbf{r}}'}}_{2}} \right)$ associated with the different realizations of the measurement process. And as result any relations between values $t$ and ${t}'$ cannot be found, because such relations can be determined only between the time coordinates of the same event measured relative to various reference frames. Accordingly it is impossible to establish a tie between the values of ${{\mathbf{r}}_{1}}$ and ${{\mathbf{r}}_{2}}$ on the one hand and of ${{{\mathbf{r}}'}_{1}}$, ${{{\mathbf{r}}'}_{2}}$ on the other hand. 
In our case, the probability amplitudes $\Psi \left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$ and ${\Psi }'\left( {t}',{{{{\mathbf{r}}'}}_{1}},{{{{\mathbf{r}}'}}_{2}} \right)$ are associated with the different realizations of the measurement process. 
As a result, one cannot conclude any kind of relation between the magnitudes $t$ and ${t}'$, since such a relations could be established only between the time coordinates of the same event, measured with respect to different reference frames. Accordingly, it is impossible to establish a tie between the values of ${{\mathbf{r}}_{1}}$ and ${{\mathbf{r}}_{2}}$, as well as between ${{{\mathbf{r}}'}_{1}}$ and ${{{\mathbf{r}}'}_{2}}$. 
%So there is no relations like Lorentz transformations between the arguments of the probability amplitudes in both reference frames. Therefore conclusions about the lengths contraction or on increasing of time intervals, which are consequences of the Lorentz transformations, become invalid.
Hence, there is no relations similar to Lorentz transformations between the arguments of the probability amplitudes in both reference frames. Therefore, the notions of length contraction and time dilation, which are the consequences of Lorentz transformations, are not applicable in our case. 

%Thus considering the problem of the probability amplitude transformation in the transition from one inertial reference frame to another we should establish a relations not between the values of the function (the probability amplitude) which correspond to the same events in the different reference frames, but between its values for the same arguments as this is done for internal symmetries. Therefore the probability amplitude in the \textit{primed} reference frame we will denote as ${\Psi }'\left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$.
Thus, in the problem of state transformation at the transition from one inertial reference frame to another, it is improper to consider connections between the values of probability amplitudes, corresponding to the same events in different reference frames. Rather, one should examine the relation between the values of probability amplitude in the different reference frames, corresponding to the same values of arguments, similar to that in dealing with the internal symmetries. Taking this into account, we will denote the probability amplitude in the \textit{primed} reference frame through ${\Psi }'\left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$. 


%A principled method for the solution of the problem of the state transformation in the transition from one inertial reference frame to another is given by the field quantization postulate, adopted in \cite{Bogolubov}. 
The principle method to solve the state transformation problem at the transition from one inertial reference frame to another is provided by the field quantization postulate, established in \cite{Bogolubov}. 
However that postulate is applied to the case of the relativistic quantum system state, which cause problem. This means, for instance, that an internal state of the meson should be described not by the two-particle probability amplitude $\Psi\left(t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$, but by the Fock space element, i.e. by the set of probability amplitudes of the type:
%======================================
\begin{equation}
%\left| \Psi  \right\rangle =\left\{ {{\Psi }_{{{n}_{q}},{{n}_{g}}}}\left( t,{{{\vec{r}}}_{1}},{{{\vec{r}}}_{2}};{{{\vec{q}}}_{1}},\ldots ,{{{\vec{q}}}_{{{n}_{q}}}};{{{\vec{g}}}_{1}},\ldots ,{{{\vec{g}}}_{{{n}_{g}}}} \right) \right\},
\left| \Psi  \right\rangle =\left\{ {{\Psi }_{{{n}_{q}},{{n}_{g}}}}\left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}};{{{\mathbf{q}}}_{1}},\ldots ,{{{\mathbf{q}}}_{{{n}_{q}}}};{{{\mathbf{g}}}_{1}},\ldots ,{{{\mathbf{g}}}_{{{n}_{g}}}} \right) \right\},
\label{eq:Fokstan}
\end{equation}
%======================================
which take into account that system besides the constituent quark and antiquark (the values ${{\mathbf{r}}_{1}},{{\mathbf{r}}_{2}}$ can be obtained during measurements of the coordinates) can contain an arbitrary number $n_{q}$ of current  quarks and antiquarks (obtained by measuring their coordinates and denoted ${{\mathbf{q}}_{1}},\ldots ,{{\mathbf{q}}_{{{n}_{q}}}}$). The notations $n_{g}$ and ${{\mathbf{g}}_{1}},\ldots ,{{\mathbf{g}}_{{{n}_{g}}}}$ have a similar meaning but for gluons.

To avoid difficulties connected with the necessity to define a large number of unknown functions we want to use advantage of the above circumstance, which consists in the fact that hadron internal state in the rest frame can be described in the nonrelativistic approximation. This in turn mean that there is a possibility to transform operators and the state vector Eq.\ref{eq:Fokstan} thus that all observed values remained unchanged. While all functions included in Eq.\ref{eq:Fokstan}, except \mbox{ $\Psi_{n_{q}=0, n_{g}=0}\left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)\equiv \Psi \left( t,{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$} were taking small values and may be set equal to zero with a high accuracy.
%It means that there is a possibility to transform operators and the state vector Eq.\ref{eq:Fokstan} thus that all observed values remained unchanged and while all the functions included in Eq.\ref{eq:Fokstan}, except for \mbox{ $\Psi_{\overset{n_{q}=0}{}\underset{n_{g}=0}{}}=0\left( t,{{{\vec{r}}}_{1}},{{{\vec{r}}}_{2}} \right)\equiv \Psi \left( t,{{{\vec{r}}}_{1}},{{{\vec{r}}}_{2}} \right)$}, turned out to take small values and they can be put equal to zero with great accuracy.

But then we also have to build a corresponding nonrelativistic approximation for the generators, which are included in the unitary operator with help of which the state can be transformed in the transition to the new reference frame according to the quantization postulate \cite{Bogolubov}. Our own formulation of such approximations is the content of the next part of the work. It is well known that the arbitrary Lorentz transformation can be made as the product of rotations and boost. Obviously, the above mentioned problems concern exactly to the boost. Therefore further we will consider only state transformations at the boost. At the same time we will consider that the boost is realized along $z$ axis. 

Before we proceed to the resolving of the mentioned problem, let's turn ours attention to the further simplification of this problem. 
%Let's consider the hadron in its rest frame. In this system the state Eq.\ref{eq:Fokstan} must be eigenstate for total momentum $\hat{\mathbf{P}}$ of the all particles that make up this system. And at the same time this state must correspond to zero eigenvalue. Before the transition to nonrelativistic approximation the temporal progress of state Eq.\ref{eq:Fokstan} of the particles system that make up hadron could be represented in next form as well 
Let us consider the hadron in its rest frame. In this system the state $\left| \Psi \right\rangle$ must be eigenstate for total momentum $\hat{\mathbf{P}}$ of the all particles that make up this system. And at the same time this state must correspond to zero eigenvalue. Before the transition to non-relativistic approximation the temporal progress of state $\left| \Psi \right\rangle$ of the particles system, that make up the hadron, can be represented in next form
%======================================
\begin{equation}
\left| \Psi \left( t \right) \right\rangle =\exp \left( -i\hat{H}t \right)\left| \Psi \left( t=0 \right) \right\rangle ,
\end{equation}
%======================================
where $\hat{H}$ is the relativistic Hamiltonian of a system of fields, whose quants are making up the hadron. In the reference frame that obtained from outcome system by applying the boost, according to \cite{Bogolubov} we will obtain:
%======================================
\begin{equation}
\left| {\Psi }'\left( t \right) \right\rangle =\hat{U}\left( Y \right)\left( \exp \left( -i\hat{H}t \right)\left| \Psi \left( t=0 \right) \right\rangle  \right).
\label{eq:peretvorenna}
\end{equation}
%======================================
%Here $\hat{U}\left( Y \right)$ is unitary state transformation operator from \cite{Bogolubov} as a result of boost with rapidity $Y$. Taking into consideration, that we dealing with the eigenstate of total momentum of the system and this state corresponds the zero eigenvalue we can write relation Eq.\ref{eq:peretvorenna} in the next form:
Here $\hat{U}\left( Y \right)$ is the unitary state transformation operator of Eq.\ref{eq:Fokstan} from \cite{Bogolubov} as a result of boost with rapidity $Y$. Taking into account that we are dealing with the eigenstate of total momentum of the system and this state corresponds the zero eigenvalue we can write relation Eq.\ref{eq:peretvorenna} in the following form:
%======================================
\begin{eqnarray}
%& \left| {\Psi }'\left( t \right) \right\rangle =\hat{U}\left( Y \right)\left( \exp \left( -i\left( \hat{H}t-\left( \hat{\vec{P}}\cdot \hat{\vec{R}} \right) \right) \right)\left| \Psi \left( t=0 \right) \right\rangle  \right),
& \left| {\Psi }'\left( t \right) \right\rangle =\hat{U}\left( Y \right)\left( \exp \left( -i\left( \hat{H}t-\left( \hat{\mathbf{P}}\cdot \hat{\mathbf{R}} \right) \right) \right)\left| \Psi \left( t=0 \right) \right\rangle  \right),
\label{eq:peretvorenna1}	%\times\nonumber  \\
\end{eqnarray}
%======================================
%where $\mathbf{R}$ is complex of the three arbitrary coordinates. The specific choice of these coordinates have no matter, because the eigenvalue of total momentum operator $\hat{\mathbf{P}}$ is equal to zero.
where $\mathbf{R}$ is a set of the three arbitrary coordinates. The specific choice of these coordinates is not sufficient, because the eigenvalue of total momentum operator $\hat{\mathbf{P}}$ is zero.


Now we rewrite expression Eq.\ref{eq:peretvorenna1} in another form:
%======================================
\begin{eqnarray}
& \left| {\Psi }'\left( t \right) \right\rangle =\hat{U}\left( Y \right)\hat{u}\left( x \right){{{\hat{U}}}^{-1}}\left( Y \right)\hat{U}\left( Y \right)\left| \Psi \left( t=0 \right) \right\rangle,
\label{eq:peretvorenna2}	
\end{eqnarray}
%======================================
where we use the following notations:
%======================================
\begin{eqnarray}
x\equiv \left( t,{{R}_{x}},{{R}_{y}},{{R}_{z}} \right), ~~ \hat{u}\left( x \right)\equiv \exp \left( -i\left( \hat{H}t-\left( \hat{\mathbf{P}}\cdot \hat{\mathbf{R}} \right) \right) \right).
\label{eq:poznachenna_x_u}	
\end{eqnarray}
%======================================

%Expression $\hat{U}\left( Y \right)\hat{u}\left( x \right){{\hat{U}}^{-1}}\left( Y \right)$ in Eq.\ref{eq:peretvorenna2} formally matches  with the expression which appears at transformation of operator field functions \cite{Bogolubov}. Therefore when we denote matrix of boost along $Oz$ direction by ${{\Lambda }^{\left( 0 \right)}}\left( Y \right)$ we get 
Expression $\hat{U}\left( Y \right)\hat{u}\left( x \right){{\hat{U}}^{-1}}\left( Y \right)$ in Eq.\ref{eq:peretvorenna2} formally coincides with the expression which appears at transformation of operator field functions \cite{Bogolubov}. Therefore, denoting the matrix of boost along $z$ direction by ${{\Lambda }^{\left( 0 \right)}}\left( Y \right)$ we get 
%======================================
\begin{eqnarray}
\hat{U}\left( Y \right)\hat{u}\left( x \right){{\hat{U}}^{-1}}\left( Y \right)=\hat{u}\left( {{\Lambda }^{\left( 0 \right)}}\left( Y \right)x \right).
\label{eq:poznachenna_u1}	
\end{eqnarray}
%======================================

%Then instead of Eq.\ref{eq:peretvorenna2}, we can write:
Then, instead of Eq.\ref{eq:peretvorenna2}, we can write:
%======================================
\begin{eqnarray}
& \left| {\Psi }'\left( t \right) \right\rangle =\exp \left( -it\left( \operatorname{ch}\left( Y \right)\hat{H}+\operatorname{sh}\left( Y \right){{{\hat{P}}}_{z}} \right) \right)\exp \left( i{{R}_{z}}\left( \operatorname{sh}\left( Y \right)\hat{H}+\operatorname{ch}\left( Y \right){{{\hat{P}}}_{z}} \right) \right)\times\nonumber  \\
& \times \exp \left( i\left( {{R}_{x}}{{{\hat{P}}}_{x}}+{{R}_{y}}{{{\hat{P}}}_{y}} \right) \right)\hat{U}\left( Y \right)\left| \Psi \left( t=0 \right) \right\rangle .
\label{eq:peretvorenna3}	
\end{eqnarray}
%======================================

%Pay attention, that operators $\hat{H}$ and $\hat{\mathbf{P}}$ included in relation Eq.\ref{eq:peretvorenna3} that belongs to original reference frame, in which we can apply nonrelativistic approximation. Using this approximation these operators may be replaced by nonrelativistic internal Hamiltonian of the quarks system which make up hadron, and nonrelativistic momentum operator of this system  accordingly. The quantity $\left| \Psi \left( t=0 \right) \right\rangle $ in such nonrelativistic approximation may be replaced by coordinate part of the probability amplitude of energy eigenstate for two-particle system (quark and antiquark). In addition, if we consider a borderline case of a small rapidities $Y$, on may note that we must choose a coordinates of the center of mass as an arbitrary coordinates of the vector $\hat{\mathbf{R}}$:
Pay attention, that operators $\hat{H}$ and $\hat{\mathbf{P}}$ are included in relation Eq.\ref{eq:peretvorenna3} that belongs to the original reference frame, in which we can apply the nonrelativistic approximation. Using this approximation these operators may be replaced by nonrelativistic internal Hamiltonian of the quarks system which make up hadron, and nonrelativistic momentum operator of this system  accordingly. The quantity $\left| \Psi \left( t=0 \right) \right\rangle $ in such nonrelativistic approximation may be replaced by coordinate part of the probability amplitude of energy eigenstate for two-particle system (quark and antiquark). In addition, if we consider a extreme case of small rapidities $Y$, one may note that we must choose a coordinates of the center of mass as an arbitrary coordinates of the vector $\hat{\mathbf{R}}$:
%======================================
\begin{align}
%\vec{R}=\vec{R}\left( {{{\vec{r}}}_{1}},{{{\vec{r}}}_{2}} \right)=\frac{{{m}_{1}}{{{\vec{r}}}_{1}}+{{m}_{2}}{{{\vec{r}}}_{2}}}{{{m}_{1}}+{{m}_{2}}}.	
\mathbf{R}=\mathbf{R}\left( {{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)=\frac{{{m}_{1}}{{{\mathbf{r}}}_{1}}+{{m}_{2}}{{{\mathbf{r}}}_{2}}}{{{m}_{1}}+{{m}_{2}}}.
%$\boldmath{NTn} + \mathbf{NTn} + \boldsymbol{\pi} + \pmb{\pi} $ = Number of lines.	
\label{eq:centr_mas}
\end{align}
%======================================
%Thus, simplification, which we have been talked about, is achieved by Eq.\ref{eq:peretvorenna1}-\ref{eq:peretvorenna3}. The simplification means that we don`t need to describe the transformation of the whole probability amplitude in an eigenstate for energy during a transition from the center-of-mass (quark and antiquark) system to another reference system, but we can limit only by transformation of the coordinate part of this probability amplitude.
Thus, the simplification discussed above and achieved throughout the Eq.\ref{eq:peretvorenna1}-\ref{eq:peretvorenna3} means that we don't need to describe the transformation of the whole probability amplitude in the energy eigenstate during the transition from the center-of-mass (quark and antiquark) system to another reference system, but we can limit that calculation only to the transformation of the coordinate part of this probability amplitude.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%    
%%%%%%%%%%%%%%%%%%%%%%%%%%%%  
%\section {Approximation of Lorentz transformation generators by differential operators}
\section {Approximation of Lorentz transformation generators using differential operators}

As it is known from \cite{Bogolubov} the component ${{\hat{M}}_{03}}$ of the angular momentum tensor is the generator of transformation $\hat{U}\left( Y \right)$. That is
%======================================
\begin{align}
\hat{U}\left( Y \right)=\exp \left( i{{{\hat{M}}}_{03}}Y \right).
\label{eq:generator}
\end{align}
%======================================

%According to \cite{Bogolubov} the representation of ${{\hat{M}}_{03}}$ by the differential operators has next form:
According to \cite{Bogolubov} the representation of ${{\hat{M}}_{03}}$ by the differential operators looks as follows:
%======================================
\begin{align}
{{\hat{M}}_{03}}=i\left( t\frac{\partial }{\partial z}+z\frac{\partial }{\partial t} \right).
\label{eq:dif1particl}
\end{align}
%======================================
%Therefore, we note that the generator representations by the differential operators can be obtained by considering a some function of coordinates and time with a corresponding substitution of the independent variables in the function. But, as was mentioned in the previous section, in our case the replacement of the independent variables is impossible. Therefore, relation Eq.\ref{eq:dif1particl} can be understood only as a limit to which the ``proper" relativistic operator ${{\hat{M}}_{03}}$ is approached in the transition to a nonrelativistic approximation. The question then arises: "To what limit should approach this operator in case of many-particle system?". Given that the spatial components of the momentum are presented as a sum of the corresponding single-particle operators, we can make an assumption that the components in which one of indices is equal to zero are also additive. Then, for the two-particle system we have:
Therefore, we note that the representation of generators through differential operators can be obtained by considering some function of coordinates and time with a corresponding substitution of the independent variables in the function. But, as was mentioned in the previous section, in our case the replacement of the independent variables is impossible. Therefore, relation Eq.\ref{eq:dif1particl} can be understood only as a limit to which the ``proper" relativistic operator ${{\hat{M}}_{03}}$ is approached in the transition to a nonrelativistic approximation. The question then arises: "To what limit should this operator approach in case of many-particle system?". Given that the spatial components of the momentum are presented as a sum of the corresponding single-particle operators, we can make an assumption that the components in which one of indices is equal to zero are also additive. Then, for the two-particle system we have:
%======================================
\begin{align}
{{\hat{M}}_{03}}=i\left( t\left( \frac{\partial }{\partial {{z}_{1}}}+\frac{\partial }{\partial {{z}_{2}}} \right)+\left( {{z}_{1}}+{{z}_{2}} \right)\frac{\partial }{\partial t} \right).
\label{eq:pripucenna}
\end{align}
%======================================
%As already noted, a quantity $\left| \Psi \left( t=0 \right) \right\rangle $, which is included in Eq.\ref{eq:peretvorenna3}, in our case can be replaced by coordinates part of energy eigenstate, which we denote as $\psi \left( {{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$ in the transition to a nonrelativistic approximation. This function does not depend on time and is the operator eigenfunction of the total momentum of system, that corresponds to zero eigenvalue. If we consider that operator Eq.\ref{eq:pripucenna} can be written as:
As already noted, a quantity $\left| \Psi \left( t=0 \right) \right\rangle $, which is included in Eq.\ref{eq:peretvorenna3}, can be replaced in our case by coordinates part of energy eigenstate, which we denote as $\psi \left( {{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$, at the transition to a nonrelativistic approximation. This function does not depend on time and is the operator eigenfunction of the total momentum of system, that corresponds to zero eigenvalue. If we consider that operator Eq.\ref{eq:pripucenna} can be written as:
%======================================
\begin{align}
{\hat M_{03}} =  - t{\hat P_z} + \left( {{z_1} + {z_2}} \right)i\frac{\partial }{{\partial t}},
\label{eq:pripucenna1}
\end{align}
%======================================
%we will come to the conclusion, that function $\psi \left( {{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$ is also the eigenfunction of operator ${{\hat{M}}_{03}}$, which corresponds to zero eigenvalue. 
Hence, we will come to the conclusion that function $\psi \left( {{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)$ is also the eigenfunction of operator ${{\hat{M}}_{03}}$, which corresponds to zero eigenvalue.

%This can be explained by the following considerations. Since the original reference frame is the center of mass frame of quark and antiquark, we have: 
This can be explained by the following reflections. Since the original reference frame is the center of mass frame of quark and antiquark, we have: 
%======================================
\begin{align}
\psi \left( {{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right)=\psi \left( {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right).
\label{eq:cm_koord_chast}
\end{align}
%======================================
If in the expression 
%======================================
\begin{align}
i\left( t\left( \frac{\partial }{\partial {{z}_{1}}}+\frac{\partial }{\partial {{z}_{2}}} \right)+\left( {{z}_{1}}+{{z}_{2}} \right)\frac{\partial }{\partial t} \right)\psi \left( {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right),
\label{eq:M03psi}
\end{align}
%======================================
from ${{\mathbf{r}}_{1}}$ and ${{\mathbf{r}}_{2}}$ we move to the new variables
%======================================
\begin{align}
{\mathbf r_ + } = {\mathbf r_1} + {\mathbf r_2},~~~{\mathbf r_ - } = {\mathbf{r_1}} - {\mathbf{r_2}},
\label{eq:novi_rplus_rminus}
\end{align}
%======================================
%then operator in Eq.\ref{eq:M03psi} will only depend on ${{z}_{+}}$, and function on which operator acts will only depend on ${{z}_{-}}$.
%in this way operator in Eq.\ref{eq:M03psi} will depends only on $\mathbf{r_ {+}}$, and function on which this operator acts will depend only on $\mathbf{r_ {-}}$.
in this way, operator in Eq.\ref{eq:M03psi} will depend only on $z$ component of vector $\mathbf{r_ {+}}$ as ${{z}_{+}}$, and function on which this operator acts will depends only on $z$ component of vector $\mathbf{r_ {-}}$ as ${{z}_{-}}$.


%Thereby we can make next conclusion that
Thereby we can make the next conclusion
%======================================
\begin{align}
 \exp \left( i{{{\hat{M}}}_{03}}Y \right)\psi \left( {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right)=\psi \left( {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right).
\label{eq:visnovok}
\end{align}
%======================================
%Namely the internal state of meson does not change in the transition to a new reference frame.
Namely, the internal state of meson does not change in the transition to a new reference frame.

In all previous discussion we have considered generator ${{\hat{M}}_{03}}$ as a component of the orbital momentum tensor. We note that for free bispinor field with explicit appearance of component spin in tensor of angular momentum \cite{Bogolubov} it is seen that this components equal to zero for tensor's components which have at least one zero index. The operator of interaction of bispinor field with gauge field does not contain derivatives from components of bispinor field, thus does not give contribution to the spin angular momentum tensor. Therefore spin component is equal to zero for ``proper" relativistic operator ${{\hat{M}}_{03}}$. It means that in the transition to the nonrelativistic bound, we can consider only the orbital contribution to the ${{\hat{M}}_{03}}$.

Hence all considerations are applicable not only for mesons but for baryons, because the presence of this non-zero spins does not change anything. In the case of baryons, taking assumption that all components of momentum are additive, instead of Eq.\ref{eq:pripucenna} we obtain:
%======================================
\begin{equation}
{{{\hat{M}}}_{03}}=i\left( t\left( \frac{\partial }{\partial {{z}_{1}}}+\frac{\partial }{\partial {{z}_{2}}}+\frac{\partial }{\partial {{z}_{3}}} \right) +\left( {{z}_{1}}+{{z}_{2}}+{{z}_{3}} \right)\frac{\partial }{\partial t} \right).
\label{eq:pripucenna3}
\end{equation}
%======================================
%This operator is expressed through operator $z$, is the component of the total momentum of the system. Therefore when we act by this operator on eigenfunction of the total moment momentum operator that corresponds to zero eigenvalue we will get zero.
This operator is expressed through the operator of $z$-component of total momentum of the system. Therefore, when we act by this operator on the eigenfunction of the total moment momentum operator that corresponds to zero eigenvalue we get zero.


%Two considerations given in this section have essential shortcomings. First, the ``proper" relativistic operator ${{\hat{M}}_{03}}$ is not realized by differential operators, but it is given in the second quantization representation. Therefore, it is more convenient to look for the nonrelativistic bound of this operator in this representation. In addition, we have essentially used the assumptions Eq.\ref{eq:pripucenna} and Eq.\ref{eq:pripucenna3}. 
The considerations in this section suffer two essential shortcomings. First, the ``proper" relativistic operator ${{\hat{M}}_{03}}$ is not realized through differential operators, but it is given in the second quantization representation. Therefore, it is more convenient to look for the nonrelativistic limit of this operator in this representation. In addition, we have essentially used the assumptions Eq.\ref{eq:pripucenna} and Eq.\ref{eq:pripucenna3}. 
%These assumptions are not required in the second quantization representation, owing to the fact that the expression for the operators do not depend either on single-particle or many-particle space these operators are set by. %, the following assumptions are not required. 
These assumptions are not required in the second quantization representation, owing to the fact that the expression for the operators does not depend on whether these operators are set on the single-particle or on the many-particle space. 

%Therefore, the arguments in this section may be considered only as auxiliary. In the next section we will show that considering the problem in the secondary quantization representation we can get the same result.
Therefore, the arguments in this section may be considered only as auxiliary. In the next section we will demonstrate that considering this problem in the secondary quantization representation one can get the same result.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%    
%%%%%%%%%%%%%%%%%%%%%%%%%%%%    
%\section{Approximation of Lorentz transformation generators in second quantization representation}
\section{Approximation of Lorentz transformation generators in the second quantization representation}

%We denote ${{\hat{q}}^{+}}\left( f,\nu ,c,\mathbf{r} \right)$ as nonrelativistic creation quark operator in the coordinate representation of the second quantization. Indexes $f,\nu,c$ set respectively flavor, spin and color state of quarks, which comes to their radius-eigenvector state corresponding to the eigenvalue $\mathbf{r}$. Creation antiquark operator in the same state denoted as ${{\hat{\bar{q}}}^{+}}\left( f,\nu ,c,\mathbf{r} \right)$ and annihilation operators as ${{\hat{q}}^{-}}\left( f,\nu ,c,\mathbf{r} \right)$ and ${{\hat{\bar{q}}}^{-}}\left( f,\nu ,c,\mathbf{r} \right)$ respectively.
We denote ${{\hat{q}}^{+}}\left( f,\nu ,c,\mathbf{r} \right)$ as nonrelativistic quark creation operator in the coordinate representation of the second quantization. Indices $f,\nu,c$ set flavor, spin and color state of quarks, respectively, where quarks are created in the radius eigenvector state corresponding to the eigenvalue $\mathbf{r}$. Creation antiquark operator in the same state denoted as ${{\hat{\bar{q}}}^{+}}\left( f,\nu ,c,\mathbf{r} \right)$ and annihilation operators as ${{\hat{q}}^{-}}\left( f,\nu ,c,\mathbf{r} \right)$ and ${{\hat{\bar{q}}}^{-}}\left( f,\nu ,c,\mathbf{r} \right)$ respectively.


%In these notations, the coordinate of the internal state of mesons as quark and antiquark system can be represented in the form:
In these notations, the coordinate of the internal state of meson as quark-antiquark system can be represented in the form:
%======================================
\begin{equation}
\begin{split}
\left| \mu  \right\rangle =\int{d{{{\mathbf{r}}}_{2}}d{{{\mathbf{r}}}_{1}}\psi \left( \left| {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right| \right)}s\left( {{\nu }_{1}},{{\nu }_{2}} \right)c\left( {{c}_{1}},{{c}_{2}} \right)a\left( {{f}_{1}},{{f}_{2}} \right)\times  \\
\times {{{\hat{q}}}^{+}}\left( {{f}_{1}},{{\nu }_{1}},{{c}_{1}},{{{\mathbf{r}}}_{1}} \right){{{\hat{\bar{q}}}}^{+}}\left( {{f}_{2}},{{\nu }_{2}},{{c}_{2}},{{{\mathbf{r}}}_{2}} \right)\left| 0 \right\rangle .
\end{split}
\label{eq:coordinatna_castina}
\end{equation}
%======================================
%In this relation we have denoted $s\left( {{\nu }_{1}},{{\nu }_{2}} \right)c\left( {{c}_{1}},{{c}_{2}} \right)a\left( {{f}_{1}},{{f}_{2}} \right)$ respectively as the spin, color and flavor of probability amplitudes, whereas the function $\psi \left( \left| {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right| \right)$ is describing the coordinate dependence of amplitude probability in the center of mass frame of quark and antiquark. 
In this relation have denoted the spin, color and flavor of probability amplitudes through $s\left( {{\nu }_{1}},{{\nu }_{2}} \right)c\left( {{c}_{1}},{{c}_{2}} \right)a\left( {{f}_{1}},{{f}_{2}} \right)$, respectively, whereas the function $\psi \left( \left| {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right| \right)$ describes the coordinate dependence of probability amplitude in the center of mass frame of quark and antiquark. 
%Since we consider the coordinate part of the energy eigenstate as $\psi \left( \left| {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right| \right)$, we should consider the eigenfunctions of the nonrelativistic Hamiltonian of quark and antiquark system. Besides, as is usually assumed the summation goes over repeated indices. We also used the usual notation for vacuum state $\left| 0 \right\rangle$. 
Since we consider the coordinate part of the energy eigenstate as $\psi \left( \left| {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right| \right)$, we should consider the eigenfunctions of the nonrelativistic Hamiltonian of quark and antiquark system. Besides, as is usually assumed the summation goes over repeated indices. Also was used the usual notation for vacuum state $\left| 0 \right\rangle$. 

%Whereas the dependence of all quantities on internal indices for us in this section will be insignificant. Hereinafter, the set of indices $\left\{ \nu ,c,f \right\}$ denote as $\xi$, and the dependence of probability amplitude on internal indices as: 
Since the dependence of all quantities on internal indices is insufficient, we hereinafter denote the set of indices $\left\{ \nu ,c,f \right\}$ by $\xi$, and the dependence of probability amplitude on internal indices as: 
%======================================
\begin{equation}
s\left( {{\nu }_{1}},{{\nu }_{2}} \right)c\left( {{c}_{1}},{{c}_{2}} \right)a\left( {{f}_{1}},{{f}_{2}} \right)\equiv F\left( {{\xi }_{1}},{{\xi }_{2}} \right).
\label{eq:poznachennaF}	
\end{equation}
%======================================
Notably, instead of Eq.\ref{eq:coordinatna_castina} we can write
%======================================
\begin{equation}
\begin{split}
 \left| \mu  \right\rangle =F\left( {{\xi }_{1}},{{\xi }_{2}} \right)\int{d{{{\mathbf{r}}}_{2}}d{{{\mathbf{r}}}_{1}}\psi \left( \left| {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right| \right)}{{{\hat{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}} \right){{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}} \right)\left| 0 \right\rangle.
\end{split}
\label{eq:coord_cast}
\end{equation}
%======================================
As it is known, in the field theory the operator ${{\hat{M}}_{03}}$ is presented in the form: 
%======================================
\begin{equation}
{{\hat{M}}_{03}}=\int{d\mathbf{r}}\left( {{x}_{3}}{{{\hat{T}}}_{00}}\left( {\mathbf{r}} \right)-{{x}_{0}}{{{\hat{T}}}_{30}}\left( {\mathbf{r}} \right) \right),
\label{eq:vt_kv_generator}	
\end{equation}
%======================================
%where ${{\hat{T}}_{00}}\left( {\mathbf{r}} \right)$ and ${{\hat{T}}_{30}}\left( {\mathbf{r}} \right)$ are operators of corresponding component of the energy-momentum tensor, ${{x}_{0}}\equiv t$ is the time component of the coordinate 4-vector, and ${{x}_{3}}\equiv \left( -z \right)$ $\textendash$ component of the coordinate 4-vector along covariant axis $Oz$. Hence, the relation Eq.\ref{eq:vt_kv_generator} clearly can be rewritten in the form:
where ${{\hat{T}}_{00}}\left( {\mathbf{r}} \right)$ and ${{\hat{T}}_{30}}\left( {\mathbf{r}} \right)$ are the operators of the corresponding components of the energy-momentum tensor, ${{x}_{0}}\equiv t$ is the time component of the coordinate 4-vector, and ${{x}_{3}}\equiv \left( -z \right)$ $\textendash$ is covariant component of the coordinate 4-vector along the $z$-axis. 
Hence, the relation Eq.\ref{eq:vt_kv_generator} clearly can be obviously rewritten in the form:
%======================================
\begin{equation}
{{\hat{M}}_{03}}=-t{{\hat{P}}_{z}}+\int{d\mathbf{r}}\left( {{x}_{3}}{{{\hat{T}}}_{00}}\left( {\mathbf{r}} \right) \right),
\label{eq:M03_t}	
\end{equation}
%======================================
where ${{\hat{P}}_{z}}$ is operator of the $z$ component of the total momentum of the system.

%Please note, that relations Eq.\ref{eq:vt_kv_generator} and Eq.\ref{eq:M03_t} are accurate and do not require any assumptions and approximations. Herewith dependence on $t$ in Eq.\ref{eq:M03_t} coincides with Eq.\ref{eq:pripucenna1}, while Eq.\ref{eq:pripucenna1} is a consequence of the assumptions Eq.\ref{eq:pripucenna} and Eq.\ref{eq:pripucenna3}. Thus, we can conclude that Eq.\ref{eq:M03_t} proves the validity of these assumptions.
Note, that relations Eq.\ref{eq:vt_kv_generator} and Eq.\ref{eq:M03_t} are accurate and do not require any assumptions and approximations. Herewith, the dependence on $t$ in Eq.\ref{eq:M03_t} coincides with Eq.\ref{eq:pripucenna1}, while Eq.\ref{eq:pripucenna1} is a consequence of the assumptions Eq.\ref{eq:pripucenna} and Eq.\ref{eq:pripucenna3}. Thus, we can conclude that Eq.\ref{eq:M03_t} proves the validity of these assumptions.


%State Eq.\ref{eq:coord_cast} is an eigenstate for the total momentum of the system, which corresponds to the zero eigenvalue, so the action of the first summand of Eq.\ref{eq:M03_t} on this condition trivially gives zero. Therefore we will consider the second summand of Eq.\ref{eq:M03_t} using the following notation for it 
The state Eq.\ref{eq:coord_cast} is an eigenstate for the total momentum of the system, which corresponds to the zero eigenvalue, so the action of the first summand of Eq.\ref{eq:M03_t} on this state trivially gives zero. Therefore, we will represent the second summand of Eq.\ref{eq:M03_t} as follows 
%======================================
\begin{equation}
{{\hat{M}}_{03}}\left( {{{\hat{T}}}_{00}} \right)=\int{{{x}_{3}}}{{\hat{T}}_{00}}\left( {\mathbf{r}} \right)d\mathbf{r}.
\label{eq:M03_ot_T00}	
\end{equation}
%======================================
%To act by this operator on the state of two-particle system Eq.\ref{eq:coord_cast}, we shall construct a nonrelativistic approximation for the energy density ${{T}_{00}}\left( {\mathbf{r}} \right)$. For the solution of this problem it is the most convinient to use representation of second quantization, because in this representation the Hamiltonian is presented by the integral from some operator-significant functions, which can be taken as nonrelativistic bound of the energy density.
In order to act by this operator on the state of two-particle system Eq.\ref{eq:coord_cast}, we shall construct a nonrelativistic approximation for the energy density ${{T}_{00}}\left( {\mathbf{r}} \right)$. For the solution of this problem it is most convenient to use the representation of second quantization, because in this representation the Hamiltonian is written as an integral from some operator-valued function, which can be taken as nonrelativistic limit of the energy density.


%Nonrelativistic Hamiltonian of the quark-antiquark which in second quantization representation can be written in the form:
Nonrelativistic Hamiltonian of the quark-antiquark system in the second quantization representation can be written in the form:
%======================================
\begin{equation}
\begin{split}
& \hat{H}={{{\hat{H}}}^{\left( 0 \right)}}+{{{\hat{H}}}^{\left( V \right)}}, \\
& {{{\hat{H}}}^{\left( 0 \right)}}=\int{d\mathbf{r}}\left( {{{\hat{\bar{q}}}}^{+}}\left( \xi ,\mathbf{r} \right)\left( -\frac{1}{2m}\Delta  \right){{{\hat{q}}}^{-}}\left( \xi ,\mathbf{r} \right) \right)+\int{d\mathbf{r}}\left( {{{\hat{q}}}^{+}}\left( \xi ,\vec{r} \right)\left( -\frac{1}{2m}\Delta  \right){{{\hat{\bar{q}}}}^{-}}\left( \xi ,\mathbf{r} \right) \right), \\
& {{{\hat{H}}}^{\left( V \right)}}=\frac{1}{2}\int{d{{{\mathbf{r}}}_{1}}d{{{\mathbf{r}}}_{2}}}V\left( {{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}} \right){{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}} \right),\\
\end{split}
\label{eq:H}
\end{equation}
%======================================
%where $V\left({{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}}\right)$ is a potential energy of interaction between the quark and antiquark, $m$ is a mass of the quark and antiquark, which is approximately independent of the flavor, because the bound state exists due to the strong interaction and other types of interaction we neglect.
where $V\left({{{\mathbf{r}}}_{2}}-{{{\mathbf{r}}}_{1}}\right)$ is the potential energy of the quark-antiquark interaction, $m$ is the mass of quark or antiquark, which is approximately independent of the flavor, because the bound state exists due to the strong interaction and other types of interactions are neglected.


%As we know when two-particle Hamiltonian representation by differential operators is considered by introducing Jacobi coordinates
%As well known in consideration of the two-particle Hamiltonian representation by differential operators one may introduce the Jacobi coordinates
As is well known, in the representation of the two-particle Hamiltonian through the differential operators is considered by introducing Jacobi coordinates
%======================================
\begin{equation}
\mathbf{R}=\frac{1}{2}\left( {{{\mathbf{r}}}_{1}}+{{{\mathbf{r}}}_{2}} \right),~~~\mathbf{r}={{\mathbf{r}}_{2}}-{{\mathbf{r}}_{1}},
\label{eq:Jacobi}	
\end{equation}
%======================================
%which helps to represent the Hamiltonian as a sum of two commuting operators $\textendash$ center of mass Hamiltonian, which depends only on $\mathbf{R}$ and internal Hamiltonian which depends only on $\mathbf{r}$. We want to achieve a similar representation in case when the Hamiltonian is written by creation and annihilation operators. And for this purpose, it is convenient to rewrite one-particle part of Hamiltonian ${{\hat{H}}^{\left( 0 \right)}}$ in the two-particle operator form. As we examine the nonrelativistic approximation, all operators can be considered on the subspace of the Fock space with a fixed number and composition of particles. In our case, we consider the subspace of states which contain one quark and one antiquark. The basic states of this subspace can be written in the form:
which helps to represent the Hamiltonian as a sum of two commuting operators $\textendash$ center of mass Hamiltonian, which depends only on $\mathbf{R}$ and internal Hamiltonian which depends only on $\mathbf{r}$. We want to achieve a similar representation in case when the Hamiltonian is written through the creation and annihilation operators. For this purpose, it is convenient to rewrite one-particle part of Hamiltonian ${{\hat{H}}^{\left( 0 \right)}}$ in the two-particle operator form. As we examine the nonrelativistic approximation, all operators can be considered on the subspace of the Fock space with a fixed number and composition of particles. In our case, we consider the subspace of states which contain one quark and one antiquark. The basis states of this subspace can be written in the form:
%======================================
\begin{equation}
\left| {{\xi }_{1}},{{\xi }_{2}},{{{\mathbf{r}}}_{1}},{{{\mathbf{r}}}_{2}} \right\rangle ={{\hat{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}} \right){{\hat{q}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}} \right)\left| 0 \right\rangle.
\label{eq:bazis}	
\end{equation}
%======================================
If one will act on arbitrary linear combination of states Eq.\ref{eq:bazis} by the operator
%======================================
\begin{equation}
\hat{E}=\int{d\mathbf{r}}{{\hat{\bar{q}}}^{+}}\left( \xi ,\mathbf{r} \right){{\hat{q}}^{-}}\left( \xi ,\mathbf{r} \right),
\label{eq:edinichnij}	
\end{equation}
%======================================
%one may note that on the subspace of states which contains one quark and one antiquark the Eq.\ref{eq:edinichnij} act as unit operator. The same property corresponds to operator
one may note that on the subspace of states which contain one quark and one antiquark the operator in Eq.\ref{eq:edinichnij} acts as unity operator. The same holds for the operator
%======================================
\begin{equation}
{\hat{E}}'=\int{d\mathbf{r}}{{\hat{q}}^{+}}\left( \xi ,\mathbf{r} \right){{\hat{\bar{q}}}^{-}}\left( \xi ,\mathbf{r} \right).
\label{eq:edinichnij1}	
\end{equation}
%======================================
which act on the same subspace. If the first summand in the one-particle part ${{\hat{H}}^{\left( 0 \right)}}$ of Hamiltonian Eq.\ref{eq:H} multiply by the unit operator Eq.\ref{eq:edinichnij1}, and the second summand multiply by Eq.\ref{eq:edinichnij} we will get the expression of the one-particle part in two-particle effective form
%======================================
\begin{equation}
\begin{split}
& {{{\hat{H}}}^{\left( 0 \right)}}=\left( -\frac{1}{2m} \right)\int{d{{{\mathbf{r}}}_{1}}d{{{\mathbf{r}}}_{2}}}\left( {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}} \right){{\Delta }_{1}}{{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}} \right)+ \right. \\
& \left. +{{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}} \right){{\Delta }_{2}}{{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}} \right) \right).\\
\end{split}
\label{eq:odnodvux}	
\end{equation}
%======================================

%Now we are replacing one-particle part of Hamiltonian Eq.\ref{eq:H} by Eq.\ref{eq:odnodvux} and after this replacement we are going to the variables Eq.\ref{eq:Jacobi}. We also introduce the following denotations
Next, let us replace the one-particle part of the Hamiltonian Eq.\ref{eq:H} by Eq.\ref{eq:odnodvux} and, after this replacement, pass to the variables Eq.\ref{eq:Jacobi}. We also introduce the following notations
%======================================
\begin{equation}
\begin{split}
& {{{\mathbf{r}}}_{1}}\left( \mathbf{R},\mathbf{r} \right)=\mathbf{R}-\frac{1}{2}\mathbf{r},~~~{{{\mathbf{r}}}_{2}}\left( \mathbf{R},\mathbf{r} \right)=\mathbf{R}+\frac{1}{2}\mathbf{r}, \\
& {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}}\left( \mathbf{R},\mathbf{r} \right) \right)={{{\hat{\bar{q}}}}^{+}_{1}},~~~{{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}}\left( \mathbf{R},\mathbf{r} \right) \right)={{{\hat{q}}}^{+}_{2}}, \\
%& {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\vec{r}}}_{2}}\left( \vec{R},\vec{r} \right) \right)={{{\hat{\bar{q}}}}^{-}}\left( 2 \right),~~~{{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\vec{r}}}_{1}}\left( \vec{R},\vec{r} \right) \right)={{{\hat{q}}}^{-}}\left( 1 \right). \\
& {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{r}}}_{2}}\left( \mathbf{R},\mathbf{r} \right) \right)={{{\hat{\bar{q}}}}^{-}_{2}},~~~{{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{r}}}_{1}}\left( \mathbf{R},\mathbf{r} \right) \right)={{{\hat{q}}}^{-}}_{1}. \\
 \end{split}
\label{eq:pozn27_06}	
\end{equation}
%======================================

Then, instead of Hamiltonian Eq.\ref{eq:H} we get
%======================================
\begin{equation}
\begin{split}
& \hat{H}={{{\hat{H}}}^{\left( {\mathbf{R}} \right)}}+{{{\hat{H}}}^{\left( \mathbf{r},V \right)}}, \\ 
& {{{\hat{H}}}^{\left( {\mathbf{R}} \right)}}=\left( -\frac{1}{4m} \right)\int{d\mathbf{R}d\mathbf{r}}{{{\hat{\bar{q}}}}^{+}_{1}}{{{\hat{q}}}^{+}_{2}}{{\Delta }_{{\mathbf{R}}}}{{{\hat{\bar{q}}}}^{-}_{2}}{{{\hat{q}}}^{-}_{1}}, \\ 
%& {{{\hat{H}}}^{\left( \vec{r},V \right)}}=\int{d\vec{R}d\vec{r}}{{{\hat{\bar{q}}}}^{+}}\left( 1 \right){{{\hat{q}}}^{+}}\left( 2 \right)\left( -\frac{1}{m}{{\Delta }_{{\vec{r}}}}+V\left( {\vec{r}} \right) \right){{{\hat{\bar{q}}}}^{-}}\left( 2 \right){{{\hat{q}}}^{-}}\left( 1 \right) .\\
& {{{\hat{H}}}^{\left( \mathbf{r},V \right)}}=\int{d\mathbf{R}d\pmb{r}}{{{\hat{\bar{q}}}}^{+}_{1}}{{{\hat{q}}}^{+}}_{2}\left( -\frac{1}{m}{{\Delta }_{{\pmb{r}}}}+V\left( {\pmb{r}} \right) \right){{{\hat{\bar{q}}}}^{-}_{2}} {{{\hat{q}}}^{-}}_{1} .\\
\end{split}
\label{eq:Hvnut_vnech}
\end{equation}
%======================================

%The operator ${\hat{H}}^{\left( {\mathbf{R}} \right)}$ we define as the center of mass Hamiltonian, and the operator ${{\hat{H}}^{\left( \mathbf{r},V \right)}}$ define as internal Hamiltonian of system. Thus the Hamiltonian Eq.\ref{eq:Hvnut_vnech} can be written as
The operator ${\hat{H}}^{\left( {\mathbf{R}} \right)}$ we define as the center of mass Hamiltonian, and the operator ${{\hat{H}}^{\left( \mathbf{r},V \right)}}$ we define as the internal Hamiltonian of the system. Thus the Hamiltonian Eq.\ref{eq:Hvnut_vnech} can be written as
%======================================
\begin{eqnarray}
\hat{H}=\int{{{{\hat{T}}}_{00}}\left( {\mathbf{R}} \right)d\mathbf{R}},
\label{eq:oprT00} 
\end{eqnarray}
%======================================
where the energy density operator ${{\hat{T}}_{00}}\left( {\mathbf{R}} \right)$ can be written in the form:
%======================================
\begin{equation}
{{\hat{T}}_{00}}\left( {\mathbf{R}} \right)=T_{00}^{\left( {\mathbf{R}} \right)}\left( {\mathbf{R}} \right)+T_{00}^{\left( {\mathbf{r}} \right)}\left( {\mathbf{R}} \right)+T_{00}^{\left( V \right)}\left( {\mathbf{R}} \right),
\label{eq:T00RrV}
\end{equation}
%======================================
with the help of the following denotations:
%======================================
\begin{equation}
\begin{split}
& T_{00}^{\left( {\mathbf{R}} \right)}\left( {\mathbf{R}} \right)=\left( -\frac{1}{4m} \right)\int{d\mathbf{r}}{{{\hat{\bar{q}}}}^{+}_{1}}{{{\hat{q}}}^{+}_{2}}{{\Delta }_{{\mathbf{R}}}}{{{\hat{\bar{q}}}}^{-}_{2}}{{{\hat{q}}}^{-}_{1}}, \\ 
& T_{00}^{\left( {\mathbf{r}} \right)}\left( {\mathbf{R}} \right)=\left( -\frac{1}{m} \right)\int{d\mathbf{r}}{{{\hat{\bar{q}}}}^{+}_{1}}{{{\hat{q}}}^{+}_{2}}{{\Delta }_{{\vec{r}}}}{{{\hat{\bar{q}}}}^{-}_{2}}{{{\hat{q}}}^{-}_{1}}, \\ 
& T_{00}^{\left( V \right)}\left( {\mathbf{R}} \right)=\int{d\mathbf{r}}{{{\hat{\bar{q}}}}^{+}_{1}}{{{\hat{q}}}^{+}_{2}}V\left( {\mathbf{r}} \right){{{\hat{\bar{q}}}}^{-}_{2}}{{{\hat{q}}}^{-}_{1}}.
\label{eq:T00RrVoboznacenija}
\end{split}
\end{equation}
%======================================

%Relations in Eqs.\ref{eq:oprT00}-\ref{eq:T00RrVoboznacenija} define the nonrelativistic approximation for the operator ${{\hat{T}}_{00}}\left( {\mathbf{r}} \right)$, which can be used to construct nonrelativistic approximation for generator Eq.\ref{eq:M03_ot_T00}. Using these expressions, we have:
Relations in Eqs.\ref{eq:oprT00}-\ref{eq:T00RrVoboznacenija} define the nonrelativistic approximation for the operator ${{\hat{T}}_{00}}\left( {\mathbf{r}} \right)$ in the representation of second quantization on Fock subspace. That approximation can be used to construct nonrelativistic approximation for generator Eq.\ref{eq:M03_ot_T00}. Using these expressions, we have:
%======================================
\begin{equation}
{{\hat{M}}_{03}}\left( {{{\hat{T}}}_{00}} \right)=\hat{M}_{03}^{\left( {\mathbf{R}} \right)}+\hat{M}_{03}^{\left( {\mathbf{r}} \right)}+\hat{M}_{03}^{\left( V \right)},
\label{eq:oprM03}
\end{equation}
%======================================
where
%======================================
\begin{equation}
\hat{M}_{03}^{\left( a \right)}=\int{d\mathbf{R}}\left( {{R}_{3}}T_{00}^{\left( a \right)}\left( {\mathbf{R}} \right) \right),
\label{eq:M03a}
\end{equation}
%======================================
and index $a$ takes three possible values $a=\mathbf{R},\mathbf{r},V$.

Having a nonrelativistic approximation for the generator Eq.\ref{eq:vt_kv_generator} we can act by the associated operator Eq.\ref{eq:coord_cast} on the nonrelativistic approximation for state Eq.\ref{eq:generator} and obtain the probability amplitude of this state in the new reference frame. 
In order to simplify the action of operator's exponent Eq.\ref{eq:generator} on Eq.\ref{eq:coord_cast} we consider that state Eq.\ref{eq:coord_cast} is the eigenstate of internal Hamiltonian ${\hat{H}}^{\left( \mathbf{r},V \right)}$. 
%We are interested in a main state of the bound quark's system that in their intrinsic center of mass  system corresponds to the eigenvalue, which is equal to the mass of hadron. But for the main state this eigenvalue is non-degenerate. Therefore, if we prove that the generator Eq.\ref{eq:vt_kv_generator} commutes with the internal Hamiltonian, it will mean that the state Eq.\ref{eq:coord_cast} is the eigenstate for the generator Eq.\ref{eq:vt_kv_generator} and for the operator in Eq.\ref{eq:generator}.
We are interested in the ground state of the bound system of quarks, that in center of mass system of quarks corresponds to the eigenvalue which equals the mass of hadron. However, for the aforementioned ground state this eigenvalue is not degenerate. 
Therefore, since that the generator Eq.\ref{eq:vt_kv_generator} commutes with the internal Hamiltonian, we get that the state Eq.\ref{eq:coord_cast} is the eigenstate for both the boost generator Eq.\ref{eq:vt_kv_generator} and the operator Eq.\ref{eq:generator}.

So, our next goal is to prove that the operators ${{\hat{M}}_{03}}\left( {{{\hat{T}}}_{00}} \right)$ and ${{\hat{H}}^{\left( \mathbf{r},V \right)}}$ commute.

For this purpose it is better to use the momentum representation for differentiation operator functions of the coordinates, which are included in the Laplace operator in the expression for the Hamiltonian Eq.\ref{eq:Hvnut_vnech}. For transition to the momentum representation we will write the operators Eq.\ref{eq:pozn27_06} in form:
%======================================
\begin{equation}
\begin{split}
& {{{\hat{\bar{q}}}}^{+}_{1}}=\frac{1}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{1}}}{{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right)\exp \left( i{{{\mathbf{p}}}_{1}}\left( \mathbf{R}-\frac{1}{2}\mathbf{r} \right) \right), \\ 
& {{{\hat{q}}}^{+}_{2}}=\frac{1}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{2}}}{{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right)\exp \left( i{{{\mathbf{p}}}_{2}}\left( \mathbf{R}+\frac{1}{2}\mathbf{r} \right) \right), \\ 
& {{{\hat{\bar{q}}}}^{-}_{2}}=\frac{1}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{3}}}{{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}} \right)\exp \left( -i{{{\mathbf{p}}}_{3}}\left( \mathbf{R}+\frac{1}{2}\mathbf{r} \right) \right), \\ 
& {{{\hat{q}}}^{-}_{1}}=\frac{1}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{4}}}{{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}} \right)\exp \left( -i{{{\mathbf{p}}}_{4}}\left( \mathbf{R}-\frac{1}{2}\mathbf{r} \right) \right), \\
\end{split} 	
\label{eq:impulspred}
\end{equation}
%======================================
where ${{\hat{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right),{{\hat{q}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right),{{\hat{\bar{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}} \right),{{\hat{q}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}} \right)$ are creation and annihilation operators of quarks in the states, which are eigenstates of momentum. Analogously, we present potential energy
%======================================
\begin{equation}
V\left( {\mathbf{r}} \right)=\frac{1}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d\mathbf{k}}V\left( {\mathbf{k}} \right)\exp \left( i\mathbf{k}\mathbf{r} \right).
\label{eq:potential}
\end{equation}
%======================================

If we substitute these expressions into Eq.\ref{eq:T00RrVoboznacenija} we will get the integrals over the variables ${{\mathbf{p}}_{1}},{{\mathbf{p}}_{2}},{{\mathbf{p}}_{3}},{{\mathbf{p}}_{4}}$, in which is convenient to make the following replacements:
%======================================
\begin{align}
{{{\mathbf{p}}}_{1}}+{{{\mathbf{p}}}_{2}}={{{\mathbf{P}}}_{12}},~~~\frac{{{{\mathbf{p}}}_{2}}-{{{\mathbf{p}}}_{1}}}{2}={{{\mathbf{p}}}_{12}},~~~{{{\mathbf{p}}}_{3}}+{{{\mathbf{p}}}_{4}}={{{\mathbf{P}}}_{34}}, ~~~\frac{{{{\mathbf{p}}}_{4}}-{{{\mathbf{p}}}_{3}}}{2}={{{\mathbf{p}}}_{34}}, 
\label{eq:Pip}	
\end{align}
%======================================
Jacobian of which is equal to 1. Considering this substitution, we will obtain:
%======================================
\begin{equation}
\begin{split}
& T_{00}^{\left( {\mathbf{R}} \right)}\left( {\mathbf{R}} \right)=\frac{1}{{{\left( 2\pi  \right)}^{3}}}\int{d{{{\mathbf{P}}}_{12}}d{{{\mathbf{P}}}_{34}}}d{{{\mathbf{p}}}_{12}}d{{{\mathbf{p}}}_{34}} \left( \frac{{{\left( {{{\mathbf{P}}}_{34}} \right)}^{2}}}{4m} \right)\\
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}-{{{\mathbf{p}}}_{12}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}+{{{\mathbf{p}}}_{12}} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}-{{{\mathbf{p}}}_{34}} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}+{{{\mathbf{p}}}_{34}} \right) \\ 
& \times \exp \left( i\left( {{{\mathbf{P}}}_{12}}-{{{\mathbf{P}}}_{34}} \right)\mathbf{R} \right)\delta \left( {{{\mathbf{p}}}_{12}}+{{{\mathbf{p}}}_{34}} \right). \\ 
\end{split}
\label{eq:T00Rimp}
\end{equation}
%======================================
       
Here $\delta \left( {{{\mathbf{p}}}_{12}}+{{{\mathbf{p}}}_{34}} \right)$ is denotation of Dirac $\delta $-function. Due to this function we may do ${{\mathbf{p}}_{34}}$ integration. Suppose that
%======================================
\begin{equation}
{{\mathbf{p}}_{12}}=\mathbf{p},~~~{{\mathbf{p}}_{34}}=-\mathbf{p},
\label{eq:p12_p}
\end{equation}
%======================================
we will get
%======================================
\begin{equation}
\begin{split}
& T_{00}^{\left( {\mathbf{R}} \right)}\left( {\mathbf{R}} \right)=\frac{1}{{{\left( 2\pi  \right)}^{3}}}\int{d{{{\mathbf{P}}}_{12}}d{{{\mathbf{P}}}_{34}}}d\mathbf{p}\left( \frac{{{\left( {{{\mathbf{P}}}_{34}} \right)}^{2}}}{4m} \right)\\
& \times {{{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}-\mathbf{p} \right)  {{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}+\mathbf{p} \right)\\
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}+\mathbf{p} \right) {{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}-\mathbf{p} \right)\exp \left( i\left( {{{\mathbf{P}}}_{12}}-{{{\mathbf{P}}}_{34}} \right)\mathbf{R} \right). \\ 
\end{split}
\label{eq:T00Rimp1}
\end{equation}
%======================================

The expression for $T_{00}^{\left( {\mathbf{r}} \right)}\left( {\mathbf{R}} \right)$ is analogous to $T_{00}^{\left( {\mathbf{R}} \right)}\left( {\mathbf{R}} \right)$, but considering that operator ${{\Delta }_{{\mathbf{r}}}}$ provides the other variable differentiation of exponent:
%======================================
\begin{equation}
\begin{split}
& T_{00}^{\left( {\mathbf{r}} \right)}\left( {\mathbf{R}} \right)=\frac{1}{{{\left( 2\pi  \right)}^{3}}}\int{d{{{\mathbf{P}}}_{12}}d{{{\mathbf{P}}}_{34}}}d\mathbf{p}\left( \frac{{{{\mathbf{p}}}^{2}}}{m} \right)\\
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}-\mathbf{p} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}+\mathbf{p} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}+\mathbf{p} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}-\mathbf{p} \right)\exp \left( i\left( {{{\mathbf{P}}}_{12}}-{{{\mathbf{P}}}_{34}} \right)\mathbf{R} \right). \\  
\end{split}
%\label{eq:T00Rimp1}
\end{equation}
%======================================


Considering Eq.\ref{eq:potential} for $T_{00}^{\left( V \right)}\left( {\mathbf{R}} \right)$ we have
%======================================
\begin{equation}
\begin{split}
& T_{00}^{\left( V \right)}\left( {\mathbf{R}} \right)=\frac{1}{{{\left( 2\pi  \right)}^{{9}/{2}\;}}}\int{d{{{\mathbf{P}}}_{12}}d{{{\mathbf{P}}}_{34}}}d{{{\mathbf{p}}}_{12}}d{{{\mathbf{p}}}_{34}}d\mathbf{k}  \cdot V ({\mathbf{k}})\\
& \times{{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}-{{{\mathbf{p}}}_{12}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}+{{{\mathbf{p}}}_{12}} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}-{{{\mathbf{p}}}_{34}} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}+{{{\mathbf{p}}}_{34}} \right) \\ 
& \times \exp \left( i\left( {{{\mathbf{P}}}_{12}}-{{{\mathbf{P}}}_{34}} \right)\mathbf{R} \right)\delta \left( {{{\mathbf{p}}}_{12}}+\mathbf{k}+{{{\mathbf{p}}}_{34}} \right). \\  
\end{split}
\label{eq:T00Vimp}
\end{equation}
%======================================


Now to ${{\mathbf{p}}_{34}}$ integrate taking into account $\delta$-function we will suppose
%======================================
\begin{equation}
{{\mathbf{p}}_{12}}=\mathbf{p},~~~{{\mathbf{p}}_{34}}=-\mathbf{p}-\mathbf{k}.
\label{eq:drugajazamena}
\end{equation}
%======================================

Then
%======================================
\begin{equation}
\begin{split}
& T_{00}^{\left( V \right)}\left( {\mathbf{R}} \right)=\frac{1}{{{\left( 2\pi  \right)}^{{9}/{2}\;}}}\int{d{{{\mathbf{P}}}_{12}}d{{{\mathbf{P}}}_{34}}}d\mathbf{p}d\mathbf{k} \cdot V ( {\mathbf{k}})\\
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}-\mathbf{p} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}+\mathbf{p} \right)\\ 
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}+\mathbf{p}+\mathbf{k} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}-\mathbf{p}-\mathbf{k} \right)\exp \left( i\left( {{{\mathbf{P}}}_{12}}-{{{\mathbf{P}}}_{34}} \right)\mathbf{R} \right). \\ 
\end{split}
\label{eq:}
\end{equation}
%======================================

Substituting the expressions for $T_{00}^{\left( {\mathbf{r}} \right)}\left( {\mathbf{R}} \right) $ and $T_{00}^{\left( V \right)}\left( {\mathbf{R}} \right)$ into Eq.\ref{eq:T00RrV} considering Eq.\ref{eq:oprT00} and Eq.\ref{eq:Hvnut_vnech} we will obtain the expression for internal Hamiltonian $\hat{H}\left( {\mathbf{r}} \right)$. Along with this, $\mathbf{R}$ integration appears, which leads again to appearance of Dirac $\delta$-function. This enable us to integrate over ${\mathbf{P}}_{34}$. Besides we set
%======================================
\begin{equation}
{{\mathbf{P}}_{12}}=\mathbf{P},~~~{{\mathbf{P}}_{34}}=\mathbf{P}.
\label{eq:poklademo}
\end{equation}
%======================================

As a result we will obtain: 
%======================================
\begin{equation}
\begin{split}
& {{{\hat{H}}}^{\left( {\mathbf{r}} \right)}}=\int{T_{00}^{\left( {\mathbf{r}} \right)}\left( {\mathbf{R}} \right)d\mathbf{R}}=\int{d\mathbf{P}}d\mathbf{p}\left( \frac{{{{\mathbf{p}}}^{2}}}{m} \right)\\
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}\mathbf{P}-\mathbf{p} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}\mathbf{P}+\mathbf{p} \right) {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}\mathbf{P}+\mathbf{p} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}\mathbf{P}-\mathbf{p} \right), \\ 
& {{{\hat{H}}}^{\left( V \right)}}=\int{T_{00}^{\left( V \right)}\left( {\mathbf{R}} \right)d\mathbf{R}}=\frac{1}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d\mathbf{P}}d\mathbf{p}d\mathbf{k}\cdot V({\mathbf{k}})\\
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}\mathbf{P}-\mathbf{p} \right)  {{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}\mathbf{P}+\mathbf{p} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}\mathbf{P}+\mathbf{p}+\mathbf{k} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}\mathbf{P}-\mathbf{p}-\mathbf{k} \right). \\  
\end{split}
\label{eq:vnutrH}
\end{equation}
%======================================

The internal Hamiltonian of the system is represented by this operators in next way
%======================================
\begin{equation}
\begin{split}
{{\hat{H}}^{\left( \mathbf{r},V \right)}}={{\hat{H}}^{\left( {\mathbf{r}} \right)}}+{{\hat{H}}^{\left( V \right)}}.
\end{split}
\label{eq:vnutrH1}
\end{equation}
%======================================

Now consider the expression for the momentum representation of generator Eq.\ref{eq:oprM03}. For $\hat{M}_{03}^{\left( {\mathbf{R}} \right)}$ we have
%======================================
\begin{equation}
\begin{split}
& \hat{M}_{0,3}^{\left( {\mathbf{R}} \right)}=\frac{1}{{{\left( 2\pi  \right)}^{3}}}\int{d{{{\mathbf{P}}}_{12}}d{{{\mathbf{P}}}_{34}}}d\mathbf{p}\left( \frac{{{\left( {{{\mathbf{P}}}_{34}} \right)}^{2}}}{4m} \right) \\ 
& \times{{{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}-\mathbf{p} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}{{{\mathbf{P}}}_{12}}+\mathbf{p} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}+\mathbf{p} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}{{{\mathbf{P}}}_{34}}-\mathbf{p} \right)\int{{{R}_{3}}\exp \left( i\left( {{{\mathbf{P}}}_{12}}-{{{\mathbf{P}}}_{34}} \right)\mathbf{R} \right)d\mathbf{R}}. \\ 
\end{split}
\label{eq:M03Rimp}
\end{equation}
%======================================

Integration variable ${{\mathbf{P}}_{12}}$ will be represented as  $\mathbf{P}$, and instead of ${{\mathbf{P}}_{34}}$ we will enter a new variable of integration $\pmb{\varepsilon}$ with the help of this expression:
%======================================
\begin{equation}
{{\mathbf{P}}_{34}}=\mathbf{P}-\pmb{\varepsilon }.
\label{eq:eps}
\end{equation}
%======================================
After this transformations:
%======================================
\begin{equation}
\begin{split}
& \hat{M}_{03}^{\left( {\mathbf{R}} \right)}=-i\int{d\mathbf{P}d\pmb{\varepsilon }}d\mathbf{p}\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right)\left( \frac{{{\left( \mathbf{P}-\pmb{\varepsilon } \right)}^{2}}}{4m} \right)\\
& \times {{{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}\mathbf{P}-\mathbf{p} \right) {{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}\mathbf{P}+\mathbf{p} \right)\\
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}\left( \mathbf{P}-\pmb{\varepsilon } \right)+\mathbf{p} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}\left( \mathbf{P}-\pmb{\varepsilon } \right)-\mathbf{p} \right). 
\end{split}
\label{eq:M03Rimp1}
\end{equation}
%======================================
Analogously,
%======================================
\begin{equation}
\begin{split}
& \hat{M}_{03}^{\left( {\mathbf{r}} \right)}=-i\int{d\mathbf{P}d\pmb{\varepsilon }}d\mathbf{p}\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right)\left( \frac{{{{\mathbf{p}}}^{2}}}{m} \right)\\
& \times {{{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}\mathbf{P}-\mathbf{p} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}\mathbf{P}+\mathbf{p} \right) \\
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}\left(\mathbf{P}-\pmb{\varepsilon } \right)+\mathbf{p} \right) {{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}\left(\mathbf{P}-\pmb{\varepsilon } \right)-\mathbf{p} \right). \\ 
& \hat{M}_{03}^{\left( V \right)}=\frac{-i}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d\mathbf{P}d\pmb{\varepsilon }}d\mathbf{p}d\mathbf{k}\frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}}V\left( {\mathbf{k}} \right)\\
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}\mathbf{P}-\mathbf{p} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}\mathbf{P}+\mathbf{p} \right)\\
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}\left( \mathbf{P}-\pmb{\varepsilon } \right)+\mathbf{p}+\mathbf{k} \right) {{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}\left( \mathbf{P}-\pmb{\varepsilon } \right)-\mathbf{p}-\mathbf{k} \right).
\end{split}
\label{eq:M03Rimp1mal}
\end{equation}
%======================================

Now we have the expressions for generator ${\hat{M}}_{03}$ and for internal Hamiltonian we are able to calculate their commutator. Due to the fact that each of the operators is composed of a few components, we will consider the commutators between this components. To simplify the calculating of commutators by Wick's theorem we will lead the products of operators in different order to the normal form.

Previously, we have used the fact that all operators are considered on Fock subspace state which contains one quark and one antiquark. Therefore operators, which in normal form containing two or more quark/antiquark operators (creation or annihilation) will have zero matrix elements for all basic elements on the Fock subspace, which we actually consider. 
Therefore, we will discard such operators. Given that each of the operators, which we consider contain one quark and one antiquark operators, their product will contain two quark operators and two antiquarks operators. So this product will contain ``extra" operators. By applying Wick's theorem to this product, we find that nonzero matrix elements on the considered space will have only those summands for which the ``extra" operators are paired together.

Let's consider the product ${{\hat{H}}^{\left( {\mathbf{r}} \right)}}\hat{M}_{03}^{\left( {\mathbf{R}} \right)}$. After reduction to normal form and discarding of terms with zero matrix elements, this product can be written in the form
%======================================
\begin{equation}
\begin{split}
& {{{\hat{H}}}^{\left( {\mathbf{r}} \right)}}\hat{M}_{0,3}^{\left( {\mathbf{R}} \right)}=-i\int{d\mathbf{P}}d\mathbf{p}d{\mathbf{P}}'d{\mathbf{p}}'d\pmb{\varepsilon }\left( \frac{{{{\mathbf{p}}}^{2}}}{m} \right)\left( \frac{{{\left( {\mathbf{P}}'-\pmb{\varepsilon } \right)}^{2}}}{4m} \right)\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right) \\ 
& \times \delta \left( \left( \frac{1}{2}\mathbf{P}+\mathbf{p} \right)-\left( \frac{1}{2}{\mathbf{P}}'+{\mathbf{p}}' \right) \right)\delta \left( \left( \frac{1}{2}\mathbf{P}-\mathbf{p} \right)-\left( \frac{1}{2}{\mathbf{P}}'-{\mathbf{p}}' \right) \right) \\ 
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}=\frac{1}{2}\mathbf{P}-\mathbf{p} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}=\frac{1}{2}\mathbf{P}+\mathbf{p} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\frac{1}{2}\left( {\mathbf{P}}'-\pmb{\varepsilon } \right)+{\mathbf{p}}' \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}}=\frac{1}{2}\left( {\mathbf{P}}'-\pmb{\varepsilon } \right)-{\mathbf{p}}' \right). \\ 
\end{split}
\label{eq:Rr}
\end{equation}    
%======================================

Here we have already considered the subsum on the internal indexes $\xi$ with the Kronecker $\delta$-symbols that appear in paired operators. Please note that during the integration we returned back to the old variables
%======================================
\begin{equation}
\begin{split}
& {{{\mathbf{p}}}_{1}}=\frac{1}{2}\mathbf{P}-\mathbf{p},~~~~{{{\mathbf{p}}}_{2}}=\frac{1}{2}\mathbf{P}+\mathbf{p}, \\ 
& {{{\mathbf{p}}}_{3}}=\frac{1}{2}{\mathbf{P}}'+{\mathbf{p}}',~~~~{{{\mathbf{p}}}_{4}}=\frac{1}{2}{\mathbf{P}}'-{\mathbf{p}}'. \\ 
\end{split}
\label{eq:starzmin}
\end{equation}
%======================================

Afterwards, due to the presence of $\delta $-functions one may easily perform the integration of components momenta ${{\mathbf{p}}_{3}}$ and ${{\mathbf{p}}_{4}}$. And as a result, obtain
%======================================
\begin{equation}
\begin{split}
& {{{\hat{H}}}^{\left( {\mathbf{r}} \right)}}\hat{M}_{03}^{\left( {\mathbf{R}} \right)}=-i\int{d{{{\mathbf{p}}}_{1}}}d{{{\mathbf{p}}}_{2}}d\pmb{\varepsilon }\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{2}}-{{{\mathbf{p}}}_{1}} \right)}^{2}}}{4m} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{1}}+{{{\mathbf{p}}}_{2}}-\pmb{\varepsilon } \right)}^{2}}}{4m} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}}-\frac{1}{2}\pmb{\varepsilon } \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}}-\frac{1}{2}\pmb{\varepsilon } \right). \\ 
\end{split}
\label{eq:Rrupr}
\end{equation} 
%======================================

In the case of multiplication of the same operators in reverse order after the analogous transformations we have
%======================================
\begin{equation}
\begin{split}
& \hat{M}_{03}^{\left( {\mathbf{R}} \right)}{{{\hat{H}}}^{\left( {\mathbf{r}} \right)}}=-i\int{d{{{\mathbf{p}}}_{1}}d{{{\mathbf{p}}}_{2}}}d{{{\mathbf{p}}}_{3}}d{{{\mathbf{p}}}_{4}}d\pmb{\varepsilon } \left( \frac{{{\left( {{{\mathbf{p}}}_{3}}-{{{\mathbf{p}}}_{4}} \right)}^{2}}}{4m} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{1}}+{{{\mathbf{p}}}_{2}}-\pmb{\varepsilon } \right)}^{2}}}{4m} \right)\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right) \\ 
& \times \delta \left( \left( {{{\vec{p}}}_{2}}-\frac{1}{2}\pmb{\varepsilon } \right)-{{{\mathbf{p}}}_{3}} \right)\delta \left( \left( {{{\mathbf{p}}}_{1}}-\frac{1}{2}\pmb{\varepsilon } \right)-{{{\mathbf{p}}}_{4}} \right){{{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}} \right). \\  
\end{split}
\label{eq:Rrobrporadok}
\end{equation}
%======================================

If we execute this expression and integrating the components ${{\mathbf{p}}_{3}}$ and ${{\mathbf{p}}_{4}}$, we will get the result that equal to Eq.\ref{eq:Rrupr}. Hence we conclude that the operators $\hat{M}_{03}^{\left( {\mathbf{R}} \right)} $ and ${{\hat{H}}^{\left( {\mathbf{r}} \right)}}$ commute with each other.

Now we consider the commutator $\left[ \hat{M}_{03}^{\left( {\mathbf{R}} \right)},{{{\hat{H}}}^{\left( V \right)}} \right]$. After reduction to normal form and return to the ``old" variables we have
%======================================
\begin{equation}
\begin{split}
& \hat{M}_{03}^{\left( {\mathbf{R}} \right)}{{{\hat{H}}}^{\left( V \right)}}=-i\frac{1}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{1}}d{{{\mathbf{p}}}_{2}}}d\mathbf{k}d\pmb{\varepsilon }\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{2}}+{{{\mathbf{p}}}_{1}}-\pmb{\varepsilon } \right)}^{2}}}{4m} \right)V\left( {\mathbf{k}} \right) \\ 
& \times {{{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}=\left( {{{\mathbf{p}}}_{2}}-\frac{1}{2}\pmb{\varepsilon } \right)+\mathbf{k} \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}}=\left( {{{\mathbf{p}}}_{1}}-\frac{1}{2}\pmb{\varepsilon } \right)-\mathbf{k} \right). \\ 
\end{split}
\label{eq:MRHVpra}
\end{equation}
%======================================

Calculation of product ${{\hat{H}}^{\left( V \right)}}\hat{M}_{03}^{\left( {\mathbf{R}} \right)}$ after similar transformations leads to the expression which coincides with the right-hand side of equality Eq.\ref{eq:MRHVpra}. Hence, operator $\hat{M}_{03}^{\left( {\mathbf{R}} \right)}$ commutes with each one of the two summands of internal Hamiltonian and thus commutes with the Hamiltonian of the bound quarks system.

Let us now calculate the commutators of the operator $\hat{M}_{03}^{\left( {\mathbf{r}} \right)}$ with the summands of internal Hamiltonian. With the help of transformations considered above a product ${\hat{H}}^{\left( \mathbf{r} \right)} \hat{M}_{03}^{\left( {\mathbf{r}} \right)}$ is reduced to the form:
%======================================
\begin{equation}
\begin{split}
& {{{\hat{H}}}^{\left( {\mathbf{r}} \right)}}\hat{M}_{03}^{\left( {\mathbf{r}} \right)}=-i\int{d{{{\mathbf{p}}}_{1}}d{{{\mathbf{p}}}_{2}}d{{{\mathbf{p}}}_{3}}d{{{\mathbf{p}}}_{4}}d\pmb{\varepsilon }}\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{2}}-{{{\mathbf{p}}}_{1}} \right)}^{2}}}{4m} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{2}}-{{{\mathbf{p}}}_{1}} \right)}^{2}}}{4m} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}={{{\mathbf{p}}}_{2}}-\frac{1}{2}\pmb{\varepsilon } \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}}={{{\mathbf{p}}}_{1}}-\frac{1}{2}\pmb{\varepsilon } \right). \\  
\end{split}
\label{eq:MrHrpra}
\end{equation}
%======================================

The same result is obtained if one construct the normal form from these operators product but in inverse order.  Thus $\hat{M}_{03}^{\left( \mathbf{r} \right)}$ and $\hat{H}^{\left( {\mathbf{r}} \right)}$ commutes with each other. 

Calculation of the product ${{\hat{H}}^{\left( V \right)}}\hat{M}_{03}^{\left( {\mathbf{r}} \right)}$ leads to result
%======================================
\begin{equation}
\begin{split}
& {{{\hat{H}}}^{\left( V \right)}}\hat{M}_{03}^{\left( {\mathbf{r}} \right)}=\frac{-i}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{1}}d{{{\mathbf{p}}}_{2}}d\pmb{\varepsilon }}d\mathbf{k}\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right)V\left( {\mathbf{k}} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{2}}-{{{\mathbf{p}}}_{1}}+2\mathbf{k} \right)}^{2}}}{4m} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}={{{\mathbf{p}}}_{2}}+\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}}={{{\mathbf{p}}}_{1}}-\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right). \\ 
\end{split}
\label{eq:HVMrpram}
\end{equation}
%======================================

Calculating the product of these operators in inverse order $\hat{M}_{03}^{\left( {\mathbf{r}} \right)}{{\hat{H}}^{\left( V \right)}}$ gives the result that does not coincide with Eq.\ref{eq:HVMrpram}, therefore operators $\hat{M}_{03}^{\left( {\mathbf{r}} \right)}$ and ${\hat{H}}^{\left( V \right)}$ do not commute with each other:
%======================================
\begin{equation}
\begin{split}
& \hat{M}_{03}^{\left( {\mathbf{r}} \right)}{{{\hat{H}}}^{\left( V \right)}}=\frac{-i}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{1}}d{{{\mathbf{p}}}_{2}}d\pmb{\varepsilon }}d\mathbf{k}V\left( {\mathbf{k}} \right)\left( \frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{2}}-{{{\mathbf{p}}}_{1}} \right)}^{2}}}{4m} \right) \\ 
& {{{\bar{q}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}={{{\mathbf{p}}}_{2}}+\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right){{{\hat{q}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{4}}={{{\mathbf{p}}}_{1}}-\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right). \\ 
\end{split}
\label{eq:HVMrobr}
\end{equation}
%======================================
But
%======================================
\begin{equation}
\begin{split}
& \hat{M}_{03}^{\left( V \right)}{{{\hat{H}}}^{\left( {\mathbf{r}} \right)}}=\frac{-i}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{1}}d{{{\mathbf{p}}}_{2}}d{{{\mathbf{p}}}_{3}}d{{{\mathbf{p}}}_{4}}d\pmb{\varepsilon }}d\mathbf{k}\frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}}V\left( {\mathbf{k}} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{2}}-{{{\mathbf{p}}}_{1}}+2\mathbf{k} \right)}^{2}}}{4m} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}={{{\mathbf{p}}}_{2}}+\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}}={{{\mathbf{p}}}_{1}}-\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right). \\ 
\end{split}
\label{eq:MVHrpram}
\end{equation}
%======================================
and 
%======================================
\begin{equation}
\begin{split}
& {{{\hat{H}}}^{\left( {\mathbf{r}} \right)}}\hat{M}_{03}^{\left( V \right)}=\frac{-i}{{{\left( 2\pi  \right)}^{{3}/{2}\;}}}\int{d{{{\mathbf{p}}}_{1}}d{{{\mathbf{p}}}_{2}}d{{{\mathbf{p}}}_{3}}d{{{\mathbf{p}}}_{4}}}d\pmb{\varepsilon }d\mathbf{k}\frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}}V\left( {\mathbf{k}} \right)\left( \frac{{{\left( {{{\mathbf{p}}}_{2}}-{{{\mathbf{p}}}_{1}} \right)}^{2}}}{m} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}={{{\mathbf{p}}}_{2}}+\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}}={{{\mathbf{p}}}_{1}}-\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right). \\ 
\end{split}
\label{eq:MVHobr}
\end{equation}
%======================================

Given that the expressions ${{\hat{H}}^{\left( V \right)}}\hat{M}_{03}^{\left( {\mathbf{r}} \right)}$ in Eq.\ref{eq:HVMrpram} and $\hat{M}_{03}^{\left( V \right)}{{\hat{H}}^{\left( {\mathbf{r}} \right)}}$ (see Eq.\ref{eq:MVHrpram}) are included in the general expression for commutator $\left[ {{{\hat{H}}}^{\left( \mathbf{r},V \right)}},{{{\hat{M}}}_{03}} \right]$ with opposite signs. And, since, same thing can be said about Eq.\ref{eq:HVMrobr} and Eq.\ref{eq:MVHobr}, we conclude:
%======================================
\begin{equation}
\left[ {{{\hat{H}}}^{\left( {\mathbf{r}} \right)}},\hat{M}_{03}^{\left( V \right)} \right]+\left[ {{{\hat{H}}}^{\left( V \right)}},\hat{M}_{03}^{\left( {\mathbf{r}} \right)} \right]=0.
\label{eq:CommutatorVrrV}
\end{equation}
%======================================
Finally, computation of products $\hat{M}_{03}^{\left( V \right)}{{\hat{H}}^{\left( V \right)}} $ and ${{\hat{H}}^{\left( V \right)}}\hat{M}_{03}^{\left( V \right)}$ give us the same result:
%======================================
\begin{equation}
\begin{split}
& \hat{M}_{03}^{\left( V \right)}{{{\hat{H}}}^{\left( V \right)}}={{{\hat{H}}}^{\left( V \right)}}\hat{M}_{03}^{\left( V \right)}=\frac{-i}{{{\left( 2\pi  \right)}^{3}}}\int{d{{{\mathbf{p}}}_{1}}d{{{\mathbf{p}}}_{2}}}d\pmb{\varepsilon }d{\mathbf{p}}'d{\mathbf{k}}'\frac{\partial \delta \left( {\pmb{\varepsilon }} \right)}{\partial {{\varepsilon }_{3}}}V\left( {\mathbf{k}} \right)V\left( {{\mathbf{k}}'} \right) \\ 
& \times {{{\hat{\bar{q}}}}^{+}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{1}} \right){{{\hat{q}}}^{+}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{2}} \right){{{\hat{\bar{q}}}}^{-}}\left( {{\xi }_{2}},{{{\mathbf{p}}}_{3}}={{{\mathbf{p}}}_{2}}+{\mathbf{k}}'+\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right){{{\hat{q}}}^{-}}\left( {{\xi }_{1}},{{{\mathbf{p}}}_{4}}={{{\mathbf{p}}}_{1}}-{\mathbf{k}}'-\mathbf{k}-\frac{1}{2}\pmb{\varepsilon } \right). \\ 
\end{split}
\label{eq:MVHVpramobr}
\end{equation}
%======================================

Therefore, if we decompose the commutator $\left[ {{{\hat{H}}}^{\left( \mathbf{r},V \right)}},{{{\hat{M}}}_{03}} \right]$ on the terms that corresponds to the summands of internal Hamiltonian and generator, then the sum of all these summands is equal to zero, hence
%======================================
\begin{equation}
\left[ {{{\hat{M}}}_{03}},{{{\hat{H}}}^{\left( \mathbf{r},V \right)}} \right]=0.
\label{eq:polnijcommutator}
\end{equation}
%======================================
 
So, as noted early in this paper, the state Eq.\ref{eq:coord_cast} is the eigenstate for the internal Hamiltonian, which corresponds to non-degenerate eigenstate. As consequence of Eq.\ref{eq:polnijcommutator} the state Eq.\ref{eq:coord_cast} must be the eigenstate for the generator of boost:
%======================================
\begin{equation}
{{\hat{M}}_{03}}\left| \mu  \right\rangle ={{m}_{03}}\left| \mu  \right\rangle,
\label{eq:eigenval}
\end{equation}
%======================================
where ${{m}_{03}}$ is the eigenvalue of generator ${{\hat{M}}_{03}}$ that correspond to the state $\left| \mu  \right\rangle$.

%If we want to determine the eigenvalue ${{m}_{03}}$, we need to use symmetry features of the eigenstate $\left| \mu  \right\rangle$. Moreover, this state should be transmitted into itself with the arbitrary inverse of axes coordinates. While, if we suppose that the potential of the quarks and antiquarks interaction is spherically symmetric, then the main state of this system should also be spherically symmetric, that is, the state that turn into itself with an arbitrary rotation. If we note the unitary operator, that represents an inversion or a rotation on the Fock sub-space, which is denoted as ${{\hat{U}}^{\left( I,R \right)}}$, then we have
In order to determine the eigenvalue ${{m}_{03}}$, we make use of symmetry features of the eigenstate $\left| \mu  \right\rangle$. That is, in particular, this state should transform into itself upon an arbitrary inversion of coordinate exes. Besides, if we suppose that the potential of the quarks and antiquarks interaction is spherically symmetric, then the ground state of this system should also be spherically symmetric, i.e. the state that turn into itself upon an arbitrary rotation. If we denote the unitary operator that represents an inversion or a rotation on the Fock subspace by as ${{\hat{U}}^{\left( I,R \right)}}$, then we have
%======================================
\begin{equation}
{{\hat{U}}^{\left( I,R \right)}}\left| \mu  \right\rangle =\left| \mu  \right\rangle .
\label{eq:UIR}
\end{equation}
%======================================
Hence, Eq.\ref{eq:eigenval} can be written as
%======================================
\begin{equation}
{{\hat{M}}_{03}}{{\hat{U}}^{\left( I,R \right)}}\left| \mu  \right\rangle ={{m}_{03}}{{\hat{U}}^{\left( I,R \right)}}\left| \mu  \right\rangle,
\label{eq:eigenval1}
\end{equation}
%======================================
or 
%======================================
\begin{equation}
{{\left( {{{\hat{U}}}^{\left( I,R \right)}} \right)}^{-1}}{{\hat{M}}_{03}}{{\hat{U}}^{\left( I,R \right)}}\left| \mu  \right\rangle ={{m}_{03}}\left| \mu  \right\rangle.
\label{eq:eigenval-1}
\end{equation} 
%======================================
%Operator $\left( \hat{U}^{\left( I,R \right)} \right)^{-1}{{\hat{M}}_{03}}{{\hat{U}}^{\left( I,R \right)}}$ associated with ${\hat{M}}_{03}$ by the tensor's transformation rule. It means, that if one chooses an inverse or a rotation, that is changing direction of $Oz$ axis on the opposite one, then we have
Operator $\left( \hat{U}^{\left( I,R \right)} \right)^{-1}{{\hat{M}}_{03}}{{\hat{U}}^{\left( I,R \right)}}$ associated with ${\hat{M}}_{03}$ by the tensor transformation rules. It means that we choose an inverse or a rotation, that changes the orientation of $z$ axis to the opposite one, then, we have
%======================================
\begin{equation}
 {{\left( {{{\hat{U}}}^{\left( I,R \right)}} \right)}^{-1}}{{\hat{M}}_{03}}{{\hat{U}}^{\left( I,R \right)}}=-{{\hat{M}}_{03}}.
\label{eq:minusM03}
\end{equation}
%======================================

But then, inserting Eq.\ref{eq:minusM03} into Eq.\ref{eq:eigenval-1} with account of Eq.\ref{eq:eigenval}, we get 
%======================================
\begin{equation}
{{m}_{03}}=0.
\label{eq:ravnonulu}
\end{equation}
%======================================

%So, if we note that $\left| {{\mu }'} \right\rangle $ is the state of two bounded particles system relative to the reference frame, which is obtained from the c.m.s. of these particles with boost along $Oz$ axis with rapidity $Y$, then we have
So, if we note that $\left| {{\mu }'} \right\rangle$ is the bound state of two-particle system in the reference frame, which is obtained from the c.m.s. of these particles by the boost transformation along $z$ axis with rapidity $Y$, then we have
%======================================
\begin{equation}
\left| {{\mu }'} \right\rangle =\exp \left( i{{{\hat{M}}}_{03}}Y \right)\left| \mu  \right\rangle. 
\label{eq:mustrih}
\end{equation}
%======================================
%Given the Eq.\ref{eq:eigenval} and Eq.\ref{eq:ravnonulu}, we see that from the whole series, that represent operator's exponent $\exp \left( i{{{\hat{M}}}_{03}}Y \right)$ after the operation on the state $\left|\mu\right\rangle$, the non zero result can be obtained only with the help of the term that has the identity operator. So, we have
Given Eq.\ref{eq:eigenval}, \ref{eq:ravnonulu}, we see that from the whole series, that represents the result of acting by the operator's exponent $\exp \left( i{{{\hat{M}}}_{03}}Y \right)$ on the state $\left|\mu\right\rangle$, there is the unity-operator summand only, which yields a non-zero result. So, we have
%======================================
\begin{equation}
\left| {{\mu }'} \right\rangle =\left| \mu  \right\rangle. 
\label{eq:mustrihravnomu}
\end{equation}
%======================================
%This result matches with the result shown in Eq.\ref{eq:visnovok}, which was obtained in the representation of the differential operators. Hence, we can make the conclusion, that the internal state of nonrelativistic frame of bound particles does not change, in the case of transmithion with boost to the reference frame relative to which this bound frame has relativistic energy-momentum.
This result matches with the result shown in Eq.\ref{eq:visnovok}, which was obtained in the representation of the differential operators. Hence, we draw the conclusion, that the internal state of nonrelativistic system of bound particles does not change
upon the boost transformation with boost to the reference frame, in which this bound system holds a relativistic energy-momentum.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%    
%%%%%%%%%%%%%%%%%%%%%%%%%%%%    
\section {The state transformation problem treatment within the common group-theoretical consideration}

%In the previous sections we considered two different representation of the boost generator ${{\hat{M}}_{03}}$ and got the identical results. The question therefore arises whether these results can be generalized. This generalization could be reached while reviewing this problem within the common group-theoretical consideration. 
In the previous sections we have considered two different representations of the boost generator ${{\hat{M}}_{03}}$ which have led us to the identical results. The question therefore arises whether these results can be generalized. This generalization can be achieved if we review the problem using the general group-theoretical concerns. 

%Let us consider the generators of Poincare group. We have four generators of space-time translations ${{\hat{P}}_{a}}$, where $a=0,1,2,3$ and six Lorentz generators ${{\hat{M}}_{ab}}=-{{\hat{M}}_{ba}}$. Commutation relations between these generators depend only on the group multiplication law, and those states transformation features of the quantum systems of interacting particles which were mentioned above have no affect on these commutation relations. In addition, these commutation relations do not depend on the generators representation, so we can examine them without using the explicit form of these generators. 
Let us consider the generators of Poincare group. We have four generators of space-time translations ${{\hat{P}}_{a}}$, where $a=0,1,2,3$ and six Lorentz generators ${{\hat{M}}_{ab}}=-{{\hat{M}}_{ba}}$. Commutation relations between these generators depend only on the group multiplication law, whereas the above mentioned state-transformation features of quantum systems of interacting particles do not affect these commutation relations. In addition, these commutation relations do not depend on the exact choice of the representation of these generators, so we can examine them without using the explicit form of these generators. 
As it is known from \cite{Raider} the operator ${{g}^{ab}}{{\hat{P}}_{a}}{{\hat{P}}_{b}}$ commutes with all generators of the Poincare group and particularly with the generator ${{\hat{M}}_{03}}$ that we are interested in. 
%Taking into account that in accordance with the field quantization postulate \cite{Bogolubov} the generator ${{\hat{P}}_{0}}$ should agree with the total Hamiltonian of the system, and the operators ${{\hat{P}}_{b}}$, where $b=1,2,3$, should agree with the operators of the momentum components, then one may note that operator ${{g}^{ab}}{{\hat{P}}_{a}}{{\hat{P}}_{b}}$ agrees with the square of the internal Hamiltonian of the system because all eigenvalues of this operator are equal to the squared eigenvalues of the internal energy of the considered particles system.
Take into account also that in accordance with the field quantization postulate \cite{Bogolubov} the generator ${{\hat{P}}_{0}}$ should match with the total Hamiltonian of the system, and the operators ${{\hat{P}}_{b}}$, where $b=1,2,3$, should match with the operators of the momentum components. Then, one may note that the operator ${{g}^{ab}}{{\hat{P}}_{a}}{{\hat{P}}_{b}}$ matches with the square of the internal Hamiltonian of the system, since all eigenvalues of this operator are equal to the corresponding squared eigenvalues of the internal energy of the considered particle system.


%So, the boost generator commutes with the square of the internal Hamiltonian irrespective of the representation of these operators and irrespective of the opportunity to apply the nonrelativistic approximation in the c.m.s of the examined particles. 
As a result, the boost generator commutes with the square of the internal Hamiltonian irrespective of the possibility to apply the nonrelativistic approximation in the c.m.s of the examined particle system. 
If such approximation is possible as in the case of interest, then the eigenstate that corresponds to the smallest eigenvalue of the square of the internal Hamiltonian is such that corresponds to the non-degenerate eigenvalue. %Following due to the commutativity of the operators ${{g}^{a,b}}{{\hat{P}}_{a}}{{\hat{P}}_{b}}$ and ${{\hat{M}}_{0,3}}$, we find out this eigenstate for ${{\hat{M}}_{0,3}}$ too, this implies as we have seen that the state does not change with the boost. Thus as we see the most significant factor is the non-degeneracy of the ground state of the bound system.
Then, because of the commutativity of the operators ${{g}^{ab}}{{\hat{P}}_{a}}{{\hat{P}}_{b}}$ and ${{\hat{M}}_{03}}$, we get that this state is eigenstate also for ${{\hat{M}}_{03}}$. It implies as we have noticed that this state does not change with the boost. Thus as seen the most significant factor is the non-degeneracy of the ground state of the bound system.


Therefore we can say that from the commutation relations between the Poincare group generators it implies that if the internal state of the interacting particles system (i.e. the eigenstate of the squared internal Hamiltonian) is non-degenerate, then it will not change with the boost.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%    
%%%%%%%%%%%%%%%%%%%%%%%%%%%%    
\section{Discussion of Results and Conclusions}

%It is known that the ground state of the system with the central interaction is spherically symmetric. From the obtained result one may conclude that this state will not change in the transition to the new frame. So it remains spherically symmetric and resists to the Lorentz contraction.
It is known that the ground state of the system with the central interaction is spherically symmetric. From the obtained result we conclude that this state will not change during the transition to the new reference frame. In other words, the state remains spherically symmetric and does not undergo to the Lorentz contraction.


%Thus the Lorentz contraction in the bound quantum systems does not exist. This conclusion can be very important, since the description of the elastic scattering of hadron-hadron processes used geometric models of discs \cite{Dremin:2013}, which are based on the assumption that the reduction should take place.
Thus the Lorentz contraction in the bound quantum systems does not exist. This conclusion can be very important, since the description of the elastic scattering of hadron-hadron processes rely on the geometric models of discs \cite{Dremin:2013}, which are based on the assumption that the contraction takes place.

%Physical reason for this follows from the previous discussion. Indeed, when we consider the problem of the rod contraction in the rod rest frame it doesn't matter the coordinates of its ends are measured simultaneously or not. It is sufficient only to ensure the simultaneity of the measurement of the rod ends coordinates in the reference frame relative to which it is moving. Therefore, the problem of the rod contraction can be considered in terms of the relations between the coordinates and time of the same events but measured in the various reference frames. In the case which was discussed in this paper the situation is different. While measuring the coordinates of interacting particles in an arbitrary reference frame we have to ensure the measurement simultaneity relative to that reference frame relative to which the measurement is taken. This is a significant difference from the problem of rod contraction.
The physical reasoning for this follows from the previous discussion. Indeed, when we consider the problem of the rod contraction in the rod's rest frame it is not sufficient whether the coordinates of its ends are measured simultaneously or not. Rather, it is sufficient only to ensure the simultaneity of measurements of the rod's ends coordinates in the reference frame relative to which it is moving. Therefore, the problem of rod contraction can be considered in terms of the relations between the coordinates and time of the same events, measured in the two different reference frames. This is not the case for the problem considered in the paper. In measuring the coordinates of interacting particles in an arbitrary reference frame we have to ensure that the measurements are simultaneous with respect to the reference frame, relative to which the measurements are taken. This makes an essential difference to the problem of rod contraction.
% in which in rod's rest reference frame the measurement  of its ends can be performed not simultaneously. 
%And unlike the rod case, there is no such reference frame, or one may say that it is does not exist at all, where the measurements of particles coordinates can be taken not at the same time. 
And unlike the rod case, there is no such reference frame, where the measurements of particle coordinates could be taken not at the same time. 
%And unlike the rod case, there is no such reference frame, or one may say that is does not exist, where the measurements of particles coordinates, considered in this work, can be taken not at the same time. 
%And unlike the rod problem there is no such reference frame where the measurements can be taken not at the same time. 
%This leads to that we should talk about different measurements events by various observers. 
%This leads to the fact that we should talk about different measurements events by various observers. Then the coordinates and time of those events are not connected with Lorentz transformation because these transformations bind the coordinates and time of the same event in the different reference frames. Taking into account that the Lorentz contraction is the exactly result of the Lorentz transformation formulas, which in our case are not occurred, there is no wonder that we have obtained the result that such contraction does not exist.
%This leads to the fact that we should talk about different measurement events held by different observers. Then the coordinates and time of those events are not connected with Lorentz transformation because these transformations related to each other the coordinates and time of the same event in the different reference frames. Taking into account that via the Lorentz contraction is the exact result of the Lorentz transformation formulas, which do not hold in our case, there is no wonder that we have obtained the result that such a contraction does not take place in the considered problem.
This leads to the fact that we should talk about different measurement events held by different observers. Then the coordinates and time of those events are not connected via Lorentz transformation because these transformations connect the coordinates and time of the same event measured in the different reference frames. Taking into account that via the Lorentz contraction is the exact result of the Lorentz transformation formulas, which do not hold in our case, there is no wonder that we have obtained the result that such a contraction does not take place in the considered problem.



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\normalsize {\textbf{REFERENCES}}
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