%Title: Equations for Relativistic Particles, Binaries and Ternaries
%Author: Harmut Pilkuhn
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\begin{document}

\title{\vskip-3cm{\baselineskip14pt \begin{flushright}
\normalsize TTP/06-29\\ \normalsize November 2006\\
\end{flushright}} \vskip0.5cm
Equations for Relativistic Particles, Binaries and Ternaries}
\author{Hartmut Pilkuhn}
\affiliation{Institut f\"ur Theoretische Teilchenphysik, Universit\"at,
D-76128 Karlsruhe, Germany, email hp@particle.uni-karlsruhe.de}
\begin{abstract} Relativistic differential equations for bound states of $n$ Dirac
particles ($n=1,2,3$) are put into forms that depend only on the
square of the total cms energy. Instead of coupling $4^n$ Dirac components,
they decompose into two equivalent equations for $4^n/2$ components $\psi$ and $\chi$
of total chiralities $+1$ and $-1$, respectively. Their time-dependent
versions are of the Klein-Gordon type; they
reproduce the relativistic kinematics for the emission of photons or pions.
Although the equations are presently extended to mesons and baryons within
perturbative QCD only, the necessity of free Klein-Gordon equations for closed
systems implies $E^2$-spectra not only for mesons, but also for baryons.
For binaries, there exists an intricate transformation which turns the
equation for $\psi$ into an effective one-body Dirac equation. The
corresponding transformation for $\chi$ is derived here.

\hfill PACS numbers: 03.65Pm, 12.39.Ki\hfill
\end{abstract}
\maketitle
\begin{section}{Introduction}
Some years ago, an eight-component equation has been derived for the boud
states of two fermions, in the first place for fermions with small
anomalous magnetic moments \cite{1}. Similar equations have recently been
found for $n$ fermions, which have $4^n/2$ components instead of the $4^n$
components of Dirac-Breit and Bethe-Salpeter equations \cite{2}.
By eliminating half of the Dirac components, one finds that the equations
actually have the square of the total cms energy $E$ as eigenvalues, not $E$
itself. Finally, time-dependent forms have been found in which $E^2$ is
replaced by $-\partial_t^2$ ($\partial_t=\partial/\partial_t$, units
$\hbar=c=1$), precisely as in a Klein-Gordon equation \cite{3}. It has
also been shown that $-\partial_t^2$ is necessary for the relativistic recoil
in radiative decays \cite{4}. The energy of a photon emitted in the transition
from an atomic state of cms energy $E$ to the ground state of energy $E_{gr}$ in
its own cms (!) is
\beg{1}\omega=(E^2-E_{gr}^2)/2E.\en
The deviations from the recoil-free $\omega=E-E_{gr}$ are so small in atoms
that they can be treated as nonrelativistic corrections. Although this paper
uses atomic notation, the $E^2$-spectrum must apply also to mesons
and baryons. In Sect.4, (\ref{1}) will be
generalized to pion emission by the baryon ``resonances'', which via unitarity
and analyticity also affects the baryon mass splittings themselves, 
as a kind of ``pionic
Lamb shift''. 

The linear, $4^n$-component equations $i\partial_t\psi=H\psi$ contradict relativity 
except for $n=1$, where the Dirac equation has an equivalent two-component
Kramers version \cite{5,6,7}. As the present formalism assumes the
conservation of energy and momentum, the only relevant equation is that of a
free electron, which is known to imply a free Klein-Gordon equation. However,
a discrete electron spectrum does appear in Landau levels, where the energies of
the emitted photons also follow (\ref{1}) \cite{3} (the relativistic Landau
levels are equidistant in $E^2$).

The Kramers equation is recalled here because it displays an unusual property
of all equations with only $4^n/2$ components: it is not hermitian. For $n=1$,
this follows simply from the elimination of components of the Dirac
equation. Decomposition of the Dirac spinor $\psi_D$ into components $\psi_r$
and $\psi_l$ of chirality $+ 1$ and $-1$ (eigenstates of the chirality matrix $\gamma^5$)
produces two coupled equations. With the momentum operators $\pi^\mu=p^\mu+eA^\mu$,
the Dirac equation assumes the form
\beg{2}(\pi^0-\bld{\sigma\pi})\psi_r=m\psi_l,\quad(\pi^0+\bld{\sigma\pi})\psi_l=m\psi_r,\en
where the $\bld\sigma$ are Pauli matrices. Using the first equation for the
elimination of $\psi_l$, one obtains a second-order equation for $\psi_r$ alone:
\beg{3}(\pi^0+\bld{\sigma\pi})(\pi^0-\bld{\sigma\pi})\psi_r=m^2\psi_r.\en
The product of operators in  (\ref{1}) is not hermitian; the hermitian conjugate operator 
appears in the equation for $\psi_l$. The expectation value of its
antihermitian component vanishes.

For $n>1$, one may similarly derive the second-order equations from the
Dirac-Breit equations \cite{8}. However, these derivations are only
approximate. They fail for the inclusion of vacuum polatization and contain incorrect
squares of Breit operators which are normally eliminated by ``positive-energy projectors''.
The correct derivation of the interaction uses instead the Born series
of quantum electrodynamics (QED). It produces a non-hermitian operator $M^2$,
which occurs in two equivalent equations,
\beg{4}\Box\psi=M^2\psi,\quad\Box\chi=(M^2)^\dagger\chi,
\quad\Box=\nabla_R^2-\partial_{t,lab}^2.\en
$\psi$ and $\chi$ have total chirality $+1$ and $-1$, but $M^2$ does not
factorize as in (\ref{3}). 

The Kramers equation (\ref{3}) is invariant under a somewhat complicated
parity transformation, $\psi_r'=m^{-1} (\pi^0{}'+\bld{\sigma\pi{}'})\psi_r(-\bld
r)$, which is another way of exchanging $\psi_r(\bld r)$ with $\psi_l(-\bld r)$
(in Dirac notation, this is simply $\psi_D'=\beta\psi_D(-\bld r)$, where
$\beta=\gamma^0$ is the matrix which exchanges $\psi_r$ with $\psi_l$).
As $M^2$ does not factorize for $n=3$, that equation is not parity
invariant: $\psi$ must be exchanged with $\chi$. For
even $n$, however, the total chirality operator $\gamma^5_{tot}=\gamma^5_1
\dots\gamma^5_n$ commutes with the parity operator
$\beta=\beta_1\dots\beta_n$, such that the separation into $\psi$ and $\chi $
is parity invariant. The binary equation ($n=2$) is rederived in the next section.
The derivation of the corresponding equation for $\chi$ is new.

Difficulties with the $4^n$-component formalism have promoted a very different
approach for $n=2$, namely nonrelativistic expansions of the QED and QCD (quantum
chromodynamics) field couplings (``NRQED'',  ``NRQCD''), by which one can
calculate the energy levels
to a certain order in $\alpha=e^2$, using perturbation theory of the two-body
Schr\"odinger equation, with the (nonrelativistic) reduced mass, $H=\bld
p^2/2\mu_{nr}+V,\;V=-\alpha/r_{12}$
\cite{9}. The method ends presently at second order perturbation theory, which
gives the energy levels to the order $\alpha^6$. Encouraged by this success,
an old claim \cite{10} has been revived that relativistic equations for fixed numbers of
particles cannot exist. It is true that the form (\ref{2}) of the Dirac equation
is not exact, but there exists a form with a more precise $g$-factor,
$g_e=2+\alpha/\pi+\dots$. The corrections to $g_e=2$ are calculated by the
relativistic Feynman rules with a precision that remains to be reached by
NRQED. The leading $\alpha^8$-term of the muonium (e$^-\mu^+$)
hyperfine structure follows from first order perturbation theory of the Dirac equation,
and for uranium atoms, the equation can be solved numerically if necessary.
\end{section}
 \begin{section}{The binary equation}
The equation is best constructed by first adding the Dirac Hamiltonians for
two free fermions, then by reducing the result in the cms ($\bld p_1+\bld
p_2=0$) to an eight-component equation, and finally by adding the interaction
as the Fourier transform of the Born series of elastic cms scattering, after
it has also been reduced to an $8\times8$ matrix \cite{1}. However, this
interaction has not yet been found from the scattering of three particles. We
therefore use the less precise Dirac-Breit equation as a starting point
\cite{8}. It also contains the sum $H_0$ of the free Dirac Hamiltonians. With
$\bld \alpha_i=\gamma^5_i\bld\sigma_i$,
\beg{5}H_0=m_++\Sigma_i\gamma^5_i\bld\sigma_i\bld p_i,\quad m_+=
\Sigma_im_i\beta_i.\en
The zero-component of the total four-momentum operator enters as
\beg{6}\pi^0_{tot}=i\Sigma_i\partial t_i-\Sigma_{i<j}V_{ij},\quad
V_{ij}=q_iq_j/r_{ij},\quad q_1=-e.\en
The only other interaction is provided by the Breit operator $H_B$, to be
specified below:
\beg{7}(\pi^0_{tot}-\Sigma_i\gamma^5_i\bld\sigma_i\bld p_i-H_B)\psi_{DB}=m_+\psi_{DB}.\en
The equation is now split into two equations for the components $\psi^{(+)}$
and $\psi^{(-)}$ of $\psi_{DB}$ which have total chirality $+1$ and $-1$, respectively:
\beg{8}(\pi^0_{tot}-\Sigma_i\gamma^5_i\bld\sigma_i\bld p_i-H_B)\psi^{(+)}=m_+\psi^{(-)},\en
\beg{9}(\pi^0_{tot}-\Sigma_i\gamma^5_i\bld\sigma_i\bld p_i-H_B)\psi^{(-)}=m_+\psi^{(+)},\en
In (\ref{8}), $\Pi_i\gamma^5_i=1$ is understood, and in (\ref{9}), $\Pi_i\gamma^5_i=-1$.

For $n=2$, one may define $\gamma^5_1=\gamma^5$, which implies $\gamma^5_2
=+\gamma^5$ and $-\gamma^5$ in (\ref{8}) and (\ref{9}), respectively. The
Breit operator may be written as 
\beg{10}H_B=\gamma^5_1\gamma^5_2b,\quad b=-\kalb V_{12}(\bld\sigma_1\bld\sigma_2+
\sigma_{1r}\sigma_{2r}),\en
with $\sigma_{ir}=\bld{r\sigma_i}/r$ and $\bld r=\bld r_1-\bld r_2$. This
leads to the following forms in (\ref{8}) and (\ref{9}):
\beg{11}[\pi^0_{tot}-\gamma^5(\bld\sigma_1\bld p_1+\bld\sigma_2\bld p_2)-b]
\psi^{(+)}=m_+\psi^{(-)},\en
\beg{12}[\pi^0_{tot}-\gamma^5(\bld\sigma_1\bld p_1-\bld\sigma_2\bld p_2)+b]
\psi^{(-)}=m_+\psi^{(+)}.\en
Elimination of $\psi^{(-)}$ by means of the first equation gives an
equation for $\psi^{(+)}$ alone.  With the abbreviations
\beg{13}\bld\sigma_1+\bld\sigma_2=\bld\sigma,\quad
\bld\sigma_1-\bld\sigma_2=\Delta\bld\sigma,\quad\bld p=\kalb(\bld p_1-\bld p_2),
\quad\bld P=\bld p_1+\bld p_2,\en
the equation becomes
\beg{14}[\pi^0_{tot}-\gamma^5(\bld\sigma\bld p+\kalb\Delta\bld{\sigma P})+b]
m_+^{-1}[\pi^0_{tot}-\gamma^5(\Delta\bld\sigma\bld p+\kalb\bld{\sigma P})-b]
\psi^{(+)}=m_+\psi^{(+)}.\en
For $\bld p_1^2=\bld p_2^2$, the product of the two inner brackets
follows from the original expressions as
\beg{14a}(\bld\sigma_1\bld p_1)^2-(\bld\sigma_2\bld p_2)^2
=\bld p_1^2-\bld p_2^2=0.\en  
This cancellation reduces the second-order differential operator to a first-order
one, which has no longer the product form. The stationary solution 
have $\pi_{tot}^0=E-V,\;V=V_{12}$. The
$V^2$ of $(E-V)^2$ is eliminated by a shift of the distance,
$r_{12}=r+\alpha/2E$ \cite{11}. This entails a change in the angular part of
$b$, which is canceled by the ``retardation'' part of the Breit
operator. There remains the unretarded Gaunt interaction,
\beg{15}b_G=-V\bld\sigma_1\bld\sigma_2.\en
The $b^2$ does not cancel, but it must be omitted anyway.
One reason is that $b$ is expected to be
valid as a first-order perturbation only. The decisive fact, however, is that $b^2$ is
as large as $V^2$ and leads to wrong results \cite{12}.
In the standard Dirac-Breit equation,  $b^2$ is eliminated by
``positive-energy projectors''. In the QED derivation, the interaction of
(\ref{14}) arises from the first Born term of elastic scattering,
\beg{16}T_{if}^{(1)}=-4\pi q_1q_2j_1^\mu g_{\mu\nu}j_2^\nu/(-\bld q^2),\quad 
j_i^\mu=\overline\psi_i'\gamma^\mu_i\psi_i,\en
after its reduction to an $8\times 8$ matrix. The second Born term
$T_{if}^{(2)}$ does not
contribute to $V^2$, i.e. all $V^2$-terms including $b^2$ are absent.
The consistent equation derived from (\ref{14}) is thus
\beg{17}[E^2-2EV-(\bld\sigma\bld p+\kalb\Delta\bld\sigma\bld P)
(E-V_-)\tilde\gamma^5-(E-V_+)\gamma^5(\Delta\bld\sigma\bld p+\kalb\bld{\sigma P})
]\psi^{(+)}=m_+^2\psi^{(+)},\en
\beg{18}V_\pm=V(1\pm\bld\sigma_1\bld\sigma_2),\quad\tilde\gamma^5
=m_+\gamma^5/m_+.\en
The algebra of Pauli matrices gives
\beg{19}(\bld\sigma_1\bld\sigma_2)\bld\sigma_1=\bld\sigma_2+i\bld\sigma^\times,\quad
(\bld\sigma_1\bld\sigma_2)\bld\sigma_2=\bld\sigma_1-i\bld\sigma^\times,\quad
\bld\sigma^\times=\bld\sigma_1\times\bld\sigma_2.\en   
Next, we use the cms, $\bld P\psi=0$.
The remaining combinations of Pauli matrices are
\beg{20}\bld\sigma(1-\bld\sigma_1\bld\sigma_2)=0,\quad
(1+\sigma_1\bld\sigma_2)\Delta\bld\sigma=2i\bld\sigma^\times.\en
As a result, (\ref{17}) is simplified as follows:
\beg{21}[E^2-2EV-E\bld p(\tilde\gamma^5\bld\sigma+\gamma^5\Delta\bld\sigma)
+2i\gamma^5V\bld{p\sigma}^\times]\psi^{(+)}=m_+^2\psi^{(+)}.\en

The $\tilde\gamma^5$ of (\ref{17}) satisfies the same Dirac algebra as $\gamma^5$,
but it is not hermitian:
\beg{22}(\tilde\gamma^5)^2=1,\quad\{\tilde\gamma^5,\beta\}=0,\quad
\beta\equiv\beta_1\beta_2.\en
Notice that $\beta_1$ and $\beta_2$ occur in $m_+^2$ in the combination
\beg{23}m_+^2=m_1^2+m_2^2+2m_1m_2\beta.\en

There exists a very special transformation which makes $\tilde\gamma^5$
hermitian,
\beg{24}\psi^{(+)}=C_1\psi,\quad C_1=(m_+m_-)^{-1/2}(m_2\beta_2+m_1\beta_1
P_{spin}),\en
\beg{25}m_\pm=m_1\beta_1\pm m_2\beta_2,\quad P_{spin}=\kalb(1+
\bld\sigma_1\bld\sigma_2).\en
The spin exchange operator $P_{spin}$ is diagonal in the triplet-singlet spin
basis, with eigenvalues $+1$ and $-1$, respectively. As $\bld\sigma$ operates
only between triplet states, one may here set $P_{spin}=1,\; C_1=(m_+/m_-)^{1/2}$.
Both $\Delta\bld\sigma$ and $\bld\sigma^\times$ are antidiagonal in this basis,
which implies opposite signs of $P_{spin}$ to their right and left,
$C_1\Delta\bld\sigma =\Delta\bld\sigma C_1^{-1}$.
Multiplication by $C_1^{-1}$ from the left and using $C^{-1}_1\gamma^5=\gamma^5C_1$
 thus removes the tilde from 
$\tilde\gamma^5\bld\sigma$, without affecting the other operators.
The result is a one-body Dirac equation with somewhat different units,
\beg{26}[E^2-m_1^2-m_2^2-2EV-2E\gamma^5\bld {p\sigma}_1+
2i\gamma^5V\bld p\bld\sigma^\times]\psi=2m_1m_2\beta\psi.\en
It may be compared with the nonrelativistic one-body reduction, which uses
$\bld p_1^2/2m_1+
\bld p_2^2/2m_2=\bld p^2/2\mu_{nr}$. In both cases, the reduction breaks down in
the presence of a magnetic field, but not for an electric dipole field. After
division by $2E$, (\ref{26}) takes a familiar form,
\beg{26a}(\epsilon-V-\gamma^5\bld {p\sigma}_1-V_{hf})\psi=\mu\beta\psi,
\quad\mu=m_1m_2/E,\en
where $V_{hf}=-i\gamma^5V\bld p\bld\sigma^\times/E$ is a hyperfine operator.
The solutions of (\ref{26}) have been discussed extensively  \cite{8}. They
are symmetric under the exchange $m_1\leftrightarrow m_2$ and contain the
signs of $m_1$ and $m_2$ only in the combination $m_1m_2$. The sum
$m_{12}=m_1+m_2$ which occurs in the ²''external field approximation'' \cite{13}
gets replaced by $E$, as in the reduced mass $\mu$ of (\ref{26a}). The only
exception is another factor $(m_1-m_2)$, in which case even powers of $m_1^2-m_2^2$
are allowed, as in the ``Salpeter shift''. The first-order perturbative energy
shift $\delta E^2=2E\delta E$  contains only even powers of $E$.


An extra factor $P_{12}$
in front of $C_1$ in (\ref{24}) exchanges $\bld\sigma_1$ with $\bld\sigma_2$. 
The $E^2$-dependence appears after a rescaling of $\bld r=\bld r_1-\bld r_2$:
\beg{27}\bld r_E=\bld r/E,\quad \bld p_E=E\bld p=-i\nabla_{\bldi rE},\quad V_E=-\alpha/r_E.\en
The time-independent version of $M^2$ in (\ref{4}) is thus
\beg{28}M^2_0=m_+^2+2V_E+2\gamma^5\bld  p_E\bld\sigma_1+2i\gamma^5V_E\bld p_E
\bld\sigma^\times/E^2.\en
The spinor notation has been modified in this paper; the $\psi^{(+)}$
and $\psi^{(-)}$ of (\ref{8}) and (\ref{9}) were originally called $\psi$ and $\chi$; the
$\psi$ of (\ref{26}) was called $\psi_1$. This entailed an error in the
definition of the $\chi$ entering (\ref{4}) \cite{3}, which will now be
corrected:

As in the Kramers equation for $\psi_l$, the equation for $\psi^{(-)}$ follows
from (\ref{14}) by exchange of the two brackets. The resulting changes in
(\ref{4}) are the replacement of $V\bld p$ by $\bld p V$ and the exchange of
$\gamma^5$ with $\tilde\gamma^5$. As explained above, the tilde can only be
removed from the combination $\tilde\gamma^5\bld\sigma$. This requires an
extra transformation,
\beg{29}\psi^{(-)}=m_+\tilde\psi^{(-)}.\en
Multiplication by $m_+^{-1}$ from the left now replaces $\gamma^5\bld\sigma$
by $m_+^{-1}\gamma^5m_+\bld{\sigma}=(\tilde\gamma^5)^\dagger\bld\sigma$.
The $(\tilde\gamma^5)^\dagger$ is tranformed to $\gamma^5$ by the inverse
of (\ref{24}),
\beg{30}\tilde\psi^{(-)}=(m_+m_-)^{-1/2}(m_2\beta_2-m_1\beta_1P_{spin})\chi=
C_1^{-1}\chi.\en
As a  result, $\chi$ now obeys the hermitian conjugate of (\ref{26})
\beg{31}[E^2-m_1^2-m_2^2-2EV-2E\gamma^5\bld {p\sigma}_1-
2i\gamma^5\bld p\bld\sigma^\times V]\chi=2m_1m_2\beta\chi.\en
This guarantees identical eigenvalues $E^2$ of (\ref{26}) and (\ref{31}).

The QED Born series is now brought into the form of an $8\times8$ matrix,
\beg{32}T_{if}=\tilde\psi^{(-)}_f{}^\dagger T_8\psi^{(+)}_i
=\psi^{(-)}_f{}^\dagger m_+^{-1}T_8\psi^{(+)}_i,\en
where the indices $i$ and $f$ refer to in- and outgoing free two-body states
in the cms. The $m_+^{-1}$ arises from (\ref{29}); it was previously justified
only by the successful construction of the interaction from $T_8$ \cite{1}.

Turning now to the quarkonium model of heavy mesons, it seems clear that the
calculation of energy levels from perturbative QCD should also profit from the
binary formulation. A major improvement is the strict absence of special
retardation operators. In principle, the equation could be solved numerically
when the coupling becomes too strong, as in the case of uranium atoms.
However, QCD calculations on a space-time lattice \cite{14} show essential
deviations from this picture.
For the light mesons, they indicate an $E^2$-spectrum which is
linear in $m_i$ in the limit $m_i\to0$, $E^2=B_0(m_1+m_2)$. 
As the invariance of (\ref{31})
under the sign change of any of the $m_i$ follows directly from the QCD
Lagrangian, the terms $B_{0i}m_i$  require that $B_{0i}$ automatically changes sign
with $m_i$. Simply adding such terms in (\ref{26}) is forbidden. A valid
extension of the quarkonium model to the light mesons seems not possible.

On the other hand, one may obtain terms like $B_{0i}m_i$
by using different right- and lefthanded masses in the free single-particle Dirac equations,
$m_i^2=m_{ir}m_{il}$. Taking for example $m_{1r}=m_{2r}=
B_0\gg m_{il}$, one obtains $m_+^2=B_0[m_{1l}+m_{2l}+2(m_{1l}m_{2l})^{1/2}\beta]$.
This gives the desired result only for $\langle\beta\rangle=0$. If on
the other hand one adds the Dirac operators of particles 1 and 2 without using
the explicitly parity-invariant forms \cite{3,8}, $m_+^2$ becomes
\beg{32a}m_+^2=B_0(m_{1l}+m_{2l})(1+2\beta).\en
A more explicit connection of this trick with broken chiral symmetry is still
missing, however.
\end{section}

\begin{section}{Ternary equations}
The derivation of equations for $n=3$ uses again the Dirac-Breit equation
(\ref{7}), for which there is presently no alternative. The $H_B$ of
(\ref{10}) is replaced by three Breit operators,
\beg{33}H_{Bi}=\gamma^5_i\gamma^5_jb_{ij},\quad b_{ij}=
-\kalb V_{ij}(\bld\sigma_i\bld\sigma_j+\sigma_{ir}\sigma_{jr}),\en
In (\ref{8}) and (\ref{9}), one may then replace $\gamma^5_i\gamma^5_j$ by 
$\gamma^5_k$ and $-\gamma^5_k$ respectively, with $k\neq i\neq j$. The product (\ref{14}) 
is replaced by
\beg{34}[\pi^0_{tot}-\Sigma_i\tilde\gamma^5_i\bld\sigma_i\bld p_i+\Sigma_{i<j}
\tilde\gamma^5_k
b_{ij}][\pi^0_{tot}-\Sigma_i\gamma^5_i\bld\sigma_i\bld p_i-\Sigma_{i<j}\gamma^5_k
b_{ij}]\psi^{(+)}=m_+\psi^{(+)}.\en
\beg{35}\tilde\gamma^5_i=m_+\gamma^5_i/m_+=\gamma^5_i(1-2m_i\beta_i/m_+),\en
A useful combination is 
\beg{36}\tilde\gamma^5_i+\gamma^5_i=2\gamma^5_im_{jk+}/m_+,\quad m_{jk+}=
m_j\beta_j+m_k\beta_k.\en
Several simplifications arise from $[\bld p_k, V_{ij}]=0$, for example
\beg{37}[E-V_{12}+\tilde\gamma^5_3b_{12}]\gamma^5_3\bld\sigma_3\bld p_3+
\tilde\gamma^5_3\bld\sigma_3\bld p_3[E-V_{12}-\gamma^5_3b_{12}]=2
(E-V_{12})\gamma^5_3\bld\sigma_3\bld p_3m_{12+}/m_+.\en

The generalization of (\ref{28}) to $\bld r_{ij}$ brings also (\ref{34})
into a form with only even powers of $E$.

The number of remaining products is still large and becomes familiar only 
in one of the static limits, for example for $m_3\gg m_1+m_2$. This limit has
\beg{37a}\tilde\gamma^5_1=\gamma^5_1,\quad\tilde\gamma^5_2=\gamma^5_2,
\quad\tilde\gamma^5_3=-\gamma^5_3.\en
The operator (\ref{37}) vanishes in this limit, and the Dirac-Breit operator
for helium-like systems appears as expected.
\end{section}

\begin{section}{The energies of emitted pions and photons}
The energy of a particle emitted from a closed system of total cms energy $E$
follows from the conservation of momentum and energy alone, without any
knowledge of the interaction. The particle will be called ``pion'', with
$E_\pi^2-\bld P_\pi^2=m_\pi^2$. Photons are included by setting $m_\pi=0$.

The final state of the emitting system will be called ``ground state'', with
$E_{gr,lab}^2-\bld P_{gr}^2=E_{gr}^2$. The decay leaves the ground state with
a recoiling momentum $\bld P_{gr}=-\bld P_\pi$, such that $E_{gr,lab}$ exceeds
$E_{gr}$ which refers to the new cms. With $\bld P_{gr}^2=\bld P_\pi^2$,
one finds $E_{gr,lab}^2-E_{gr}^2=E_\pi^2-m_\pi^2$, or $(
E_{gr,lab}-E_\pi)(E_{gr,lab}+E_\pi)=E_{gr}^2-m_\pi^2$. Energy conservation requires
$E_{gr,lab}+E_\pi=E$, which then gives $E_{gr,lab}$ and $E_\pi$ separately:
\beg{38}E_{gr,lab}=(E^2+E_{gr}^2-m_\pi^2)/2E,\quad E_\pi=
(E^2-E_{gr}^2+m_\pi^2)/2E.\en
The second equation reduces to (\ref{1}) for $m_\pi=0$.

It will now be shown that this consequence of momentum and energy
conservation follows from (\ref{4}) using first order time-dependent Dirac
perturbation theory. $\Box$ is Lorentz transformed to the cms, where it
reduces to $-\partial_t^2$ ($t=$ cms time). Separation of a factor $e^{-iEt}$
from $\psi$ and $\chi$ gives
\beg{39}-\partial_t^2e^{-iEt}=e^{-iEt}(E^2+2iE\partial_t-\partial_t^2).\en 
The second-order derivative is neglected, and after a time scaling,
\beg{40}t=E\tau,\quad E\partial_t=\partial_\tau,\en
Dirac's perturbation theory for the $\tau$ dependence can be adopted. 
The result is $2EE_\pi=E^2-E_{gr}^2$, which gives the factor $(2E)^{-1}$ in (\ref{1}).
As both equations
(\ref{38}) contain $E_{gr}^2-m_\pi^2$, the time-independent equation for the
final state must be generalized to
\beg{41}M_0^2\psi_{gr}=(E_{gr}^2-m_\pi^2)\psi_{gr}.\en
Re-absorption of the pion or photon is treated accordingly. The residual
$E^2$-dependence of $M_0^2$ is not converted to $\partial_t^2$. In the
hyperfine operator of (\ref{28}) and in the running coupling constant
$\alpha_s(E^2)$ of QCD, one may take $E^2=E_{gr}^2$.
\end{section}
\begin{section}{Summary}
The relativistic two-body equation can be reduced to a one-body
equation, under the same conditions as the nonrelativistic one. In first-order
perturbation theory, the antihermitian part of its differential operator must be included
only for degenerate  states. The free eight-component spinors $\psi^{(\pm)}_i$ of the
Dirac equation are both needed for the construction of the interaction; the general
relation between between $\psi^{(-)}$ and the $\chi$ of (\ref{4}) is derived
in (\ref{29}) and (\ref{30}) as $\psi^{(-)}=m_+C_1^{-1}\chi$.
The statinary ternary equation has also $E^2$ as eigenvalues. This implies
that symmetry relations within baryon multiplets contain the squares of
masses, just as in meson mulitplets. The equation remains
complicated but becomes manageable in the static limit of helium-like states. 
The extension to states containing light quarks is questionable but perhaps
not impossible. However, the $E^2$-result follows already from Lorentz
invariance and should apply to all states.

For binaries, the absence of energy transfer between the
external fermion legs in the cms eliminates the extra retardation operators of
Breit \cite{14}. Corresponding simplifications are expected for the ternary
equation, but they remain to be derived. Finally, first-order perturbation theory
of the time-dependent equations (\ref{4}) gives simple expressions for the
energies of emitted particles which cannot be derived from first-order
time-dependent equations of the type $i\partial_t\psi=H\psi$.

The author would like to thank U. Nierste and A.G. Grozin for helpful comments.
\end{section} 
\begin{thebibliography}{99}
\bibitem{1}R. H\"ackl, V. Hund, and H. Pilkuhn, Phys. Rev. A 57, 3268 (1998)\\
\bibitem{2} H. Pilkuhn, Phys. Lett. B 614, 62 (2005)\\
\bibitem{3} H. Pilkuhn, Contemp. Physics 47, 11 (2006)\\
\bibitem{4} H. Pilkuhn, Phys. Rev. A 70, 044701 (2004)\\
\bibitem{5} H.A. Kramers, {\it The Foundations of Quantum Mechanics},
North-Holland, Amsterdam 1957\\
\bibitem{6} R.P. Feynman and M. Gell-Mann, Phys. Rev. 109, 193 (1958)\\
\bibitem{7} P.A. Chalmers and W. Siegel, Phys. Rev. D 59, 45012, 45013
(1999)\\
\bibitem{8} H. Pilkuhn, {\it Relativistic Quantum Mechanics} (Springer,
Berlin, 2. ed. 2005)\\
\bibitem{9} N. Brambilla et al., Rev. Mod. Physics 77, 1423 (2005)\\
\bibitem{10} D.G. Currie, T.F. Jordan and E.C.G. Sudarshan, Rev. Mod. Phys. 35,
350 (1963)\\
\bibitem{11} J. Schwinger, {\it Sources and Fields}. Addison-Wesley, Boston
1973.\\
\bibitem{12} G. Breit, Phys. Rev. 34, 553 (1929); 36, 383 (1930)\\ 
\bibitem{13} M.I. Eides et al., Physics reports 342, 63 (2001)\\
\bibitem{14} I. Montvay and G. M\"unster, {\it Quantum Fields on a Lattice},
Cambridge University Press (1994)\\ 

\end{thebibliography}
\end{document}
