%Title: Energy-square equations for relativistic closed systems.
%Author: Hartmut Pilkuhn
%Published: <EM> Physics Letters B </EM> <B> 614 </B> (2005) 62-65. 
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\begin{document}
\title{Energy-square equations for relativistic closed systems}
\author{Hartmut Pilkuhn}
\affiliation{Institut f\"ur Theoretische Teilchenphysik, Universit\"at,
D-76128 Karlsruhe, Germany}
\begin{abstract}{The most powerful relativistic two-body equations
have only half as many components as their Dirac-Breit and
Bethe-Salpeter counterparts and contain only the square of the total
cms energy. In this letter, the "energy-square" theorem is extended to
systems of $n>2$ Dirac particles. The case $n=3$ is discussed for baryons.}
\begin{center}{PACS numbers: 11.10.St, 11.30.Cp, 30.30.Ce, 31.30.Jv.}
\end{center}\end{abstract}\maketitle
\vspace{0.8cm}%----------------------------------------------------------------------
%\newpage
\begin{section}{Introduction}\label{ch1}
According to Einstein, a closed system of total energy $K^0$ moving
with total momentum $\bld K$ satisfies
$K^0{}^2-\bld K^2=constant$ (units $c=\hbar=1$ will be used). 
For a single atom in its ground state, the $constant$ is $M^2$, $M$
being the atomic mass. For all other atomic states, $M^2$ is better
called $E^2$, $E$ being the total cms energy. In relativistic quantum
mechanics, it is customary to calculate $E$ as the eigenvalue of a
differential equation $H_D\psi_D=E\psi_D$. The index $D$ stands for
Dirac. $H_D$ is the relativistic "Hamiltonian" for the system's
constituents, which for a closed system includes the nucleus.
It has been derived from quantum electrodynamics (QED) with ever
increasing precision. Its crudest version is the Dirac-Coulomb Hamiltonian
$H_{\rm DC}$, next comes the Dirac-Breit Hamiltonian $H_{\rm DB}$.
The method is presently extended to mesons as quark-antiquark bound states
and baryons as three-quark bound states, replacing QED by quantum
chromodynamics (QCD). The results for the short-range interaction are
encouraging. 

Mainly with this extension in mind it is pointed out here that the above
Einstein relation indicates a better approach to the fully
relativistic quantum mechanics
of closed systems. By reducing the S-matrix for the scattering of $n$ fermions
from $4^n\times4^n$ components to $4^n\times4^n/4$ components before
including the interaction, one derives equations $h\psi=\kalb
E^2\psi$ of a new "hamiltonian" $h$, 
with a small residual $E^2$-dependence in some terms of
$h$ [1]. In the residual terms, one may approximate $E^2$ by $M^2$.
From the point of view of QED and QCD, this "$E^2$-theorem" follows from
CPT invariance. For every atom in a level $E$, there should exist an
antiatom of exactly the same energy. In a formulation that dispenses with
creation and annihilation operators for atoms and antiatoms, the
antiatomic state appears as an atomic state of energy
$-E$. Multilocal fields for composite systems are presently used in
situations where their nonlocality is negligible, for example in the
derivation of Bose-Einstein and Fermi-Dirac statistics for mesons and
baryons. In particle theory, the theorem appeared first
in a current algebra derivation of the Gell-Mann -Okubo mass formula for
the light pseudoscalar mesons [2].

The problem is of course the construction of $h$ from the reduced S-matrix.
It has been solved for leptonium (e$^-$e$^+$ and e$^-\mu^+$)
[3,4]. For hydrogen, the inclusion of the proton's large anomalous
magnetic moment in the relativistic proton current $J_p^\mu$ is a mere
technicality. In this letter, we consider $n$ point
Dirac particles, including a "nucleus" of mass $m_n$ and kinetic
energy operator $\bld\alpha_n\bld p_n=\gamma^5_n\bld\sigma_n\bld p_n$, where
$\bld\sigma_i$ denotes the $2\times2$ Pauli matrix of particle $i$. We 
first show how $H_{\rm DC}$
can be converted to an equivalent $h_C$, for which the $E^2$-theorem
is proven. The corresponding conversion of $H_{\rm DB}$ to $h_B$
is straightforward, but the result is uninteresting in view of the
cancellations  expected from the reduced S-matrix approach. 
We shall thus only need the Gaunt part of the Breit operator between
particles $i$ and $j$, which combines with their Coulomb potential $V_{ij}$
into $V_{ij}(1-\gamma^5_i\gamma^5_j\bld\sigma_i\bld\sigma_j)$.
We shall also
include a vector potential $\bld A(\bld r)$, by taking the individual
momentum operators as
\beg{1}\bld\pi_i=\bld p_i+q_i\bld A(t,\,\bld r_i),\quad\bld p_i=-i\nabla_i,\en
with $q_i=-e$ for electrons and $+Ze$ for the "nucleus". The extension
to the radiation field $\bld
A(t,\bld r)=e^{i\omega t}\bld A(\bld r)$ shows that the time-dependent
perturbation theory based on
\beg{2}i\partial_\tau\psi= h\psi,\quad \tau=t/E\equiv t_E\en
reproduces the relativistic atomic recoil in radiative decays, while
the traditional  $i\partial_t\psi_D=H_D\psi_D$ fails to do so
[5]. After a summary of the two-fermion equation in section 3, the
case $n=3$ will be discussed in section 4.

The Dirac-Coulomb equation for $n$ idealized Dirac particles including
the nucleus can be put into the form
\beg{3}[\pi^0-\sum_{i=1}^n(m_i\beta_i+\gamma^5_i\bld\sigma_i\bld\pi_i)]
\psi_D=0,\quad \pi^0=E-V_{tot},\en
\beg{4}V_{tot}=\Sigma_{i<j}V_{ij},\quad V_{ij}=q_iq_j/|\bld
r_i-\bld r_j|,\en
with $\beta_i=\gamma^0_i$. For $n=1$, one has $V_{tot}=0$, and the cms
condition $\Sigma_i\bld p_i=0$ (for $\bld A=0$) leads to $E_1\psi_D=m_1\beta_1
\psi_D$. The construction of $h$ uses the chiral basis, in which
$\gamma^5$ and $\beta$ are diagonal and antidiagonal, respectively:
\beg{5}\psi_D=\left(\begin{array}{c}\psi_r\\ \psi_l\end{array}\right),\quad
\gamma^5=\left(\begin{array}{cc}1&0\\0&-1\end{array}\right),\quad
\beta=\left(\begin{array}{cc}0&1\\1&0\end{array}\right).\en
For a single free electron, elimination of the "lefthanded"
two-component spinor $\psi_l$ yields
$E_1^2\psi_r=m_1^2\psi_r$. This trivial example shows how the
construction of $h$ halves the number of components of $\psi_D$.
\end{section}
\begin {section}{The energy-square form of the Dirac-Coulomb equation}
\label{ch2}
The spinor $\psi_D$ of (\ref{3}) may be decomposed into two spinors
$\psi$ and $\chi$ which are eigenstates of the total chirality operator:
\beg{6}\gamma^5_{tot}=\Pi_{i=1}^n\gamma^5_i.\en
$(\gamma^5_i)^2=1$ implies $(\gamma^5_{tot})^2=1$, such that its
eigenvalues are $+1$ and $-1$:
\beg{7}\psi_D=\left(\begin{array}{c}\psi\\ \chi\end{array}\right),
\quad\gamma^5_{tot}\psi=\psi,\quad\gamma^5_{tot}\chi=-\chi.\en
As each $\beta_i$ anticommutes with $\gamma^5_{tot}$, it has only nondiagonal
elements in this basis. With the abbreviation $m_+=\Sigma_im_i\beta_i$, 
(\ref{3}) becomes
\beg{8}(\pi^0-\Sigma_i\gamma^5_i\bld\sigma_i\bld\pi_i
)\psi=m_+\chi,\quad(\pi^0-\Sigma_i\gamma^5_i\bld\sigma_i\bld\pi_i
)\chi=m_+\psi.\en
One of the $\gamma^5_i$ may be eliminated using (\ref{7}), say $\gamma^5_n$:
\beg{9}\gamma^5_n=\gamma^5_{tot}\gamma^5_e,\quad\gamma^5_e=
\Pi_{i\ne n}^n\gamma^5_i.\en
Choosing $n$ as the "nucleus" makes $\gamma^5_e$ the total electronic
chirality. Using now the first equation to eliminate $\chi$, the
second one produces
\beg{10}(\pi^0-\Sigma\gamma^5_i\bld\sigma_i\bld
\pi_i+\gamma^5_e\bld\sigma_n\bld\pi_n)m_+^{-1}(\pi^0-\Sigma\gamma^5_i 
\bld\sigma_i\bld\pi_i-\gamma^5_e\bld\sigma_n\bld\pi_n)\psi=m_+\psi,\en
where the summation symbol refers only to the electrons.

The factor $m_+^{-1}$ is removed by multiplying (\ref{10}) by $m_+$,
\beg{11}(\pi^0-\Sigma\gamma^5_i{}'\bld\sigma_i\bld
\pi_i+\gamma^5_e{}'\bld\sigma_n\bld\pi_n)(\pi^0-\Sigma\gamma^5_i 
\bld\sigma_i\bld\pi_i-\gamma^5_e\bld\sigma_n\bld\pi_n)\psi=m_+^2\psi,\en
\beg{12}\gamma^5{}'=m_+\gamma^5/m_+,\quad(\gamma^5{}')^2=1,\quad
\gamma_i^5{}'\beta_i+\beta_i\gamma_i^5{}'=0.\en
After a rescaling of variables, each bracket in (\ref{11})
contains only odd powers of $E$:
\beg{13}\bld r_i=E\bld r_{iE},\quad\bld p_i=\bld p_{iE}/E,\quad V_{tot}=E^{-1}
V_{tot}(\bld r_{iE}).\en
The new variables $\bld r_{iE}$ will also appear in orthogonality
relations. A constant magnetic field $\bld B$
provides $\bld A=\kalb B\times\bld r=\kalb E\bld
B\times\bld r_E$. Clearly, the product of the two brackets contains
only even powers of $E$, which proves the $E^2$-theorem for
$H_{\rm DC}$. The addition of Breit operators does not alter the conclusion.
In the new "hamiltonian" $h$ of (\ref{11}),
terms of order $V^2$ will now also be neglected: 
\beg{16}2h_C=m_+^2+2V_{tot,E}+\Sigma(\gamma^5_i+\gamma^5_i{}')\bld\sigma_i
\bld\pi_{iE}+(\gamma^5_e-\gamma^5_e{}')\bld\sigma_n\bld
\pi_{nE}+2h(\bld\pi_i,\bld\pi_j,\bld\pi_n).\en
The last term is not written down here because of its length.
It vanishes for $n=2$ and $\bld A=0$.

When particle $n$ is a nucleus with mass $m_n\gg m_e$, one defines $E^2=(m_n+
E_e)^2$, $E_e$ being the "electronic energy". Inserting
\beg{17}m_+^2=m_n^2+\Sigma_im_i^2+2m_n\Sigma_im_i\beta_n\beta_i+2\Sigma_{i<j}
m_im_j\beta_i\beta_j,\en
one may cancel the $m_n^2$ in the equation $(2h-E^2)\psi=0$,
and often also neglect $m_im_j=m_e^2$ relative to $m_nm_e$. The phase between
$\psi$ and $\chi$ in (\ref{7}) may be fixed by $\beta_n\psi=\chi$, $\beta_n
\chi=\psi$, which implies $\beta_n=1$ in (\ref{17}).
The resulting $(m_+^2-m_n^2)/2m_n$ is $\Sigma_im_e\beta_i$, in
agreement with the mass operator of the recoilless Dirac equation.

In the radiative decays of atoms or hadrons, the relativistic recoil appears
automatically in the time-dependent perturbation theory based on
(\ref{2}) with $\bld A=e^{i\omega t}\bld A(\bld r)$, using the
unperturbed solution [5] $h^0(\tau)=exp\{iE^2\tau\}h(\tau=0)$. If one
wants to use standard orthogonality relations for $h$, one must replace
$E^2$ by $M^2$ in the residual terms of $h$.

Relativistic recoil is small in atomic radiation but becomes
appreciable in the hyperon decay $\Sigma^0\to\Lambda\gamma$, and even
enormous in the meson decay $\omega\to\pi^0\gamma$. It shows that 
the energy-square theorem must apply even when the interaction
considered here is incomplete.
\end{section}

\begin{section}{Hydrogenic atoms}\label{ch3}
For $n=2$, $q_1=-e,\; q_2=Ze$ gives $V_{tot}=V(r)=-Ze^2/r$, $r
=|\bld r_1-\bld r_2|$.
One may also abbreviate $\gamma^5_e=\gamma^5_1\equiv\gamma^5$, and 
$\bld\sigma_1\bld\pi_1\pm\bld\sigma_2\bld\pi_2=\pi_\pm$. The resulting
$h$ is
\beg{18}2h=m_+^2+2EV+
[E-V(1+\bld\sigma_1\bld\sigma_2)]\gamma^5\pi_++\gamma^5{}'\pi_-[E-
V(1-\bld\sigma_1\bld\sigma_2)]+\gamma^5{}'\gamma^5\pi^+\pi^-.\en
For $\bld A=0$, the cms condition $\bld p_1+\bld p_2=0$ allows one to put
$\bld p_1=\bld p,\;\bld p_2=-\bld p$, which results in $\pi_+\pi_-=0$.
The term $\pi_-V(1-\bld\sigma_1\bld\sigma_2)=\bld p(\bld\sigma_1+\bld\sigma_2)
(1-\bld\sigma_1\bld\sigma_2)V$ vanishes then also because of 
$(\bld\sigma_1+\bld\sigma_2)(1-\bld\sigma_1\bld\sigma_2)=0$.
The remainder gives, after a transformation $\psi=C_1\psi_1$,
\beg{19}h_1(\bld A=0)=\kalb m_+^2+\gamma^5\bld{\sigma}_1\bld
p_E+V_E+h_{hf},\en
\beg{20}h_{hf}=i\gamma^5\bld\sigma^\times V_E\bld p_E/E^2,
\quad\sigma^\times=\bld\sigma_1\times\bld\sigma_2.\en
An equivalent transformation $\psi=C_2\psi_2$ produces an $h_2$ which
has $\bld\sigma_1$ and $\bld\sigma_2$ exchanged [6].
(\ref{19}) is the complete hamiltonian to the order $\alpha^4$. The
$V^2$ expected from $\pi^{02}$ in (\ref{11}) is excluded, and so is the 
retardation
part of the Breit operator. The remaining Gaunt interaction
$-V\bld\alpha_1\bld\alpha_2$ produces $h_{hf}$. It may be remembered
that $V(1-\bld\alpha_1\bld\alpha_2)$ is the Fourier transform of the one-photon
exchange amplitude in the cms, $T^{(1)}=\gamma^\mu_1g_{\mu\nu}\gamma^\nu_2/\bld
q^2$, in which retardation is excluded by $q^0=0$. There exist of
course two-photon exchange potentials and Bethe logarithms
(calculated from time-ordered
products of fields as usual), their largest contributions to $E^2$
being of the order $\alpha^5\log\alpha$ and $\alpha^5$.
 
Finally, a word about parity invariance. For $n=2$, the transformation 
$\psi_D^{\it P}(\bld r_1,\bld r_2)=\beta_1\beta_2\psi_D(-\bld
r_1,-\bld r_2)$ 
is consistent with the decomposition (\ref{7}), as $\beta_1\beta_2$
commutes with $\gamma^5_1\gamma^5_2$. The decomposition is also parity
invariant for $n=4$, but not for $n=1$ and $n=3$. For such cases, the notation
$\psi=\psi_r,\,\chi=\psi_l$ of the single-particle Dirac equation in
the chiral basis (\ref{5}) is appropriate. For $n=1$, the parity
transformation uses the linear relation between $\psi_r$ and $\psi_l$. The
corresponding relation exists also for the n-body Dirac-Coulomb
equation (\ref{8}) and its second-order form (\ref{11}), 
but it disappears after the cancellation
of $[V(...)]^2$. The parity transformation involves then two separate
spinors $\psi$ and $\chi$, which are related only by the fact that $\chi$
satisfies the hermitian adjoint equation of $\psi$, with the same real
eigenvalue. This feature appears also for $n=2$ with a spinless
particle 2; it caused considerable concern in an early paper [7].
\end{section}
\begin {section}{baryons}\label{ch4}
In the quark model, baryons are taken as three-fermion bound
states. Their shortrange interactions should follow the QED formalism,
with a one-gluon exchange potential V (including vacuum polarization)
and the corresponding Gaunt interaction dominating at short
distances. A more complete shortrange interaction will have to use the
three-body S-matrix [8]. As $V+V_G$ is a sum of two-fermion operators, we
assume instead the type of cancellations that occur for $n=2$: No retardation,
and no terms of order $V^2$. The Gaunt interaction may be included in
the definition of $\pi^0$, similar to (\ref{16}). A separate treatment
of a "nucleus" is not necessary. With $\gamma^5_i\gamma^5_j\psi=\gamma^5_k\psi
\;(i\ne j\ne k)$ and $\gamma^5_i\gamma^5_j\chi=-\gamma^5_k\chi$, one
must distinguish between a $\pi^0_r$ in the right bracket of
(\ref{10}), and a $\pi^0_l$ in the left one:
\beg{20a}(\pi^0_l-\Sigma_i\gamma^5_i\bld\sigma_i\bld\pi_i)m_+^{-1}(
\pi^0_r-\Sigma_i\gamma^5_i\bld\sigma_i\bld\pi_i)\psi=m_+\psi,\en
\beg{21}\pi^0_{r,l}=E-\Sigma_{i<j}V_{ij}(1\mp \gamma^5_k\bld\sigma_i
\bld\sigma_j).\en
In the form (\ref{11}), $\pi^0_l$ has $\gamma^5_k$ replaced by $\gamma^5_k{}'$.
With the above cancellations assumed, one obtains the "Coulomb-Gaunt"
hamiltonian,
\beg{22}h_{CG}=h_C+\kalb\Sigma_{i<j}[V_{ij,E}\bld\sigma_i\bld\sigma_j
\gamma^5_k{}'(1-\Sigma_i\gamma^5_i\bld\sigma_i\bld\pi_{iE}/E^2)-(1-
\Sigma_i\gamma^5_i{}'\bld\sigma_i\bld\pi_{iE}/E^2)V_{ij}\bld\sigma_i\bld
\sigma_j\gamma^5_k].\en
For $n=2$, the transformation $\psi=C_1\psi_1$ was necessary to remove
the pole of $m_+^{-1}$ at $m_2=m_1$, which occurs at $\beta=-1$. For
$n=3$, a corrsponding transformation is unnecessary, as $m_1(\beta_1+\beta_2
+\beta_3)\ne 0$. The second operator in each bracket produces new
hyperfine operators, which are not written out here:
\beg{23}h_{CG}=h_C+h_G'+h_{hf},\quad h_G'=\Sigma_{i<j}V_{ij,E}
\bld\sigma_i\bld\sigma_j\kalb(\gamma^5_k{}'-\gamma^5_k),\en
\beg{24}\kalb(\gamma^5_k{}'-\gamma^5_k)=\gamma^5_km_k\beta_k(m_1\beta_1+
m_2\beta_2+m_3\beta_3)^{-1}.\en
When $k$ represents a heavy quark ($m_k\gg m_i,\,m_j$), $h_G'$ is 
particularly large; it replaces essentially the Breit
operator of the two light quarks.
\end{section}
\begin {section}{References}\label{ch5}
1. H. Pilkuhn , {\it Relativistic Quantum Mechanics} 
(Springer, Berlin 2003)\\
2. M. Gell-Mann, R. Oakes and B. Renner, Phys. Rev. 175 (1968), 2195\\
3. R. H\"ackl, V. Hund and H. Pilkuhn, Phys. Rev. A 57 (1998), 3268 \\
4. V. Hund and H. Pilkuhn, J. Phys. B 33 (2000), 1617 \\
5. H. Pilkuhn, Phys. Rev. A 70 (2004), 044701\\
6. H. Pilkuhn, Phys. Letters B 512 (2001), 412\\
7. H. Pilkuhn, J. Phys. B 16 (1984), 4061\\
8. R. J. Eden et al, {\it The Analytic S-Matrix} (Cambridge Univ.
Press, Cambridge 1965)\\
\end{section}
\end{document}