$\displaystyle\langle{0}|\bar{q}(x)\gamma_{5}{\textstyle\frac{1}{2}}\tau^{i}q(x)\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{q}({\boldsymbol{x}}_{1})\Phi_{\pi}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\tau^{j}q({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
$\displaystyle=-iN\mathrm{Tr}\,(\gamma_{5}{\textstyle\frac{1}{2}}\tau^{i}\gamma^{0}\gamma_{5}\tau^{j}\gamma^{0})=4iN\,\delta^{ij}=-i\frac{M_{\pi}^{2}}{2m}\,f_{\pi}\,\delta^{ij}$

This determines the normalization of the pion wave function,

$\displaystyle\Phi_{\pi}({\boldsymbol{x}})=i\,\frac{M_{\pi}^{2}}{8m}\,f_{\pi}\big{[}J_{0}({\textstyle\frac{1}{4}}V^{\prime}r^{2})+i\,{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}\,J_{1}({\textstyle\frac{1}{4}}V^{\prime}r^{2})\big{]}\gamma_{5}$
(520)

####  VIII.6.2 Finite quark mass $m_{u}=m_{d}=m\neq 0$

The pion becomes massive ($M_{\pi}>0$) through the explicit breaking of chiral
symmetry by a non-vanishing quark mass $m\neq 0$. The $0^{++}$ $\sigma$ must
remain massless ($M_{\sigma}=0$) to ensure that its mixing with the vacuum
does not break Poincaré invariance. The wave function
$\Phi_{\sigma}({\boldsymbol{x}})$ which solves the BSE (506) with $M=0$ but
$m\neq 0$ has the structure of (510) with

$\displaystyle H_{1}(r)$
$\displaystyle=Ne^{-iV^{\prime}r^{2}/4}\Big{\\{}\Big{[}1-\frac{2m^{2}}{(V^{\prime}r)^{2}}\Big{]}L_{{\textstyle\frac{im^{2}}{2V^{\prime}}}-{\textstyle\frac{1}{2}}}\big{(}{\textstyle\frac{1}{2}}iV^{\prime}r^{2}\big{)}+\frac{2m^{2}}{(V^{\prime}r)^{2}}L_{{\textstyle\frac{im^{2}}{2V^{\prime}}}+{\textstyle\frac{1}{2}}}\big{(}{\textstyle\frac{1}{2}}iV^{\prime}r^{2}\big{)}\Big{\\}}=N\big{[}J_{0}({\textstyle\frac{1}{4}}V^{\prime}r^{2})+{\cal
O}\left(m^{2}\right)\big{]}$ $\displaystyle H_{2}(r)$
$\displaystyle=Ne^{-iV^{\prime}r^{2}/4}\Big{\\{}\Big{[}\frac{2}{V^{\prime}r^{2}}-i\Big{]}L_{{\textstyle\frac{im^{2}}{2V^{\prime}}}-{\textstyle\frac{1}{2}}}\big{(}{\textstyle\frac{1}{2}}iV^{\prime}r^{2}\big{)}-\frac{2}{V^{\prime}r^{2}}L_{{\textstyle\frac{im^{2}}{2V^{\prime}}}+{\textstyle\frac{1}{2}}}\big{(}{\textstyle\frac{1}{2}}iV^{\prime}r^{2}\big{)}\Big{\\}}=N\big{[}J_{1}({\textstyle\frac{1}{4}}V^{\prime}r^{2})+{\cal
O}\left(m^{2}\right)\big{]}$ $\displaystyle H_{3}(r)$
$\displaystyle=-\frac{2m}{V^{\prime}r}\,H_{2}(r)$ (521)

where the $L_{n}(x)$ are Laguerre functions.

Exercise A.24: Verify the expressions (VIII.6.2) for radial functions
$H_{1}(r),H_{2}(r)$ and $H_{3}(r)$.

The pion state in the rest frame is

$\displaystyle\left|{\pi,i}\right\rangle=\hat{\pi}_{i}\left|{0}\right\rangle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{q}({\boldsymbol{x}}_{1})\Phi_{\pi}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\tau^{i}q({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(522)

The pion wave function has the form given in (415), where $F_{3}=0$ for $j=0$.
With the notational change $F_{2}\to F_{2}/r$ this means

$\displaystyle\Phi_{\pi}({\boldsymbol{x}})=\Big{[}F_{1}(r)+i\,{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}\,F_{2}(r)+\gamma^{0}\,F_{4}(r)\Big{]}\gamma_{5}$
(523)

Using this in the bound state equation (506) and collecting terms with the
same Dirac structure we get the conditions:

$\displaystyle\gamma_{5}:\hskip 14.22636pt$
$\displaystyle-\frac{2}{r}F_{2}-F_{2}^{\prime}+mF_{4}={\textstyle\frac{1}{2}}(M_{\pi}-V)F_{1}$
$\displaystyle
i\,{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}\,\gamma_{5}:\hskip
14.22636pt$ $\displaystyle
F_{1}^{\prime}={\textstyle\frac{1}{2}}(M_{\pi}-V)F_{2}$
$\displaystyle\gamma^{0}\gamma_{5}:\hskip 14.22636pt$ $\displaystyle
mF_{1}={\textstyle\frac{1}{2}}(M_{\pi}-V)F_{4}$ (524)

Eliminating $F_{2}$ and $F_{4}$ gives the radial equation

$\displaystyle
F_{1}^{\prime\prime}+\Big{(}\frac{2}{r}+\frac{V^{\prime}}{M_{\pi}-V}\Big{)}F_{1}^{\prime}+\big{[}{\textstyle\frac{1}{4}}(M_{\pi}-V)^{2}-m^{2}\big{]}F_{1}=0$
(525)

For the regular solution $F_{1}(r\to 0)\sim r^{0}[1+{\cal
O}\left(r^{2}\right)]$, $F_{2}(r\to 0)\sim r$ and
$F_{4}(0)=F_{1}(0)\,2m/M_{\pi}$. Thus

$\displaystyle\Phi_{\pi}^{(0)}({\boldsymbol{x}}=0)=F_{1}(0)\Big{(}1+\frac{2m}{M_{\pi}}\gamma^{0}\Big{)}\gamma_{5}$
(526)

The superscript on $\Phi_{\pi}$ reminds that this is the rest frame wave
function. In a frame with ${\boldsymbol{P}}\neq 0$ the pion state takes the
form (455) at $t=0$. For a general time, suppressing color and Dirac indices,

$\displaystyle\left|{M_{\pi},i,{\boldsymbol{P}},t}\right\rangle=e^{-iP^{0}t}\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{q}(t,{\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\pi}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\tau^{i}\,q(t,{\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(527)

where $P^{0}=\sqrt{{\boldsymbol{P}}^{2}+M_{\pi}^{2}}$. The axial current
identities (518) and (519) probe the pion state at $t=x^{0}$ and
${\boldsymbol{x}}_{1}={\boldsymbol{x}}_{2}={\boldsymbol{x}}$, involving
$\Phi_{\pi}^{({\boldsymbol{P}})}(0)$. Since $V^{\prime}r=0$ at $r=0$ the frame
dependence of the wave function at the origin is that of a non-interacting
state, given by (459). The wave function of a pion with momentum
${\boldsymbol{P}}={\boldsymbol{\hat{\xi}}}\,M_{\pi}\sinh|\boldsymbol{\xi}|$ is
then, at the origin,

$\displaystyle\Phi_{\pi}^{({\boldsymbol{P}})}({\boldsymbol{x}}=0)$
$\displaystyle=e^{-\boldsymbol{\xi}\cdot{\boldsymbol{\alpha}}/2}\Phi_{\pi}^{(0)}({\boldsymbol{x}}=0)e^{\boldsymbol{\xi}\cdot{\boldsymbol{\alpha}}/2}=F_{1}(0)\Big{(}1+\frac{2m}{M_{\pi}}\gamma^{0}e^{\boldsymbol{\xi}\cdot{\boldsymbol{\alpha}}}\Big{)}\gamma_{5}=F_{1}(0)\Big{[}1+\frac{2m}{M_{\pi}^{2}}(\gamma^{0}P^{0}+{\boldsymbol{P}}\cdot\boldsymbol{\gamma})\Big{]}\gamma_{5}$
$\displaystyle\gamma^{0}\Phi_{\pi}^{({\boldsymbol{P}})}(0)\gamma^{0}$
$\displaystyle=-F_{1}(0)\Big{[}1+\frac{2m}{M_{\pi}^{2}}\not{P}\Big{]}\gamma_{5}$
(528)

The CSB matrix elements become

$\displaystyle\langle{0}|\bar{q}(x)\gamma^{\mu}\gamma_{5}{\textstyle\frac{1}{2}}\tau^{i}q(x)\,\left|{M_{\pi},j,{\boldsymbol{P}},x^{0}}\right\rangle$
$\displaystyle=\delta_{ij}e^{-iP\cdot
x}\,\mathrm{Tr}\,\big{\\{}\gamma^{\mu}\gamma_{5}\gamma^{0}\Phi_{\pi}^{({\boldsymbol{P}})}(0)\gamma^{0}\big{\\}}=\delta_{ij}e^{-iP\cdot
x}F_{1}(0)\,\frac{2m}{M_{\pi}^{2}}\,4P^{\mu}$
$\displaystyle\langle{0}|\bar{q}(x)\gamma_{5}{\textstyle\frac{1}{2}}\tau^{i}q(x)\,\left|{M_{\pi},j,{\boldsymbol{P}},x^{0}}\right\rangle$
$\displaystyle=\delta_{ij}e^{-iP\cdot
x}\,\mathrm{Tr}\,\big{\\{}\gamma_{5}\gamma^{0}\Phi_{\pi}^{({\boldsymbol{P}})}(0)\gamma^{0}\big{\\}}=\delta_{ij}e^{-iP\cdot
x}(-4)F_{1}(0)$ (529)

Comparing with the rhs. of relations (518) and (519) gives in both cases

$\displaystyle F_{1}(0)=i\,\frac{M_{\pi}^{2}}{8m}\,f_{\pi}$ (530)

in agreement with our previous result (520) for $m=0$.

I leave to the future a more comprehensive study of spontaneous chiral
symmetry breaking in the present context.

##  IX Bound state epilogue

I conclude with several remarks on bound states. The more subjective ones may
serve to stimulate further discussion on these topics.

Confinement is an essential aspect of hadrons, and has been demonstrated in
lattice QCD. Analytic approaches to confinement are often formulated in terms
of quark and gluon Green functions, aiming to show that colored states are
unphysical. This deals directly with the fundamental fields of the theory, and
benefits from the accumulated experience with (non-)perturbative methods for
local fields. A drawback is that one needs to prove that something does not
exist, with little guidance from data.

Here I try to approach confinement from the opposite direction, in terms of
the color singlet bound states that are the asymptotic states of QCD. A main
challenge is that bound states are extended objects, and more difficult to
deal with than the pointlike quarks and gluons. The experience gained from QED
atoms is valuable, even though it does not address confinement. Experts regard
atoms as “non-perturbative” Bodwin  _et al._ (1985), yet use PQED for precise
evaluations of binding energies. This cautions not to rush to judgement based
on the non-perturbative nature of hadrons.

The hadron spectrum has surprising “atomic” features, despite their large
binding energies. Why can hadrons be classified by their valence quarks and
$J^{PC}$ quantum numbers only Zyla  _et al._ (2020)? Their gluon and sea quark
constituents do not feature in the spectrum as clearly as they do in deep
inelastic scattering. Why do hadron decays obey the OZI rule Okubo (1963);
Zweig (1964); Iizuka (1966), e.g., favor $\phi(1020)\to K\bar{K}$ over
$\phi(1020)\to\pi\pi\pi$? What causes quark-hadron duality, which in various
guises pervades hadron dynamics? These features are pictured by dual diagrams
(Harari (1969); Rosner (1969); Zweig (2015) and Fig. 19 a) which show only
valence quarks, no gluons. We lack understanding based on QCD.

Simple features are precious: experience shows that they often have
correspondingly simple explanations. Taking the data at face value limits the
options in choosing the approach. An explanation based on QCD needs to use
perturbation theory, which is our only general analytic method. Perturbative
methods are successful for the atomic spectrum as well as for hard scattering
in QCD. Addressing bound states in motion with a perturbative expansion is
possible in QED and conceivable in QCD. Imposing restrictive conditions on an
explicit yet formally exact method may reveal the QCD solution, or the
theoretical inconsistency of such conditions.

Quarks and gluons do not move faster than light. This implies retarded
interactions between bound state constituents. In a Fock state expansion the
retardation is described by higher Fock states, with gluons “on their way”
between valence quarks. For the non-relativistic constituents of atoms photon
exchange is almost instantaneous, so retardation is a higher order,
relativistic correction. For light, relativistic quarks retarded interactions
would be expected to be prominent and higher Fock states significant. Yet data
suggests that most hadrons may be viewed as $q\bar{q}$ and $qqq$ states. Is it
conceivable that the valence quark Fock states dominate also for light
hadrons?

Gauge theories can, depending on the choice of gauge, have instantaneous
interactions. The absence of the $\partial_{t}A^{0}$ and
$\boldsymbol{\nabla}\cdot{\boldsymbol{A}}_{L}$ terms in the action means that
$A^{0}$ and the longitudinal ${\boldsymbol{A}}_{L}$ fields do not propagate in
time and space. Their values are determined by the gauge, which can be fixed
over all space at an instant of time. This is illustrated by a comparison of
the $A^{0}$ propagator in the Coulomb and Landau gauges,

$\displaystyle D_{C}^{00}(q^{0},{\boldsymbol{q}})$
$\displaystyle=i\,\frac{1}{{\boldsymbol{q}}^{2}}\hskip 73.97733pt\mbox{Coulomb
gauge:}\ \boldsymbol{\nabla}\cdot{\boldsymbol{A}}=0$ $\displaystyle
D_{L}^{00}(q^{0},{\boldsymbol{q}})$
$\displaystyle=-i\frac{1-(q^{0})^{2}/q^{2}}{q^{2}+i\varepsilon}\hskip
28.45274pt\mbox{Landau gauge:}\ \ \partial_{\mu}A^{\mu}=0$ (531)

The Coulomb gauge propagator is independent of $q^{0}$ and thus
$\propto\delta(t)$ after a Fourier transform $q^{0}\to t$. The gauge fixing
term $\propto(\partial_{\mu}A^{\mu})^{2}$ of Landau gauge adds the missing
derivatives $\partial_{t}A^{0}$ and
$\boldsymbol{\nabla}\cdot{\boldsymbol{A}}_{L}$ to the action, allowing all
components of $A^{\mu}$ to propagate. The free field boundary condition of the
perturbative $S$-matrix at $t=\pm\infty$ is also covariant, resulting in an
explicitly Poincaré invariant expansion of scattering amplitudes in terms of
Feynman diagrams in Landau gauge.

Bound states defined at an instant of time retain explicit symmetry only under
space translations and rotations (in the rest frame). Coulomb gauge maintains
these symmetries, sets ${\boldsymbol{A}}_{L}=0$ and determines $A^{0}$ through
Gauss’ law,

$\displaystyle\frac{\delta\mathcal{S}_{QED}}{\delta
A^{0}(x)}=-\boldsymbol{\nabla}^{2}A^{0}(t,{\boldsymbol{x}})-e\psi^{\dagger}\psi(t,{\boldsymbol{x}})=0$
(532)

The positions of the charges at any time $t$ determine $A^{0}$ instantaneously
at all positions ${\boldsymbol{x}}$. This gives the Coulomb potential
$V(r)=-\alpha/r$ for Positronium, which is the dominant interaction in atomic
rest frames. The $e^{-}e^{+}$ Fock state wave function is determined by the
Schrödinger equation, and other Fock states are suppressed by powers of
$\alpha$. These features make Coulomb gauge a common choice for bound state
calculations.

Quantization in Coulomb gauge is complicated by the absence of a conjugate
field for $A^{0}$, see Feinberg (1978); Christ and Lee (1980); Weinberg
(2005). Gauss’ law (532) is an operator relation which defines $A^{0}$ in
terms of $\psi^{\dagger}\psi$ as a non-local quantum field. The classical
potential of the Schrödinger equation given by $A^{0}$ in Coulomb gauge is
more simply realized in temporal gauge, $A^{0}=0$. Quantization is
straightforward in temporal gauge, since ${\boldsymbol{A}}_{L}$ does have a
conjugate field, ${\boldsymbol{E}}_{L}$. Gauss’ law is no longer an operator
equation of motion since $A^{0}$ is fixed. Physical Fock states are defined to
be invariant under time independent gauge transformations (which preserve
$A^{0}=0$). This constrains the value of ${\boldsymbol{E}}_{L}$ for each
physical state to be such that Gauss’ law is satisfied (Willemsen (1978);
Bjorken (1979); Christ and Lee (1980); Leibbrandt (1987); Strocchi (2013) and
section V), giving rise to the classical potential.

In a perturbative approach the classical potential is weak, being proportional
to $\alpha$ or ${\alpha_{s}}$. The QCD action has no parameter like
$\Lambda_{QCD}\sim 1$ fm-1 for the confinement scale. However, Gauss’ law
determines ${\boldsymbol{E}}_{L}^{a}$ only up to a boundary condition. In
section V.3.2 I consider adding a homogeneous solution (162) to the gluon
exchange term. This gives rise to a spatially constant gluon field energy
density (365) for each color component of a Fock state. Being a color octet,
${\boldsymbol{E}}_{L}^{a}$ cancels in the sum over the color components of
singlet Fock states, avoiding long range effects.

The homogeneous solution generates a confining potential of ${\cal
O}\left(\alpha_{s}^{0}\right)$. In section VIII I determined the potential for
various Fock states and made first checks of the ${\cal
O}\left(\alpha_{s}^{0}\right)$ dynamics. Essential and non-trivial features
include the gauge invariance of electromagnetic form factors and the correct
dependence of the bound state energies on the CM momentum,
$E({\boldsymbol{P}})$. String breaking and duality can arise through an
overlap between single and multiple bound states as suggested by dual diagrams
(Fig. 19 a). There is a $J^{PC}=0^{++}$ solution with $P^{\mu}=0$ in all
frames whose mixing with the vacuum allows solutions with spontaneously broken
chiral symmetry.

A non-vanishing field energy density would explain why confinement is not seen
in Feynman diagrams, which expand around free states. However, the homogeneous
solution (162) represents a major departure from previous experience. It
requires much further study, both of theoretical consistency and
phenomenological relevance.

###### Acknowledgements.

During my work on the topics presented here I have benefited from the
collaboration and advice of many colleagues. Particular thanks are due to
Jean-Paul Blaizot, Stan Brodsky, Dennis D. Dietrich, Matti Järvinen, Stephane
Peigné and Johan Rathsman. I am grateful for the hospitality of the University
of Pavia, and the stimulating response to my lectures there in early 2020.
During earlier stages of this project I have enjoyed visits of a month or more
to ECT* (Trento), CERN-TH (Geneva), CP3 (Odense), NIKHEF (Amsterdam), IPhT
(Saclay) and GSI (Darmstadt). I have the privileges of Professor Emeritus at
the Physics Department of Helsinki University. Travel grants from the Magnus
Ehrnrooth Foundation have allowed me to maintain contacts and present my
research to colleagues.

##  Appendix A Solutions to exercises

###  A.1 Order of box diagram

The leading contribution comes from the range of the loop integral $\int
d^{4}\ell$ where $\ell^{0}\sim\alpha^{2}m$ and
$|{\boldsymbol{\ell}}|\sim\alpha m$. Thus $\int d^{4}\ell\sim\alpha^{5}$. The
fermions are off-shell similarly to $E_{p_{1}}-m\sim\alpha^{2}$, as are the
photons,
$q^{2}=(q^{0})^{2}-{\boldsymbol{q}}^{2}\simeq-{\boldsymbol{q}}^{2}\sim\alpha^{2}$.
Considering also the factor $e^{4}\sim\alpha^{2}$ from the vertices the power
of $\alpha$ is altogether $5+4\times(-2)+2=-1$.

###  A.2 Contribution of the diagrams in Fig. 11(b,c)

As in Ex. A.1 the leading contribution comes from the range of the loop
integral $\int d^{4}\ell$ where $\ell$ is of the order of the bound state
momenta. Then the Dirac structures simplify as in $A_{1}$ of (27), giving
$(2m)^{4}$. The two photon propagators reduce to the $\ell^{0}$-independent
Coulomb exchanges contained in $A_{1}({\boldsymbol{p}},{\boldsymbol{\ell}})$
and $V({\boldsymbol{\ell}}-{\boldsymbol{q}})$ of (28). The fermion propagators
with momenta $\ell$ and $\ell-p_{1}-p_{2}\equiv\ell-P$ may be expressed as in
(16), keeping only the terms with the electron and positron poles. The
relevant part of the $\ell^{0}$ integral is then

$\displaystyle\int\frac{d\ell^{0}}{2\pi}\,\frac{1}{\ell^{0}-E_{\ell}+i\varepsilon}\,\frac{1}{\ell^{0}-P^{0}+E_{\ell}-i\varepsilon}=\frac{i}{P^{0}-2E_{\ell}+i\varepsilon}$
(A.533)

Noting that the factor of $1/2E_{\ell}$ in each fermion propagator (16) gives
$1/(2m)^{2}$ we arrive at (28).

In the crossed diagram (c) of Fig. 11 the second factor in (A.533) is
$1/(q_{1}^{0}-p_{2}^{0}-\ell^{0}+E-i\varepsilon)$. The pole in $\ell^{0}$ now
has ${\rm Im}\,\ell^{0}<0$ as in the first factor of (A.533). This allows
closing the contour in the ${\rm Im}\,\ell^{0}>0$ hemisphere giving no leading
order contribution. In fact only the ladder diagrams shown in Fig. 10
contribute at leading ${\cal O}\left(1/\alpha\right)$. They generate the
classical field of the bound state.

###  A.3 Derivation of (47)

For any (finite) momentum exchange the antifermion energy in the loops of Fig.
11(b,c) is $\bar{E}=m_{T}+{\cal O}\left(1/m_{T}\right)$. Since
$p_{2}^{0}=m_{T}$ we need only retain the negative energy pole term in the
antifermion propagators (cf. (16)), in which $-p_{2}^{0}+\bar{E}$ is of ${\cal
O}\left(1/m_{T}\right)$. For the electron lines to go on-shell requires non-
vanishing energy transfer $\ell^{0}-p_{1}^{0}\neq 0$, which causes the
antifermion propagator to be off-shell by ${\cal O}\left(m_{T}\right)$. The
photon propagators are thus independent of $\ell^{0}$, e.g., $D^{00}(\ell-
p_{1})=i/({\boldsymbol{\ell}}-{\boldsymbol{p}}_{1})^{2}$. The only relevant
poles of the $\ell^{0}$ integration are in the antifermion propagators,

$\displaystyle\int\frac{d\ell^{0}}{2\pi}\Big{[}\frac{i}{\ell^{0}-p_{1}^{0}-i\varepsilon}+\frac{i}{q_{1}^{0}-\ell^{0}-i\varepsilon}\Big{]}=i\int\frac{d\ell^{0}}{2\pi}\,2\pi
i\delta(\ell^{0}-p_{1}^{0})=-1$ (A.534)

The factor $-i$ at the antifermion vertices cancels with the $i$ of the
Coulomb photon propagators. The standard rules for the electron line then
gives (47).

In the diagram with three uncrossed photon exchanges with momenta
$\ell_{1}-p_{1},\ \ell_{2}-\ell_{1}$ and $q_{1}-\ell_{2}$ the two antifermion
propagators similarly give

$\displaystyle\int\frac{d\ell_{1}^{0}\,d\ell_{2}^{0}}{(2\pi)^{2}}\,\frac{i^{2}}{(\ell_{1}^{0}-p_{1}^{0}-i\varepsilon)(\ell_{2}^{0}-q_{1}^{0}-i\varepsilon)}$
(A.535)

The five other diagrams with crossed photons ensure the convergence of the
integrals similarly as in (A.534), and do not contribute when the integration
contours of $\ell_{1}^{0}$ and $\ell_{2}^{0}$ are closed in the upper half
plane. The full result is then due to (A.535), which equals 1 (given the
convergence). If the contours are chosen differently then the same result will
come from diagrams with crossed photons. The factors associated with the
electron line are again given by the standard rules, making the three photon
contribution equal to scattering from an external potential.

###  A.4 Derivation of (64)

Inserting the completeness condition for the Dirac wave functions,

$\sum_{n}\big{[}\Psi_{n,\alpha}({\boldsymbol{x}})\Psi_{n,\beta}^{\dagger}({\boldsymbol{y}})+\overline{\Psi}_{n,\alpha}({\boldsymbol{x}})\overline{\Psi}_{n,\beta}^{\dagger}({\boldsymbol{y}})\big{]}=\delta_{\alpha\beta}\delta^{3}({\boldsymbol{x}}-{\boldsymbol{y}})$
(A.536)

into the Dirac Hamiltonian (49) gives, recalling that the wave functions
satisfy (44) and (45),

$\displaystyle H_{D}$ $\displaystyle=$ $\displaystyle\int
d{\boldsymbol{x}}\,d{\boldsymbol{y}}\,\bar{\psi}_{\alpha^{\prime}}({\boldsymbol{x}})\big{[}-i\boldsymbol{\nabla}_{\boldsymbol{x}}\cdot\boldsymbol{\gamma}+m+e\not{A}({\boldsymbol{x}})\big{]}_{\alpha^{\prime}\alpha}\sum_{n}\big{[}\Psi_{n,\alpha}({\boldsymbol{x}})\Psi_{n,\beta}^{\dagger}({\boldsymbol{y}})+\overline{\Psi}_{n,\alpha}({\boldsymbol{x}})\overline{\Psi}_{n,\beta}^{\dagger}({\boldsymbol{y}})\big{]}\psi_{\beta}({\boldsymbol{y}})$
(A.537) $\displaystyle=$ $\displaystyle\sum_{n}\int
d{\boldsymbol{x}}\,d{\boldsymbol{y}}\,\psi_{\alpha}^{\dagger}({\boldsymbol{x}})\big{[}M_{n}\Psi_{n,\alpha}({\boldsymbol{x}})\Psi_{n,\beta}^{\dagger}({\boldsymbol{y}})-\bar{M}_{n}\overline{\Psi}_{n,\alpha}({\boldsymbol{x}})\overline{\Psi}_{n,\beta}^{\dagger}({\boldsymbol{y}})\big{]}\psi_{\beta}({\boldsymbol{y}})$
$\displaystyle=$
$\displaystyle\sum_{n}\big{[}M_{n}c_{n}^{\dagger}c_{n}-\bar{M}_{n}\bar{c}_{n}\bar{c}_{n}^{\dagger}\big{]}\to\sum_{n}\big{[}M_{n}c_{n}^{\dagger}c_{n}+\bar{M}_{n}\bar{c}_{n}^{\dagger}\bar{c}_{n}\big{]}$

In the last step I normal-ordered the operators, neglecting the zero-point
energies according to (57).

###  A.5 The expressions (65) for vacuum state

(a) The $B$ and $D$ coefficients defined in (IV.3) and (62) satisfy

$B_{mp}\overline{B}_{np}+D_{mp}\overline{D}_{np}=\sum_{\boldsymbol{p}}\Psi_{m,\alpha}^{\dagger}({\boldsymbol{p}})\big{[}u_{\alpha}({\boldsymbol{p}},\lambda)u_{\beta}^{\dagger}({\boldsymbol{p}},\lambda)+v_{\alpha}(-{\boldsymbol{p}},\lambda)v_{\beta}^{\dagger}(-{\boldsymbol{p}},\lambda)\big{]}\overline{\Psi}_{n,\beta}({\boldsymbol{p}})=\int\frac{d{\boldsymbol{p}}}{(2\pi)^{3}}\Psi_{m,\alpha}^{\dagger}({\boldsymbol{p}})\overline{\Psi}_{n,\alpha}({\boldsymbol{p}})=0$
(A.538)

Multiplying by $(B^{-1})_{qm}({\overline{D}}^{\,-1})_{rn}$ and summing over
$m,n$ gives

$-(B^{-1})_{qm}D_{mr}=({\overline{D}}^{\,-1})_{rn}{\overline{B}}_{nq}$ (A.539)

Using also $b^{\dagger}_{q}d^{\dagger}_{r}=-d^{\dagger}_{r}b^{\dagger}_{q}$
shows the equivalence of the two expressions for the vacuum state in (65).

(b) In order to verify that $c_{n}\left|{\Omega}\right\rangle=0$ we note that
since $b_{p}$ essentially differentiates the exponent in
$\left|{\Omega}\right\rangle$,

$B_{nq}b_{q}\left|{\Omega}\right\rangle=-B_{nq}\big{(}B^{-1}\big{)}_{qm}D_{mr}d_{r}^{\dagger}\left|{\Omega}\right\rangle=-D_{nr}d_{r}^{\dagger}\left|{\Omega}\right\rangle$
(A.540)

This cancels the contribution of the second term in the definition (IV.3) of
$c_{n}$. The demonstration that $\bar{c}_{n}$ annihilates the vacuum is
similar. Thus

$c_{n}\left|{\Omega}\right\rangle=\bar{c}_{n}\left|{\Omega}\right\rangle=H_{D}\left|{\Omega}\right\rangle=0$
(A.541)

###  A.6 Derivation of the identities (IV.4)

We may start by evaluating (here I make no difference between lower and upper
indices)

$\displaystyle(\hat{\boldsymbol{x}}\times{\boldsymbol{L}})^{i}=\epsilon_{ijk}\hat{x}^{j}\epsilon_{kln}x^{l}(-i\partial_{n})=(\delta_{il}\delta_{jn}-\delta_{in}\delta_{jl})\hat{x}^{j}x^{l}(-i\partial_{n})=-i\hat{x}^{i}r\partial_{r}+ir\partial_{i}$
(A.542)

Multiplying by $\alpha^{i}/r$ we have the first relation,

$\displaystyle\frac{1}{r}{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}\times{\boldsymbol{L}}=-i{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}\,\partial_{r}+i{\boldsymbol{\alpha}}\cdot\boldsymbol{\nabla}$
(A.543)

The second identity:

$\displaystyle-i({\boldsymbol{\alpha}}\cdot\boldsymbol{\nabla})i({\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}})=\alpha^{i}\alpha^{j}\partial_{i}\frac{x^{j}}{r}=(\delta_{ij}+i\gamma_{5}\epsilon_{ijk}\alpha^{k})\Big{(}\frac{\delta^{ij}}{r}-\frac{x^{i}x^{j}}{r^{3}}+\frac{x^{j}}{r}\partial_{i}\Big{)}=\frac{2}{r}+\partial_{r}+\frac{1}{r}\gamma_{5}{\boldsymbol{\alpha}}\cdot{\boldsymbol{L}}$
(A.544)

###  A.7 Derivation of (127)

The momentum space wave function (60) for $j={\textstyle\frac{1}{2}}$ and
positive parity is

$\displaystyle\Psi_{1/2,\lambda,+}=2\pi\int_{0}^{\infty}dr\int_{-1}^{1}d\cos\theta\,r^{2}e^{-ipr\cos\theta}\big{[}F(r)+iG(r)\,\alpha\cdot\hat{\boldsymbol{p}}\,\cos\theta\big{]}\left(\begin{array}[]{c}\chi_{\lambda}\\\\[5.69054pt]
0\end{array}\right)$ (A.547)

I chose the $z$-axis of the ${\boldsymbol{x}}$-integration along
${\boldsymbol{p}}$. The integration over the azimuthal angle $\varphi$ leaves
only the $\alpha_{z}$ component of
${\boldsymbol{\alpha}}\cdot{\boldsymbol{x}}$. Expressing the factor
$\cos\theta$ as a derivative of $\exp(-ipr\cos\theta)$ and using $G=iF$ (which
holds at large $r$, where the stationary point is located for large $p$) we
have

$\displaystyle\Psi_{1/2,\lambda,+}=2\pi\int_{0}^{\infty}dr\,r^{2}\,F(r)\big{(}1-\frac{i}{r}\,\alpha\cdot\hat{\boldsymbol{p}}\,\partial_{p}\big{)}\frac{i}{pr}\big{(}e^{-ipr}-e^{ipr}\big{)}\left(\begin{array}[]{c}\chi_{\lambda}\\\\[5.69054pt]
0\end{array}\right)$ (A.550)

For $p\to\infty$ the phase of $e^{\pm ipr}$ is rapidly oscillating, so the
leading contribution comes from the region of the $r$-integration where the
phase of the integrand is stationary. The stationary phase approximation is

$\displaystyle\int dr\,f(r)e^{i\varphi(r)}\simeq
e^{\varepsilon(\varphi^{\prime\prime}(r_{s}))i\pi/4}\,f(r_{s})\,e^{i\varphi(r_{s})}$
(A.551)

The function $f(r)$ is assumed to be varying slowly compared to the phase
$\exp[i\varphi(r)]$. The phase is stationary at $\varphi^{\prime}(r_{s})=0$
and $\varepsilon(x)=1\ (-1)$ for $x>0\ (x<0)$. According to (126) we have in
the integral of (A.550) $\varphi(r)=V^{\prime}r^{2}/2\mp pr$. There is a
stationary phase for $r>0$ only with the $\exp(-ipr)$ term, giving
$r_{s}=p/V^{\prime}$, $\varphi(r_{s})=-p^{2}/2V^{\prime}$ and
$\varphi^{\prime\prime}(r_{s})>0$. From (126) the contribution proportional to
the unit Dirac matrix in (A.550) is then

$\displaystyle\big{(}\Psi_{1/2,\lambda,+}\big{)}_{F}=\frac{2\pi
iN}{p}\Big{(}\frac{p}{V^{\prime}}\Big{)}^{\beta+1}\,e^{-ip^{2}/2V^{\prime}}\left(\begin{array}[]{c}\chi_{\lambda}\\\\[5.69054pt]
0\end{array}\right)$ (A.554)

The leading term in the contribution $\propto\alpha\cdot\hat{\boldsymbol{p}}$
has $\partial_{p}\to-ip/V^{\prime}=-ir_{s}$, giving the full result

$\displaystyle\Psi_{1/2,\lambda,+}=\frac{2\pi
iN}{p}\Big{(}\frac{p}{V^{\prime}}\Big{)}^{\beta+1}(1-\alpha\cdot\hat{\boldsymbol{p}})\,e^{-ip^{2}/2V^{\prime}}\left(\begin{array}[]{c}\chi_{\lambda}\\\\[5.69054pt]
0\end{array}\right)$ (A.557)

Consider now the expressions for the $u$ and $v$ spinors for
$|{\boldsymbol{p}}|\to\infty$,

$\displaystyle u({\boldsymbol{p}},\lambda)$
$\displaystyle\equiv\frac{{\not{p}}+m}{\sqrt{E_{p}+m}}\left(\begin{array}[]{c}\chi_{\lambda}\\\\[5.69054pt]
0\end{array}\right)\simeq\sqrt{|{\boldsymbol{p}}|}\,(1+\alpha\cdot\hat{\boldsymbol{p}})\left(\begin{array}[]{c}\chi_{\lambda}\\\\[5.69054pt]
0\end{array}\right)\ \ \ (|{\boldsymbol{p}}|\to\infty)$ (A.562) $\displaystyle
v({\boldsymbol{p}},\lambda)$
$\displaystyle\equiv\frac{-{\not{p}}+m}{\sqrt{E_{p}+m}}\left(\begin{array}[]{c}0\\\\[5.69054pt]
\bar{\chi}_{\lambda}\end{array}\right)\simeq\sqrt{|{\boldsymbol{p}}|}\,(1+\alpha\cdot\hat{\boldsymbol{p}})\left(\begin{array}[]{c}0\\\\[5.69054pt]
\bar{\chi}_{\lambda}\end{array}\right)\ \ \ (|{\boldsymbol{p}}|\to\infty)$
(A.567)

Consequently, omitting a common factor in the distributions
$e^{\pm}({\boldsymbol{p}},s)$ (IV.3),

$\displaystyle e^{-}(p,s)$
$\displaystyle=u^{\dagger}({\boldsymbol{p}},s)\Psi_{1/2,\lambda,+}\sim\left(\begin{array}[]{cc}\chi_{s}^{\dagger}&0\end{array}\right)(1+\alpha\cdot\hat{\boldsymbol{p}})(1-\alpha\cdot\hat{\boldsymbol{p}})\left(\begin{array}[]{c}\chi_{\lambda}\\\\[5.69054pt]
0\end{array}\right)=0$ (A.571) $\displaystyle e^{+}(p,s)$
$\displaystyle=v^{\dagger}(-{\boldsymbol{p}},s)\Psi_{1/2,\lambda,+}\sim\left(\begin{array}[]{cc}0&\bar{\chi}_{s}^{\dagger}\end{array}\right)(1-\alpha\cdot\hat{\boldsymbol{p}})(1-\alpha\cdot\hat{\boldsymbol{p}})\left(\begin{array}[]{c}\chi_{\lambda}\\\\[5.69054pt]
0\end{array}\right)=-2\bar{\chi}_{s}^{\dagger}\,\boldsymbol{\sigma}\cdot\hat{\boldsymbol{p}}\,\chi_{\lambda}$
(A.575)

which establishes (127).

###  A.8 Gauge transformations generated by Gauss operator

The unitary operator defined in (140),

$\displaystyle U(t)=1+i\int
d{\boldsymbol{y}}\,G(t,{\boldsymbol{y}})\delta\Lambda({\boldsymbol{y}})=1+i\int
d{\boldsymbol{y}}\,\big{[}\partial_{i}E^{i}(t,{\boldsymbol{y}})-e\psi^{\dagger}\psi(t,{\boldsymbol{y}})\big{]}\delta\Lambda({\boldsymbol{y}})$
(A.576)

transforms $A^{j}(t,{\boldsymbol{x}})$ as,

$\displaystyle\delta A^{j}(t,{\boldsymbol{x}})$ $\displaystyle\equiv
U(t)A^{j}(t,{\boldsymbol{x}})U^{-1}(t)-A^{j}(t,{\boldsymbol{x}})=i\left[{\int
d{\boldsymbol{y}}\,\big{[}\partial_{i}E^{i}(t,{\boldsymbol{y}})-e\psi^{\dagger}\psi(t,{\boldsymbol{y}})\big{]}\delta\Lambda({\boldsymbol{y}})},{A^{j}(t,{\boldsymbol{x}})}\right]$
$\displaystyle=-i\left[{\int
d{\boldsymbol{y}}\,E^{i}(t,{\boldsymbol{y}})\partial_{i}^{y}\delta\Lambda({\boldsymbol{y}})},{A^{j}(t,{\boldsymbol{x}})}\right]=\partial_{j}\delta\Lambda({\boldsymbol{x}})$

Similarly for the electron field,

$\displaystyle\delta\psi(t,{\boldsymbol{x}})$ $\displaystyle\equiv
U(t)\psi(t,{\boldsymbol{x}})U^{-1}(t)-\psi(t,{\boldsymbol{x}})=-ie\left[{\int
d{\boldsymbol{y}}\,\psi^{\dagger}\psi(t,{\boldsymbol{y}})\delta\Lambda({\boldsymbol{y}})},{\psi(t,{\boldsymbol{x}})}\right]=ie\,\delta\Lambda({\boldsymbol{x}})\psi(t,{\boldsymbol{x}})$
(A.578)

###  A.9  Derive (172).

The anticommutation relation of the electron fields gives

$\displaystyle\boldsymbol{\mathcal{J}}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1})\Big{[}{\boldsymbol{J}}{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})-{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2}){\boldsymbol{J}}\Big{]}\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(A.579)

${\boldsymbol{J}}$ commutes with the $\Lambda_{\pm}$ projectors (VI.1.1),
which have a rotationally invariant form. It is instructive to see how this
works out explicitly. The commutator
$\big{[}{{\overset{\rightarrow}{\Lambda}}_{-}},{{\boldsymbol{L}}+{\boldsymbol{S}}}\big{]}$
gets contributions from the derivatives in
${\overset{\rightarrow}{\Lambda}}_{-}$ differentiating ${\boldsymbol{x}}$ in
${\boldsymbol{L}}=-i{\boldsymbol{x}}\times\boldsymbol{\nabla}$, and from the
commutators of the Dirac matrices in ${\overset{\rightarrow}{\Lambda}}_{-}$
with
${\boldsymbol{S}}={\textstyle\frac{1}{2}}\gamma_{5}{\boldsymbol{\alpha}}$.

The $\boldsymbol{\nabla}^{2}$ in $E=\sqrt{-\boldsymbol{\nabla}^{2}+m^{2}}$ of
${\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}})$ commutes with
$L^{i}$,

$\displaystyle\left[{\partial_{j}\partial_{j}},{\varepsilon_{ik\ell}x^{k}\partial_{\ell}}\right]=\partial_{j}\varepsilon_{ij\ell}\partial_{\ell}+\varepsilon_{ij\ell}\partial_{\ell}\partial_{j}=0$
(A.580)

since $\partial_{j}\partial_{\ell}=\partial_{\ell}\partial_{j}$, whereas
$\varepsilon_{ij\ell}=-\varepsilon_{i\ell j}$. The commutator of
$i{\boldsymbol{\alpha}}\cdot\boldsymbol{\nabla}$ with $L^{i}$ (in my notation
$\alpha_{j}=\alpha^{j}$),

$\displaystyle\left[{i\alpha_{j}\partial_{j}},{-i\varepsilon_{ik\ell}x^{k}\partial_{\ell}}\right]=\varepsilon_{ij\ell}\alpha_{j}\partial_{\ell}$
(A.581)

is cancelled by its commutator with $S^{i}$. Using
$\alpha_{i}\alpha_{j}=\delta_{ij}+i\varepsilon_{ijk}\alpha_{k}\gamma_{5}$,

$\displaystyle\left[{i\alpha_{j}\partial_{j}},{{\textstyle\frac{1}{2}}\gamma_{5}\alpha_{i}}\right]=-\varepsilon_{jik}\alpha_{k}\partial_{j}$
(A.582)

Finally, $\left[{\gamma^{0}},{{\boldsymbol{S}}}\right]=0$. We see that the
commutator
$\big{[}{{\overset{\rightarrow}{\Lambda}}_{-}},{{\boldsymbol{J}}}\big{]}=0$
requires contributions from both ${\boldsymbol{L}}$ and ${\boldsymbol{S}}$.
Similarly ${\boldsymbol{J}}$ commutes with other rotationally invariant
structures. Bringing ${\boldsymbol{J}}$ through the $\Lambda_{\pm}$ projectors
gives (172).

The commutator with the orbital angular momentum ${\boldsymbol{L}}$ in (172)
arises from

$\displaystyle{\boldsymbol{x}}_{1}\times(-i{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1})\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})-\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\boldsymbol{x}}_{2}\times(i{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2})=({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\times(-i{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1})\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$
(A.583)

Hence in
$\left[{{\boldsymbol{L}}},{\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})}\right]={\boldsymbol{x}}\times[-i\boldsymbol{\nabla}\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})]$
the ${\boldsymbol{x}}$-derivatives of ${\boldsymbol{L}}$ apply only to
$\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})$.

###  A.10 Derivation of (181)

The transformation (179) of the fields under charge conjugation gives

$\displaystyle\mathcal{C}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle=-\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\psi({\boldsymbol{x}}_{1})^{T}{\alpha_{2}}{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2}){\alpha_{2}}\bar{\psi}^{T}({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(A.584)

Take the transpose on the rhs., recalling that the anticommutation of the
fields gives a minus sign,

$\displaystyle\mathcal{C}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$
$\displaystyle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{2}){\alpha_{2}}^{T}\frac{1}{2E_{2}}(E_{2}+i{\boldsymbol{\alpha}}^{T}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2}-\gamma^{0}m){\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}}^{T}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$
$\displaystyle\times\frac{1}{2E_{1}}(E_{1}-i{\boldsymbol{\alpha}}^{T}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}+\gamma^{0}m){\alpha_{2}}^{T}\psi({\boldsymbol{x}}_{1})\left|{0}\right\rangle$
(A.585)

Recalling that
${\alpha_{2}}{\boldsymbol{\alpha}}^{T}{\alpha_{2}}=-{\boldsymbol{\alpha}}$ and
changing integration variables
${\boldsymbol{x}}_{1}\leftrightarrow{\boldsymbol{x}}_{2}$ we get

$\displaystyle\mathcal{C}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1}){\alpha_{2}}{\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}}^{T}({\boldsymbol{x}}_{2}-{\boldsymbol{x}}_{1}){\alpha_{2}}{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(A.586)

Comparing with the definition (165) of
$\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$ we see that
(180) implies (181).

###  A.11  Verify (186).

At lowest order in $\alpha$ the projectors $\Lambda_{\pm}$ in the definition
(165) of the Positronium state may be expressed in terms of the CM momentum
${\boldsymbol{P}}$, by partial integration and ignoring the ${\cal
O}\left(\alpha\right)$ contributions from differentiating
$\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$,

$\displaystyle\Lambda_{\pm}({\boldsymbol{P}})=\frac{1}{2E_{P}}\big{(}E_{P}\mp{\boldsymbol{\alpha}}\cdot{\boldsymbol{P}}\pm
2m\gamma^{0}\big{)}=\Lambda_{\pm}^{\dagger}({\boldsymbol{P}})=\big{[}\Lambda_{\pm}({\boldsymbol{P}})\big{]}^{2}\hskip
56.9055pt\Lambda_{+}({\boldsymbol{P}})\Lambda_{-}({\boldsymbol{P}})=0$ (A.587)

Anticommutating the fields in
$\langle{e^{-}e^{+};\mathcal{B}^{\prime},{\boldsymbol{P}}^{\prime}}|$ with
those in $\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$ gives
using (183),

$\displaystyle\langle{e^{-}e^{+};\mathcal{B}^{\prime},{\boldsymbol{P}}^{\prime}}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$
$\displaystyle=N_{\mathcal{B}^{\prime}}^{({\boldsymbol{P}}^{\prime})*}N_{\mathcal{B}}^{(P)}\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,e^{i({\boldsymbol{P}}-{\boldsymbol{P}}^{\prime})\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}{F^{({\boldsymbol{P}}^{\prime})}}^{*}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})F^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\mathrm{Tr}\,_{\mathcal{B},\mathcal{B}^{\prime}}$
$\displaystyle=\big{|}N_{\mathcal{B}}^{(P)}\big{|}^{2}(2\pi)^{3}\delta({\boldsymbol{P}}-{\boldsymbol{P}}^{\prime})\int
d{\boldsymbol{x}}\,|F^{({\boldsymbol{P}})}({\boldsymbol{x}})|^{2}\,\mathrm{Tr}\,_{\mathcal{B},\mathcal{B}^{\prime}}$
(A.588) $\displaystyle\mathrm{Tr}\,_{\mathcal{B},\mathcal{B}^{\prime}}$
$\displaystyle=\mathrm{Tr}\,\big{\\{}\Gamma_{\mathcal{B}^{\prime}}^{\dagger}\Lambda_{+}({\boldsymbol{P}})\Gamma_{\mathcal{B}}\Lambda_{-}({\boldsymbol{P}})\big{\\}}$

Commuting $\Lambda_{-}({\boldsymbol{P}})$ through $\Gamma_{\mathcal{B}}$ gives
for $\Gamma_{\mathcal{B}}=\gamma_{5}$ (Parapositronium) and
$\Gamma_{\mathcal{B}}={\alpha_{3}}$ (Orthopositronium with $\lambda=0$),

$\displaystyle\mathrm{Tr}\,_{\mathcal{B},\mathcal{B}^{\prime}}=\frac{2m}{E_{P}}\,\mathrm{Tr}\,\big{\\{}\Gamma_{\mathcal{B}^{\prime}}^{\dagger}\Lambda_{+}({\boldsymbol{P}})\gamma^{0}\Gamma_{\mathcal{B}}\big{\\}}=\frac{8m^{2}}{E_{P}^{2}}\delta_{\mathcal{B},\mathcal{B}^{\prime}}$
(A.589)

while for
$\Gamma_{\mathcal{B}}={\boldsymbol{e}}_{\pm}\cdot{\boldsymbol{\alpha}}$
(Orthopositronium with $\lambda=\pm 1$),

$\displaystyle\mathrm{Tr}\,_{\mathcal{B},\mathcal{B}^{\prime}}=\mathrm{Tr}\,\big{\\{}\Gamma_{\mathcal{B}^{\prime}}^{\dagger}\Lambda_{+}({\boldsymbol{P}})\Gamma_{\mathcal{B}}\big{\\}}=2\delta_{\mathcal{B},\mathcal{B}^{\prime}}$
(A.590)

Using these expressions for $\mathrm{Tr}\,_{\mathcal{B},\mathcal{B}^{\prime}}$
and the normalization (183) of $F^{({\boldsymbol{P}})}({\boldsymbol{x}})$ in
(A.11) the normalization (168) of the state implies (186).

###  A.12 Derive the expression for
$\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ in (VI.3.2).

The contribution to $\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$
from $\left[{\mathcal{H}_{int}},{\bar{\psi}({\boldsymbol{x}}_{1})}\right]$ is

$\displaystyle e\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1}){\boldsymbol{\alpha}}\cdot{\boldsymbol{\varepsilon}}_{s}^{*}({\boldsymbol{q}})e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{1}}a^{\dagger}({\boldsymbol{q}},s){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(A.591)

where I inserted ${\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})$
to select the $b^{\dagger}$ contribution in $\bar{\psi}({\boldsymbol{x}}_{1})$
as in (167). We have then

$\displaystyle{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\alpha^{j}{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})$
$\displaystyle={\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\frac{1}{2E_{1}}\Big{[}(E_{1}+i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}-m\gamma^{0})\alpha^{j}+\big{\\{}{\alpha^{j}},{-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}}\big{\\}}\Big{]}={\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\frac{-2i{\buildrel\leftarrow\over{\partial}}_{1,j}}{2E_{1}}$
$\displaystyle\to{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\frac{-P^{j}}{E_{P}}$
(A.592)

The first term in the square bracket vanishes when multiplied by
${\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})$. The final result
follows after partial integration, with the leading order term due to
$i{\buildrel\rightarrow\over{\partial}}_{1,j}\exp[i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2]$
in (A.591).

The contribution to $\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$
from $\left[{\mathcal{H}_{int}},{\psi({\boldsymbol{x}}_{2})}\right]$ is
similarly

$\displaystyle e\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2}){\boldsymbol{\alpha}}\cdot{\boldsymbol{\varepsilon}}_{s}^{*}({\boldsymbol{q}})e^{-{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{2}}a^{\dagger}({\boldsymbol{q}},s){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(A.593)

where now

$\displaystyle{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\alpha^{j}{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})$
$\displaystyle=\alpha^{j}\frac{1}{2E_{2}}\big{[}(E_{2}-i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}+m\gamma^{0})+\big{\\{}{\alpha^{j}},{i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}}\big{\\}}\big{]}{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})=\frac{2i{\buildrel\rightarrow\over{\partial}}_{2,j}}{2E_{2}}{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})$
$\displaystyle\to\frac{P^{j}}{E_{P}}{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})$
(A.594)

Using (A.12) in (A.591) and (A.12) in (A.593) and adding the two contributions
gives (VI.3.2).

###  A.13 Derive the expression (309).

The frame dependence of functions like $\phi_{0}(\tau)$ and $\phi_{1}(\tau)$
that do not explicitly depend on $P$ or $E$ arises only due to the
$P$-dependence of $\tau(x)$.

$\displaystyle\left.\frac{\partial\tau}{\partial
P}\right|_{x}=\frac{\partial}{\partial
P}\big{[}M^{2}-2EV+V^{2}\big{]}/V^{\prime}=-\frac{2xP}{E}$ (A.595)

Recalling also that for functions that depend on $x$ only via $\tau$,

$\displaystyle\left.\frac{\partial}{\partial
x}\right|_{P}=\left.\frac{\partial\tau}{\partial
x}\right|_{P}\frac{\partial}{\partial\tau}=-2(E-V)\frac{\partial}{\partial\tau}$
(A.596)

we have

$\displaystyle\left.\frac{\partial\phi_{0,1}}{\partial\xi}\right|_{x}=\left.E\frac{\partial\phi_{0,1}}{\partial
P}\right|_{x}=\left.\frac{xp}{E-V}\partial_{x}\phi_{0,1}\right|_{P}$ (A.597)

Applying this to $e^{\sigma_{1}\zeta/2}\Phi^{(P)}e^{-\sigma_{1}\zeta/2}$ gives

$\displaystyle\left.\frac{\partial}{\partial\xi}\Big{[}e^{\sigma_{1}\zeta/2}\Phi^{(P)}e^{-\sigma_{1}\zeta/2}\Big{]}\right|_{x}$
$\displaystyle=\left.e^{\sigma_{1}\zeta/2}\bigg{\\{}\frac{1}{2}\,\frac{\partial\zeta}{\partial\xi}\right|_{x}\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}+\left.\frac{\partial\Phi^{(P)}}{\partial\xi}\right|_{x}\bigg{\\}}e^{-\sigma_{1}\zeta/2}$
(A.598)
$\displaystyle=\frac{xp}{E-V}\,\partial_{x}\Big{[}e^{\sigma_{1}\zeta/2}\Phi^{(P)}e^{-\sigma_{1}\zeta/2}\Big{]}\Big{|}_{\xi}=\left.\frac{xp}{E-V}\,e^{\sigma_{1}\zeta/2}\bigg{\\{}\frac{1}{2}\,\frac{\partial\zeta}{\partial
x}\right|_{\xi}\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}+\left.\frac{\partial\Phi^{(P)}}{\partial
x}\right|_{\xi}\bigg{\\}}e^{-\sigma_{1}\zeta/2}$

which implies

$\displaystyle\frac{\partial\Phi^{(P)}}{\partial\xi}\Big{|}_{x}=\frac{xp}{E-V}\,\frac{\partial\Phi^{(P)}}{\partial
x}\Big{|}_{\xi}+\frac{1}{2}\Big{(}\frac{xp}{E-V}\,\frac{\partial\zeta}{\partial
x}\Big{|}_{\xi}-\frac{\partial\zeta}{\partial\xi}\Big{|}_{x}\Big{)}\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}$
(A.599)

It remains to work out the derivatives of $\zeta$. From its definition (306),

$\displaystyle\partial_{x}(\sinh\zeta)\big{|}_{\xi}$
$\displaystyle=\partial_{x}\zeta\big{|}_{\xi}\,\cosh\zeta=\frac{V^{\prime}P(E-V)}{(V^{\prime}\tau)^{3/2}}\hskip
61.17325pt\Longrightarrow\hskip 28.45274pt\frac{\partial\zeta}{\partial
x}\Big{|}_{\xi}=\frac{P}{\tau}$
$\displaystyle\partial_{\xi}(\sinh\zeta)\big{|}_{x}$
$\displaystyle=\partial_{\xi}\zeta\big{|}_{x}\,\cosh\zeta\,=\frac{(E-V)(M^{2}-EV)}{(V^{\prime}\tau)^{3/2}}\hskip
28.45274pt\Longrightarrow\hskip
28.45274pt\frac{\partial\zeta}{\partial\xi}\Big{|}_{x}=\frac{M^{2}-EV}{V^{\prime}\tau}$
(A.600)

Using these in (A.599) gives (309).

###  A.14 Derive the expression (VII.3.1).

The expressions (317) for $\phi_{1}(\tau)$ and $\phi_{0}(\tau)$ at large
$\tau$ are,

$\displaystyle\phi_{1}(|\tau|\to\infty)$
$\displaystyle=\frac{4V^{\prime}}{\sqrt{\pi}\,m}\,\sqrt{e^{\pi
m^{2}/V^{\prime}}-1}\,e^{-\theta(-\tau)\pi
m^{2}/2V^{\prime}}\sin\big{[}{\textstyle\frac{1}{4}}\tau-{\textstyle\frac{1}{2}}m^{2}/V^{\prime}\log({\textstyle\frac{1}{2}}|\tau|)+\arg\Gamma(1+im^{2}/2V^{\prime})\big{]}$
$\displaystyle\phi_{0}(|\tau|\to\infty)$
$\displaystyle=-i\frac{4V^{\prime}}{\sqrt{\pi}\,m}\,\sqrt{e^{\pi
m^{2}/V^{\prime}}-1}\,e^{-\theta(-\tau)\pi
m^{2}/2V^{\prime}}\cos\big{[}{\textstyle\frac{1}{4}}\tau-{\textstyle\frac{1}{2}}m^{2}/V^{\prime}\log({\textstyle\frac{1}{2}}|\tau|)+\arg\Gamma(1+im^{2}/2V^{\prime})\big{]}$
(A.601)

Since state $B$ has positive parity
$\phi_{B1}[\tau_{B}(x=0)]=\phi_{B1}(\tau_{B}=M_{B}^{2}/V^{\prime})=0$
according to (305). This determines the masses $M_{Bn}$ in the Bj limit,

$\displaystyle{\textstyle\frac{1}{4}}M_{Bn}^{2}/V^{\prime}-{\textstyle\frac{1}{2}}m^{2}/V^{\prime}\log({\textstyle\frac{1}{2}}M_{Bn}^{2}/V^{\prime})+\arg\Gamma(1+im^{2}/2V^{\prime})=n\cdot\pi$
(A.602)

where $n$ is a large positive integer. Subtracting the lhs. from the arguments
of the sin and cos functions in (A.14) gives a sign $(-1)^{n}$ to $\phi_{B0}$
and $\phi_{B1}$. Using also $-2E_{B}\,x=2P_{B}^{1}\,x-2P_{A}^{+}(1-{x_{Bj}})x$
from (346) gives

$\displaystyle\phi_{B1}(|\tau_{B}|\to\infty)=$
$\displaystyle\frac{4V^{\prime}(-1)^{n}}{\sqrt{\pi}\,m}\,\sqrt{e^{\pi
m^{2}/V^{\prime}}-1}\,\exp\big{[}-\theta(-\tau_{B})\pi
m^{2}/2V^{\prime}\big{]}$
$\displaystyle\times\sin\Big{[}{\textstyle\frac{1}{2}}P_{B}^{1}x-{\textstyle\frac{1}{2}}P_{A}^{+}(1-{x_{Bj}})x+{\textstyle\frac{1}{4}}V^{\prime}x^{2}-\frac{m^{2}}{2V^{\prime}}\log\Big{|}1-\frac{V^{\prime}x}{P_{A}^{+}(1-{x_{Bj}})}\Big{|}\Big{]}$
$\displaystyle\phi_{B0}(|\tau_{B}|\to\infty)=$
$\displaystyle-i\frac{4V^{\prime}(-1)^{n}}{\sqrt{\pi}\,m}\,\sqrt{e^{\pi
m^{2}/V^{\prime}}-1}\,\exp\big{[}-\theta(-\tau_{B})\pi
m^{2}/2V^{\prime}\big{]}$
$\displaystyle\times\cos\Big{[}{\textstyle\frac{1}{2}}P_{B}^{1}x-{\textstyle\frac{1}{2}}P_{A}^{+}(1-{x_{Bj}})x+{\textstyle\frac{1}{4}}V^{\prime}x^{2}-\frac{m^{2}}{2V^{\prime}}\log\Big{|}1-\frac{V^{\prime}x}{P_{A}^{+}(1-{x_{Bj}})}\Big{|}\Big{]}$
(A.603)

The asymptotically large phase ${\textstyle\frac{1}{2}}P_{B}^{1}x$ must cancel
the one in the factor
$\sin\big{(}{\textstyle\frac{1}{2}}q^{1}x\big{)}=\sin\big{[}{\textstyle\frac{1}{2}}(P_{B}^{1}-P_{A}^{1})x\big{]}$
of (VII.3.1) to give a non-vanishing result. Using

$\displaystyle\sin\alpha\sin\beta$
$\displaystyle={\textstyle\frac{1}{2}}\big{[}\cos(\alpha-\beta)-\cos(\alpha+\beta)\big{]}$
$\displaystyle\sin\alpha\cos\beta$
$\displaystyle={\textstyle\frac{1}{2}}\big{[}\sin(\alpha+\beta)+\sin(\alpha-\beta)\big{]}$
(A.604)

and defining the angle

$\displaystyle\varphi_{B}(x)\equiv{\textstyle\frac{1}{2}}\big{[}P_{A}^{+}(1-{x_{Bj}})-P_{A}^{1}\big{]}x+\frac{m^{2}}{2V^{\prime}}\log\Big{|}1-\frac{V^{\prime}x}{P_{A}^{+}(1-{x_{Bj}})}\Big{|}-{\textstyle\frac{1}{4}}V^{\prime}x^{2}$
(A.605)

the expression (VII.3.1) for $G_{AB}$ gives (VII.3.1) when terms with
asymptotically large phases are neglected,

###  A.15 Do the $x$-integral in (VII.3.1) numerically for the parameters in
Fig. 16, and compare.

Separate the integral in the form factor (VII.3.1) into three parts,

$\displaystyle E_{B}G(q^{2})$
$\displaystyle=(-1)^{n}\frac{16iV^{\prime}}{\sqrt{\pi}m}\,\sqrt{e^{\pi
m^{2}/V^{\prime}}-1}\,(I_{1}+I_{2}+I_{3})$ (A.606) $\displaystyle I_{1}$
$\displaystyle=\int_{0}^{\infty}dx\big{[}i\sin\varphi_{B}\,\phi_{A0}(\tau_{A})+\cos\varphi_{B}\,\phi_{A1}(\tau_{A})\big{]}e^{-\theta(x-x_{0})\pi
m^{2}/2V^{\prime}}$ $\displaystyle I_{2}$
$\displaystyle=\int_{0}^{x_{0}}dx\,\cos\varphi_{B}\,\phi_{A1}(\tau_{A})\,\frac{2m^{2}}{V^{\prime}(x_{0}-x)(P_{A}^{+}-V^{\prime}x)}$
$\displaystyle I_{3}$
$\displaystyle=-\int_{x_{0}}^{\infty}dx\,\cos\varphi_{B}\,\phi_{A1}(\tau_{A})\,\frac{2m^{2}}{V^{\prime}(x-x_{0})(P_{A}^{+}-V^{\prime}x)}e^{-\pi
m^{2}/2V^{\prime}}$ (A.607)

The $I_{1}$ integrand oscillates with constant amplitude at large $x$, where
the approximation (A.14) for $\phi_{A}(\tau_{A}\to\infty)$ applies. $I_{1}$ is
further divided into three parts. In $I_{1a}$ the range is $0<x<x_{1}$, where
$x_{1}>x_{0}$ and $\tau_{A}(x_{1})>0$, ensuring that $\theta(-\tau_{A})=0,\
\theta(-\tau_{B})=1$ for $x>x_{1}$. $I_{1b}$ integrates over $x_{1}\leq
x\leq\infty$ with $\phi_{A0}$ and $\phi_{A1}$ replaced by the difference with
their large $x$ approximations, $\phi_{A}-\phi_{A}^{as}$. Finally $I_{1c}$
integrates $\phi_{A}^{as}$over $x_{1}\leq x\leq\infty$:

$\displaystyle I_{1a}$
$\displaystyle=\int_{0}^{x_{1}}dx\big{[}i\sin\varphi_{B}\,\phi_{A0}(\tau_{A})+\cos\varphi_{B}\,\phi_{A1}(\tau_{A})\big{]}e^{-\theta(x-x_{0})\pi
m^{2}/2V^{\prime}}$ $\displaystyle I_{1b}$
$\displaystyle=\int_{x_{1}}^{\infty}dx\big{\\{}i\sin\varphi_{B}[\phi_{A0}(\tau_{A})-\phi_{A0}^{as}(\tau_{A})]+\cos\varphi_{B}[\phi_{A1}(\tau_{A})-\phi_{A1}^{as}(\tau_{A})]\big{\\}}e^{-\pi
m^{2}/2V^{\prime}}$ (A.608) $\displaystyle I_{1c}$
$\displaystyle=\int_{x_{1}}^{\infty}dx\big{\\{}i\sin\varphi_{B}\,\phi_{A0}^{as}(\tau_{A})+\cos\varphi_{B}\,\phi_{A1}^{as}(\tau_{A})\big{\\}}e^{-\pi
m^{2}/2V^{\prime}}$

The oscillations in $I_{1b}$ at large $x$ are damped, allowing a numerical
integration. The phase in $\phi_{A}^{as}$ (A.14) is

$\displaystyle\varphi_{A}={\textstyle\frac{1}{4}}\tau_{A}-{\textstyle\frac{1}{2}}m^{2}/V^{\prime}\log({\textstyle\frac{1}{2}}|\tau_{A}|)+\arg\Gamma(1+im^{2}/2V^{\prime})$
(A.609)

The $I_{1c}$ integral reduces to

$\displaystyle I_{1c}$
$\displaystyle=-\frac{4V^{\prime}}{\sqrt{\pi}\,m}\sqrt{1-e^{-\pi
m^{2}/V^{\prime}}}\int_{x_{1}}^{\infty}dx\,\sin\varphi_{C}$
$\displaystyle\varphi_{C}$
$\displaystyle\equiv-(\varphi_{A}+\varphi_{B})={\textstyle\frac{1}{2}}P_{A}^{+}{x_{Bj}}x-{\textstyle\frac{1}{4}}M_{A}^{2}/V^{\prime}-\arg\Gamma(1+im^{2}/2V^{\prime})+\frac{m^{2}}{2V^{\prime}}\big{[}\log({\textstyle\frac{1}{2}}\tau_{A})-\log(x/x_{0}-1)\big{]}$
(A.610)

This integral is evaluated by rotating the contour in $u\equiv x-x_{1}$ by
$\pi/2$ to ensure exponential convergence,

$\displaystyle\int_{x_{1}}^{\infty}dx\,\sin\varphi_{C}={\rm
Im}\int_{x_{1}}^{\infty}dx\,e^{i\varphi_{C}(x)}={\rm
Im}\int_{0}^{i\infty}du\,e^{i\varphi_{C}(u+x_{1})}$ (A.611)

In $I_{2}$ the range $0<x<x_{0}$ is transformed into $0<y<\infty$ through

$\displaystyle y=-\log(1-x/x_{0})\hskip 56.9055ptx=x_{0}(1-e^{-y})\hskip
56.9055pt\frac{dx}{x_{0}-x}=dy$ (A.612)

The $y$-integration is further split into $0<y<y_{2}$ and $y_{2}<y<\infty$.
The first path is finite and readily integrated numerically. The latter path
is rotated by $\pi/2$, giving exponential convergence in
$y_{2}<y<y_{2}+i\infty$ when using $\cos\varphi_{B}={\rm
Re}\exp(-i\varphi_{B})$, due to the term $ym^{2}/2V^{\prime}$ in
$-\varphi_{B}$ (VII.3.1). $x\simeq x_{0}$ is constant on the complex path when
$y_{2}$ is large, allowing the integral to be be evaluated analytically. The
result for $I_{2}$ should be independent of $y_{2}$.

In $I_{3}$ the integration contour is split into $x_{0}<x<2x_{0}$ and
$2x_{0}<x<\infty$. In the first range the integration variable is changed to

$\displaystyle y=-\log(x/x_{0}-1)\hskip 56.9055ptx=x_{0}(1+e^{-y})\hskip
56.9055pt\frac{dx}{x_{0}-x}=-dy$ (A.613)

The $y$-range $0<y<\infty$ is further split into $0<y<y_{3}$ and
$y_{3}<y<\infty$. The first is integrated numerically, and in the second the
path is rotated by $\pi/2$, ranging over $y_{3}<y<y_{3}+i\infty$. For large
$y_{3}$ the value of $x\simeq x_{0}$ is constant on the complex contour,
allowing an analytic integration. The integration over $2x_{0}<x<\infty$ is
numerically stable, as the oscillations at large $x$ are damped.

###  A.16 Derive the $q\bar{q}g$ potential (VIII.1.4)

Using the commutators in (VIII.1.1) and (VIII.1.3) the operation of
$\mathcal{E}_{a}({\boldsymbol{y}})$ (VIII.1) on the
$\left|{q\bar{q}g}\right\rangle$ state (379) gives

$\displaystyle\mathcal{E}_{a}({\boldsymbol{y}})\left|{q\bar{q}g}\right\rangle$
$\displaystyle=\big{\\{}\bar{\psi}_{A^{\prime}}({\boldsymbol{x}}_{1})T_{A^{\prime}A}^{a}A_{b}^{i}({\boldsymbol{x}}_{g})T_{AB}^{b}\psi_{B}({\boldsymbol{x}}_{2})\delta({\boldsymbol{y}}-{\boldsymbol{x}}_{1})$
$\displaystyle+\bar{\psi}_{A}({\boldsymbol{x}}_{1})if_{abc}A_{c}^{i}({\boldsymbol{x}}_{g})T_{AB}^{b}\psi_{B}({\boldsymbol{x}}_{2})\delta({\boldsymbol{y}}-{\boldsymbol{x}}_{g})$
$\displaystyle-\bar{\psi}_{A}({\boldsymbol{x}}_{1})A_{b}^{i}({\boldsymbol{x}}_{g})T_{AB}^{b}T_{BB^{\prime}}^{a}\psi_{B^{\prime}}({\boldsymbol{x}}_{2})\delta({\boldsymbol{y}}-{\boldsymbol{x}}_{2})\big{\\}}\left|{0}\right\rangle$
(A.614)

When $\mathcal{E}_{a}({\boldsymbol{y}})$ and
$\mathcal{E}_{a}({\boldsymbol{z}})$ in $\mathcal{H}_{V}^{(0)}$ (V.3.2) act on
the same (quark or gluon) constituent we may use the previous results
(VIII.1.1) and (376) showing that the coefficients of
${\boldsymbol{x}}_{1}^{2}$ and ${\boldsymbol{x}}_{2}^{2}$ are $C_{F}$ while
that of ${\boldsymbol{x}}_{g}^{2}$ is $N_{c}$, multiplied by the common factor
$\big{(}{\textstyle\frac{1}{2}}\kappa^{2}{\textstyle\int}d{\boldsymbol{x}}+g\kappa\big{)}$.
The new contributions are

$\displaystyle{\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{2}:$
$\displaystyle\hskip
28.45274pt-2\,\bar{\psi}_{A^{\prime}}({\boldsymbol{x}}_{1})T^{a}_{A^{\prime}A}\,A_{b}^{i}({\boldsymbol{x}}_{g})T^{b}_{AB}\,T^{a}_{BB^{\prime}}\psi_{B^{\prime}}({\boldsymbol{x}}_{2})\left|{0}\right\rangle=\frac{1}{N_{c}}\left|{q\bar{q}g}\right\rangle$
$\displaystyle{\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{g}:$
$\displaystyle\hskip
28.45274pt2\,\bar{\psi}_{A^{\prime}}({\boldsymbol{x}}_{1})T^{a}_{A^{\prime}A}\,if_{abc}A_{c}^{i}({\boldsymbol{x}}_{g})\,T^{b}_{AB}\psi_{B}({\boldsymbol{x}}_{2})\left|{0}\right\rangle=-N_{c}\left|{qg\bar{q}}\right\rangle$
$\displaystyle{\boldsymbol{x}}_{2}\cdot{\boldsymbol{x}}_{g}:$
$\displaystyle\hskip
28.45274pt-2\bar{\psi}_{A}({\boldsymbol{x}}_{1})\,if_{abc}A_{c}^{i}({\boldsymbol{x}}_{g})\,T^{b}_{AB}\,T^{a}_{BB^{\prime}}\psi_{B^{\prime}}({\boldsymbol{x}}_{2})\left|{0}\right\rangle=-N_{c}\left|{qg\bar{q}}\right\rangle$
(A.615)

Altogether,

$\displaystyle\mathcal{H}_{V}^{(0)}\left|{q\bar{q}g}\right\rangle$
$\displaystyle=\big{(}{\textstyle\frac{1}{2}}\kappa^{2}{\textstyle\int}d{\boldsymbol{x}}+g\kappa\big{)}\big{[}C_{F}({\boldsymbol{x}}_{1}^{2}+{\boldsymbol{x}}_{2}^{2})+N_{c}{\boldsymbol{x}}_{g}^{2}-N_{c}({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})\cdot{\boldsymbol{x}}_{g}+{\textstyle\frac{1}{N_{c}}}{\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{2}\big{]}\left|{q\bar{q}g}\right\rangle$
$\displaystyle=\big{(}{\textstyle\frac{1}{2}}\kappa^{2}{\textstyle\int}d{\boldsymbol{x}}+g\kappa\big{)}\big{[}d_{q\bar{q}g}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2},{\boldsymbol{x}}_{g})\big{]}^{2}\left|{q\bar{q}g}\right\rangle$
$\displaystyle
d_{q\bar{q}g}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2},{\boldsymbol{x}}_{g})$
$\displaystyle\equiv\sqrt{{\textstyle\frac{1}{4}}(N_{c}-{\textstyle\frac{2}{N_{c}}})({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})^{2}+N_{c}({\boldsymbol{x}}_{g}-{\textstyle\frac{1}{2}}{\boldsymbol{x}}_{1}-{\textstyle\frac{1}{2}}{\boldsymbol{x}}_{2})^{2}}$
(A.616)

For the ${\cal O}\left(\kappa^{2}\right)$ term to give the universal energy
$E_{\Lambda}$ (365) we need to choose the normalization of the homogeneous
solution as

$\displaystyle\kappa_{q\bar{q}g}=\frac{\Lambda^{2}}{g\sqrt{C_{F}}}\,\frac{1}{d_{q\bar{q}g}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{g},{\boldsymbol{x}}_{2})}$
(A.617)

The ${\cal O}\left(g\kappa\right)$ contribution to $\mathcal{H}_{V}$ gives the
potential,

$\displaystyle
V_{q\bar{q}g}^{(0)}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2},{\boldsymbol{x}}_{g})=g\kappa_{q\bar{q}g}\,\big{[}d_{q\bar{q}g}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2},{\boldsymbol{x}}_{g})\big{]}^{2}=\frac{\Lambda^{2}}{\sqrt{C_{F}}}\,d_{q\bar{q}g}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2},{\boldsymbol{x}}_{g})$
(A.618)

When the self-energies are subtracted $\mathcal{H}_{V}^{(1)}$ has
contributions only from the three terms in (A.16),

$\displaystyle
V_{q\bar{q}g}^{(1)}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2},{\boldsymbol{x}}_{g})={\textstyle\frac{1}{2}}\,{\alpha_{s}}\Big{[}\frac{1}{N_{c}}\,\frac{1}{|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}|}-N_{c}\Big{(}\frac{1}{|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{g}|}+\frac{1}{|{\boldsymbol{x}}_{2}-{\boldsymbol{x}}_{g}|}\Big{)}\Big{]}$
(A.619)

###  A.17 Derive the expression for $\mathcal{H}_{V}^{(0)}\left|{8\otimes
8}\right\rangle$ in (VIII.1.7)

Recall from (391),

$\displaystyle\left|{8\otimes
8}\right\rangle=\bar{\psi}_{A}({\boldsymbol{x}}_{1})T_{AB}^{b}\psi_{B}({\boldsymbol{x}}_{2})\,\bar{\psi}_{C}({\boldsymbol{x}}_{3})T_{CD}^{b}\psi_{D}({\boldsymbol{x}}_{4})\left|{0}\right\rangle$
$\displaystyle\bar{\psi}_{A}({\boldsymbol{x}}_{1})\psi_{B}({\boldsymbol{x}}_{2})\,\bar{\psi}_{B}({\boldsymbol{x}}_{3})\psi_{A}({\boldsymbol{x}}_{4})\left|{0}\right\rangle=2\left|{8\otimes
8}\right\rangle+{\textstyle\frac{1}{N_{c}}}\left|{1\otimes 1}\right\rangle$
(A.620)

In the following I leave out the common factor
$\big{(}{\textstyle\frac{1}{2}}\kappa^{2}{\textstyle\int}d{\boldsymbol{x}}+g\kappa\big{)}$
in $\mathcal{H}_{V}^{(0)}$ (VIII.1),

$\displaystyle\mathcal{H}_{V}^{(0)}$
$\displaystyle=\big{(}{\textstyle\frac{1}{2}}\kappa^{2}{\textstyle\int}d{\boldsymbol{x}}+g\kappa\big{)}\int
d{\boldsymbol{y}}\,d{\boldsymbol{z}}\,{\boldsymbol{y}}\cdot{\boldsymbol{z}}\,\mathcal{E}_{a}({\boldsymbol{y}})\mathcal{E}_{a}({\boldsymbol{z}})$
(A.621)

and make use of the commutation relations (VIII.1.1),

$\displaystyle\big{[}{\mathcal{E}_{a}({\boldsymbol{x}})},{\bar{\psi}_{A}({\boldsymbol{x}}_{1})}\big{]}=\bar{\psi}_{A^{\prime}}({\boldsymbol{x}}_{1})T_{A^{\prime}A}^{a}\delta({\boldsymbol{x}}-{\boldsymbol{x}}_{1})\hskip
56.9055pt\big{[}{\mathcal{E}_{a}({\boldsymbol{x}})},{\psi_{A}({\boldsymbol{x}}_{2})}\big{]}=-T_{AA^{\prime}}^{a}\psi_{A^{\prime}}({\boldsymbol{x}}_{2})\delta({\boldsymbol{x}}-{\boldsymbol{x}}_{2})$
(A.622)

and of the SU($N_{c}$) generator relations (VIII.1.1).

The commutators of $\mathcal{E}_{a}({\boldsymbol{y}})$ and
$\mathcal{E}_{a}({\boldsymbol{z}})$ with the same quark at
${\boldsymbol{x}}_{i}\ (i=1,\ldots 4)$ in $\left|{8\otimes 8}\right\rangle$
gives ${\boldsymbol{y}}\cdot{\boldsymbol{z}}={\boldsymbol{x}}_{i}^{2}$, color
factor $T^{a}T^{a}=C_{F}\,I$ and state $\left|{8\otimes 8}\right\rangle$. The
commutators with $\bar{\psi}({\boldsymbol{x}}_{1})$ and
$\psi({\boldsymbol{x}}_{2})$ gives

$\displaystyle{\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{2}:$
$\displaystyle\hskip
28.45274pt-2\bar{\psi}_{A^{\prime}}({\boldsymbol{x}}_{1})T_{A^{\prime}A}^{a}T_{AB}^{b}T_{BB^{\prime}}^{a}\psi_{B^{\prime}}({\boldsymbol{x}}_{2})\bar{\psi}_{C}({\boldsymbol{x}}_{3})T_{CD}^{b}\psi_{D}({\boldsymbol{x}}_{4})\left|{0}\right\rangle={\textstyle\frac{1}{N_{c}}}\left|{8\otimes
8}\right\rangle$ (A.623)

The commutators with $\bar{\psi}({\boldsymbol{x}}_{1})$ and
$\bar{\psi}({\boldsymbol{x}}_{3})$ give

$\displaystyle{\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{3}:$
$\displaystyle\hskip
28.45274pt2\bar{\psi}_{A^{\prime}}({\boldsymbol{x}}_{1})T_{A^{\prime}A}^{a}T_{AB}^{b}\psi_{B}({\boldsymbol{x}}_{2})\bar{\psi}_{C^{\prime}}({\boldsymbol{x}}_{3})T_{C^{\prime}C}^{a}T_{CD}^{b}\psi_{D}({\boldsymbol{x}}_{4})\left|{0}\right\rangle$
(A.624)

The color factors

$\displaystyle{\textstyle\frac{1}{2}}(\delta_{A^{\prime}C}\delta_{AC^{\prime}}-{\textstyle\frac{1}{N_{c}}}\delta_{A^{\prime}A}\delta_{C^{\prime}C})T_{AB}^{b}T_{CD}^{b}={\textstyle\frac{1}{4}}\delta_{A^{\prime}C}\delta_{AC^{\prime}}(\delta_{AD}\delta_{BC}-{\textstyle\frac{1}{N_{c}}}\delta_{AB}\delta_{CD})-{\textstyle\frac{1}{2N_{c}}}\delta_{A^{\prime}A}\delta_{C^{\prime}C}T_{AB}^{b}T_{CD}^{b}$
(A.625)

give for the coefficient of ${\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{3}$ in
(A.624),

$\displaystyle\big{[}{\textstyle\frac{1}{2}}\bar{\psi}_{B}({\boldsymbol{x}}_{1})\psi_{B}({\boldsymbol{x}}_{2})\bar{\psi}_{D}({\boldsymbol{x}}_{3})\psi_{D}({\boldsymbol{x}}_{4})\left|{0}\right\rangle-{\textstyle\frac{1}{2N_{c}}}\bar{\psi}_{D}({\boldsymbol{x}}_{1})\psi_{B}({\boldsymbol{x}}_{2})\bar{\psi}_{B}({\boldsymbol{x}}_{3})\psi_{D}({\boldsymbol{x}}_{4})\big{]}\left|{0}\right\rangle-{\textstyle\frac{1}{N_{c}}}\left|{8\otimes
8}\right\rangle={\textstyle\frac{C_{F}}{N_{c}}}\left|{1\otimes
1}\right\rangle-{\textstyle\frac{2}{N_{c}}}\left|{8\otimes 8}\right\rangle$
(A.626)

The commutators with $\bar{\psi}({\boldsymbol{x}}_{1})$ and
$\bar{\psi}({\boldsymbol{x}}_{4})$ give

$\displaystyle{\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{4}:$
$\displaystyle\hskip
28.45274pt-2\bar{\psi}_{A^{\prime}}({\boldsymbol{x}}_{1})T_{A^{\prime}A}^{a}T_{AB}^{b}\psi_{B}({\boldsymbol{x}}_{2})\bar{\psi}_{C}({\boldsymbol{x}}_{3})T_{CD}^{b}T_{DD^{\prime}}^{a}\psi_{D^{\prime}}({\boldsymbol{x}}_{4})\left|{0}\right\rangle$
(A.627)

Now the color factors

$\displaystyle{\textstyle\frac{1}{2}}(\delta_{A^{\prime}D^{\prime}}\delta_{AD}-{\textstyle\frac{1}{N_{c}}}\delta_{A^{\prime}A}\delta_{D^{\prime}D})T_{AB}^{b}T_{CD}^{b}={\textstyle\frac{1}{4}}\delta_{A^{\prime}D^{\prime}}\delta_{AD}(\delta_{AD}\delta_{BC}-{\textstyle\frac{1}{N_{c}}}\delta_{AB}\delta_{CD})-{\textstyle\frac{1}{2N_{c}}}\delta_{A^{\prime}A}\delta_{DD^{\prime}}T_{AB}^{b}T_{CD}^{b}$
(A.628)

give for the coefficient of ${\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{4}$ in
(A.627),

$\displaystyle\big{(}-{\textstyle\frac{N_{c}}{2}}+{\textstyle\frac{1}{2N_{c}}}\big{)}\bar{\psi}_{A^{\prime}}({\boldsymbol{x}}_{1})\psi_{B}({\boldsymbol{x}}_{2})\bar{\psi}_{B}({\boldsymbol{x}}_{3})\psi_{A^{\prime}}({\boldsymbol{x}}_{4})\left|{0}\right\rangle+{\textstyle\frac{1}{N_{c}}}\left|{8\otimes
8}\right\rangle=-{\textstyle\frac{C_{F}}{N_{c}}}\left|{1\otimes
1}\right\rangle-{\textstyle\frac{N_{c}^{2}-2}{N_{c}}}\left|{8\otimes
8}\right\rangle$ (A.629)

The coefficients of ${\boldsymbol{x}}_{2}\cdot{\boldsymbol{x}}_{3},\
{\boldsymbol{x}}_{2}\cdot{\boldsymbol{x}}_{4}$ and
${\boldsymbol{x}}_{3}\cdot{\boldsymbol{x}}_{4}$ are the same as those of
${\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{4},\
{\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{3}$ and
${\boldsymbol{x}}_{1}\cdot{\boldsymbol{x}}_{2}$, respectively. Altogether,

$\displaystyle\mathcal{H}_{V}^{(0)}\left|{8\otimes 8}\right\rangle$
$\displaystyle=\big{(}{\textstyle\frac{1}{2}}\kappa^{2}{\textstyle\int}d{\boldsymbol{x}}+g\kappa\big{)}\Big{\\{}\big{[}{\textstyle\frac{1}{2}}N_{c}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{4})^{2}+{\textstyle\frac{1}{2}}N_{c}({\boldsymbol{x}}_{2}-{\boldsymbol{x}}_{3})^{2}-{\textstyle\frac{1}{2N_{c}}}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})^{2}$
$\displaystyle-{\textstyle\frac{1}{2N_{c}}}({\boldsymbol{x}}_{3}-{\boldsymbol{x}}_{4})^{2}-{\textstyle\frac{2}{N_{c}}}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\cdot({\boldsymbol{x}}_{3}-{\boldsymbol{x}}_{4})\big{]}\left|{8\otimes
8}\right\rangle+{\textstyle\frac{C_{F}}{N_{c}}}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\cdot({\boldsymbol{x}}_{3}-{\boldsymbol{x}}_{4})\left|{1\otimes
1}\right\rangle\Big{\\}}$ (A.630)

When expressed in terms of the separations (393) this gives (VIII.1.7).

###  A.18 Verify that the expression (VIII.2.3) for
$\Phi_{-+}({\boldsymbol{x}})$ satisfies the bound state equation (405) given
the radial equation (417).

The BSE as in (405) applied to $\Phi_{-+}({\boldsymbol{x}})$ in the
alternative forms of (VIII.2.3),

$\displaystyle{\overset{\rightarrow}{\mathfrak{h}}}_{-}\Phi({\boldsymbol{x}})+\Phi({\boldsymbol{x}}){\overset{\leftarrow}{\mathfrak{h}}}_{-}=0$
$\displaystyle\Phi_{-+}({\boldsymbol{x}})={\overset{\rightarrow}{\mathfrak{h}}}_{+}\gamma_{5}\,F_{1}(r)Y_{j\lambda}(\hat{\boldsymbol{x}})=F_{1}(r)Y_{j\lambda}(\hat{\boldsymbol{x}})\,\gamma_{5}{\overset{\leftarrow}{\mathfrak{h}}}_{+}$
(A.631)

allows the use of

$\displaystyle{\overset{\rightarrow}{\mathfrak{h}}}_{-}{\overset{\rightarrow}{\mathfrak{h}}}_{+}$
$\displaystyle=\frac{4}{(M-V)^{2}}(-{\overset{\rightarrow}{\boldsymbol{\nabla}}}^{2}+m^{2})-1+\frac{4iV^{\prime}}{r(M-V)^{3}}\,{\boldsymbol{\alpha}}\cdot{\boldsymbol{x}}\,(i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}+m\gamma^{0})$
$\displaystyle{\overset{\leftarrow}{\mathfrak{h}}}_{+}{\overset{\leftarrow}{\mathfrak{h}}}_{-}$
$\displaystyle=(-{\overset{\leftarrow}{\boldsymbol{\nabla}}}^{2}+m^{2})\frac{4}{(M-V)^{2}}-1+(i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}-m\gamma^{0})\,{\boldsymbol{\alpha}}\cdot{\boldsymbol{x}}\,\frac{4iV^{\prime}}{r(M-V)^{3}}$
(A.632)

Moving the $\gamma_{5}$ to the right in the BSE,

$\displaystyle{\overset{\rightarrow}{\mathfrak{h}}}_{-}{\overset{\rightarrow}{\mathfrak{h}}}_{+}\gamma_{5}F_{1}(r)Y_{j\lambda}(\hat{\boldsymbol{x}})$
$\displaystyle+F_{1}(r)Y_{j\lambda}(\hat{\boldsymbol{x}})\gamma_{5}{\overset{\leftarrow}{\mathfrak{h}}}_{+}{\overset{\leftarrow}{\mathfrak{h}}}_{-}$
$\displaystyle=\Big{[}\frac{8}{(M-V)^{2}}(-\boldsymbol{\nabla}^{2}+m^{2})-2+\frac{4iV^{\prime}x^{j}}{r(M-V)^{3}}\big{\\{}{\alpha_{j}},{i{\buildrel\rightarrow\over{\partial}}_{k}\alpha_{k}+m\gamma^{0}}\big{\\}}\Big{]}F_{1}(r)Y_{j\lambda}(\hat{\boldsymbol{x}})\gamma_{5}$
(A.633)

Using $\left\\{{\alpha_{j}},{\alpha_{k}}\right\\}=2\delta_{jk}$,
$\left\\{{\alpha_{j}},{\gamma^{0}}\right\\}=0$,
$x^{j}{\buildrel\rightarrow\over{\partial}}_{j}=r{\overset{\rightarrow}{\partial}_{r}}$
and
$\boldsymbol{\nabla}^{2}=(1/r^{2})\partial_{r}(r^{2}\partial_{r})-{\boldsymbol{L}}^{2}/r^{2}$
with
${\boldsymbol{L}}^{2}Y_{j\lambda}(\hat{\boldsymbol{x}})=j(j+1)Y_{j\lambda}(\hat{\boldsymbol{x}})$
gives the radial equation (417).

###  A.19 Derive the coupled equations (VIII.3.1) from the bound state
equation (456).

I make use of commutator identities such as,

$\displaystyle\left[{A},{BC}\right]$ $\displaystyle=$
$\displaystyle\left[{A},{B}\right]C+B\left[{A},{C}\right]$ (A.634)
$\displaystyle\left\\{{A},{BC}\right\\}$ $\displaystyle=$
$\displaystyle\left[{A},{B}\right]C+B\left\\{{A},{C}\right\\}=\left\\{{A},{B}\right\\}C-B\left[{A},{C}\right]$
(A.635) $\displaystyle\left\\{{A},{\left\\{{B},{C}\right\\}}\right\\}$
$\displaystyle=$ $\displaystyle-\left[{B},{\left[{A},{C}\right]}\right]\hskip
32.72049pt{\rm when}\ \ \big{\\{}{A},{B}\big{\\}}=0$ (A.636)
$\displaystyle\left\\{{A},{\left[{B},{C}\right]}\right\\}$ $\displaystyle=$
$\displaystyle-\left\\{{B},{\left[{A},{C}\right]}\right\\}\hskip
28.45274pt{\rm when}\ \ \big{\\{}{A},{B}\big{\\}}=0$ (A.637)
$\displaystyle\left[{A},{\left\\{{B},{C}\right\\}}\right]$ $\displaystyle=$
$\displaystyle-\left[{B},{\left\\{{A},{C}\right\\}}\right]\hskip
28.45274pt{\rm when}\ \ \big{\\{}{A},{B}\big{\\}}=0$ (A.638)
$\displaystyle\left\\{{A},{\left[{A},{C}\right]}\right\\}$ $\displaystyle=$
$\displaystyle\left[{A},{\left\\{{A},{C}\right\\}}\right]=\left[{A^{2}},{C}\right]$
(A.639) $\displaystyle\left\\{{A},{\left\\{{A},{C}\right\\}}\right\\}$
$\displaystyle=$ $\displaystyle 2A\left\\{{A},{C}\right\\}\hskip
42.67912pt{\rm when}\ \ A^{2}=1$ (A.640)
$\displaystyle\left[{A},{\left[{A},{C}\right]}\right]$ $\displaystyle=$
$\displaystyle 2A\left[{A},{C}\right]\hskip 46.94687pt{\rm when}\ \ A^{2}=1$
(A.641)

Taking the commutator
$i\boldsymbol{\nabla}\cdot\left[{{\boldsymbol{\alpha}}},{\rm BSE}\right]$ of
the bound state equation (456) gives

$\displaystyle\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{(E-V)\Phi^{(P)}}\right]$
$\displaystyle=\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\big{\\{}{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\big{\\}}}\right]-{\textstyle\frac{1}{2}}\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\big{[}{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\big{]}}\right]+m\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\big{[}{\gamma^{0}},{\Phi^{(P)}}\big{]}}\right]$
(A.642)

The first term on the rhs. vanishes due to the commutator identity (A.639),
when we recall that $\boldsymbol{\nabla}$ in the BSE always operates on
$\Phi^{(P)}$. The identity (A.636) implies for the third term on the rhs. of
(A.642),

$\displaystyle
m\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\big{[}{\gamma^{0}},{\Phi^{(P)}}\big{]}}\right]=-m\left\\{{\gamma^{0}},{\big{\\{}{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\big{\\}}}\right\\}$
(A.643)

Using the original BSE (456) on the rhs. of (A.643) we get

$\displaystyle
m\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\big{[}{\gamma^{0}},{\Phi^{(P)}}\big{]}}\right]$
$\displaystyle=-m\left\\{{\gamma^{0}},{{\textstyle\frac{1}{2}}\left[{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\right]}\right\\}+m^{2}\left\\{{\gamma^{0}},{\left[{\gamma^{0}},{\Phi^{(P)}}\right]}\right\\}-m\left\\{{\gamma^{0}},{(E-V)\Phi^{(P)}}\right\\}$
$\displaystyle={\textstyle\frac{1}{2}}m\left\\{{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\left[{\gamma^{0}},{\Phi^{(P)}}\right]}\right\\}-m(E-V)\left\\{{\gamma^{0}},{\Phi^{(P)}}\right\\}$
$\displaystyle={\textstyle\frac{1}{2}}\left\\{{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{-\big{\\{}{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\big{\\}}+{\textstyle\frac{1}{2}}\big{[}{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\big{]}+(E-V)\Phi^{(P)}}\right\\}-m(E-V)\left\\{{\gamma^{0}},{\Phi^{(P)}}\right\\}$
(A.644)

where I used (A.637), (A.639) and in the last step expressed
$m\left[{\gamma^{0}},{\Phi_{P}}\right]$ using the BSE (456). The second term
on the rhs. of (A.19) vanishes according to (A.639). Inserting this result in
(A.642) we have

$\displaystyle\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{(E-V)\Phi^{(P)}}\right]$
$\displaystyle=$
$\displaystyle-{\textstyle\frac{1}{2}}\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\left[{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\right]}\right]-{\textstyle\frac{1}{2}}\left\\{{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\left\\{{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\right\\}}\right\\}+{\textstyle\frac{1}{2}}(E-V)\left[{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\right]-m(E-V)\left\\{{\gamma^{0}},{\Phi^{(P)}}\right\\}$
(A.645)

The sum of the first two terms on the rhs. simplifies. With
$\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}=\alpha^{i}\partial_{i}$ and
${\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}=P^{j}\alpha^{j}$,

$\displaystyle\left[{\alpha^{i}},{\left[{\alpha^{j}},{\partial_{i}\Phi^{(P)}}\right]}\right]$
$\displaystyle=\alpha^{i}(\alpha^{j}\partial_{i}\Phi^{(P)}-\partial_{i}\Phi^{(P)}\alpha^{j})-(\alpha^{j}\partial_{i}\Phi^{(P)}-\partial_{i}\Phi^{(P)}\alpha^{j})\alpha^{i}$
$\displaystyle\left\\{{\alpha^{j}},{\left\\{{\alpha^{i}},{\partial_{i}\Phi^{(P)}}\right\\}}\right\\}$
$\displaystyle=\alpha^{j}(\alpha^{i}\partial_{i}\Phi^{(P)}+\partial_{i}\Phi^{(P)}\alpha^{i})+(\alpha^{i}\partial_{i}\Phi^{(P)}+\partial_{i}\Phi^{(P)}\alpha^{i})\alpha^{j}$
(A.646)

so that

$\displaystyle\left[{\alpha^{i}},{\left[{\alpha^{j}},{\partial_{i}\Phi^{(P)}}\right]}\right]+\left\\{{\alpha^{j}},{\left\\{{\alpha^{i}},{\partial_{i}\Phi^{(P)}}\right\\}}\right\\}$
$\displaystyle=(\alpha^{i}\alpha^{j}+\alpha^{j}\alpha^{i})\partial_{i}\Phi^{(P)}+\partial_{i}\Phi^{(P)}(\alpha^{j}\alpha^{i}+\alpha^{i}\alpha^{j})=4\partial_{j}\Phi^{(P)}$
(A.647)

Using this in (A.19) and dividing by $E-V$ gives

$\displaystyle\frac{1}{E-V}\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{(E-V)\Phi^{(P)}}\right]-{\textstyle\frac{1}{2}}\left\\{{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\right\\}+m\left\\{{\gamma^{0}},{\Phi^{(P)}}\right\\}=-\frac{2i}{E-V}{\boldsymbol{P}}\cdot\boldsymbol{\nabla}\Phi^{(P)}$
(A.648)

For a linear potential
$i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}\,V^{\prime}|{\boldsymbol{x}}|=iV^{\prime}{\boldsymbol{\alpha}}\cdot{\boldsymbol{x}}/r$,
where $r=|{\boldsymbol{x}}|$. Bringing this derivative to the rhs. in (A.648),

$\displaystyle\left[{i\boldsymbol{\nabla}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\right]-{\textstyle\frac{1}{2}}\left\\{{{\boldsymbol{P}}\cdot{\boldsymbol{\alpha}}},{\Phi^{(P)}}\right\\}+m\left\\{{\gamma^{0}},{\Phi^{(P)}}\right\\}=\frac{1}{E-V}\Big{(}-2i{\boldsymbol{P}}\cdot\boldsymbol{\nabla}\Phi^{(P)}+\frac{V^{\prime}}{r}\left[{i{\boldsymbol{\alpha}}\cdot{\boldsymbol{x}}},{\Phi^{(P)}}\right]\Big{)}$
(A.649)

The lhs. is now the same as in the original BSE (456), with commutators and
anticommutators interchanged. Adding and subtracting the two equations and
dividing by $E-V$ we get equations (VIII.3.1).

###  A.20 Derive the frame dependence (459) of
$\Phi^{(P)}_{V=0}({\boldsymbol{x}})$ using the boost generator
$\mathcal{K}_{0}^{z}$.

The action of $\mathcal{K}_{0}^{z}(t=0)$ (464) on the state (455) with
${\boldsymbol{P}}=(0,0,P)$ and $V=0$,

$\displaystyle\left|{M,P}\right\rangle_{0}=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1})e^{iP(z_{1}+z_{2}/2}\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(A.650)

is determined by

$\displaystyle\left[{\mathcal{K}_{0}^{z}},{\bar{\psi}({\boldsymbol{x}}_{1})}\right]$
$\displaystyle=\psi^{\dagger}({\boldsymbol{x}}_{1})\big{[}z_{1}(-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}-m\gamma^{0})+{\textstyle\frac{1}{2}}i{\alpha_{3}}\big{]}\gamma^{0}=\bar{\psi}({\boldsymbol{x}}_{1})\big{[}z_{1}(i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}-m\gamma^{0})-{\textstyle\frac{1}{2}}i{\alpha_{3}}\big{]}$
$\displaystyle\left[{\mathcal{K}_{0}^{z}},{\psi({\boldsymbol{x}}_{2})}\right]$
$\displaystyle=-\big{[}z_{2}(i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}-m\gamma^{0})+{\textstyle\frac{1}{2}}i{\alpha_{3}}\big{]}\psi({\boldsymbol{x}}_{2})$
(A.651)

Making the derivatives act on the wave function through partial integration,

$\displaystyle\mathcal{K}^{z}_{0}\left|{M,P}\right\rangle_{0}=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1})\Big{\\{}$
$\displaystyle\big{[}z_{1}(-i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}-m\gamma^{0})-{\textstyle\frac{1}{2}}i{\alpha_{3}}\big{]}e^{iP(z_{1}+z_{2}/2}\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$
$\displaystyle-e^{iP(z_{1}+z_{2}/2}\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\big{[}z_{2}(-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2}-m\gamma^{0})+{\textstyle\frac{1}{2}}i{\alpha_{3}}\big{]}\Big{\\}}\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(A.652)

Using
$z_{2}(-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2})+{\textstyle\frac{1}{2}}i{\alpha_{3}}=(-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2})z_{2}-{\textstyle\frac{1}{2}}i{\alpha_{3}}$
and then expressing
$z_{1,2}={\textstyle\frac{1}{2}}(z_{1}+z_{2})\pm{\textstyle\frac{1}{2}}(z_{1}-z_{2})$,
the BSE (457) satisfied by $\Phi^{(P)}_{V=0}$,

$\displaystyle\big{(}i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}-{\textstyle\frac{1}{2}}P{\alpha_{3}}+m\gamma^{0}\big{)}\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})+\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\big{(}-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2}+{\textstyle\frac{1}{2}}P{\alpha_{3}}-m\gamma^{0}\big{)}=E\,\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$
(A.653)

reduces the coefficient of ${\textstyle\frac{1}{2}}(z_{1}+z_{2})$ to $-E$,
with $E=M\cosh\xi=\sqrt{P^{2}+M^{2}}$. We have then

$\displaystyle\mathcal{K}^{z}_{0}\left|{M,P}\right\rangle_{0}=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1})e^{iP(z_{1}+z_{2}/2}\Big{\\{}-{\textstyle\frac{1}{2}}(z_{1}+z_{2})E\,\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})-{\textstyle\frac{1}{2}}i\big{[}{{\alpha_{3}}},{\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}\big{]}$
(A.654)
$\displaystyle-{\textstyle\frac{1}{2}}(z_{1}-z_{2})\big{[}(i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}-{\textstyle\frac{1}{2}}P{\alpha_{3}}+m\gamma^{0})\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})+\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})(i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2}-{\textstyle\frac{1}{2}}P{\alpha_{3}}+m\gamma^{0})\big{]}\Big{\\}}\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$

where ${\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}$ and
${\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2}$ only differentiate
$\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$. Subtracting the
two BSE equations (VIII.3.1) gives

$\displaystyle\big{(}i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}-{\textstyle\frac{1}{2}}P{\alpha_{3}}+m\gamma^{0}\big{)}\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})+\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\big{(}i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2}-{\textstyle\frac{1}{2}}P{\alpha_{3}}+m\gamma^{0}\big{)}=-2i\frac{P}{E}\,{\buildrel\rightarrow\over{\partial}}_{z_{1}}\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$
(A.655)

Thus

$\displaystyle-id\xi\mathcal{K}^{z}_{0}\left|{M,P}\right\rangle_{0}$
$\displaystyle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1})e^{iP(z_{1}+z_{2}/2}\Big{\\{}{\textstyle\frac{1}{2}}id\xi\,E(z_{1}+z_{2})\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})-{\textstyle\frac{1}{2}}d\xi\big{[}{{\alpha_{3}}},{\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}\big{]}$
$\displaystyle+d\xi(z_{1}-z_{2})\frac{P}{E}\,\partial_{z_{1}}\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\Big{\\}}\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(A.656)

Let us now assume that the frame dependence (459) holds, and show that it
agrees with the change in the wave function (A.20) caused by the infinitesimal
boost. According to (459),

$\displaystyle\Phi^{(P+dP)}_{V=0}({\boldsymbol{x}})$
$\displaystyle=e^{-(\xi+d\xi){\alpha_{3}}/2}\Phi^{(P=0)}_{V=0}({\boldsymbol{x}}_{R})e^{(\xi+d\xi){\alpha_{3}}/2}$
$\displaystyle{\boldsymbol{x}}_{R}$
$\displaystyle=(x,y,z\cosh(\xi+d\xi))=(x,y,z\cosh\xi)+(0,0,d\xi z\sinh\xi)$
(A.657)

The first term within the $\\{\ \\}$ of (A.20) reflects the change in the
plane wave phase of $\left|{M,P}\right\rangle_{0}$,

$\displaystyle e^{i(P+d\xi
E)(z_{1}+z_{2})/2}=e^{i(P(z_{1}+z_{2})/2}\big{[}1+{\textstyle\frac{1}{2}}i\,d\xi
E(z_{1}+z_{2})\big{]}$ (A.658)

The second term is due to the $\exp[\mp(\xi+d\xi){\alpha_{3}}/2]$ factors in
$\Phi^{(P+dP)}_{V=0}({\boldsymbol{x}})$,

$\displaystyle\exp[-(\xi+d\xi){\alpha_{3}}/2]\Phi^{(P=0)}_{V=0}({\boldsymbol{x}}_{1R}-{\boldsymbol{x}}_{2R})=(1-{\textstyle\frac{1}{2}}d\xi{\alpha_{3}})\exp(-\xi{\alpha_{3}}/2)\Phi^{(P=0)}_{V=0}({\boldsymbol{x}}_{1R}-{\boldsymbol{x}}_{2R})$
(A.659)

and similarly for
$\Phi^{(P=0)}_{V=0}({\boldsymbol{x}}_{1R}-{\boldsymbol{x}}_{2R})\exp[(\xi+d\xi){\alpha_{3}}/2]$.
The third term in (A.20) relates to the Lorentz contraction, i.e., the change
in ${\boldsymbol{x}}_{R}$ (A.20),

$\displaystyle
e^{-\xi{\alpha_{3}}/2}d\xi(z_{1}-z_{2})\sinh\xi\frac{\partial}{\partial
z_{1R}}\Phi^{(P=0)}_{V=0}({\boldsymbol{x}}_{1R}-{\boldsymbol{x}}_{2R})e^{\xi{\alpha_{3}}/2}=d\xi(z_{1}-z_{2})\frac{\sinh\xi}{\cosh\xi}\,\frac{\partial}{\partial
z_{1}}\Phi^{(P)}_{V=0}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$ (A.660)

This confirms that the frame dependence of the state
$\left|{M,P}\right\rangle_{0}$ (A.650) implied by (A.20) agrees with the
transformation of a boost.

###  A.21 Show that $\Phi^{(P)}(\tau)$ given by (471) satisfies the BSE (457)
at $\boldsymbol{x}_{\perp}=0$.

Denoting by $B$ the lhs. of (457) at $\boldsymbol{x}_{\perp}=0$ when the wave
function $\Phi^{(P)}(\tau)$ is given by (471),

$\displaystyle e^{\zeta{\alpha_{3}}/2}\,B\,e^{-\zeta{\alpha_{3}}/2}$
$\displaystyle=e^{\zeta{\alpha_{3}}/2}\big{[}i{\overset{\rightarrow}{\boldsymbol{\nabla}}}\cdot{\boldsymbol{\alpha}}-{\textstyle\frac{1}{2}}(E-V+P{\alpha_{3}})+m\gamma^{0}\big{]}e^{-\zeta{\alpha_{3}}/2}\Phi^{(0)}(\tau)$
$\displaystyle+\Phi^{(0)}(\tau)e^{\zeta{\alpha_{3}}/2}\big{[}i{\overset{\leftarrow}{\boldsymbol{\nabla}}}\cdot{\boldsymbol{\alpha}}-{\textstyle\frac{1}{2}}(E-V-P{\alpha_{3}})-m\gamma^{0}\big{]}e^{-\zeta{\alpha_{3}}/2}$
(A.661)

We need to show that $B=0$. For the $i\partial_{z}$ terms,

$\displaystyle
e^{\zeta{\alpha_{3}}/2}i{\buildrel\rightarrow\over{\partial}}_{z}{\alpha_{3}}e^{-\zeta{\alpha_{3}}/2}$
$\displaystyle=i{\buildrel\rightarrow\over{\partial}}_{z}{\alpha_{3}}-{\textstyle\frac{1}{2}}i({\buildrel\rightarrow\over{\partial}}_{z}\zeta)$
$\displaystyle
e^{\zeta{\alpha_{3}}/2}i{\buildrel\leftarrow\over{\partial}}_{z}{\alpha_{3}}e^{-\zeta{\alpha_{3}}/2}$
$\displaystyle=i{\buildrel\leftarrow\over{\partial}}_{z}{\alpha_{3}}+{\textstyle\frac{1}{2}}i({\buildrel\rightarrow\over{\partial}}_{z}\zeta)$
(A.662)

The contributions $\propto\partial_{z}\zeta$ cancel in (A.21). Transforming
$\partial_{z}=-2(E-V)\partial_{\tau}=-2\sqrt{V^{\prime}\tau}\cosh\zeta\,\partial_{\tau}$
(which requires both a linear potential and $\boldsymbol{x}_{\perp}=0$),

$\displaystyle i{\buildrel\rightarrow\over{\partial}}_{z}{\alpha_{3}}$
$\displaystyle=-2\sqrt{V^{\prime}\tau}e^{\zeta{\alpha_{3}}}{\alpha_{3}}\,i{\buildrel\rightarrow\over{\partial}}_{\tau}+2\sqrt{V^{\prime}\tau}\sinh\zeta\,i{\buildrel\rightarrow\over{\partial}}_{\tau}$
$\displaystyle i{\buildrel\leftarrow\over{\partial}}_{z}{\alpha_{3}}$
$\displaystyle=-2i{\buildrel\leftarrow\over{\partial}}_{\tau}{\alpha_{3}}\sqrt{V^{\prime}\tau}e^{-\zeta{\alpha_{3}}}-2i{\buildrel\leftarrow\over{\partial}}_{\tau}\sqrt{V^{\prime}\tau}\sinh\zeta$
(A.663)

The terms $\propto\sinh\zeta$ cancel in (A.21). Expressing $E-V\pm
P{\alpha_{3}}=\sqrt{V^{\prime}\tau}\exp(\pm\zeta{\alpha_{3}})$, commuting the
$\exp(\pm\zeta{\alpha_{3}}/2)$ factors using
$i\boldsymbol{\nabla}_{\perp}\cdot{\boldsymbol{\alpha}_{\perp}}\,\exp(-\zeta{\alpha_{3}}/2)=\exp(\zeta{\alpha_{3}}/2)i\boldsymbol{\nabla}_{\perp}\cdot{\boldsymbol{\alpha}_{\perp}}$
(since $\boldsymbol{\nabla}_{\perp}\zeta=0$) and similarly for the
$m\gamma^{0}$ terms we get,

$\displaystyle e^{\zeta{\alpha_{3}}/2}\,B\,e^{-\zeta{\alpha_{3}}/2}$
$\displaystyle=e^{\zeta{\alpha_{3}}}\big{[}-2\sqrt{V^{\prime}\tau}\,i{\buildrel\rightarrow\over{\partial}}_{\tau}{\alpha_{3}}+i{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{\perp}\cdot{\boldsymbol{\alpha}_{\perp}}-{\textstyle\frac{1}{2}}\sqrt{V^{\prime}\tau}+m\gamma^{0}\big{]}\Phi^{(0)}(\tau)$
$\displaystyle+\Phi^{(0)}(\tau)\big{[}-2\,i{\buildrel\leftarrow\over{\partial}}_{\tau}{\alpha_{3}}\sqrt{V^{\prime}\tau}+i{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{\perp}\cdot{\boldsymbol{\alpha}_{\perp}}-{\textstyle\frac{1}{2}}\sqrt{V^{\prime}\tau}-m\gamma^{0}\big{]}e^{-\zeta{\alpha_{3}}}$
(A.664)

The terms in [ ] depend only on $\tau$, i.e., they are as in the $\zeta=0$ BSE
of $\Phi^{(0)}(\tau)$. Expressing
$\exp(\pm\zeta{\alpha_{3}})=\cosh\zeta\pm{\alpha_{3}}\sinh\zeta$ the
coefficent of $\cosh\zeta$ is the rest frame BSE, which $\Phi^{(0)}(\tau)$
satisfies by definition. Consequently the two terms in [ ] give equal and
opposite contributions to the $\zeta=0$ BSE. Using this leaves an
anticommutator with ${\alpha_{3}}$. Transforming back $\tau\to z$ at $\zeta=0$
allows to identify the ${\overset{\rightarrow}{\mathfrak{h}}}_{-}$ operator
(403),

$\displaystyle e^{\zeta{\alpha_{3}}/2}\,B\,e^{-\zeta{\alpha_{3}}/2}$
$\displaystyle=\sinh\zeta\big{\\{}{{\alpha_{3}}},{\big{(}-2\sqrt{V^{\prime}\tau}\,i{\buildrel\rightarrow\over{\partial}}_{\tau}{\alpha_{3}}+i{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{\perp}\cdot{\boldsymbol{\alpha}_{\perp}}-{\textstyle\frac{1}{2}}\sqrt{V^{\prime}\tau}+m\gamma^{0}\big{)}\Phi^{(0)}(\tau)}\big{\\}}$
(A.665)
$\displaystyle=\sinh\zeta\big{\\{}{{\alpha_{3}}},{\Big{[}i{\overset{\rightarrow}{\boldsymbol{\nabla}}}\cdot{\boldsymbol{\alpha}}-{\textstyle\frac{1}{2}}\big{(}M-V)+m\gamma^{0}\Big{]}\Phi^{(0)}(0,0,z)}\big{\\}}={\textstyle\frac{1}{2}}(M-V)\sinh\zeta\big{\\{}{{\alpha_{3}}},{{\overset{\rightarrow}{\mathfrak{h}}}_{-}\Phi^{(0)}(0,0,z)}\big{\\}}$

The expressions in (419), (430) and (439) show that
$\big{\\{}{{\alpha_{3}}},{{\overset{\rightarrow}{\mathfrak{h}}}_{-}\Phi^{(0)}(0,0,z)}\big{\\}}=0$
for all wave functions. Hence $B=0$ and $\Phi^{(P)}(\tau)$ given by (471)
solves the BSE for all $P$ at $\boldsymbol{x}_{\perp}=0$.

###  A.22 Prove the orthogonality relation (473) for states with wave
functions satisfying the BSE (456).

I follow the proof presented in Dietrich  _et al._ (2013). From the expression
(455) for the states in terms of their wave functions,

$\displaystyle\langle{M_{B},{\boldsymbol{P}}_{B}}|M_{A},{\boldsymbol{P}}_{A}\rangle$
$\displaystyle=\int
d{\boldsymbol{x}}_{1B}d{\boldsymbol{x}}_{2B}d{\boldsymbol{x}}_{1A}d{\boldsymbol{x}}_{2A}\langle{0}|\psi^{\dagger}({\boldsymbol{x}}_{2B})e^{-i{\boldsymbol{P}}_{B}\cdot({\boldsymbol{x}}_{1B}+{\boldsymbol{x}}_{2B})/2}\Phi_{B}^{({\boldsymbol{P}}_{B}){\dagger}}({\boldsymbol{x}}_{1B}-{\boldsymbol{x}}_{2B})\gamma^{0}\psi({\boldsymbol{x}}_{1B})$
$\displaystyle\times\bar{\psi}({\boldsymbol{x}}_{1A})e^{i{\boldsymbol{P}}_{A}\cdot({\boldsymbol{x}}_{1A}+{\boldsymbol{x}}_{2A})/2}\Phi_{A}^{({\boldsymbol{P}}_{A})}({\boldsymbol{x}}_{1A}-{\boldsymbol{x}}_{2A})\psi({\boldsymbol{x}}_{1A})\left|{0}\right\rangle$
(A.666)

The field contractions set
${\boldsymbol{x}}_{1A}={\boldsymbol{x}}_{1B}\equiv{\boldsymbol{x}}_{1}$ and
${\boldsymbol{x}}_{2A}={\boldsymbol{x}}_{2B}\equiv{\boldsymbol{x}}_{2}$. Then
$\int d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}=\int
d[({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2]d({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$
and the integral over ${\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2}$ sets
${\boldsymbol{P}}_{A}={\boldsymbol{P}}_{B}\equiv{\boldsymbol{P}}$,

$\displaystyle\langle{M_{B},{\boldsymbol{P}}_{B}}|M_{A},{\boldsymbol{P}}_{A}\rangle$
$\displaystyle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,e^{i({\boldsymbol{P}}_{A}-{\boldsymbol{P}}_{B})\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\mathrm{Tr}\,\Big{[}\Phi_{B}^{({\boldsymbol{P}}_{B}){\dagger}}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\Phi_{A}^{({\boldsymbol{P}}_{A})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})\Big{]}$
$\displaystyle=(2\pi)^{3}\delta^{3}({\boldsymbol{P}}_{A}-{\boldsymbol{P}}_{B})\int
d{\boldsymbol{x}}\,\mathrm{Tr}\,\Big{[}\Phi_{B}^{({\boldsymbol{P}}){\dagger}}({\boldsymbol{x}})\Phi_{A}^{({\boldsymbol{P}})}({\boldsymbol{x}})\Big{]}$
(A.667)

The BSE (456) for $\Phi_{A}^{({\boldsymbol{P}})}$ and
$\Phi_{B}^{({\boldsymbol{P}}){\dagger}}$ are

$\displaystyle
i\boldsymbol{\nabla}\cdot\big{\\{}{{\boldsymbol{\alpha}}},{\Phi_{A}^{({\boldsymbol{P}})}({\boldsymbol{x}})}\big{\\}}-{\textstyle\frac{1}{2}}{\boldsymbol{P}}\cdot\big{[}{{\boldsymbol{\alpha}}},{\Phi_{A}^{({\boldsymbol{P}})}({\boldsymbol{x}})}\big{]}+m\big{[}{\gamma^{0}},{\Phi_{A}^{({\boldsymbol{P}})}({\boldsymbol{x}})}\big{]}$
$\displaystyle=\big{[}E_{A}-V({\boldsymbol{x}})\big{]}\Phi_{A}^{({\boldsymbol{P}})}({\boldsymbol{x}})$
$\displaystyle-i\boldsymbol{\nabla}\cdot\big{\\{}{{\boldsymbol{\alpha}}},{\Phi_{B}^{({\boldsymbol{P}}){\dagger}}({\boldsymbol{x}})}\big{\\}}+{\textstyle\frac{1}{2}}{\boldsymbol{P}}\cdot\big{[}{{\boldsymbol{\alpha}}},{\Phi_{B}^{({\boldsymbol{P}}){\dagger}}({\boldsymbol{x}})}\big{]}-m\big{[}{\gamma^{0}},{\Phi_{B}^{({\boldsymbol{P}}){\dagger}}({\boldsymbol{x}})}\big{]}$
$\displaystyle=\big{[}E_{B}-V({\boldsymbol{x}})\big{]}\Phi_{B}^{({\boldsymbol{P}}){\dagger}}({\boldsymbol{x}})$
(A.668)

Multiplying the first equation by $\Phi_{B}^{(P){\dagger}}({\boldsymbol{x}})$
from the left, the second by $\Phi^{(P)}_{A}({\boldsymbol{x}})$ from the right
and taking the trace of their difference the terms
$\propto{\textstyle\frac{1}{2}}{\boldsymbol{P}}$, $m$ and $V$ cancel, giving

$\displaystyle
2i\mathrm{Tr}\,\Big{[}{\boldsymbol{\alpha}}\cdot\boldsymbol{\nabla}\big{\\{}{\Phi_{B}^{({\boldsymbol{P}}){\dagger}}({\boldsymbol{x}})},{\Phi_{A}^{({\boldsymbol{P}})}({\boldsymbol{x}})}\big{\\}}\Big{]}=\big{[}E_{A}-E_{B}\big{)}\mathrm{Tr}\,\Big{[}\Phi_{B}^{({\boldsymbol{P}}){\dagger}}({\boldsymbol{x}})\Phi_{A}^{({\boldsymbol{P}})}({\boldsymbol{x}})\Big{]}$
(A.669)

Integrating both sides $\int_{-\infty}^{\infty}d{\boldsymbol{x}}$ the lhs.
vanishes due to the substitution at (one component of)
${\boldsymbol{x}}=\pm\infty$. The vanishing of the rhs. implies (for
$M_{A}\neq M_{B}$) the orthogonality of the states according to (A.22).

###  A.23 Verify the expression (476) for the global norm of
$\Phi_{-+}({\boldsymbol{x}})$ in terms of $F_{1}(r)$.

According to (A.22) the bound state norm is proportional to the trace on the
lhs. of (476). The expression (VIII.2.3) of $\Phi_{-+}({\boldsymbol{x}})$
implies

$\displaystyle N$ $\displaystyle\equiv\int
d{\boldsymbol{x}}\,\mathrm{Tr}\,\big{\\{}\Phi_{-+}^{\dagger}({\boldsymbol{x}})\Phi_{-+}({\boldsymbol{x}})\big{\\}}$
$\displaystyle=\int
dr\,r^{2}d\Omega\,\mathrm{Tr}\,\Big{\\{}Y_{j\lambda}^{*}(\Omega)F_{1}^{*}(r)\gamma_{5}\Big{[}(-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}+m\gamma^{0})\frac{2}{M-V}+1\Big{]}\Big{[}\frac{2}{M-V}(i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}+m\gamma^{0})+1\Big{]}\gamma_{5}F_{1}(r)Y_{j\lambda}(\Omega)\Big{\\}}$
$\displaystyle=4\int
dr\,r^{2}d\Omega\,Y_{j\lambda}^{*}F_{1}^{*}\Big{[}{\buildrel\leftarrow\over{\partial}}_{j}\frac{4}{(M-V)^{2}}{\buildrel\rightarrow\over{\partial}}_{j}+\frac{4m^{2}}{(M-V)^{2}}+1\Big{]}F_{1}Y_{j\lambda}$
$\displaystyle=4\int
dr\,r^{2}d\Omega\,Y_{j\lambda}^{*}F_{1}^{*}\Big{[}-\frac{4}{(M-V)^{2}}{\overset{\rightarrow}{\boldsymbol{\nabla}}}^{2}-\frac{8V^{\prime}}{(M-V)^{3}}\partial_{r}+\frac{4m^{2}}{(M-V)^{2}}+1\Big{]}F_{1}Y_{j\lambda}$
(A.670)

Expressing ${\overset{\rightarrow}{\boldsymbol{\nabla}}}^{2}$ in spherical
coordinates and using the radial equation (417),

$\displaystyle{\overset{\rightarrow}{\boldsymbol{\nabla}}}^{2}F_{1}(r)Y_{j\lambda}(\Omega)=\Big{[}F^{\prime\prime}_{1}+\frac{2}{r}F^{\prime}_{1}-\frac{j(j+1)}{r^{2}}F_{1}\Big{]}Y_{j\lambda}=\Big{[}-\frac{V^{\prime}}{M-V}F^{\prime}_{1}-{\textstyle\frac{1}{4}}(M-V)^{2}F_{1}+m^{2}F_{1}\Big{]}Y_{j\lambda}$
(A.671)

Substituting this into (A.23) and using $\int
d\Omega\,|Y_{j\lambda}(\Omega)|^{2}=1$ gives (476).

###  A.24 Verify the expressions (VIII.6.2) for radial functions
$H_{1}(r),H_{2}(r)$ and $H_{3}(r)$.

The structure (510) of the $J^{PC}=0^{++}$ wave function,

$\displaystyle\Phi_{\sigma}({\boldsymbol{x}})=H_{1}(r)+i\,{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}\,H_{2}(r)+i\,\gamma^{0}{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}\,H_{3}(r)$
(A.672)

follows from its parity and charge conjugation quantum numbers, as listed in
(413). The three Dirac structures involving the orbital angular momentum
operator ${\boldsymbol{L}}$ do not contribute for $j=0$, since
${\boldsymbol{L}}$ acts on $Y_{00}$ in (406), which has no angular dependence.
The BSE (506) involves the (anti)commutators

$\displaystyle{\textstyle\frac{1}{2}}\left\\{{{\boldsymbol{\alpha}}},{\Phi_{\sigma}({\boldsymbol{x}})}\right\\}$
$\displaystyle={\boldsymbol{\alpha}}H_{1}(r)+i\hat{\boldsymbol{x}}H_{2}(r)+\gamma^{0}\hat{\boldsymbol{x}}\times{\boldsymbol{\alpha}}\,\gamma_{5}H_{3}(r)$
$\displaystyle{\textstyle\frac{1}{2}}\left[{\gamma^{0}},{\Phi_{\sigma}({\boldsymbol{x}})}\right]$
$\displaystyle=i\gamma^{0}{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}H_{2}(r)+i{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}H_{3}(r)$
(A.673)

Inserting the expression (A.672) into the BSE (506) with $M=0$ gives, using
$\boldsymbol{\nabla}f(r)=\hat{\boldsymbol{x}}f^{\prime}(r)$,
$\partial_{i}x^{j}=\delta_{ij}$ and
$\boldsymbol{\nabla}\cdot\hat{\boldsymbol{x}}\times{\boldsymbol{\alpha}}\,f(r)=0$,

$\displaystyle
i\,{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}H_{1}^{\prime}-\frac{2}{r}H_{2}-H_{2}^{\prime}+im\gamma^{0}{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}H_{2}+im{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}H_{3}+{\textstyle\frac{1}{2}}V^{\prime}r\,\Phi_{\sigma}({\boldsymbol{x}})=0$
(A.674)

The coefficients of the Dirac structures $1,\
i{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}$ and
$i\gamma^{0}{\boldsymbol{\alpha}}\cdot\hat{\boldsymbol{x}}$ impose

$\displaystyle-\frac{2}{r}H_{2}-H_{2}^{\prime}+{\textstyle\frac{1}{2}}V^{\prime}r\,H_{1}$
$\displaystyle=0$ $\displaystyle
H_{1}^{\prime}+mH_{3}+{\textstyle\frac{1}{2}}V^{\prime}r\,H_{2}$
$\displaystyle=0$ $\displaystyle
mH_{2}+{\textstyle\frac{1}{2}}V^{\prime}r\,H_{3}$ $\displaystyle=0$ (A.675)

The first and third equations allow to express $H_{1}$ and $H_{3}$,
respectively, in terms of $H_{2}$. Substituting them into the second equation
gives the differential equation

$\displaystyle
H_{2}^{\prime\prime}+\frac{1}{r}H_{2}^{\prime}+\Big{[}\frac{1}{4}(V^{\prime}r)^{2}-m^{2}-\frac{4}{r^{2}}\Big{]}H_{2}=0$
(A.676)

It is straightforward to verify that the expressions for $H_{i}(r)$ given in
(VIII.6.2) satisfy the above equations. Their properties $H_{1}(r\to 0)\sim
r^{0}$, $H_{2}(r\to 0)\sim r^{2}$ and $H_{3}(r\to 0)\sim r^{1}$ ensure that
the wave function is locally normalizable at $r=0$.

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