d{\boldsymbol{x}}d{\boldsymbol{y}}d{\boldsymbol{z}}\Big{[}\partial_{i}^{x}\frac{e}{4\pi|{\boldsymbol{x}}-{\boldsymbol{y}}|}\psi^{\dagger}\psi({\boldsymbol{y}})\Big{]}\Big{[}\partial_{i}^{x}\frac{e}{4\pi|{\boldsymbol{x}}-{\boldsymbol{z}}|}\psi^{\dagger}\psi({\boldsymbol{z}})\Big{]}\left|{phys}\right\rangle$
$\displaystyle={\textstyle\frac{1}{2}}\int
d{\boldsymbol{x}}d{\boldsymbol{y}}\,\frac{e^{2}}{4\pi|{\boldsymbol{x}}-{\boldsymbol{y}}|}\big{[}\psi^{\dagger}\psi({\boldsymbol{x}})\big{]}\big{[}\psi^{\dagger}\psi({\boldsymbol{y}})\big{]}\left|{phys}\right\rangle$
(145)

$\mathcal{H}_{V}\left|{phys}\right\rangle$ contributes a potential which
depends only on the instantaneous positions of the electrons and positrons,
regardless of their momenta (which may be relativistic). The other terms of
$\mathcal{H}$ determine the propagation of the transverse photons and fermions
in time, as well as the transitions between them.

This method can be applied to atoms in any frame. Given that non-valence Fock
states are suppressed by powers of $\alpha$, calculations with a given degree
of precision require to include a limited number of terms in the Fock
expansion. In section VI I illustrate the method by considering several
aspects of Positronium at rest and in motion. In section VII I study the
strongly bound states of QED in $D=1+1$ dimensions.

###  V.3 Temporal gauge in QCD

####  V.3.1 Canonical quantization

The canonical quantization of QCD in temporal gauge $A^{0}_{a}=0$ proceeds as
in QED Willemsen (1978); Bjorken (1979); Christ and Lee (1980); Leibbrandt
(1987); Strocchi (2013). The QCD action is

$\displaystyle\mathcal{S}_{QCD}$ $\displaystyle=\int
d^{4}x\big{[}-{\textstyle\frac{1}{4}}F_{\mu\nu}^{a}F^{\mu\nu}_{a}+\bar{\psi}(i\not{\partial}-m-g\not{A}_{a}T^{a})\psi\big{]}$
$\displaystyle
F_{\mu\nu}^{a}=\partial_{\mu}A_{\nu}^{a}-\partial_{\nu}A_{\mu}^{a}-gf_{abc}A_{\mu}^{b}A_{\nu}^{c}$
(146)

The electric field $E_{a}^{i}=F_{a}^{i0}=-\partial_{0}A_{a}^{i}$ is conjugate
to $A_{i}^{a}=-A_{a}^{i}$, giving the equal-time commutation relations

$\displaystyle\left[{E_{a}^{i}(t,{\boldsymbol{x}})},{A_{b}^{j}(t,{\boldsymbol{y}})}\right]$
$\displaystyle=i\delta_{ab}\delta^{ij}\delta({\boldsymbol{x}}-{\boldsymbol{y}})$
$\displaystyle\left\\{{\psi^{A\,{\dagger}}_{\alpha}(t,{\boldsymbol{x}})},{\psi_{\beta}^{B}(t,{\boldsymbol{y}})}\right\\}=\delta^{AB}\delta_{\alpha\beta}\,\delta({\boldsymbol{x}}-{\boldsymbol{y}})$
(147)

The $a,b$ ($A,B$) are color indices in the adjoint (fundamental)
representation of SU(3). The Hamiltonian is

$\displaystyle\mathcal{H}_{QCD}$ $\displaystyle=\int
d{\boldsymbol{x}}\big{[}E_{a}^{i}\partial_{0}A_{i}^{a}+i\psi_{A}^{\dagger}\partial_{0}\psi_{A}-\mathcal{L}_{QCD}\big{]}=\int
d{\boldsymbol{x}}\big{[}{\textstyle\frac{1}{2}}E_{a}^{i}E_{a}^{i}+{\textstyle\frac{1}{4}}F_{a}^{ij}F_{a}^{ij}+\psi^{\dagger}(-i{\boldsymbol{\alpha}}\cdot\boldsymbol{\nabla}+m\gamma^{0}-g{\boldsymbol{\alpha}}\cdot{\boldsymbol{A}}_{a}T^{a})\psi\big{]}$
(148)

where

$\displaystyle\int
d{\boldsymbol{x}}\,{\textstyle\frac{1}{4}}F_{a}^{ij}F_{a}^{ij}=\int
d{\boldsymbol{x}}\big{[}{\textstyle\frac{1}{2}}A_{a}^{i}(-\delta_{ij}\boldsymbol{\nabla}^{2}+\partial_{i}\partial_{j})A_{a}^{j}+gf_{abc}(\partial_{i}A_{a}^{j})A_{b}^{i}A_{c}^{j}+{\textstyle\frac{1}{4}}g^{2}f_{abc}f_{ade}A_{b}^{i}A_{c}^{j}A_{d}^{i}A_{e}^{j}\big{]}$
(149)

has both longitudinal and transverse gluon fields.

Gauss’ operator

$\displaystyle
G_{a}(x)\equiv\frac{\delta\mathcal{S}_{QCD}}{\delta{A_{a}^{0}(x)}}=\partial_{i}E_{a}^{i}(x)+gf_{abc}A_{b}^{i}E_{c}^{i}-g\psi^{\dagger}T^{a}\psi(x)$
(150)

generates time-independent gauge transformations similarly as in QED (140),
which leave the gauge condition $A^{0}_{a}=0$ invariant. The longitudinal
electric field ${\boldsymbol{E}}_{L}^{a}$ is fixed by constraining physical
states to be invariant under the gauge transformations generated by
$G_{a}(x)$,

$\displaystyle G_{a}(x)\left|{phys}\right\rangle=0$ (151)

This constraint is independent of time since Gauss’ operator commutes with the
Hamiltonian, $\left[{G_{a}(t,{\boldsymbol{x}})},{\mathcal{H}(t)}\right]=0$. It
constrains the longitudinal electric field for physical states,

$\displaystyle\boldsymbol{\nabla}\cdot
E_{L}^{a}({\boldsymbol{x}})\left|{phys}\right\rangle=g\big{[}-f_{abc}A_{b}^{i}E_{c}^{i}+\psi^{\dagger}T^{a}\psi({\boldsymbol{x}})\big{]}\left|{phys}\right\rangle$
(152)

We may solve for ${\boldsymbol{E}}_{L}^{a}$ analogously222At higher orders in
$g$ one needs to take into account the contribution of ${\boldsymbol{E}}_{L}$
on the rhs. of (152). For large gauge fields this leads to the issue of Gribov
copies Gribov (1978), but they do not appear in a perturbative expansion. as
for QED in section V.2.3,

$\displaystyle{\boldsymbol{E}}_{L}^{a}({\boldsymbol{x}})\left|{phys}\right\rangle$
$\displaystyle=-\boldsymbol{\nabla}_{x}\int
d{\boldsymbol{y}}\,\frac{g}{4\pi|{\boldsymbol{x}}-{\boldsymbol{y}}|}\,\mathcal{E}_{a}({\boldsymbol{y}})\left|{phys}\right\rangle$
$\displaystyle\mathcal{E}_{a}({\boldsymbol{y}})$
$\displaystyle=-f_{abc}A_{b}^{i}E_{c}^{i}({\boldsymbol{y}})+\psi^{\dagger}T^{a}\psi({\boldsymbol{y}})$
(153)

The contribution of the longitudinal electric field to the QCD Hamiltonian
(148) is then

$\displaystyle\mathcal{H}_{V}^{QCD}\left|{phys}\right\rangle$
$\displaystyle\equiv{\textstyle\frac{1}{2}}\int
d{\boldsymbol{x}}\,\big{(}{\boldsymbol{E}}_{L}^{a}\big{)}^{2}\left|{phys}\right\rangle={\textstyle\frac{1}{2}}\int
d{\boldsymbol{y}}d{\boldsymbol{z}}\,\frac{{\alpha_{s}}}{|{\boldsymbol{y}}-{\boldsymbol{z}}|}\,\mathcal{E}_{a}({\boldsymbol{y}})\mathcal{E}_{a}({\boldsymbol{z}})\left|{phys}\right\rangle$
(154)

####  V.3.2 Specification of temporal gauge in QCD

There is a relevant difference between QED and QCD which needs to be
considered when determining the longitudinal electric field from the QCD gauge
constraint (152). To illustrate, compare the expectation value of the field in
an $e^{-}e^{+}$ Fock component of Positronium and in an analogous color
singlet $q\bar{q}$ component of a meson at $t=0$,

$\displaystyle\left|{e^{-}e^{+}}\right\rangle$
$\displaystyle\equiv\bar{\psi}_{\alpha}({\boldsymbol{x}}_{1})\psi_{\beta}({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(155) $\displaystyle\left|{q\,\bar{q}}\right\rangle$
$\displaystyle\equiv\sum_{A}\bar{\psi}_{\alpha}^{A}({\boldsymbol{x}}_{1})\psi_{\beta}^{A}({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(156)

The Dirac components $\alpha,\beta$ are irrelevant here and will be
suppressed. Repeated color indices are summed. Note that “color singlet”
refers to global gauge transformations333 The global SU(3) transformations
should not be regarded as a subgroup of the local ones, see Ch. 7 of Strocchi
(2013)., the local temporal gauge being fixed by (143) and (V.3.1).

The expectation values of the QED (143) and QCD (V.3.1) longitudinal electric
fields in these states are, using the canonical commutation relations for the
fermions and recalling that ${\boldsymbol{E}}_{L}\left|{0}\right\rangle=0$,

$\displaystyle\langle{e^{-}e^{+}}|{\boldsymbol{E}}_{L}({\boldsymbol{x}})\left|{e^{-}e^{+}}\right\rangle$
$\displaystyle=-\boldsymbol{\nabla}_{x}\Big{(}\frac{e}{4\pi|{\boldsymbol{x}}-{\boldsymbol{x}}_{1}|}-\frac{e}{4\pi|{\boldsymbol{x}}-{\boldsymbol{x}}_{2}|}\Big{)}\langle{e^{-}e^{+}}|e^{-}e^{+}\rangle$
(157)
$\displaystyle\langle{q\,\bar{q}}|{\boldsymbol{E}}_{L}^{a}({\boldsymbol{x}})\left|{q\,\bar{q}}\right\rangle$
$\displaystyle=-\boldsymbol{\nabla}_{x}\Big{(}\frac{g}{4\pi|{\boldsymbol{x}}-{\boldsymbol{x}}_{1}|}-\frac{g}{4\pi|{\boldsymbol{x}}-{\boldsymbol{x}}_{2}|}\Big{)}\langle{q\,\bar{q}}|\bar{\psi}_{A}({\boldsymbol{x}}_{1})T_{AB}^{a}\psi_{B}({\boldsymbol{x}}_{2})\left|{0}\right\rangle\propto\mathrm{Tr}\,T^{a}=0$
(158)

In QED the charges of $e^{-}$ and $e^{+}$ give rise to the expected dipole
electric field, while in QCD the expectation value of an octet field in a
singlet state vanishes everywhere. Comparing similarly444The singular “self-
energy” contributions $\propto 1/0$ are independent of
${\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2}$ and subtracted. the instantaneous
potentials $\mathcal{H}_{V}^{QED}$ (V.2.3) and $\mathcal{H}_{V}^{QCD}$ (154),

$\displaystyle\langle{e^{-}e^{+}}|\mathcal{H}_{V}^{QED}\left|{e^{-}e^{+}}\right\rangle$
$\displaystyle=-\frac{\alpha}{|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}|}\,\langle{e^{-}e^{+}}|e^{-}e^{+}\rangle$
(159)
$\displaystyle\langle{q\,\bar{q}}|\mathcal{H}_{V}^{QCD}\left|{q\,\bar{q}}\right\rangle$
$\displaystyle=-\frac{\alpha_{s}}{|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}|}\,\langle{q\,\bar{q}}|\bar{\psi}_{A}({\boldsymbol{x}}_{1})T_{AB}^{a}T_{BC}^{a}\psi_{C}({\boldsymbol{x}}_{2})\left|{0}\right\rangle=-C_{F}\frac{\alpha_{s}}{|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}|}\,\langle{q\,\bar{q}}|q\,\bar{q}\rangle$
(160)

These are the Coulomb potentials of QED and QCD, again as expected. The
electron feels only the positron field, and each quark of a given color
interacts with its antiquark of opposite color. The sum over the potential
energies of all color-anticolor components $A$ in (156) gives the Casimir
$C_{F}=(N_{c}^{2}-1)/2N_{c}$ of the fundamental representation.

The solution of the QED gauge constraint (141), the longitudinal electric
field (143), is determined using the physical boundary condition that the
electric field vanishes at spatial infinity. This boundary condition is no
longer evident for QCD, since the expectation value (158) of the color
electric field in any case vanishes at all ${\boldsymbol{x}}$, due to the sum
over quark colors. There seems to be no compelling reason to require that the
gauge field of each color-anticolor component $A$ of the state (156) should
vanish at spatial infinity.

The gauge constraint (152) fully determines ${\boldsymbol{E}}_{L}^{a}$ only
given a boundary condition at spatial infinity. ${\boldsymbol{E}}_{L}^{a}$ may
be specified by the particular solution (V.3.1) and a homogeneous solution
${\boldsymbol{E}}_{H}^{a}$ which satisfies

$\displaystyle\boldsymbol{\nabla}\cdot{\boldsymbol{E}}_{H}^{a}\left|{phys}\right\rangle=0$
(161)

There is apparently only one homogeneous solution which is invariant under
translations and rotations,

$\displaystyle{\boldsymbol{E}}^{a}_{H}({\boldsymbol{x}})\left|{phys}\right\rangle$
$\displaystyle=-\kappa\,\boldsymbol{\nabla}_{x}\int
d{\boldsymbol{y}}\,{\boldsymbol{x}}\cdot{\boldsymbol{y}}\,\mathcal{E}_{a}({\boldsymbol{y}})\left|{phys}\right\rangle$
(162)

where $\mathcal{E}_{a}({\boldsymbol{y}})$ is defined in (V.3.1) and the
normalization $\kappa$ is independent of ${\boldsymbol{x}}$, but may depend on
the state $\left|{phys}\right\rangle$. The complete longitudinal electric
field is then

$\displaystyle{\boldsymbol{E}}_{L}^{a}({\boldsymbol{x}})\left|{phys}\right\rangle$
$\displaystyle=-\boldsymbol{\nabla}_{x}\int
d{\boldsymbol{y}}\Big{[}\kappa\,{\boldsymbol{x}}\cdot{\boldsymbol{y}}+\frac{g}{4\pi|{\boldsymbol{x}}-{\boldsymbol{y}}|}\Big{]}\mathcal{E}_{a}({\boldsymbol{y}})\left|{phys}\right\rangle$
(163)

and its contribution to the Hamiltonian (148) is

$\displaystyle\mathcal{H}_{V}$ $\displaystyle\equiv{\textstyle\frac{1}{2}}\int
d{\boldsymbol{x}}\,E_{a,L}^{i}E_{a,L}^{i}={\textstyle\frac{1}{2}}\int
d{\boldsymbol{x}}\Big{\\{}\partial_{i}^{x}\int
d{\boldsymbol{y}}\Big{[}\kappa\,{\boldsymbol{x}}\cdot{\boldsymbol{y}}+\frac{g}{4\pi|{\boldsymbol{x}}-{\boldsymbol{y}}|}\Big{]}\mathcal{E}_{a}({\boldsymbol{y}})\Big{\\}}\Big{\\{}\partial_{i}^{x}\int
d{\boldsymbol{z}}\Big{[}\kappa\,{\boldsymbol{x}}\cdot{\boldsymbol{z}}+\frac{g}{4\pi|{\boldsymbol{x}}-{\boldsymbol{z}}|}\Big{]}\mathcal{E}_{a}({\boldsymbol{z}})\Big{\\}}$
$\displaystyle=\int
d{\boldsymbol{y}}d{\boldsymbol{z}}\Big{\\{}\,{\boldsymbol{y}}\cdot{\boldsymbol{z}}\Big{[}{\textstyle\frac{1}{2}}\kappa^{2}{\textstyle\int}d{\boldsymbol{x}}+g\kappa\Big{]}+{\textstyle\frac{1}{2}}\frac{{\alpha_{s}}}{|{\boldsymbol{y}}-{\boldsymbol{z}}|}\Big{\\}}\mathcal{E}_{a}({\boldsymbol{y}})\mathcal{E}_{a}({\boldsymbol{z}})$
(164)

where the terms of ${\cal O}\left(g\kappa,g^{2}\right)$ were integrated by
parts. The term of ${\cal O}\left(\kappa^{2}\right)$ is due to an
${\boldsymbol{x}}$-independent field energy density. It is $\propto\int
d{\boldsymbol{x}}$ but irrelevant provided it is universal, i.e., the same for
all Fock components of all bound states. This determines the normalization
$\kappa$ in (163) for each state $\left|{phys}\right\rangle$, up to a
universal scale $\Lambda$.

The scale $\Lambda$ is unrelated to the coupling $g$, so the $g\kappa$ term in
(V.3.2) may be viewed as an instantaneous ${\cal
O}\left(\alpha_{s}^{0}\right)$ potential. All relevant symmetries, in
particular exact Poincaré invariance, must appear at each order of
${\alpha_{s}}$. The boost covariance of Positronia in QED is ensured by a
combination of the Coulomb potential and ${\cal O}\left(\alpha\right)$
transverse photon exchange (section VI). The boost covariance of QCD bound
states must at ${\cal O}\left(\alpha_{s}^{0}\right)$ be achieved by the
instantaneous potential alone, akin to QED in $D=1+1$ (section VII). This
appears to be satisfied (section VIII.3).

##  VI Applications to Positronium atoms

I now illustrate the approach to QED bound states described above with several
applications to Positronium atoms. The expansion starts with valence Fock
states, here $\left|{e^{-}e^{+}}\right\rangle$, with higher Fock states
included perturbatively. The first task is then to define the valence Fock
states and determine the constraints on their wave functions imposed by the
symmetries of translations and rotations, as well as parity and charge
conjugation (section VI.1). I express the valence Fock states using field
operators (here the electron field $\psi(t,{\boldsymbol{x}})$), similarly as
in the representations (53), (54) of the Dirac bound states.

In section VI.2 I determine the wave functions of Para- and Orthopositronium
atoms at lowest ${\cal O}\left(\alpha^{2}\right)$ in their binding energy
$E_{b}$. The rest frame wave functions then satisfy the Schrödinger equation.
For atomic CM momentum ${\boldsymbol{P}}\neq 0$ one needs to include the Fock
state with one transverse photon $\left|{e^{-}e^{+}\gamma}\right\rangle$
(VI.3). The hyperfine splitting between Ortho- and Parapositronium at rest is
calculated at ${\cal O}\left(\alpha^{4}\right)$ in the rest frame (VI.4),
taking into account the transverse photon state
$\left|{e^{-}e^{+}\gamma}\right\rangle$ and Orthopositronium annihilation into
a virtual photon, $e^{-}e^{+}\to\gamma\to e^{-}e^{+}$. Finally I calculate the
electromagnetic form factor of Positronium (VI.5) and deep inelastic
scattering on Positronia (VI.6) in a general frame.

###  VI.1 The $\left|{e^{-}e^{+}}\right\rangle$ Fock states of Para- and
Orthopositronium atoms

####  VI.1.1 Definition of the Fock states

The $\left|{e^{-}e^{+}}\right\rangle$ Fock states of Parapositronium
($J^{PC}=0^{-+}$) and Orthopositronium ($J^{PC}=1^{--}$) atoms, jointly
denoted by $\mathcal{B}$, may be expressed in terms of two electron fields,

$\displaystyle\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle\equiv\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(165)

where $\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}$ is a $4\times 4$ wave function
of the atom $\mathcal{B}$ with CM momentum ${\boldsymbol{P}}$. The
$\Lambda_{\pm}$ are Dirac projection operators,

$\displaystyle{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})$
$\displaystyle\equiv\frac{1}{2E_{1}}\Big{[}E_{1}-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}+\gamma^{0}m\Big{]}=\big{[}{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\big{]}^{2}$
$\displaystyle{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})$
$\displaystyle\equiv\frac{1}{2E_{2}}\Big{[}E_{2}+i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}-\gamma^{0}m\Big{]}=\big{[}{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\big{]}^{2}$
(166)

where $E=\sqrt{-\boldsymbol{\nabla}^{2}+m^{2}}$. The projectors select the
$b^{\dagger}$ operator in $\bar{\psi}({\boldsymbol{x}}_{1})$ and $d^{\dagger}$
in $\psi({\boldsymbol{x}}_{2})$, defined as in (51),

$\displaystyle\bar{\psi}({\boldsymbol{x}}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}})$
$\displaystyle=\int\frac{d{\boldsymbol{k}}}{(2\pi)^{3}2E_{k}}\sum_{\lambda}\bar{u}({\boldsymbol{k}},\lambda)\,e^{-i{\boldsymbol{k}}\cdot{\boldsymbol{x}}}\,b_{{\boldsymbol{k}},\lambda}^{\dagger}\hskip
56.9055pt{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}})\psi({\boldsymbol{x}})=\int\frac{d{\boldsymbol{k}}}{(2\pi)^{3}2E_{k}}\sum_{\lambda}v({\boldsymbol{k}},\lambda)\,e^{-i{\boldsymbol{k}}\cdot{\boldsymbol{x}}}\,d_{{\boldsymbol{k}},\lambda}^{\dagger}$
(167)

Since $b\left|{0}\right\rangle=d\left|{0}\right\rangle=0$ in (165) the
projectors actually have no effect. However, in operations on the states they
allow to use the anticommutation relations
$\left\\{{\psi^{\dagger}_{\alpha}(t,{\boldsymbol{x}})},{\psi_{\beta}(t,{\boldsymbol{y}})}\right\\}=\delta_{\alpha,\beta}\,\delta^{3}({\boldsymbol{x}}-{\boldsymbol{y}})$,
to which the $b$ and $d$ operators contribute. The projectors ensure that the
coefficients of $b$ and $d$ in (165) vanish, so that no spurious contributions
arise. I assume the normalization

$\displaystyle\langle{e^{-}e^{+};\mathcal{B}^{\prime},{\boldsymbol{P}}^{\prime}}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle=2E_{P}(2\pi)^{3}\delta({\boldsymbol{P}}-{\boldsymbol{P}}^{\prime})\delta_{\mathcal{B},\mathcal{B}^{\prime}}\hskip
56.9055ptE_{P}=\sqrt{{\boldsymbol{P}}^{2}+4m^{2}}$ (168)

where $E_{P}$ is the energy of the atom at ${\cal O}\left(\alpha^{0}\right)$.

The Hamiltonian (138) is symmetric under translations, rotations, parity and
charge conjugation. The states may be classified by their transformation under
those symmetries, giving constraints on the wave functions
$\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}})$.

####  VI.1.2 Translations

Under space translations
${\boldsymbol{x}}\to{\boldsymbol{x}}+{\boldsymbol{\ell}}$ the electron field
is transformed by the operator

$U(\boldsymbol{\ell})=\exp[-i{\boldsymbol{\ell}}\cdot\boldsymbol{\mathcal{P}}]\hskip
28.45274pt{\rm where}\hskip 28.45274pt\boldsymbol{\mathcal{P}}=\int
d{\boldsymbol{x}}\,\psi^{\dagger}({\boldsymbol{x}})(-i{\overset{\rightarrow}{\boldsymbol{\nabla}}})\psi({\boldsymbol{x}})$
(169)

The momentum operator satisfies

$\displaystyle\left[{\boldsymbol{\mathcal{P}}},{\psi({\boldsymbol{x}})}\right]=i{\overset{\rightarrow}{\boldsymbol{\nabla}}}\psi({\boldsymbol{x}})\hskip
56.9055pt\left[{\boldsymbol{\mathcal{P}}},{\bar{\psi}({\boldsymbol{x}})}\right]=\bar{\psi}({\boldsymbol{x}})i{\overset{\leftarrow}{\boldsymbol{\nabla}}}$
(170)

With $\boldsymbol{\mathcal{P}}\left|{0}\right\rangle=0$ we have
$\boldsymbol{\mathcal{P}}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle={\boldsymbol{P}}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$.

####  VI.1.3 Rotations

Rotations are generated by the angular momentum operator
$\boldsymbol{\mathcal{J}}$, which was defined already in (70) for the Dirac
equation. With ${\boldsymbol{\alpha}}\equiv\gamma^{0}\boldsymbol{\gamma}$,

$\displaystyle\boldsymbol{\mathcal{J}}$ $\displaystyle=\int
d{\boldsymbol{x}}\,\psi^{\dagger}({\boldsymbol{x}})\,\boldsymbol{J}\,\psi({\boldsymbol{x}})\hskip
82.51282pt\boldsymbol{J}={\boldsymbol{L}}+{\boldsymbol{S}}={\boldsymbol{x}}\times(-i\boldsymbol{\nabla})+{\textstyle\frac{1}{2}}\gamma_{5}{\boldsymbol{\alpha}}$
(171)

Rest frame states are taken to be eigenstates of
$\boldsymbol{\mathcal{J}}^{2}$ and $\mathcal{J}^{z}$. For a
${\boldsymbol{P}}=0$ Positronium state expressed as in (165),

$\displaystyle\boldsymbol{\mathcal{J}}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\left[{{\boldsymbol{J}}},{\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}\right]{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(172)

Exercise A.9: Derive (172).

For the state to have total angular momentum $j$ and $j^{z}=\lambda$ in the
rest frame,

$\displaystyle\boldsymbol{\mathcal{J}}^{2}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle=j(j+1)\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle\hskip
56.9055pt\mathcal{J}^{z}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle=\lambda\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle$
(173)

the wave function should satisfy

$\displaystyle\left[{J^{i}},{\left[{J^{i}},{\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})}\right]}\right]=j(j+1)\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})\hskip
56.9055pt\left[{J^{z}},{\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})}\right]=\lambda\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})$
(174)

####  VI.1.4 Parity $\eta_{P}$

The parity operator $\mathbb{P}$ transforms the electron field as

$\mathbb{P}\psi(t,{\boldsymbol{x}})\mathbb{P}^{\dagger}=\gamma^{0}\psi(t,-{\boldsymbol{x}})\hskip
56.9055pt\mathbb{P}\bar{\psi}(t,{\boldsymbol{x}})\mathbb{P}^{\dagger}=\bar{\psi}(t,-{\boldsymbol{x}})\gamma^{0}$
(175)

Changing the integration variables
${\boldsymbol{x}}_{1,2}\to-{\boldsymbol{x}}_{1,2}$ in (165) and noting that
$\gamma^{0}\Lambda_{\pm}({\boldsymbol{x}})=\Lambda_{\pm}(-{\boldsymbol{x}})\gamma^{0}$,

$\mathbb{P}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\gamma^{0}e^{-i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}(-{\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})\gamma^{0}{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})=\eta_{P}\left|{e^{-}e^{+};\mathcal{B},-{\boldsymbol{P}}}\right\rangle$
(176)

if the wave function satisfies

$\gamma^{0}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}(-{\boldsymbol{x}})\gamma^{0}=\eta_{P}\Phi_{\mathcal{B}}^{(-{\boldsymbol{P}})}({\boldsymbol{x}})\hskip
56.9055pt(\eta_{P}=\pm 1)$ (177)

Note that parity reverses the CM momentum ${\boldsymbol{P}}$ of the state.

####  VI.1.5 Charge conjugation $\eta_{C}$

The charge conjugation operator $\mathbb{C}$ transforms particles into
antiparticles.

$\mathbb{C}b({\boldsymbol{p}},\lambda)\mathbb{C}^{\dagger}=d({\boldsymbol{p}},\lambda)\hskip
56.9055pt\mathbb{C}d({\boldsymbol{p}},\lambda)\mathbb{C}^{\dagger}=b({\boldsymbol{p}},\lambda)$
(178)

In the Dirac representation of the $\gamma$ matrices (here $T$ indicates
transpose and ${\alpha_{2}}\equiv\gamma^{0}\gamma^{2}$)

$\mathbb{C}\psi(t,{\boldsymbol{x}})\mathbb{C}^{\dagger}=-i\gamma^{2}\psi^{*}(t,{\boldsymbol{x}})=i{\alpha_{2}}\bar{\psi}^{T}(t,{\boldsymbol{x}})\hskip
56.9055pt\mathbb{C}\bar{\psi}(t,{\boldsymbol{x}})\mathbb{C}^{\dagger}=i\psi^{T}(t,{\boldsymbol{x}}){\alpha_{2}}$
(179)

This implies
$v({\boldsymbol{k}},\lambda)=-i\gamma^{2}u^{*}({\boldsymbol{k}},\lambda)$ and
thus $\bar{\chi}_{\lambda}=i\sigma_{2}\chi_{\lambda}$ in (21). For a
Positronium state to be an eigenstate of charge conjugation,

$\mathbb{C}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle=\eta_{C}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$
(180)

its wave function should satisfy

${\alpha_{2}}\big{[}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}(-{\boldsymbol{x}})\big{]}^{T}{\alpha_{2}}=\eta_{C}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}})\hskip
56.9055pt(\eta_{C}=\pm 1)$ (181)

Exercise A.10: Derive (181).

####  VI.1.6 Wave functions of Para - and Orthopositronium

Non-relativistic Para- and Orthopositronium have zero orbital angular momentum
in the rest frame,
$\big{[}{{\boldsymbol{L}}},{\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})}\big{]}=0$.
Hence their wave functions have no angular dependence. The radial dependence
factorizes from the Dirac structure since the spin, parity and charge
conjugation constraints are independent of the radial coordinate
$r=|{\boldsymbol{x}}|$,

$\displaystyle\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})=N_{\mathcal{B}}^{(0)}\,\Gamma_{\mathcal{B}}\,F^{(0)}(r)$
(182)

where $\Gamma_{\mathcal{B}}$ is an ${\boldsymbol{x}}$-independent $4\times 4$
Dirac matrix. Para- and Orthopositronia have the same radial function $F(r)$
and the same binding energy $E_{b}$ at ${\cal O}\left(\alpha^{2}\right)$. The
energy degeneracy holds for all ${\boldsymbol{P}}$, indicating that the
factorization (182) holds in any frame,

$\displaystyle\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}})=N_{\mathcal{B}}^{(P)}\,\Gamma_{\mathcal{B}}\,F^{({\boldsymbol{P}})}({\boldsymbol{x}})\hskip
56.9055pt\int
d{\boldsymbol{x}}\,|F^{({\boldsymbol{P}})}({\boldsymbol{x}})|^{2}=E_{P}$ (183)

I verify this in section VI.3. The boosted radial function
$F^{({\boldsymbol{P}})}({\boldsymbol{x}})$ is angular dependent for
${\boldsymbol{P}}\neq 0$ due to Lorentz contraction in the
${\boldsymbol{P}}$-direction. With its normalization fixed as in (183) the
constants $N_{\mathcal{B}}^{(P)}$ are determined by the normalization (168) of
the state.

In the following I take ${\boldsymbol{P}}=(0,0,P)$ in the $z$-direction, and
consider Orthopositronium with $j^{z}=\lambda$. The following Dirac structures
$\Gamma_{\mathcal{B}}$ in (182) give the correct $J^{PC}$ quantum numbers of
the Positronia:

$\displaystyle\textbf{Parapositronium:}\ J^{PC}=0^{-+}\hskip
28.45274pt\Gamma_{Para}=\gamma_{5}$ (184)

  * •

Spin:
$\left[{{\boldsymbol{S}}},{\gamma_{5}}\right]=\left[{{\textstyle\frac{1}{2}}\gamma_{5}{\boldsymbol{\alpha}}},{\gamma_{5}}\right]=0$,
hence $j=s=0$.

  * •

Parity: $\gamma^{0}\gamma_{5}\gamma^{0}=-\gamma_{5}$, hence $\eta_{P}=-1$.

  * •

Charge conjugation: ${\alpha_{2}}\gamma_{5}^{T}{\alpha_{2}}=\gamma_{5}$, hence
$\eta_{C}=+1$.

$\displaystyle\textbf{Orthopositronium:}\ J^{PC}=1^{--}\hskip
28.45274pt\Gamma_{Ortho}^{\lambda}={\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}\hskip
28.45274pt{\boldsymbol{e}}_{\pm 1}=-\frac{1}{\sqrt{2}}(\pm 1,i,0)\hskip
28.45274pt{\boldsymbol{e}}_{0}=(0,0,1)$ (185)

  * •

Spin:
$\left[{S^{z}},{{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}}\right]=\lambda\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}$,
hence $j^{z}=\lambda$, and
$\sum_{i}\left[{S^{i}},{\left[{S^{i}},{{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}}\right]}\right]=2\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}$,
hence $j=1$.

  * •

Parity:
$\gamma^{0}\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}\,\gamma^{0}=-{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}$,
hence $\eta_{P}=-1$.

  * •

Charge conjugation:
${\alpha_{2}}\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}^{T}\,{\alpha_{2}}=-{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}$,
hence $\eta_{C}=-1$.

The constants $N_{\mathcal{B}}^{(P)}$ of (183) are then determined by the
state normalization (168) to be, at ${\cal O}\left(\alpha^{0}\right)$,

$\displaystyle
N_{Para}^{({\boldsymbol{P}})}=N_{Ortho}^{({\boldsymbol{P}})}(\lambda=0)=\frac{E_{P}}{2m}\hskip
56.9055ptN_{Ortho}^{({\boldsymbol{P}})}(\lambda=\pm 1)=1$ (186)

Exercise A.11: Verify (186).

###  VI.2 The Schrödinger equation for Positronium at ${\boldsymbol{P}}=0$

The Schrödinger equation for the rest frame wave function follows from the
condition that the (Para- or Ortho-) Positronium state (165) be an eigenstate
of the Hamiltonian (138),

$\displaystyle\mathcal{H}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle=\big{[}\mathcal{H}_{0}(f)+\mathcal{H}_{V}\big{]}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle=(2m+E_{b})\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle$
(187)

Transverse photons do not contribute to $E_{b}$ at ${\cal
O}\left(\alpha^{2}\right)$. Their coupling to electrons is proportional to the
3-momentum of the electron, which in the rest frame is of ${\cal
O}\left(\alpha m\right)$.

The free fermion Hamiltonian acting on the electron fields gives (note that
$\left\\{{\psi},{\bar{\psi}}\right\\}$ leaves a $\gamma^{0}$, and
${\boldsymbol{\alpha}}\gamma^{0}=-\gamma^{0}{\boldsymbol{\alpha}}$)

$\displaystyle\big{[}{\mathcal{H}_{0}(f)},{\bar{\psi}({\boldsymbol{x}}_{1})}\big{]}$
$\displaystyle=\int
d{\boldsymbol{x}}\,\psi^{\dagger}({\boldsymbol{x}})(-i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}+m\gamma^{0})\big{\\{}{\psi({\boldsymbol{x}})},{\bar{\psi}({\boldsymbol{x}}_{1})}\big{\\}}=\bar{\psi}({\boldsymbol{x}}_{1})(-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}+m\gamma^{0})$
$\displaystyle\big{[}{\mathcal{H}_{0}(f)},{\psi({\boldsymbol{x}}_{2})}\big{]}$
$\displaystyle=-(-i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}+m\gamma^{0})\psi({\boldsymbol{x}}_{2})$
(188)

Together with the projection operators $\Lambda_{\pm}$ in
$\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle$ these give
energies $E=\sqrt{-\boldsymbol{\nabla}^{2}+m^{2}}$,

$\displaystyle(-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}+m\gamma^{0})\frac{1}{2E_{1}}(E_{1}-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}+m\gamma^{0})$
$\displaystyle=\frac{1}{2E_{1}}\big{[}E_{1}(-i{\boldsymbol{\alpha}}\cdot{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}+m\gamma^{0})-{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1}^{2}+m^{2}\big{]}=\Lambda_{+}(i{\overset{\leftarrow}{\boldsymbol{\nabla}}}_{1})E_{1}$
$\displaystyle-\frac{1}{2E_{2}}(E_{2}+i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}-m\gamma^{0})(-i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}+m\gamma^{0})$
$\displaystyle=\frac{1}{2E_{2}}\big{[}E_{2}(i{\boldsymbol{\alpha}}\cdot{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}-m\gamma^{0})-{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2}^{2}+m^{2}\big{]}=E_{2}\Lambda_{-}(i{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2})$
(189)

Through a partial integration the $\boldsymbol{\nabla}^{2}$ in the energies
acts on the wave function. At ${\cal O}\left(\alpha^{2}\right)$ we have
$E\simeq m-\boldsymbol{\nabla}^{2}/2m$, giving the kinetic contribution of the
Schrödinger equation with reduced mass $m/2$,

$\displaystyle\Big{(}2m-\frac{\boldsymbol{\nabla}_{1}^{2}}{m}\Big{)}\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$
(190)

The potential energy arises from the instantaneous part (V.2.3) of the
Hamiltonian,

$\displaystyle\mathcal{H}_{V}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle$
$\displaystyle=\frac{1}{2}\int
d{\boldsymbol{x}}d{\boldsymbol{y}}\,\frac{e^{2}}{4\pi|{\boldsymbol{x}}-{\boldsymbol{y}}|}\big{[}\psi^{\dagger}\psi({\boldsymbol{x}})\big{]}\big{[}\psi^{\dagger}\psi({\boldsymbol{y}})\big{]}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}=0}\right\rangle$
(191)

Because Gauss’ law in temporal gauge is imposed as a constraint on the
physical states we have $\psi^{\dagger}\psi\left|{0}\right\rangle=0$ as in
(144). The effect of $\mathcal{H}_{V}$ is then to multiply the wave function
by the Coulomb potential,

$\displaystyle-\frac{\alpha}{|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}|}\,\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$
(192)

Combining (190) and (192) the eigenstate condition (187) implies the
Schrödinger equation for the wave function,

$\displaystyle\Big{(}2m-\frac{\boldsymbol{\nabla}^{2}}{m}-\frac{\alpha}{|{\boldsymbol{x}}|}\Big{)}\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})=(2m+E_{b})\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}})$
(193)

Since the Schrödinger equation is a Dirac scalar it gives the usual $\ell=0$
radial equation for $F^{(0)}(r)$ in (182),

$\displaystyle\frac{1}{m}\Big{[}{F^{(0)}}^{\prime\prime}(r)+\frac{2}{r}F^{(0)^{\prime}}(r)\Big{]}+\Big{[}\frac{\alpha}{r}+E_{b}\Big{]}F^{(0)}(r)=0\
\ \ \ \
{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\Longrightarrow}\
\ \ \ \ F^{(0)}(r)=\frac{\alpha^{3/2}\,m^{2}}{\sqrt{4\pi}}\exp(-\alpha
mr/2)\hskip 28.45274ptE_{b}=-{\textstyle\frac{1}{4}}m\alpha^{2}$ (194)

where the normalization of $F^{(0)}(r)$ is determined by (183) ($E_{P=0}=2m$).

###  VI.3 Positronium with momentum ${\boldsymbol{P}}$

Fock states quantized at equal time transform non-covariantly under boosts
since the definition of time is frame dependent. The Poincaré invariance of
the QED action nevertheless guarantees that measurables will be Lorentz
covariant. In this section I demonstrate this for the binding energy of
Positronia at ${\cal O}\left(\alpha^{2}\right)$. Lorentz covariance determines
the momentum dependence of the binding,

$\displaystyle\Delta
E({\boldsymbol{P}})\equiv\sqrt{{\boldsymbol{P}}^{2}+(2m+E_{b})^{2}}-\sqrt{{\boldsymbol{P}}^{2}+4m^{2}}=\frac{2mE_{b}}{E_{P}}+{\cal
O}\left(\alpha^{4}\right)\hskip 56.9055pt\Delta E({\boldsymbol{P}}=0)=E_{b}$
(195)

where $E_{P}=\sqrt{{\boldsymbol{P}}^{2}+4m^{2}}$. In section VI.4 I evaluate
the hyperfine splitting between Ortho- and Parapositronia at ${\cal
O}\left(\alpha^{4}\right)$ for ${\boldsymbol{P}}=0$. In sections VI.5 and VI.6
I consider the covariance of form factors and deep inelastic scattering.

The importance of properly taking into account the momentum dependence of
bound state wave functions was emphasized in Brodsky and Primack (1969). The
frame dependence of atomic wave functions is of general interest, since it
shows how the classical concept of Lorentz contraction is realized for quantum
bound states. Surprisingly, there appears to be only one study Järvinen (2005)
of atoms with general CM momenta ${\boldsymbol{P}}$, even at leading order.
The following analysis is equivalent to that one, but is formulated in terms
of Fock states in temporal gauge.

In atoms with CM momentum ${\boldsymbol{P}}$ transverse photons contribute at
leading order to the binding energy $E_{b}$, since they couple at ${\cal
O}\left(\alpha^{0}\right)$ to electrons whose momenta
$\propto{\boldsymbol{P}}$. I consider first the kinetic and potential energies
of the $\left|{e^{-}e^{+}}\right\rangle$ Fock state (165), and then determine
the wave function of the $\left|{e^{-}e^{+}\gamma}\right\rangle$ state in
Positronium. Only terms which contribute to $E_{b}$ at ${\cal
O}\left(\alpha^{2}\right)$ are retained.

####  VI.3.1 Kinetic and potential energy

The above relations (VI.2) and (VI.2) are valid for all ${\boldsymbol{P}}$.
After a partial integration the derivatives in $E_{1}$ and $E_{2}$ operate in
(165) on
$\exp\big{[}i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2\big{]}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})$.
Let
${\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}={\textstyle\frac{1}{2}}({\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}+{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2})+{\textstyle\frac{1}{2}}({\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}-{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2})$,
and similarly for ${\overset{\leftarrow}{\boldsymbol{\nabla}}}_{2}$. Then the
first term gives $i{\boldsymbol{P}}/2$ while the second, denoted
$\boldsymbol{\nabla}\equiv{\textstyle\frac{1}{2}}({\overset{\rightarrow}{\boldsymbol{\nabla}}}_{1}-{\overset{\rightarrow}{\boldsymbol{\nabla}}}_{2})$,
gives ${\cal O}\left(\alpha\right)$ derivatives of
$\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}})$,

$\displaystyle
E_{1}=\sqrt{({\textstyle\frac{1}{2}}{\boldsymbol{P}}-i\boldsymbol{\nabla})^{2}+m^{2}}\hskip
56.9055ptE_{2}=\sqrt{({\textstyle\frac{1}{2}}{\boldsymbol{P}}+i\boldsymbol{\nabla})^{2}+m^{2}}$
(196)

At ${\cal O}\left(\alpha^{2}\right)$ we need to keep two powers of
$\boldsymbol{\nabla}$. Using
$\sqrt{1+x}=1+{\textstyle\frac{1}{2}}x-\frac{1}{8}x^{2}+{\cal
O}\left(x^{3}\right)$ and denoting

$\displaystyle E_{P}\equiv\sqrt{{\boldsymbol{P}}^{2}+4m^{2}}\hskip
56.9055pt\gamma\equiv\frac{E_{P}}{2m}\hskip
56.9055pt\beta\equiv\frac{P}{E_{P}}$ (197)

we get

$\displaystyle\mathcal{H}_{0}(f)\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle:\hskip
28.45274ptE_{1}+E_{2}\simeq
E_{P}-\frac{2}{E_{P}}\Big{(}\boldsymbol{\nabla}_{\perp}^{2}-\frac{1}{\gamma^{2}}\boldsymbol{\nabla}_{\|}^{2}\Big{)}=E_{P}-\frac{1}{m\gamma}\Big{(}\boldsymbol{\nabla}_{\perp}^{2}-\frac{1}{\gamma^{2}}\boldsymbol{\nabla}_{\|}^{2}\Big{)}$
(198)

where $\perp$ and $\|$ refer to the ${\boldsymbol{P}}$-direction, here taken
to be along the $z$-axis.

The potential energy depends only on the instantaneous positions of the
fermions and thus gives the same result as in (192) for ${\boldsymbol{P}}=0$,
the wave functions is multiplied by

$\displaystyle\mathcal{H}_{V}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle:\hskip
42.67912pt-\frac{\alpha}{|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}|}=-e^{2}\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}}\,\frac{1}{{\boldsymbol{q}}^{2}}e^{-i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}$
(199)

This already signals the importance of transverse photon exchange, since the
potential energy should be commensurate with $\Delta E({\boldsymbol{P}})$ in
(195), which is ${\boldsymbol{P}}$-dependent.

####  VI.3.2 The transverse photon Fock state

For ${\boldsymbol{P}}\neq 0$ the transverse photon vertices
$\propto{\boldsymbol{P}}$ are of ${\cal O}\left(\alpha^{0}\right)$, so
transverse and Coulomb photon exchanges contribute at the same order in
$\alpha$. The transverse photon and its conjugate electric field in temporal
gauge $(A^{0}=0)$ may at $t=0$ be expanded in photon creation and annihilation
operators $a^{\dagger}$ and $a$, with polarization 3-vectors
${\boldsymbol{\varepsilon}}_{s}$, $s=\pm 1$,

$\displaystyle{\boldsymbol{A}}_{T}({\boldsymbol{x}})=\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s=\pm
1}\Big{[}{\boldsymbol{\varepsilon}}_{s}({\boldsymbol{q}})\,e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}}a({\boldsymbol{q}},s)+{\boldsymbol{\varepsilon}}_{s}^{*}({\boldsymbol{q}})\,e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}}a^{\dagger}({\boldsymbol{q}},s)\Big{]}$
$\displaystyle{\boldsymbol{E}}_{T}({\boldsymbol{x}})=i\int\frac{d{\boldsymbol{q}}}{2(2\pi)^{3}}\sum_{s=\pm
1}\Big{[}{\boldsymbol{\varepsilon}}_{s}({\boldsymbol{q}})\,e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}}a({\boldsymbol{q}},s)-{\boldsymbol{\varepsilon}}_{s}^{*}({\boldsymbol{q}})\,e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}}a^{\dagger}({\boldsymbol{q}},s)\Big{]}$
(200)
$\displaystyle\left[{a({\boldsymbol{q}},s)},{a^{\dagger}({\boldsymbol{q}}^{\prime},s^{\prime})}\right]=(2\pi)^{3}2|{\boldsymbol{q}}|\delta({\boldsymbol{q}}-{\boldsymbol{q}}^{\prime})\delta_{s,s^{\prime}}$
$\displaystyle{\boldsymbol{q}}\cdot{\boldsymbol{\varepsilon}}_{s}({\boldsymbol{q}})=0\hskip
28.45274pt{\boldsymbol{\varepsilon}}_{s}^{*}({\boldsymbol{q}})\cdot{\boldsymbol{\varepsilon}}_{s^{\prime}}({\boldsymbol{q}})=\delta_{s,s^{\prime}}\hskip
28.45274pt\sum_{s=\pm
1}\varepsilon_{s}^{i}({\boldsymbol{q}}){\varepsilon_{s}^{j}}^{*}({\boldsymbol{q}})=\delta^{ij}-\frac{q^{i}q^{j}}{{\boldsymbol{q}}^{2}}$

The interaction between the electron and the transverse photon fields in the
Hamiltonian (138) is given by

$\displaystyle\mathcal{H}_{int}({\boldsymbol{A}}_{T})=-e\int
d{\boldsymbol{x}}\,\psi^{\dagger}({\boldsymbol{x}})\,{\boldsymbol{\alpha}}\cdot{\boldsymbol{A}}_{T}({\boldsymbol{x}})\psi({\boldsymbol{x}})$
(201)

This creates $\left|{e^{-}e^{+}\gamma}\right\rangle$ states with a photon of
momentum ${\boldsymbol{q}}$ and polarization $s$. At leading order,

$\displaystyle\mathcal{H}_{int}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle=e\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})$
$\displaystyle\times\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s}\frac{1}{E_{P}}{\boldsymbol{P}}\cdot\varepsilon_{s}^{*}({\boldsymbol{q}})\,a^{\dagger}({\boldsymbol{q}},s)\big{(}-e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{1}}+e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{2}}\big{)}\left|{0}\right\rangle\equiv\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s}\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$
(202)

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

Operating with $\mathcal{H}_{int}$ a second time and retaining only the terms
where the transverse photon is absorbed, giving an
$\left|{e^{-}e^{+}}\right\rangle$ Fock state,

$\displaystyle\mathcal{H}_{int}^{2}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$
$\displaystyle=e^{2}\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})$
$\displaystyle\times\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s}\frac{1}{E_{P}^{2}}\big{[}{\boldsymbol{P}}\cdot\varepsilon_{s}^{*}({\boldsymbol{q}})\big{]}\big{[}{\boldsymbol{P}}\cdot\varepsilon_{s}({\boldsymbol{q}})\big{]}\big{|}-e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{1}}+e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{2}}\big{|}^{2}\left|{0}\right\rangle$
(203) $\displaystyle\mbox{where}\ \ \
\big{|}-e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{1}}+e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{2}}\big{|}^{2}=2-e^{-i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}-e^{i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}$
(204)

The ${\boldsymbol{x}}_{1,2}$-independent term in (204) corresponds to the
absorption of the photon on the same fermion from which it was emitted. This
loop contribution gives a multiplicative renormalization of the state and does
not contribute to the eigenstate condition at lowest order. Neglecting this
term and summing over the photon polarization $s$ using (VI.3.2) we have

$\displaystyle\mathcal{H}_{int}^{2}\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$
$\displaystyle=-e^{2}\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})$
$\displaystyle\times\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\,\beta^{2}\,\frac{{\boldsymbol{q}}_{\perp}^{2}}{{\boldsymbol{q}}^{2}}\,\big{[}e^{-i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}+e^{i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}\big{]}\left|{0}\right\rangle$
(205)

where $\beta=|{\boldsymbol{P}}|/E_{P}$ and ${\boldsymbol{q}}_{\perp}$ is the
component of ${\boldsymbol{q}}$ orthogonal to ${\boldsymbol{P}}$ (i.e., to the
$z$-axis).

####  VI.3.3 The bound state condition

The Positronium state, including the $\left|{e^{-}e^{+}\gamma}\right\rangle$
Fock component, can be expressed as the superposition

$\displaystyle\left|{\mathcal{B},{\boldsymbol{P}}}\right\rangle=\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle+\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s}C_{\gamma}({\boldsymbol{q}})\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$
(206)

where $\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ is defined in
(VI.3.2) and I anticipated that its relative weight
$C_{\gamma}({\boldsymbol{q}})$ is independent of $s$.
$C_{\gamma}({\boldsymbol{q}})$ should be determined so that
$\left|{\mathcal{B},{\boldsymbol{P}}}\right\rangle$ is an eigenstate of
$\mathcal{H}$,

$\displaystyle\mathcal{H}\left|{\mathcal{B},{\boldsymbol{P}}}\right\rangle$
$\displaystyle=(\mathcal{H}_{0}(f)+\mathcal{H}_{V})\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle+\mathcal{H}_{int}\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s}C_{\gamma}({\boldsymbol{q}})\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$
$\displaystyle+\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s}\Big{[}1+\big{(}\mathcal{H}_{0}(f)+\mathcal{H}_{0}(A)\big{)}C_{\gamma}({\boldsymbol{q}})\Big{]}\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle=E_{\mathcal{B}}^{({\boldsymbol{P}})}\left|{\mathcal{B},{\boldsymbol{P}}}\right\rangle$
(207)

$\mathcal{H}_{0}(f)+\mathcal{H}_{V}$ modify
$\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}$ by the factors in (198) and (199).
The $\mathcal{H}_{int}$ term is given by (VI.3.2), adding the factor
$C_{\gamma}({\boldsymbol{q}})$ to the integrand of the
${\boldsymbol{q}}$-integration. The first contribution to
$\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ follows from
(VI.3.2). The action of $\mathcal{H}_{0}(f)$ is similar to that on
$\left|{e^{-}e^{+};\mathcal{B},{\boldsymbol{P}}}\right\rangle$ in (198), but
now the additional ${\boldsymbol{x}}_{1,2}$ dependence in
$\exp(-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{1,2})$ leaves an ${\cal
O}\left(\alpha\right)$ contribution. Since
$\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ is already
suppressed by a factor $e$ we may here neglect the ${\cal
O}\left(\alpha^{2}\right)$ terms, including its potential energy
($\mathcal{H}_{V}$). Recalling that $E_{i}=$
${\scriptsize\sqrt{-\boldsymbol{\nabla}_{i}^{2}+m^{2}}}$ and separating the
${\cal O}\left(\alpha^{0}\right)$ contribution through,

$\displaystyle(E_{1}+E_{2})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}$
$\displaystyle\simeq
e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Big{(}\sqrt{{\textstyle\frac{1}{4}}E_{P}^{2}-i{\boldsymbol{P}}\cdot\boldsymbol{\nabla}_{1}}+\sqrt{{\textstyle\frac{1}{4}}E_{P}^{2}-i{\boldsymbol{P}}\cdot\boldsymbol{\nabla}_{2}}\,\Big{)}$
$\displaystyle\simeq
e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Big{[}E_{P}-\frac{i}{E_{P}}\,{\boldsymbol{P}}\cdot(\boldsymbol{\nabla}_{1}+\boldsymbol{\nabla}_{2})\Big{]}$
(208)

gives the sum of the fermion kinetic energies as

$\displaystyle\mathcal{H}_{0}(f)\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle:\hskip
28.45274ptE_{1}+E_{2}=E_{P}-\beta q_{\|}+{\cal O}\left(\alpha^{2}\right)$
(209)

where $\beta=P/E_{P}$ and ${\boldsymbol{P}}\cdot{\boldsymbol{q}}=Pq_{\|}$.

The kinetic energy of the photon follows from
$\mathcal{H}_{0}{(A)}=\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}\,2|{\boldsymbol{q}}|}|{\boldsymbol{q}}|\sum_{s=\pm
1}a^{\dagger}({\boldsymbol{q}},s)a({\boldsymbol{q}},s)$,

$\displaystyle\mathcal{H}_{0}(A)\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle:\hskip
28.45274ptE_{\gamma}=|{\boldsymbol{q}}|$ (210)

Comparing the coefficients of
$\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ in the eigenstate
condition (VI.3.3) gives, since
$E_{\mathcal{B}}^{({\boldsymbol{P}})}=E_{P}+{\cal O}\left(\alpha^{2}\right)$,

$\displaystyle 1+C_{\gamma}({\boldsymbol{q}})(E_{P}+|{\boldsymbol{q}}|-\beta
q_{\|})=C_{\gamma}({\boldsymbol{q}})E_{P}+{\cal
O}\left(\alpha^{2}\right)\hskip
28.45274pt{\color[rgb]{1,0,0}\definecolor[named]{pgfstrokecolor}{rgb}{1,0,0}\Longrightarrow}\hskip
28.45274ptC_{\gamma}({\boldsymbol{q}})=-\frac{1}{|{\boldsymbol{q}}|-\beta
q_{\|}}=-\frac{|{\boldsymbol{q}}|+\beta
q_{\|}}{{\boldsymbol{q}}_{\perp}^{2}+q_{\|}^{2}/\gamma^{2}}$ (211)

Including $C_{\gamma}({\boldsymbol{q}})$ in (VI.3.2) we have in (VI.3.3),

$\displaystyle\mathcal{H}_{int}\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s}C_{\gamma}({\boldsymbol{q}})\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$
$\displaystyle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})$
$\displaystyle\times
e^{2}\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\beta^{2}\,\frac{{\boldsymbol{q}}_{\perp}^{2}}{{\boldsymbol{q}}^{2}}\frac{|{\boldsymbol{q}}|+\beta
q_{\|}}{{\boldsymbol{q}}_{\perp}^{2}+q_{\|}^{2}/\gamma^{2}}\,\big{[}e^{-i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}+e^{i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}\big{]}\left|{0}\right\rangle$
(212)

The $\beta q_{\|}$ term in the numerator does not contribute since the
remaining integrand is symmetric under ${\boldsymbol{q}}\to-{\boldsymbol{q}}$.
We may then use this symmetry to set
$\exp[i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})]\to\exp[-i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})]$
and cancel a factor $2|{\boldsymbol{q}}|$. Combining this transverse photon
contribution with the Coulomb one (199) the integral over ${\boldsymbol{q}}$
becomes (with ${\boldsymbol{x}}={\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}$),

$\displaystyle-e^{2}$
$\displaystyle\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}}\,\frac{1}{{\boldsymbol{q}}^{2}}\Big{[}1-\frac{\beta^{2}q_{\perp}^{2}}{{\boldsymbol{q}}_{\perp}^{2}+q_{\|}^{2}/\gamma^{2}}=\frac{{\boldsymbol{q}}^{2}/\gamma^{2}}{{\boldsymbol{q}}_{\perp}^{2}+q_{\|}^{2}/\gamma^{2}}\Big{]}e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}}=-\frac{e^{2}}{\gamma^{2}}\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}}\,\frac{e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}}}{{\boldsymbol{q}}_{\perp}^{2}+q_{\|}^{2}/\gamma^{2}}=-\frac{\alpha}{\gamma\sqrt{{\boldsymbol{x}}_{\perp}^{2}+\gamma^{2}x_{\|}^{2}}}$
(213)

Adding the kinetic energy (198) the eigenstate condition (VI.3.3) imposes the
bound state condition on $\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}$, with the
required eigenvalue (195),

$\displaystyle\Big{[}E_{P}-\frac{1}{m\gamma}\Big{(}\boldsymbol{\nabla}_{\perp}^{2}-\frac{1}{\gamma^{2}}\boldsymbol{\nabla}_{\|}^{2}\Big{)}-\frac{\alpha}{\gamma\sqrt{{\boldsymbol{x}}_{\perp}^{2}+\gamma^{2}x_{\|}^{2}}}\Big{]}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}})=\Big{(}E_{P}+\frac{1}{\gamma}\,E_{b}\Big{)}\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}})$
(214)

A comparison with the ${\boldsymbol{P}}=0$ equation (193) shows that up to a
normalization we have

$\displaystyle\Phi_{\mathcal{B}}^{({\boldsymbol{P}})}({\boldsymbol{x}})=\Phi_{\mathcal{B}}^{(0)}({\boldsymbol{x}}_{\perp},\gamma
x_{\|})$ (215)

i.e., standard Lorentz contraction. The contraction is the same for all Dirac
components, justifying the factorization (183). According to (194) the
contracted radial function is

$\displaystyle
F^{({\boldsymbol{P}})}({\boldsymbol{x}})=\gamma\,F^{(0)}(r_{P})=\gamma\,\frac{\alpha^{3/2}m^{2}}{\sqrt{4\pi}}\,e^{-\alpha
mr_{P}/2}\hskip
56.9055ptr_{P}\equiv\sqrt{{\boldsymbol{x}}_{\perp}^{2}+\gamma^{2}x_{\|}^{2}}\hskip
28.45274pt\gamma=\frac{E_{P}}{2m}=\frac{\sqrt{{\boldsymbol{P}}^{2}+4m^{2}}}{2m}$
(216)

The Lorentz contraction of $F^{({\boldsymbol{P}})}({\boldsymbol{x}})$ agrees
with the classical result. Recall, however, that the
$\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ Fock component
(206) also contributes. The kinetic energies of its constituents (209), (210)
are of ${\cal O}\left(\alpha\right)$, i.e., large compared to the ${\cal
O}\left(\alpha^{2}\right)$ binding energy of
$\left|{\mathcal{B},{\boldsymbol{P}}}\right\rangle$. By the uncertainty
principle $\left|{\mathcal{B},{\boldsymbol{P}}}\right\rangle$ fluctuates into
$\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ only a fraction
$\alpha$ of the time. Equivalently, the norm of the
$\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ Fock component is
of ${\cal O}\left(\alpha\right)$.

###  VI.4 * Hyperfine splitting of Positronium at ${\boldsymbol{P}}=0$

Hyperfine splitting (hfs) is defined as the difference between the Ortho- and
Parapositronium binding energies, $E_{hfs}=E_{b}(Ortho)-E_{b}(Para)$. The hfs
(I.1) is known to high accuracy, with current work addressing the ${\cal
O}\left(\alpha^{7}m\right)$ contribution using the methods of NRQED. Here I
shall illustrate the Fock state method in temporal gauge by evaluating the
${\cal O}\left(\alpha^{4}m\right)$ contribution. At this order the hfs arises
from transverse photon exchange between the $e^{-}$ and $e^{+}$, as well as
from the annihilation contribution $e^{-}e^{+}\to\gamma\to e^{-}e^{+}$ (for
Orthopositronium only). In the Fock state approach this means considering the
$\left|{e^{-}e^{+}\gamma}\right\rangle$ and $\left|{\gamma}\right\rangle$ Fock
states, respectively.

####  VI.4.1 The transverse photon Fock state
$\left|{e^{-}e^{+}\gamma}\right\rangle$ contribution

In section VI.3 I evaluated the transverse photon exchange contribution to
Positronium with ${\boldsymbol{P}}\neq 0$ at leading order. Here I consider
transverse photon exchange for Positronium at rest (${\boldsymbol{P}}=0$). The
two electron-photon vertices are then proportional to ${\cal
O}\left(\alpha\right)$ Bohr momenta, which makes the transverse photon
contribution to be of ${\cal O}\left(\alpha^{4}\right)$ in the rest frame. I
discuss only photon emission from the $e^{-}$ and absorption by the $e^{+}$.
The converse contribution is identical and is taken into account by a factor 2
in the final result. Photons both emitted and absorbed on the $e^{-}$ do not
contribute to the spin correlation (hfs) between the $e^{-}$ and $e^{+}$.

The Positronium state including the Fock state with a transverse photon is
(206)

$\displaystyle\left|{\mathcal{B}}\right\rangle=\left|{e^{-}e^{+};\mathcal{B}}\right\rangle+\left|{e^{-}e^{+}\gamma;\mathcal{B}}\right\rangle$
(217)

where $\left|{e^{-}e^{+};\mathcal{B}}\right\rangle$ is defined in (165) with
${\boldsymbol{P}}=0$ and the wave function
$\Phi_{\mathcal{B}}({\boldsymbol{x}})=\Gamma_{\mathcal{B}}\,F(r)$ according to
(182) and (186). Its Dirac structures are $\Gamma_{Para}=\gamma_{5}$,
$\Gamma_{Ortho}={\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}$ as in
(184), (185) and the radial function $F(r)$ is given in (194). The transverse
photon state is as in (206) with $C_{\gamma}=-1/|{\boldsymbol{q}}|$ from (211)
with ${\boldsymbol{P}}=0$,

$\displaystyle\left|{e^{-}e^{+}\gamma;\mathcal{B}}\right\rangle=\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2|{\boldsymbol{q}}|}\sum_{s}\frac{-1}{|{\boldsymbol{q}}|}\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$
(218)

The $\left|{e^{-}e^{+}\gamma;{\boldsymbol{q}},s}\right\rangle$ state created
by $\mathcal{H}_{int}({\boldsymbol{A}}_{T})$ (201) acting on
$\left|{e^{-}e^{+};\mathcal{B}}\right\rangle$ with $e^{-}\to e^{-}\gamma$ is
given in (A.591).

When the Hamiltonian acts on its eigenstate $\left|{\mathcal{B}}\right\rangle$
the energy eigenvalue may be read off from the coefficient of the
$\left|{e^{-}e^{+};\mathcal{B}}\right\rangle$ Fock state. Projecting on this
state,

$\displaystyle\mathcal{H}\left|{\mathcal{B}}\right\rangle=\Big{[}2m-{\textstyle\frac{1}{4}}m\alpha^{2}+{\cal
O}\left(\alpha^{4}\right)+\frac{\langle{e^{-}e^{+};\mathcal{B}}|\mathcal{H}_{int}\left|{e^{-}e^{+}\gamma;\mathcal{B}}\right\rangle}{\langle{e^{-}e^{+};\mathcal{B}}|e^{-}e^{+};\mathcal{B}\rangle}\Big{]}\left|{e^{-}e^{+};\mathcal{B}}\right\rangle+\ldots$
(219)

The ${\cal O}\left(\alpha^{4}\right)$ term represents higher order corrections
to the eigenvalue from
$(\mathcal{H}_{0}+\mathcal{H}_{V})\left|{e^{-}e^{+};\mathcal{B}}\right\rangle$,
e.g., due to the Taylor expansion of the energies $E_{i}$ in (VI.2). They are
the same for Para- and Orthopositronium and do not affect the hfs. I keep only
terms which are spin-, i.e., $\Gamma_{\mathcal{B}}$-dependent.

The absorption of the photon on the $e^{+}$ is given by
$\left[{\mathcal{H}_{int}},{\psi({\boldsymbol{x}}_{2})}\right]$ analogously as
the emission from $e^{-}$ in (A.591),

$\displaystyle\mathcal{H}_{int}\left|{e^{-}e^{+}\gamma;\mathcal{B}}\right\rangle=-e^{2}\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}2{\boldsymbol{q}}^{2}}\sum_{s}\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1})$
$\displaystyle{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1}){\boldsymbol{\alpha}}\cdot{\boldsymbol{\varepsilon}}_{s}^{*}({\boldsymbol{q}})e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{1}}{\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})\Gamma_{\mathcal{B}}F(r)$
$\displaystyle\times{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2}){\boldsymbol{\alpha}}\cdot{\boldsymbol{\varepsilon}}_{s}({\boldsymbol{q}})e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{2}}{\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(220)

where $r=|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}|$ and the extra pair of
$\Lambda_{\pm}$ project $b^{\dagger}$ from $\bar{\psi}({\boldsymbol{x}}_{1})$
and $d^{\dagger}$ from $\psi({\boldsymbol{x}}_{2})$ (167). At ${\cal
O}\left(\alpha^{4}\right)$ two momenta can contribute, one each from the
photon vertices. The identities in (A.12) and (A.12) show that
${\boldsymbol{\alpha}}$ bracketed by two projectors becomes a derivative. This
contribution reduces the Dirac structure of (VI.4.1) so that the overlap in
(219) is independent of $\Gamma_{\mathcal{B}}$, i.e., it does not contribute
to the hfs. Hence we need only consider the contributions

$\displaystyle
e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{1}}{\overset{\leftarrow}{\Lambda}}({\boldsymbol{x}}_{1})\to-\frac{1}{2m}{\boldsymbol{\alpha}}\cdot{\boldsymbol{q}}\,e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{1}}\hskip
56.9055pt{\overset{\rightarrow}{\Lambda}}({\boldsymbol{x}}_{2})e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{2}}\to-\frac{1}{2m}{\boldsymbol{\alpha}}\cdot{\boldsymbol{q}}\,e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}_{2}}$
(221)

At ${\cal O}\left(\alpha^{4}\right)$ the remaining projectors can be set to
lowest order, $\Lambda_{\pm}=(1\pm\gamma^{0})/2$. The products of
${\boldsymbol{\alpha}}$-matrices may be reduced through
$\alpha_{i}\alpha_{j}=\delta_{ij}+i\varepsilon_{ijk}\alpha_{k}\gamma_{5}$. The
$\delta_{ij}$-function does not contribute here since
${\boldsymbol{q}}\cdot{\boldsymbol{\varepsilon}}_{s}({\boldsymbol{q}})=0$,

$\displaystyle{\varepsilon_{s}^{i}}^{*}q^{j}\alpha_{i}\alpha_{j}=i\varepsilon_{ijk}{\varepsilon_{s}^{i}}^{*}q^{j}\alpha_{k}\gamma_{5}\hskip
56.9055ptq^{\ell}\varepsilon_{s}^{m}\alpha_{\ell}\alpha_{m}=i\varepsilon_{\ell
mn}q^{\ell}\varepsilon_{s}^{m}\alpha_{n}\gamma_{5}$ (222)

Since $\left[{\gamma_{5}},{\Gamma_{\mathcal{B}}}\right]=0$ the $\gamma_{5}$’s
cancel, $\gamma_{5}^{2}=1$. The sum over photon polarizations (VI.3.2) gives
$\sum_{s}{\varepsilon_{s}^{i}}^{*}({\boldsymbol{q}})\varepsilon_{s}^{m}({\boldsymbol{q}})\to\delta^{im}$,
since the $-q^{i}q^{m}/{\boldsymbol{q}}^{2}$ term vanishes. Then
$i^{2}\varepsilon_{ijk}\varepsilon_{in\ell}q^{j}q^{\ell}={\boldsymbol{q}}^{2}\delta^{kn}-q^{k}q^{n}$.

Writing
$\alpha_{k}\Gamma_{\mathcal{B}}=\left[{\alpha_{k}},{\Gamma_{\mathcal{B}}}\right]+\Gamma_{\mathcal{B}}\alpha_{k}$
the second term does not give an hfs, while the commutator vanishes for
$\Gamma_{Para}=\gamma_{5}$. Hence is suffices to consider the Orthopositronium
$(j^{z}=\lambda)$ contribution,
$\left[{\alpha_{k}},{{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}}\right]=-2i\varepsilon_{jkp}e_{\lambda}^{j}\alpha_{p}\gamma_{5}$.
This is multiplied by the $\alpha_{n}$ in (222), giving
$-2i\varepsilon_{jkp}e_{\lambda}^{j}\alpha_{p}\alpha_{n}\gamma_{5}=2\varepsilon_{jkp}\varepsilon_{nip}e_{\lambda}^{j}\alpha_{i}=2(e_{\lambda}^{n}\alpha_{k}-\delta^{kn}{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}})$.
Combined with the ${\boldsymbol{q}}$-dependence found above we have

$\displaystyle({\boldsymbol{q}}^{2}\delta^{kn}-q^{k}q^{n})2(e_{\lambda}^{n}\alpha_{k}-\delta^{kn}{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}})=-2\big{[}{\boldsymbol{q}}^{2}{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}+({\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{q}})({\boldsymbol{\alpha}}\cdot{\boldsymbol{q}})\big{]}$
(223)

The contribution to (VI.4.1) that is relevant for the hfs is thus

$\displaystyle\mathcal{H}_{int}\left|{e^{-}e^{+}\gamma;Ortho}\right\rangle=\frac{e^{2}}{4m^{2}}\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}{\boldsymbol{q}}^{2}}\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1})\Lambda_{+}F(r)e^{-i{\boldsymbol{q}}\cdot({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2})}\big{[}{\boldsymbol{q}}^{2}{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}+({\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{q}})({\boldsymbol{\alpha}}\cdot{\boldsymbol{q}})\big{]}\Lambda_{-}\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(224)

where $\Lambda_{\pm}=(1\pm\gamma^{0})/2$. In the matrix elements of (219) both
electron fields are annihilated and the integral $\int
d({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2=(2\pi)^{3}\delta(\boldsymbol{0})$
cancels with the norm (168) in the denominator up to a factor $4m$. Recalling
that we only considered photon emission from $e^{-}$ and absorption on $e^{+}$
and so are getting half of the hfs,

$\displaystyle{\textstyle\frac{1}{2}}E_{hfs}^{T}$
$\displaystyle=\frac{\langle{e^{-}e^{+};\mathcal{B}}|\mathcal{H}_{int}\left|{e^{-}e^{+}\gamma;\mathcal{B}}\right\rangle}{\langle{e^{-}e^{+};\mathcal{B}}|e^{-}e^{+};\mathcal{B}\rangle}$
$\displaystyle=\frac{e^{2}}{16m^{3}}\int
d{\boldsymbol{x}}\,\frac{d{\boldsymbol{q}}}{(2\pi)^{3}{\boldsymbol{q}}^{2}}\,|F(r)|^{2}e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}}\mathrm{Tr}\,\big{\\{}{\textstyle\frac{1}{2}}(1-\gamma^{0}){\boldsymbol{e}}_{\lambda}^{*}\cdot{\boldsymbol{\alpha}}{\textstyle\frac{1}{2}}(1+\gamma^{0})\big{[}{\boldsymbol{q}}^{2}{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}+{\boldsymbol{e}}_{\lambda}^{i}\alpha^{j}q^{i}q^{j}\big{]}\big{\\}}$
(225)

The factor multiplying $q^{i}q^{j}$ in the integrand is symmetric under
$q^{i}\to-q^{i},\ x^{i}\to-x^{i}$, allowing
$q^{i}q^{j}\to{\textstyle\frac{1}{3}}{\boldsymbol{q}}^{2}\delta^{ij}$. The
trace factor becomes
${\textstyle\frac{2}{3}}{\boldsymbol{q}}^{2}\mathrm{Tr}\,\big{\\{}({\boldsymbol{e}}_{\lambda}^{*}\cdot{\boldsymbol{\alpha}})({\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}})\big{\\}}={\textstyle\frac{8}{3}}{\boldsymbol{q}}^{2}$.
With $F(r)$ given by (194),

$\displaystyle{\textstyle\frac{1}{2}}E_{hfs}^{T}=\frac{e^{2}}{6m^{3}}\int
d{\boldsymbol{x}}\,|F(r)|^{2}\int\frac{d{\boldsymbol{q}}}{(2\pi)^{3}}e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}}=\frac{4\pi\alpha}{6m^{3}}|F(0)|^{2}=\frac{1}{6}m\alpha^{4}$
(226)

The contribution to the hfs from transverse photon exchange is thus as
expected Penin (2014),

$\displaystyle E_{hfs}^{T}=\frac{1}{3}m\alpha^{4}$ (227)

####  VI.4.2 Hyperfine splitting from annihilation: $e^{-}e^{+}\to\gamma\to
e^{-}e^{+}$

The
$\left|{e^{-}e^{+}}\right\rangle\to\left|{\gamma}\right\rangle\to\left|{e^{-}e^{+}}\right\rangle$
transition is proportional to the square $|\Phi_{\mathcal{B}}(0)|^{2}$ of the
${\boldsymbol{P}}=0$ Positronium (165) wave function at the origin.
$\Phi_{\mathcal{B}}({\boldsymbol{x}})=\Gamma_{\mathcal{B}}F(r)$ where
$\Gamma_{\mathcal{B}}$ for Para- and Orthopositronium are in (184), (185) and
their common radial function $F(r)$ is in (194). Counting also the vertex
couplings the transition is $\propto e^{2}\,|F(0)|^{2}\propto\alpha^{4}$.
Hence we may neglect the ${\cal O}\left(\alpha m\right)$ relative (Bohr)
momenta in evaluating the hfs at ${\cal O}\left(\alpha^{4}\right)$. The
projectors $\Lambda_{\pm}$ in the state (165) may then be replaced with
${\textstyle\frac{1}{2}}(1\pm\gamma^{0})$. Annihilating both the $e^{-}$ and
$e^{+}$ in the state (165) by $\mathcal{H}_{int}$ gives

$\displaystyle\mathcal{H}_{int}\left|{e^{-}e^{+};\mathcal{B}}\right\rangle$
$\displaystyle=-e\int
d{\boldsymbol{x}}\,\bar{\psi}({\boldsymbol{x}}){\boldsymbol{\alpha}}\cdot{\boldsymbol{A}}({\boldsymbol{x}})\psi({\boldsymbol{x}})\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1}){\textstyle\frac{1}{2}}(1+\gamma^{0})\Phi_{\mathcal{B}}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\textstyle\frac{1}{2}}(1-\gamma^{0})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(228) $\displaystyle=-e\int
d{\boldsymbol{x}}\,\mathrm{Tr}\,\big{\\{}{\boldsymbol{\alpha}}\cdot{\boldsymbol{A}}({\boldsymbol{x}})\gamma^{0}{\textstyle\frac{1}{2}}(1+\gamma^{0})\Gamma_{\mathcal{B}}F(0){\textstyle\frac{1}{2}}(1-\gamma^{0})\big{\\}}\left|{0}\right\rangle=-{\textstyle\frac{1}{2}}eF(0)\int
d{\boldsymbol{x}}\,\mathrm{Tr}\,[\alpha^{i}\Gamma_{\mathcal{B}}]\,A^{i}({\boldsymbol{x}})\left|{0}\right\rangle$

As expected due to charge conjugation invariance, this vanishes for
Parapositronium, $\Gamma_{Para}=\gamma_{5}$. Hence the annihilation
contribution to the hfs arises only from Orthopositronium,
$\Gamma_{Ortho}^{\lambda}={\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}$
for states with $j^{z}=\lambda$:

$\displaystyle\mathcal{H}_{int}\left|{e^{-}e^{+};\mathcal{O}_{\lambda}}\right\rangle$
$\displaystyle=-2eF(0)\int
d{\boldsymbol{x}}\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{A}}({\boldsymbol{x}})\left|{0}\right\rangle\equiv-2eF(0)\left|{{\boldsymbol{A}},\lambda}\right\rangle$
(229)

The relevant action of the Hamiltonian (138) on this state is given by the
canonical commutation relations (137),

$\displaystyle\mathcal{H}\left|{{\boldsymbol{A}},\lambda}\right\rangle\to\int
d{\boldsymbol{y}}\,{\textstyle\frac{1}{2}}{\boldsymbol{E}}^{2}({\boldsymbol{y}})\int
d{\boldsymbol{x}}\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{A}}({\boldsymbol{x}})\left|{0}\right\rangle=i\int
d{\boldsymbol{x}}\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{E}}({\boldsymbol{x}})\left|{0}\right\rangle\equiv
i\left|{{\boldsymbol{E}},\lambda}\right\rangle$ (230)

$\mathcal{H}\left|{{\boldsymbol{E}},\lambda}\right\rangle$ has an overlap
$C_{\mathcal{O}}$ with Orthopositronium. Neglecting the other states which do
not contribute here,

$\displaystyle\mathcal{H}_{int}\left|{{\boldsymbol{E}},\lambda}\right\rangle=-e\int
d{\boldsymbol{y}}\,\psi^{\dagger}({\boldsymbol{y}})\,{\boldsymbol{\alpha}}\cdot{\boldsymbol{A}}({\boldsymbol{y}})\psi({\boldsymbol{y}})\int
d{\boldsymbol{x}}\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{E}}({\boldsymbol{x}})\left|{0}\right\rangle=ie\int
d{\boldsymbol{x}}\,\psi^{\dagger}({\boldsymbol{x}})\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}\,\psi({\boldsymbol{x}})\left|{0}\right\rangle=C_{\mathcal{O}}\left|{e^{-}e^{+};\mathcal{O}_{\lambda}}\right\rangle+\ldots$
(231)

With the normalization (168) the overlap $C_{\mathcal{O}}$ is

$\displaystyle
C_{\mathcal{O}}=\frac{\langle{e^{-}e^{+};\mathcal{O}_{\lambda}}|\mathcal{H}\left|{{\boldsymbol{E}},\lambda}\right\rangle}{\langle{e^{-}e^{+};\mathcal{O}_{\lambda}}|e^{-}e^{+};\mathcal{O}_{\lambda}\rangle}=$
$\displaystyle=\frac{1}{4m(2\pi)^{3}\delta^{3}(0)}\langle{0}|\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\psi^{\dagger}({\boldsymbol{x}}_{2})\,{\boldsymbol{e}}_{\lambda}^{*}\cdot{\boldsymbol{\alpha}}\,F^{*}(|{\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}|){\textstyle\frac{1}{2}}(1+\gamma^{0})\gamma^{0}\psi({\boldsymbol{x}}_{1})$
$\displaystyle\times ie\int
d{\boldsymbol{x}}\,\psi^{\dagger}({\boldsymbol{x}})\,{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}\,\psi({\boldsymbol{x}})\left|{0}\right\rangle=\frac{ie}{2m}\,F^{*}(0)$
(232)

The Orthopositronium state may at ${\cal O}\left(\alpha^{4}\right)$ be
considered as a superposition of the three states involved,

$\displaystyle\left|{\mathcal{O}_{\lambda}}\right\rangle=\left|{e^{-}e^{+};\mathcal{O}_{\lambda}}\right\rangle+C_{A}\left|{{\boldsymbol{A}},\lambda}\right\rangle+C_{E}\left|{{\boldsymbol{E}},\lambda}\right\rangle$
(233)

and should be an eigenstate of $\mathcal{H}$ with eigenvalue
$E_{\mathcal{O}}$. Using (229), (230) and (VI.4.2),

$\displaystyle\mathcal{H}\left|{\mathcal{O}_{\lambda}}\right\rangle$
$\displaystyle=(2m-{\textstyle\frac{1}{4}}m\alpha^{2}+C_{\mathcal{O}}C_{E})\left|{e^{-}e^{+};\mathcal{O}_{\lambda}}\right\rangle+C_{A}\,i\left|{{\boldsymbol{E}},\lambda}\right\rangle-2eF(0)\left|{{\boldsymbol{A}},\lambda}\right\rangle$
$\displaystyle=E_{\mathcal{O}}\,\Big{[}\left|{e^{-}e^{+};\mathcal{O}_{\lambda}}\right\rangle-\frac{2eF(0)}{E_{\mathcal{O}}}\left|{{\boldsymbol{A}},\lambda}\right\rangle+\frac{iC_{A}}{E_{\mathcal{O}}}\left|{{\boldsymbol{E}},\lambda}\right\rangle\Big{]}=E_{\mathcal{O}}\left|{\mathcal{O}_{\lambda}}\right\rangle$
(234)

The eigenstate constraint requires (at leading order) $C_{A}=-2eF(0)/2m$ and
$C_{E}=iC_{A}/2m=-2ieF(0)/4m^{2}$. With the value of $F(0)$ in (194) this
gives the hfs term in $E_{\mathcal{O}}$ as

$\displaystyle
C_{\mathcal{O}}C_{E}=\frac{ie}{2m}\,\frac{-2ie}{4m^{2}}|F(0)|^{2}={\textstyle\frac{1}{4}}m\alpha^{4}$
(235)

as quoted in Penin (2014).

###  VI.5 * Electromagnetic form factor of Positronium atoms in an arbitrary
frame

In this section I evaluate the electromagnetic form factors of Positronium.
The elastic form factor is evaluated with leading order wave functions in an
arbitrary frame, demonstrating the Lorentz covariance of the result. The
transition form factor from Para- to Orthopositronium is calculated in the
rest frame only.

The electromagnetic current $j^{\mu}(z)$ may be translated to the origin
($z=0$) using the four-momentum operator $\hat{P}$,

$\displaystyle j^{\mu}(z)=\bar{\psi}(z)\gamma^{\mu}\psi(z)=e^{i\hat{P}\cdot
z}j^{\mu}(0)e^{-i\hat{P}\cdot z}$ (236)

The EM form factor $F_{AB}^{\mu}$ is the expectation value of the current
between atoms $A,B$ of three-momenta
${\boldsymbol{P}_{A}},{\boldsymbol{P}_{B}}$ whose four-momenta satisfy
$P_{A}^{2}=M_{A}^{2}$, $P_{B}^{2}=M_{B}^{2}$,

$\displaystyle F_{AB}^{\mu}(z)$
$\displaystyle=\langle{B,{\boldsymbol{P}_{B}}}|j^{\mu}(z)\left|{A,{\boldsymbol{P}_{A}}}\right\rangle=e^{i(P_{B}-P_{A})\cdot
z}\langle{B,{\boldsymbol{P}_{B}}}|j^{\mu}(0)\left|{A,{\boldsymbol{P}_{A}}}\right\rangle$
$\displaystyle F_{AB}^{\mu}(q)$ $\displaystyle=\int d^{4}z\,e^{-iq\cdot
z}\,F^{\mu}_{AB}(z)=(2\pi)^{4}\delta^{4}(P_{b}-P_{a}-q)G^{\mu}_{AB}(q)$ (237)

In the following I consider $G^{\mu}_{AB}(q)$, keeping in mind the four-
momentum constraint.

The Positronium state is defined in (165). With a short-hand notation
$\Psi_{A}$ the wave function of the incoming state is

$\displaystyle\left|{A,{\boldsymbol{P}}}\right\rangle$ $\displaystyle=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\bar{\psi}({\boldsymbol{x}}_{1})\Psi_{A}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
$\displaystyle\Psi_{A}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2})$
$\displaystyle={\overset{\leftarrow}{\Lambda}}_{+}({\boldsymbol{x}}_{1})e^{i{\boldsymbol{P}}_{A}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2}\Phi^{({\boldsymbol{P}}_{A})}_{A}({\boldsymbol{x}}_{1}-{\boldsymbol{x}}_{2}){\overset{\rightarrow}{\Lambda}}_{-}({\boldsymbol{x}}_{2})$
(238)

where the projectors $\Lambda_{\pm}$ are defined in (VI.1.1). The same
notation for the final state $\langle{B,{\boldsymbol{P}}}|$ gives

$\displaystyle G_{AB}^{\mu}(q)=\int
d{\boldsymbol{y}}_{1}d{\boldsymbol{y}}_{2}\,d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\langle{0}|\psi^{\dagger}({\boldsymbol{y}}_{2})\Psi_{B}^{\dagger}({\boldsymbol{y}}_{1},{\boldsymbol{y}}_{2})\gamma^{0}\psi({\boldsymbol{y}}_{1})\,\bar{\psi}(0)\gamma^{\mu}\psi(0)\,\bar{\psi}({\boldsymbol{x}}_{1})\Psi_{A}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(239)

The contraction of $\psi(0)$ with $\bar{\psi}({\boldsymbol{x}}_{1})$
corresponds to $j^{\mu}$ interacting with the $e^{-}$. This sets
${\boldsymbol{x}}_{1}={\boldsymbol{y}}_{1}=0$ and
${\boldsymbol{y}}_{2}={\boldsymbol{x}}_{2}$. I denote this contribution
$G_{AB}^{\mu}(q,e^{-})$. Interaction with $e^{+}$ corresponds to $\psi(0)$
contracting with $\psi^{\dagger}({\boldsymbol{y}}_{2})$ and is denoted
$G_{AB}^{\mu}(q,e^{+})$. It has a minus sign due to anticommutation and sets
${\boldsymbol{x}}_{2}={\boldsymbol{y}}_{2}=0$ and
${\boldsymbol{y}}_{1}={\boldsymbol{x}}_{1}$. Thus

$\displaystyle G_{AB}^{\mu}(q,e^{-})$ $\displaystyle=\int
d{\boldsymbol{x}}\,\mathrm{Tr}\,\big{\\{}\Psi_{A}(0,-{\boldsymbol{x}})\Psi_{B}^{\dagger}(0,-{\boldsymbol{x}})\gamma^{\mu}\gamma^{0}\big{\\}}$
(240) $\displaystyle G_{AB}^{\mu}(q,e^{+})$ $\displaystyle=-\int
d{\boldsymbol{x}}\,\mathrm{Tr}\,\big{\\{}\Psi_{B}^{\dagger}(-{\boldsymbol{x}},0)\Psi_{A}(-{\boldsymbol{x}},0)\gamma^{0}\gamma^{\mu}\big{\\}}$

The two contributions are related by charge conjugation. Using (181) and
recalling that $\alpha_{2}\gamma^{\mu}\alpha_{2}=-(\gamma^{\mu})^{T}$ and
$\alpha_{2}\,{\boldsymbol{\alpha}}\,\alpha_{2}=-{\boldsymbol{\alpha}}^{T}$ we
have

$\displaystyle{\alpha_{2}}\Psi^{T}({\boldsymbol{x}}_{2},{\boldsymbol{x}}_{1}){\alpha_{2}}=\eta_{C}\,\Psi({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2})$
(241)

Multiplying the argument of the $\mathrm{Tr}\,$ in $G_{AB}^{\mu}(q,e^{-})$ by
${\alpha_{2}}$ from the left and right and taking its transpose shows that

$\displaystyle
G_{AB}^{\mu}(q,e^{+})=-\eta^{A}_{C}\eta^{B}_{C}G_{AB}^{\mu}(q,e^{-})$ (242)

As expected the photon ($\eta_{C}^{\gamma}=-1$) requires
$\eta^{A}_{C}=-\eta^{B}_{C}$ when $A$ and $B$ are eigenstates of charge
conjugation,

$\displaystyle G_{AB}^{\mu}(q)=(1-\eta^{A}_{C}\eta^{B}_{C})\int
d{\boldsymbol{x}}\,\mathrm{Tr}\,\big{\\{}\Psi_{A}(0,-{\boldsymbol{x}})\Psi_{B}^{\dagger}(0,-{\boldsymbol{x}})\gamma^{\mu}\gamma^{0}\big{\\}}$
(243)

####  VI.5.1 Parapositronium form factor

I take both $A$ and $B$ to be Parapositronium and consider only
$G_{AB}^{\mu}(q,e^{-})$. This is relevant for states which are not eigenstates
of charge conjugation, e.g., a hypothetical $\mu^{-}e^{+}$ atom where the muon
and electron have the same mass. Even for standard Positronium
$G_{AB}^{\mu}(q,e^{-})\neq 0$ and should have the form required by Lorentz
covariance,

$\displaystyle G^{\mu}(q,e^{-})=(P_{A}+P_{B})^{\mu}F(q^{2})$ (244)

After partial integrations in the state (VI.5) we need consider only the
projector derivatives acting on the ${\cal O}\left(\alpha^{0}\right)$ phase
$\exp[i{\boldsymbol{P}}\cdot({\boldsymbol{x}}_{1}+{\boldsymbol{x}}_{2})/2]$,
since $\boldsymbol{\nabla}\Phi^{(P)}_{A,B}({\boldsymbol{x}})$ is of ${\cal
O}\left(\alpha\right)$. The projectors then become,

$\displaystyle\Lambda_{\pm}({\boldsymbol{P}})\equiv\frac{1}{2E_{P}}\big{(}E_{P}\mp{\boldsymbol{\alpha}}\cdot{\boldsymbol{P}}\pm
M\gamma^{0}\big{)}=\Lambda_{\pm}^{\dagger}({\boldsymbol{P}})=\big{[}\Lambda_{\pm}({\boldsymbol{P}})\big{]}^{2}\hskip
56.9055pt\Lambda_{+}({\boldsymbol{P}})\Lambda_{-}({\boldsymbol{P}})=0$ (245)

where $E_{P}=\sqrt{{\boldsymbol{P}}^{2}+M^{2}}$ and $M=2m$ at leading order.

The Parapositronium states are relativistically normalized (168), with the
Lorentz contracted wave function $\Phi^{(P)}({\boldsymbol{x}})$ given in
(183), (184), (186) and (216). Taking
${\boldsymbol{P}}=(0,0,P)=(0,0,M\sinh\xi)$ along the $z$-axis,

$\displaystyle\Phi^{(P)}({\boldsymbol{x}})=\frac{E_{P}}{M}\,\gamma_{5}F^{({\boldsymbol{P}})}({\boldsymbol{x}})\hskip
56.9055ptF^{({\boldsymbol{P}})}({\boldsymbol{x}})$
$\displaystyle=\frac{E_{P}}{M}\,F^{(0)}(r_{P})=\frac{E_{P}}{M}\,\frac{\alpha^{3/2}m^{2}}{\sqrt{4\pi}}\,e^{-\alpha
mr_{P}/2}$ (246) $\displaystyle r_{P}$
$\displaystyle\equiv\sqrt{x^{2}+y^{2}+z^{2}\cosh^{2}\xi}$ (247)

The photon momentum ${\boldsymbol{q}}$ must be of ${\cal
O}\left(\alpha\right)$ for a leading order overlap between $\Psi_{A}$ and
$\Psi_{B}$. Hence we may set
${\boldsymbol{P}}_{B}={\boldsymbol{P}}_{A}+{\boldsymbol{q}}={\boldsymbol{P}}+{\cal
O}\left(\alpha\right)$ in $G_{AB}^{\mu}(q,e^{-})$ (240). However, we need to
retain the ${\boldsymbol{q}}$-dependence of the
$\Psi_{A}(0,-{\boldsymbol{x}})\Psi_{B}^{\dagger}(0,-{\boldsymbol{x}})$ phase
factor
$\exp[i({\boldsymbol{P}}_{B}-{\boldsymbol{P}}_{A})\cdot{\boldsymbol{x}}/2]=\exp(i{\boldsymbol{q}}\cdot{\boldsymbol{x}}/2)$,
which reflects the photon wave function. The expression for
$G_{AB}^{\mu}(q,e^{-})$ is then

$\displaystyle G^{\mu}(q,e^{-})$
$\displaystyle=\Big{(}\frac{E_{P}}{M}\Big{)}^{2}\int
d{\boldsymbol{x}}\,|F^{({\boldsymbol{P}})}({\boldsymbol{x}})|^{2}e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}/2}\mathrm{Tr}\,\big{\\{}\Lambda_{+}({\boldsymbol{P}})\gamma_{5}\Lambda_{-}({\boldsymbol{P}})\Lambda_{-}({\boldsymbol{P}})\gamma_{5}\Lambda_{+}({\boldsymbol{P}})\gamma^{\mu}\gamma^{0}\big{\\}}$
(248)

From the definitions (245) of $\Lambda_{\pm}({\boldsymbol{P}})$ follows that

$\displaystyle\Lambda_{+}({\boldsymbol{P}})\gamma_{5}\Lambda_{-}({\boldsymbol{P}})=\frac{M}{E_{P}}\,\Lambda_{+}({\boldsymbol{P}})\gamma^{0}\gamma_{5}$
(249)

Using this the trace in (248) becomes

$\displaystyle\mathrm{Tr}\,^{\mu}=\Big{(}\frac{M}{E_{P}}\Big{)}^{2}\mathrm{Tr}\,\big{\\{}\Lambda_{+}({\boldsymbol{P}})\gamma^{\mu}\gamma^{0}\big{\\}}=\Big{(}\frac{M}{E_{P}}\Big{)}^{2}\,\frac{2P^{\mu}}{E_{P}}$
(250)

so that

$\displaystyle G^{\mu}(q,e^{-})$ $\displaystyle=\frac{2P^{\mu}}{E_{P}}\,\int
d{\boldsymbol{x}}\,|F^{({\boldsymbol{P}})}({\boldsymbol{x}})|^{2}e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}/2}$
(251)

Changing the integration variable to
${\boldsymbol{x}}_{R}\equiv(x,y,z\cosh\xi)$ gives
$d{\boldsymbol{x}}=d{\boldsymbol{x}}_{R}/\cosh\xi$ and
$r_{P}=|{\boldsymbol{x}}_{R}|$ in (246). The photon four-momentum $q$ is
constrained by kinematics,

$\displaystyle M^{2}=(P+q)^{2}=M^{2}+2P\cdot q+{\cal
O}\left(\alpha^{2}\right)=M^{2}$ (252)

which at ${\cal O}\left(\alpha\right)$ implies $P\cdot
q=E_{P}q^{0}-{\boldsymbol{P}}\cdot{\boldsymbol{q}}=0$. In the Positronium rest
frame (${\boldsymbol{P}}=0$) this means $q^{0}_{R}=0$. Thus
$q^{z}=q^{z}_{R}\cosh\xi$, where $q^{z}_{R}$ is the $z$-component of the
photon momentum in the rest frame. Hence
${\boldsymbol{q}}\cdot{\boldsymbol{x}}={\boldsymbol{q}}_{R}\cdot{\boldsymbol{x}}_{R}$.
Recalling that $\cosh\xi=E_{P}/M$ and using the expression for
$F^{({\boldsymbol{P}})}({\boldsymbol{x}})$ in (246) gives

$\displaystyle G^{\mu}(q,e^{-})$
$\displaystyle=\frac{2P^{\mu}}{E_{P}}\,\frac{E_{P}}{M}\,\frac{\alpha^{3}m^{4}}{4\pi}\int
d{\boldsymbol{x}}_{R}\,\exp\big{[}(-\alpha
M|{\boldsymbol{x}}_{R}|+i{\boldsymbol{q}}_{R}\cdot{\boldsymbol{x}}_{R})/2\big{]}$
(253)

The integral is as in the rest frame, i.e., it is
${\boldsymbol{P}}$-independent, so the result is covariant and agrees with
(244),

$\displaystyle G^{\mu}(q,e^{-})$ $\displaystyle=2P^{\mu}\,\frac{(\alpha
M)^{4}}{(Q^{2}+\alpha^{2}M^{2})^{2}}$ (254)

where $2P^{\mu}=(P_{A}+P_{B})^{\mu}+{\cal O}\left(\alpha\right)$ and
$Q^{2}={\boldsymbol{q}}_{R}^{2}=-q^{2}$.

####  VI.5.2 Positronium transition form factor

The $\gamma^{*}(q)$ \+ Parapositronium $\to$ Orthopositronium transition
electromagnetic form factor has the structure

$\displaystyle
G_{\lambda}^{\mu}(q)=i\varepsilon^{\mu\nu\rho\sigma}P_{\nu}q_{\rho}e_{\sigma}^{\lambda}F(q^{2})$
(255)

where $P$ is the four-momentum of one of the Positronia, $e^{\lambda}$ is the
polarization vector (185) of the Orthopositronium (with
$e^{\lambda}_{\sigma=0}=0$) and $q$ is the photon momentum. Symmetries and
gauge invariance force the kinematic factor to be of ${\cal O}\left(q\right)$,
i.e., of ${\cal O}\left(\alpha\right)$. This reflects the spin flip, from
$S=0$ for Parapositronium to $S=1$ for Orthopositronium.

In section VI.3 I demonstrated that transverse photon exchange contributes to
the binding of Positronium at leading order for ${\boldsymbol{P}}\neq 0$. The
photon exchange may at ${\cal O}\left(q\right)$ involve a spin flip at one of
its vertices, whereupon the transition to Orthopositronium proceeds at ${\cal
O}\left(\alpha^{0}\right)$. I shall not here work out the ${\cal
O}\left(\alpha\right)$ corrections to the wave function of Positronium in
motion, but limit myself to evaluating the transition form factor in the rest
frame of the target, ${\boldsymbol{P}}_{A}=0$.

Using the expressions (245) for the $\Lambda_{\pm}({\boldsymbol{P}})$
projectors (with ${\boldsymbol{P}}_{A}=0,\
{\boldsymbol{P}}_{B}={\boldsymbol{q}},\ E_{B}\simeq M$) the Positronium wave
functions given in (183), (184), (185) and (186) are,

$\displaystyle\Psi_{A}(0,-{\boldsymbol{x}})$
$\displaystyle={\textstyle\frac{1}{2}}(1+\gamma^{0})\gamma_{5}\,F(r)$
$\displaystyle\Psi_{B}(0,-{\boldsymbol{x}})$
$\displaystyle=N_{\lambda}\Lambda_{+}({\boldsymbol{q}}){\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}\Lambda_{-}({\boldsymbol{q}})F(r)e^{-i{\boldsymbol{q}}\cdot{\boldsymbol{x}}/2}$
(256)

The expression for $\Psi_{B}(0,-{\boldsymbol{x}})$ may be simplified using
$\Lambda_{+}({\boldsymbol{q}})\Lambda_{-}({\boldsymbol{q}})=0$,

$\displaystyle\Lambda_{+}({\boldsymbol{q}}){\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}\,\Lambda_{-}({\boldsymbol{q}})=\left\\{{\Lambda_{+}({\boldsymbol{q}})},{{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}}\right\\}\Lambda_{-}({\boldsymbol{q}})=\big{(}{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{\alpha}}-\frac{1}{M}{\boldsymbol{e}}_{\lambda}\cdot{\boldsymbol{q}}\big{)}\Lambda_{-}({\boldsymbol{q}})$
(257)

The form factor (243) is then in the target rest frame,

$\displaystyle G_{\lambda}^{\mu}(q)$
$\displaystyle=\frac{{N_{\lambda}}^{*}}{2M}\int
d{\boldsymbol{x}}\,|F(r)|^{2}e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}/2}\mathrm{Tr}\,_{\lambda}^{\mu}$
$\displaystyle\mathrm{Tr}\,_{\lambda}^{\mu}$
$\displaystyle=\mathrm{Tr}\,\big{\\{}(1+\gamma^{0})\gamma_{5}(M+{\boldsymbol{\alpha}}\cdot{\boldsymbol{q}}-M\gamma^{0})\big{(}{\boldsymbol{e}}_{\lambda}^{*}\cdot{\boldsymbol{\alpha}}-\frac{1}{M}{\boldsymbol{e}}_{\lambda}^{*}\cdot{\boldsymbol{q}}\big{)}\gamma^{\mu}\gamma^{0}\big{\\}}=\mathrm{Tr}\,\big{\\{}\gamma_{5}\,{\boldsymbol{\alpha}}\cdot{\boldsymbol{q}}\,{\boldsymbol{e}}_{\lambda}^{*}\cdot{\boldsymbol{\alpha}}\,\gamma^{\mu}\gamma^{0}\big{\\}}$
$\displaystyle=-4i\,\delta^{\mu,i}\varepsilon_{ijk}q^{j}{e_{\lambda}^{k}}^{*}$
(258)

This agrees with the kinematic factor in (255) for ${\boldsymbol{P}}=0$. The
Orthopositronium is transversely polarized because
${\boldsymbol{e}}_{\lambda=0}\parallel{\boldsymbol{P}}={\boldsymbol{q}}$ gives
$\mathrm{Tr}\,_{0}^{\mu}=0$. The normalization $N_{\lambda=\pm 1}=1$ (186).
The invariant form factor has the same integral as in (253),

$\displaystyle F(q^{2})=\frac{2}{M^{2}}\int
d{\boldsymbol{x}}\,|F(r)|^{2}e^{i{\boldsymbol{q}}\cdot{\boldsymbol{x}}/2}=\frac{2}{M}\,\frac{(\alpha
M)^{4}}{(Q^{2}+\alpha^{2}M^{2})^{2}}$ (259)

where $Q^{2}=-q^{2}$.

I leave it as an exercise (without a worked-out solution) to show that the
transition form factor agrees with (255) in a general frame.

###  VI.6 * Deep inelastic scattering on Parapositronium in a general frame

The target $A$ vertex of Deep Inelastic Scattering (DIS),
$\gamma^{*}(q)+A(P_{A})\to X$, is as in a transition form factor (VI.5) for
each final state $X$, with the Parapositronium state $A$ defined in (VI.5).
Now the photon is taken to have an asymptotically large momentum $q$, and the
squared vertex is summed over the final states $X$. In the absence of
radiative effects we may describe $X$ in the basis of free $e^{-}e^{+}$
states,

$\displaystyle\left|{X}\right\rangle=b^{\dagger}_{{\boldsymbol{k}}_{1},\lambda_{1}}\,d^{\dagger}_{{\boldsymbol{k}}_{2},\lambda_{2}}\left|{0}\right\rangle$
(260)

constrained by momentum conservation, $q+P_{A}=k_{1}+k_{2}$. In the Bjorken
limit

$\displaystyle{x_{Bj}}=\frac{Q^{2}}{2P_{A}\cdot q}$ (261)

is fixed as $q\to\infty$, with $Q^{2}=-q^{2}>0$. The frame is defined by
keeping the target 4-momentum $P_{A}^{\mu}$ fixed in the Bj limit, with the
three-momenta of $q^{\mu}=(q^{0},0,0,-|{\boldsymbol{q}}|)$ and
$P_{A}^{\mu}=(E_{A},0,0,P_{A})$ along the $z$-axis. The target mass
$M=2m+{\cal O}\left(\alpha^{2}\right)$.

The amplitude for $\gamma^{*}A\to X$ corresponding to (239) is, suppressing
the momentum conserving $\delta$-function as in (VI.5),

$\displaystyle G^{\mu}_{AX}(q)=\int
d{\boldsymbol{x}}_{1}d{\boldsymbol{x}}_{2}\,\langle{0}|d_{{\boldsymbol{k}}_{2},\lambda_{2}}b_{{\boldsymbol{k}}_{1},\lambda_{1}}\,\bar{\psi}(0)\gamma^{\mu}\psi(0)\bar{\psi}({\boldsymbol{x}}_{1})\Psi_{A}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2})\psi({\boldsymbol{x}}_{2})\left|{0}\right\rangle$
(262)

I consider only scattering from the electron, hence the contractions

$\displaystyle\left\\{{b_{{\boldsymbol{k}}_{1},\lambda_{1}}},{\bar{\psi}(0)}\right\\}$
$\displaystyle=\bar{u}({\boldsymbol{k}}_{1},\lambda_{1})$
$\displaystyle\left\\{{d_{{\boldsymbol{k}}_{2},\lambda_{2}}},{\psi({\boldsymbol{x}}_{2})}\right\\}$
$\displaystyle=e^{-i{\boldsymbol{k}}_{2}\cdot{\boldsymbol{x}}_{2}}v({\boldsymbol{k}}_{2},\lambda_{2})$
(263)

resulting in

$\displaystyle G^{\mu}_{AX}(q)$ $\displaystyle=\int
d{\boldsymbol{x}}_{2}\,\bar{u}({\boldsymbol{k}}_{1},\lambda_{1})\gamma^{\mu}\gamma^{0}\Psi_{A}(0,{\boldsymbol{x}}_{2})v({\boldsymbol{k}}_{2},\lambda_{2})e^{-i{\boldsymbol{k}}_{2}\cdot{\boldsymbol{x}}_{2}}$
(264)

According to (167) the $\Lambda_{-}$ projector in (VI.5) reduces to unity when
acting on
$e^{-i{\boldsymbol{k}}_{2}\cdot{\boldsymbol{x}}_{2}}v({\boldsymbol{k}}_{2},\lambda_{2})$.
After a partial integration of the $\gamma^{0}\Lambda_{+}$ projector in the
target state (VI.5) it acts on
$\Psi_{A}({\boldsymbol{x}}_{1},{\boldsymbol{x}}_{2})$ and gives
$(\not{P}_{A}+M)$ when the ${\cal O}\left(\alpha\right)$ contribution from
differentiating the radial function is neglected. The radial function $F(r)$
is as in (246) evaluated with Lorentz contracted argument. Thus

$\displaystyle\gamma^{0}\Psi_{A}(0,{\boldsymbol{x}}_{2})$
$\displaystyle=\frac{E_{A}}{2M^{2}}(\not{P}_{A}+M)e^{i{\boldsymbol{P}_{A}}\cdot{\boldsymbol{x}}_{2}/2}\gamma_{5}\,F(|{\boldsymbol{x}}_{2P}|)$
$\displaystyle{\boldsymbol{x}}_{2P}^{\perp}$
$\displaystyle={\boldsymbol{x}}_{2}^{\perp}\hskip
56.9055ptz_{2P}=z_{2}\cosh\xi=z_{2}\,E_{A}/M$ (265)

The integral over ${\boldsymbol{x}}_{2}$ in $G^{\mu}_{AX}(q)$ can be done by a
change of integration variable,
${\boldsymbol{x}}_{2}\to{\boldsymbol{x}}_{2P}$, giving

$\displaystyle G^{\mu}_{AX}(q)$
$\displaystyle=\frac{\sqrt{\pi}\,\alpha^{5/2}M^{2}}{8\big{(}\alpha^{2}M^{2}/16+{\boldsymbol{P}}_{2}^{2}\big{)}^{2}}\,\bar{u}({\boldsymbol{k}}_{1},\lambda_{1})\gamma^{\mu}(\not{P}_{A}+M)\gamma_{5}\,v({\boldsymbol{k}}_{2},\lambda_{2})$
$\displaystyle{\boldsymbol{P}}_{2}$
$\displaystyle\equiv\big{(}-{\boldsymbol{k}}_{2}^{\perp},({\textstyle\frac{1}{2}}P_{A}^{z}-k_{2}^{z})/\cosh\xi\big{)}$
(266)

The denominator of $G^{\mu}_{AX}(q)$ shows that ${\boldsymbol{P}}_{2}$, i.e.,
${\boldsymbol{k}}_{2}^{\perp}$ and
${\textstyle\frac{1}{2}}P_{A}^{z}-k_{2}^{z}$ must be of ${\cal
O}\left(\alpha\right)$ for a leading order contribution.

Squaring the form factor and summing over the helicities
$\lambda_{1},\lambda_{2}$,

$\displaystyle\sum_{\lambda_{1},\lambda_{2}}G^{\mu}_{AX}(q){G^{\nu}_{AX}}^{\dagger}(q)$
$\displaystyle=\frac{\pi\alpha^{5}M^{4}}{64\big{(}\alpha^{2}M^{2}/16+{\boldsymbol{P}}_{2}^{2}\big{)}^{4}}\,\mathrm{Tr}\,^{\mu\nu}$
$\displaystyle\mathrm{Tr}\,^{\mu\nu}$
$\displaystyle=\mathrm{Tr}\,\big{\\{}\gamma^{\mu}(\not{P}_{A}+M)(\not{k}_{2}+m)(\not{P}_{A}+M)\gamma^{\nu}(\not{k}_{1}+m)\big{\\}}$
$\displaystyle=2M^{2}\mathrm{Tr}\,\big{\\{}\gamma^{\mu}(\not{P}_{A}+M)\gamma^{\nu}(\not{k}_{1}+m)\big{\\}}=8M^{2}\big{[}P_{A}^{\mu}k_{1}^{\nu}+P_{A}^{\nu}k_{1}^{\mu}-g^{\mu\nu}(P_{A}\cdot
k_{1}-{\textstyle\frac{1}{2}}M^{2})\big{\\}}$ (267)

where I used $k_{2}={\textstyle\frac{1}{2}}P_{A}+{\cal O}\left(\alpha\right)$.
We may check gauge invariance at leading order,

$\displaystyle q_{\mu}\mathrm{Tr}\,^{\mu\nu}$
$\displaystyle=2M^{2}\mathrm{Tr}\,\big{\\{}(\not{k}_{1}+\not{k}_{2}-\not{P}_{A})(\not{P}_{A}+M)\gamma^{\nu}(\not{k_{1}}+m)\big{\\}}=2M^{2}\mathrm{Tr}\,\big{\\{}(m-{\textstyle\frac{1}{2}}\not{P}_{A})(\not{P}_{A}+M)\gamma^{\nu}(\not{k_{1}}+m)\big{\\}}=0$
(268)

In the notation Peskin and Schroeder (1995)

$\displaystyle\textrm{Im}\,W^{\mu\nu}(P_{a},q)$
$\displaystyle=\frac{1}{2}\int\frac{d{\boldsymbol{k}}_{1}d{\boldsymbol{k}}_{2}}{(2\pi)^{6}4E_{1}E_{2}}\,(2\pi)^{4}\delta^{4}(q+P_{a}-k_{1}-k_{2})G^{\mu}(q){G^{\nu}}^{\dagger}(q)$
$\displaystyle W^{\mu\nu}$
$\displaystyle=\Big{(}-g^{\mu\nu}+\frac{q^{\mu}q^{\nu}}{q^{2}}\Big{)}W_{1}+\Big{(}P_{a}^{\mu}-q^{\mu}\frac{P_{a}\cdot
q}{q^{2}}\Big{)}\Big{(}P_{a}^{\nu}-q^{\nu}\frac{P_{a}\cdot
q}{q^{2}}\Big{)}W_{2}$ (269)

Identifying $W_{1}$ from the coefficient of $-g^{\mu\nu}$ in (VI.6)
($P_{A}\cdot k_{1}\gg{\textstyle\frac{1}{2}}M^{2}$) we get for the electron
distribution,

$\displaystyle f_{e/A}({x_{Bj}})$
$\displaystyle=\frac{1}{\pi}\textrm{Im}\,W_{1}=\int\frac{d{\boldsymbol{k}}_{2}}{(2\pi)^{2}4E_{1}E_{2}}\delta(q^{0}+E_{A}-E_{1}-E_{2})\frac{\alpha^{5}M^{6}\,P_{A}\cdot
k_{1}}{16\big{(}\alpha^{2}M^{2}/16+{\boldsymbol{P}}_{2}^{2}\big{)}^{4}}$ (270)

As $q=(q^{0},0,0,-|{\boldsymbol{q}}|)\to\infty$ at fixed ${x_{Bj}}$ and
$P_{A}$,

$\displaystyle
E_{1}=\sqrt{({\boldsymbol{q}}+{\boldsymbol{P}}_{A}-{\boldsymbol{k}}_{2})^{2}+m^{2}}\simeq\sqrt{{\boldsymbol{q}}^{2}-2|{\boldsymbol{q}}|(P_{A}^{z}-k_{2}^{z})}\simeq|{\boldsymbol{q}}|-(P_{A}-k_{2})^{z}$
(271)

Defining the light-front notation by $q^{\pm}\equiv q^{0}\pm q^{z}$ we have,

$\displaystyle Q^{2}=-q^{+}q^{-}=2{x_{Bj}}P_{A}\cdot
q\simeq{x_{Bj}}P_{A}^{+}q^{-}\ \ \ \ \Longrightarrow\ \
q^{+}=-{x_{Bj}}P_{A}^{+}$ (272)

The energy constraint in (270) becomes

$\displaystyle q^{0}+E_{A}-E_{1}-E_{2}$
$\displaystyle=q^{0}-|{\boldsymbol{q}}|+(P_{A}-k_{2})^{z}+E_{A}-E_{2}=q^{+}+P_{A}^{+}-k_{2}^{+}=P_{A}^{+}(1-{x_{Bj}})-k_{2}^{+}=0$
(273)

Recalling that $k_{2}={\textstyle\frac{1}{2}}P_{A}$ at ${\cal
O}\left(\alpha^{0}\right)$ I denote the ${\cal O}\left(\alpha\right)$
difference ${x_{Bj}}-{\textstyle\frac{1}{2}}$ as

$\displaystyle{x_{Bj}}={\textstyle\frac{1}{2}}(1+\alpha\,\tilde{x}_{B})\hskip
56.9055ptk_{2}^{+}={\textstyle\frac{1}{2}}P_{A}^{+}(1-\alpha\tilde{x}_{B})=m\,e^{\xi}(1-\alpha\tilde{x}_{B})$
(274)

Neglecting terms of ${\cal O}\left(\alpha^{2}\right)$ we have in
${\boldsymbol{P}}_{2}$ (VI.6),

$\displaystyle
k_{2}^{z}-{\textstyle\frac{1}{2}}P_{a}^{z}={\textstyle\frac{1}{2}}(k_{2}^{+}-k_{2}^{-})-{\textstyle\frac{1}{2}}P_{a}^{z}={\textstyle\frac{1}{2}}me^{\xi}(1-\alpha\tilde{x}_{B})-\frac{m^{2}e^{-\xi}}{2m(1-\alpha\tilde{x}_{B})}-m\sinh\xi=-m\alpha\tilde{x}_{B}\cosh\xi$
(275)

With the change of variables $dk_{2}^{z}/E_{2}=dk_{2}^{+}/k_{2}^{+}$ the
$\delta$-function in (270) may be integrated using (273). Substituting
$P_{A}\cdot k_{1}/E_{1}k_{2}^{+}\simeq 2$ and the result (275) we have the
frame independent result

$\displaystyle f_{e/A}({x_{Bj}})$
$\displaystyle=\frac{\alpha^{5}M^{6}}{32}\int\frac{d{\boldsymbol{k}}_{2}^{\perp}}{(2\pi)^{2}}\frac{1}{\big{(}{{\boldsymbol{k}}_{2}^{\perp}}^{2}+\alpha^{2}M^{2}/16+\alpha^{2}M^{2}{\tilde{x}_{B}}^{2}/4\big{]}^{4}}=\frac{1}{6\pi\,\alpha}\,\frac{1}{({\tilde{x}_{B}}^{2}+{\textstyle\frac{1}{4}})^{3}}\hskip
56.9055pt\tilde{x}_{B}=\frac{2}{\alpha}({x_{Bj}}-{\textstyle\frac{1}{2}})$
(276)

We may check that there is a single $e^{-}$ in the bound state,

$\displaystyle\int_{0}^{1}d{x_{Bj}}\,f_{e/A}({x_{Bj}})={\textstyle\frac{1}{2}}\alpha\int_{-\infty}^{\infty}\frac{d\tilde{x}_{B}}{6\pi\alpha({\tilde{x}_{B}}^{2}+{\textstyle\frac{1}{4}})^{3}}=1$
(277)

##  VII QED in $D=1+1$ dimensions

In this section I apply the perturbative bound state method described in
chapter V to QED in $D=1+1$ dimensions (QED2), also known as the “Massive
Schwinger model” Schwinger (1962); Coleman  _et al._ (1975); Coleman (1976).
QED (and QCD) in two dimensions is often considered as a model for
confinement, since the Coulomb potential is linear. The coupling $e$ has
dimension of mass, so the dimensionless parameter relevant for dynamics is its
ratio $e/m$ to the electron mass. For $e/m\ll 1$ the fermions are weakly bound
and their wave function satifies the Schrödinger equation. For $e/m\gg 1$ the
spectrum is that of weakly interacting bosons: The strong coupling locks the
fermion degrees of freedom into compact neutral bound states. In the limit of
the massless Schwinger model ($e/m\to\infty$) QED2 reduces to a free theory,
with only a pointlike, non-interacting massive ($M=e/\sqrt{\pi}$) boson field.

A perturbative approach requires the coupling to be small, $e/m<1$. Highly
excited states are nevertheless strongly bound due to the linear potential.
Several features are similar to those of the Dirac equation in a linear
potential discussed in section IV.6. There are also important differences,
first of all because translation invariance allows to define the bound state
momentum. The peculiar feature of a constant (local) norm for wave functions
at large separations of the charges occurs here as well, but now it does not
imply a continuous spectrum. Highly excited states have features of duality
similar to those observed for hadrons.

In section VI we saw that transverse photon exchange contributes to the
binding of Positronium atoms even at leading order for non-vanishing atomic
momentum $P$. In $D=1+1$ there are no transverse photons. Boost covariance is
realized differently, and requires a linear potential. I shall verifty that
form factors and deep inelastic scattering are frame independent.

###  VII.1 QED2 bound states in $A^{0}=0$ gauge

####  VII.1.1 Temporal gauge in $D=1+1$

Quantization in temporal gauge proceeds as in section V.2.3, adapted to
$D=1+1$. The QED2 action is

$\displaystyle\mathcal{S}=\int
d^{2}x\big{[}-{\textstyle\frac{1}{2}}F_{10}F^{10}+\bar{\psi}(\not{\partial}-m-e{\not{A}})\psi\big{]}$
(278)

The electric field $E^{1}=F^{10}=-\partial_{0}A^{1}$ is conjugate to the
photon field $A_{1}$, hence

$\displaystyle\left[{E^{1}(t,x)},{A^{1}(t,y)}\right]=i\delta(x-y)$ (279)

The Hamiltonian is

$\displaystyle\mathcal{H}$ $\displaystyle=\int
dx\big{[}E^{1}\partial_{0}A_{1}+i\psi^{\dagger}\partial_{0}\psi-\mathcal{L}\big{]}=\int
dx\big{[}{\textstyle\frac{1}{2}}(E^{1})^{2}+\psi^{\dagger}(-i\alpha^{1}\partial_{1}+m\gamma^{0}-e\alpha^{1}A^{1})\psi\big{]}\equiv\mathcal{H}_{V}+\mathcal{H}_{0}+\mathcal{H}_{int}$
(280) $\displaystyle\mathcal{H}_{V}(t)$ $\displaystyle=\int
dx\,{\textstyle\frac{1}{2}}\big{[}E^{1}(t,x)\big{]}^{2}\hskip
28.45274pt\mathcal{H}_{0}(t)=\int
dx\big{[}\bar{\psi}(-i\alpha^{1}\partial_{1}+m\gamma^{0}\psi)\big{]}\hskip
28.45274pt\mathcal{H}_{int}(t)=-e\int
dx\big{[}\psi^{\dagger}\,\alpha^{1}A^{1}(t,x)\psi\big{]}$

Gauss’ operator is

$\displaystyle
G(t,x)\equiv\frac{\delta\mathcal{S}}{\delta{A^{0}(t,x)}}=\partial_{1}E^{1}(t,x)-e\psi^{\dagger}\psi(t,x)$
(281)

$G(t,x)=0$ (Gauss’ law) is imposed as a constraint on physical states, fixing
the remaining gauge degrees of freedom and defining the value of $E^{1}$ for
those states,

$\displaystyle G(t,x)\left|{phys}\right\rangle$
$\displaystyle=\big{[}\partial_{1}E^{1}(x)-e\psi^{\dagger}\psi(x)\big{]}\left|{phys}\right\rangle=0$
(282)

Solving for $E^{1}$, using $\partial_{x}^{2}|x-y|=2\delta(x-y)$,

$\displaystyle E^{1}(t,x)\left|{phys}\right\rangle$
$\displaystyle=\partial_{x}\int
dy\,{\textstyle\frac{1}{2}}e|x-y|\psi^{\dagger}\psi(t,y)\left|{phys}\right\rangle$
(283)

The vacuum $\left|{0}\right\rangle$ is a physical state with locally vanishing
charge distribution,

$\displaystyle E^{1}(t,x)\left|{0}\right\rangle=0$ (284)

The $\mathcal{H}_{V}$ part of the Hamiltonian (280) generates an instantaneous
linear potential,

$\displaystyle\mathcal{H}_{V}\left|{phys}\right\rangle$
$\displaystyle=\frac{1}{2}\int
dx\,\big{[}E^{1}(x)\big{]}^{2}\left|{phys}\right\rangle=\frac{e^{2}}{8}\int
dxdydz\big{[}\partial_{x}|x-y|\psi^{\dagger}\psi(y)\big{]}\big{[}\partial_{x}|x-z|\psi^{\dagger}\psi(z)\big{]}\left|{phys}\right\rangle$
$\displaystyle=-\frac{e^{2}}{4}\int
dxdy\,\psi^{\dagger}\psi(x)|x-y|\psi^{\dagger}\psi(y)\left|{phys}\right\rangle$
(285)

####  VII.1.2 States and wave functions of QED2

An $e^{-}e^{+}$ valence Fock state with CM momentum $P$ is defined analogously
to Positronium (165),

$\displaystyle\left|{M,P}\right\rangle=\int
dx_{1}dx_{2}\,\bar{\psi}(x_{1})e^{iP(x_{1}+x_{2})/2}\Phi^{(P)}(x_{1}-x_{2})\psi(x_{2})\left|{0}\right\rangle$
(286)

When bound by $\mathcal{H}_{V}$ (VII.1.1) this is taken as the lowest order
contribution of a bound state expansion, where higher orders are
perturbatively generated by $\mathcal{H}_{int}$. Hence the lowest order wave
functions $\Phi^{(P)}(x_{1}-x_{2})$ and energy eigenvalues $E(P)$ are
determined by the eigenstate condition

$\displaystyle(\mathcal{H}_{0}+\mathcal{H}_{V})\left|{M,P}\right\rangle=E(P)\left|{M,P}\right\rangle$
(287)

Each order of the perturbative expansion should be Poincaré covariant, which
implies $E(P)=\sqrt{P^{2}+M^{2}}$. This will be seen to be satisfied, and the
$P$-dependence of the wave function $\Phi^{(P)}(x)$ determined. I do not here
consider the higher order corrections defined by $\mathcal{H}_{int}$ (280).

At large values of the linear potential generated by $\mathcal{H}_{V}$ the
state $\left|{M,P}\right\rangle$ has contributions from virtual $e^{\pm}$
pairs, as in (IV.3) and (65) for the Dirac case. In terms of Feynman diagrams
these effects are due to $Z$-diagrams (Fig. 13(b)), and they give rise to
negative energy components of the wave function. The virtual pairs are
implicitly included by omitting from (286) the energy projectors
$\Lambda_{\pm}$ (VI.1.1) used for Positronium in (165).

Applying the free Hamiltonian to the state (286),

$\displaystyle\mathcal{H}_{0}\left|{M,P}\right\rangle=\int
dx_{1}dx_{2}\big{[}$
$\displaystyle\bar{\psi}(x_{1})(-i\alpha^{1}{\buildrel\leftarrow\over{\partial}}_{1}+m\gamma^{0})e^{iP(x_{1}+x_{2})/2}\Phi^{(P)}(x_{1}-x_{2})\psi(x_{2})$
$\displaystyle-\bar{\psi}(x_{1})e^{iP(x_{1}+x_{2})/2}\Phi^{(P)}(x_{1}-x_{2})(-i\alpha^{1}{\buildrel\rightarrow\over{\partial}}_{2}+m\gamma^{0})\psi(x_{2})\big{]}\left|{0}\right\rangle$
(288)

Partially integrating the derivatives, so that they act on the wave function
instead of on the electron fields,

$\displaystyle\mathcal{H}_{0}\left|{M,P}\right\rangle=\int
dx_{1}dx_{2}\big{[}$
$\displaystyle\bar{\psi}(x_{1})e^{iP(x_{1}+x_{2})/2}(i\alpha^{1}{\buildrel\rightarrow\over{\partial}}_{1}-{\textstyle\frac{1}{2}}\alpha^{1}P+m\gamma^{0})\Phi^{(P)}(x_{1}-x_{2})\psi(x_{2})$
$\displaystyle+\bar{\psi}(x_{1})\Phi^{(P)}(x_{1}-x_{2})(-i\alpha^{1}{\buildrel\leftarrow\over{\partial}}_{2}+{\textstyle\frac{1}{2}}\alpha^{1}P-m\gamma^{0})e^{iP(x_{1}+x_{2})/2}\psi(x_{2})\big{]}\left|{0}\right\rangle$
(289)

The instantaneous potential generated by $\mathcal{H}_{V}$ (VII.1.1) is seen
from

$\displaystyle\mathcal{H}_{V}\left|{M,P}\right\rangle=\int
dx_{1}dx_{2}\,\bar{\psi}(x_{1})e^{iP(x_{1}+x_{2})/2}{\textstyle\frac{1}{2}}e^{2}|x_{1}-x_{2}|\Phi^{(P)}(x_{1}-x_{2})\psi(x_{2})\left|{0}\right\rangle$
(290)

In $D=1+1$ the Dirac matrices can be represented by the $2\times 2$ Pauli
matrices. I shall use

$\displaystyle\gamma^{0}=\sigma_{3}\hskip
56.9055pt\gamma^{1}=i\sigma_{2}\hskip
56.9055pt\alpha^{1}=\alpha_{1}=\gamma^{0}\gamma^{1}=\sigma_{1}$ (291)

With this notation the bound state condition (287) implies for the wave
function

$\displaystyle
i\partial_{x}\big{\\{}{\sigma_{1}},{\Phi^{(P)}(x)}\big{\\}}-{\textstyle\frac{1}{2}}P\big{[}{\sigma_{1}},{\Phi^{(P)}(x)}\big{]}+m\big{[}{\sigma_{3}},{\Phi^{(P)}(x)}\big{]}$
$\displaystyle=\big{[}E-V(x)\big{]}\Phi^{(P)}(x)$ $\displaystyle
V(x)={\textstyle\frac{1}{2}}e^{2}\,|x|\equiv V^{\prime}|x|=V^{\prime}x\ \ \
(x\geq 0)$ (292)

In the following I assume that $x\geq 0$. The wave function for $x<0$ is then
determined by its parity $\eta_{P}$ as in (177),

$\sigma_{3}\,\Phi^{(P)}(-x)\sigma_{3}=\eta_{P}\Phi^{(-P)}(x)\hskip
56.9055pt(\eta_{P}=\pm 1)$ (293)

It can be shown (see Exercise A.19 for the derivation in $D=3+1$) that
(VII.1.2) is equivalent to the two coupled equations

$\displaystyle\Big{[}\frac{2}{E-V}\big{(}i\sigma_{1}{\buildrel\rightarrow\over{\partial}}_{x}+m\sigma_{3}-{\textstyle\frac{1}{2}}\sigma_{1}P\big{)}-1\Big{]}\Phi^{(P)}$
$\displaystyle=-\frac{2i}{(E-V)^{2}}P\partial_{x}\Phi^{(P)}+\frac{iV^{\prime}}{(E-V)^{2}}\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}$
$\displaystyle\Phi^{(P)}\Big{[}\big{(}i\sigma_{1}{\buildrel\leftarrow\over{\partial}}_{x}-m\sigma_{3}+{\textstyle\frac{1}{2}}\sigma_{1}P\big{)}\frac{2}{E-V}-1\Big{]}$
$\displaystyle=\frac{2i}{(E-V)^{2}}P\partial_{x}\Phi^{(P)}-\frac{iV^{\prime}}{(E-V)^{2}}\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}$
(294)

The $2\times 2$ wave function may be expanded in Pauli matrices,

$\displaystyle\Phi^{(P)}(x)=\phi^{\scriptscriptstyle{(P)}}_{0}(x)\,I+\phi^{\scriptscriptstyle{(P)}}_{1}(x)\,\sigma_{1}+\phi^{\scriptscriptstyle{(P)}}_{2}(x)\,i\sigma_{2}+\phi^{\scriptscriptstyle{(P)}}_{3}(x)\,\sigma_{3}$
(295)

where $I$ stands for the unit $2\times 2$ matrix. The coefficients of the
Pauli matrices in the bound state equation (VII.1.2) give four conditions,

$\displaystyle I$ $\displaystyle:\hskip
28.45274pt2i\partial_{x}\phi^{\scriptscriptstyle{(P)}}_{1}(x)=(E-V)\phi^{\scriptscriptstyle{(P)}}_{0}(x)$
$\displaystyle\sigma_{1}$ $\displaystyle:\hskip
28.45274pt2i\partial_{x}\phi^{\scriptscriptstyle{(P)}}_{0}(x)+2m\phi^{\scriptscriptstyle{(P)}}_{2}(x)=(E-V)\phi^{\scriptscriptstyle{(P)}}_{1}(x)$
$\displaystyle i\sigma_{2}$ $\displaystyle:\hskip
28.45274ptP\phi^{\scriptscriptstyle{(P)}}_{3}(x)+2m\phi^{\scriptscriptstyle{(P)}}_{1}(x)=(E-V)\phi^{\scriptscriptstyle{(P)}}_{2}(x)$
$\displaystyle\sigma_{3}$ $\displaystyle:\hskip
28.45274ptP\phi^{\scriptscriptstyle{(P)}}_{2}(x)=(E-V)\phi^{\scriptscriptstyle{(P)}}_{3}(x)$
(296)

####  VII.1.3 Rest frame and non-relativistic limit

Consider first the rest frame, $P=0,\ E=M$. The conditions (VII.1.2) give

$\displaystyle\phi^{\scriptscriptstyle(0)}_{0}(x)=\frac{2i}{M-V}\,\partial_{x}\phi^{\scriptscriptstyle(0)}_{1}(x)$
$\displaystyle\hskip
56.9055pt\phi^{\scriptscriptstyle(0)}_{2}(x)=\frac{2m}{M-V}\,\phi^{\scriptscriptstyle(0)}_{1}(x)\hskip
56.9055pt\phi^{\scriptscriptstyle(0)}_{3}(x)=0$
$\displaystyle\partial_{x}^{2}\phi^{\scriptscriptstyle(0)}_{1}(x)$
$\displaystyle+\frac{V^{\prime}}{M-V}\,\partial_{x}\phi^{\scriptscriptstyle(0)}_{1}(x)+\big{[}{\textstyle\frac{1}{4}}(M-V)^{2}-m^{2}\big{]}\phi^{\scriptscriptstyle(0)}_{1}(x)=0$
(297)

In the NR limit, with $e\ll m$ and $V(x)\ll m$, the equation for
$\phi^{\scriptscriptstyle(0)}_{1}(x)$ reduces to the Schrödinger equation with
binding energy $E_{b}=M-2m$,

$\displaystyle\Big{[}-\frac{1}{m}\partial_{x}^{2}+V(x)\Big{]}\phi^{\scriptscriptstyle(0)}_{1}(x)=E_{b}\phi^{\scriptscriptstyle(0)}_{1}(x)$
(298)

The normalizable solution is given by the Airy function,

$\displaystyle\phi^{\scriptscriptstyle(0)}_{1}(x)=N\mbox{Ai}\big{[}m(V-E_{b})/(mV^{\prime})^{2/3}\big{]}\hskip
56.9055pt(x\geq 0)$ (299)

The coefficient $N$ may be chosen to be real, with size fixed by the
normalization (168) of the state. The energy eigenvalues are determined by
continuity at $x=0$. From (293) and (295) follows
$\phi^{\scriptscriptstyle(0)}_{1}(-x)=-\eta_{P}\phi^{\scriptscriptstyle(0)}_{1}(x)$,
so that

$\displaystyle\phi^{\scriptscriptstyle(0)}_{1}(x=0)=0\ \ \ (\eta_{P}=+1)\hskip
56.9055pt\partial_{x}\phi^{\scriptscriptstyle(0)}_{1}(x=0)=0\ \ \
(\eta_{P}=-1)$ (300)

The relations (VII.1.3) reduce in the NR limit to
$\phi^{\scriptscriptstyle(0)}_{2}(x)=\phi^{\scriptscriptstyle(0)}_{1}(x)$,
$\phi^{\scriptscriptstyle(0)}_{0}(x)=\phi^{\scriptscriptstyle(0)}_{3}(x)=0$.
Hence the $2\times 2$ wave function has the structure

$\displaystyle\Phi^{(0)}_{NR}(x)=(\sigma_{1}+i\sigma_{2})\phi^{\scriptscriptstyle(0)}_{1}(x)$
(301)

The projectors (VI.1.1) in the NR limit are
$\Lambda_{\pm}={\textstyle\frac{1}{2}}(1\pm\sigma_{3})$. The wave function
satisfies
$\Lambda_{+}\Phi^{(0)}_{NR}(x)=\Phi^{(0)}_{NR}(x)\Lambda_{-}=\Phi^{(0)}_{NR}(x)$,
showing that it has no negative energy components. Hence there are no virtual
$e^{-}e^{+}$ pairs in the NR bound state (286).

####  VII.1.4 Solution for any $M$ and $P$

Consider now the bound state conditions (VII.1.2) for arbitrary momenta $P$,
without assuming $V\ll E$. The last two relations allow to express
$\phi^{\scriptscriptstyle{(P)}}_{2}(x)$ and
$\phi^{\scriptscriptstyle{(P)}}_{3}(x)$ in terms of
$\phi^{\scriptscriptstyle{(P)}}_{1}(x)$,

$\displaystyle\phi^{\scriptscriptstyle{(P)}}_{2}(x)=\frac{E-V}{(E-V)^{2}-P^{2}}\,2m\phi^{\scriptscriptstyle{(P)}}_{1}(x)\hskip
56.9055pt\phi^{\scriptscriptstyle{(P)}}_{3}(x)=\frac{P}{(E-V)^{2}-P^{2}}\,2m\phi^{\scriptscriptstyle{(P)}}_{1}(x)$
(302)

The denominators are the square of the kinetic 2-momentum
$\Pi(x)\equiv(E-V,P)$. This motivates changing the variables $x$ into the
“Lorentz invariant” $\tau(x)$, defined as

$\displaystyle\tau(x)\equiv\big{[}(E-V)^{2}-P^{2}\big{]}/V^{\prime}\hskip
56.9055pt\partial_{x}=-2(E-V)\partial_{\tau}$ (303)

The relation between $\partial_{x}$ and $\partial_{\tau}$ is crucial in the
following, and is valid only for a linear potential, $V(x)=V^{\prime}x$
($x\geq 0$). When the equations (VII.1.2) for
$\phi^{\scriptscriptstyle{(P)}}_{0}(x)$ and
$\phi^{\scriptscriptstyle{(P)}}_{1}(x)$ are expressed in terms of $\tau$
rather than $x$ they turn out to be frame independent, i.e., no factors of $E$
or $P$ appear Hoyer (1986). With the shorthand notation
$\phi_{0,1}(\tau)\equiv\phi^{\scriptscriptstyle{(P)}}_{0,1}\big{[}x(\tau)\big{]}$,

$\displaystyle\partial_{\tau}\phi_{1}(\tau)=\frac{i}{4}\,\phi_{0}(\tau)\hskip
56.9055pt\partial_{\tau}\phi_{0}(\tau)=\frac{i}{4}\Big{(}1-\frac{4m^{2}}{V^{\prime}\tau}\Big{)}\phi_{1}(\tau)$
(304)

The superscript $(P)$ on $\phi_{0,1}(\tau)$ is omitted since as functions of
$\tau$ they are the same in all frames. The $P$-dependence of
$\phi^{\scriptscriptstyle{(P)}}_{0,1}\big{[}x(\tau)\big{]}$ as functions of
$x$ arises only from the mapping $x(\tau)$ defined by (303). The equivalence
of this with the $P$-dependence induced by actually boosting the state was
verified in Dietrich  _et al._ (2012) (in $A^{1}=0$ gauge).

The parity constraint (293) on $\Phi^{(P)}(x)$ implies, in view of the
expansion (295), the relations

$\displaystyle\phi^{\scriptscriptstyle{(P)}}_{0,3}(x)=\eta_{P}\phi^{\scriptscriptstyle{(-P)}}_{0,3}(-x)\hskip
56.9055pt\phi^{\scriptscriptstyle{(P)}}_{1,2}(x)=-\eta_{P}\phi^{\scriptscriptstyle{(-P)}}_{1,2}(-x)$
(305)

Consider first $\phi_{0}$ and $\phi_{1}$, which for $x\geq 0$ are functions
only of $\tau$ in (304). Since $\tau(x)$ is invariant under $P\to-P$ we need
not be concerned with the sign change of $P$ under parity. Continuity at $x=0$
requires for $\eta_{P}=+1$ that $\phi_{1}\big{[}\tau(x=0)\big{]}=0$ and for
$\eta_{P}=-1$ that $\phi_{0}\big{[}\tau(x=0)\big{]}=0$. The relations (304)
ensure that $\partial_{\tau}\phi_{1}(\tau)=0$ when $\phi_{0}(\tau)=0$ and vice
versa, as required by the opposite parities of $\phi_{0}$ and $\phi_{1}$.

For $P=0$ (303) gives $V^{\prime}\tau(x=0)=M^{2}$. Hence the condition
$\phi_{1}(\tau=M^{2}/V^{\prime})=0$ determines the masses $M$ of the bound
states with $\eta_{P}=+1$. Similarly, the zeros of $\phi_{0}(\tau)$ determine
the $\eta_{P}=-1$ masses. When $P\neq 0$ we have
$V^{\prime}\tau(x=0)=E^{2}-P^{2}$, whereas the zeros of the functions
$\phi_{0,1}(\tau)$ are independent of $P$. Satisfying the parity constraint
for all $P$ then requires the energies $E$ to satisfy $E^{2}-P^{2}=M^{2}$, as
expected from Lorentz covariance. This allows to express $\tau(x)$ in (303) as
$\tau(x)=\big{[}M^{2}-2EV+V^{2}\big{]}/V^{\prime}$.

The function $\phi^{\scriptscriptstyle{(P)}}_{2}(x)$ given by (302) has the
same $x\to-x$ symmetry as $\phi^{\scriptscriptstyle{(P)}}_{1}(x)$ as required
by (305). $\phi^{\scriptscriptstyle{(P)}}_{3}(x)$ is likewise related to
$\phi^{\scriptscriptstyle{(P)}}_{1}(x)$ by a coefficient which is symmetric
under $x\to-x$, but antisymmetric under $P\to-P$. Hence
$\phi^{\scriptscriptstyle{(P)}}_{3}(x)$ has opposite parity constraint
compared to $\phi^{\scriptscriptstyle{(P)}}_{1}(x)$, which is again consistent
with (305).

Defining the $x$-dependent “boost parameter” $\zeta(x)$ by

$\displaystyle\cosh\zeta=\frac{E-V}{\sqrt{V^{\prime}\tau}}\hskip
56.9055pt\sinh\zeta=\frac{P}{\sqrt{V^{\prime}\tau}}$ (306)

the full wave function (295) may be expressed using (302) as

$\displaystyle\Phi^{(P)}=\phi_{0}+\phi_{1}\Big{[}\sigma_{1}+\frac{2m(E-V)}{V^{\prime}\tau}\,i\sigma_{2}+\frac{2mP}{V^{\prime}\tau}\,\sigma_{3}\Big{]}=e^{-\sigma_{1}\zeta/2}\Big{(}\phi_{0}+\phi_{1}\sigma_{1}+\frac{2m}{\sqrt{V^{\prime}\tau}}\phi_{1}\,i\sigma_{2}\Big{)}e^{\sigma_{1}\zeta/2}$
(307)

In the latter expression the term in ( ) depends on $\tau$ only, whereas
$\zeta$ depends also explicitly on $P$. In the weak coupling limit $(V\ll m)$
$\zeta(x)$ reduces to the standard boost parameter $\xi$,

$\displaystyle\cosh\xi=\frac{E}{M}\hskip 56.9055pt\sinh\xi=\frac{P}{M}$ (308)

The expression (307) allows to determine the frame dependence of
$\Phi^{(P)}(x)$ at constant $x$,

$\displaystyle\left.\frac{\partial\Phi^{(P)}(x)}{\partial\xi}\right|_{x}=\left.\frac{xP}{E-V}\partial_{x}\Phi^{(P)}\right|_{\xi}-\frac{E}{2(E-V)}\,\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}$
(309)

where the $x$-derivative on the rhs. is taken at constant $P$.

Exercise A.13: Derive the expression (309). Hint: You may start from
$e^{\sigma_{1}\zeta/2}\Phi^{(P)}e^{-\sigma_{1}\zeta/2}$, which is a function
only of $\tau$.

Eliminating $\phi_{0}(\tau)$ in (304) gives

$\displaystyle\partial_{\tau}^{2}\phi_{1}(\tau)+\frac{1}{16}\Big{(}1-\frac{4m^{2}}{V^{\prime}\tau}\Big{)}\phi_{1}(\tau)=0$
(310)

The regular, arbitrarily normalized analytic solutions for $\phi_{0,1}(\tau)$
are555The factor $\sqrt{V^{\prime}}$ in the wave function gives it the correct
dimension, corresponding to a relativistically normalized state in $D=1+1$.
Analytic solutions are given in Dietrich  _et al._ (2013) also for bound
states of fermions with unequal masses. Dietrich  _et al._ (2013)

$\displaystyle\phi_{1}(\tau)$
$\displaystyle=\sqrt{V^{\prime}}\,\tau\,\exp(-i\tau/4){\,{{}_{1}}F_{1}}(1-im^{2}/2V^{\prime},2,i\tau/2)=\phi_{1}^{*}(\tau)$
$\displaystyle\phi_{0}(\tau)$
$\displaystyle=-\phi_{1}(\tau)-4i\sqrt{V^{\prime}}\,\exp(-i\tau/4){\,{{}_{1}}F_{1}}(1-im^{2}/2V^{\prime},1,i\tau/2)=-\phi_{0}^{*}(\tau)$
(311)

####  VII.1.5 Weak coupling limit

The Positronium states of QED4 were in (165) defined with the $\Lambda_{\pm}$
projectors (VI.1.1), and the ${\boldsymbol{x}}$-dependence of the
$\left|{e^{-}e^{+}}\right\rangle$ Fock state wave function
$\Phi^{(P)}({\boldsymbol{x}})$ (183) was found to Lorentz contract (216). Here
I verify that the QED2 wave functions have analogous properties in the weak
coupling limit, although there is no transverse photon contribution and the
states (286) are defined without projecting on the lowest Fock state.

According to (VII.1.2) the Dirac structure (295) of the QED2 wave function
reduces for $V\ll m$ and $M\simeq 2m$ to

$\displaystyle\Phi^{(P)}_{NR}(x)=\Big{(}\sigma_{1}+\frac{E}{M}\,i\sigma_{2}+\frac{P}{M}\,\sigma_{3}\Big{)}\phi^{\scriptscriptstyle{(P)}}_{1,NR}(x)$
(312)

and the variable $\tau$ of (303) simplifies to

$\displaystyle
V^{\prime}\tau_{NR}(x)=M^{2}-2EV(x)=M(M-2V^{\prime}x\cosh\xi)\hskip
56.9055pt\cosh\xi=\frac{E}{M}$ (313)

The dependence on $x\cosh\xi$ means that $\phi_{1}\big{[}\tau_{NR}(x)\big{]}$
Lorentz contracts similarly as $F^{({\boldsymbol{P}})}({\boldsymbol{x}})$ in
(216). The leading order expression (A.587) of the projectors $\Lambda_{\pm}$
is in $D=1+1$

$\displaystyle\Lambda_{\pm}(P)=\frac{1}{2E}(E\mp\sigma_{1}P\pm M\sigma_{3})$
(314)

The Dirac structure of the QED2 wave function is $\sigma_{1}+i\sigma_{2}$ for
$P=0$ (301). The analogy with QED4 (215) suggests that $\Phi^{(P)}_{NR}(x)$
should for general $P$ be a $2\times 2$ matrix proportional to

$\displaystyle\Lambda_{+}(P)(\sigma_{1}+i\sigma_{2})\Lambda_{-}(P)=\frac{M(E+M)}{2E^{2}}\Big{(}\sigma_{1}+\frac{E}{M}\,i\sigma_{2}+\frac{P}{M}\,\sigma_{3}\Big{)}$
(315)

which agrees with (312): The properties of the weakly bound states in QED2 and
QED4 are analogous at all $P$.

The non-relativistic limit of $\phi_{1}(\tau)$ may be determined from its
analytic expression (VII.1.4). The scaling of the coordinate $x$ in the limit
$m\to\infty$ at fixed $V^{\prime}={\textstyle\frac{1}{2}}e^{2}$ is given by
the Schrödinger equation (298) for $P=0$: $\partial_{x}^{2}\propto
mV^{\prime}x$, i.e., $x\propto(mV^{\prime})^{-1/3}$, and
$E_{b}=M-2m\propto(mV^{\prime})^{2/3}/m$. In this limit Dietrich  _et al._
(2013); Hoyer (2014)

$\displaystyle\phi_{1,NR}^{\scriptscriptstyle{(0)}}(x)=\lim_{m\to\infty}\phi_{1}(\tau)=4\sqrt{V^{\prime}}\Big{(}\frac{V^{\prime}}{m^{2}}\Big{)}^{1/3}e^{\pi
m^{2}/2V^{\prime}}\,\mbox{Ai}\big{[}m(V-E_{b})/(mV^{\prime})^{2/3}\big{]}\Big{[}1+{\cal{O}}\Big{(}m^{2}/V^{\prime}\Big{)}^{-2/3}\Big{]}$
(316)

which relates the normalization of the NR solution (299) to that of the
general solution (VII.1.4).

####  VII.1.6 Large separations between $e^{-}$ and $e^{+}$

The variable $\tau(x)$ (303) grows with the separation $x$ of the fermions.
For $|\tau|\to\infty$ the wave function $\phi_{1}(\tau)$ (VII.1.4) oscillates
with constant amplitude, up to corrections of ${\cal O}\left(1/|\tau|\right)$:

$\displaystyle\phi_{1}(|\tau|\to\infty)=\frac{4V^{\prime}}{\sqrt{\pi}\,m}\sqrt{\exp(\pi
m^{2}/V^{\prime})-1}\;e^{-\theta(-\tau)\pi
m^{2}/2V^{\prime}}\cos\Big{[}{\textstyle\frac{1}{4}}\tau-(m^{2}/2V^{\prime})\log({\textstyle\frac{1}{2}}|\tau|)+\arg\Gamma(1+im^{2}/2V^{\prime})-\pi/2\Big{]}$
(317)

where $\theta(x)=1\ (0)$ for $x>0\ (x<0)$ is the step function. From (304) we
have $\phi_{0}(\tau)=-4i\partial_{\tau}\phi_{1}(\tau)$, so that

$\displaystyle\lim_{|\tau|\to\infty}\big{[}\phi_{1}(\tau)+\phi_{0}(\tau)\big{]}=\lim_{|\tau|\to\infty}\big{[}\phi_{1}(\tau)-\phi_{0}(\tau)\big{]}^{*}$
$\displaystyle=N\exp\big{[}i\tau/4-i(m^{2}/2V^{\prime})\log(|\tau|/2)+i\arg\Gamma(1+im^{2}/2V^{\prime})-i\pi/2\big{]}$
$\displaystyle N$
$\displaystyle=\frac{4V^{\prime}}{\sqrt{\pi}\,m}\sqrt{\exp(\pi
m^{2}/V^{\prime})-1}\;e^{-\theta(-\tau)\pi m^{2}/2V^{\prime}}\hskip 56.9055pt$
(318)

has an $x$-independent local norm $N^{2}$. This $x$-dependence of the wave
function is made possible by modes with large negative kinetic energy, which
balance the linear potential to give a fixed energy eigenvalue. The asymptotic
wave function thus describes virtual $e^{-}e^{+}$ pairs, illustrated by time-
ordered $Z$-diagrams such as in Fig. 13 b. The negative energy components
created by the $b\,d$ operators in the state (286) dominate for $x\to\infty$
Dietrich  _et al._ (2013); Hoyer (2014). The pairs give rise to a sea
distribution for ${x_{Bj}}\to 0$ in deep inelastic scattering (see section
VII.3 and Fig. 17). The Dirac radial functions (126) similarly have a constant
local norm at large values of $r$.

####  VII.1.7 Bound state masses and duality

The wave function $\phi_{1}(\tau)$ in (VII.1.4) was chosen to satisfy
$\phi_{1}(\tau=0)=0$, ensuring that $\phi_{2}(0)$ and $\phi_{3}(0)$ (302) are
finite. The general solution of the differential equation (310) has
$\phi_{1}(\tau=0)\neq 0$, giving singular $\phi_{2}$ and $\phi_{3}$. The
requirement that the wave function is regular at $\tau=0$ implies a discrete
QED2 spectrum. This criterion of local normalizability is the relativistic
generalization of the requirement of a finite global norm for Schrödinger wave
functions. In fact, the non-relativistic limit of the general solution for
$\phi_{1}(\tau)$ (with singular $\phi_{2,3}(\tau=0)$) adds an Airy Bi function
to (316), which increases exponentially at large $x$ Dietrich  _et al._
(2013).

The bound state masses $M$ of locally normalizable solutions are determined by
the parity constraint (305) at $x=0$, which requires
$\phi_{1}(\tau=M^{2}/V^{\prime})=0$ for $\eta_{P}=+1$, and
$\partial_{\tau}\phi_{1}(\tau=M^{2}/V^{\prime})=0$ for $\eta_{P}=-1$. At high
masses $M$ we may use the asymptotic expression (317) for
$\phi_{1}(\tau\to\infty)$. The squared eigenvalues $M_{n}^{2}$ are then given
by integers $n$ and lie on asymptotically linear “Regge trajectories”,

$\displaystyle M_{n}^{2}=n\,2\pi V^{\prime}+{\cal O}\left(m^{2}\log
n\right)\hskip 56.9055pt\eta_{P}=(-1)^{n}$ (319)

For small electron masses $m$ the trajectory is linear down to low excitations
Dietrich  _et al._ (2013). In the range of $x$ where $V(x)\ll M$ the state
(286) is dominated by positive energy $b^{\dagger}\,d^{\dagger}$
contributions, as would be expected for parton-hadron duality Dietrich  _et
al._ (2013); Hoyer (2014). The states have an overlap with multiple bound
states generated via string breaking as shown in Fig. 19 a. I return to these
issues in the context of QCD in $D=3+1$ dimensions (section VIII.4).

Consider next a ground state of mass $M$. With increasing $x$ the first
virtual $e^{-}e^{+}$ pair (string breaking) in the $P=0$ wave function is
expected when the potential has reached twice the mass of the bound state,
i.e., at $V(x)=2M$. This energetically allows a (virtual) bound state pair to
appear (as depicted in Fig. 19 b below). I illustrate this with a numerical
example in Fig. 15(a), where $e/m=0.71$ and $M=4.86\sqrt{V^{\prime}}$. The
dynamics is nearly non-relativistic at low $x$, as evidenced by the agreement
of the blue (exact, (VII.1.4)) and red dashed (Schrödinger, (316)) wave
functions. Both are exponentially suppressed with increasing $x$. In the
relativistic range, $V(x)\gtrsim M$, the exact wave function (blue line)
begins to increase, reaching a maximum at $V(x)=2M$. The wave function is
symmetric around $V^{\prime}x=M$ because $\tau(x)$ in (303) satisfies
$\tau(x)=\tau(2M/V^{\prime}-x)$.666There is no symmetry for $V^{\prime}x>2M$
since the wave function is defined by parity for $x<0$. For $P=M\sinh\xi>0$
the virtual pair is expected at $V(x)=2E$, since each bound state has
increased energy due to the boost: $M\cosh\xi=E$. This is verified in Fig.
15(b), and is again due to the symmetry of $\tau(x)$.

Figure 15: The wave function $\phi_{1}\big{[}\tau(x)\big{]}$ of the
$\eta_{P}=-1$ ground state for $m=2\sqrt{V^{\prime}}$, i.e., $e/m=1/\sqrt{2}$.
Blue line: Relativistic expression (VII.1.4), implying a bound state mass
$M=4.86\sqrt{V^{\prime}}$. Dashed red line: Non-relativistic Airy function
(316), requiring $E_{b}=0.81\sqrt{V^{\prime}}$. Both functions are normalized
to unity at $x=0$. (a) Rest frame, $P=0$. (b) $P=5\sqrt{V^{\prime}}$, with the
non-relativistic (dashed red) line Lorentz contracted, $x\to xE/M$ as in
(313).

###  VII.2 * Bound state form factors in QED2

####  VII.2.1 Form factor definition and symmetry under parity

The electromagnetic form factor is defined as for Positronium in $D=3+1$, see
section VI.5. The form factor for $\gamma^{*}+A\to B$ is as in (VI.5) and
(240),

$\displaystyle F_{AB}^{\mu}(q)$ $\displaystyle=\int d^{2}z\,e^{-iq\cdot
z}\langle{M_{B},P_{B}}|\bar{\psi}(z)\gamma^{\mu}\psi(z)\left|{M_{A},P_{A}}\right\rangle=(2\pi)^{2}\delta^{2}(P_{B}-P_{A}-q)G_{AB}^{\mu}(q)$
$\displaystyle G_{AB}^{\mu}(q)$
$\displaystyle=\int_{-\infty}^{\infty}dx\,e^{i(P_{B}-P_{A})x/2}\,\mathrm{Tr}\,\big{[}\Phi_{B}^{\dagger}(x)\gamma^{\mu}\gamma^{0}\Phi_{A}(x)-\Phi_{B}^{\dagger}(-x)\Phi_{A}(-x)\gamma^{0}\gamma^{\mu}\big{]}$
(320)

where the two contributions to $G_{AB}^{\mu}(q)$ arise from scattering on
$e^{-}$ and $e^{+}$, respectively. The wave function
$\Phi_{A}(x)\equiv\Phi_{A}^{(P_{A})}(x)$ determines the state as in (286), and
$\gamma^{0},\,\gamma^{1}$ are defined in (291). The $e^{-}$ and $e^{+}$
contributions can be related using an analogy to charge conjugation in $D=3+1$
(181). According to the relations (302) the parity relations (305) become,
when the sign of $P$ is not reversed,

$\displaystyle\phi^{\scriptscriptstyle{(P)}}_{0}(x)=\eta_{P}\phi^{\scriptscriptstyle{(P)}}_{0}(-x)\hskip
56.9055pt\phi^{\scriptscriptstyle{(P)}}_{1,2,3}(x)=-\eta_{P}\phi^{\scriptscriptstyle{(P)}}_{1,2,3}(-x)$
(321)

This implies for the wave function expressed as in (295), in analogy to charge
conjugation

$\displaystyle\sigma_{2}\big{[}\Phi^{(P)}(-x)\big{]}^{T}\sigma_{2}=\eta_{P}{\Phi^{(P)}}(x)$
(322)

Bracketing the trace of the second term in (VII.2.1) with $\sigma_{2}$ and
transposing it,

$\displaystyle-\mathrm{Tr}\,\big{[}\sigma_{2}\Phi_{B}^{\dagger}(-x)\Phi_{A}(-x)\gamma^{0}\gamma^{\mu}\sigma_{2}\big{]}^{T}=-\eta_{P}^{A}\eta_{P}^{B}\mathrm{Tr}\,\big{[}\gamma^{\mu}\gamma^{0}\Phi_{A}(x)\Phi_{B}^{\dagger}(x)\big{]}$
(323)

where
$\sigma_{2}(\gamma^{0}\gamma^{\mu})^{T}\sigma_{2}=\gamma^{\mu}\gamma^{0}$ for
$\mu=0,\,1$. Hence the $e^{-}$ and $e^{+}$ contributions are related similarly
as in (242) and

$\displaystyle G_{AB}^{\mu}(q)$
$\displaystyle=(1-\eta_{P}^{A}\eta_{P}^{B})\int_{-\infty}^{\infty}dx\,e^{i(P_{B}-P_{A})x/2}\,\mathrm{Tr}\,\big{[}\Phi_{B}^{\dagger}(x)\gamma^{\mu}\gamma^{0}\Phi_{A}(x)\big{]}$
(324)

vanishes unless $\eta_{P}^{A}=-\eta_{P}^{B}$.

####  VII.2.2 Gauge invariance

I follow the derivation in Dietrich  _et al._ (2013) and consider only
scattering from $e^{-}$, i.e., leave out the factor
$1-\eta_{P}^{A}\eta_{P}^{B}$ of (324). Denoting $E_{A,B}=P_{A,B}^{\,0}$ and
$P_{A,B}=P_{A,B}^{1}$, gauge invariance requires that

$\displaystyle
q_{\mu}G^{\mu}_{AB}=(P_{B}-P_{A})_{\mu}G^{\mu}_{AB}=\int_{-\infty}^{\infty}dx\,e^{i(P_{B}-P_{A})x/2}\,\mathrm{Tr}\,\big{\\{}\Phi_{B}^{\dagger}\big{[}(E_{B}-E_{A})+(P_{B}-P_{A})\sigma_{1}\big{]}\Phi_{A}\big{\\}}=0$
(325)

The bound state equation (VII.1.2) for $\Phi_{A}$ and $\Phi_{B}^{\dagger}$
are, with $V=V^{\prime}x$ and $x\geq 0$,

$\displaystyle(E_{A}-V)\Phi_{A}$
$\displaystyle=i\partial_{x}\left\\{{\sigma_{1}},{\Phi_{A}}\right\\}-{\textstyle\frac{1}{2}}P_{A}\left[{\sigma_{1}},{\Phi_{A}}\right]+m\left[{\sigma_{3}},{\Phi_{A}}\right]$
$\displaystyle\Big{|}\ -\Phi_{B}^{\dagger}\,\times$
$\displaystyle\Phi_{B}^{\dagger}(E_{B}-V)$
$\displaystyle=-i\partial_{x}\big{\\{}{\sigma_{1}},{\Phi_{B}^{\dagger}}\big{\\}}+{\textstyle\frac{1}{2}}P_{B}\big{[}{\sigma_{1}},{\Phi_{B}^{\dagger}}\big{]}-m\big{[}{\sigma_{3}},{\Phi_{B}^{\dagger}}\big{]}$
$\displaystyle\Big{|}\ \times\,\Phi_{A}\hskip 8.5359pt$ (326)

Multiplying the equations as indicated in the margin their sum becomes

$\displaystyle\Phi_{B}^{\dagger}(E_{B}-E_{A})\Phi_{A}=$
$\displaystyle-i\Phi_{B}^{\dagger}\partial_{x}\left\\{{\sigma_{1}},{\Phi_{A}}\right\\}-i\partial_{x}\big{\\{}{\sigma_{1}},{\Phi_{B}^{\dagger}}\big{\\}}\Phi_{A}+{\textstyle\frac{1}{2}}P_{B}\big{[}{\sigma_{1}},{\Phi_{B}^{\dagger}}\big{]}\Phi_{A}+{\textstyle\frac{1}{2}}P_{A}\Phi_{B}^{\dagger}\left[{\sigma_{1}},{\Phi_{A}}\right]$
$\displaystyle-m\left[{\sigma_{3}},{\Phi_{B}^{\dagger}}\right]\Phi_{A}-m\Phi_{B}^{\dagger}\left[{\sigma_{3}},{\Phi_{A}}\right]$
(327)

When the trace is taken the terms with $\partial_{x}$ form a total derivative,

$\displaystyle-i\mathrm{Tr}\,\big{\\{}\Phi_{B}^{\dagger}(\sigma_{1}\partial_{x}\Phi_{A}+\partial_{x}\Phi_{A}\sigma_{1})+(\sigma_{1}\partial_{x}\Phi_{B}^{\dagger}+\partial_{x}\Phi_{B}^{\dagger}\sigma_{1})\Phi_{A}\big{\\}}=-i\mathrm{Tr}\,\big{\\{}\partial_{x}\big{(}\Phi_{B}^{\dagger}\sigma_{1}\Phi_{A}+\Phi_{B}^{\dagger}\Phi_{A}\sigma_{1}\big{)}\big{\\}}$
(328)

Partially integrating this term in (325) it becomes

$\displaystyle-{\textstyle\frac{1}{2}}(P_{B}-P_{A})\mathrm{Tr}\,\big{\\{}\Phi_{B}^{\dagger}\sigma_{1}\Phi_{A}+\Phi_{B}^{\dagger}\Phi_{A}\sigma_{1}\big{\\}}$
(329)

The ${\cal O}\left(P\right)$ terms in (VII.2.2) may in the trace be expressed
as

$\displaystyle{\textstyle\frac{1}{2}}P_{B}\mathrm{Tr}\,\big{\\{}(\sigma_{1}\Phi_{B}^{\dagger}-\Phi_{B}^{\dagger}\sigma_{1})\Phi_{A}\big{\\}}+{\textstyle\frac{1}{2}}P_{A}\mathrm{Tr}\,\big{\\{}\Phi_{B}^{\dagger}(\sigma_{1}\Phi_{A}-\Phi_{A}\sigma_{1})\big{\\}}={\textstyle\frac{1}{2}}(P_{B}-P_{A})\mathrm{Tr}\,\big{\\{}\Phi_{B}^{\dagger}\Phi_{A}\sigma_{1}-\Phi_{B}^{\dagger}\sigma_{1}\Phi_{A}\big{\\}}$
(330)

The ${\cal O}\left(m\right)$ term vanishes when traced,

$\displaystyle-m\mathrm{Tr}\,\big{\\{}(\sigma_{3}\Phi_{B}^{\dagger}-\Phi_{B}^{\dagger}\sigma_{3})\Phi_{A}+\Phi_{B}^{\dagger}(\sigma_{3}\Phi_{A}-\Phi_{A}\sigma_{3})\big{\\}}=0$
(331)

The sum of (329) and (330) gives

$\displaystyle-(P_{B}-P_{A})\mathrm{Tr}\,\big{\\{}\Phi_{B}^{\dagger}\sigma_{1}\Phi_{A}\big{\\}}$
(332)

This cancels the second term in (325), thus ensuring gauge invariance.

####  VII.2.3 Lorentz covariance

Lorentz covariance requires that the form factor can be written, given the
gauge invariance (325),

$\displaystyle G^{\mu}_{AB}(q)$
$\displaystyle=\epsilon^{\mu\nu}(P_{B}-P_{A})_{\nu}G_{AB}(q^{2})$
$\displaystyle\epsilon^{01}=-\epsilon^{10}=1$ (333)

With $E_{A,B}=M_{A,B}\cosh\xi$ and $P_{A,B}=M_{A,B}\sinh\xi$ we have

$\displaystyle\frac{\delta E_{A,B}}{\delta\xi}=P_{A,B}\hskip
85.35826pt\frac{\delta P_{A,B}}{\delta\xi}=E_{A,B}$ (334)

Recalling that $P_{0}=P^{0}$ whereas $P_{1}=-P^{1}$ we should have

$\displaystyle\frac{\delta G_{AB}^{0}(q)}{\delta\xi}=G_{AB}^{1}(q)\hskip
85.35826pt\frac{\delta G_{AB}^{1}(q)}{\delta\xi}=G_{AB}^{0}(q)$ (335)

Subtracting the coupled bound state equations in (VII.1.2) gives an expression
for

$\displaystyle\frac{P}{E-V}\partial_{x}\Phi^{(P)}=-{\textstyle\frac{1}{2}}\partial_{x}\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}-{\textstyle\frac{1}{4}}iP\big{\\{}{\sigma_{1}},{\Phi^{(P)}}\big{\\}}+{\textstyle\frac{1}{2}}im\big{\\{}{\sigma_{3}},{\Phi^{(P)}}\big{\\}}+\frac{V^{\prime}}{2(E-V)}\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}$
(336)

Using this in (309) gives

$\displaystyle\frac{\partial\Phi^{(P)}}{\partial\xi}$
$\displaystyle=-{\textstyle\frac{1}{2}}\partial_{x}\Big{(}x\big{[}{\sigma_{1}},{\Phi^{(P)}}\big{]}\Big{)}+{\textstyle\frac{1}{2}}ix\Big{(}-{\textstyle\frac{1}{2}}P\big{\\{}{\sigma_{1}},{\Phi^{(P)}}\big{\\}}+m\big{\\{}{\sigma_{3}},{\Phi^{(P)}}\big{\\}}\Big{)}$
$\displaystyle\frac{\partial\Phi^{(P){\dagger}}}{\partial\xi}$
$\displaystyle={\textstyle\frac{1}{2}}\partial_{x}\Big{(}x\big{[}{\sigma_{1}},{\Phi^{(P){\dagger}}}\big{]}\Big{)}+{\textstyle\frac{1}{2}}ix\Big{(}{\textstyle\frac{1}{2}}P\big{\\{}{\sigma_{1}},{\Phi^{(P){\dagger}}}\big{\\}}-m\big{\\{}{\sigma_{3}},{\Phi^{(P){\dagger}}}\big{\\}}\Big{)}$
(337)

According to the expression (324) for $G_{AB}^{\mu}$ (without the factor
$1-\eta_{P}^{A}\eta_{P}^{B}$),

$\displaystyle\frac{\partial
G_{AB}^{\mu}}{\partial\xi}=\int_{-\infty}^{\infty}dx\,e^{i(P_{B}-P_{A})x/2}\Big{(}{\textstyle\frac{1}{2}}ix(E_{B}-E_{A})\mathrm{Tr}\,\big{\\{}\Phi_{B}^{\dagger}(x)\gamma^{\mu}\gamma^{0}\Phi_{A}(x)\big{\\}}+\frac{\partial}{\partial\xi}\mathrm{Tr}\,\big{\\{}\Phi_{B}^{\dagger}(x)\gamma^{\mu}\gamma^{0}\Phi_{A}(x)\big{\\}}\Big{)}$
(338)

The second term has the contributions

$\displaystyle\mathrm{Tr}\,\Big{\\{}\frac{\partial\Phi_{B}^{\dagger}}{\partial\xi}\gamma^{\mu}\gamma^{0}\Phi_{A}\Big{\\}}$
$\displaystyle={\textstyle\frac{1}{2}}\mathrm{Tr}\,\Big{\\{}\big{[}{\sigma_{1}},{\partial_{x}(x\Phi_{B}^{\dagger})}\big{]}\gamma^{\mu}\gamma^{0}\Phi_{A}+ix\Big{(}{\textstyle\frac{1}{2}}P_{B}\big{\\{}{\sigma_{1}},{\Phi_{B}^{\dagger}}\big{\\}}\gamma^{\mu}\gamma^{0}\Phi_{A}-m\big{\\{}{\sigma_{3}},{\Phi_{B}^{\dagger}}\big{\\}}\Big{)}\gamma^{\mu}\gamma^{0}\Phi_{A}\Big{\\}}$
$\displaystyle\mathrm{Tr}\,\Big{\\{}\Phi_{B}^{\dagger}\gamma^{\mu}\gamma^{0}\frac{\partial\Phi_{A}}{\partial\xi}\Big{\\}}$
$\displaystyle={\textstyle\frac{1}{2}}\mathrm{Tr}\,\Big{\\{}-\Phi_{B}^{\dagger}\gamma^{\mu}\gamma^{0}\left[{\sigma_{1}},{\partial_{x}(x\Phi_{A})}\right]+ix\Big{(}-{\textstyle\frac{1}{2}}P_{A}\Phi_{B}^{\dagger}\gamma^{\mu}\gamma^{0}\left\\{{\sigma_{1}},{\Phi_{A}}\right\\}+m\Phi_{B}^{\dagger}\gamma^{\mu}\gamma^{0}\left\\{{\sigma_{3}},{\Phi_{A}}\right\\}\Big{)}\Big{\\}}$
(339)