c}{4}\bar{\psi}(\reflectbox{$\vec{\reflectbox{$D$}}$}\boldsymbol{\gamma}_{\mathrm{B}}^{5}\boldsymbol{\gamma}_{\mathrm{B}}^{\nu}\mathbf{t}^{\mu}\boldsymbol{\gamma}_{\mathrm{F}}-\bar{\boldsymbol{\gamma}}_{\mathrm{F}}\boldsymbol{\gamma}_{\mathrm{B}}^{5}\boldsymbol{\gamma}_{\mathrm{B}}^{\nu}\mathbf{t}^{\mu}\vec{D})\psi$
$\displaystyle\hskip
11.38092pt+\bar{\Psi}_{\Re}\mathbf{t}^{\mu}\boldsymbol{\gamma}_{\mathrm{B}}^{\nu}\boldsymbol{\gamma}_{\mathrm{B}}^{5}\Psi_{\Re}.$
(141)

The first term of Eq. (141) is equal to the trace-reversed SEM tensor of Dirac
field in an external electromagnetic field. The second term is equal to the
SEM tensor of the electromagnetic field. Since the SEM tensor of the
electromagnetic field is traceless, the trace reversion has no effect on it.
The conservation law for the SEM tensor is given by
$\partial_{\nu}T^{\mu\nu}=0$. That Eq. (141) leads to the well-known
expressions of the symmetric SEM tensors of the Dirac and electromagnetic
fields is revealed in the subsections below.

####  VII.3.1 SEM tensor of the Dirac field

Using Eq. (141), for the trace-reversed SEM tensor of the Dirac field, we
obtain

$\displaystyle\bar{T}_{\mathrm{D}}^{\mu\nu}$ $\displaystyle=\frac{i\hbar
c}{4}\bar{\psi}(\reflectbox{$\vec{\reflectbox{$D$}}$}\boldsymbol{\gamma}_{\mathrm{B}}^{5}\boldsymbol{\gamma}_{\mathrm{B}}^{\nu}\mathbf{t}^{\mu}\boldsymbol{\gamma}_{\mathrm{F}}-\bar{\boldsymbol{\gamma}}_{\mathrm{F}}\boldsymbol{\gamma}_{\mathrm{B}}^{5}\boldsymbol{\gamma}_{\mathrm{B}}^{\nu}\mathbf{t}^{\mu}\vec{D})\psi$
$\displaystyle=\frac{i\hbar
c}{4}\bar{\psi}(\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\vec{D}^{\nu}+\boldsymbol{\gamma}_{\mathrm{F}}^{\nu}\vec{D}^{\mu}-\reflectbox{$\vec{\reflectbox{$D$}}$}^{\nu}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}-\reflectbox{$\vec{\reflectbox{$D$}}$}^{\mu}\boldsymbol{\gamma}_{\mathrm{F}}^{\nu}$
$\displaystyle\hskip
14.22636pt-\eta^{\mu\nu}\boldsymbol{\gamma}_{\mathrm{F}}^{\rho}\vec{D}_{\rho}+\reflectbox{$\vec{\reflectbox{$D$}}$}_{\rho}\boldsymbol{\gamma}_{\mathrm{F}}^{\rho}\eta^{\mu\nu})\psi.$
(142)

In the last equality, we have used the mathematical identities
$\bar{\boldsymbol{\gamma}}_{\mathrm{F}}\boldsymbol{\gamma}_{\mathrm{B}}^{5}\boldsymbol{\gamma}_{\mathrm{B}}^{\nu}\mathbf{t}^{\mu}\vec{D}=-\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\vec{D}^{\nu}-\boldsymbol{\gamma}_{\mathrm{F}}^{\nu}\vec{D}^{\mu}+\eta^{\mu\nu}\boldsymbol{\gamma}_{\mathrm{F}}^{\rho}\vec{D}_{\rho}$
and
$\reflectbox{$\vec{\reflectbox{$D$}}$}\boldsymbol{\gamma}_{\mathrm{B}}^{5}\boldsymbol{\gamma}_{\mathrm{B}}^{\nu}\mathbf{t}^{\mu}\boldsymbol{\gamma}_{\mathrm{F}}=-\reflectbox{$\vec{\reflectbox{$D$}}$}^{\mu}\boldsymbol{\gamma}_{\mathrm{F}}^{\nu}-\reflectbox{$\vec{\reflectbox{$D$}}$}^{\nu}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}+\reflectbox{$\vec{\reflectbox{$D$}}$}_{\rho}\boldsymbol{\gamma}_{\mathrm{F}}^{\rho}\eta^{\mu\nu}$.
In terms of the trace-reversed SEM tensor in Eq. (142), the actual SEM tensor
$T_{\mathrm{D}}^{\mu\nu}$ of the Dirac field is given by

$\displaystyle T_{\mathrm{D}}^{\mu\nu}$
$\displaystyle=\bar{T}_{\mathrm{D}}^{\mu\nu}-\frac{1}{2}\eta^{\mu\nu}\bar{T}_{\mathrm{D}}=\left[\begin{array}[]{cccc}W_{\mathrm{D}}&c\mathbf{G}_{\mathrm{D}}\\\
c\mathbf{G}_{\mathrm{D}}&\boldsymbol{\mathcal{T}}_{\mathrm{D}}\end{array}\right]$
(145) $\displaystyle=\frac{i\hbar
c}{4}\bar{\psi}(\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\vec{D}^{\nu}+\boldsymbol{\gamma}_{\mathrm{F}}^{\nu}\vec{D}^{\mu}-\reflectbox{$\vec{\reflectbox{$D$}}$}^{\nu}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}-\reflectbox{$\vec{\reflectbox{$D$}}$}^{\mu}\boldsymbol{\gamma}_{\mathrm{F}}^{\nu})\psi.$
(146)

This is the well-known result for the symmetric SEM tensor of the Dirac field
in an external electromagnetic field Peskin2018 . In general, all four terms
on the last line of Eq. (146) are necessary for the SEM tensor to be symmetric
and its components real valued. The energy density, momentum density, and
stress tensor components of the Dirac field SEM tensor are given by

$\displaystyle W_{\mathrm{D}}$ $\displaystyle=\rho_{\hat{H}}^{\mathrm{(D)}},$
$\displaystyle\mathbf{G}_{\mathrm{D}}$
$\displaystyle=\boldsymbol{\rho}_{\hat{\mathbf{p}}-q_{\mathrm{e}}\mathbf{A}_{\Re}}^{\mathrm{(D)}}+\frac{1}{2}\nabla\times\boldsymbol{\rho}_{\hat{\mathbf{S}}}^{\mathrm{(D)}},$
$\displaystyle\boldsymbol{\mathcal{T}}_{\mathrm{D}}$
$\displaystyle=\frac{1}{2}\big{(}\boldsymbol{\rho}_{\hat{\mathbf{v}}\otimes(\hat{\mathbf{p}}-q_{\mathrm{e}}\mathbf{A}_{\Re})}^{\mathrm{(D)}}+\boldsymbol{\rho}_{(\hat{\mathbf{p}}-q_{\mathrm{e}}\mathbf{A}_{\Re})\otimes\hat{\mathbf{v}}}^{\mathrm{(D)}}\big{)}.$
(147)

Here $\hat{\mathbf{v}}=c\boldsymbol{\alpha}_{\mathrm{F}}$ is the effective
velocity operator for the Dirac field, obtained in the Heisenberg picture from
the time dependence of the position operator as
$\hat{\mathbf{v}}=\frac{\partial\mathbf{r}}{\partial
t}=\frac{i}{\hbar}[\hat{H},\mathbf{r}]=c\boldsymbol{\alpha}_{\mathrm{F}}$. The
Dirac field Hamiltonian operator $\hat{H}$ used here is given in Eq. (136).

####  VII.3.2 SEM tensor of the electromagnetic field

Using Eq. (141) and the traceless property of the SEM tensor of the
electromagnetic field, we obtain the SEM tensor of the electromagnetic field,
given by

$\displaystyle T_{\mathrm{M}}^{\mu\nu}$
$\displaystyle=\bar{\Psi}_{\Re}\mathbf{t}^{\mu}\boldsymbol{\gamma}_{\mathrm{B}}^{\nu}\boldsymbol{\gamma}_{\mathrm{B}}^{5}\Psi_{\Re}=\left[\begin{array}[]{cccc}W_{\mathrm{M}}&c\mathbf{G}_{\mathrm{M}}\\\
c\mathbf{G}_{\mathrm{M}}&\boldsymbol{\mathcal{T}}_{\mathrm{M}}\end{array}\right]$
(150)
$\displaystyle=\frac{1}{\mu_{0}}\Big{(}F_{\;\,\rho}^{\mu}F^{\rho\nu}+\frac{1}{4}\eta^{\mu\nu}F_{\rho\sigma}F^{\rho\sigma}\Big{)}.$
(151)

In terms of the electric and magnetic fields, the components of the SEM tensor
$T_{\mathrm{M}}^{\mu\nu}$ in Eq. (151) are given by

$\displaystyle W_{\mathrm{M}}$
$\displaystyle=\frac{1}{2}\Big{(}\varepsilon_{0}\mathbf{E}_{\Re}^{2}+\frac{1}{\mu_{0}}\mathbf{B}_{\Re}^{2}\Big{)},$
$\displaystyle\mathbf{G}_{\mathrm{M}}$
$\displaystyle=\varepsilon_{0}\mathbf{E}_{\Re}\times\mathbf{B}_{\Re},$
$\displaystyle\boldsymbol{\mathcal{T}}_{\mathrm{M}}$
$\displaystyle=W_{\mathrm{M}}\mathbf{I}_{3}-\varepsilon_{0}\mathbf{E}_{\Re}\otimes\mathbf{E}_{\Re}-\frac{1}{\mu_{0}}\mathbf{B}_{\Re}\otimes\mathbf{B}_{\Re}.$
(152)

The SEM tensor $T_{\mathrm{M}}^{\mu\nu}$ in Eq. (151) is equal to the well-
known symmetric SEM tensor of the electromagnetic field Jackson1999 ;
Landau1989 .

For time-harmonic fields, the time averages of the energy and momentum density
components $W_{\mathrm{M}}$ and $\mathbf{G}_{\mathrm{M}}$ in Eq. (152) over
the harmonic cycle, denoted by the brackets with a subscript $\omega$, are
related to the expectation value densities in Eqs. (97), (98), and (100) as
$\langle W_{\mathrm{M}}\rangle_{\omega}=\rho_{\hat{H}}^{\mathrm{(M)}}$ and
$\langle\mathbf{G}_{\mathrm{M}}\rangle_{\omega}=\boldsymbol{\rho}_{\hat{\mathbf{p}}}^{\mathrm{(M)}}+\frac{1}{2}\nabla\times\boldsymbol{\rho}_{\hat{\mathbf{S}}}^{\mathrm{(M)}}$.
Apart from the need of the harmonic cycle time average on the left-hand sides,
these equations are analogous to those in the case of the Dirac field in Eq.
(147).

###  VII.4 Summary of the generating Lagrangian density of gravity and its
special unitary symmetry

We expect that the eight-spinor formulation of the standard model is necessary
for the unification of gravity into it. This conclusion is based on the
surprising simplicity of how all the fermionic and bosonic fields of the
standard model can be coupled to the gravity in the gauge theory Partanen2023c
arising from the SU(8)4D symmetry presented in this work. The decades of work
on this problem without an unambiguous solution also suggest that the
formulation of the quantum field theory of gravity using the standard model
without eight-spinor structures is extremely difficult if not impossible.

In the derivation of the SEM tensor using the generating Lagrangian density of
gravity, the trace-reversed symmetric SEM tensor follows directly from the
internal symmetry of the generating Lagrangian density of gravity. This is in
accordance with Noether’s theorem, which states that each generator of a
continuous symmetry is associated with a conserved current Noether1918 ;
Zee2010 . In the present case, the conserved currents of the four symmetry
generators $\mathbf{t}^{\mu}$ combine to a single SEM tensor and the trace-
reverse is related to how the source term appears in the Yang-Mills gauge
theory of gravity Partanen2023c . Using the internal special unitary symmetry
SU(8)4D instead of the external space-time symmetry differs from the
conventional Lagrangian derivation of the SEM tensor Landau1989 ; Bliokh2014b
. In the conventional derivation Landau1989 , the SEM tensor follows from the
external space-time symmetry of the action and it is asymmetric if no
additional symmetrization procedures are introduced, such as the Belinfante-
Rosenfeld symmetrization Schroder1968 ; Ramos2015 ; Belinfante1940 ;
Rosenfeld1940 ; Weinberg1995 . This observation is one of the foundations for
the development of the Yang-Mills gauge theory of gravity in Ref.
Partanen2023c .

We have shown how the SEM tensor acting as the source term in the Yang-Mills
gauge theory of gravity arises from the special unitary symmetry, but have
left the description of the tensor gauge field to a separate work
Partanen2023c . Here we briefly note that the tensor gauge field is a Lorentz-
invariant tensor whose representation as a $4\times 4$ matrix is invariant in
the Lorentz transformation of second-rank tensors, and whose representation in
terms of $8\times 8$ matrices Partanen2023c  is invariant in the Lorentz
transformation of spin-2 fields, given in Sec III.5. The different Lorentz
transformation properties of vector and tensor gauge fields are one indication
why the Yang-Mills gauge theory of gravity cannot be derived from the
conventional Lagrangian density, but the generating Lagrangian density of
gravity is needed. The conventional Lagrangian density follows from the
gravitational-gauge-invariant form of the Lagrangian density of the Yang-Mills
gauge theory of gravity in the Minkowski metric limit Partanen2023c . Now, we
end our discussion on the generating Lagrangian density of gravity and its
special unitary symmetry and move to a brief representation of the second
quantization of the electromagnetic field in the present eight-spinor
formalism.

##  VIII Second quantization of spinorial electromagnetic field

The goal of this section is to give the reader a concise idea of how the
definitions of the key quantum operators and their matrix elements in the
spinorial form transform into the pertinent quantities in the conventional
QED. The quantization of the photon spinor field form $\Psi_{\Re}$, made of
real-valued fields, follows trivially from the conventional QED Loudon2000 .
It is obtained by substituting the conventional Hermitian electric and
magnetic field operators in places of the electric and magnetic fields in the
photon spinor in Eq. (1). Similarly, the charge-current spinor $\Phi_{\Re}$ is
quantized by substituting the conventional Dirac field operator into Eq. (124)
or the corresponding components of the electric four-current density into Eq.
(1).

This section is devoted to the quantization of the complex-valued photon
spinor field form $\Psi$, which has the role of the photon wavefunction in the
present theory. Since $\Psi$ is complex valued, the corresponding operator is
non-Hermitian in analogy to the Dirac field operator. We demonstrate the
usability of the photon-spinor-field operator in the calculation of the
correct second-quantized operators of physical quantities in a way analogous
to the Dirac theory. We also compare the single-photon states in the second
quantization to the first-quantized single-photon states presented in Sec. V
and in Appendix A.

###  VIII.1 Photon-spinor-field operator

Following the conventional quantization procedure of fields in the plane-wave
basis Peskin2018 ; Loudon2000 , the photon-spinor-field operator is given in
the Heisenberg picture by

$\hat{\Psi}=\sum_{q}\int\frac{V}{(2\pi)^{3}}\frac{1}{\sqrt{2\hbar\omega_{\mathbf{k}}}}\Psi_{\mathbf{k},q}(\hat{a}_{\mathbf{k},q}+\hat{a}_{\mathbf{k},q}^{\dagger})d^{3}k.$
(153)

Here $\hat{a}_{\mathbf{k},q}$ and $\hat{a}_{\mathbf{k},q}^{\dagger}$ are the
conventional photon annihilation and creation operators, respectively, and $V$
is the quantization volume. The operators $\hat{a}_{\mathbf{k},q}$ and
$\hat{a}_{\mathbf{k},q}^{\dagger}$ satisfy the well-known bosonic commutation
relations, given by

$\displaystyle[\hat{a}_{\mathbf{k},q},\hat{a}_{\mathbf{k}^{\prime},q^{\prime}}^{\dagger}]$
$\displaystyle=\frac{(2\pi)^{3}}{V}\delta_{q,q^{\prime}}\delta(\mathbf{k}-\mathbf{k}^{\prime}),$
$\displaystyle[\hat{a}_{\mathbf{k},q},\hat{a}_{\mathbf{k}^{\prime},q^{\prime}}]$
$\displaystyle=[\hat{a}_{\mathbf{k},q}^{\dagger},\hat{a}_{\mathbf{k}^{\prime},q^{\prime}}^{\dagger}]=0.$
(154)

For the action of the photon annihilation and creation operators on photon
states of the second quantization, see Sec. VIII.3 below.

We define the commutator of two field operators of the form in Eq. (153) so
that there is a scalar product between the spinor terms $\Psi_{\mathbf{k},q}$,
whence the commutator applies to the annihilation and creation operators only.
Consequently, since the sum
$\hat{a}_{\mathbf{k},q}+\hat{a}_{\mathbf{k},q}^{\dagger}$ is a Hermitian
operator, the equal-time commutation relations for $\hat{\Psi}$ and
$\hat{\Psi}^{\dagger}$ are all zero as

$[\hat{\Psi}(\mathbf{r}),\hat{\Psi}^{\dagger}(\mathbf{r}^{\prime})]=[\hat{\Psi}(\mathbf{r}),\hat{\Psi}(\mathbf{r}^{\prime})]=[\hat{\Psi}^{\dagger}(\mathbf{r}),\hat{\Psi}^{\dagger}(\mathbf{r}^{\prime})]=0.\\!\\!$
(155)

It also follows that the operators $\hat{\Psi}^{(+)}$ and $\hat{\Psi}^{(-)}$,
which are the annihilation operator part of $\hat{\Psi}$ and the creation
operator part of $\hat{\Psi}^{\dagger}$, respectively, satisfy the canonical
equal-time commutation relations, given by

$\displaystyle[\hat{\Psi}^{(+)}(\mathbf{r}),\hat{\Psi}^{(-)}(\mathbf{r}^{\prime})]$
$\displaystyle=\delta(\mathbf{r}-\mathbf{r}^{\prime}),$
$\displaystyle[\hat{\Psi}^{(+)}(\mathbf{r}),\hat{\Psi}^{(+)}(\mathbf{r}^{\prime})]$
$\displaystyle=[\hat{\Psi}^{(-)}(\mathbf{r}),\hat{\Psi}^{(-)}(\mathbf{r}^{\prime})]=0.$
(156)

The photon-spinor-field operator in Eq. (153) can also be expressed in the
spherical state basis, presented in Appendix A. In this basis, it is written
as

$\hat{\Psi}=\\!\sum_{J,M,\eta}\int_{0}^{\infty}\\!\\!\\!\\!\frac{2V\omega^{2}}{\pi
c^{3}\sqrt{2\hbar\omega}}\Psi_{\omega,J,M}^{(\eta)}(\hat{a}_{\omega,J,M}^{(\eta)}+\hat{a}_{\omega,J,M}^{(\eta){\dagger}})d\omega.$
(157)

Here the quantum number $J$ ranges from 1 to infinity, $M$ ranges from $-J$ to
$J$, and the state parameter $\eta$ has values $\mathrm{e}$ and $\mathrm{m}$
for the states associated with the electric and magnetic transitions,
respectively.

The bosonic commutation relations of the photon annihilation operators
$\hat{a}_{\omega,J,M}^{(\eta)}$ and creation operators
$\hat{a}_{\omega,J,M}^{(\eta){\dagger}}$ in the spherical basis are given by

$\displaystyle[\hat{a}_{\omega,J,M}^{(\eta)},\hat{a}_{\omega^{\prime},J^{\prime},M^{\prime}}^{(\eta^{\prime}){\dagger}}]$
$\displaystyle=\frac{\pi
c^{3}}{2V\omega^{2}}\delta_{\eta,\eta^{\prime}}\delta_{J,J^{\prime}}\delta_{M,M^{\prime}}\delta(\omega-\omega^{\prime}),$
$\displaystyle[\hat{a}_{\omega,J,M}^{(\eta)},\hat{a}_{\omega^{\prime},J^{\prime},M^{\prime}}^{(\eta^{\prime})}]$
$\displaystyle=[\hat{a}_{\omega,J,M}^{(\eta){\dagger}},\hat{a}_{\omega^{\prime},J^{\prime},M^{\prime}}^{(\eta^{\prime}){\dagger}}]=0.$
(158)

Another standard approach for the quantization of the electromagnetic field is
to use wavepacket states and the corresponding wavepacket creation and
annihilation operators Loudon2000 . This approach to the quantization of
$\Psi$ is also possible.

###  VIII.2 Second-quantized operators

Next, we define quantum operators for the electromagnetic field in the second-
quantized multi-photon picture. In distinction to the operators in the first
quantization, we denote the second-quantized operators with an underline.
Then, the second-quantized operator $\underline{\hat{O}}$, which corresponds
to an arbitrary first-quantized operator $\hat{O}$, is given by

$\underline{\hat{O}}=\int:\\!\hat{\Psi}^{\dagger}\hat{O}\hat{\Psi}\\!:d^{3}r.$
(159)

Here the colons denote the conventional normal ordering of operators, which
removes the infinite constant term that otherwise arises from the commutator
of the annihilation and creation operators Loudon2000 .

Second-quantized operators obtain simple forms in the eigenbasis of the
pertinent first-quantized operators. Therefore, as examples of second-
quantized operators, we present selected operators using the pertinent
eigenbases. The operators are obtained by substituting the first-quantized
pertinent operators from Sec. IV and the photon-spinor-field operator from
either Eq. (153) or (157) into Eq. (159). Using the normal ordering relations
of bosonic operators, given as an example in the plane-wave basis by
$:\hat{a}_{\mathbf{k},q}^{\dagger}\hat{a}_{\mathbf{k},q}:=\hat{a}_{\mathbf{k},q}^{\dagger}\hat{a}_{\mathbf{k},q}=\hat{n}_{\mathbf{k},q}$
and
$:\hat{a}_{\mathbf{k},q}\hat{a}_{\mathbf{k},q}^{\dagger}:=\hat{a}_{\mathbf{k},q}^{\dagger}\hat{a}_{\mathbf{k},q}=\hat{n}_{\mathbf{k},q}$,
where
$\hat{n}_{\mathbf{k},q}=\hat{a}_{\mathbf{k},q}^{\dagger}\hat{a}_{\mathbf{k},q}$
is the single-mode number operator, after some algebra, we obtain the
conventional results

$\underline{\hat{n}}=\sum_{q}\int\frac{V}{(2\pi)^{3}}\hat{n}_{\mathbf{k},q}d^{3}k=\sum_{J,M,\eta}\int_{0}^{\infty}\frac{2V\omega^{2}}{\pi
c^{3}}\hat{n}_{\omega,J,M}^{(\eta)}d\omega,$ (160)
$\displaystyle\underline{\hat{H}}$
$\displaystyle=\sum_{q}\int\frac{V}{(2\pi)^{3}}\hbar\omega_{\mathbf{k}}\hat{n}_{\mathbf{k},q}d^{3}k$
$\displaystyle=\sum_{J,M,\eta}\int_{0}^{\infty}\frac{2V\omega^{2}}{\pi
c^{3}}\hbar\omega\hat{n}_{\omega,J,M}^{(\eta)}d\omega,$ (161)
$\underline{\hat{\mathbf{p}}}=\sum_{q}\int\frac{V}{(2\pi)^{3}}\hbar\mathbf{k}\hat{n}_{\mathbf{k},q}d^{3}k,$
(162) $\underline{\hat{\mathfrak{h}}}=\sum_{q}\int\frac{V}{(2\pi)^{3}}\hbar
q\hat{n}_{\mathbf{k},q}d^{3}k.$ (163)
$\underline{\hat{J}^{2}}=\sum_{J,M,\eta}\int_{0}^{\infty}\frac{2V\omega^{2}}{\pi
c^{3}}\hbar^{2}J(J+1)\hat{n}_{\omega,J,M}^{(\eta)}d\omega,$ (164)
$\underline{\hat{J}_{z}}=\sum_{J,M,\eta}\int_{0}^{\infty}\frac{2V\omega^{2}}{\pi
c^{3}}\hbar M\hat{n}_{\omega,J,M}^{(\eta)}d\omega,$ (165)

The second-quantized square of the spin operator is trivially given by
$\underline{\hat{S}^{2}}=\hbar^{2}S(S+1)\underline{\hat{n}}$. The conventional
results in Eqs. (160)–(165) strongly indicate the correctness of the present
formalism since the results predicted by these conventional operators agree
with ample experimental evidence.

Note that first-quantized density operators defined through Eq. (92) can also
be used in place of $\hat{O}$ in Eq. (159). Thus, Eq. (159) also defines the
density operators in the second quantization picture. For example, using the
first-quantized number-density operator
$\hat{\rho}_{\hat{n}}(\mathbf{r},\mathbf{r}^{\prime})=\delta(\mathbf{r}-\mathbf{r}^{\prime})$,
following from Eqs. (70) and (92), we then obtain the second-quantized number
density operator as

$\underline{\hat{\rho}_{\hat{n}}}^{\\!\\!(M)}=:\\!\hat{\Psi}^{\dagger}\hat{\Psi}\\!:.$
(166)

In analogy to the Dirac theory, $\underline{\hat{\rho}_{\hat{n}}}^{\\!\\!(M)}$
is equal to the time component of the number-density-current four-vector
operator, which is given for photons by

$\underline{\hat{\mathcal{I}}_{\mathrm{M}}^{\mu}}=:\\!\hat{\bar{\Psi}}\boldsymbol{\gamma}_{\mathrm{B}}^{\mu}\hat{\Psi}\\!:.$
(167)

Using Eq. (159) with the pertinent first-quantized density operators obtaned
through Eq. (92), we obtain the second-quantized Hamiltonian, momentum,
orbital angular momentum, spin angular momentum, and helicity density
operators of the electromagnetic field as

$\underline{\hat{\rho}_{\hat{H}}}^{\\!\\!(M)}=:\\!\hat{\Psi}^{\dagger}\hat{H}\hat{\Psi}\\!:,$
(168)
$\textstyle\underline{\hat{\boldsymbol{\rho}}_{\hat{\mathbf{p}}}}^{\\!\\!(M)}=:\\!\frac{1}{2}[\hat{\Psi}^{\dagger}\hat{\mathbf{p}}\hat{\Psi}-(\hat{\mathbf{p}}\hat{\Psi}^{\dagger})\hat{\Psi}]\\!:,$
(169)
$\textstyle\underline{\hat{\boldsymbol{\rho}}_{\hat{\mathbf{L}}}}^{\\!\\!(M)}=:\\!\frac{1}{2}[\hat{\Psi}^{\dagger}\hat{\mathbf{L}}\hat{\Psi}-(\hat{\mathbf{L}}\hat{\Psi}^{\dagger})\hat{\Psi}]\\!:,$
(170)
$\textstyle\underline{\hat{\boldsymbol{\rho}}_{\hat{\mathbf{S}}}}^{\\!\\!(M)}=:\\!\hat{\Psi}^{\dagger}\hat{\mathbf{S}}\hat{\Psi}\\!:,$
(171)
$\underline{\hat{\rho}_{\hat{\mathfrak{h}}}}^{\\!\\!(M)}=:\\!\hat{\Psi}^{\dagger}\hat{\mathfrak{h}}\hat{\Psi}\\!:.$
(172)

The expectation values of these operators for second-quantized single-photon
states can be compared to the first-quantized density expectation values in
Eqs. (97)–(101). This comparison is discussed in Sec. VIII.3 below.

###  VIII.3 Photon states in the second quantization

The single-photon plane-wave Fock state is the momentum eigenstate that is a
continuum state written in the Heisenberg picture in terms of the plane-wave
spinorial wavefunctions $\Psi_{\mathbf{k},q}(t,\mathbf{r})$, presented in Sec.
V and in Appendix A, as

$|1_{\mathbf{k},q}\rangle=\int\Psi_{\mathbf{k},q}(0,\mathbf{r})|\mathbf{r}\rangle
d^{3}r.$ (173)

Similarly, we can write the single-photon spherical Fock state in terms of the
spherical state wavefunctions $\Psi_{\omega,J,M}^{(\eta)}(t,\mathbf{r})$,
presented in Appendix A, as

$|1_{\omega,J,M}^{(\eta)}\rangle=\int\Psi_{\omega,J,M}^{(\eta)}(0,\mathbf{r})|\mathbf{r}\rangle
d^{3}r.$ (174)

As conventional, many-photon states are formed as symmetric product states of
the single-photon states. The direct sum of all product states of different
numbers of photons then forms a Fock space.

In analogy to the standard QED, operating on a single-mode plane-wave vacuum
state $|0_{\mathbf{k},q}\rangle$, the creation operator
$\hat{a}_{\mathbf{k},q}^{\dagger}$ creates a single-photon Fock state
$|1_{\mathbf{k},q}\rangle$ with a wave vector $\mathbf{k}$ and the helicity
quantum number $q$ as
$|1_{\mathbf{k},q}\rangle=\sqrt{\hbar\omega_{\mathbf{k}}}\hat{a}_{\mathbf{k},q}^{\dagger}|0_{\mathbf{k},q}\rangle$,
and the annihilation operator $\hat{a}_{\mathbf{k},q}$ annihilates this state
as
$\hat{a}_{\mathbf{k},q}|1_{\mathbf{k},q}\rangle=\sqrt{\hbar\omega_{\mathbf{k}}}|0_{\mathbf{k},q}\rangle$.
Correspondingly, the operator $\hat{\Psi}^{(-)}$ creates a single-photon state
at position $\mathbf{r}$ as
$\hat{\Psi}^{(-)}|0_{\mathbf{r}}\rangle=|1_{\mathbf{r}}\rangle$. The operator
$\hat{\Psi}^{(+)}$ annihilates this state as
$\hat{\Psi}^{(+)}|1_{\mathbf{r}}\rangle=|0_{\mathbf{r}}\rangle$. Similar
relations apply to spherical photon states. The completeness relation for
single-particle states is given in the plane-wave and spherical state bases by
Peskin2018

$\displaystyle\hat{\mathbf{1}}_{\mathrm{1}}$
$\displaystyle=\sum_{q}\int\frac{V}{(2\pi)^{3}}\frac{1}{\hbar\omega_{\mathbf{k}}}|1_{\mathbf{k},q}\rangle\langle
1_{\mathbf{k},q}|d^{3}k$
$\displaystyle=\sum_{J,M,\eta}\int_{0}^{\infty}\frac{2V\omega^{2}}{\pi
c^{3}}\frac{1}{\hbar\omega}|1_{\omega,J,M}^{(\eta)}\rangle\langle
1_{\omega,J,M}^{(\eta)}|d\omega.$ (175)

This operator is an identity operator within the subspace of single-particle
states, and it is zero in the rest of the Hilbert space.

It is interesting to compare the single-photon expectation values of the
second-quantized density operators in Eqs. (168)–(172) to the first-quantized
density expectation values in Eqs. (97)–(101). For simple comparison, we do
not form narrow-frequency-band wavepacket states here but use the delta-
function-normalized Fock states. As an example, for the Hamiltonian density
operator expectation value for the single-photon plane-wave Fock state, we
obtain

$\frac{\langle
1_{\mathbf{k},q}|\underline{\hat{\rho}_{\hat{H}}}|1_{\mathbf{k},q}\rangle}{\langle
1_{\mathbf{k},q}|1_{\mathbf{k},q}\rangle}=\frac{\Psi_{\mathbf{k},q}^{\dagger}\hat{H}\Psi_{\mathbf{k},q}}{\int\Psi_{\mathbf{k},q}^{\dagger}\Psi_{\mathbf{k},q}d^{3}r}=\hbar\omega_{\mathbf{k}}\frac{\Psi_{\mathbf{k},q}^{\dagger}\Psi_{\mathbf{k},q}}{\int\Psi_{\mathbf{k},q}^{\dagger}\Psi_{\mathbf{k},q}d^{3}r}.$
(176)

In the calculation, we have used $\langle
1_{\mathbf{k},q}|\hat{a}_{\mathbf{k}^{\prime},q^{\prime}}^{\dagger}\hat{a}_{\mathbf{k}^{\prime\prime},q^{\prime\prime}}|1_{\mathbf{k},q}\rangle=\hbar\omega_{\mathbf{k}}[(2\pi)^{3}/V]^{2}\delta_{q,q^{\prime}}\delta_{q^{\prime},q^{\prime\prime}}\delta(\mathbf{k}-\mathbf{k}^{\prime})\delta(\mathbf{k}^{\prime}-\mathbf{k}^{\prime\prime})$.
The result of Eq. (176) corresponds to the first-quantized Hamiltonian density
expectation value in Eq. (97) if $\Psi_{\mathbf{k},q}$ is used in Eq. (97).
Corresponding relations apply to other density operators. Thus, the second-
quantized expectation value densities for single-photon states are equal to
the first-quantized expectation value densities as expected. The comparison of
the first-quantization and second-quantization density expectation values
shows that the spinorial photon wavefunction $\Psi$ is a rigorous physical
concept in understanding single-photon states.

##  IX Conclusions

In conclusion, we have presented QED based on an eight-dimensional spinorial
photon equation. The spinorial photon equation is equivalent to the full set
of Maxwell’s equations. Consequently, it is equivalent to several formulations
of electrodynamics, such as the most conventional three-vector-calculus,
electromagnetic-tensor-field representation, exterior algebra of differential
forms, and the space-time algebra Dressel2015 . It provides an elegant
representation of classical electrodynamics and QED, but it does not produce
new physics if no other elements are added to the theory.

In comparison to other known formulations of electrodynamics, the spinorial
photon equation is the most analogous to the Dirac equation. Instead of the
conventional $4\times 4$ Dirac gamma matrices, it is given in terms of
$8\times 8$ gamma matrices satisfying the Dirac algebra. In contrast to
$6\times 6$ gamma matrices that have been studies in some previous works
Mohr2010 ; Barnett2014 , the gamma matrices of the present theory allow the
description the physics of all Maxwell’s equations by a single equation, and
accordingly, _complete reformulation_ of QED using eight-spinors. Many
properties of the Dirac field are directly transferable to the eight-spinor
photon theory. For example, we have formulated the Lorentz transformations of
eight-spinors in analogy to the Lorentz transformation of Dirac spinors. As a
result, the relativistic quantum spin operators of light emerge naturally
through the generators of Lorentz transformations on eight-component photon
spinors. Therefore, our work provides a well-defined photon wavefunction and
field operator representations with a natural emergence of the relativistic
quantum spin structure of light.

The spinorial photon equation leads to the formulation of the Lagrangian
density of QED using eight-spinors. It enables the generating Lagrangian
density of gravity with internal special-unitary-symmetry-based coupling of
the quantum fields of the standard model to gravity. The internal special
unitary symmetry of the generating Lagrangian density of gravity was shown to
enable an elegant derivation of the symmetric SEM tensors of the
electromagnetic and Dirac fields. Due to the analogous forms of the underlying
theories, the present eight-spinor formalism can be extended to describe the
other fundamental interactions and the related fermionic and bosonic fields of
the standard model. The related fundamental consequence, the Yang-Mills gauge
theory of gravity, is studied in a separate work Partanen2023c .

###### Acknowledgements.

This work has been funded by the Academy of Finland under Contract No. 349971.
Wolfram Mathematica has been extensively used to verify the equations of the
present work.

##  Appendix A Photon spinor wavefunctions for plane-wave and spherical states

###  A.1 Plane wave wavefunctions

The photon spinor wavefunctions for the plane-wave states corresponding to the
wave vector $\mathbf{k}$ and the helicity quantum number $q$ are given by

$\Psi_{\mathbf{k},q}=\left[\begin{array}[]{c}0\\\
\boldsymbol{\mathcal{E}}_{\mathbf{k},q}\\\ 0\\\
i\boldsymbol{\mathcal{B}}_{\mathbf{k},q}\end{array}\right].$ (177)

Here the electric and magnetic three-vector plane-wave state components
$\boldsymbol{\mathcal{E}}_{\mathbf{k},q}$ and
$\boldsymbol{\mathcal{B}}_{\mathbf{k},q}$ are defined to absorb the
normalizing prefactors of Eq. (1), and they are written as

$\boldsymbol{\mathcal{E}}_{\mathbf{k},q}=\sqrt{\frac{\hbar\omega_{\mathbf{k}}}{2V}}\mathbf{e}_{\mathbf{k}}^{(q)}e^{i(\mathbf{k}\cdot\mathbf{r}-\omega_{\mathbf{k}}t)},$
(178)
$\boldsymbol{\mathcal{B}}_{\mathbf{k},q}=\frac{\nabla\times\boldsymbol{\mathcal{E}}_{\mathbf{k},q}}{ik}=\frac{\mathbf{k}}{k}\times\boldsymbol{\mathcal{E}}_{\mathbf{k},q}.$
(179)

Here the helicity quantum number $q=\pm 1$ and the wave vector $\mathbf{k}$
appearing as indices in the spherical polarization vector
$\mathbf{e}_{\mathbf{k}}^{(q)}$ correspond to a fixed polarization and wave
vector state of a photon. We define the spherical polarization vectors as
$\mathbf{e}_{\mathbf{k}}^{(0)}=\mathbf{R}_{\mathbf{k}}\cdot\hat{\mathbf{z}}$
and $\mathbf{e}_{\mathbf{k}}^{(\pm
1)}=\mathbf{R}_{\mathbf{k}}\cdot\frac{1}{\sqrt{2}}(\hat{\mathbf{x}}\pm
i\hat{\mathbf{y}})$, where $\hat{\mathbf{x}}$, $\hat{\mathbf{y}}$, and
$\hat{\mathbf{z}}$ are unit vectors parallel to the $x$, $y$ and $z$ axes,
respectively, and $\mathbf{R}_{\mathbf{k}}$ is a rotation matrix that rotates
$\hat{\mathbf{z}}$ parallel to the wave vector $\mathbf{k}$. The rotation
matrix $\mathbf{R}_{\mathbf{k}}$ is given by
$\mathbf{R}_{\mathbf{k}}=\mathbf{I}_{3}\cos\theta_{\mathbf{k}}+(\mathbf{s}\cdot\hat{\mathbf{a}})\sin\theta_{\mathbf{k}}+(\hat{\mathbf{a}}\otimes\hat{\mathbf{a}})(1-\cos\theta_{\mathbf{k}})$,
where $\theta_{\mathbf{k}}$ is the polar angle of the wave vector,
$\hat{\mathbf{a}}=(\hat{\mathbf{z}}\times\mathbf{k})/|\hat{\mathbf{z}}\times\mathbf{k}|$,
and $\mathbf{s}=(\mathbf{s}^{x},\mathbf{s}^{y},\mathbf{s}^{z})$ is a vector
made of the $3\times 3$ rotation generator matrices, given by

$\displaystyle\mathbf{s}^{x}=\left[\begin{array}[]{ccc}0&0&0\\\ 0&0&-1\\\
0&1&0\end{array}\right]\\!\\!,\hskip
2.84544pt\mathbf{s}^{y}=\left[\begin{array}[]{ccc}0&0&1\\\ 0&0&0\\\
-1&0&0\end{array}\right]\\!\\!,$ (186) $\displaystyle\hskip
42.67912pt\mathbf{s}^{z}=\left[\begin{array}[]{ccc}0&-1&0\\\ 1&0&0\\\
0&0&0\end{array}\right]\\!\\!.$ (190)

The plane-wave states are eigenstates of the helicity operator satisfying
$(\hat{\mathfrak{h}}/\hbar)[0,\boldsymbol{\mathcal{E}}_{\mathbf{k},q},0,i\boldsymbol{\mathcal{B}}_{\mathbf{k},q}]^{T}=q[0,\boldsymbol{\mathcal{E}}_{\mathbf{k},q},0,i\boldsymbol{\mathcal{B}}_{\mathbf{k},q}]^{T}=[0,i\boldsymbol{\mathcal{B}}_{\mathbf{k},q},0,\boldsymbol{\mathcal{E}}_{\mathbf{k},q}]^{T}$.
In the last form of this equation, we have pointed out that the action of the
helicity operator in Eq. (78) and the division by $\hbar$ is equivalent to
swapping of the upper and lower three-vector components in the spinorial
photon state. This result is in agreement with the helicity-chirality
equivalence for massless particles,
$\hat{\mathfrak{h}}/\hbar\leftrightarrow\boldsymbol{\gamma}_{\mathrm{B}}^{5}$,
discussed in Sec. II.5. From the point of view of the light source, the
helicity quantum number $q=1$ corresponds to clockwise or right-handed
circular polarization, while $q=-1$ corresponds to counter-clockwise or left-
handed circular polarization.

The normalization condition for the plane-wave photon states is written as

$\int\Psi_{\mathbf{k},q}^{\dagger}\Psi_{\mathbf{k}^{\prime},q^{\prime}}d^{3}r=\hbar\omega_{\mathbf{k}}\frac{(2\pi)^{3}}{V}\delta_{q,q^{\prime}}\delta(\mathbf{k}-\mathbf{k}^{\prime}).$
(191)

This kind of a normalization of one-particle states is known to be Lorentz
invariant Peskin2018 . The plane-wave photon states satisfy the space and time
reversal symmetry relation, given by

$\Psi_{\mathbf{k},q}(-t,-\mathbf{r})=-\boldsymbol{\gamma}_{\mathrm{B}}^{0}\Psi_{\mathbf{k},-q}^{*}(t,\mathbf{r}).$
(192)

This relation is useful in checking that the CPT symmetry is satisfied as
discussed in Sec. V.4.

###  A.2 Spherical state wavefunctions

The photon spinor wavefunctions for the spherical states are given by

$\Psi_{\omega,J,M}^{\mathrm{(e)}}=\left[\begin{array}[]{c}0\\\
\boldsymbol{\mathcal{E}}_{\omega,J,M}\\\ 0\\\
i\boldsymbol{\mathcal{B}}_{\omega,J,M}\end{array}\right],\hskip
11.38092pt\Psi_{\omega,J,M}^{\mathrm{(m)}}=\left[\begin{array}[]{c}0\\\
i\boldsymbol{\mathcal{B}}_{\omega,J,M}\\\ 0\\\
\boldsymbol{\mathcal{E}}_{\omega,J,M}\end{array}\right].$ (193)

The spinors $\Psi_{\omega,J,M}^{\mathrm{(e)}}$ and
$\Psi_{\omega,J,M}^{\mathrm{(m)}}$ are typically associated with the electric
and magnetic atomic transitions, respectively. These spinors are found to be
related to each other by the action of the helicity operator in Eq. (78) as
$\Psi_{\omega,J,M}^{\mathrm{(m)}}=(\hat{\mathfrak{h}}/\hbar)\Psi_{\omega,J,M}^{\mathrm{(e)}}$
and
$\Psi_{\omega,J,M}^{\mathrm{(e)}}=(\hat{\mathfrak{h}}/\hbar)\Psi_{\omega,J,M}^{\mathrm{(m)}}$.
This corresponds to the helicity-chirality equivalence for massless particles,
$\hat{\mathfrak{h}}/\hbar\leftrightarrow\boldsymbol{\gamma}_{\mathrm{B}}^{5}$,
discussed in Sec. II.5. The normalized electric and magnetic three-vector
spherical state components $\boldsymbol{\mathcal{E}}_{\omega,J,M}$ and
$\boldsymbol{\mathcal{B}}_{\omega,J,M}$ in Eq. (193) are written as

$\displaystyle\boldsymbol{\mathcal{E}}_{\omega,J,M}$
$\displaystyle=\sqrt{\frac{\hbar\omega}{2V}}\,e^{-i\omega
t}\Big{[}\sqrt{\frac{J}{2J+1}}j_{J+1}\Big{(}\frac{\omega
r}{c}\Big{)}\boldsymbol{Y}_{J,J+1,M}(\theta_{\mathbf{r}},\phi_{\mathbf{r}})$
$\displaystyle\hskip
11.38092pt-\sqrt{\frac{J+1}{2J+1}}j_{J-1}\Big{(}\frac{\omega
r}{c}\Big{)}\boldsymbol{Y}_{J,J-1,M}(\theta_{\mathbf{r}},\phi_{\mathbf{r}})\Big{]},$
(194) $\displaystyle\boldsymbol{\mathcal{B}}_{\omega,J,M}$
$\displaystyle=\frac{\nabla\times\boldsymbol{\mathcal{E}}_{\omega,J,M}}{i\omega/c}$
$\displaystyle=\sqrt{\frac{\hbar\omega}{2V}}\,e^{-i\omega
t}j_{J}\Big{(}\frac{\omega
r}{c}\Big{)}\boldsymbol{Y}_{J,J,M}(\theta_{\mathbf{r}},\phi_{\mathbf{r}})$
(195)

Here $j_{J}(kr)$ are the spherical Bessel functions of the first kind and
$\mathbf{Y}_{J,L,M}(\theta_{\mathbf{r}},\phi_{\mathbf{r}})$ are the vector
spherical harmonic functions, defined as Edmonds1974

$\displaystyle\mathbf{Y}_{J,L,M}(\theta_{\mathbf{r}},\phi_{\mathbf{r}})$
$\displaystyle=\sum_{n=-1}^{1}\langle L,M-n,1,n|J,M\rangle
Y_{L,M-n}(\theta_{\mathbf{r}},\phi_{\mathbf{r}})\mathbf{u}^{(n)}.$ (196)

The terms of this series are formed from the well-known Clebsch-Gordan
coefficients $\langle L,M-n,1,n|J,M\rangle$ Rose1957 , the scalar spherical
harmonic functions $Y_{L,M}(\theta_{\mathbf{r}},\phi_{\mathbf{r}})$, and the
spherical unit vector $\mathbf{u}^{(n)}$. The spherical unit vectors
$\mathbf{u}^{(n)}$ are defined as $\mathbf{u}^{(0)}=\hat{\mathbf{z}}$,
$\mathbf{u}^{(-1)}=\frac{1}{\sqrt{2}}(\hat{\mathbf{x}}-i\hat{\mathbf{y}})$,
and
$\mathbf{u}^{(+1)}=-\frac{1}{\sqrt{2}}(\hat{\mathbf{x}}+i\hat{\mathbf{y}})$.
For the spherical harmonic functions, we use the definition written in terms
of the associated Legendre polynomials $P_{L,M}(x)$ as

$Y_{L,M}(\theta_{\mathbf{r}},\phi_{\mathbf{r}})=\sqrt{\frac{2L+1}{4\pi}\frac{(L-M)!}{(L+M)!}}\,P_{L,M}(\cos\theta_{\mathbf{r}})e^{iM\phi_{\mathbf{r}}}.$
(197)

For the associated Legendre polynomials, we use the Condon-Shortley phase
convention Condon1970 . Thus, the Condon-Shortley phase is also included in
our definition of the spherical harmonic functions. The normalization
condition for the spherical photon states is written as

$\displaystyle\int\Psi_{\omega,J,M}^{(\eta){\dagger}}\Psi_{\omega^{\prime},J^{\prime},M^{\prime}}^{(\eta^{\prime})}d^{3}r$
$\displaystyle=\hbar\omega\frac{\pi
c^{3}}{2V\omega^{2}}\delta_{\eta,\eta^{\prime}}\delta_{J,J^{\prime}}\delta_{M,M^{\prime}}\delta(\omega-\omega^{\prime}).$
(198)

The spherical photon states satisfy the space and time reversal symmetry
relations, given by

$\Psi_{\omega,J,M}^{(\eta)}(t,-\mathbf{r})=(-1)^{J+1+\delta_{\eta,m}}\boldsymbol{\gamma}_{\mathrm{B}}^{0}\Psi_{\omega,J,M}^{(\eta)}(t,\mathbf{r}),$
(199)
$\Psi_{\omega,J,M}^{(\eta)}(-t,\mathbf{r})=(-1)^{M+1}\Psi_{\omega,J,-M}^{(\eta)*}(t,\mathbf{r}).$
(200)

These relations are useful in checking that the CPT symmetry is satisfied as
discussed in Sec. V.4.

That the two linearly independent vectors in Eq. (193) are needed for the
spherical states is manifested by the fact that the plane-wave states
$\Psi_{\mathbf{k},q}$ can be written as a sum over spherical photon states,
where both $\Psi_{\omega,J,M}^{\mathrm{(e)}}$ and
$\Psi_{\omega,J,M}^{\mathrm{(m)}}$ are needed, as Edmonds1974 ; Rose1957

$\displaystyle\Psi_{\mathbf{k},q}$
$\displaystyle=-i\sqrt{2\pi}\sum_{J=1}^{\infty}\sum_{M=-J}^{J}i^{J}\sqrt{2J+1}$
$\displaystyle\hskip 14.22636pt\times
D_{M,q}^{(J)}(\phi_{\mathbf{k}},\theta_{\mathbf{k}},-\phi_{\mathbf{k}})(\Psi_{\omega,J,M}^{(m)}+q\Psi_{\omega,J,M}^{\mathrm{(e)}}).$
(201)

Here $D_{M,q}^{(J)}(\phi_{\mathbf{k}},\theta_{\mathbf{k}},-\phi_{\mathbf{k}})$
are Wigner D-matrix elements corresponding to the three Eulerian angles that
are in our case formed from the components of the wave vector $\mathbf{k}$ in
the spherical coordinates, i.e.,
$\mathbf{k}=(k,\theta_{\mathbf{k}},\phi_{\mathbf{k}})$.

##  Appendix B Derivation of the electric four-current density in QED

Here we briefly present the derivation of the electromagnetic four-current
density using the free Dirac field term
$\mathcal{L}_{\mathrm{QED,0}}=\mathcal{L}_{\mathrm{QED,D}}$ of the
conventional QED Lagrangian density in Eq. (120). The Lagrangian density
$\mathcal{L}_{\mathrm{QED,0}}$ can be called the generating Lagrangian density
of QED. The derivation of the electric four-current density using
$\mathcal{L}_{\mathrm{QED,0}}$ highlights the complete analogy with the
derivation of the symmetric SEM tensor using the generating Lagrangian density
of gravity in Sec. VII. Thus, as the starting point, the generating Lagrangian
density of QED is given by

$\mathcal{L}_{\mathrm{QED,0}}=\textstyle\frac{i}{2}\hbar
c\bar{\psi}(\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\vec{\partial}_{\mu}-\reflectbox{$\vec{\reflectbox{$\partial$}}$}_{\mu}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu})\psi-
m_{\mathrm{e}}c^{2}\bar{\psi}\psi.$ (202)

The generating Lagrangian density of QED satisfies the global unitary symmetry
U(1). The unitary transformation associated with this symmetry is given by

$\psi\rightarrow U_{\mathrm{e}}\psi,\hskip 14.22636pt\text{where
}U_{\mathrm{e}}=e^{i\theta}.$ (203)

The infinitesimal variations of the Dirac field $\psi$ and its adjoint
$\bar{\psi}$ with respect to the symmetry transformation parameter $\theta$
are given by

$\delta\psi=i\psi\delta\theta,\hskip
14.22636pt\delta\bar{\psi}=-i\bar{\psi}\delta\theta.$ (204)

Using these infinitesimal variations, the variation of the generating
Lagrangian density $\mathcal{L}_{\mathrm{QED,0}}$ in Eq. (202) is written as

$\displaystyle\delta\mathcal{L}_{\mathrm{QED,0}}$
$\displaystyle=\textstyle\frac{i}{2}\hbar
c(\delta\bar{\psi})(\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\vec{\partial}_{\mu}-\reflectbox{$\vec{\reflectbox{$\partial$}}$}_{\mu}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu})\psi-
m_{\mathrm{e}}c^{2}(\delta\bar{\psi})\psi$ $\displaystyle\hskip
14.22636pt+\frac{i}{2}\hbar
c\bar{\psi}(\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\vec{\partial}_{\mu}-\reflectbox{$\vec{\reflectbox{$\partial$}}$}_{\mu}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu})(\delta\psi)-m_{\mathrm{e}}c^{2}\bar{\psi}(\delta\psi)$
$\displaystyle=-\hbar
c\bar{\psi}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\psi\partial_{\mu}\delta\theta$
$\displaystyle=-\frac{\hbar}{q_{\mathrm{e}}}J_{\mathrm{e\Re}}^{\mu}\partial_{\mu}\theta.$
(205)

In the last equality of Eq. (205), we have defined the electric four-current
density $J_{\mathrm{e\Re}}^{\mu}$ as

$J_{\mathrm{e\Re}}^{\mu}=q_{\mathrm{e}}c\bar{\psi}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\psi.$
(206)

The derivation of the electric four-current density above using Eqs.
(202)–(205) can be seen to be completely analogous to the derivation of the
trace-reversed symmetric SEM tensor using Eqs. (137)–(140).

For completeness, next we describe the derivation of the electromagnetic-
gauge-invariant Lagrangian density of QED from the generating Lagrangian
density of QED in Eq. (202) using the conventional gauge theory. In the case
of the gravitational coupling of the present theory, analogous steps are not
described in the present work, but they are left as a topic of the Yang-Mills
gauge theory of gravity in Ref. Partanen2023c .

The variation of the generating Lagrangian density of QED in Eq. (205) is
generally nonzero for a position- and time-dependent $\theta$, for which
$\partial_{\mu}\theta\neq 0$. Therefore, the generating Lagrangian density of
QED in Eq. (202) is not invariant with respect to the local form of the
symmetry transformation in Eq. (203) with $\theta$ depending on the position
and time. To promote the global symmetry of constant $\theta$ to a local
symmetry of space-time dependent $\theta(t,\mathbf{r})$, the partial
derivatives $\vec{\partial}_{\mu}$ and
$\reflectbox{$\vec{\reflectbox{$\partial$}}$}_{\mu}$ in the generating
Lagrangian density of QED in Eq. (202) are replaced by electromagnetic-gauge-
covariant derivatives, given by Eq. (122) as
$\vec{D}_{\mu}=\vec{\partial}_{\mu}+i\frac{q_{\mathrm{e}}}{\hbar}A_{\Re\mu}$
and
$\reflectbox{$\vec{\reflectbox{$D$}}$}_{\mu}=\reflectbox{$\vec{\reflectbox{$\partial$}}$}_{\mu}-i\frac{q_{\mathrm{e}}}{\hbar}A_{\Re\mu}$.
These derivatives bring the electromagnetic gauge potential $A_{\Re\mu}$ to
the theory.

The electromagnetic-gauge-covariant derivative transforms by the same unitary
transformation as the Dirac field itself in Eq. (203) as
$\vec{D}_{\mu}\psi\rightarrow U_{\mathrm{e}}\vec{D}_{\mu}\psi$. The
corresponding transformation for the Dirac adjoint is given by
$\bar{\psi}\reflectbox{$\vec{\reflectbox{$D$}}$}_{\mu}\rightarrow\bar{\psi}\reflectbox{$\vec{\reflectbox{$D$}}$}_{\mu}U_{\mathrm{e}}^{*}$.
That these transformations are satisfied requires that the gauge field
$A_{\Re\mu}$ transforms as
$A_{\Re\mu}\rightarrow(U_{\mathrm{e}}A_{\Re\mu}+\frac{i\hbar}{q_{\mathrm{e}}}\partial_{\mu}U_{\mathrm{e}})U_{\mathrm{e}}^{*}=A_{\Re\mu}-\frac{\hbar}{q_{\mathrm{e}}}\partial_{\mu}\theta$.
Replacing the partial derivative operators $\vec{\partial}_{\mu}$ and
$\reflectbox{$\vec{\reflectbox{$\partial$}}$}_{\mu}$ in the generating
Lagrangian density of QED in Eq. (202) by the electromagnetic-gauge-covariant
derivative operators $\vec{D}_{\mu}$ and
$\reflectbox{$\vec{\reflectbox{$D$}}$}_{\mu}$ makes the Lagrangian density
invariant with respect to the local space-time-dependent form of the U(1)
symmetry transformation in Eq. (203).

To construct the complete electromagnetic-gauge-invariant Lagrangian density,
one must also add an electromagnetic-gauge-invariant term that depends on the
gauge field $A_{\Re\mu}$ only. Utilizing the gauge theory, this is obtained
from the commutator of the electromagnetic-gauge-covariant derivatives. The
commutator
$[\vec{D}_{\mu},\vec{D}_{\nu}]=\frac{iq_{\mathrm{e}}}{\hbar}F_{\mu\nu}$
defines an antisymmetric field strength tensor $F_{\mu\nu}$, which corresponds
to Eq. (129), as

$F_{\mu\nu}=\partial_{\mu}A_{\Re\nu}-\partial_{\nu}A_{\Re\mu}.$ (207)

In the U(1) gauge symmetry transformation in Eq. (138), the field strength
tensor $F_{\mu\nu}$ transforms as $F_{\mu\nu}\rightarrow
U_{\mathrm{e}}F_{\mu\nu}U_{\mathrm{e}}^{*}=F_{\mu\nu}$. Using the gauge
theory, we obtain an electromagnetic-gauge-invariant Lagrangian density term,
which depends on the gauge field $A_{\Re\mu}$ only, as
$\mathcal{L}_{\mathrm{QED,M}}=-\frac{1}{4\mu_{0}}F_{\mu\nu}F^{\mu\nu}$, in
agreement with Eq. (120). The prefactor of $\mathcal{L}_{\mathrm{QED,M}}$ is
found from comparison of the resulting Euler-Lagrange equations of the
electromagnetic gauge field to Maxwell’s equations. The complete
electromagnetic-gauge-invariant generalization of the generating Lagrangian
density of QED in Eq. (202) is then given by

$\mathcal{L}_{\mathrm{QED}}=\frac{i\hbar
c}{2}\bar{\psi}(\boldsymbol{\gamma}_{\mathrm{F}}^{\mu}\vec{D}_{\mu}-\reflectbox{$\vec{\reflectbox{$D$}}$}_{\mu}\boldsymbol{\gamma}_{\mathrm{F}}^{\mu})\psi-
m_{\mathrm{e}}c^{2}\bar{\psi}\psi-\frac{1}{4\mu_{0}}F_{\mu\nu}F^{\mu\nu}.$
(208)

This equation is equivalent to Eq. (120). The electromagnetic gauge invariance
means that the Lagrangian density in Eq. (208) satisfies
$\delta\mathcal{L}_{\mathrm{QED}}=0$, when varied with respect to the symmetry
transformation parameter $\theta$. Thus, the addition of the gauge field
cancels the variation of the generating Lagrangian density of QED obtained
without the gauge field in Eq. (205).

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