# Nonlinear chiral kinetic theory

Kazuya Mameda  Department of Physics, Tokyo University of Science, Tokyo
162-8601, Japan  RIKEN iTHEMS, RIKEN, Wako 351-0198, Japan

###### Abstract

From quantum field theory, we derive the chiral kinetic theory involving
nonlinear quantum corrections coupled with spacetime-dependent electromagnetic
fields and fluid velocity gradients. An equilibrium Wigner function determined
by the kinetic equation verifies the nondissipativeness of the charge induced
by the magneto-vortical coupling. We reveal that this nonlinear chiral kinetic
theory is consistent with the one-loop Euler–Heisenberg effective theory,
indicating an indirect evidence of the trace anomaly in the kinetic theory. We
also argue a potential issue on the regularization, and demonstrate the
availability of the point-splitting regularization in the nonlinear chiral
kinetic theory.

††preprint: RIKEN-iTHEMS-Report-23

##  I Introduction

The chiral kinetic theory (CKT) is one of the prominent theoretical tools to
describe transport phenomena of massless degrees of freedom. In this
framework, a lot of transport phenomena are displayed with the Berry monopole
[1, 2, 3], as in the electron transport theory [4]. A significant advantage of
the CKT is the versatile applicability not only to heavy-ion collisions [5,
6], Weyl semimetal [7, 8] and neutrino physics [9, 10, 11], but also to the
photonic transport [12, 13, 14, 15]. The CKT has also inspired us to elucidate
many aspects in relativistic quantum transport, such as the Lorentz covariance
[16, 17, 18], collisional effects [18, 19, 20, 21], the mass corrections [22,
23, 24], the strong magnetic field limit [25, 26, 27, 28], the different
derivations [29, 30, 31, 32, 33, 34], and gravitational contributions [35, 36,
37, 38] (see also Ref. [39] and reference therein).

In spite of various developments, the usual CKT includes only the linear
quantum correction. One limitation of this linear CKT is found in the
transport phenomena induced by the nonlinear coupling of background fields. A
particular example belonging to this category is the charge density of chiral
fermions under external magnetic field and vortical field. Such an induced
charge is originally discovered from the diagrammatic computation based on the
linear response theory [40], and the agreement is found from the Dirac theory
of a rotating fermions (for instance, see Ref. [41]). Importantly, this charge
generation is believed to be originated from quantum anomaly, and thus to be
nondissipative [42]. Nevertheless, the nondissipativeness cannot be verified
within thermal field theory, including the linear response theory. Indeed, the
equilibration under magnetic field and rotation is subtle, since the
coexistence of these external fields generates the drift force playing a role
of an effective electric field. The kinetic theory based on the Wigner
function [43] would provide a field-theoretical manifestation of the
nondissipativeness, and thus the anomalous nature. In this direction, the off-
equilibrium formulation of the kinetic theory is required, beyond the near-
equilibrium studies [44, 28].

Another limitation of the linear CKT is uncovered in the trace anomaly of
quantum electrodynamics (QED), which is also the nonlinear quantum effect in
the kinetic theory. While the chiral anomaly is well known as a consequence of
the Berry curvature, it is unobvious how the trace anomaly is interpreted in
the kinetic description. An important clue to answer this question is the
consistency of the kinetic theory and quantum field theory. Particularly, the
CKT and the Euler–Heisenberg effective theory [45, 46] should inherit the same
QED properties, since both theories describe fermionic dynamics under
background electromagnetic fields. Such a consistency is also a guiding
principle in developing the CKT with nonlinear quantum corrections.

In this paper, based on quantum field theory, we formulate the nonlinear CKT,
i.e., the CKT involving the nonlinear quantum correction coupled with
spacetime-dependent electromagnetic and fluid velocity fields. For this
purpose, we derive the off-equilibrium Wigner function [43] in the
collisionless limit as a simple attempt. Although the equilibrium state is not
completely determined in the collisionless case, the frame-independence of the
Wigner function provides a strong constraint for the equilibrium [37]. From an
equilibrium Wigner function found in this way, we show the nondissipativeness
of the magneto-vortical transport found in Ref. [40]. We also find that the
nonlinear CKT yields transport phenomena consistent with the Euler–Heisenberg
effective theory. This consistency further elucidates the kinetic encoding of
the charge renormalization and the QED $\beta$-function, which is an indirect
evidence of the trace anomaly in the CKT.

As a striking difference from the linear CKT, the nonlinear CKT bears an
inevitable ultraviolet divergence to be properly regularized. In this paper,
we pose a potential issue on this regularization; the competent techniques,
such as Pauli–Villars regularization and dimensional regularization, are
incompatible with the CKT. Instead, we implements the point-splitting
regularization [47] in the nonlinear CKT. Despite the violation of the
translational invariance, this scheme is not only compatible with the Wigner
function, but also helpful in elucidating the consistency with the
Euler–Heisenberg theory.

This paper is organized as follows. In Sec. II, we derive the off-equilibrium
Wigner function at $O(\hbar^{2})$, except for the distribution function. In
Sec. III, analyzing the frame-dependence of the nonlinear CKT, we identify an
equilibrium Wigner function. In Sec. IV, we demonstrate the computational
manner of the momentum integral in the CKT, including the implementation of
the point-splitting regularization. In Sec. V, we evaluate the $O(\hbar^{2})$
contributions to the equilibrium charge current and energy-momentum tensor. In
Sec. VI, we show the consistency of the nonlinear CKT and the Euler–Heisenberg
theory. Section VII is devoted to the summary of this paper. We set $e=1$ in
this paper unless otherwise stated, and use the mostly negative Minkowski
metric.

##  II Nonlinear chiral kinetic theory

###  II.1 Transport equations

Based on quantum field theory, the transport theory is constructed from the
Dyson-Schwinger equation for the Green’s function. When we consider virtual
gauge fields, the corresponding equation for Dirac propagators yields the
collisional kinetic theory. This is important for pursuing the dynamical
evolution in practical systems. Nevertheless, since our present interest is to
formulate the kinetic theory with nonlinear quantum corrections, through this
paper, we only focus on the collisionless limit.

We consider the Dirac theory of fermion fields $\psi$ and $\bar{\psi}$ coupled
with an external electromagnetic field $A_{\mu}$. The two-point correlation
functions
$S_{\alpha\beta}^{<}(x,y):=\langle\bar{\psi}_{\beta}(y)\psi_{\alpha}(x)\rangle$
and
$S_{\alpha\beta}^{>}(x,y):=\langle\psi_{\alpha}(x)\bar{\psi}_{\beta}(y)\rangle$
obey

$D_{x,\mu}S^{<}(x,y)=S^{>}(x,y)\overleftarrow{D}_{x,\mu}=0$ (1)

with $D_{\mu}\psi(x):=(\partial_{\mu}+{\mathrm{i}}A_{\mu}/\hbar)\psi(x)$ and
$\bar{\psi}(x)\overleftarrow{D}_{\mu}:=\psi(x)(\overleftarrow{\partial}_{\mu}-{\mathrm{i}}A_{\mu}/\hbar)$.
Note that here we implicitly enclosed the Wilson line, which ensures the gauge
covariance of $S^{\gtrless}$. This is equivalent to define the gauge covariant
translation operator as $\psi(x+y):={\mathrm{e}}^{y\cdot D}\psi(x)$. Fourier-
transforming Eq. (1), we get the transport equation of the Wigner function

$W^{\gtrless}(x,p):=\int_{y}{\mathrm{e}}^{-{\mathrm{i}}p\cdot
y/\hbar}S^{\gtrless}(x-y/2,x+y/2)$ (2)

with $\int_{y}:=\int{\mathrm{d}}^{4}y$. The original transport equation of
$W^{\gtrless}(x,p)$ contains the full quantum effect, and can be expanded in
terms of $\hbar$ [43]. This expansion is the same as that in terms of the
spacetime gradient since $\hbar$ always accompanies a spacetime derivative.
The first nonlinear terms of $O(\hbar^{2})$ thus emerge together with the
second power of background electromagnetic fields and vortical fields, and
their derivatives. In the following analysis, we discuss only the lesser part
$W(x,p):=W^{<}(x,p)$, which describes the kinetic theory of fermions.

In four-dimensional spacetime, the Wigner function can be decomposed with the
basis of the Clifford algebra as

$W=\mathcal{F}+i\gamma^{5}\mathcal{P}+\gamma^{\mu}\mathcal{V}_{\mu}+\gamma^{5}\gamma^{\mu}\mathcal{A}_{\mu}+\tfrac{1}{2}\sigma^{\mu\nu}{\mathcal{S}}_{\mu\nu},$
(3)

where $\mathcal{F}$, $\mathcal{P}$, $\mathcal{V}_{\mu}$ $\mathcal{A}_{\mu}$
and $\mathcal{S}_{\mu\nu}$ are some coefficient fields dependent on $x^{\mu}$
and $p_{\mu}$. For the transport equation of chiral fermions, the right-handed
projection of $W(x,p)$ is decoupled (and so is the left-handed one) from other
channels. We denote this by

$\mathcal{R}(x,p):=\frac{1}{2}\mathrm{tr}[\gamma^{\mu}P_{\mathrm{R}}W(x,p)]$
(4)

with $P_{\mathrm{R}}:=\frac{1}{2}(1+\gamma^{5})$ and the trace is for the
spinor indices. The equations of motion for $\mathcal{R}^{\mu}$ are derived as
follows:

$\displaystyle(\Delta_{\mu}+\hbar^{2}P_{\mu})\mathcal{R}^{\mu}=0,$ (5)
$\displaystyle(p_{\mu}+\hbar^{2}Q_{\mu})\mathcal{R}^{\mu}=0,$ (6)
$\displaystyle\hbar\varepsilon_{\mu\nu\rho\sigma}\Delta^{\rho}\mathcal{R}^{\sigma}+4\Bigl{[}p_{[\mu}+\hbar^{2}Q_{[\mu}\Bigr{]}\mathcal{R}_{\nu]}=0.$
(7)

Here we defined
$X_{[\mu}Y_{\nu]}:=\frac{1}{2}(X_{\mu}Y_{\nu}-X_{\nu}Y_{\mu})$, the Levi-
Civita tensor with $\varepsilon^{0123}=1$ and the following differential
operators:

$\Delta_{\mu}=\partial_{\mu}-F_{\mu\lambda}\partial_{p}^{\lambda},\quad
P_{\mu}=\frac{1}{24}(\partial_{p}\cdot\partial)^{2}F_{\mu\nu}\partial_{p}^{\nu},\quad
Q_{\mu}=-\frac{1}{12}\partial_{p}\cdot\partial F_{\mu\nu}\partial^{\nu}_{p}.$
(8)

Contracting Eq. (7) with $p^{\nu}$ and using Eq. (6), we get the useful
equation

$p^{2}\mathcal{R}_{\mu}=\frac{\hbar}{2}\varepsilon_{\mu\nu\rho\sigma}p^{\nu}\Delta^{\rho}\mathcal{R}^{\sigma}+2\hbar^{2}p^{\nu}Q_{[\mu}\mathcal{R}_{\nu]}-\hbar^{2}p_{\mu}Q\cdot\mathcal{R}.$
(9)

Once $\mathcal{R}^{\mu}$ is determined from the above equations of motion, we
can compute physical quantities. By implementing the inverse Wigner
transformation of two point functions, the charge current, energy-momentum
tensor and spin tensor are expressed with $\mathcal{R}^{\mu}$, as follows:

$\displaystyle J^{\mu}(x,y)$ $\displaystyle=$ $\displaystyle
2\int_{p}{\mathrm{e}}^{{\mathrm{i}}p\cdot y/\hbar}\mathcal{R}^{\mu}(x,p),$
(10) $\displaystyle T^{\mu\nu}(x,y)$ $\displaystyle=$ $\displaystyle
2\int_{p}{\mathrm{e}}^{{\mathrm{i}}p\cdot
y/\hbar}\Bigl{[}p^{(\mu}\mathcal{R}^{\nu)}(x,p)+\hbar^{2}Q^{(\mu}\mathcal{R}^{\nu)}(x,p)\Bigr{]},$
(11) $\displaystyle S^{\mu\nu\rho}(x,y)$ $\displaystyle=$
$\displaystyle-2\hbar\,\varepsilon^{\mu\nu\rho\sigma}\int_{p}{\mathrm{e}}^{{\mathrm{i}}p\cdot
y/\hbar}\mathcal{R}_{\sigma}(x,p)$ (12)

with $\int_{p}:=\int\frac{{\mathrm{d}}^{4}p}{(2\pi)^{4}}$ and
$X_{(\mu}Y_{\nu)}:=\frac{1}{2}(X_{\mu}Y_{\nu}+X_{\nu}Y_{\mu})$. In Appendix A,
we derive Eqs. (10)-(12) from the two-point functions. In the usual analysis
with the Wigner function approach, the above quantities are defined in the
$y\to 0$ limit. However, this parameter $y$ plays a role of the ultraviolet
regulator when we implement the point-splitting regularization. For this
reason, hereafter we keep $y$ finite.

From these expressions (10)-(12), it is manifested that Eqs. (5)-(7)
correspond to the Ward identities which massless fermions should respect. The
first equation (5) is related to charge conservation, and thus interpreted as
the kinetic equation, which determines the distribution function in
$\mathcal{R}^{\mu}$. The latter two (6) and (7) imply the conformal invariance
and the Lorentz invariance (i.e., angular momentum conservation),
respectively. These two determine the off-equilibrium Wigner function, except
for the distribution function.

###  II.2 Solution up to $O(\hbar^{2})$

In the following, we look for the solution of Eqs. (6)-(7) and (9), with the
parametrization:

$\mathcal{R}^{\mu}=\mathcal{R}^{\mu}_{(0)}+\hbar\mathcal{R}^{\mu}_{(1)}+\hbar^{2}\mathcal{R}_{(2)}^{\mu}.$
(13)

For the latter computation of the nonlinear solution
$\mathcal{R}^{\mu}_{(2)}$, let us first briefly review the $O(\hbar^{0})$ and
$O(\hbar)$ parts [18]. The $O(\hbar^{0})$ solution is readily found from Eqs.
(6) and (9) as

$\mathcal{R}^{\mu}_{(0)}=2\pi\delta(p^{2})p^{\mu}f_{(0)},$ (14)

where $f_{(0)}$ is a function that satisfies $\delta(p^{2})p^{2}f_{(0)}=0$.
The delta function $\delta(p^{2})$ represents the on-shell condition of the
chiral fermion: $p^{2}=(p_{0})^{2}-|{\boldsymbol{p}}|^{2}=0$. This $f_{(0)}$
has both particle and antiparticle contributions. At equilibrium, $f_{(0)}$ is
the Fermi distribution function, with which the Wigner function
$\mathcal{R}^{\mu}_{(0)}$ reproduces the usual lesser propagator [48].

Let us solve the first-order part. Inserting the zeroth-order solution (14)
into Eq. (9), we get the first-order correction as

$\mathcal{R}^{\mu}_{(1)}=2\pi\delta(p^{2})\biggl{[}\widetilde{\mathcal{R}}^{\mu}_{(1)}-\frac{1}{p^{2}}\tilde{F}^{\mu\nu}p_{\nu}f_{(0)}\biggr{]}$
(15)

with
$\tilde{F}_{\mu\nu}=\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}F^{\rho\sigma}$.
The second term is apparently singular, but it accounts for the chiral anomaly
in the CKT. Also, we emphasize the existence of the first term, which is
admitted as long as it satisfies
$\delta(p^{2})p^{2}\widetilde{\mathcal{R}}^{\mu}_{(1)}=0$ and
$\delta(p^{2})p\cdot\widetilde{\mathcal{R}}_{(1)}=0$. This extra term is
determined from Eq. (7) at $O(\hbar)$, as follows:

$\begin{split}\widetilde{\mathcal{R}}_{\mu}^{(1)}\delta(p^{2})=\delta(p^{2})\biggl{[}p_{\mu}\frac{n\cdot\widetilde{\mathcal{R}}_{(1)}}{p\cdot
n}+\frac{\varepsilon_{\mu\nu\rho\sigma}p^{\rho}n^{\sigma}}{2p\cdot
n}\Delta^{\nu}f_{(0)}\biggr{]},\end{split}$ (16)

where we introduce an arbitrary vector field $n^{\mu}(x)$. Thus, the first
correction part is given by

$\mathcal{R}^{\mu}_{(1)}=2\pi\delta(p^{2})\biggl{[}p^{\mu}f_{(1)}+\biggl{(}\Sigma_{n}^{\mu\nu}\Delta_{\nu}-\frac{1}{p^{2}}\tilde{F}^{\mu\nu}p_{\nu}\biggr{)}f_{(0)}\biggr{]},$
(17)

where we define

$f_{(1)}:=\frac{n\cdot\widetilde{\mathcal{R}}^{(1)}}{p\cdot
n},\quad\Sigma_{n}^{\mu\nu}:=\frac{\varepsilon^{\mu\nu\rho\sigma}p_{\rho}n_{\sigma}}{2p\cdot
n}.$ (18)

This tensor $\Sigma_{n}^{\mu\nu}$ corresponds to the spin of chiral fermions
and $n^{\mu}$ is the degrees of freedom for the frame choice of the spin [16,
17]. It is worth mentioning that
$\delta(p^{2})p^{2}\widetilde{\mathcal{R}}^{\mu}_{(1)}=0$ implies
$\delta(p^{2})p^{2}f_{(1)}=0$. Such a nonsingular condition for $f_{(1)}$ is
important, in particular, when we determine the equilibrium form of $f_{(1)}$.
Also, $\delta(p^{2})p^{2}f_{(1)}=0$ ensures that the above solution (17)
fulfills Eqs. (6) and (9).

In a totally parallel manner, we can solve the second-order part
$\mathcal{R}^{\mu}_{(2)}$. The derivation is shown in Appendix B (see also
Ref. [37]). The result is

$\begin{split}\mathcal{R}_{\mu}^{(2)}&=2\pi\delta(p^{2})\biggl{[}p_{\mu}f_{(2)}+\biggl{(}\Sigma_{\mu\nu}^{u}\Delta^{\nu}-\frac{1}{p^{2}}\tilde{F}_{\mu\nu}p^{\nu}\biggr{)}f_{(1)}-\Sigma_{\mu\nu}^{u}\varepsilon^{\nu\rho\sigma\lambda}\Delta_{\rho}\frac{n_{\sigma}}{2p\cdot
n}\Delta_{\lambda}f_{(0)}\biggr{]}\\\
&\quad+\frac{2\pi}{p^{2}}\biggl{[}-p_{\mu}Q\cdot
p+2p^{\nu}Q_{[\mu}p_{\nu]}\biggr{]}f_{(0)}\delta(p^{2})\\\
&\quad+2\pi\frac{\delta(p^{2})}{p^{2}}\biggl{(}\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}p^{\nu}\Delta^{\rho}+\frac{p_{\mu}p^{\nu}}{p^{2}}\tilde{F}_{\nu\sigma}-\tilde{F}_{\mu\sigma}\biggr{)}\biggl{(}\Sigma^{\sigma\lambda}_{n}\Delta_{\lambda}-\frac{1}{p^{2}}\tilde{F}^{\sigma\lambda}p_{\lambda}\biggr{)}f_{(0)}\\\
&\quad+2\pi\frac{\delta(p^{2})}{p^{2}}\Sigma_{\mu\nu}^{u}\biggl{[}\Delta_{\alpha}\Sigma^{\alpha\nu}_{n}+\frac{n_{\alpha}}{p\cdot
n}\tilde{F}^{\alpha\nu}+\frac{1}{p^{2}}\tilde{F}^{\nu\lambda}p_{\lambda}\biggr{]}p\cdot\Delta
f_{(0)},\end{split}$ (19)

where $\Delta_{\mu}$ and $Q_{\mu}$ operate all on the right. Here, another
vector field $u^{\mu}$ and spin tensor $\Sigma^{\mu\nu}_{u}$ are introduced,
similarly to $n^{\mu}$ and $\Sigma^{\mu\nu}_{n}$ in $\mathcal{R}^{\mu}_{(1)}$.
The new factor $f_{(2)}$ is the second-order counterpart of $f_{(1)}$, and is
required to satisfy the nonsingular condition $\delta(p^{2})p^{2}f_{(2)}=0$.
For $u^{\mu}=n^{\mu}$, the above solution
$\mathcal{R}^{\mu}=\mathcal{R}^{\mu}_{(0)}+\hbar\mathcal{R}^{\mu}_{(1)}+\hbar^{2}\mathcal{R}^{\mu}_{(2)}$
can be recast in a simpler form. Then, $f_{{(0)}}$, $f_{{(1)}}$ and
$f_{{(2)}}$ in $\mathcal{R}^{\mu}$ are totally combined as the single function
$f:=f_{(0)}+\hbar f_{(1)}+\hbar^{2}f_{(2)}$, as are so in the gravitational
case [37]. Inserting this $\mathcal{R}^{\mu}$ into Eq. (5), we get the
$n^{\mu}$-dependent nonlinear chiral kinetic equation to determine the single
distribution function $f$. Such a structure is the same as the linear chiral
kinetic equation. For this reason, $u^{\mu}$ could be regarded as the degrees
of freedom for the Lorentz transformation. On the other hand, the above
interpretation of $u^{\mu}$ is inapplicable for $u^{\mu}\neq n^{\mu}$, and
thus the physical meaning of $u^{\mu}$ is not completely identified. To
address this problem, we should study the Lorentz transformation up to
$O(\hbar^{2})$ in quantum field theory [18]. Although this is an important
task to manifest the nonlinear-order side-jump effect [16, 17], we will
analyze it in a future publication. Hereafter, we call both $n^{\mu}$ and
$u^{\mu}$ the frame vectors.

##  III Equilibrium

###  III.1 Frame-dependence

As is well known, the CKT depends on the frame vectors $n^{\mu}$ and
$u^{\mu}$. Since the frames are auxiliary fields to obtain the solutions (17)
and (19), however, physical quantities should be independent of the frames,
and so is $\mathcal{R}^{\mu}$. On the other hand, the distribution function
depends on the frame [17]. In the linear CKT, the frame transformation law of
$f_{(1)}$ is determined by imposing $\mathcal{R}^{\mu}_{(1)}$ keeps frame-
independent [18]. Similarly, in the nonlinear CKT, we can compute the
transformation law of $f_{(2)}$ from the frame-independence of
$\mathcal{R}^{\mu}_{(2)}$ [37]. Let us first focus on the variation in terms
of $n^{\mu}$. Suppose that we take the transformation of the frame vector as
$n^{\mu}\to n^{\prime\mu}$. Then the corresponding transformation of the
distribution function is written as $f_{(1)}\to f_{(1)}+\delta_{n}f_{(1)}$,
$f_{(2)}\to f_{(2)}+\delta_{n}f_{(2)}$. It is worthwhile to mention that the
variations $\delta_{n}f_{{(1)},{(2)}}$ should be nonsingular because so are
$f_{{(1)},{(2)}}$. That is, we impose
$\delta(p^{2})p^{2}\delta_{n}f_{(1)}=\delta(p^{2})p^{2}\delta_{n}f_{(2)}=0$.

The frame-independence of $\mathcal{R}_{(1)}^{\mu}$ is represented as
$\mathcal{R}_{(1)}^{\mu}|_{n^{\prime}}-\mathcal{R}_{(1)}^{\mu}|_{n}=0$, where
$\mathcal{R}_{(1)}^{\mu}|_{n}$ is the Wigner function in Eq. (17) with a frame
$n^{\mu}$. From this equation, we determine the transformation law of
$f_{(1)}$, as follows: [17, 18]

$\begin{split}\delta_{n}f_{(1)}&=-\frac{n^{\mu}}{p\cdot
n}\Sigma_{\mu\nu}^{n^{\prime}}\Delta^{\nu}f_{(0)}+p^{2}\delta_{n}g_{(1)},\end{split}$
(20)

where $\delta_{n}g_{(1)}$ is a nonsingular scalar fulfills
$\delta(p^{2})p^{2}\delta_{n}g_{(1)}=0$. In the linear CKT, this
$\delta_{n}g_{(1)}$ can be ignored; such a term does not affect
$\mathcal{R}^{\mu}_{(1)}$. This is, however, not the case in the nonlinear
CKT. Indeed, from a similar but more complicated evaluation for
$\mathcal{R}^{\mu}_{(2)}$, we obtain the variation of $f_{(2)}$ as

$\begin{split}\delta_{n}f_{(2)}&=\Sigma_{\mu\nu}^{u}\biggl{[}\Delta^{\mu}\frac{\varepsilon^{\nu\rho\alpha\beta}n_{\alpha}n^{\prime}_{\beta}}{2p\cdot
n\,p\cdot
n^{\prime}}\Delta_{\rho}f_{(0)}-F^{\mu\nu}\delta_{n}g_{(1)}\biggr{]}\\\
&\quad+\frac{1}{p^{2}}\biggl{[}\Sigma^{u}_{\mu\nu}\Delta^{\mu}-\tilde{F}_{\mu\nu}\biggl{(}\frac{p^{\mu}}{p^{2}}-\frac{u^{\mu}}{p\cdot
u}\biggr{)}\biggr{]}\Sigma^{\nu\lambda}_{n^{\prime}}\frac{n_{\lambda}}{p\cdot
n}p\cdot\Delta f_{(0)}.\end{split}$ (21)

which involves $\delta_{n}g_{(1)}$. The same analysis can be performed for the
variation in terms of $u^{\mu}$. Then, we find $\delta_{u}f_{(1)}=0$ and

$\delta_{u}f_{(2)}=-\frac{u^{\mu}}{p\cdot
u}\Sigma^{u^{\prime}}_{\mu\nu}\biggl{[}\Delta^{\nu}f_{(1)}-\varepsilon^{\nu\rho\sigma\lambda}\Delta_{\rho}\frac{n_{\sigma}}{2p\cdot
n}\Delta_{\lambda}f_{(0)}+\frac{1}{p^{2}}\biggl{(}\Delta_{\alpha}\Sigma_{n}^{\alpha\nu}+\frac{n_{\alpha}}{p\cdot
n}\tilde{F}^{\alpha\nu}+\frac{1}{p^{2}}\tilde{F}^{\nu\lambda}p_{\lambda}\biggr{)}p\cdot\Delta
f_{(0)}\biggr{]}.$ (22)

###  III.2 Equilibrium Wigner function

Let us apply the above argument to the equilibrium solution of the nonlinear
CKT. In the collisionless case, the kinetic theory itself cannot generally
determine equilibrium. The frame transformation laws (20)-(22) however provide
strong constraints to fix the equilibrium distribution functions. To
illustrate this fact, let us here employ the equilibrium distribution function
so that the classical Wigner function (14) is reproduced as the well-known
form of the lesser Green’s function of free fermions, that is,

$\displaystyle
f_{(0)}=\epsilon(p_{0})\,n_{F}(-\mu+p\cdot\xi),\quad\partial_{\mu}\alpha-
F_{\mu\nu}\beta^{\nu}=0,\quad\partial_{\mu}\beta_{\nu}+\partial_{\nu}\beta_{\mu}=0,$
(23)

where we define $\epsilon(x):=\theta(x)-\theta(-x)$ with the step function
$\theta(x)$, and the Fermi distribution function
$n_{F}(x):=({\mathrm{e}}^{\beta x}+1)^{-1}$. The parameters $\alpha$ and
$\beta^{\mu}$ are defined as $\alpha=-\beta\mu$, $\beta^{\mu}=\beta\xi^{\mu}$
and $\xi\cdot\xi=1$ with chemical potential $\mu$, inverse temperature $\beta$
and fluid velocity $\xi^{\mu}$. The Wigner function $\mathcal{R}^{\mu}_{(0)}$
with this $f_{(0)}$ in fact solves the classical kinetic equation (5):
$\Delta\cdot\mathcal{R}_{(0)}=2\pi\delta(p^{2})f^{\prime}_{(0)}p^{\mu}(\partial_{\mu}\alpha+p^{\nu}\partial_{\mu}\beta_{\nu}-F_{\mu\nu}\beta^{\nu})=0$
with $f^{\prime}_{(0)}={\mathrm{d}}f_{(0)}(x)/{\mathrm{d}}x$ and
$x=\alpha+\beta\cdot p$.

Then, using the above $f_{(0)}$, we compute the transformation law of
$f_{(1)}$ and $f_{(2)}$. From Eq. (20), we obtain

$\begin{split}\delta_{n}f_{(1)}&=f^{\prime}_{(0)}\frac{1}{2}(\Sigma^{\nu\rho}_{n^{\prime}}-\Sigma^{\nu\rho}_{n})\partial_{\nu}\beta_{\rho}+p^{2}\biggl{[}\delta_{n}g_{(1)}-f_{(0)}^{\prime}\frac{\varepsilon^{\mu\nu\alpha\beta}n_{\alpha}n_{\beta}^{\prime}}{4p\cdot
np\cdot n^{\prime}}\partial_{\nu}\beta_{\rho}\biggr{]}.\end{split}$ (24)

The above equation holds when we choose

$f_{{(1)}}=f_{(0)}^{\prime}\frac{1}{2}\Sigma_{n}^{\mu\nu}\partial_{\mu}\beta_{\nu},\quad\delta_{n}g_{(1)}=f_{(0)}^{\prime}\frac{\varepsilon^{\mu\nu\alpha\beta}n_{\alpha}n_{\beta}^{\prime}}{4p\cdot
np\cdot n^{\prime}}\partial_{\nu}\beta_{\rho}.$ (25)

Similarly, the variations of $f_{(2)}$ are calculated as follows:

$\begin{split}\delta_{n}f_{(2)}&=\frac{1}{4}\Sigma_{\mu\nu}^{u}\Delta^{\mu}\varepsilon^{\nu\beta\rho\lambda}\,\biggl{(}\frac{n^{\prime}_{\beta}}{p\cdot
n^{\prime}}-\frac{n_{\beta}}{p\cdot
n}\biggr{)}\partial_{\rho}\beta_{\lambda}f^{\prime}_{(0)},\\\
\delta_{u}f_{(2)}&=\frac{1}{4}(\Sigma_{\mu\nu}^{u^{\prime}}-\Sigma_{\mu\nu}^{u})\Delta^{\mu}\varepsilon^{\nu\beta\rho\lambda}\frac{n_{\beta}}{p\cdot
n}\partial_{\rho}\beta_{\lambda}f^{\prime}_{(0)}.\end{split}$ (26)

We note that all singular terms with $(p^{2})^{-1}$ or $(p^{2})^{-2}$ in Eqs.
(21) and (22) disappear, thanks to $p\cdot\Delta f_{(0)}=0$. The above
equations indicate that the second-order quantum correction $f_{(2)}$ may be
deduced as

$f_{(2)}=\Sigma^{u}_{\mu\nu}\Delta^{\mu}\biggl{(}f^{\prime}_{(0)}\frac{\varepsilon^{\nu\rho\sigma\lambda}}{4\,p\cdot
n}n_{\rho}\partial_{\sigma}\beta_{\lambda}\biggr{)}+\phi_{(2)}.$ (27)

Here $\phi_{(2)}$ is a frame-independent term in the equilibrium distribution
function. Such an ambiguity in $f_{(2)}$ cannot be determined in the present
framework, which ignore the collisional effect.

At the equilibrium we found above, the Wigner function (19) is reduced. First,
we assume $\phi_{(2)}=0$ for simplicity. Plugging Eqs. (25) and (27) into
Eq.(19), one can show that the frame-dependence of $\mathcal{R}^{\mu}_{(2)}$
is totally compensated, as it should. Eventually, Eq. (19) is recast into the
four different pieces as $\mathcal{R}^{(2)}_{\mu}=\mathcal{R}^{(\partial
F)}_{\mu}+\mathcal{R}^{(FF)}_{\mu}+\mathcal{R}^{(F\omega)}_{\mu}+\mathcal{R}^{(\omega\omega)}_{\mu}$
with

$\displaystyle\mathcal{R}^{(\partial F)}_{\mu}$
$\displaystyle=2\pi\frac{\delta(p^{2})}{p^{2}}\cdot\frac{1}{12}\Biggl{[}p_{\mu}f_{(0)}\biggl{(}-\frac{8}{p^{2}}\biggr{)}p^{\rho}\partial^{\lambda}F_{\rho\lambda}+p_{\mu}f_{(0)}^{\prime}\biggl{(}\partial^{\rho}F_{\rho\lambda}\beta^{\lambda}-\frac{4}{p^{2}}p^{\rho}p\cdot\partial
F_{\rho\lambda}\beta^{\lambda}\biggr{)}$
$\displaystyle\qquad\qquad\quad+p_{\mu}f_{(0)}^{\prime\prime}\biggl{(}2p^{\rho}\beta\cdot\partial
F_{\rho\lambda}\beta^{\lambda}\biggr{)}+f_{(0)}\biggl{(}8\partial^{\lambda}F_{\mu\lambda}-\frac{8}{p^{2}}p\cdot\partial
F_{\mu\lambda}p^{\lambda}\biggr{)}$
$\displaystyle\qquad\qquad\quad+f^{\prime}_{(0)}\biggl{(}p\cdot\partial
F_{\mu\lambda}\beta^{\lambda}+p^{\nu}\partial_{\mu}F_{\nu\lambda}\beta^{\lambda}\biggr{)}+f_{(0)}^{\prime\prime}\biggl{(}-p^{2}\beta\cdot\partial
F_{\mu\lambda}\beta^{\lambda}\biggr{)}\Biggr{]}$ (28)
$\displaystyle\mathcal{R}^{(FF)}_{\mu}$
$\displaystyle=2\pi\frac{\delta(p^{2})}{(p^{2})^{2}}\,\cdot
2\biggl{(}-\frac{p_{\mu}p^{\nu}}{p^{2}}F_{\nu\sigma}+F_{\mu\sigma}\biggr{)}F^{\sigma\lambda}p_{\lambda}f_{(0)},$
(29) $\displaystyle\mathcal{R}^{(F\omega)}_{\mu}$
$\displaystyle=2\pi\frac{\delta(p^{2})}{p^{2}}\biggl{(}-p_{\mu}\frac{p^{\nu}p_{\rho}}{p^{2}}\omega_{\nu\sigma}\tilde{F}^{\sigma\rho}+\frac{3}{4}\omega_{\mu\sigma}\tilde{F}^{\sigma\nu}p_{\nu}+\frac{1}{4}\tilde{F}_{\mu\sigma}\omega^{\sigma\nu}p_{\nu}\biggr{)}f_{(0)},$
(30) $\displaystyle\mathcal{R}^{(\omega\omega)}_{\mu}$
$\displaystyle=2\pi\delta(p^{2})\cdot\frac{1}{4}\biggl{(}p_{\mu}\frac{p^{\nu}p^{\rho}}{p^{2}}\omega_{\nu\sigma}{\omega_{\rho}}^{\sigma}-\omega_{\mu\sigma}{\omega_{\nu}}^{\sigma}p^{\nu}\biggr{)}f_{(0)}^{\prime\prime},$
(31)

where we introduce
$\omega^{\mu\nu}:=\frac{\beta^{-1}}{2}\varepsilon^{\mu\nu\rho\sigma}\partial_{\rho}\beta_{\sigma}$.
We also note that the derivative of vorticity disappears, i.e.,
$\mathcal{R}^{\mu}_{(\partial\omega)}=0$, owing to the identity
$\partial_{\mu}\partial_{\nu}\beta_{\rho}=0$ for the Killing vector
$\beta_{\rho}$.

At this point, it is not guaranteed that the above $\mathcal{R}^{\mu}$ is
really an equilibrium Wigner function, because we have not yet analyzed the
$O(\hbar^{2})$ part of the kinetic equation (5) 111One can readily check that
the $O(\hbar)$ part of Eq. (5) holds for the linear-order solution (17). .
Plugging Eqs. (14) and (III.2)-(31) to the kinetic equation (5) and carrying
out a tedious computation, we arrive at

$\begin{split}\delta(p^{2})\biggl{[}\biggl{(}\frac{f^{\prime\prime}_{(0)}}{6p^{2}}\partial^{\mu}\beta^{\rho}p^{\nu}p_{\rho}-\frac{f^{\prime\prime}_{(0)}}{8}\partial^{\mu}\beta^{\nu}-\frac{f_{(0)}^{\prime\prime\prime}}{12}\partial^{\mu}\beta^{\rho}\beta^{\nu}p_{\rho}\biggr{)}\beta\cdot\partial+\frac{f_{(0)}^{\prime\prime}}{24}p^{\mu}\beta^{\nu}\beta^{\rho}\beta^{\sigma}\partial_{\rho}\partial_{\sigma}\biggr{]}F_{\mu\nu}=0.\end{split}$
(32)

Using
$\beta^{\rho}\beta^{\sigma}\partial_{\rho}\partial_{\sigma}F_{\mu\nu}=\beta\cdot\partial(\beta\cdot\partial
F_{\mu\nu})-(\beta\cdot\partial\beta_{\sigma})\partial^{\sigma}F_{\mu\nu}$, we
find that all the terms in the above kinetic equation contain
$\beta\cdot\partial F_{\mu\nu}$ or $\beta\cdot\partial\beta_{\mu}$. As long as
we consider a finite $F_{\mu\nu}$, hence, the above reduced kinetic equation
implies that either of the following conditions should be fulfilled: 222Note
that $\partial_{\mu}\beta_{\nu}=0$ is an equilibrium condition. In this case,
$\beta\cdot\partial F_{\mu\nu}=0$ automatically holds because of
$0=\partial_{[\mu}\partial_{\nu]}\alpha=\partial_{[\mu}(F_{\nu]\lambda}\beta^{\lambda})$.
This condition is however a special case of the condition (33a).

$\displaystyle 1)\quad\beta\cdot\partial
F_{\mu\nu}=0,\quad\beta\cdot\partial\beta_{\mu}=0,$ (33a) $\displaystyle
2)\quad\partial_{\lambda}F_{\mu\nu}=0.$ (33b)

These are the additional equilibrium conditions on top of those in Eq. (23).
The meaning of the condition (33a) is understandable when we take
$\xi^{\mu}=(1,\boldsymbol{0})$. The first equation in Eq. (33a) implies the
time-independence of background electromagnetic fields. The second means that
the background fluid has no acceleration, or equivalently, there is no
temperature gradient:
$0=\beta\cdot\partial\beta_{\mu}=-\beta\partial_{\mu}\beta$ with
$\beta:=\sqrt{\beta\cdot\beta}$. On the other hand, the acceleration term is
admitted under the condition (33b), where electromagnetic fields are constant.
This is the case employed in Ref. [44].

We here discuss the case with $\phi_{(2)}\neq 0$ in Eq. (27). One can readily
check that in this case the extra term $\delta(p^{2})p\cdot\Delta\phi_{(2)}$
emerges in the kinetic equation (32). However, the singular term with $p^{-2}$
cannot be eliminated by the $\phi_{(2)}$ term, since
$\delta(p^{2})\phi_{(2)}=0$ is required from the nonsingular condition
$\delta(p^{2})f_{(2)}=0$. Moreover, the other terms in Eq. (32) are not
canceled by the $\phi_{(2)}$ term. Hence,
$\delta(p^{2})p\cdot\Delta\phi_{(2)}=0$ is demanded. As the simplest choice,
we may take $\phi_{(2)}=0$ hereafter. This is a difference from the CKT in
curved spacetime; under a weak static gravitational field, a finite
$\phi_{(2)}$ is required for the realization of an equilibrium [37].

##  IV Momentum integral

###  IV.1 Regularization

The equilibrium physical quantities are computed as the momentum integral with
the Winger function in Eqs. (III.2)-(31) with the distribution function (23)
under the condition (33). Before the computation, we demonstrate how to
evaluate the momentum integrals. The integrals that we encounter in the
following section are generally written as

$\int_{p}2\pi\frac{{\mathrm{d}}^{l}\delta(p^{2})}{({\mathrm{d}}p^{2})^{l}}p^{\mu_{1}}\cdots
p^{\mu_{j}}\frac{{\mathrm{d}}^{k}f_{(0)}(p_{0})}{{\mathrm{d}}p_{0}^{k}}{\mathrm{e}}^{{\mathrm{i}}p\cdot
y/\hbar}$ (34)

with $f_{(0)}$ given by Eq. (23). Here we replaced the singular factor
$(p^{2})^{-l}$ in the Wigner functions with the derivative of $\delta(p^{2})$,
through the identity
$l!\delta(x)=(-x)^{l}{\mathrm{d}}^{l}\delta(x)/{\mathrm{d}}x^{l}$.

For the latter convenience, we here decompose Eq. (23) into the vacuum and
matter parts as
$f_{(0)}(p_{0})=f_{{(0)}\mathrm{vac}}(p_{0})+f_{{(0)}\mathrm{mat}}(p_{0})$
with $f_{{(0)}\mathrm{vac}}(p_{0}):=-\theta(-p_{0})$ and
$f_{{(0)}\mathrm{mat}}(p_{0}):=\theta(p_{0})n_{F}(p_{0}-\mu)+\theta(-p_{0})n_{F}(-p_{0}+\mu)$.
In Eq. (34) the former may result in the divergence at the ultraviolet regime
$p_{0}\sim-\infty$ unless $k\geq 1$. For this divergence, the parameter
$y^{\mu}$ plays a role of the cutoff scale. This is nothing but the point-
splitting regularization. On the other hand, the latter does not require such
a regulation. Therefore, in the following, we evaluate these two contributions
in different ways; for the vacuum contributions, we keep $y$ finite so that
the point-splitting regularization is implemented, but for the matter part we
take $y\to 0$ before integration 333The point-splitting regularization with
$n_{F}(p_{0}\mp\mu)$ would in principle be possible, but is not so easy as
that of the vacuum; due to the pole at $p_{0}=\pm\mu+{\mathrm{i}}(2n+1)\pi T$
for $n=0,\pm 1,\cdots$, it is nontrivial to perform the Wick rotation, which
is required in implementing the point-splitting regularization.. It should
also be emphasized that we face no infrared divergence in Eq. (34), thanks to
the cancellation of those from the vacuum and matter parts.

We comment on the regularization in the CKT. In usual quantum field theory,
when we regularize a divergent integral, it is preferred to choose a
regularization scheme to respect the gauge, Lorentz, and translational
invariances. It is, however, not so easy to find out such an appropriate
scheme for Eq. (34). For instance, the Pauli–Villars scheme is obviously
unsuitable, since the CKT possesses no mass parameter; a Pauli–Villars
regulator would be useful for the kinetic theory of massive fermions [22, 23,
24]. Dimensional regularization is also incompatible with the CKT, since
$\varepsilon^{\mu\nu\rho\sigma}$ and $\gamma^{5}$ cannot be extended
straightforwardly in a general $d$-dimensional spacetime [49]. Indeed, the
Wigner functions derived in Secs. II-III are no longer correct in $d\neq 4$
dimensions, for the following two reasons. First, the Clifford basis
decomposition (3) is unjustified in $d\neq 4$ dimensions. This implies that
our starting point at Eqs. (5)-(7) is modified. Second, the uselessness of the
Schouten identity in $d\neq 4$ dimensions brings a lot of extra singular terms
with $p^{-2}$ in intermediate steps of calculation. Then we would not derive
the solution that satisfies appropriate conditions, such as
$\delta(p^{2})p^{2}f_{(2)}=0$.

The above circumstance compels us to choose a regularization scheme that
sacrifices at least one symmetry. Among such schemes, the point-splitting
regularization is compatible with the Wigner function because the point-
splitting parameter is naturally introduced as $y^{\mu}$, as shown in the
charge current (10) and the energy-momentum tensor (11). This is the reason
why we employ the point-splitting regularization in this paper. Although this
scheme in general violates the translational invariance (namely,
$\partial_{\mu}T^{\mu\nu}+F^{\mu\nu}J_{\mu}\neq 0$), it can reveal the
consistency with the Euler–Heisenberg theory, as we discuss later. The
analysis with a more appropriate regularization will be shown in feature
publication.

###  IV.2 Matter part

We demonstrate how to compute the matter part in Eq. (34). We perform first
the integral in terms of $p_{0}$ and then $p_{i}$. In this way, by decomposing
each $p^{\mu}$ into the transverse component to
$\xi^{\mu}:=(1,\boldsymbol{0})$ and the longitudinal one, we can replace
integrands with nonvanishing tensor form; for instance, $p_{\alpha}\to
p_{0}\xi_{\alpha}$ and
$p_{\alpha}p_{\beta}\to(p_{0})^{2}\xi_{\alpha}\xi_{\beta}+\frac{{\boldsymbol{p}}^{2}}{3}\Delta_{\alpha\beta}$
with the transverse projector
$\Delta^{\mu\nu}:=\xi^{\mu}\xi^{\nu}-g^{\mu\nu}$. Performing the tensor
decomposition of the integrands, we express Eq. (34) as the linear combination
of

$\mathcal{I}^{l}_{n,m,k}:=\int_{p}2\pi\frac{{\mathrm{d}}^{l}\delta(p^{2})}{({\mathrm{d}}p^{2})^{l}}(p_{0})^{n}|{\boldsymbol{p}}|^{m-n}\frac{{\mathrm{d}}^{k}f_{{(0)}\mathrm{mat}}}{{\mathrm{d}}p_{0}^{k}}.$
(35)

In order to handle the derivative on $\delta(p^{2})$, we use the chain rule;
e.g.
$\frac{{\mathrm{d}}}{{\mathrm{d}}p^{2}}\delta(p^{2})=\frac{1}{2p_{0}}\frac{{\mathrm{d}}}{{\mathrm{d}}p_{0}}\delta(p^{2})$.
Then, the integration by parts in terms of $p_{0}$ removes the derivative on
$\delta(p^{2})$. It is worthwhile to notice that this step generates no
surface term because of $f_{{(0)}\mathrm{mat}}(p_{0}\to\pm\infty)=0$. In
Appendix C, we show the detailed evaluation. After this step, the integral
$\mathcal{I}^{l}_{n,m,k}$ is written as the linear combination of another
integral sequence

$\begin{split}\mathcal{J}_{m,k}&:=\int_{0}^{\infty}{\mathrm{d}}p\,p^{m}\frac{{\mathrm{d}}^{m}}{{\mathrm{d}}p^{m}}\Bigl{[}n_{F}(p-\mu)-(-1)^{a+b}n_{F}(p+\mu)\Bigr{]}.\end{split}$
(36)

There is an important remark about the above computation manner. In Eq. (35),
we have only the matter part $f_{{(0)}\mathrm{mat}}$ since the vacuum part is
evaluated with the point-splitting regularization. In some regularization
scheme, it is in principle possible to evaluate Eq. (35) including the vacuum
contribution. In this case, we replace $f_{{(0)}\mathrm{mat}}$ with
$f_{(0)}=f_{{(0)}\mathrm{vac}}+f_{{(0)}\mathrm{mat}}$ in the integrand and
evaluate the integral in the almost same manner. Only one difference is that
we carefully take into account the surface term contributions from the
$p_{0}$-integral. Such contributions always appear for $k=0$ due to the vacuum
contribution at ultraviolet regime: $f_{{(0)}}(p_{0}\to+\infty)=0$ but
$f_{{(0)}}(p_{0}\to-\infty)=-1$. Although Ref. [44] performs a similar
integration by parts, the above surface terms are missing.

###  IV.3 Vacuum part

Now we compute the vacuum contribution of Eq. (34) with the point-splitting
regularization. What we need to evaluate is

$\mathcal{K}_{n}^{\mu_{1}\cdots\mu_{m}}(y):=\int_{p}2\pi\frac{{\mathrm{d}}^{n}\delta(p^{2})}{({\mathrm{d}}p^{2})^{n}}p^{\mu_{1}}\cdots
p^{\mu_{m}}\bigl{[}-\theta(-p_{0})\bigr{]}{\mathrm{e}}^{{\mathrm{i}}p\cdot
y/\hbar}.$ (37)

It is efficient to first evaluate $\mathcal{K}_{1}$,
$\mathcal{K}_{2}^{\mu\nu}$ and $\mathcal{K}_{3}^{\mu\nu\rho\sigma}$, which
would lead to the logarithmic ultraviolet divergence without the point-
splitting. After the contour deformation to obtain an integral on the
Euclidean momentum phase space, we can evaluate these three integrals. As
shown in Appendix D, the result is as follows:

$\begin{split}\mathcal{K}_{1}(y)=-\frac{\mathcal{J}(y)}{8\pi^{2}},\quad\mathcal{K}_{2}^{\mu\nu}(y)=\frac{\mathcal{J}(y)}{16\pi^{2}}g^{\mu\nu},\quad\mathcal{K}_{3}^{\mu\nu\rho\sigma}(y)=-\frac{\mathcal{J}(y)}{32\pi^{2}}(g^{\mu\nu}g^{\rho\sigma}+g^{\mu\rho}g^{\nu\sigma}+g^{\mu\sigma}g^{\nu\rho}),\end{split}$
(38)

with the regularized integral:

$\mathcal{J}(y):=\int_{0}^{y^{-1}}\frac{{\mathrm{d}}p}{p}.$ (39)

We again emphasize that the infrared divergence at $p\sim 0$ are completely
canceled by those of the matter part.

All other types of integrals in Eq. (37) are generated by the derivative of
Eq. (38) with respect to $y^{\mu}$. It is important to remind then that in the
point-splitting regularization, we take the limit of $y\to 0$ symmetrically at
the end of evaluation, as follows [50]:

$\underset{y\to
0}{\mathrm{symm\,lim}}\,\frac{y^{\mu}}{y^{2}}=0,\qquad\underset{y\to
0}{\mathrm{symm\,lim}}\,\frac{y^{\mu}y^{\nu}}{y^{2}}=\frac{g^{\mu\nu}}{4}.$
(40)

Thanks to the first equation, for example, we readily find
$\mathcal{K}^{\mu}_{1}=-{\mathrm{i}}\hbar\partial^{\mu}_{y}\mathcal{K}_{1}\propto
y^{\mu}/y^{2}\to 0$ in this limit. Eventually, the integrals (37) other than
the three in Eq. (38) vanish in the following section.

##  V Equilibrium transport

We can now evaluate the charge current (10) and the energy-momentum tensor
(11), from the momentum integral (34). It is then convenient to introduce the
following four-vector fields:

$\begin{split}&B^{\mu}:=\tilde{F}^{\mu\nu}\xi_{\nu},\quad
E^{\mu}:=F^{\mu\nu}\xi_{\nu},\\\
&\omega^{\mu}:=\omega^{\mu\nu}\xi_{\nu}=\frac{1}{2}\beta^{-1}\varepsilon^{\mu\nu\rho\sigma}\xi_{\nu}\partial_{\rho}\beta_{\sigma},\quad
a^{\mu}:=\beta^{-1}\xi_{\nu}\partial^{\mu}\beta^{\nu}=\beta^{-1}\partial^{\mu}\beta.\end{split}$
(41)

Hereafter, we focus on the equilibrium cases described by either the condition
(33a) or (33b), on top of those in Eq. (23). Therefore, in the following
analysis, either $a_{\mu}$ or $\partial_{\mu}F_{\mu\nu}$ should vanish
depending on the choice of Eq. (33a) or (33b).

The classical and the first-order contributions can be evaluated with the
integral formulas in Appendix C. As derived in many literatures, we get [17]:

$\begin{split}J^{\mu}_{(1)}&=\frac{\mu}{4\pi^{2}}B^{\mu}+\biggl{(}\frac{\mu^{2}}{4\pi^{2}}+\frac{T^{2}}{12}\biggr{)}\omega^{\mu},\\\
T^{\mu\nu}_{(1)}&=\biggl{(}\frac{\mu^{2}}{4\pi^{2}}+\frac{T^{2}}{12}\biggr{)}B^{(\mu}\xi^{\nu)}+\biggl{(}\frac{\mu^{3}}{3\pi^{2}}+\frac{\mu
T^{2}}{3}\biggr{)}\omega^{(\mu}\xi^{\nu)},\end{split}$ (42)

which represent the chiral magnetic effect [51, 52, 53] and the chiral
vortical effect [54, 55, 56].

For the nonlinear-order contributions to Eqs. (10) and (11), we differently
evaluate the matter and vacuum part, with the help of the integral formulas in
Appendices. C and D, respectively. Since $\mathcal{R}^{\mu}_{(2)}$ is
decomposed into the four different pieces (III.2)-(31), so are the
corresponding charge current $J^{\mu}_{(2)}$ and energy-momentum tensor
$T^{\mu\nu}_{(2)}$. The resulting expressions are as follows:

$\begin{split}&J_{(\partial
F)}^{\mu}=-\frac{\mathcal{J}_{-1,0}-\mathcal{J}}{12\pi^{2}}\partial_{\lambda}F^{\mu\lambda},\\\
&J_{(FF)}^{\mu}=\frac{\mathcal{J}_{-1,1}}{12\pi^{2}}\biggl{[}\frac{1}{2}\xi^{\mu}(E^{2}+B^{2})+\varepsilon^{\mu\nu\rho\sigma}\xi_{\nu}E_{\rho}B_{\sigma}\biggr{]},\\\
&J_{(F\omega)}^{\mu}=\frac{1}{8\pi^{2}}\Bigl{[}-\xi^{\mu}(B\cdot\omega+E\cdot
a)+\varepsilon^{\mu\nu\rho\sigma}\xi_{\nu}B_{\rho}a_{\sigma}\Bigr{]},\\\
&J^{\mu}_{(\omega\omega)}=-\frac{\mu}{4\pi^{2}}\xi^{\mu}(\omega^{2}+a^{2}),\end{split}$
(43) $\begin{split}T^{\mu\nu}_{(\partial
F)}&=\frac{\mu}{24\pi^{2}}\biggl{[}-\xi^{(\mu}\partial_{\lambda}F^{\nu)\lambda}+2\xi^{\mu}\xi^{\nu}\xi^{\lambda}\partial^{\rho}F_{\rho\lambda}-g^{\mu\nu}\xi_{\lambda}\partial_{\rho}F^{\rho\lambda}+\xi_{\lambda}\partial^{(\mu}F^{\nu)\lambda}\biggr{]},\\\
T^{\mu\nu}_{(FF)}&=\frac{\mathcal{J}_{-1,0}-\mathcal{J}}{12\pi^{2}}\biggl{[}{F^{\mu}}_{\sigma}F^{\nu\sigma}-\frac{1}{4}g^{\mu\nu}F_{\alpha\beta}^{2}\biggr{]},\\\
T^{\mu\nu}_{(F\omega)}&=\frac{\mu}{8\pi^{2}}\biggl{[}-\xi^{\mu}\xi^{\nu}(\omega\cdot
B+a\cdot
E)+\omega^{(\mu}B^{\nu)}+a^{(\mu}E^{\nu)}+2\xi^{(\mu}\varepsilon^{\nu)\rho\sigma\lambda}a_{\rho}B_{\sigma}\xi_{\lambda}\biggr{]},\\\
T^{\mu\nu}_{(\omega\omega)}&=\biggl{(}\frac{\mu^{2}}{2\pi^{2}}+\frac{T^{2}}{6}\biggr{)}\biggl{[}\biggl{(}\frac{1}{4}g^{\mu\nu}-\xi^{\mu}\xi^{\nu}\biggr{)}(\omega^{2}+a^{2})+\xi^{(\mu}\varepsilon^{\nu)\rho\sigma\lambda}a_{\rho}\omega_{\sigma}\xi_{\lambda}\biggr{]}.\end{split}$
(44)

Here, the longitudinal component of the derivative disappears, i.e.,
$\xi\cdot\partial F_{\mu\nu}=0$, due to the equilibrium condition (33). For
the energy-momentum tensor, the term with $Q^{\mu}$ in Eq. (11) yields no
contribution, as is readily checked. We again emphasize that either
$\partial_{\lambda}F_{\mu\nu}$ or $a^{\mu}$ is admitted to survive due to the
conditions (33a) and (33b), respectively.

We should make a comparison with Ref. [44]. The authors derived almost the
same transport as above, except for $J^{\mu}_{(\partial F)}$,
$T^{\mu\nu}_{(\partial F)}$ and $T^{\mu\nu}_{(FF)}$. The first two were not
computed since the authors focused only on constant background fields. The
stark difference from Ref. [44] is found in $T^{\mu\nu}_{(FF)}$. The two
underlying reasons of this difference are elucidated by recalling the
arguments in Sec. IV. First, the authors did not take into account finite
surface terms because of the vacuum contribution in Eq. (34). Second, they
implemented dimensional regularization, without caring about the modification
on the Clifford algebra in $d\neq 4$ dimensions. As a result, while our
energy-momentum tensor agrees with that from the Euler–Heisenberg effective
theory, that derived in Ref. [44] does not (see Sec.VI).

An important observation in Eqs. (LABEL:eq:J2) and (44) is the finite
contributions from the magneto-vortical terms $J^{\mu}_{(F\omega)}$ and
$T^{\mu\nu}_{(F\omega)}$. In particular, the charge density
$J^{0}_{(F\omega)}\sim B\cdot\omega$ agree with that derived in Refs. [40, 41,
44, 28]. There is, however, a crucial contrast with them in terms of the
derivations. On the one hand, the above early studies implicitly assume an
equilibrium under magnetic field and vorticity, despite the subtlety of this
assumption; the interplay of magnetic field and rotation classically generates
an effective electric field, which in general prohibits the equilibration. On
the other hand, our $J^{0}_{(F\omega)}\sim B\cdot\omega$ is derived from the
equilibrium Wigner function, which is determined by the kinetic equation. We
hence verifies the nondissipativeness of the above magneto-vortical effect,
based on quantum field theory. This is one of the main findings in this paper.

However, the above result does not reproduce the induced current $\sim
B\cdot\omega B^{\mu}/|B|$, which is discovered in Ref. [40]. This is because
our classical Wigner function $\mathcal{R}^{\mu}_{(0)}$ is independent of
$B^{\mu}$. Contrary, if $\mathcal{R}^{\mu}_{(0)}$ depends on $B^{\mu}$, there
emerges $B^{\mu}/|B|$ as a possible tensorial basis of
$\mathcal{R}^{\mu}_{(2)}$, similarly to the fluid velocity $\xi^{\mu}$. This
is in fact the case of the CKT in the strong magnetic field [28]. Hence,
although the magneto-vortical coupling generates both the charge $\sim
B\cdot\omega$ and the current $\sim B\cdot\omega B^{\mu}/|B|$, they are
qualitatively different. Such a difference would be related to their anomalous
nature [57].

Let us argue the conservation laws for the transport in Eqs. (LABEL:eq:J2) and
(44). One can compute the divergences $\partial_{\mu}J^{\mu}_{(2)}$ with the
help of Eq. (84) and the formulas in Appendix E. We then observe
$\partial_{\mu}J^{\mu}_{(F\omega)}=\partial_{\mu}J^{\mu}_{(\omega\omega)}=\partial_{\mu}(J^{\mu}_{(FF)}+J^{\mu}_{(\partial
F)})=0$. Therefore, the nonlinear contribution of the charge current is
conserved:

$\partial_{\mu}J^{\mu}_{(2)}=0,$ (45)

where we impose $a_{\mu}\partial_{\nu}F_{\rho\sigma}=0$ because of the
equilibrium condition (33). This relation holds under both the conditions
(33a) and (33b). The divergence $\partial_{\mu}T^{\mu\nu}_{(2)}$ is computed
in a similar manner. We find
$\partial_{\mu}T^{\mu\nu}_{(F\omega)}+F^{\mu\nu}J_{\mu}^{(F\omega)}=\partial_{\mu}T^{\mu\nu}_{(\omega\omega)}+F^{\mu\nu}J_{\mu}^{(\omega\omega)}=0$,
but

$\begin{split}&\partial_{\mu}T^{\mu\nu}_{(FF)}+F^{\mu\nu}(J_{\mu}^{(FF)}+J_{\mu}^{(\partial
F)})=\frac{1}{48\pi^{2}}\Bigl{[}a^{\nu}F_{\alpha\beta}F^{\alpha\beta}-4a^{\mu}F_{\mu\sigma}F^{\nu\sigma}\Bigr{]},\\\
&\partial_{\mu}T^{\mu\nu}_{(\partial
F)}=\frac{1}{48\pi^{2}}\Bigl{[}-\xi^{\nu}E_{\mu}\partial_{\lambda}F^{\lambda\mu}+2\xi_{\mu}E^{\nu}\partial_{\lambda}F^{\lambda\mu}-2\xi_{\lambda}E_{\mu}\partial^{(\mu}F^{\nu)\lambda}\Bigr{]},\end{split}$
(46)

where we again drop the product terms $\sim
a_{\mu}\partial_{\lambda}F_{\nu\rho}$. Thus, we arrive at

$\partial_{\mu}T^{\mu\nu}_{(2)}+F^{\mu\nu}J_{\mu}^{(2)}\neq 0.$ (47)

This violation of the translational invariance is a compensation of the point-
splitting regularization.

Lastly, we look at the trace of the energy-momentum tensors in Eq. (44). We
first notice that $T^{\mu\nu}_{(\partial F)}$, $T^{\mu\nu}_{(F\omega)}$ and
$T^{\mu\nu}_{(\omega\omega)}$ are traceless irrelevantly to the regularization
scheme. The same is true for $T^{\mu\nu}_{(FF)}$, as long as we utilize the
point-splitting regularization. Eventually, no trace anomaly is reproduced:

${T^{\mu}}_{\mu{(2)}}=0.$ (48)

This is another compensation of the point-splitting regularization; the
energy-momentum conservation and the tracelessness do not simultaneously hold.
We emphasize that the QED trace anomaly stems from the fermion loop
corrections, regardless of whether electromagnetic fields are background or
not [58, 59, 60]; it is generally inevitable to introduce some regularization
scale. Hence, Eq. (48) is just a consequence of our regularization.

##  VI Consistency with Euler–Heisenberg effective theory

For consistency check, let us make a comparison with the Euler–Heisenberg
effective theory, which is described by the following effective Lagrangian:

$\begin{split}\mathcal{L}_{\mathrm{EH}}&=-\mathcal{F}-\frac{e^{2}}{8\pi^{2}}\int_{s_{0}}^{\infty}\frac{{\mathrm{d}}s}{s}\,{\mathrm{e}}^{-sm^{2}}\frac{\operatorname{Re}\cosh\Bigl{[}\hbar\,es\sqrt{2(\mathcal{F}+{\mathrm{i}}\mathcal{G})}\Bigr{]}}{\operatorname{Im}\cosh\Bigl{[}\hbar\,es\sqrt{2(\mathcal{F}+{\mathrm{i}}\mathcal{G})}\Bigr{]}}\,\mathcal{G}\end{split}$
(49)

with $\mathcal{F}:=F_{\alpha\beta}^{2}/4$,
$\mathcal{G}:=F^{\alpha\beta}\tilde{F}_{\alpha\beta}/4$ and $m$ being the
fermion mass. Here $\hbar$ and $e$ are explicitly written. In the above
Lagrangian, we do not put the conventional counterterms $\sim 1/s^{3}$ and
$\sim e^{2}\mathcal{F}/s$, which accounts for the vacuum energy
renormalization and the charge renormalization. Instead of this minimal
subtraction, we introduced the ultraviolet cutoff parameter $s_{0}$, which
plays the similar role to $y^{-1}$ in the point-splitting regularization. The
charge current and the energy-momentum tensor are obtained from the derivative
of the corresponding action with respect to gauge field $A_{\mu}$ and metric
tensor $g_{\mu\nu}$, respectively. We define them as

$\begin{split}J^{\mu}_{\mathrm{EH}}:=-\frac{\delta}{\delta
A_{\mu}}\int{\mathrm{d}}^{4}x\,(\hbar^{2}\mathcal{L}_{\mathrm{EH}}),\qquad
T^{\mu\nu}_{\mathrm{EH}}:=\frac{2}{\sqrt{-g}}\frac{\delta}{\delta
g^{\mu\nu}}\int{\mathrm{d}}^{4}x\sqrt{-g}\,(\hbar^{2}\mathcal{L}_{\mathrm{EH}}).\end{split}$
(50)

For the energy-momentum tensor $T^{\mu\nu}_{\mathrm{EH}}$, we utilized the
effective action in a general curved spacetime with $g:=\det(g_{\mu\nu})$. We
note that the factor $\hbar^{2}$ is from our convention for the comparison
with the CKT analysis; on top of $\hbar^{-1}$ by definition of action, the
extra $\hbar^{3}$ is multiplied because we abbreviate the $\hbar^{-3}$ in the
momentum phase space, following the usual convention in the CKT.

The above Lagrangian can be expanded in terms of power of $\hbar$. This is
generally written as follows:

$\hbar^{2}\mathcal{L}_{\mathrm{EH}}=-\hbar^{2}\mathcal{F}+\mathcal{L}_{\mathrm{EH{(0)}}}+\hbar^{2}\mathcal{L}_{\mathrm{EH{(2)}}}+\hbar^{4}\mathcal{L}_{\mathrm{EH(4)}}+\cdots.$
(51)

For the latter convenience, here we multiplied $\hbar^{2}$ by both sides. In
Eq. (51), what we are now interested in is

$\begin{split}\mathcal{L}_{\mathrm{EH}(2)}=-\frac{e^{2}}{48\pi^{2}}F_{\mu\nu}^{2}\int_{s_{0}}^{\infty}\frac{{\mathrm{d}}s}{s}\,{\mathrm{e}}^{-sm^{2}}=-\frac{e^{2}}{24\pi^{2}}F_{\mu\nu}^{2}\int_{0}^{s_{0}^{-1/2}}\frac{{\mathrm{d}}p}{p}\,{\mathrm{e}}^{-m^{2}/p^{2}}.\end{split}$
(52)

The mass parameter $m$ is the convergence factor of the infrared regime at
$s\to\infty$ or $p\to 0$. In our CKT analysis at equilibrium, we do not care
about the infrared divergence, thanks to the cancellation by the matter part
of $f_{(0)}$. For comparison with the CKT, we consider the limit of $m\to 0$
444Even if we take the massless limit after performing the integration in
$\mathcal{L}_{\mathrm{EH{(2)}}}$, the logarithmically divergent behavior in
terms of $s_{0}$ is unchanged. Thus, the order of taking the limit is
irrelevant to the present discussion. . By replacing $s_{0}^{-1/2}$ with
$y^{-1}$, we reduce Eq. (52) to

$\begin{split}\mathcal{L}_{\mathrm{EH}(2)}\bigl{|}_{m\to
0}=-\frac{e^{2}\mathcal{J}}{24\pi^{2}}F_{\mu\nu}^{2},\end{split}$ (53)

where $\mathcal{J}$ is given by Eq. (39). Inserting Eq. (53) into Eq. (50) and
setting $e=1$ as we do in the CKT, we arrive at the following relations:

$\begin{split}J^{\mu}_{\mathrm{EH}{(2)}}\bigl{|}_{m\to
0}&=\frac{\mathcal{J}}{6\pi^{2}}\partial_{\lambda}F^{\mu\lambda}=2J^{\mu}_{(\partial
F)\,\mathrm{vac}},\\\ T^{\mu\nu}_{\mathrm{EH}{(2)}}\bigl{|}_{m\to
0}&=-\frac{\mathcal{J}}{6\pi^{2}}\biggl{[}{F^{\mu}}_{\sigma}F^{\nu\sigma}-\frac{1}{4}g^{\mu\nu}F_{\alpha\beta}^{2}\biggr{]}=2T^{\mu\nu}_{(FF)\,\mathrm{vac}},\end{split}$
(54)

where ‘$\mathrm{vac}$’ denotes the vacuum contribution. The factor $2$ on the
right-hand sides is understood as the degrees of freedom of chirality. These
relations guarantee the correctness of $J^{\mu}_{(\partial F)}$ and
$T^{\mu\nu}_{(FF)}$ in Eqs. (LABEL:eq:J2) and (44). We note that the matter
part and the vortical terms in Eqs. (LABEL:eq:J2) and (44) are not included
here, as they are not enclosed in Eq. (49).

There is an important remark about the spacetime-dependence of electromagnetic
fields. Since the original Euler–Heisenberg effective theory is for a constant
$F_{\mu\nu}$, one might be skeptical that the above comparison is meaningful
for $J^{\mu}_{\mathrm{EH{(2)}}}\sim\partial_{\lambda}F^{\mu\lambda}$. If we
take into account the coordinate-dependence of $F_{\mu\nu}$, the effective
Lagrangian acquires the derivative corrections. However, the leading
derivative correction is of $O\bigl{(}(\partial F)^{2}\bigr{)}$ or of
$O\bigl{(}F\partial^{2}F\bigr{)}$ [61, 62]. In the power counting of $\hbar$,
such a term is the fourth-order term, as it contains four derivatives of gauge
field. For this reason, even when $F_{\mu\nu}$ is spacetime-dependent, the
Lagrangian $\mathcal{L}_{\mathrm{EH{(2)}}}$ is unmodified and thus so is Eq.
(54). Therefore, we conclude that the nonlinear CKT is consistent with the
Euler–Heisenberg effective theory.

The Euler–Heisenberg effective theory also reveals underlying physics of the
logarithmic behavior of $\mathcal{J}$ in Eqs. (LABEL:eq:J2) and (44). To
illustrate it, we write the Lagrangian (49) in the following form:

$\hbar^{2}\mathcal{L}_{\mathrm{EH}}=-\frac{\hbar^{2}}{4}F_{\mu\nu}^{2}\biggl{(}1+\frac{e^{2}}{12\pi^{2}}\log\frac{s_{0}^{-1}}{m^{2}}\biggr{)}+\mathrm{const.}+O(\hbar^{4}),$
(55)

where the constant is the term without $F_{\mu\nu}$. The logarithmic behavior
is the same as that found in the vacuum polarization of QED. From the above
Lagrangian, hence, we read off the effective charge
$e^{2}_{\mathrm{eff}}(M):=e^{2}(1+\frac{e^{2}}{12\pi^{2}}\log\frac{M^{2}}{m^{2}})$,
and the $\beta$-function
$\beta(e_{\mathrm{eff}}):=M{\mathrm{d}}e_{\mathrm{eff}}(M)/{\mathrm{d}}M=e^{3}_{\mathrm{eff}}(M)/(12\pi^{2})$
[63]. Equation (54) shows that this characteristic of the charge
renormalization is inherited not only in the Euler–Heisenberg theory, but also
in the nonlinear CKT through the same logarithm $\mathcal{J}\sim\log y^{-1}$.
At the same time, in spite of Eq. (48), we find that the logarithm
$\mathcal{J}$ in Eq. (44) is an indirect evidence of the trace anomaly, which
is determined by the QED $\beta$-function.

##  VII Summary

In this paper, we formulated the nonlinear CKT under arbitrary background
electromagnetic fields. We derived the off-equilibrium Wigner function for
arbitrary frame vectors. Imposing the frame-independence of this Wigner
function, we identified an equilibrium Wigner function, which solves the
kinetic equation. As an application, we compute the transport phenomena at the
equilibrium. We then found that the charge induced by the interplay of
magnetic field and vorticity [40] are permitted at the equilibrium of the
nonlinear CKT. This analysis based on the Wigner function is, to the best of
our knowledge, the first field-theoretical verification of the
nondissipativeness of the above charge generation. Besides, as an important
finding, we also showed that the nonlinear CKT and the Euler–Heisenberg
effective theory share equivalent transport phenomena. The ultraviolet
logarithmic behavior in the nonlinear CKT is not only a kinetic encoding of
the charge renormalization but also an indirect signature of the trace anomaly
in the kinetic description.

Also, we posed the potential issue that the prominent schemes, i.e.,
Pauli–Villars regularization and dimensional regularization, are incompatible
with the CKT. The incompatibility of the latter scheme is one reason of the
fact that the energy-momentum tensor in Ref. [44] disagree with that derived
from the Euler–Heisenberg effective theory. For this reason, we employed the
point-splitting regularization, which is much more compatible with the Wigner
function but cannot directly reproduce the trace anomaly. For the complete
reproduction of the trace anomaly in the kinetic description, we should find
out an appropriate regularization in CKT, or rely on frameworks other than the
CKT. For the latter option, the kinetic theory of massive fermions involving
the $O(\hbar^{2})$ correction is one of the candidates, since Pauli–Villars
regularization could be applicable.

Several potential developments are invoked from the nonlinear CKT. First, the
nonlinear transport phenomena is one of the pivotal research fields in
condensed matter physics [64]. Also, the merit of the nonlinear CKT could be
found in, for instance, the so-called nonlinear Hall effect [65, 66], which
originates from the Berry curvature dipole. In the nonlinear CKT, such
contribution would be hidden (see also Refs. [67, 68, 69], which argue
nonlinear corrections of the Berry curvature). Besides, it is straightforward
but complicated to extend the present nonlinear CKT to the collisional case by
starting from the Kadanoff–Baym equation with fermionic self-energy [70, 71].
In the nonlinear CKT, it is also interesting to take into account dynamical
gauge fields, which bring the chiral plasma instabilities [72]. These
applications will be discussed elsewhere.

###### Acknowledgements.

The author thanks Yoshimasa Hidaka for giving valuable comments.

##  Appendix A Charge current, energy-momentum tensor and spin tensor at
$O(\hbar^{2})$

In this Appendix, we derive the charge current and energy-momentum tensor with
the Wigner function. The former for the right-handed massless fermions is
defined as

$\begin{split}J^{\mu}(x,y):=\mathrm{tr}\Bigl{\langle}\bar{\psi}_{+}\gamma^{\mu}P_{\mathrm{R}}\psi_{-}\Bigr{\rangle},\end{split}$
(56)

where we define $P_{\mathrm{R}}:=\frac{1}{2}(1+\gamma^{5})$,
$O_{-}:={\mathrm{e}}^{-y\cdot D/2}O(x)$ and
$O_{+}:=O(x){\mathrm{e}}^{y\cdot\overleftarrow{D}/2}$ with
$D_{\mu}:=\partial_{\mu}+{\mathrm{i}}A_{\mu}/\hbar$,
$\overleftarrow{D}_{\mu}:=\overleftarrow{\partial}_{\mu}-{\mathrm{i}}A_{\mu}/\hbar$.
Here, the operators ${\mathrm{e}}^{-y\cdot D/2}$ and
${\mathrm{e}}^{y\cdot\overleftarrow{D}/2}$ represent the covariant
translation, and thus their insertion is equivalent to enclosing the Wilson
line [43]. In the $y\to 0$ limit, the above current is reduced to the usual
definition in quantum field theory. Let us here recall that the Wigner
function is defined as

$\mathcal{R}^{\mu}(x,p):=\frac{1}{2}\mathrm{tr}\Bigl{[}\gamma^{\mu}P_{\mathrm{R}}W(x,p)\Bigr{]},\quad
W_{ab}(x,p):=\int_{y}{\mathrm{e}}^{-{\mathrm{i}}p\cdot
y/\hbar}\,\mathrm{tr}\Bigl{\langle}(\bar{\psi}_{+})_{b}(\psi_{-})_{a}\Bigr{\rangle}.$
(57)

Then, performing the inverse Wigner transformation, we write the above current
as Eq. (10):

$J^{\mu}(x,y)=2\int_{p}{\mathrm{e}}^{{\mathrm{i}}p\cdot
y/\hbar}\mathcal{R}^{\mu}(x,p).$ (58)

For the spin tensor, the inverse Wigner transformation of the standard field-
theoretical definition yields Eq. (12):

$\begin{split}S^{\mu\nu\rho}(x,y)&:=\frac{\hbar}{4}\,\mathrm{tr}\Bigl{\langle}\bar{\psi}_{+}\bigl{\\{}\gamma^{\mu},\sigma^{\nu\rho}\bigr{\\}}P_{\mathrm{R}}\psi_{-}\Bigr{\rangle}=-2\hbar\varepsilon^{\mu\nu\rho\sigma}\int_{p}{\mathrm{e}}^{{\mathrm{i}}p\cdot
y/\hbar}\mathcal{R}_{\sigma}(x,p)\end{split}$ (59)

with $\sigma^{\mu\nu}=\frac{{\mathrm{i}}}{2}[\gamma^{\mu},\gamma^{\nu}]$.

Let us derive the kinetic expression of the energy-momentum tensor. Unlike the
charge current and spin tensor, the definition of the energy-momentum tensor
is ambiguous due to the derivative operator. We here employ the canonical
energy-momentum tensor defined as follows:

$\begin{split}T_{\text{can}}^{\mu\nu}(x,y)&:=\frac{{\mathrm{i}}\hbar}{2}(t^{\mu\nu}-g^{\mu\nu}{t^{\lambda}}_{\lambda}),\quad
t^{\mu\nu}=\mathrm{tr}\Bigl{\langle}\bar{\psi}_{+}\gamma^{\mu}P_{\mathrm{R}}(D^{\nu}\psi)_{-}-(\bar{\psi}\overleftarrow{D}^{\nu})_{+}\gamma^{\mu}P_{\mathrm{R}}\psi_{-}\Bigr{\rangle}.\end{split}$
(60)

Note that $(D^{\nu}\psi)_{-}={\mathrm{e}}^{-y\cdot D/2}D_{\mu}\psi$ is
inequivalent to $D_{\mu}\psi_{-}=D_{\mu}{\mathrm{e}}^{-y\cdot D/2}\psi$ when
electromagnetic fields are spacetime-dependence [see Eq. (62)]. In the limit
of $y\to 0$, this definition is consistent with the classical canonical
momentum tensor

$\Theta^{\mu\nu}(x)=\partial^{\mu}\psi\frac{\partial\mathcal{L}}{\partial\partial_{\nu}\psi}+\partial^{\mu}\bar{\psi}\frac{\partial\mathcal{L}}{\partial\partial_{\nu}\bar{\psi}}-g^{\mu\nu}\mathcal{L}$
(61)

with
$\mathcal{L}=\frac{{\mathrm{i}}\hbar}{2}\Bigl{[}\bar{\psi}(x)\gamma^{\lambda}P_{\mathrm{R}}D_{\lambda}\psi(x)-\bar{\psi}(x)\overleftarrow{D}_{\lambda}\gamma^{\lambda}P_{\mathrm{R}}\psi(x)\Bigr{]}$.
In Eq. (60), the last term with $g^{\mu\nu}$ vanishes due to the Dirac
equation. To reduce the first two terms, we prepare the following identities:

$\begin{split}D_{\mu}{\mathrm{e}}^{y\cdot
D}\psi(x)&=\biggl{[}{\mathrm{e}}^{y\cdot
D}D_{\mu}+\frac{{\mathrm{i}}y^{\lambda}}{\hbar}\mathcal{F}_{\mu\lambda}(x,y){\mathrm{e}}^{y\cdot
D}\biggr{]}\psi(x),\\\ \partial_{\mu}^{y}{\mathrm{e}}^{y\cdot
D}\psi(x)&=\biggl{[}D_{\mu}{\mathrm{e}}^{y\cdot
D}-\frac{{\mathrm{i}}y^{\lambda}}{\hbar}\mathcal{G}_{\mu\lambda}(x,y){\mathrm{e}}^{y\cdot
D}\biggr{]}\psi(x)\end{split}$ (62)

with

$\mathcal{F}_{\mu\lambda}(x,y)=\sum_{n=0}^{\infty}\frac{(y\cdot\partial)^{n}}{(n+1)!}F_{\mu\lambda}(x),\quad\mathcal{G}_{\mu\lambda}(x,y)=\sum_{n=0}^{\infty}\frac{(y\cdot\partial)^{n}}{(n+2)!}F_{\mu\lambda}(x).$
(63)

These are derived from
${\mathrm{e}}^{Y}X{\mathrm{e}}^{-Y}={\mathrm{e}}^{\mathcal{C}(Y)}X$ with
$\mathcal{C}(Y)X:=[Y,X]$. Performing the inverse Wigner transformation, we
rewrite Eq. (60) as

$\begin{split}T^{\mu\nu}_{\text{can}}(x,y)&=\biggl{[}-{\mathrm{i}}\hbar\partial^{\nu}_{y}+\frac{1}{12}y\cdot\partial
F^{\nu\lambda}y_{\lambda}\biggr{]}\mathrm{tr}\Bigl{\langle}\bar{\psi}_{+}\gamma^{\mu}P_{\mathrm{R}}\psi_{-}\Bigr{\rangle}\\\
&=2\int_{p}{\mathrm{e}}^{ip\cdot
y/\hbar}p^{\nu}\mathcal{R}^{\mu}(x,p)+\frac{1}{12}y\cdot\partial
F^{\nu\lambda}y_{\lambda}\cdot 2\int_{p}{\mathrm{e}}^{ip\cdot
y/\hbar}\mathcal{R}^{\mu}(x,p)\\\ \end{split}$ (64)

up to $O(\hbar^{2})$. In the second line, we need carefully to perform the
integral by parts because the surface terms in general are generated. At least
at the equilibrium described by Eq. (23), however, we can show that no surface
term appears, as follows. The second term can be decomposed into the
contributions from the vacuum and the matter parts, namely,
$f_{{(0)}\mathrm{vac}}(p_{0})=-\theta(-p_{0})$ and
$f_{{(0)}\mathrm{mat}}(p_{0})=\theta(p_{0})n_{F}(p_{0}-\mu)+\theta(-p_{0})n_{F}(-p_{0}+\mu)$.
The former should be proportional to $y^{\lambda}y^{\rho}y^{-3}$ for the
dimensional reason, and thus vanishes in the symmetric limit of $y\to 0$. The
latter yields no surface term because of
$f_{{(0)}\mathrm{mat}}(p_{0})\delta(p^{2})\to 0$ for $p_{\mu}\to 0$.
Therefore, at the equilibrium, performing the integral by parts leads to

$\begin{split}T^{\mu\nu}_{\text{can}}(x,y)&=2\int_{p}{\mathrm{e}}^{{\mathrm{i}}p\cdot
y/\hbar}(p^{\nu}+\hbar^{2}Q^{\nu})\mathcal{R}^{\mu}(x,p).\end{split}$ (65)

Its symmetric part is given by Eq. (11).

##  Appendix B Solution at $O(\hbar^{2})$

In this Appendix, we derive the second-order solution
$\mathcal{R}^{\mu}_{(2)}$. The basic step of the following calculation is
parallel to Ref. [37]. Inserting $\mathcal{R}_{(0)}^{\mu}$ and
$\mathcal{R}_{(1)}^{\mu}$ into Eq. (9), we find

$\mathcal{R}_{\mu}^{(2)}=2\pi\delta(p^{2})\widetilde{\mathcal{R}}_{\mu}^{(2)}+\frac{2\pi}{p^{2}}\biggl{[}-p_{\mu}Q\cdot
pf_{(0)}+p^{\nu}\mathcal{D}_{\mu\nu}-p^{\nu}\tilde{F}_{\mu\nu}f_{(1)}\biggr{]}\delta(p^{2}),$
(66)

with $\widetilde{\mathcal{R}}_{\mu}^{(2)}$ satisfying
$p^{2}\delta(p^{2})\widetilde{\mathcal{R}}^{\mu}_{(2)}=\delta(p^{2})p\cdot\widetilde{\mathcal{R}}_{(2)}=0$.
Here, we have introduced

$\mathcal{D}_{\mu\nu}\delta(p^{2}):=\frac{1}{2}\varepsilon_{\mu\nu\rho\sigma}\Delta^{\rho}\biggl{(}\Sigma^{\sigma\lambda}_{n}(\Delta_{\lambda}f_{(0)})-\frac{1}{p^{2}}\tilde{F}^{\sigma\lambda}p_{\lambda}f_{(0)}\biggr{)}\delta(p^{2})+2Q_{[\mu}p_{\nu]}f_{(0)}\delta(p^{2}),$
(67)

where $(\Delta_{\lambda}f_{(0)})$ represents the derivative operation acting
only on $f_{(0)}$, but others operate on all on the right. For this
$\mathcal{R}^{\mu}_{(2)}$, Eq. (7) yields

$\begin{split}\widetilde{\mathcal{R}}_{\mu}^{(2)}\delta(p^{2})&=\delta(p^{2})\Bigl{[}p_{\mu}f_{(2)}+\Sigma_{\mu\nu}^{u}\Delta^{\nu}f_{(1)}\Bigr{]}+\frac{1}{p^{2}}\varepsilon^{\alpha\beta\gamma\nu}\Sigma_{\mu\nu}^{u}p_{\alpha}\mathcal{D}_{\beta\gamma}\delta(p^{2}),\end{split}$
(68)

where we introduce a vector $u^{\nu}$, and define
$f_{(2)}:=u\cdot\widetilde{\mathcal{R}}_{(2)}/(p\cdot u)$ and
$\Sigma_{u}^{\mu\nu}:=\varepsilon^{\mu\nu\rho\sigma}p_{\rho}u_{\sigma}/(2p\cdot
u)$, similarly to Eq. (18). The last term with complicated structure due to
the Levi-Civita symbols can be reduced with the Schouten identity:
$\varepsilon^{\mu\nu\rho\sigma}p^{\lambda}+\varepsilon^{\nu\rho\sigma\lambda}p^{\mu}+\varepsilon^{\rho\sigma\lambda\mu}p^{\nu}+\varepsilon^{\sigma\lambda\mu\nu}p^{\rho}+\varepsilon^{\lambda\mu\nu\rho}p^{\sigma}=0$
and

$\begin{split}&\Sigma_{n}^{\lambda[\mu}p^{\nu]}=-\frac{1}{2}\Sigma_{n}^{\mu\nu}p^{\lambda}-\frac{1}{4}\varepsilon^{\mu\nu\lambda\rho}p_{\rho}+\frac{1}{4}\varepsilon^{\mu\nu\lambda\rho}\frac{n_{\rho}p^{2}}{p\cdot
n}.\end{split}$ (69)

In the CKT at $O(\hbar^{2})$, Eq. (69) is quite helpful in the sense that the
frame-independent part and the $p^{2}$ term can be extracted. After
straightforward computation with these relations, we arrive at

$\begin{split}\frac{1}{p^{2}}\varepsilon^{\alpha\beta\gamma\nu}\Sigma_{\mu\nu}^{u}p_{\alpha}\mathcal{D}_{\beta\gamma}\delta(p^{2})&=-\delta(p^{2})\Sigma_{\mu\nu}^{u}\varepsilon^{\nu\rho\sigma\lambda}\Delta_{\rho}\frac{n_{\sigma}}{2p\cdot
n}\Delta_{\lambda}f_{(0)}\\\
&\quad+\frac{\delta(p^{2})}{p^{2}}\Sigma_{\mu\nu}^{u}\biggl{[}\Delta_{\alpha}\Sigma^{\alpha\nu}_{n}+\frac{n_{\alpha}}{p\cdot
n}\tilde{F}^{\alpha\nu}+\frac{1}{p^{2}}\tilde{F}^{\nu\lambda}p_{\lambda}\biggr{]}p\cdot\Delta
f_{(0)}.\end{split}$ (70)

It should be mentioned that the singular factors $(p^{2})^{-1}$ and
$(p^{2})^{-2}$ in Eq. (70) does not conflict with the nonsingular condition
$\delta(p^{2})p^{2}\widetilde{\mathcal{R}}^{\mu}_{(2)}=0$. One can show this
by noting $(p^{2})^{-n}\delta(p^{2})p\cdot\Delta f_{(0)}\neq 0$ for $n\geq 1$
but $\delta(p^{2})p\cdot\Delta f_{(0)}=0$, which follows from the classical
kinetic equation (5). Plugging Eqs. (68) and (70) into Eq. (66), and
proceeding computation, we obtain the second-order solution
$\mathcal{R}^{\mu}_{(2)}$ in Eq. (19).

##  Appendix C Integral formulas for matter contribution

In this Appendix, we derive the integral formulas for the matter contribution.
At equilibrium, the matter contribution in Eq. (34) is the following form:

$\int_{p}2\pi\frac{\delta(p^{2})}{(p^{2})^{l}}p^{\mu_{1}}\cdots
p^{\mu_{j}}\frac{{\mathrm{d}}^{k}f_{{(0)}\mathrm{mat}}}{{\mathrm{d}}p_{0}^{k}}$
(71)

with
$f_{{(0)}\mathrm{mat}}=\theta(p_{0})n_{F}(p_{0}-\mu)+\theta(-p_{0})n_{F}(-p_{0}+\mu)$
and $n_{F}(x)=({\mathrm{e}}^{\beta x}+1)^{-1}$. In the integrands, we can
implement the following replacement:

$\begin{split}p_{\alpha}&\to p_{0}\xi_{\alpha},\\\
p_{\alpha}p_{\beta}&\to(p_{0})^{2}\xi_{\alpha}\xi_{\beta}+\frac{{\boldsymbol{p}}^{2}}{3}\Delta_{\alpha\beta},\\\
p_{\alpha}p_{\beta}p_{\gamma}&\to(p_{0})^{3}\xi_{\alpha}\xi_{\beta}\xi_{\gamma}+\frac{p_{0}{\boldsymbol{p}}^{2}}{3}(\xi_{\alpha}\Delta_{\beta\gamma}+\xi_{\beta}\Delta_{\gamma\alpha}+\xi_{\gamma}\Delta_{\alpha\beta}),\\\
p_{\alpha}p_{\beta}p_{\gamma}p_{\delta}&\to(p_{0})^{4}\xi_{\alpha}\xi_{\beta}\xi_{\gamma}\xi_{\delta}\\\
&\quad+\frac{(p_{0})^{2}{\boldsymbol{p}}^{2}}{3}(\xi_{\alpha}\xi_{\beta}\Delta_{\gamma\delta}+\xi_{\alpha}\xi_{\gamma}\Delta_{\beta\delta}+\xi_{\alpha}\xi_{\delta}\Delta_{\beta\gamma}+\xi_{\beta}\xi_{\gamma}\Delta_{\alpha\delta}+\xi_{\beta}\xi_{\delta}\Delta_{\alpha\gamma}+\xi_{\gamma}\xi_{\delta}\Delta_{\alpha\beta})\\\
&\quad+\frac{|{\boldsymbol{p}}|^{4}}{15}(\Delta_{\alpha\beta}\Delta_{\gamma\delta}+\Delta_{\alpha\gamma}\Delta_{\beta\delta}+\Delta_{\alpha\delta}\Delta_{\beta\gamma}),\end{split}$
(72)

with $\xi^{\mu}:=(1,\boldsymbol{0})$ and the transverse projector
$\Delta^{\mu\nu}:=\xi^{\mu}\xi^{\nu}-g^{\mu\nu}$. Then, the above integral is
represented as a linear combination of

$\mathcal{I}^{l}_{n,m,k}=\int_{p}2\pi\frac{{\mathrm{d}}^{l}\delta(p^{2})}{({\mathrm{d}}p^{2})^{l}}F_{n,m,k},\quad
F_{n,m,k}:=(p_{0})^{n}|{\boldsymbol{p}}|^{n-m}\frac{{\mathrm{d}}^{k}f_{{(0)}\mathrm{mat}}}{{\mathrm{d}}p_{0}^{k}},$
(73)

with $\int_{p}=\int\frac{{\mathrm{d}}^{4}p}{(2\pi)^{4}}$. We start from

$\delta(p^{2})=\frac{1}{2|{\boldsymbol{p}}|}(\delta_{+}+\delta_{-}),\quad\delta_{\pm}:=\delta(p_{0}\mp|{\boldsymbol{p}}|).$
(74)

Then the first, second, and third derivatives are computed as

$\begin{split}\delta^{\prime}(p^{2})&=\frac{1}{4p_{0}|{\boldsymbol{p}}|}(\delta^{\prime}_{+}+\delta^{\prime}_{-}),\\\
\delta^{\prime\prime}(p^{2})&=-\frac{1}{8p_{0}^{3}|{\boldsymbol{p}}|}(\delta^{\prime}_{+}+\delta^{\prime}_{-})+\frac{1}{8p_{0}^{2}|{\boldsymbol{p}}|}(\delta^{\prime\prime}_{+}+\delta^{\prime\prime}_{-}),\\\
\delta^{\prime\prime\prime}(p^{2})&=\frac{3}{16p_{0}^{5}|{\boldsymbol{p}}|}(\delta^{\prime}_{+}+\delta^{\prime}_{-})-\frac{3}{16p_{0}^{4}|{\boldsymbol{p}}|}(\delta^{\prime\prime}_{+}+\delta^{\prime\prime}_{-})+\frac{1}{16p_{0}^{3}|{\boldsymbol{p}}|}(\delta^{\prime\prime\prime}_{+}+\delta^{\prime\prime\prime}_{-}).\end{split}$
(75)

where the primes on $\delta_{\pm}$ denote the derivative with respect to
$p_{0}$. For $l=0$, we readily compute Eq. (73) by using Eq. (74).

For $l\geq 1$, performing the integration by parts, we can replace
${\mathrm{d}}/{\mathrm{d}}p_{0}$ in Eq. (75) with that on $F_{n,m,k}$. For
instance, the integral for $l=1$ reads

$\begin{split}\mathcal{I}^{1}_{n,m,k}&=-\int_{\boldsymbol{p}}\frac{1}{4|{\boldsymbol{p}}|}\int{\mathrm{d}}p_{0}(\delta_{+}+\delta_{-})\frac{{\mathrm{d}}}{{\mathrm{d}}p_{0}}\frac{F_{n,m,k}}{p_{0}}\end{split}$
(76)

with $\int_{\boldsymbol{p}}=\int\frac{{\mathrm{d}}^{3}p}{(2\pi)^{3}}$. In a
similar manner, we obtain

$\begin{split}\mathcal{I}^{2}_{n,m,k}&=\int_{\boldsymbol{p}}\frac{1}{8|{\boldsymbol{p}}|}\int{\mathrm{d}}p_{0}(\delta_{+}+\delta_{-})\biggl{[}\frac{{\mathrm{d}}}{{\mathrm{d}}p_{0}}\frac{F_{n,m,k}}{p_{0}^{3}}+\frac{{\mathrm{d}}^{2}}{{\mathrm{d}}p_{0}^{2}}\frac{F_{n,m,k}}{p_{0}^{2}}\biggr{]},\end{split}$
(77)
$\begin{split}\mathcal{I}^{3}_{n,m,k}&=-\int_{\boldsymbol{p}}\frac{1}{16|{\boldsymbol{p}}|}\int{\mathrm{d}}p_{0}(\delta_{+}+\delta_{-})\biggl{[}\frac{{\mathrm{d}}}{{\mathrm{d}}p_{0}}\frac{3F_{n,m,k}}{p_{0}^{5}}+\frac{{\mathrm{d}}^{2}}{{\mathrm{d}}p_{0}^{2}}\frac{3F_{n,m,k}}{p_{0}^{4}}+\frac{{\mathrm{d}}^{3}}{{\mathrm{d}}p_{0}^{3}}\frac{F_{n,m,k}}{p_{0}^{3}}\biggr{]}.\end{split}$
(78)

Carrying out the momentum integration in Eqs. (76), (77) and (78), we finally
derive

$\displaystyle\mathcal{I}_{n,m,k}^{0}$ $\displaystyle=$
$\displaystyle\frac{1}{4\pi^{2}}\mathcal{J}_{m+1,k},$ (79)
$\displaystyle\mathcal{I}_{n,m,k}^{1}$ $\displaystyle=$
$\displaystyle\frac{-1}{8\pi^{2}}\biggl{[}(n-1)\mathcal{J}_{m-1,k}+\mathcal{J}_{m,k+1}\biggr{]},$
(80) $\displaystyle\mathcal{I}_{n,m,k}^{2}$ $\displaystyle=$
$\displaystyle\frac{1}{16\pi^{2}}\biggl{[}(n-1)(n-3)\mathcal{J}_{m-3,k}+(2n-3)\mathcal{J}_{m-2,k+1}+\mathcal{J}_{m-1,k+2}\biggr{]},$
(81) $\displaystyle\mathcal{I}_{n,m,k}^{3}$ $\displaystyle=$
$\displaystyle\frac{-1}{32\pi^{2}}\biggl{[}(n-1)(n-3)(n-5)\mathcal{J}_{m-5,k}+3(n^{2}-5n+5)\mathcal{J}_{m-4,k+1}$
(82)
$\displaystyle\qquad\quad+3(n-2)\mathcal{J}_{m-3,k+2}+\mathcal{J}_{m-2,k+3}\biggr{]},$

where the integral sequence $\mathcal{J}_{m,k}$ is given by

$\mathcal{J}_{m,k}:=\int_{0}^{\infty}{\mathrm{d}}p\,p^{m}\frac{{\mathrm{d}}^{k}}{{\mathrm{d}}p^{k}}\Bigl{[}n_{F}(p-\mu)-(-1)^{m+k}n_{F}(p+\mu)\Bigr{]}.$
(83)

One can show the following recursion equations for $m\geq 0$ and $k\geq 0$:

$\mathcal{J}_{m+1,k+1}=-(m+1)\mathcal{J}_{m,k},\quad\partial_{\mu}\mathcal{J}_{m,k}=E_{\mu}\mathcal{J}_{m,k+1}+a_{\mu}(\mathcal{J}_{m+1,k+1}+k\mathcal{J}_{m,k}).$
(84)

The former is useful to reduce Eqs. (80)-(82), and the latter is helpful to
compute the divergence of $J^{\mu}$ and $T^{\mu\nu}$.

##  Appendix D Point-splitting regularization

In this Appendix, we demonstrate the evaluation of $\mathcal{K}_{1}$,
$\mathcal{K}^{\mu\nu}_{2}$ and $\mathcal{K}^{\mu\nu\rho\sigma}_{3}$, where

$\mathcal{K}_{n}^{\mu_{1}\cdots\mu_{m}}(y):=\int_{p}2\pi\frac{{\mathrm{d}}^{n}\delta(p^{2})}{({\mathrm{d}}p^{2})^{n}}p^{\mu_{1}}\cdots
p^{\mu_{m}}\bigl{[}-\theta(-p_{0})\bigr{]}{\mathrm{e}}^{{\mathrm{i}}p\cdot
y/\hbar}.$ (85)

As usual, the point-splitting regularization is implemented with the Euclidean
momentum integral. For the above integral, the simple Wick rotation with
$p_{0}\to-{\mathrm{i}}p_{4}$ cannot be admitted due to the delta function and
step function, which are defined on real space. For this reason, we first
write them as

$\delta(x)=\frac{1}{\pi}{\rm
Im}\frac{1}{x-{\mathrm{i}}\epsilon},\quad\theta(x)=\frac{1}{2\pi{\mathrm{i}}}\int_{-\infty}^{\infty}{\mathrm{d}}\tau\frac{{\mathrm{e}}^{{\mathrm{i}}x\tau}}{\tau-{\mathrm{i}}\eta}$
(86)

with positive infinitesimals $\epsilon$ and $\eta$.

Let us first compute $\mathcal{K}_{1}(y)$, which can be expressed as

$\mathcal{K}_{1}(y)=\frac{1}{2{\mathrm{i}}}(\mathcal{K}_{+}-\mathcal{K}_{-}),\quad\mathcal{K}_{\pm}:=\frac{-2}{2\pi{\mathrm{i}}}\int_{-\infty}^{\infty}\frac{{\mathrm{d}}\tau}{\tau+{\mathrm{i}}\eta}\int_{p}\frac{{\mathrm{e}}^{{\mathrm{i}}p_{0}(\tau+y_{0}/\hbar)-{\mathrm{i}}{\boldsymbol{p}}\cdot\boldsymbol{y}/\hbar}}{(p^{2}\mp{\mathrm{i}}\epsilon)^{2}}.$
(87)

We can now deform the contours of $p_{0}$-integral, together with that of
$\tau$-integral, along the imaginary axis. Introducing another positive
infinitesimal $\delta$, we compute

$\begin{split}\mathcal{K}_{\pm}&=\frac{-2}{2\pi{\mathrm{i}}}\int_{{\mathrm{i}}\infty+\delta}^{-{\mathrm{i}}\infty+\delta}\frac{{\mathrm{d}}\tau}{\tau+{\mathrm{i}}\eta}\int_{\boldsymbol{p}}\int_{{\mathrm{i}}\infty}^{-{\mathrm{i}}\infty}(\pm
1)\frac{{\mathrm{d}}p_{0}}{2\pi}\frac{{\mathrm{e}}^{{\mathrm{i}}p_{0}(\tau+y_{0}/\hbar)-{\mathrm{i}}{\boldsymbol{p}}\cdot\boldsymbol{y}/\hbar}}{(p^{2}\mp{\mathrm{i}}\epsilon)^{2}}\\\
&=(\pm
1)\cdot\frac{-2}{2\pi{\mathrm{i}}}\cdot\frac{1}{{\mathrm{i}}}\int_{-\infty}^{\infty}\frac{{\mathrm{d}}\tau_{E}}{-{\mathrm{i}}\tau_{E}+\delta+{\mathrm{i}}\eta}\int_{\boldsymbol{p}}\int_{-\infty}^{\infty}\frac{{\mathrm{d}}p_{4}}{2\pi{\mathrm{i}}}\frac{{\mathrm{e}}^{p_{4}(-{\mathrm{i}}\tau_{E}+\delta+y_{0}/\hbar)-{\mathrm{i}}{\boldsymbol{p}}\cdot\boldsymbol{y}/\hbar}}{(-p^{2}_{E})^{2}}\\\
&=(\pm
1)\cdot\frac{-2}{2\pi{\mathrm{i}}}\cdot\frac{1}{{\mathrm{i}}}\int_{-\infty}^{\infty}\frac{{\mathrm{d}}\tau_{E}}{\tau_{E}+{\mathrm{i}}\delta}\int_{p_{E}}\frac{{\mathrm{e}}^{-{\mathrm{i}}p_{4}(\tau_{E}+{\mathrm{i}}\delta)-{\mathrm{i}}p_{E}\cdot
y_{E}/\hbar}}{(p^{2}_{E})^{2}}\\\ &=(\pm
1)\cdot(-2{\mathrm{i}})\int_{p_{E}}\theta(p_{4})\frac{{\mathrm{e}}^{-{\mathrm{i}}p_{E}\cdot
y_{E}/\hbar}}{(p^{2}_{E})^{2}},\end{split}$ (88)

where we denote the Euclidean splitting parameter as
$y_{E}=\sqrt{y_{4}^{2}+\boldsymbol{y}^{2}}$ with $y_{4}:=iy_{0}$ and inner
product as $p_{E}\cdot y_{E}=\boldsymbol{p}\cdot\boldsymbol{y}+p_{4}y_{4}$. In
the following, we suppress the subscript $E$. Due to this contour deformation,
the integral (87) is represented as the momentum integral in the Euclidean
four-dimensional half hypersphere:

$\begin{split}\mathcal{K}_{1}(y)&=-2\int_{p}\theta(p_{4})\frac{{\mathrm{e}}^{-{\mathrm{i}}p\cdot
y/\hbar}}{p^{4}}=-\frac{1}{4\pi^{2}}\int_{0}^{\infty}\frac{{\mathrm{d}}p}{p}\int_{p_{4}>0}\frac{{\mathrm{d}}\Omega}{2\pi^{2}}{\mathrm{e}}^{-{\mathrm{i}}py\cos\omega/\hbar},\end{split}$
(89)

where $\omega$ is the angular valuable defined by $p\cdot y=py\cos\omega$.

To proceed, it is useful to introduce two integral sequences. The first one is

$\mathcal{Z}_{n}(x):=\int_{0}^{1}\frac{{\mathrm{d}}\zeta}{\pi/2}\zeta^{n}\sqrt{1-\zeta^{2}}\,{\mathrm{e}}^{-{\mathrm{i}}x\zeta}.$
(90)

This $\mathcal{Z}_{n}(x)$ can be written with the Bessel function of the first
kind $J_{n}(x)$ and the Struve function ${\bf H}_{n}(x)$, as follows:

$\begin{split}\mathcal{Z}_{1}(x)&=\frac{2}{3\pi}-{\mathrm{i}}\frac{J_{2}(x)}{x}-\frac{{\bf
H}_{2}(x)}{x},\\\
\mathcal{Z}_{2}(x)&=\frac{-2{\mathrm{i}}x}{15\pi}+\frac{J_{2}(x)-xJ_{3}(x)}{x^{2}}+{\mathrm{i}}\frac{-{\bf
H}_{2}(x)+x{\bf H}_{3}(x)}{x^{2}},\\\
\mathcal{Z}_{3}(x)&=\frac{4}{15\pi}+{\mathrm{i}}\frac{-3J_{3}(x)+xJ_{4}(x)}{x^{2}}+\frac{3{\bf
H}_{3}(x)-x{\bf H}_{2}(x)}{x^{2}},\\\
\mathcal{Z}_{4}(x)&=-\frac{2{\mathrm{i}}x}{21\pi}+\frac{2xJ_{4}(x)-(x^{2}-3)J_{3}(x)}{x^{3}}+{\mathrm{i}}\frac{-2x{\bf
H}_{4}(x)+(x^{2}-3){\bf H}_{3}(x)}{x^{3}},\\\
\mathcal{Z}_{5}(x)&=-\frac{2(x^{2}-8)}{105\pi}+{\mathrm{i}}\frac{(x^{2}-15)J_{4}(x)}{x^{3}}+\frac{-(x^{2}-15){\bf
H}_{4}(x)}{x^{3}}.\end{split}$ (91)

One important property of $\mathcal{Z}_{n}(x)$ is that the integral from $0$
to $\infty$ are analytically evaluated as:

$\begin{split}\int_{0}^{\infty}{\mathrm{d}}z\,\mathcal{Z}_{1}(z)&=-\frac{{\mathrm{i}}}{2},\quad\int_{0}^{\infty}{\mathrm{d}}z\,\mathcal{Z}_{2}(z)=-\frac{2{\mathrm{i}}}{3\pi},\quad\int_{0}^{\infty}{\mathrm{d}}z\,\mathcal{Z}_{3}(z)=-\frac{{\mathrm{i}}}{8},\\\
\int_{0}^{\infty}{\mathrm{d}}z\,\mathcal{Z}_{4}(z)&=-\frac{4i}{15\pi},\quad\int_{0}^{\infty}{\mathrm{d}}z\,\mathcal{Z}_{5}(z)=-\frac{{\mathrm{i}}}{16}.\end{split}$
(92)

Another property is the recurrence relation

$\mathcal{Z}^{\prime}_{n}(x)=-{\mathrm{i}}\mathcal{Z}_{n+1}(x),\quad\mathcal{Z}_{n}(x)=-{\mathrm{i}}\int_{\infty}^{x}{\mathrm{d}}z\,\mathcal{Z}_{n+1}(z),$
(93)

where the latter follow from
$\mathcal{Z}_{n}(x)\underset{x\to\infty}{\longrightarrow}0$.

The second useful integral is

$\mathcal{A}^{\mu_{1}\cdots\mu_{n}}(x):=\int_{p_{4}>0}\frac{{\mathrm{d}}\Omega}{2\pi^{2}}\hat{p}^{\mu_{1}}\cdots\hat{p}^{\mu_{n}}{\mathrm{e}}^{-{\mathrm{i}}x\cos\omega}.$
(94)

This tensor is decomposed into the longitudinal component to
$\hat{y}^{\mu}:=y^{\mu}/y$ and transverse one with the projector
$\tilde{\Delta}^{\mu\nu}:=\delta^{\mu\nu}-\hat{y}^{\mu}\hat{y}^{\nu}$. The
coefficients of them is determined by $\mathcal{Z}_{n}$, as follows:

$\begin{split}\mathcal{A}&=\mathcal{Z}_{0},\quad\mathcal{A}^{\mu}=\hat{y}^{\mu}\mathcal{Z}_{1},\quad\mathcal{A}^{\mu\nu}=\hat{y}^{\mu}\hat{y}^{\nu}\mathcal{Z}_{2}+\tilde{\Delta}^{\mu\nu}\frac{1}{3}\Bigl{[}\mathcal{Z}_{0}-\mathcal{Z}_{2}\Bigr{]},\\\
\mathcal{A}^{\mu\nu\rho}&=\hat{y}^{\mu}\hat{y}^{\nu}\hat{y}^{\rho}\mathcal{Z}_{3}+\left(\hat{y}^{\mu}\tilde{\Delta}^{\nu\rho}+\hat{y}^{\nu}\tilde{\Delta}^{\rho\mu}+\hat{y}^{\rho}\tilde{\Delta}^{\mu\nu}\right)\frac{1}{3}\Bigl{[}\mathcal{Z}_{1}-\mathcal{Z}_{3}\Bigr{]},\\\
\mathcal{A}^{\mu\nu\rho\sigma}&=\hat{y}^{\mu}\hat{y}^{\nu}\hat{y}^{\rho}\hat{y}^{\sigma}\mathcal{Z}_{4}\\\
&\quad+\left(\tilde{\Delta}^{\mu\nu}\hat{y}^{\rho}\hat{y}^{\sigma}+\tilde{\Delta}^{\nu\rho}\hat{y}^{\sigma}\hat{y}^{\mu}+\tilde{\Delta}^{\rho\sigma}\hat{y}^{\mu}\hat{y}^{\nu}+\tilde{\Delta}^{\mu\sigma}\hat{y}^{\nu}\hat{y}^{\rho}+\tilde{\Delta}^{\mu\rho}\hat{y}^{\nu}\hat{y}^{\sigma}+\tilde{\Delta}^{\nu\sigma}\hat{y}^{\mu}\hat{y}^{\rho}\right)\frac{1}{3}\Bigl{[}\mathcal{Z}_{2}-\mathcal{Z}_{4}\Bigr{]},\\\
&\quad+\left(\tilde{\Delta}^{\mu\nu}\tilde{\Delta}^{\rho\sigma}+\tilde{\Delta}^{\mu\rho}\tilde{\Delta}^{\nu\sigma}+\tilde{\Delta}^{\mu\sigma}\tilde{\Delta}^{\nu\rho}\right)\frac{1}{15}\Bigl{[}\mathcal{Z}_{0}-2\mathcal{Z}_{2}+\mathcal{Z}_{4}\Bigr{]},\end{split}$
(95)

where we abbreviate the argument $x$ on $\mathcal{A}^{\mu_{1}\cdots\mu_{n}}$
and $\mathcal{Z}_{n}$.

Let us come back to the evaluation of $\mathcal{K}_{1}$ With the help of
$\mathcal{A}$ and $\mathcal{Z}_{0}$, we get

$\begin{split}\mathcal{K}_{1}(y)&=-\frac{1}{4\pi^{2}}\int_{0}^{\infty}\frac{{\mathrm{d}}p}{p}\mathcal{A}(py/\hbar)=-\frac{1}{4\pi^{2}}\int_{0}^{\infty}\frac{{\mathrm{d}}p}{p}\mathcal{Z}_{0}(py/\hbar)\\\
&=-\frac{(-{\mathrm{i}}/\hbar)}{4\pi^{2}}\int_{0}^{\infty}{\mathrm{d}}p\int_{\infty}^{y}{\mathrm{d}}z\,\mathcal{Z}_{1}(pz/\hbar)\\\
&=-\frac{(-{\mathrm{i}})}{4\pi^{2}}\int_{\infty}^{y}\frac{{\mathrm{d}}z}{z}\cdot\frac{-{\mathrm{i}}}{2}=-\frac{\mathcal{J}}{8\pi^{2}}\end{split}$
(96)

with

$\mathcal{J}:=\int_{0}^{y^{-1}}\frac{{\mathrm{d}}p}{p}.$ (97)

Here we interchanged the order of integration, as do in the point-splitting
regularization for axial anomaly in two dimensions [50]. The ultraviolet
logarithmic divergence is now regularized by the splitting parameter $y$. The
infrared divergence ($z=0$) is to be canceled by the matter part.

In the same manner, we can calculate $\mathcal{K}^{\mu\nu}_{2}$ and
$\mathcal{K}^{\mu\nu\rho\sigma}_{3}$. The only extra relation to be utilized
is

$\frac{{\mathrm{d}}^{n}\delta(p^{2})}{({\mathrm{d}}p^{2})^{n}}=\frac{(-1)^{n}\,n!}{\pi}\mathrm{Im}\frac{1}{\,(p^{2}-{\mathrm{i}}\epsilon)^{n+1}}.$
(98)

Finally, we get the following expressions:

$\mathcal{K}_{2}^{\mu\nu}(y)=\frac{\mathcal{J}}{16\pi^{2}}g^{\mu\nu},\quad\mathcal{K}_{3}^{\mu\nu\rho\sigma}(y)=-\frac{\mathcal{J}}{32\pi^{2}}(g^{\mu\nu}g^{\rho\sigma}+g^{\mu\rho}g^{\nu\sigma}+g^{\mu\sigma}g^{\nu\rho}),$
(99)

where we perform the analytic continuation to Minkowski spacetime after
integration.

##  Appendix E Formulas of background fields

In this Appendix, we show several formulas of electromagnetic field and fluid
vorticity fields, which are defined in Eq. (LABEL:eq:Fomega). The two rank
tensors $F_{\mu\nu}$, $\beta^{-1}\partial_{\mu}\beta_{\nu}$ and their duals
are expanded with $E_{\mu},B_{\mu},a_{\mu}$ and $\omega_{\mu}$ as follows:

$\begin{split}&F_{\mu\nu}=E_{\mu}\xi_{\nu}-E_{\nu}\xi_{\mu}-\varepsilon_{\mu\nu\rho\sigma}B^{\rho}\xi^{\sigma},\quad\tilde{F}_{\mu\nu}=B_{\mu}\xi_{\nu}-B_{\nu}\xi_{\mu}+\varepsilon_{\mu\nu\rho\sigma}E^{\rho}\xi^{\sigma},\\\
&\beta^{-1}\partial_{\mu}\beta_{\nu}=a_{\mu}\xi_{\nu}-a_{\nu}\xi_{\mu}-\varepsilon_{\mu\nu\rho\sigma}\omega^{\rho}\xi^{\sigma},\quad\omega_{\mu\nu}=\omega_{\mu}\xi_{\nu}-\omega_{\nu}\xi_{\mu}+\varepsilon_{\mu\nu\rho\sigma}a^{\rho}\xi^{\sigma}.\end{split}$
(100)

Thanks to them, one can show
$\varepsilon_{\mu\nu\rho\sigma}\omega^{\rho}E^{\sigma}=-\varepsilon_{\mu\nu\rho\sigma}a^{\rho}B^{\sigma}$
or equivalently,

$\omega_{[\mu}E_{\nu]}=-a_{[\mu}B_{\nu]}.$ (101)

Using the equilibrium conditions $\partial_{\mu}\alpha=F_{\mu\nu}\beta^{\nu}$
in Eq. (23) and $\beta\cdot\partial F_{\mu\nu}=0$ in Eq. (33), we find

$0=\partial_{[\mu}\partial_{\nu]}\alpha=F_{\lambda[\mu}\partial_{\nu]}\beta^{\lambda}.$
(102)

Combined with Eqs. (100) and (101), this leads to

$a_{[\mu}E_{\nu]}=\omega_{[\mu}B_{\nu]}=0.$ (103)

Besides, by using the expanded form (100), we express the derivatives of the
background fields as

$\begin{split}&\partial_{\mu}E_{\nu}=\xi_{\mu}\xi_{\nu}(E\cdot
a+B\cdot\omega)-g_{\mu\nu}B\cdot\omega+B_{\mu}\omega_{\nu}+2\xi_{(\mu}\varepsilon_{\nu)\rho\sigma\lambda}a^{\rho}B^{\sigma}\xi^{\sigma}+\xi^{\rho}\partial_{\mu}F_{\nu\rho},\\\
&\partial_{\mu}B_{\nu}=\xi_{\mu}\xi_{\nu}(a\cdot B-\omega\cdot
E)+g_{\mu\nu}\omega\cdot
E-E_{\mu}\omega_{\nu}-2\xi_{(\mu}\varepsilon_{\nu)\rho\sigma\lambda}a^{\rho}E^{\sigma}\xi^{\lambda}+\xi^{\lambda}\partial_{\mu}\tilde{F}_{\nu\lambda},\\\
&\partial_{\mu}a_{\nu}=\xi_{\mu}\xi_{\nu}(a^{2}+\omega^{2})-g_{\mu\nu}\omega^{2}+\omega_{\mu}\omega_{\nu}-a_{\mu}a_{\nu}+2\xi_{(\mu}\varepsilon_{\nu)\rho\sigma\lambda}a^{\rho}\omega^{\sigma}\xi^{\lambda},\\\
&\partial_{\mu}\omega_{\nu}=g_{\mu\nu}a\cdot\omega-2a_{\mu}\omega_{\nu}.\end{split}$
(104)

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