# Universal conductivity at a 2d superconductor-insulator transition: the
effects of quenched disorder and Coulomb interaction

Chao-Jung Lee  Department of Physics, California Institute of Technology,
Pasadena, CA 91125, USA  Michael Mulligan  Department of Physics and
Astronomy, University of California, Riverside, CA 92521, USA

###### Abstract

We calculate the zero-temperature universal electrical conductivity at a
superconductor-insulator transition in two spatial dimensions. We focus on
transitions in the universality class of the dirty 3d XY model. We use a dual
model consisting of a single Dirac fermion at zero density coupled to a Chern-
Simons gauge field in the presence of a quenched random mass, with or without
an unscreened Coulomb interaction. Our calculation is performed in a $1/N_{f}$
expansion, where $N_{f}$ is the number of Dirac fermions. At zeroth order, the
model exhibits particle-vortex self-dual electrical transport with
$\sigma_{xx}\lesssim(2e)^{2}/h$ and small, but finite $\sigma_{xy}$.
Corrections of ${\cal O}(1/N_{f})$ due to fluctuations in the Chern-Simons
gauge field and disorder produce violations of self-duality. We find these
violations to be milder when the Coulomb interaction is present.

###### Contents

  1. I Introduction
  2. II The Model
    1. II.1 The effective action
    2. II.2 The effective gauge field propagator
  3. III Composite fermion conductivity
    1. III.1 Definition of the composite fermion conductivity
    2. III.2 Two-loop corrections to the conductivity
  4. IV Quantum critical electrical transport
    1. IV.1 RG flow and critical composite fermion conductivity
    2. IV.2 Electrical response
  5. V Discussion
  6. A Gauge boson propagator
  7. B Random mass correction to the composite fermion conductivity
    1. B.1  $\hat{\Pi}^{(b+c)}$
    2. B.2 $\hat{\Pi}^{(a)}$
  8. C Gauge field correction to the composite fermion conductivity
    1. C.1 Useful integrals
      1. C.1.1 Basic integral building blocks
      2. C.1.2 More complicated integral formulas
    2. C.2  Anti-symmetric component of $\Pi^{(b+c)}$
    3. C.3 Anti-symmetric component of $\Pi^{(a)}$
    4. C.4 Combining the anti-symmetric components of $\Pi^{(a)}$ and $\Pi^{(b+c)}$
    5. C.5 Symmetric component of $\Pi^{(b+c)}$
    6. C.6 Symmetric component of $\Pi^{(a)}$
    7. C.7 Combining the symmetric components of $\Pi^{(a)}$ and $\Pi^{(b+c)}$

##  I Introduction

Continuous quantum phase transitions [1, 2] are one of the most intriguing
phenomena in condensed matter physics, due to the possibility of emergent
behaviors that are different from the proximate phases. Superconductor to
insulator transitions (SITs) in disordered two-dimensional thin films, the
focus of this paper, provide some of the best examples ([3], for review).
Here, as a tuning parameter, such as the charge density, disorder, or external
magnetic field, is varied, the electrons in the disordered film transition
between a superconducting and insulating ground state. At a critical value of
the tuning parameter, the electrons form a metal with finite, nonzero dc
resistance (measured as the temperature $T\rightarrow 0$) on the order of the
quantum of Cooper-pair resistance $R_{Q}=h/4e^{2}\approx 6.45\
k\Omega/\square$, consistent with the theoretical prediction [4, 5, 6] that
the critical resistivity is a universal amplitude.

This prediction is based on the hypothesis [5, 6] that such transitions are
due to the phase disordering of the superconducting order parameter, rather
than the loss of superconducting pairing amplitude [7]. The critical
properties of such a transition should then be described by a model of
interacting charge-2e Cooper-pair bosons moving in a random potential in two
spatial dimensions [5, 6]. One of the more intriguing possibilities suggested
by this model is that the transition might be self-dual, i.e., the critical
Hamiltonian for Cooper-pair bosons on the brink of localization is the same as
the dual critical Hamiltonian for vortices [8, 9, 10] on the brink of
condensation. A consequence of self-duality is the so-called semicircle law
for the dc electrical conductivity:

$\displaystyle\sigma^{2}_{xx}+\sigma^{2}_{xy}=\big{(}\frac{4e^{2}}{h}\big{)}^{2}.$
(I.1)

This relation is particularly interesting because it relates the dissipative
($\sigma_{xx}$) and nondissipative ($\sigma_{xy}$) parts of the conductivity
to a universal constant, $R_{Q}$. There is strong experimental evidence [11]
that field-tuned transitions are self-dual with $\sigma_{xx}=1/R_{Q}$ and
$\sigma_{xy}=0$. This may be surprising, given the presence of the nonzero
magnetic field, and suggests an emergent particle-hole symmetry [12].
Measurements [3] of the longitudinal resistance near $R_{Q}$ at charge density
or disorder tuned SITs are suggestive of an underlying self-duality, so long
as the impurities are nonmagnetic.

The argument for self-duality within the boson model is indirect [5, 6]. Here,
we will consider an alternative description, given in terms of 3d quantum
electrodynamics with a single Dirac fermion coupled to a Chern-Simons gauge
field [13, 14, 15, 16]. In close analogy to theories [17, 18, 19] for the
half-filled Landau level, we will refer to the Dirac fermions in this
description as composite fermions. In this composite fermion model, self-
duality is a consequence of an unbroken particle-vortex symmetry that (loosely
speaking) acts on the Hamiltonian of the model as a time-reversal symmetry and
therefore fixes the composite fermion Hall conductivity to be zero [16]. (The
composite fermion conductivity is distinct from the electrical conductivity.
We will recall the correspondence between the two in §IV.2.) This means that,
so long as particle-vortex symmetry is preserved, the composite fermion theory
will yield self-dual response. The precise manner in which self-duality (I.1)
is realized depends on the specific value of the composite fermion
longitudinal conductivity. For instance, a composite fermion longitudinal
conductivity of ${1\over 2}R^{-1}_{Q}$ reproduces (I.1) with $\sigma_{xy}=0$.

Calculations of the longitudinal conductivity are notoriously challenging
(e.g., [20, 21, 22]). To gain insight, we consider a limit of the composite
fermion theory that corresponds to lattice bosons at integer filling with
charge-conserving disorder [4]. In this limit, the critical properties are
those of the dirty 3d XY model [4], recently clarified in [23]. In the fermion
dual, the Dirac composite fermions lie at zero density in the presence of a
quenched random mass. In [24], we showed that this theory admits
renormalization group (RG) fixed points at finite disorder, with or without
the Coulomb interaction, when the mean of the random mass is tuned to zero.
(Other symmetry classes of disorder have not, as yet, yielded accessible fixed
points in this theory.) These fixed points are controlled within a $1/N_{f}$
expansion, where $N_{f}$ is the number of fermion flavors. Related works
studying the effects of quenched randomness on theories of Dirac fermions
coupled to a fluctuating boson include [25, 26, 27, 28, 29, 30].

An unscreened Coulomb interaction is believed to be relevant to the
experimentally-realized SITs [5, 6]; within the 3d XY model, the Coulomb
interaction corresponds to a relevant perturbation. General constraints on the
transport properties of a self-dual SIT, with Coulomb interactions, were
derived in [31, 32]. The stability of these results to fluctuations in the
Coulomb interaction, in the presence of disorder, is not addressed in these
papers. Here, we determine the electrical conductivity at the random-mass
fixed points of [24], with or without the unscreened Coulomb interaction, to
${\cal O}(1/N_{f})$, with our results given in Eqs. (IV.19) and (IV.20) and
Figs. 4(a) and 4(b). At ${\cal O}(1/N_{f}^{0})$, the theory exhibits self-dual
electrical transport with $\sigma_{xx}\lesssim 1/R_{Q}$ and finite, nonzero
$\sigma_{xy}$. The leading ${\cal O}(1/N_{f})$ corrections, due to
fluctuations in the Chern-Simons gauge field and disorder, violate self-
duality. These violations are found to be smaller at the fixed point with
unscreened Coulomb interaction.

There are two important things to note about our results. First, the random
mass does not preserve particle-vortex symmetry. This is not the underlying
reason why we find nonzero composite fermion Hall conductivity: The ${\cal
O}(1/N_{f})$ violation of self-duality is due to the fluctuations of the
Chern-Simons gauge field; there is a nonzero composite fermion Hall
conductivity in the pure limit without disorder. We are unaware of a
calculable model of a disordered fixed point that is self-dual. Second, we
calculate the conductivity in the phase-coherent regime
$\hbar\omega/k_{B}T\rightarrow\infty$ (where $\omega$ is the measuring
frequency). This regime gives a universal conductivity tensor that, in
general, differs from conductivity in the incoherent regime
$\hbar\omega/k_{B}T\rightarrow 0$ [22].

The structure of this paper is as follows. In §II, we define the model that we
study. In §III, we sketch the calculation of the composite fermion
conductivity for general values of the coupling constants. In §IV, we evaluate
these composite fermion conductivities at two RG fixed points, corresponding
to disordered quantum critical points with or without the unscreened Coulomb
interaction, and use the duality dictionary to translate these quantum
critical composite fermion conductivities to electrical conductivities. In §V,
we discuss our results. Three appendices contain technical details of the
calculations summarized in the main parts of the paper: Appendix A contains
the derivation of the effective gauge field propagator; Appendices B and C
contain details for the evaluation of loop diagrams.

##  II The Model

In this section, we define the Dirac composite theory of the SIT. See [24] for
additional details.

###  II.1 The effective action

The total Euclidean effective action for the Dirac composite fermion theory
has three parts:

$\displaystyle S_{\text{tot}}=S_{0}+S_{c}+S_{\text{dis}}.$ (II.1)

We will define each part in the following three paragraphs.

To begin, $S_{0}$ is a Euclidean action of quantum electrodynamics in 3d. It
consists of a two-component Dirac fermion $\psi$ that is coupled to a
dynamical Chern-Simons gauge field $a_{\mu}$ ($\hbar=2e=1$):

$\displaystyle S_{0}$ $\displaystyle=\sum_{I=1}^{N_{f}}\int
d^{2}xd\tau\,\bar{\psi}_{I}\gamma^{0}\,\,(\partial_{\tau}-i\frac{g}{\sqrt{N_{f}}}\,a_{\tau})\psi_{I}+\bar{\psi}_{I}\gamma^{j}\,(\partial_{j}-i\frac{g}{\sqrt{N_{f}}}\,a_{j})\psi_{I}$
$\displaystyle+\frac{i\,\kappa}{2}ada+\frac{i}{4\pi}(-2Ada+AdA).$ (II.2)

In (II.1), $\bar{\psi}\equiv\psi^{\dagger}\gamma^{0}$ and the $2\times 2$
gamma matrices are chosen such that the anti-commutator
$\\{\gamma^{\mu},\gamma^{\nu}\\}=2\delta^{\mu\nu}\bm{1}_{2\times 2}$, where
$\mu,\nu\in\\{0,1,2\\}=\\{\tau,x,y\\}$. We follow the Einstein summation
convention, with, for instance, the spatial indices $j\in\\{x,y\\}$ summed
over above. (Being in Euclidean signature, upper and lower indices are
equivalent and will be used interchangeably.) The Chern-Simons term
$ada\equiv\epsilon^{\mu\nu\lambda}a_{\mu}\partial_{\nu}a_{\lambda}$ (and
similarly for the other Chern-Simons terms), where the Levi-Civita symbol
$\epsilon^{012}\equiv+1$. Here, $N_{f}$ is the number of Dirac fermion
flavors. $S_{0}$ is dual to the critical 3d XY model when $N_{f}=1$ and the
Chern-Simons coefficient $\kappa=\frac{1}{4\pi}$ [13, 14, 16], with the Dirac
mass playing the role as the critical tuning parameter. We take $N_{f}>1$ in
order to study the theory within a controlled $1/N_{f}$ expansion. $A_{\mu}$
is a probe, nondynamical electromagnetic gauge potential that serves to define
the electrical current and corresponding electrical conductivity of the model.
The gauge coupling $g$ is set to unity in the infrared; we keep $g$ general
for the moment.

Next, we introduce the Coulomb interaction. Following [33, 24], the Coulomb
interaction dualizes into the composite fermion theory as

$\displaystyle
S_{c}=\frac{1}{2}\int\frac{d^{3}k}{(2\pi)^{3}}a_{T}(k)\Big{(}\,|\bm{k}|\,w_{x}\Big{)}a_{T}(-k),$
(II.3)

where we introduced the effective Coulomb coupling
$w_{x}\equiv\frac{e^{2}_{*}}{8\pi^{2}}$. Here, $e_{\ast}$ is the charge of the
Cooper-pair bosons (nominally, $e_{\ast}=2e$). The Fourier space integration
measure is $d^{3}k\equiv dk_{x}dk_{y}dk_{0}$, where $k_{0}$ is the zero-
temperature Matsubara frequency. We adopt the following notation throughout
this paper:

$\displaystyle\bm{k}=(k_{x},k_{y}),\;k_{\mu}=(k_{0},k_{x},k_{y}),\;|\bm{k}|=\sqrt{k_{x}^{2}+k_{y}^{2}},\;k=\sqrt{k_{0}^{2}+k_{x}^{2}+k_{y}^{2}}.$
(II.4)

The “$T$” subscript indicates that $a_{T}$ is the transverse component of the
Chern-Simons gauge field: $a_{T}$ is related to the Cartesian components
$(a_{x},a_{y})$ via $a_{x}(q)=i\frac{q_{y}}{|\bm{q}|}a_{T}(q)$ and
$a_{y}(q)=-i\frac{q_{x}}{|\bm{q}|}a_{T}(q)$, provided we choose Coulomb gauge,
wherein the longitudinal component of $a_{j}$ is set to zero.

Finally, we turn to the introduction of the quenched randomness. Previous
analyses [24, 29] show that if all types of disorder (quenched random
couplings that couple to composite fermion bilinears), regardless of the
symmetry they individually preserve, are added to the theory, the system flows
to strong coupling, out of reach of any analytic control. However, if charge-
conjugation symmetry is imposed, only a random mass term
$m(\bm{x})\bar{\psi}\psi$ is allowed and the theory flows to an accessible
disordered fixed point [24]. Applying the standard replica trick to disorder
average, the action picks up the term,

$\displaystyle S_{\text{dis}}=\frac{-1}{2}\int d^{2}xd\tau
d\tau^{\prime}\sum_{r,r^{\prime}=1}^{n_{r}}g_{m}\,(\bar{\psi}_{r}\psi_{r})_{x,\tau}\,(\bar{\psi}_{r^{\prime}}\psi_{r^{\prime}})_{x,\tau^{\prime}}\,,$
(II.5)

where $r,r^{\prime}$ are the replica indices and the number of replica $n_{r}$
is to be set to zero in the last step of any calculation. The replica indices
are not important in the later calculations, so we will not write them out
explicitly anymore (this is also why we didn’t include them in the earlier
parts of the total action). The parameter $g_{m}$ is the disorder strength,
which is always non-negative. The random mass disorder does not preserve
particle-vortex symmetry. This will be apparent later when we discuss the
quantum critical conductivity of the model.

###  II.2 The effective gauge field propagator

In the Cartesian basis, the one-loop gauge field self-energy induced by the
Dirac fermions is

$\displaystyle\Pi_{\mu\nu}^{1-loop}=\frac{-g^{2}}{16}\frac{\delta_{\mu\nu}k^{2}-k_{\mu}k_{\nu}}{k}.$
(II.6)

This self-energy is ${\cal O}(N_{f}^{0})$, i.e., the same order as the bare
terms in the total action that are quadratic in the gauge field. Adding this
self-energy to $S_{c}$ and the Chern-Simons term in $S_{0}$ involving
$a_{\mu}$ only, we obtain the one-loop gauge field action,

$\displaystyle
S_{\text{gauge}}=\frac{1}{2}\int\frac{d^{3}k}{(2\pi)^{3}}\begin{pmatrix}a_{0}&a_{T}\end{pmatrix}_{k}\begin{pmatrix}\frac{g_{X}^{2}\bm{k}^{2}}{k}&i\kappa\,|\bm{k}|\\\
i\kappa\,|\bm{k}|&\,g_{X}^{2}\,k+|\bm{k}|\,w_{x}\\\
\end{pmatrix}\begin{pmatrix}a_{0}\\\ a_{T}\end{pmatrix}_{-k},$ (II.7)

where $g_{X}^{2}\equiv\frac{g^{2}}{16}$. This action defines the one-loop
corrected gauge field propagator.

The computation we summarize in the next section is performed with the gauge
field, expressed in the Cartesian basis. We therefore translate the gauge
field propagator, defined by Eq. (II.7), to the Cartesian basis:

$\displaystyle
D_{00}(k_{0},\bm{k})=\frac{1}{\bm{k}^{2}}\frac{1}{\kappa^{2}+g_{X}^{4}+\frac{w_{x}g_{X}^{2}\,|\bm{k}|}{k}}(g_{X}^{2}\,k+|\bm{k}|\,w_{x})$
(II.8) $\displaystyle
D_{0i}(k_{0},\bm{k})=-D_{i0}=\frac{1}{\bm{k}^{2}}\frac{1}{\kappa^{2}+g_{X}^{4}+\frac{w_{x}g_{X}^{2}\,|\bm{k}|}{k}}\,(\frac{\kappa\,\epsilon_{ij}k_{j}}{1})$
(II.9) $\displaystyle
D_{ij}(k_{0},\bm{k})=\frac{1}{\kappa^{2}+g_{X}^{4}+\frac{w_{x}g_{X}^{2}\,|\bm{k}|}{k}}(\frac{g_{X}^{2}}{k})(\delta_{ij}-\frac{k_{i}k_{j}}{\bm{k^{2}}})$
(II.10)

To obtain $D_{\mu\nu}$ above, we have used the gauge fixing term
$\frac{(\nabla a)^{2}}{2\lambda}$. See Appendix A for details.

##  III Composite fermion conductivity

In this section, we sketch the computation of the composite fermion
conductivity, to two-loop order. The details of this calculation are in
Appendices B and C. This composite fermion conductivity is related to the
electrical conductivity in the next section.

###  III.1 Definition of the composite fermion conductivity

The composite fermion conductivity $\sigma^{\psi}_{ij}(\omega,{\bf p})$ is
determined by the real-frequency retarded gauge field self-energy
$\Pi^{R}_{ij}$, which is related to the imaginary-time self-energy $\Pi^{\rm
tot}_{ij}$ as follows:

$\displaystyle\sigma^{\psi}_{ij}(\omega,\bm{p})=\frac{i}{\omega}\Pi^{R}_{ij}(\omega+i0^{+},\bm{p})=\frac{i}{\omega}\Pi^{\rm
tot}_{ij}(ip_{0}\to\omega+i0^{+},\bm{p}).$ (III.1)

Our definition of $\sigma_{ij}^{\psi}$ includes the contributions from the
composite fermions; it does not include the tree-level Chern-Simons term. We
decompose $\Pi^{\rm tot}_{ij}$ into symmetric $\Pi^{\rm tot}_{(\text{S})}$ and
anti-symmetric $\Pi^{\rm tot}_{(\text{A})}$ components as

$\displaystyle\Pi^{\rm
tot}_{\mu\nu}(p)=(\delta_{\mu\nu}-\frac{p_{\mu}p_{\nu}}{p^{2}})\;\Pi^{\rm
tot}_{(\text{S})}(p)+\frac{p_{\kappa}}{|p|}\,\epsilon_{\kappa\mu\nu}\Pi^{\rm
tot}_{(\text{A})}(p).$ (III.2)

To extract the symmetric component of $\Pi^{\rm tot}_{ij}$, we will set
$(\mu,\nu)=(x,x)$ and evaluate:

$\displaystyle\Pi^{\rm
tot}_{(\text{S})}=\frac{1}{1-\frac{p_{x}^{2}}{p^{2}}}\Pi^{\rm tot}_{xx}.$
(III.3)

The anti-symmetric component will be obtained by contracting $\Pi^{\rm
tot}_{ij}$ with $\epsilon_{\mu\nu\lambda}p^{\lambda}$:

$\displaystyle\Pi^{\rm
tot}_{(\text{A})}=\frac{p_{\lambda}\epsilon_{\mu\nu\lambda}}{2|p|}\Pi^{\rm
tot}_{\mu\nu}.$ (III.4)

###  III.2 Two-loop corrections to the conductivity

We now calculate the composite fermion conductivity to two-loops, obtained
from the two-loop corrections to $\Pi^{\rm tot}_{ij}$ by Eq. (III.1). The one-
loop corrections have already been included in the 1-loop corrected gauge
boson self-energy. The two-loop corrections result from self-energy
corrections due to the random mass $\hat{\Pi}_{ij}$ to ${\cal O}(g_{m})$ and
the fluctuations of the gauge boson $\Pi_{ij}$ to ${\cal O}(1/N_{f})$. The
diagrams contributing to these corrections are shown in Figs. 1 and 2. In each
figure, diagrams (b) and (c) are equal.

Figure 1: Two-loop diagrams with random mass corrections, denoted as
$\hat{\Pi}_{ij}$.  Figure 2: Two-loop diagrams with gauge field corrections,
denoted as $\Pi_{ij}$.

We start with the corrections due to the random mass. The diagrams in Fig. 1
are:

$\displaystyle\hat{\Pi}^{(a)}_{\mu\nu}(p)=-(\frac{ig}{\sqrt{N_{f}}})^{2}\,N_{f}\int_{k,q}\text{Tr}[\gamma_{\mu}G(q)G(q+k)\gamma_{\nu}G(q+k+p)G(p+q)]\times
2\pi g_{m}\delta(k_{0}),\quad\quad\quad$ (III.5)
$\displaystyle\hat{\Pi}^{(b+c)}_{\mu\nu}(p)=-2(\frac{ig}{\sqrt{N_{f}}})^{2}\,N_{f}\int_{k,q}\text{Tr}[\gamma_{\mu}G(q)G(k+q)G(q)\gamma_{\nu}G(p+q)]\times
2\pi g_{m}\delta(k_{0}),$ (III.6)

where we have used the shorthand
$\int_{k,q}\equiv\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}q}{(2\pi)^{3}}$
and the trace is performed over the Dirac indices. Here, $G(k)$ is the
composite fermion propagator:

$\displaystyle-iG(k)={k_{0}\gamma^{0}+k_{i}\gamma^{i}\over
k_{0}^{2}+k_{i}k^{i}}.$ (III.7)

To evaluate the gauge field diagrams in Fig. 2, we replace the disorder-
induced four-point interaction by the gauge field propagator and disorder
strength coupling by the fermion-gauge field coupling:

$\displaystyle\Pi^{(a)}_{\mu\nu}(p)=-\frac{g^{4}}{N_{f}}\int_{k,q}\text{Tr}[\gamma_{\mu}G(q)\gamma_{\alpha}G(q+k)\gamma_{\nu}G(q+k+p)\gamma_{\beta}G(p+q)]\,D_{\alpha\beta}(k),\qquad$
(III.8)
$\displaystyle\Pi^{(b+c)}_{\mu\nu}(p)=-\frac{2g^{4}}{N_{f}}\int_{k,q}\text{Tr}[\gamma_{\mu}G(q)\gamma_{\alpha}G(k+q)\gamma_{\beta}G(q)\gamma_{\nu}G(p+q)]\,D_{\alpha\beta}(k),\qquad$
(III.9)

where $D_{\alpha\beta}$ is the gauge field propagator.

If the Coulomb interaction is present, the 3d Euclidean symmetry that rotates
the temporal and spatial directions into one another is lost in the gauge
field propagators; this symmetry is already broken by the random mass. (This
is due to the fact that the Coulomb interaction, considered as an
instantaneous interaction, is fundamentally due to an electron density-density
interaction that is mediated by the temporal component of the electromagnetic
gauge field. We have translated this interaction into the composite fermion
theory [24].) This makes the evaluation of these integrals by a naive Feynman
parameterization difficult. Instead, we re-express all terms in the integrands
involving $q$ using partial fractions in a way that allows us to carry out the
UV-finite integrals over $q$, after which we perform the $k$ integrals. The
lengthy calculations that do this are relegated to Appendices B and C. We now
summarize the results.

The only nonzero contribution of the random mass diagrams is to
$\hat{\Pi}_{xx}$:

$\displaystyle\hat{\Pi}^{(a+b+c)}_{xx}(p_{0},\bm{p}=0)=3\times(-1)(\frac{ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\times\frac{1}{2\pi}\times\frac{|p_{0}|}{96}.$
(III.10)

This result was first obtained by Thomson and Sachdev in [29].

The gauge field diagrams result in nonzero contributions to both the symmetric
and anti-symmetric components of $\Pi_{ij}$. The anti-symmetric component is

$\displaystyle\Pi^{(a+b+c)}_{(\text{A})}(p)=-\frac{1}{2|p|}\frac{g^{4}}{N_{f}}\int_{-1}^{1}dz\,p^{2}\,\frac{-z+(-1+z^{2})\text{ArcTanh}[z]}{8\pi^{2}\,z}\times\frac{1}{A_{X}+B_{X}\sqrt{1-z^{2}}},\qquad$
(III.11) $\displaystyle
A_{X}\equiv\frac{\kappa^{2}+g_{X}^{4}}{\kappa},\;\;B_{X}\equiv\frac{w_{x}g_{X}^{2}}{\kappa}.$
(III.12)

The symmetric component is

$\displaystyle\Pi^{(a+b+c)}_{(\text{S})}(p_{0},\bm{p}=0)$
$\displaystyle=-\frac{g^{4}}{N_{f}}\,|p|\int_{-1}^{1}dz\frac{f^{w_{x}}_{0}(z)+f^{w_{x}}_{0}(-z)}{2},$
(III.13) $\displaystyle\frac{f^{w_{x}}_{0}(z)+f^{w_{x}}_{0}(-z)}{2}$
$\displaystyle\equiv\frac{-5\,C_{Y}\,z(-1+z^{2})+z\sqrt{1-z^{2}}(9+z^{2})}{192\pi^{2}\,z\,[\;A_{Y}\sqrt{1-z^{2}}+B_{Y}(1-z^{2})\;\;]}$
(III.14)
$\displaystyle+\frac{6(-1+z^{2})(C_{Y}-C_{Y}\,z^{2}+2\sqrt{1-z^{2}})\,\text{arctanh}[z]}{192\pi^{2}\,z\,[\;A_{Y}\sqrt{1-z^{2}}+B_{Y}(1-z^{2})\;\;]},$
(III.15)

where

$\displaystyle A_{Y}\equiv\frac{\kappa^{2}+g_{X}^{4}}{g_{X}^{2}},\qquad
B_{Y}\equiv w_{x},\qquad C_{Y}\equiv\frac{w_{x}}{g_{X}^{2}}.$ (III.16)

As a consistency check, we note the precise agreement of our results in Eqs.
(III.11) and (III.13) for the gauge field diagrams, evaluated at $w_{x}=0$,
with the earlier computations of these same diagrams, computed in the absence
of the Coulomb interaction, in [34, 35, 36]:

$\displaystyle\Pi^{(a+b+c)}_{(\text{A})}(p;w_{x}=0)=\frac{1}{2|p|}\frac{g^{4}}{N_{f}}\frac{\kappa}{\kappa^{2}+g_{X}^{4}}\;(p^{2}\;\frac{4+\pi^{2}}{32\pi^{2}}\;),$
(III.17)
$\displaystyle\Pi^{(a+b+c)}_{(\text{S})}(p;w_{x}=0,\kappa=0)=\frac{-g^{4}}{N_{f}}\times
0.00893191.$ (III.18)

Plugging these results into the definition Eq. (III.1) using $\Pi^{\rm
tot}_{ij}=\hat{\Pi}_{ij}+\Pi_{ij}$, we find that to
$\mathcal{O}(\frac{1}{N_{f}},g_{m})$ the composite fermion conductivity equals

$\displaystyle\sigma^{\psi}_{xx}(\omega)=\frac{1}{16}-\frac{3g_{m}}{96\pi}+\frac{i}{\omega}\Pi^{(a+b+c)}_{(\text{S})}(|p_{0}|\to
i\omega,\bm{p}=0),$ (III.19)
$\displaystyle\sigma^{\psi}_{xy}(\omega)=\frac{i}{\omega}\Pi^{(a+b+c)}_{(\text{A})}(|p_{0}|\to
i\omega,\bm{p}=0).$ (III.20)

##  IV Quantum critical electrical transport

###  IV.1 RG flow and critical composite fermion conductivity

The perturbative RG flows for this system $S_{\rm tot}$ (II.1) were found in
[24]. In that study, we allowed for the possibility of a dissipative Coulomb
interaction. We will not consider this possibility here. With the Fermi
velocity set equal to unity, the beta functions
($\beta_{X}\equiv-\mu\frac{\partial X}{\partial\mu}$) are

$\displaystyle\beta_{w_{x}}=w_{x}(z-1)=w_{x}[\frac{g_{m}}{2\pi}-F_{w}(w_{x},\kappa)\;],$
(IV.1)
$\displaystyle\beta_{g_{m}}=2g_{m}(z-1-\gamma_{\bar{\psi}\psi})=-2g_{m}[\frac{g_{m}}{2\pi}+F_{m}(w_{x},\kappa)\;].$
(IV.2)

The dynamical exponent $z$, fermion anomalous dimension
$\gamma_{\bar{\psi}\psi}$, and loop-functions $F_{w},\,F_{m}$ are defined as
follows:

$\displaystyle z$ $\displaystyle=1+\frac{g_{m}}{2\pi}-F_{w}(w_{x},\kappa),$
(IV.3) $\displaystyle\gamma_{\bar{\psi}\psi}$
$\displaystyle=2\frac{g_{m}}{2\pi}+F_{m}-F_{w},$ (IV.4) $\displaystyle
F_{w}(w_{x},\kappa)$
$\displaystyle=\frac{1}{4\pi^{2}N_{f}}\int_{-\infty}^{\infty}dy\frac{g_{1}(2y^{2}-1)+w_{x}\sqrt{1+y^{2}}}{(1+y^{2})^{2}[\sqrt{1+y^{2}}(g_{1}^{2}+\kappa^{2})+g_{1}w_{x}\;]},$
(IV.5) $\displaystyle F_{m}(w_{x},\kappa)$
$\displaystyle=\frac{1}{4\pi^{2}N_{f}}\int_{-\infty}^{\infty}dy\Big{[}\frac{g_{1}(-2y^{2}-3)-w_{x}\sqrt{1+y^{2}}}{(1+y^{2})^{2}[\sqrt{1+y^{2}}(g_{1}^{2}+\kappa^{2})+g_{1}w_{x}\;]}$
(IV.6)
$\displaystyle+\frac{\sqrt{1+y^{2}}(g_{1}^{2}-\kappa^{2})+g_{1}w_{x}\;}{2(1+y^{2})[\sqrt{1+y^{2}}(g_{1}^{2}+\kappa^{2})+g_{1}w_{x}\;]^{2}}\Big{]},$
(IV.7)

where $g_{1}\equiv g^{2}_{X}=\frac{g^{2}}{16}$. The dynamical critical
exponent and fermion anomalous dimension determine the inverse correction
length exponent: $\nu^{-1}=z-\gamma_{\bar{\psi}\psi}$. Fig. 3 depicts the RG
flow diagram for these beta functions (IV.1) and (IV.2).

Figure 3: RG flow of the strength of the random mass
($g_{m}=2\pi\overline{g_{m}}$) and Coulomb coupling ($w_{x})$ evaluated at
$N_{f}=2$. The circle denotes the fixed point without the Coulomb interaction.
The star is the fixed point with finite Coulomb interaction.

As Fig. 3 indicates, a finite disorder fixed point exists whether or not the
Coulomb interaction is present. In the absence of the Coulomb interaction
($w_{x}=0$), $\beta_{w_{x}}$ in Eq. (IV.1) vanishes. At $w_{x}=0$, the $y$
integral in Eq. (IV.2) can be evaluated analytically. The solution
$g_{m}^{\ast}$ is given by

$\displaystyle\frac{g_{m}^{*}}{2\pi}\equiv-
F_{m}(0,\kappa)=\frac{g_{1}^{2}(-3+16g_{1})+(3+16g_{1})\kappa^{2}}{12\pi^{2}\,N_{f}(g_{1}^{2}+\kappa^{2})^{2}}\approx\frac{1.41091}{N_{f}},$
(IV.8)

where we plugged in $\kappa=\frac{1}{4\pi}$ in the right-most equality.
Substituting these fixed point values of $w_{x}$ and $g_{m}$ into the
expressions for the conductivity, Eqs. (III.19) and (III.20), we obtain at
$\omega=0$:

$\displaystyle\sigma^{\psi}_{xx}=\frac{1}{16}-\frac{0.084774}{N_{f}}\qquad(w_{x}=0),$
(IV.9)
$\displaystyle\sigma^{\psi}_{xy}=\frac{-0.170657}{N_{f}}\qquad(w_{x}=0).$
(IV.10)

The correlation length and dynamical critical exponents at this fixed point
are [24] $(\nu,z)=(1,\frac{1.4}{N_{f}})$. Eq. (IV.9) makes clear that we
cannot reliably extrapolate $N_{f}\rightarrow 1$, since this would violate the
general requirement that the longitudinal conductivity is non-negative.

In the presence of the Coulomb interaction ($w_{x}\neq 0$), we have to solve
the flow equations in (IV.1) and (IV.2) together. The fixed point solution is
found to lie at

$\displaystyle(w_{x}^{*},g_{m}^{*})\approx(0.184193,2\pi\,\frac{0.393787}{N_{f}}).$
(IV.11)

Using Eqs. (III.19) and (III.20), this gives the dc composite fermion
conductivity:

$\displaystyle\sigma^{\psi}_{xx}=\frac{1}{16}+\frac{0.0177677}{N_{f}}\qquad(w_{x}\neq
0),$ (IV.12)
$\displaystyle\sigma^{\psi}_{xy}=\frac{-0.0903845}{N_{f}}\qquad(w_{x}\neq 0),$
(IV.13)

The critical exponents at this fixed point are [24] $(\nu,z)=(1,1)$.

We observe that disorder, without the Coulomb interaction, suppresses the
composite fermion longitudinal conductivity (IV.9). Inclusion of the Coulomb
interaction results in an enhancement of the composite fermion longitudinal
conductivity (IV.12). Disorder and the fluctuations of the Chern-Simons gauge
field produce a nonzero composite fermion Hall conductivity, (IV.10) and
(IV.12). Note that the composite fermion Hall conductivity remains nonzero in
the pure limit at leading order in $1/N_{f}$.

###  IV.2 Electrical response

To determine the quantum critical electrical transport at the fixed points
described in the previous section, we must first recall the dictionary that
relates the composite fermion and electrical conductivities [16]. The two-loop
calculation described previously can be understood to give rise to the
following quadratic effective action, in which the composite fermion is
integrated out:

$\displaystyle
S^{\text{eff}}_{\text{DCF}}=\int\frac{d^{2}kd\omega}{(2\pi)^{3}}\frac{-i\omega}{2}\Big{(}a_{i}\,\sigma^{\psi}_{ij}\,a_{j}+{1\over
2\pi}\epsilon_{tij}(\frac{1}{2}a_{i}a_{j}-2A_{i}a_{j}+A_{i}A_{j})\Big{)}.$
(IV.14)

Here, we have used our gauge freedom to set $a_{t}=A_{t}=0$. Terms that are
higher-order in the gauge field $a_{i}$ are ignored. We integrate over $a_{i}$
to obtain the effective electrical response action:

$\displaystyle
S^{\text{eff}}_{\text{DCF}}=\int\frac{d^{2}kd\omega}{(2\pi)^{3}}\frac{-i\omega}{2}\epsilon_{tij}A_{i}\,\sigma_{ij}\,A_{j},$
(IV.15)

where the electrical conductivity is

$\displaystyle\sigma_{xx}=\frac{\sigma^{\psi}_{xx}}{(\sigma^{\psi}_{xx})^{2}+(\sigma^{\psi}_{xy}+\frac{1}{4\pi})^{2}}\cdot\big{(}{1\over
2\pi}\big{)}^{2},$ (IV.16) $\displaystyle\sigma_{xy}={1\over
2\pi}-\frac{\sigma^{\psi}_{xy}+\frac{1}{4\pi}}{(\sigma^{\psi}_{xx})^{2}+(\sigma^{\psi}_{xy}+\frac{1}{4\pi})^{2}}\cdot\big{(}{1\over
2\pi}\big{)}^{2}.$ (IV.17)

An immediate consequence of this dictionary is that

$\displaystyle\sigma_{xx}^{2}+\sigma_{xy}^{2}=\frac{(\sigma^{\psi}_{xx})^{2}+(\sigma^{\psi}_{xy}-\frac{1}{4\pi})^{2}}{(\sigma^{\psi}_{xx})^{2}+(\sigma^{\psi}_{xy}+\frac{1}{4\pi})^{2}}\cdot\big{(}{1\over
2\pi}\big{)}^{2}.$ (IV.18)

This shows that self-dual electrical transport (I.1) occurs when the composite
fermion Hall conductivity $\sigma_{xy}^{\psi}=0$. Our system, however, is not
self-dual, since $\sigma_{xy}^{\psi}\neq 0$ (see (IV.10) and (IV.13)).

We now use (IV.16) and (IV.17) and the composite fermion conductivities
computed in the previous section to obtain the electrical conductivities. We
begin with the fixed point without the Coulomb interaction. Plugging in the
two-loop composite fermion conductivities (IV.9) and (IV.10) into Eqs. (IV.16)
and (IV.17), we find the dc electrical conductivities to ${\cal O}(1/N_{f}$):

$\displaystyle\big{(}\sigma_{xx},\sigma_{xy}\big{)}=\Big{(}0.97+{2.26\over
N_{f}},-0.24-{1.91\over N_{f}}\Big{)}\cdot{1\over
2\pi},\quad\big{(}w_{x}=0\big{)}.$ (IV.19)

Similarly, for the fixed point with Coulomb interaction, we use the two-loop
composite fermion conductivities (IV.12) and (IV.13) to obtain the dc
electrical conductivities:

$\displaystyle\big{(}\sigma_{xx},\sigma_{xy}\big{)}=\Big{(}0.97+{1.43\over
N_{f}},-0.24-{0.06\over N_{f}}\Big{)}\cdot{1\over 2\pi},\quad\big{(}w_{x}\neq
0\big{)}.$ (IV.20)

At ${\cal O}(1/N_{f}^{0})$, we find that self-duality (I.1) occurs with small,
but nonzero $\sigma_{xy}$ and a longitudinal conductivity $\sigma_{xx}\lesssim
R^{-1}_{Q}$. Finite $1/N_{f}$ corrections, due to the fluctuating Chern-Simons
gauge field and disorder, lead to violations of self-duality. These violations
are milder when the Coulomb interaction is included $(w_{x}\neq 0$) than when
only short-ranged interactions are present ($w_{x}=0$). The expressions for
the quantum critical electrical conductivities, (IV.19) and (IV.20), are
reliable at large $N_{f}$. To interpolate these results to small, but finite
$N_{f}$, we use the exact expressions that obtain from substituting the
composite fermion conductivities, either with (IV.9) and (IV.10) or without
(IV.12) and (IV.13) the Coulomb interaction, into Eqs. (IV.16) and (IV.17).
The results are plotted in Figs. 4(a) and 4(b). As we saw before in (IV.9),
Fig. 4(a) indicates the extrapolation $N_{f}\rightarrow 1$ is dubious, since
it results in a negative longitudinal conductivity.

(a) Zero Coulomb interaction $(w_{x}=0)$ and finite disorder

(b) Finite Coulomb interaction $(w_{x}\neq 0)$ and finite disorder

(c) Infinite Coulomb interaction $(w_{x}=\infty)$ without disorder

Figure 4: Electrical response components for two fixed points with varying
number of fermion flavor $N_{f}$. The values in the figures are all in the
unit of $\frac{1}{2\pi}$

We end this section with a speculative comment about another way to obtain
self-duality. Examining the composite fermion Hall conductivity (III.20), we
observe that $\sigma_{xy}^{\psi}\sim{1\over\text{finite}+w_{x}}$, where
“finite” refers to an ${\cal O}(1)$ constant. Taking $w_{x}\rightarrow\infty$,
then results in $\sigma_{xy}^{\psi}=0$. We plot the associated electrical
conductivities in Fig. 4(c). We do not have an argument for why the Coulomb
coupling would flow to strong coupling; $w_{x}=\infty$ is not a fixed point of
the beta functions (IV.1) and (IV.2). We note that in [24] we studied, in
addition, the effect of a dissipative Coulomb interaction, which allowed for
fixed points with infinite strength Coulomb interaction. The gauge field
propagator at such fixed points is different from that considered in this
paper and so a determination of the electrical conductivity at such fixed
points is left for future work.

##  V Discussion

In this paper, we studied the quantum critical electrical conductivity at the
superconductor-insulator transition in two spatial dimensions. We focused on
transitions modeled by a theory of charge-2e bosons on a lattice at
commensurate filling, with charge-conserving quenched disorder. The critical
properties are those of the dirty 3d XY universality class. For our study, we
used a composite fermion theory, consisting of a single Dirac fermion coupled
to a Chern-Simons gauge field with quenched random mass disorder, that is dual
to the dirty 3d XY model. There are at least two reasons to prefer the
composite fermion description. The first is that it can exhibit a particle-
vortex symmetry at the level of its Lagrangian. A consequence of this symmetry
is the semicircle law (I.1), which appears to be realized in some experiments
[11]. A second reason to prefer the composite fermion theory is that an
unscreened Coulomb interaction is straightforwardly included in the model.
This allows for a comparison of the properties of the theory with or without
the Coulomb interaction.

Our calculation of the critical conductivity was performed in an expansion in
$1/N_{f}$, where $N_{f}$ is the number of fermion flavors. Our results are
summarized by Eqs. (IV.19) and (IV.20) and in Figs. 4(a) and 4(b). At order
$1/N_{f}^{0}$, the theory is self-dual, realizing the semicircle law (I.1)
with $\sigma_{xx}\lesssim(2e)^{2}/h$ and small, finite $\sigma_{xy}$.
Fluctuation effects due to the gauge field and the random mass contribute
corrections of order $1/N_{f}$ to the conductivity that violate self-duality.
We find these violations to be smaller when the Coulomb interaction is
present.

In the previous section, based on the leading corrections to the composite
fermion Hall conductivity, we speculated that infinite strength Coulomb
interactions of the sort we studied in [24] might produce self-dual transport.
We leave an explicit calculation of the critical conductivity at infinite
strength Coulomb interactions for future work. We note that such fixed points
have dynamical critical exponent $z>1$.

It would be interesting to carry out these electrical transport calculations
at the dirty XY model fixed point found in [23], i.e., within the usual
bosonic description, and to compare the results to those in this paper. The
fixed point in [23] does not include the unscreened Coulomb interaction. The
bosonic description of the dirty XY model fixed point gives critical exponents
that are in closer agreement with numerical experiment [37] when
$N_{f}\rightarrow 1$, than those of the composite fermion theory (at least to
${\cal O}(1/N_{f})$).

There are two aspects as to why we focused on a theory corresponding to a
model of lattice bosons with charge-conserving disorder. First, regarding the
type of disorder within this lattice boson model, we found in [24] that other
types of disorder do not yield accessible disordered fixed points with or
without a finite Coulomb interaction. These calculations were performed in a
$1/N_{f}$ expansion; it is possible that a different choice of artificial
expansion parameter (perhaps one that preserves the same symmetries as the
$N_{f}=1$ theory) may find nontrivial fixed points. Second, regarding the
commensurability constraint, the dual composite fermions acquire a finite
chemical potential if this is relaxed. It would be extremely interesting to
find disordered fixed points in this more general situation, as we believe it
more closely resembles the experimentally-realized superconductor-insulator
transitions. Recent work has shown [38, 39] how the optical conductivity is
constrained by the anomaly structure of emergent symmetries in a class of non-
Fermi liquids (so-called “Hertz-Millis” theories) consisting of a Fermi
surface coupled to a gapless bosonic order parameter. Such models are closely
related to the theory we studied (see [40] for a related theory studied
recently), when the composite fermion density is nonzero. It would be
worthwhile to understand the interplay of this emergent anomaly structure with
particle-vortex symmetry.

## Acknowledgements

We thank Hart Goldman, Yen-Wen Lu, and Sri Raghu for useful discussions. This
material is based upon work supported by the U.S. Department of Energy, Office
of Science, Office of Basic Energy Sciences under Award No. DE-SC0020007.

##  Appendix A Gauge boson propagator

Start from Eq. (II.7), with $A\equiv\frac{+g_{X}^{2}\bm{k}^{2}}{k}$, $B\equiv
i\kappa\,|\bm{k}|$, and $C\equiv g_{X}^{2}\,k+|\bm{k}|\,w_{x}$. This one-loop
effective action includes the leading-order polarization:

$\displaystyle\Pi_{00}=-g_{X}^{2}\frac{\bm{k}^{2}}{k},\;\;\;$ (A.1)
$\displaystyle\Pi_{TT}=(\delta_{ij}-\frac{k_{i}k_{j}}{\bm{k}^{2}})\Pi_{ij}=-g_{X}^{2}\,k.$
(A.2)

We work in Coulomb gauge (vanishing longitudinal component of the gauge
field). Including the gauge-fixing term,
$\int_{x}\frac{(\nabla_{j}a_{j})^{2}}{2\lambda}$, which preserves $SO(2)$
rotation invariance, the one-loop effective action is

$\displaystyle S_{\rm gauge}^{(\text{gauge-
fixed})}\equiv\frac{1}{2}\int_{k}\begin{pmatrix}a_{0}&a_{T}\end{pmatrix}_{k}\begin{pmatrix}A&B\\\
B&C\\\ \end{pmatrix}\begin{pmatrix}a_{0}\\\ a_{T}\end{pmatrix}_{-k}$ (A.3)
$\displaystyle=\lim_{\lambda\rightarrow
0}\frac{1}{2}\int_{k}\begin{pmatrix}a_{0}&a_{x}&a_{y}\end{pmatrix}_{k}(\begin{pmatrix}A&B\frac{ik_{y}}{|\bm{k}|}&B\frac{-ik_{x}}{|\bm{k}|}\\\
B\frac{-ik_{y}}{|\bm{k}|}&\frac{k_{y}^{2}}{|\bm{k}|^{2}}C&\frac{-k_{x}k_{y}}{|\bm{k}|^{2}}C\\\
B\frac{+ik_{x}}{|\bm{k}|}&\frac{-k_{x}k_{y}}{|\bm{k}|^{2}}C&\frac{k_{x}^{2}}{|\bm{k}|^{2}}C\end{pmatrix}+\frac{1}{\lambda}\begin{pmatrix}0&0&0\\\
0&k_{x}k_{x}&k_{x}k_{y}\\\
0&k_{x}k_{y}&k_{y}k_{y}\end{pmatrix})\begin{pmatrix}a_{0}\\\ a_{x}\\\
a_{y}\end{pmatrix}_{-k},$ (A.4)

where $\int_{k}\equiv\int\frac{d^{3}k}{(2\pi)^{3}}$. We obtain the gauge field
propagators (II.8) - (II.10), upon first inverting the $3\times 3$ matrix
kernel and then taking the limit $\lambda\rightarrow 0$.

##  Appendix B Random mass correction to the composite fermion conductivity

In this appendix, we detail the evaluation of Eqs. (III.5) and (III.6),
following [29].

###  B.1  $\hat{\Pi}^{(b+c)}$

We first consider the subdiagram of Eq. III.6:

$\displaystyle\begin{tabular}[]{lll}&\includegraphics[scale={0.5}]{./Subdiagram_RMass_b.png}&\end{tabular}=\int\frac{d^{3}k}{(2\pi)^{3}}\bm{1}\gamma_{\sigma}\,\bm{1}(+i)\frac{\,(q+k)_{\sigma}}{(q_{0}+k_{0})^{2}+\bm{(k+q)}^{2}}2\pi\delta(k_{0})$
(B.2)
$\displaystyle=\int\frac{d^{2}k}{(2\pi)^{2}}\gamma_{\sigma}(+i)\frac{\,(q_{0})\delta_{\sigma
0}+(q_{a}+k_{a})\delta_{\sigma a}}{(q_{0}+0)^{2}+(\bm{k+q})^{2}}$ (B.3)
$\displaystyle=\int\frac{d^{2}\bm{\ell}}{(2\pi)^{2}}\gamma_{\sigma}(+i)\frac{\,(q_{0})\delta_{\sigma
0}+(\ell_{a})\delta_{\sigma a}}{(q_{0}+0)^{2}+(\bm{\ell})^{2}}\equiv
i\,\gamma_{0}\,I_{X}(q_{0})\,\delta_{\sigma 0}+0,$ (B.4)

Then, the full diagram of Eq. III.6 reads:

$\displaystyle\Pi^{(b+c)}_{xx}(p_{0},\bm{p}=0)$ (B.5)
$\displaystyle=-2(\frac{ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,\int_{q}\text{Tr}[\gamma_{x}G(q)\Big{(}\int\frac{d^{3}k}{(2\pi)^{3}}\bm{1}G(k+q)\bm{1}\times
2\pi\delta(k_{0})g_{m}\;\Big{)}G(q)\gamma_{x}G(p+q)]\,\qquad$ (B.6)
$\displaystyle=-2(\frac{ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,\int_{q}\text{Tr}[\gamma_{x}G(q)\Big{(}i\,I_{X}\gamma_{0}+0\Big{)}G(q)\gamma_{x}G(p+q)]\times
g_{m}\,$ (B.7)
$\displaystyle=-2(\frac{ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,\int_{q}\text{Tr}[\gamma_{x}\frac{(+i)q_{\rho}\gamma_{\rho}}{q^{2}_{0}+\bm{q}^{2}}\Big{(}i\,I_{X}\gamma_{0}+0\Big{)}\frac{(i)q_{\tau}\gamma_{\tau}}{q^{2}_{0}+\bm{q}^{2}}\gamma_{x}(i)\frac{(q+p)_{\kappa}\gamma_{\kappa}}{(q_{0}+p_{0})^{2}+(\bm{q+0})^{2}}]\times
g_{m}.\,\quad\quad\,$ (B.8)

Using

$\displaystyle\text{Tr}[\gamma_{x}\gamma_{\rho}\gamma_{0}\gamma_{\tau}\gamma_{x}\gamma_{\kappa}]\,q_{\rho}\,q_{\tau}\,(q+p)_{\kappa}=2(p_{0}+q_{0})\,(\bm{q^{2}}-q_{0}^{2})$
(B.9)

we obtain

$\displaystyle\Pi^{(b+c)}_{xx}(p_{0},\bm{p}=0)=2(-1)(\frac{+ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\int\frac{d^{2}\bm{q}dq_{0}}{(2\pi)^{3}}\frac{1}{[q_{0}^{2}+\bm{q}]^{2}}\frac{2(p_{0}+q_{0})\,(\bm{q^{2}}-q_{0}^{2})}{(q_{0}+p_{0})^{2}+\bm{q}^{2}}I_{X}(q_{0})$
(B.10)
$\displaystyle=2(-1)(\frac{+ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\int\frac{d^{2}\bm{q}dq_{0}}{(2\pi)^{3}}\frac{2(p_{0}+q_{0})\,(\bm{q^{2}}-q_{0}^{2})}{[q_{0}^{2}+\bm{q}]^{2}\;[(q_{0}+p_{0})^{2}+\bm{q}^{2}]}\int^{\Lambda}\frac{d^{2}\bm{\ell}}{(2\pi)^{2}}\frac{q_{0}}{q_{0}^{2}+\bm{\ell}^{2}}$
(B.11)
$\displaystyle=2(-1)(\frac{+ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\int\frac{dq_{0}}{2\pi}\Big{(}2(p_{0}+q_{0})q_{0}$
$\displaystyle\times\frac{1}{4\pi}\frac{2q_{0}^{2}-2(p_{0}+q_{0})^{2}+[(p_{0}+q_{0})^{2}+q_{0}^{2}]\log[\frac{(q_{0}+p_{0})^{2}}{q_{0}^{2}}]}{[(p_{0}+q_{0})^{2}-q_{0}^{2}]^{2}}\Big{)}\frac{1}{4\pi}\log[\frac{\Lambda^{2}}{q_{0}^{2}}].\qquad\;$
(B.12)

Here, we used

$\displaystyle\int_{0}^{\Lambda}\frac{d^{2}\ell}{(2\pi)^{2}}\frac{1}{q_{0}^{2}+\ell^{2}}=\frac{1}{4\pi}\log[\frac{\Lambda^{2}+q_{0}^{2}}{q_{0}^{2}}]\to\frac{1}{4\pi}\log[\frac{\Lambda^{2}}{q_{0}^{2}}].$
(B.13)

We simplify the integrand by defining

$\displaystyle K(q_{0})$ (B.14)
$\displaystyle\equiv\Big{(}2(p_{0}+q_{0})q_{0}\times\frac{1}{4\pi}\frac{2q_{0}^{2}-2(p_{0}+q_{0})^{2}+[(p_{0}+q_{0})^{2}+q_{0}^{2}]\log[\frac{(q_{0}+p_{0})^{2}}{q_{0}^{2}}]}{[(p_{0}+q_{0})^{2}-q_{0}^{2}]^{2}}\Big{)}\frac{1}{4\pi}(-\log[q_{0}^{2}])\quad\qquad\;$
(B.16)
$\displaystyle=(\frac{-\log[q_{0}^{2}]}{4\pi})\frac{(p_{0}+q_{0})q_{0}}{\pi(p_{0}+2q_{0})p_{0}}\Big{(}-1+\frac{1}{2}\frac{[(p_{0}+q_{0})^{2}+q_{0}^{2}]\log[\frac{(q_{0}+p_{0})^{2}}{q_{0}^{2}}]}{(p_{0}+2q_{0})^{2}\,p_{0}^{2}}\Big{)}$

A change of variables gives

$\displaystyle\int_{-\infty}^{\infty}\frac{dq_{0}}{2\pi}K(q_{0}=xp_{0})$
(B.17)
$\displaystyle=|p_{0}|\int_{-\infty}^{\infty}\frac{dx}{2\pi}\,\frac{-1}{4\pi}(\log[x^{2}]+\log[p_{0}^{2}])\frac{x(1+x)}{1(1+2x)}\,\frac{-1}{\pi}\Big{[}1+\frac{x^{2}+(x+1)^{2}}{2(1+2x)}\log[\frac{x^{2}}{(x+1)^{2}}]\;\Big{]}\quad$
(B.18)
$\displaystyle=|p_{0}|\int_{-\infty}^{\infty}\frac{dx}{2\pi}\,\frac{-1}{4\pi}(\log[x^{2}])\frac{x(1+x)}{1(1+2x)}\,\frac{-1}{\pi}\Big{[}1+\frac{x^{2}+(x+1)^{2}}{2(1+2x)}\log[\frac{x^{2}}{(x+1)^{2}}]\;\Big{]}$
$\displaystyle\quad+|p_{0}|\int_{-\infty}^{\infty}\frac{dx}{2\pi}\,\frac{-1}{4\pi}(\log[p_{0}^{2}])\frac{x(1+x)}{1(1+2x)}\,\frac{-1}{\pi}\Big{[}1+\frac{x^{2}+(x+1)^{2}}{2(1+2x)}\log[\frac{x^{2}}{(x+1)^{2}}].$
(B.19)

The principal value of the second term, $\int dx\log[p_{0}^{2}]\times[...]$ is
zero. If we make the plot, we find it is an odd function with respect to
$x=\frac{1}{2}$, so after we shift $x\to x-\frac{1}{2}$, the integral
vanishes. For the same reason, the $\log\Lambda^{2}$ term in Eq. (B.12)
vanishes.

Hence, we symmetrize the first line of Eq. B.19, the $\log q_{0}^{2}$ term of
Eq. (B.12):

$\displaystyle\int_{-\infty}^{\infty}\frac{dq_{0}}{2\pi}\frac{K(q_{0})+K(-q_{0})}{2}=|p_{0}|\int_{-\infty}^{\infty}\frac{dx}{2\pi}\frac{\log[x^{2}]}{4\pi^{2}}\Big{[}\frac{x(1+x)}{1(1+2x)}(1+\frac{x^{2}+(x+1)^{2}}{2(1+2x)}\log[\frac{x^{2}}{(x+1)^{2}}]$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad+\frac{-x(1-x)}{1(1-2x)}(1+\frac{x^{2}+(-x+1)^{2}}{2(1-2x)}\log[\frac{x^{2}}{(-x+1)^{2}}]\;)\Big{]}$
(B.20) $\displaystyle=\frac{1}{2\pi}\times\frac{|p_{0}|}{96}.$ (B.21)

Thus, we conclude that

$\displaystyle\Pi^{(b+c)}_{xx}(p_{0},\bm{p}=0)=2(-1)(\frac{+ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\times\frac{1}{2\pi}\times\frac{|p_{0}|}{96}.$
(B.22)

###  B.2 $\hat{\Pi}^{(a)}$

Next we turn to Eq. (III.5). Again, consider the subdiagram,

$\displaystyle\begin{tabular}[]{lll}&\includegraphics[scale={0.5}]{./Subdiagram_RMass_a.png}&\end{tabular}=\int\frac{d^{3}k}{(2\pi)^{3}}i\frac{(q_{0}+k_{0}+p_{0})\gamma_{0}+\bm{(q+k+p)}\cdot\vec{\gamma}}{(q_{0}+k_{0}+p_{0})^{2}+(\bm{q+k+p})^{2}}\bm{1}\,i\frac{(q_{0}+p_{0})\gamma_{0}+\bm{(q+p)}\cdot\vec{\gamma}}{(q_{0}+p_{0})^{2}+(\bm{q+p})^{2}}$
(B.24)
$\displaystyle\qquad\qquad\qquad\qquad\times\gamma_{\mu}\,i\frac{(q_{0})\gamma_{0}+\bm{(q)}\cdot\vec{\gamma}}{(q_{0})^{2}+(\bm{q})^{2}}\bm{1}\,i\frac{(q_{0}+k_{0})\gamma_{0}+\bm{(q+k)}\cdot\vec{\gamma}}{(q_{0}+k_{0})^{2}+(\bm{q+k})^{2}}\times
2\pi\delta(k_{0})\,g_{m}$ (B.25)
$\displaystyle=g_{m}\int\frac{d^{2}\bm{k}}{(2\pi)^{2}}\gamma_{\tau}\bm{1}\gamma_{\kappa}\gamma_{\mu}\gamma_{\rho}\bm{1}\gamma_{\sigma}\frac{(q+k+p)_{\tau}}{(q+k+p)^{2}}\frac{(q+p)_{\kappa}}{(q+p)^{2}}\frac{q_{\rho}}{q^{2}}\frac{(q+k)_{\sigma}}{(q+k)^{2}}\Big{|}_{k_{0}=0}.$
(B.26)

Consider the terms involving $k$. Setting $\bm{p}=0$, we find:

$\displaystyle
I_{\tau\sigma}(q_{0},p_{0})\equiv\int\frac{d^{2}\bm{k}}{(2\pi)^{2}}\frac{(q+k+p)_{\tau}}{(q+k+p)^{2}}\frac{(q+k)_{\sigma}}{(q+k)^{2}}\Big{|}_{k_{0}=0,\bm{p}=0}$
(B.27)
$\displaystyle=\int\frac{d^{2}\bm{\ell}}{(2\pi)^{2}}\frac{(q_{0}+k_{0}+p_{0})\delta_{\tau
0}+(\bm{\ell+p})_{\tau}\delta_{\tau
a}}{(q_{0}+k_{0}+p_{0})^{2}+\bm{\ell}^{2}}\;\frac{(q_{0}+k_{0})\delta_{\sigma
0}+(\bm{\ell})_{\sigma}\delta_{\sigma
b}}{(q_{0}+k_{0})^{2}+\bm{\ell}^{2}}\Big{|}_{k_{0}=0,\bm{p}=0}$ (B.28)
$\displaystyle=\int\frac{d^{2}\bm{\ell}}{(2\pi)^{2}}\frac{(q_{0}+p_{0})\delta_{\tau
0}+(\bm{\ell})_{\tau}\delta_{\tau
a}}{(q_{0}+p_{0})^{2}+\bm{\ell}^{2}}\;\frac{(q_{0}+0)\delta_{\sigma
0}+(\bm{\ell})_{\sigma}\delta_{\sigma b}}{(q_{0}+0)^{2}+\bm{\ell}^{2}}$ (B.29)
$\displaystyle=\int\frac{d^{2}\bm{\ell}}{(2\pi)^{2}}\frac{(q_{0}+p_{0})q_{0}\,\delta_{\tau
0}\delta_{\sigma
0}}{[(q_{0}+p_{0})^{2}+\bm{\ell}^{2}]\,[q_{0}^{2}+\bm{\ell}^{2}]}+\int\frac{d^{2}\bm{\ell}}{(2\pi)^{2}}\frac{\bm{\ell}^{2}\delta_{ab}\delta_{\tau
a}\delta_{\sigma
b}}{d}\frac{1}{[(q_{0}+p_{0})^{2}+\bm{\ell}^{2}]\,[q_{0}^{2}+\bm{\ell}^{2}]}\quad\quad$
(B.30)

Now we set $\mu=\nu=x$ . First consider the temporal component of the gamma
index trace in Eq. (III.5):

$\displaystyle\text{Tr}[\gamma_{\tau}\gamma_{\kappa}\gamma_{\mu=x}\gamma_{\rho}\gamma_{\sigma}\gamma_{\nu=x}]\,\delta_{\tau
0}\delta_{\sigma
0}=\text{Tr}[\gamma_{0}\gamma_{\kappa}\gamma_{x}\gamma_{\rho}\gamma_{0}\gamma_{x}]=(-1)\text{Tr}[\gamma_{\kappa}\gamma_{x}\gamma_{\rho}\gamma_{x}].$
(B.31)

Contracting this with the momentum gives

$\displaystyle(-1)\text{Tr}[\gamma_{\kappa}\gamma_{x}\gamma_{\rho}\gamma_{x}]\,(q+p)_{\kappa}q_{\rho}=2q_{0}(p_{0}+q_{0}).$
(B.32)

Next consider the spatial components of the gamma index trace. We perform the
trace in $2-\epsilon$ spatial dimensions, rather than $2$ spatial dimensions:

$\displaystyle\text{Tr}[\gamma_{\tau}\gamma_{\kappa}\gamma_{\mu=x}\gamma_{\rho}\gamma_{\sigma}\gamma_{\nu=x}]\,\delta_{ab}\delta_{\tau
a}\delta_{\sigma b}$ (B.33) $\displaystyle=\sum_{a=2-\epsilon\text{~{}number
of
indices}}\text{Tr}[\gamma_{a}\gamma_{\kappa}\gamma_{x}\gamma_{\rho}\gamma_{a}\gamma_{x}]=\text{Tr}[\gamma_{\kappa}\gamma_{x}\gamma_{\rho}\sum_{a}(\gamma_{a}\gamma_{x}\gamma_{a})]$
(B.34)
$\displaystyle=\text{Tr}[\gamma_{\kappa}\gamma_{x}\gamma_{\rho}(2\gamma_{x}-(2-\epsilon)\gamma_{x})]=\epsilon\;\text{Tr}[\gamma_{\kappa}\gamma_{x}\gamma_{\rho}\gamma_{x}].$
(B.35)

Contracting this with the momentum yields

$\displaystyle\epsilon\;\text{Tr}[\gamma_{\kappa}\gamma_{x}\gamma_{\rho}\gamma_{x}]\,(q+p)_{\kappa}q_{\rho}=\epsilon\times(-2)q_{0}(p_{0}+q_{0}).$
(B.36)

Putting this all together, we find:

$\displaystyle\Pi^{(a)}_{\mu\nu}(p_{0},\bm{p}=0)=(-1)(\frac{+ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\int\frac{d^{3}q}{(2\pi)^{3}}\frac{2q_{0}(p_{0}+q_{0})}{[(q_{0}+p_{0})^{2}+(\bm{q+0})^{2}]\;[q_{0}^{2}+\bm{q}^{2}]}$
$\displaystyle\times\Big{(}\int\frac{d^{2}\bm{\ell}}{(2\pi)^{2}}\frac{(q_{0}+p_{0})q_{0}\,}{[(q_{0}+p_{0})^{2}+\bm{\ell}^{2}]\,[q_{0}^{2}+\bm{\ell}^{2}]}-\epsilon\int\frac{d^{2-\epsilon}\bm{\ell}}{(2\pi)^{2}}\frac{\bm{\ell}^{2}}{2}\frac{1}{[(q_{0}+p_{0})^{2}+\bm{\ell}^{2}]\,[q_{0}^{2}+\bm{\ell}^{2}]}\Big{)}.$
(B.37)

We wish to extract finite parts of this expression. The finite part of the
first $d^{2}\bm{\ell}$ integral is

$\displaystyle\int\frac{d^{2}\bm{\ell}}{(2\pi)^{2}}\frac{(q_{0}+p_{0})q_{0}\,}{[(q_{0}+p_{0})^{2}+\bm{\ell}^{2}]\,[q_{0}^{2}+\bm{\ell}^{2}]}=\frac{1}{4\pi}\frac{(q_{0}+p_{0})q_{0}}{p_{0}(p_{0}+2q_{0})}\log[\frac{(q_{0}+p_{0})^{2}}{q_{0}^{2}}].$
(B.38)

For the second integral, we need to extract the $\frac{1}{\epsilon}$ part.
Using the usual Feynman parametrization,

$\displaystyle\int\frac{d^{2-\epsilon}\bm{\ell}}{(2\pi)^{2}}\frac{\bm{\ell}^{2}}{2}\frac{1}{[(q_{0}+p_{0})^{2}+\bm{\ell}^{2}]\,[q_{0}^{2}+\bm{\ell}^{2}]}=\int\frac{d^{2-\epsilon}\bm{\ell}}{(2\pi)^{2}}\int_{0}^{1}dx\frac{1}{2}\frac{\bm{\ell}^{2}}{[\bm{\ell}^{2}+\Delta(x,q_{0},p_{0})]^{2}}$
(B.39)
$\displaystyle=\int_{0}^{1}dx\frac{1}{2}\frac{1}{2\pi\epsilon}+\mathcal{O}(\epsilon^{0})=\frac{1}{4\pi\epsilon}+\mathcal{O}(\epsilon^{0}).$
(B.40)

It remains to perform the $d^{2}\bm{q}$ integral, which has the same structure
as the one above:

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{2q_{0}(p_{0}+q_{0})}{[(q_{0}+p_{0})^{2}+\bm{q}^{2}]\;[q_{0}^{2}+\bm{q}^{2}]}=\int\frac{dq_{0}}{2\pi}\frac{1}{2\pi}\frac{(q_{0}+p_{0})q_{0}}{p_{0}(p_{0}+2q_{0})}\log[\frac{(q_{0}+p_{0})^{2}}{q_{0}^{2}}].$
(B.41)

Putting this together, we have

$\displaystyle\Pi^{(a)}_{\mu\nu}(p_{0},\bm{p}=0)=(-1)(\frac{+ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\int_{-\infty}^{\infty}\frac{dq_{0}}{2\pi}\frac{1}{2\pi}\frac{(q_{0}+p_{0})q_{0}}{p_{0}(p_{0}+2q_{0})}\log[\frac{(q_{0}+p_{0})^{2}}{q_{0}^{2}}]$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\times\Big{(}\frac{1}{4\pi}\frac{(q_{0}+p_{0})q_{0}}{p_{0}(p_{0}+2q_{0})}\log[\frac{(q_{0}+p_{0})^{2}}{q_{0}^{2}}]-\frac{1}{4\pi}\Big{)}$
(B.42)
$\displaystyle=(-1)(\frac{+ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\times\frac{1}{2\pi}\times\frac{|p_{0}|}{96}.$
(B.43)

Summarizing the results of this section, we find the random mass diagrams
equal

$\displaystyle\Pi^{(a+b+c)}_{\mu\nu}(p_{0},\bm{p}=0)=3\times(-1)(\frac{+ig}{\sqrt{N_{f}}})^{2}\,N_{f}\,g_{m}\times\frac{1}{2\pi}\times\frac{|p_{0}|}{96}.$
(B.44)

Notice there is no off-diagonal component. By Eq. (III.1), this self-energy
contributes the following to the composite fermion conductivity:

$\displaystyle\sigma^{g_{m}}_{xx}=\lim_{p_{0}\to
0}\frac{\Pi^{(\text{S})}(\bm{p}=0)}{p_{0}}=\frac{i}{\omega}\Pi^{xx}(\bm{p}=0,|p_{0}|\to
i|\omega|)=\frac{-3g_{m}}{96\times 2\pi},$ (B.45)

where we set the $g^{2}=1$ and $\omega>0$.

##  Appendix C Gauge field correction to the composite fermion conductivity

In this appendix, we detail the evaluation of Eqs. (III.8) and (III.9).

###  C.1 Useful integrals

To evaluate gauge field corrections to the composite fermion conductivity, we
need the formulas given in the next two subsections. In the expressions below,
$\Lambda$ is a momentum integral UV cutoff.

####  C.1.1 Basic integral building blocks

The simplest integrals we will need are the following:

$\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{k^{2}}=\int_{0}^{\Lambda}dk\frac{4\pi
k^{2}dk}{(2\pi)^{3}\,k^{2}}=\frac{\Lambda}{2\pi^{2}},$ (C.2)
$\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{k^{2}(k+p)^{2}}=\frac{1}{8|p|},$
$\displaystyle J_{2}(p,q)\equiv$
$\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{(k+p)^{2}(k+q)^{2}}=\int\frac{d^{3}K}{(2\pi)^{3}}\frac{1}{(K)^{2}(K-p+q)^{2}}=\frac{1}{8|p-q|},$
(C.3) $\displaystyle J_{3}(p,q)\equiv$
$\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{k^{2}(k+p)^{2}(k+q)^{2}}=\frac{1}{8|p|\,|q|\,|p-q|},$
(C.5)
$\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}\frac{k^{2}}{(k+p)^{2}(k+q)^{2}}=\frac{\Lambda}{2\pi^{2}}+\frac{p\cdot
q}{8|p-q|}.$

Next we consider the integral,

$\displaystyle I(p,q)\,$
$\displaystyle\equiv\int\frac{d^{3}k}{(2\pi)^{3}}\frac{k^{2}}{(k+p)^{2}(k+q)^{2}(k+p+q)^{2}}$
(C.7)
$\displaystyle=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{(k+q)^{2}+(k+p)^{2}-(k+p+q)^{2}+2p\cdot
q}{(k+p)^{2}(k+q)^{2}(k+p+q)^{2}}$
$\displaystyle=\frac{1}{8|q|}+\frac{1}{8|p|}-\frac{1}{8|p-q|}+\frac{2p\cdot
q}{8|p||q||p-q|}.$

It is symmetric: $I(p,q)=I(q,p)$

Next we consider the more complicated integral:

$\displaystyle{\color[rgb]{0,0,0}J_{4}(p,q)}\,$
$\displaystyle\equiv\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{(k)^{2}(k+p)^{2}(k+q)^{2}(k+p+q)^{2}}$
(C.9) $\displaystyle=\frac{1}{8|p||q|(p\cdot
q)}\bigg{(}\frac{1}{|p-q|}-\frac{1}{|p+q|}\bigg{)}.$

To prove this result, we re-express the integrand using partial fractions:

$\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{(k)^{2}(k+p)^{2}(k+q)^{2}(k+p+q)^{2}}$
(C.10)
$\displaystyle=\int\frac{d^{3}k}{(2\pi)^{3}}\Big{(}\frac{1}{(k)^{2}(k+p)^{2}(k+q)^{2}}-\frac{1}{(k)^{2}(k+p)^{2}(k+p+q)^{2}}$
$\displaystyle\qquad\qquad\qquad\qquad+\frac{1}{(k+q)^{2}(k+p+q)^{2}(k+p)^{2}}-\frac{1}{(k+q)^{2}(k+p+q)^{2}\,k^{2}}\Big{)}\frac{1}{2p\cdot
q}$ (C.11) $\displaystyle=\frac{1}{8(p\cdot
q)\;|p|\,|q|}(\frac{1}{|p-q|}-\frac{1}{|p+q|}).\qquad$ (C.12)

Now consider the three-propagator integral

$\displaystyle L(p_{5},p,k)\,$
$\displaystyle\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\frac{q\cdot
p_{5}}{(q)^{2}(q+p)^{2}(q+k)^{2}}$ (C.14)
$\displaystyle=(-1)\frac{(|k|+|p|-|k-p|)\;\Big{(}|p|\,(p_{5}\cdot
k)+|k|\,(p_{5}\cdot p)\Big{)}}{16\,|k|\,|p|\,(k\cdot p+|k|\,|p|)\,|k-p|}.$

Consider the four-propagator integral:

$\displaystyle M(k;p)\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\bm{k\cdot
q}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}.$ (C.15)

$M(k;p)$ is not symmetric with respect to $k,p$. It is difficult to evaluate
this four-propagator integral using the usual Feynman parameterization. (The
three-propagator integral (C.14) can be evaluated in this way.) Instead, we
use the partial fraction trick in Eq. (C.11) to rewrite it as multiple three-
propagator integrals, and then use the 3-propagator result in Eq. (C.14). We
find:

$\displaystyle M(k;p)=\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\bm{k\cdot
q}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.16)
$\displaystyle=\int\frac{d^{3}q}{(2\pi)^{3}}(\bm{k\cdot
q})\Big{(}\frac{1}{q^{2}(q+k)^{2}(q+p)^{2}}-\frac{1}{q^{2}(q+k)^{2}(q+k+p)^{2}}$
$\displaystyle\quad+\frac{1}{(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}-\frac{1}{q^{2}(q+k+p)^{2}(q+p)^{2}}\Big{)}\frac{1}{2(k\cdot
p)}$ (C.17) $\displaystyle=\frac{1}{2(k\cdot
p)}\Big{(}L(\bm{k},k,p)-L(\bm{k},k,k+p)+0-L(\bm{k},k+p,p)\Big{)}$ (C.18)
$\displaystyle+\frac{1}{2(k\cdot
p)}\int\frac{d^{3}\ell}{(2\pi)^{3}}\frac{\bm{(\ell-k)\cdot
k}}{\ell^{2}(\ell+p)^{2}(\ell-k+p)^{2}}$ (C.19)
$\displaystyle=\frac{1}{2(k\cdot
p)}\Big{(}L(\bm{k},k,p)-L(\bm{k},k,k+p)+L(\bm{k},p,p-k)-\frac{\bm{k}^{2}}{8|p|\,|p-k|\,|k|}$
(C.20) $\displaystyle-L(\bm{k},k+p,p)\Big{)}.\quad\quad\quad$ (C.21)

A “quadratic in $q$ integral” can be obtained from the above building blocks:

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(k\cdot q)\,(\bm{k\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.22)
$\displaystyle=\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\frac{1}{2}[(q+k)^{2}-k^{2}-q^{2}]}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}\,(\bm{k\cdot
q})\qquad\qquad$ (C.23)
$\displaystyle=\frac{1}{2}\Big{(}L(\bm{k},p,k+p)-k^{2}\,M(k;p)-L(\bm{k},p,p-k)+\bm{k}^{2}\frac{1}{8|p|\,|p-k|\,|k|}\Big{)}.$
(C.24)

####  C.1.2 More complicated integral formulas

The momenta $p_{5}=(p^{0}_{5},p^{x}_{5},p^{y}_{5})$ and
$p_{5}=(q^{0}_{6},q^{x}_{6},q^{y}_{6})$ appearing below are arbitrary external
three-momenta that are independent of any integral momentum variables. We list
the following integrals:

$\displaystyle N(p_{5};p,k)\big{|}_{p\neq k}\,$
$\displaystyle\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\frac{q\cdot
p_{5}}{(q^{2})\;(q^{2})\;(q+p)^{2}(q+k)^{2}}$ (C.26)
$\displaystyle=(-1)\frac{|p|^{2}(p_{5}\cdot k)+|k|^{2}(p_{5}\cdot
p)}{16\,|k|^{3}\,|p|^{3}\,|k-p|},$ $\displaystyle
R_{1}(p_{5};k)\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\frac{q\cdot
p_{5}}{q^{2}(q+k)^{2}q^{2}}=(-1)\frac{p_{5}\cdot k\;}{16\,|k|^{3}},$ (C.27)
$\displaystyle
R_{2}(p_{5},p_{6};k)\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(q\cdot
p_{5})\;(q\cdot p_{6})}{q^{2}(q+k)^{2}q^{2}}=\frac{k^{2}(p_{5}\cdot
p_{6})+(p_{5}\cdot k)(p_{6}\cdot k)}{32\,|k|^{3}},$ (C.28) $\displaystyle
S(p_{5};k,p)\,$ $\displaystyle\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(q\cdot
p_{5})^{2}}{q^{2}(q+k)^{2}\,(q+p)^{2}}$ (C.30)
$\displaystyle=\frac{|p|\,(\,|k|+2|p|+|k-p|\,)\,(p_{5}\cdot
k)^{2}+|k|\,(\,|p|+2|k|+|k-p|\,)\,(p_{5}\cdot
p)^{2}}{16\,|k-p|\;|k|\;|p|\;\big{(}\,|k|+|p|+|k-p|\,\big{)}^{2}}$
$\displaystyle+\frac{|k-p|\;|k|\;|p|(\,|k|+|p|+|k-p|\,)\,p_{5}^{2}+2|k|\,|p|\,(p_{5}\cdot
k)\,(p_{5}\cdot
p)}{16\,|k-p|\;|k|\;|p|\;\big{(}\,|k|+|p|+|k-p|\,\big{)}^{2}},$ $\displaystyle
R_{3}(p_{5};k,p)\,$
$\displaystyle\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(q\cdot
p_{5})^{2}}{q^{2}(q+k)^{2}q^{2}(q+p)^{2}}$ (C.32)
$\displaystyle=\frac{p^{2}\,|k-p|\,k^{2}\,p_{5}^{2}\,(|k-p|+|k|+|p|)+2k^{2}\,p^{2}\,(p_{5}\cdot
k)\,(p_{5}\cdot
p)}{16\,|k-p|\,|k|^{3}\,|p|^{3}\;\big{(}\,|k|+|p|+|k-p|\,\big{)}^{2}}$
$\displaystyle+\frac{k^{3}(|k-p|+|k|+2|p|)\,(p_{5}\cdot
p)^{2}+p^{3}(|k-p|+2|k|+|p|)\,(p_{5}\cdot
k)^{2}}{16\,|k-p|\,|k|^{3}\,|p|^{3}\;\big{(}\,|k|+|p|+|k-p|\,\big{)}^{2}}.$

Next we consider

$\displaystyle M_{2}(p_{5};k,p)\,$
$\displaystyle\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(q\cdot
p_{5})^{2}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.36)
$\displaystyle=\int\frac{d^{3}q}{(2\pi)^{3}}(q\cdot
p_{5})^{2}\Big{(}\frac{1}{q^{2}(q+k)^{2}(q+p)^{2}}-\frac{1}{q^{2}(q+k)^{2}(q+k+p)^{2}}$
$\displaystyle+\frac{1}{(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}-\frac{1}{q^{2}(q+k+p)^{2}(q+p)^{2}}\Big{)}\frac{1}{2(k\cdot
p)}$ $\displaystyle=\frac{1}{2(k\cdot p)}\Big{[}S(p_{5};k,p)-S(p_{5};k,k+p)$
$\displaystyle+\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(q\cdot
p_{5})^{2}}{(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}-S(p_{5};k+p,p)\Big{]}.$

To further evaluate, we can shift the momentum:

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(q\cdot
p_{5})^{2}}{(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}=\int\frac{d^{3}\ell}{(2\pi)^{3}}\frac{[(\ell-k)\cdot
p_{5}]^{2}}{\ell^{2}(\ell+p)^{2}(\ell-k+p)^{2}}$
$\displaystyle=\int\frac{d^{3}\ell}{(2\pi)^{3}}\frac{(\ell\cdot
p_{5})^{2}-2(\ell\cdot p_{5})(k\cdot p_{5})+(k\cdot
p_{5})^{2}}{\ell^{2}(\ell+p)^{2}(\ell+p-k)^{2}}$
$\displaystyle=S(p_{5};p,p-k)-2(k\cdot p_{5})\,L(p_{5};p,p-k)+(k\cdot
p_{5})^{2}\frac{1}{8|p|\,|p-k|\,|k|}.$ (C.37)

Thus, we have reduced $M_{2}$ to a linear combination of integrals we have
already computed:

$\displaystyle M_{2}(p_{5};k,p)\,$ $\displaystyle=\frac{1}{2(k\cdot
p)}\Big{[}S(p_{5};k,p)-S(p_{5};k,k+p)-S(p_{5};k+p,p)$ (C.38)
$\displaystyle+S(p_{5};p,p-k)-2(k\cdot p_{5})\,L(p_{5};p,p-k)+(k\cdot
p_{5})^{2}\frac{1}{8|p|\,|p-k|\,|k|}\Big{]}.\qquad$

###  C.2  Anti-symmetric component of $\Pi^{(b+c)}$

The anti-symmetric part of the gauge field propagator is

$\displaystyle
D^{\text{Anti}}_{\alpha\beta}(k)=\frac{\kappa}{\kappa^{2}+g_{X}^{4}+\frac{w_{x}g_{X}^{2}\,|\bm{k}|}{|k|}}(\frac{\epsilon_{\alpha\beta\lambda}k_{\lambda}\delta_{\lambda
j}}{\bm{k}^{2}})\equiv\tilde{C}_{A}(k)\,\frac{\epsilon_{\alpha\beta\lambda}k_{\lambda}\delta_{\lambda
j}}{\bm{k}^{2}}.$ (C.39)

It is important to keep in mind that $\tilde{C}_{A}$ is a function of the
momentum $k$ carried by the gauge field.

We want to evaluate $\Pi_{(\rm A)}^{(b+c)}$:

$\displaystyle\Pi^{(b+c)}_{(\text{A})}(p)=\frac{p_{\lambda}\epsilon_{\mu\nu\lambda}}{{\color[rgb]{0,0,0}2}|p|}\Pi^{(b+c)}_{\mu\nu}(p)$
$\displaystyle=\frac{p_{\lambda}\epsilon_{\mu\nu\lambda}}{{\color[rgb]{0,0,0}2}|p|}\times
2\frac{(-1)g^{4}}{N_{f}}\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}q}{(2\pi)^{3}}\text{Tr}[\gamma_{\mu}\frac{i\,q^{\rho}\gamma_{\rho}}{q^{2}}\gamma_{\alpha}\frac{i\,(k+q)^{\tau}\gamma_{\tau}}{(k+q)^{2}}\gamma_{\beta}\frac{i\,q^{\sigma}\gamma_{\sigma}}{q^{2}}\gamma_{\nu}\frac{i\,(p+q)^{\kappa}\gamma_{\kappa}}{(p+q)^{2}}]\,D^{\text{Anti}}_{\alpha\beta}(k).\qquad\;\;\;$
(C.40) (C.41)

The following identity is useful:

$\displaystyle\epsilon_{ijk}\epsilon_{abc}=\text{Det}\begin{pmatrix}\delta_{ia}&\delta_{ib}&\delta_{ic}\\\
\delta_{ja}&\delta_{jb}&\delta_{jc}\\\ \delta_{ka}&\delta_{kb}&\delta_{kc}\\\
\end{pmatrix}.$ (C.42)

Consider the following terms in the integrand in Eq. (C.41):

$\displaystyle\text{Tr}[\gamma_{\mu}\frac{i\,q^{\rho}\gamma_{\rho}}{q^{2}}\gamma_{\alpha}\frac{i\,(k+q)^{\tau}\gamma_{\tau}}{(k+q)^{2}}\gamma_{\beta}\frac{i\,q^{\sigma}\gamma_{\sigma}}{q^{2}}\gamma_{\nu}\frac{i\,(p+q)^{\kappa}\gamma_{\kappa}}{(p+q)^{2}}]\,\,\,(p_{\lambda}\epsilon_{\mu\nu\lambda})\,\frac{(\epsilon_{\alpha\beta\eta}\,k_{\eta}\delta_{\eta
j})}{\bm{k}^{2}}$ (C.43)
$\displaystyle=\frac{1}{q^{2}(k+q)^{2}q^{2}(p+q)^{2}}\text{Tr}[\gamma_{\mu}\gamma_{\rho}\gamma_{\alpha}\gamma_{\tau}\gamma_{\beta}\gamma_{\sigma}\gamma_{\nu}\gamma_{\kappa}]\,q^{\rho}(k+q)^{\tau}q^{\sigma}(p+q)^{\kappa}\,(p_{\lambda}\epsilon_{\mu\nu\lambda})\,\frac{(\epsilon_{\alpha\beta\eta}\,k_{\eta}\delta_{\eta
j})}{\bm{k}^{2}}\quad\quad\quad$ (C.44)
$\displaystyle=\frac{1}{q^{2}(k+q)^{2}q^{2}(p+q)^{2}\;}\times
8\,\Big{(}[q^{2}(p\cdot q)+p^{2}q^{2}]\,\bm{k}^{2}+[(p\cdot
q)\,q^{2}+p^{2}q^{2}]\,\bm{k\cdot q}\Big{)}\frac{1}{\bm{k}^{2}}.$ (C.45)

We perform the convergent $q$-integral first:

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\Big{(}\text{Eq}.~{}\eqref{tmp-bc-
integral}\Big{)}\,$
$\displaystyle=\int\frac{d^{3}q}{(2\pi)^{3}}\frac{1}{(k+q)^{2}q^{2}(p+q)^{2}\;}$
(C.47) $\displaystyle\times 8\,\Big{(}[(p\cdot q)+p^{2}]\,\bm{k}^{2}+[(p\cdot
q)\,+p^{2}]\,\bm{k\cdot q}\Big{)}\frac{1}{\bm{k}^{2}}.$

Look at

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{p\cdot
q+p^{2}}{(k+q)^{2}q^{2}(p+q)^{2}\;}=\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\frac{1}{2}[(p+q)^{2}-p^{2}-q^{2}]+p^{2}}{(k+q)^{2}q^{2}(p+q)^{2}\;}$
(C.48)
$\displaystyle=\frac{1}{2}\int\frac{d^{3}q}{(2\pi)^{3}}\Big{(}\frac{1}{(k+q)^{2}q^{2}}-\frac{1}{(k+q)^{2}(p+q)^{2}\;}+\frac{p^{2}}{(k+q)^{2}q^{2}(p+q)^{2}\;}\Big{)}$
(C.49)
$\displaystyle=\frac{1}{2}(\frac{1}{8|k|}-\frac{1}{8|p-k|}+\frac{p^{2}}{8|p||k||p-k|}),$
(C.50)

where we used a formula in Appendix C.1.1. Next consider

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(p\cdot q+p^{2})\,\bm{k\cdot
q}}{(k+q)^{2}q^{2}(p+q)^{2}\;}\,$
$\displaystyle=\frac{1}{2}\int\frac{d^{3}q}{(2\pi)^{3}}\Big{(}\frac{\bm{k\cdot
q}}{(k+q)^{2}q^{2}}-\frac{\bm{k\cdot q}}{(k+q)^{2}(p+q)^{2}\;}$ (C.52)
$\displaystyle+\frac{p^{2}\bm{k\cdot q}}{(k+q)^{2}q^{2}(p+q)^{2}\;}\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}\frac{-\bm{k}^{2}}{16|k|}+\frac{\bm{k}^{2}+\bm{k\cdot
p}}{16|k-p|}+p^{2}\,L(\bm{k},p,k)\Big{)}\qquad$

where we used Eq. (C.14). Therefore,

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\Big{(}\text{ Eq}.~{}\eqref{tmp-bc-
integral}\Big{)}$ (C.53)
$\displaystyle=\frac{8}{2}(\frac{1}{8|k|}-\frac{1}{8|p-k|}+\frac{p^{2}}{8|p||k||p-k|})+\frac{8}{\bm{k}^{2}}\frac{1}{2}\Big{(}\frac{-\bm{k}^{2}}{16|k|}+\frac{\bm{k}^{2}+\bm{k\cdot
p}}{16|k-p|}+p^{2}\,L(\bm{k},p,k)\Big{)}\quad$ (C.54)
$\displaystyle=\frac{1}{2}(\frac{1}{|k|}-\frac{1}{|p-k|}+\frac{p^{2}}{|p||k||p-k|})+\frac{1}{2}\Big{(}\frac{-1}{2|k|}+\frac{1+\frac{\bm{k\cdot
p}}{\bm{k}^{2}}}{2|k-p|}\Big{)}+\frac{4\,p^{2}}{\bm{k}^{2}}\,L(\bm{k},p,k)$
(C.55) $\displaystyle=\frac{1}{4}(\frac{1}{|k|}+\frac{-1+\frac{\bm{k\cdot
p}}{\bm{k}^{2}}}{|k-p|})+\frac{p^{2}}{2\,|p||k||p-k|}+\frac{4\,p^{2}}{\bm{k}^{2}}\,L(\bm{k},p,k),$
(C.56)

where we used

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\bm{k\cdot
q}}{(k+q)^{2}q^{2}}=\frac{-\bm{k^{2}}\,}{16|k|},$ (C.57)
$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\bm{k\cdot
q}}{(q+k)^{2}(q+p)^{2}}=-\frac{\bm{k}^{2}+\bm{k\cdot p}}{16|k-p|}.$ (C.58)

Summarizing, we find that Eq. (C.41) equals

$\displaystyle\Pi^{(b+c)}_{(\text{A})}(p)\,$
$\displaystyle=\frac{1}{{\color[rgb]{0,0,0}2}|p|}\;{\color[rgb]{0,0,0}2}\frac{(-1)g^{4}}{N_{f}}\Big{(}\int\tilde{C}_{A}(k)\frac{dk_{0}d^{2}\bm{k}}{(2\pi)^{3}}\;\frac{1}{4}(\frac{1}{|k|}+\frac{-1+\frac{\bm{k\cdot
p}}{\bm{k}^{2}}}{|k-p|})+\frac{p^{2}}{2\,|p||k||p-k|}$ (C.61)
$\displaystyle+\frac{4\,p^{2}}{\bm{k}^{2}}\,L(\bm{k},p,k)\;\;\Big{)}$
$\displaystyle=\frac{1}{{\color[rgb]{0,0,0}2}|p|}\;2\frac{(-1)g^{4}}{N_{f}}\Big{(}\int\tilde{C}_{A}(k)\frac{dk_{0}d^{2}\bm{k}}{(2\pi)^{3}}\;\frac{1}{4}\frac{\bm{k\cdot
p}}{|k-p|\;\bm{k}^{2}}+\frac{p^{2}}{2\,|p||k||p-k|}+\frac{4\,p^{2}}{\bm{k}^{2}}\,L(\bm{k},p,k)$
$\displaystyle+\frac{1}{4|k|}-\frac{1}{4|k-p|}\Big{)}.$

###  C.3 Anti-symmetric component of $\Pi^{(a)}$

We aim to evaluate

$\displaystyle\Pi^{(a)}_{(\text{A})}(p)$
$\displaystyle=\frac{p_{\lambda}\epsilon_{\mu\nu\lambda}}{{\color[rgb]{0,0,0}2}|p|}\Pi^{(a)}_{\mu\nu}(p)$
(C.64)
$\displaystyle=\frac{p_{\lambda}\epsilon_{\mu\nu\lambda}}{{\color[rgb]{0,0,0}2}|p|}\frac{(-1)g^{4}}{N_{f}}\int_{k}\int_{q}\text{Tr}[\gamma_{\mu}G(q)\gamma_{\alpha}G(q+k)\gamma_{\nu}G(q+k+p)\gamma_{\beta}G(p+q)]$
$\displaystyle\times\,\frac{\tilde{C}_{A}(k)\,\epsilon_{\alpha\beta\lambda}k_{\eta}\delta_{\eta
j}}{\bm{k^{2}}}$
$\displaystyle=\frac{1}{{\color[rgb]{0,0,0}2}|p|}\frac{(-1)g^{4}}{N_{f}}\int_{k}\tilde{C}_{A}(k)\int_{q}\Big{[}-8(k\cdot
p)\,(p\cdot q)k_{\eta}-8(p\cdot q)^{2}k_{\eta}+8(k\cdot q)p^{2}k_{\eta}$
$\displaystyle+4(p\cdot q)p^{2}k_{\eta}+12p^{2}q^{2}k_{\eta}+8(k\cdot
q)(p\cdot q)p_{\eta}+4k^{2}(p\cdot q)p_{\eta}-4(k\cdot
p)q^{2}p_{\eta}-4(k\cdot p)p^{2}q_{\eta}$ $\displaystyle-8(k\cdot
q)p^{2}q_{\eta}-4k^{2}p^{2}q_{\eta}\Big{]}\times\frac{1}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}\times\,\frac{k_{\eta}\delta_{\eta
j}}{\bm{k^{2}}}.$

The integral over $q$ in Eq. (C.64) can be decomposed as

$\displaystyle\mathcal{I_{C}}\equiv\int_{q}\frac{C_{1}(p\cdot q)+C_{2}(k\cdot
q)+C_{3}\,q^{2}+C_{4}(k\cdot q)(p\cdot q)+C_{5}\bm{(k\cdot q)}+C_{6}(k\cdot
q)\bm{(k\cdot q)}+C_{7}\,(p\cdot q)^{2}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}},$
(C.65)

for some $C_{i}$, independent of $q$. Using the building block integrals from
Appendix C.1 and writing the norms $|k|=k$, and $|p|=p$, we find

$\displaystyle\mathcal{I_{C}}$
$\displaystyle={\color[rgb]{0,0,0}(\frac{-1}{2|k|}+\frac{1}{4|k-p|}+\frac{1}{4|k+p|})}+(\frac{2p}{4|k|\,|k-p|}+\frac{3p}{4|k|\,|k+p|})+{\color[rgb]{0,0,0}\frac{-k^{2}\,p}{2|k|\,|k-p|\,(k\cdot
p)}}$ (C.68) $\displaystyle+\frac{-p^{3}}{2|k|\,|k-p|\,(k\cdot
p)}+{\color[rgb]{0,0,0}\frac{k^{2}\,p}{2|k|\,|k+p|\,(k\cdot
p)}}+\frac{p^{3}}{2|k|\,|k+p|\,(k\cdot
p)}+\frac{-p^{2}}{8|k|\,(|p|\,|k|+k\cdot p)}$
$\displaystyle+\frac{p^{2}}{8|k-p|\,(|p|\,|k|+k\cdot
p)}+\frac{p^{3}}{8\,|k|\,|k-p|\,(|p|\,|k|+k\cdot
p)}+\frac{-p}{4\big{(}k^{2}+|k|\,|k+p|+k\cdot p\big{)}}$
$\displaystyle+\frac{p^{2}}{8k\,\big{(}k^{2}+|k|\,|k+p|+k\cdot
p\big{)}}+\frac{p^{2}}{8|k+p|\,\big{(}k^{2}+|k|\,|k+p|+k\cdot p\big{)}}$
$\displaystyle+\frac{-|p|\,|k|}{8|k+p|\,\big{(}k^{2}+|k|\,|k+p|+k\cdot
p\big{)}}+\frac{-p\,|k+p|}{8|k|\,\big{(}k^{2}+|k|\,|k+p|+k\cdot p\big{)}}$
$\displaystyle+\frac{p^{2}}{8|k|\,\big{(}p^{2}+|p|\,|k-p|-k\cdot
p\big{)}}+\frac{-p^{2}}{8|k-p|\,\big{(}p^{2}+|p|\,|k-p|-k\cdot p\big{)}}$
$\displaystyle+\frac{p^{3}}{8|k|\,|k-p|\,\big{(}p^{2}+|p|\,|k-p|-k\cdot
p\big{)}}+\frac{p^{2}}{8|k|\,\big{(}p^{2}+|p|\,|k+p|\,+k\cdot p\big{)}}$
$\displaystyle+\frac{-p^{2}}{8|k+p|\,\big{(}p^{2}+|p|\,|k+p|\,+k\cdot
p\big{)}}+\frac{1}{8|k|\,|k+p|\,\big{(}p^{2}+|p|\,|k+p|+k\cdot p\big{)}}.$

Note that, throughout the calculation, we have kept $p$ general. Many terms
above are individually divergent when integrated over $k$. We cut off the
divergent $k$ integrals with the cutoff $\Lambda$: For instance,

$\displaystyle\int_{-1}^{1}d\cos\theta\int_{0}^{\Lambda}dk\frac{2\pi
k^{2}}{(2\pi)^{3}}\frac{1}{k}=\frac{\Lambda^{2}}{4\pi^{2}},$ (C.69)
$\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{|k-p|}=\frac{\Lambda^{2}}{4\pi^{2}}+\frac{-p^{2}}{12\pi^{2}},$
(C.70)
$\displaystyle{\color[rgb]{0,0,0}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{-k^{2}\,p}{2|k|\,|k-p|\,(k\cdot
p)}}=\frac{p^{2}(12+\pi^{2})}{64\pi^{2}}+0\cdot\log\Lambda+\frac{-p\Lambda}{4\pi^{2}}.$
(C.71)

We are careful not to shift the momenta arbitrarily in any divergent integral;
otherwise, we are liable to obtain an incorrect result.

###  C.4 Combining the anti-symmetric components of $\Pi^{(a)}$ and
$\Pi^{(b+c)}$

We now add together the anti-symmetric components of $\Pi^{(a)}$ and
$\Pi^{(b+c)}$ in (C.64) and (C.61) to find:

$\displaystyle\Pi^{(a+b+c)}_{(\text{A})}(p)=-\frac{1}{{\color[rgb]{0,0,0}2}|p|}\frac{g^{4}}{N_{f}}\int_{k}$
$\displaystyle\times\tilde{C}_{A}(k)\Big{[}2\times\big{(}\frac{1}{4}\frac{\bm{k\cdot
p}}{|k-p|\;\bm{k}^{2}}+\frac{p^{2}}{2\,|p||k||p-k|}+\frac{4\,p^{2}}{\bm{k}^{2}}\,L(\bm{k},p,k)+\frac{1}{4|k|}-\frac{1}{4|k-p|}\;\big{)}+\mathcal{I_{C}}\;\Big{]}.\,\qquad\;\;$
(C.72)

To perform the integral over $k$, we take $p$ to lie along the $k_{\tau}$
axis, with $k\cdot p=|k||p|\cos\theta$, so that ${\bf k}\cdot{\bf p}=0$ and

$\displaystyle\int\frac{d^{3}k}{(2\pi)^{3}}(\ldots)=\int_{-1}^{1}d\cos\theta\int_{0}^{\Lambda}dk\frac{2\pi
k^{2}}{(2\pi)^{3}}(\dots).$ (C.73)

After performing the $\int_{0}^{\Lambda}dk$ integral, we perform a
$\frac{1}{\Lambda}$ expansion, and then do the $\int_{-1}^{1}d\cos\theta$
integral. The $\log\Lambda$ divergent terms vanish after the angular
integration. Letting $\cos\theta=z$, we symmetrize with respect to $z$ to get
rid of terms that are odd in $z$ and should therefore vanish after performing
the $z$ integral. The result is

$\displaystyle\Pi^{(a+b+c)}_{(\text{A})}(p)\,$
$\displaystyle=\frac{1}{{\color[rgb]{0,0,0}2}|p|}\frac{(-1)g^{4}}{N_{f}}\int_{-1}^{1}dz\,|p|^{2}\,\frac{-z+(-1+z^{2})\text{ArcTanh}[z]}{8\pi^{2}\,z}$
(C.76) $\displaystyle\times\frac{1}{A_{X}+B_{X}\sqrt{1-z^{2}}},$
$\displaystyle\sigma^{\psi}_{xy}=\frac{i}{\omega}\Pi^{(b+c+a)}(\bm{p}=0,|p_{0}|\to
i|\omega|),$

with
$A_{X}\equiv\frac{\kappa^{2}+g_{X}^{4}}{\kappa},B_{X}\equiv\frac{w_{x}g_{X}^{2}}{\kappa}$.

As a consistency check: When $w_{x}=0$ (vanishing Coulomb interaction),

$\displaystyle\Pi^{(a+b+c)}_{(\text{A})}(p;w_{x}=0)=\frac{1}{{\color[rgb]{0,0,0}2}|p|}\frac{(-1)g^{4}}{N_{f}}\frac{\kappa}{\kappa^{2}+g_{X}^{4}}\;(-p^{2}\;\frac{4+\pi^{2}}{32\pi^{2}}+0\cdot\log\Lambda+\frac{p}{4\pi^{2}}\,\Lambda\;),$
(C.77)

which agrees with [34]. We drop the linear divergence since it is an artifact
of the (gauge-noninvariant) hard cutoff.

###  C.5 Symmetric component of $\Pi^{(b+c)}$

Now we consider the symmetric component of $\Pi^{(b+c)}$. We will need the
symmetric part of the gauge field propagator:

$\displaystyle
D^{\text{Sym}}_{00}(k_{0},\bm{k})=\frac{|k|}{\bm{k}^{2}}\times\frac{1}{\frac{\kappa^{2}+g_{X}^{4}}{g_{X}^{2}}+w_{x}\,\sqrt{1-\cos^{2}\theta}}(1+\,\frac{w_{x}}{g_{X}^{2}}\sqrt{1-\cos^{2}\theta}),$
(C.78) $\displaystyle
D^{\text{Sym}}_{ij}(k_{0},\bm{k})=\frac{1}{|k|}\;(\delta_{ij}-\frac{k_{i}k_{j}}{\bm{k^{2}}})\times\frac{1}{\frac{\kappa^{2}+g_{X}^{4}}{g_{X}^{2}}+w_{x}\sqrt{1-\cos^{2}\theta}}.$
(C.79)

Here we are denoting $\sqrt{1-\cos^{2}\theta}=|{\bf k}|/|k|$. We parameterize
the gauge field propagator as

$\displaystyle{\color[rgb]{0,0,0}D^{\text{Sym}}_{\alpha\beta}(k)}=\Big{(}\delta_{\alpha
0}\delta_{\beta 0}\,f_{A}(k)+\delta_{\alpha i}\delta_{\beta
j}\;\delta_{ij}f_{B}(k)+\delta_{\alpha i}\delta_{\beta
j}\;k_{i}k_{j}f_{C}(k)\Big{)},\;\;\;$ (C.80)

where $f_{A},f_{B},f_{C}$ are constants. Using the gauge field propagator
above, the symmetric component of $\Pi^{(b+c)}$ is

$\displaystyle\Pi^{(b+c)}_{(\text{S})}(p)\,$
$\displaystyle=\frac{1}{2}\delta_{\mu\nu}\Pi^{(b+c)}_{\mu\nu}(p)$ (C.83)
$\displaystyle=\frac{\delta_{\mu\nu}}{2}\times
2\frac{(-1)g^{4}}{N_{f}}\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}q}{(2\pi)^{3}}\text{Tr}[\gamma_{\mu}\frac{i\,q^{\rho}\gamma_{\rho}}{q^{2}}\gamma_{\alpha}\frac{i\,(k+q)^{\tau}\gamma_{\tau}}{(k+q)^{2}}\gamma_{\beta}$
$\displaystyle\times\frac{i\,q^{\sigma}\gamma_{\sigma}}{q^{2}}\gamma_{\nu}\frac{i\,(p+q)^{\kappa}\gamma_{\kappa}}{(p+q)^{2}}]\,D^{\text{Sym}}_{\alpha\beta}(k)\qquad\;\;$

The trace $\text{Tr}[\ldots]_{\mu\nu}$ in the integrand evaluates to

$\displaystyle\text{Tr}[\gamma_{\mu}\frac{i\,q^{\rho}\gamma_{\rho}}{q^{2}}\gamma_{\alpha}\frac{i\,(k+q)^{\tau}\gamma_{\tau}}{(k+q)^{2}}\gamma_{\beta}\frac{i\,q^{\sigma}\gamma_{\sigma}}{q^{2}}\gamma_{\nu}\frac{i\,(p+q)^{\kappa}\gamma_{\kappa}}{(p+q)^{2}}]\,\,\,(\frac{\delta_{\mu\nu}}{2})D^{\text{Sym}}_{\alpha\beta}(k)$
(C.84)
$\displaystyle=\frac{1}{q^{2}(k+q)^{2}q^{2}(p+q)^{2}}\Big{(}\delta_{\alpha\beta}[2(k\cdot
q)\,(p\cdot q)-(k\cdot p)\,q^{2}+(k\cdot q)\,q^{2}+(p\cdot q)\,q^{2}+q^{4}\,]$
$\displaystyle+(p_{\alpha}k_{\beta}+p_{\beta}k_{\alpha})\,q^{2}+(q_{\alpha}k_{\beta}+q_{\beta}k_{\alpha})\,[-2(p\cdot
q)-q^{2}]+(q_{\alpha}p_{\beta}+q_{\beta}p_{\alpha})\,q^{2}$ (C.85)
$\displaystyle+(q_{\alpha}q_{\beta})\,[-4(p\cdot
q)-2q^{2}]\Big{)}\,D^{\text{Sym}}_{\alpha\beta}(k)$
$\displaystyle=\frac{1}{q^{2}(k+q)^{2}q^{2}(p+q)^{2}}\Big{(}\delta_{\alpha\beta}[2(k\cdot
q)\,(p\cdot q)-(k\cdot p)\,q^{2}+(k\cdot q)\,q^{2}+(p\cdot q)\,q^{2}+q^{4}\,]$
$\displaystyle+(p_{\alpha}k_{\beta}+p_{\beta}k_{\alpha})\,q^{2}+(q_{\alpha}k_{\beta}+q_{\beta}k_{\alpha})\,[-2(p\cdot
q)-q^{2}]+(q_{\alpha}p_{\beta}+q_{\beta}p_{\alpha})\,q^{2}$ (C.86)
$\displaystyle+(q_{\alpha}q_{\beta})\,[-4(p\cdot
q)-2q^{2}]\Big{)}\,{\color[rgb]{0,0,0}D^{\text{Sym}}_{\alpha\beta}(k)}.$

Next, we decompose the terms in the integrand with different $q$ dependencies
into various partial fractions:

$\displaystyle\text{Tr}[\ldots]_{\mu\nu}(\frac{\delta_{\mu\nu}}{2})D^{\text{Sym}}_{\alpha\beta}(k)$
(C.88)
$\displaystyle=\delta_{\alpha\beta}D^{\text{Sym}}_{\alpha\beta}(k)\Big{[}\frac{1}{2(q+k)^{2}(q+p)^{2}}+\frac{1}{2q^{2}\,q^{2}}+\frac{-k^{2}}{2(q+k)^{2}q^{4}}+\frac{-p^{2}}{2(q+p)^{2}q^{4}}$
$\displaystyle+\frac{k^{2}\,p^{2}}{2(q+k)^{2}(q+p)^{2}\,q^{4}}+\frac{-k\cdot
p}{(q+k)^{2}(q+p)^{2}\,q^{2}}\Big{]}$
$\displaystyle+(p_{\alpha}k_{\beta}+p_{\beta}k_{\alpha})D^{\text{Sym}}_{\alpha\beta}(k)\frac{1}{q^{2}(q+k)^{2}(q+p)^{2}}$
$\displaystyle+\Big{(}\frac{-2f_{B}(\bm{k\cdot
q})-2f_{C}\,\bm{k}^{2}(\bm{k\cdot
q})-2f_{A}\,k_{0}\,q_{0}}{(q+k)^{2}q^{4}}+\frac{2f_{B}\,p^{2}(\bm{k\cdot
q})+2f_{C}\,p^{2}\bm{k}^{2}(\bm{k\cdot
q})+2f_{A}\,p^{2}\,k_{0}\,q_{0}}{(q+k)^{2}(q+p)^{2}\,q^{4}}\Big{)}$
$\displaystyle+\frac{2f_{B}(\bm{p\cdot
q})+2f_{A}\,p_{0}q_{0}+2f_{C}\,(\bm{k\cdot p})\,(\bm{k\cdot
q})}{(q+k)^{2}\,(q+p)^{2}\,q^{2}}$
$\displaystyle+(-2)\frac{f_{B}\bm{q}^{2}+f_{C}\,(\bm{k\cdot
q})^{2}+f_{A}\,q_{0}^{2}}{(q+k)^{2}\,q^{4}}+(+2)\frac{f_{B}\,p^{2}\bm{q}^{2}+f_{C}\,p^{2}\,(\bm{k\cdot
q})^{2}+f_{A}\,p^{2}\,q_{0}^{2}}{(q+k)^{2}(q+p)^{2}\,q^{4}}.$ (C.89)

Each of the terms in the first few lines diverge as $|q|\to 0^{+}$ in the IR,
however, their sum is IR finite. Combining some of these terms with one
another, we perform the following $q$ integral:

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}{\color[rgb]{0,0,0}\frac{1}{2}}\Big{(}\frac{1}{q^{2}\,q^{2}}+\frac{-k^{2}}{(q+k)^{2}q^{4}}+\frac{-p^{2}}{(q+p)^{2}q^{4}}+\frac{k^{2}\,p^{2}}{(q+k)^{2}(q+p)^{2}\,q^{4}}\Big{)}$
(C.90)
$\displaystyle={\color[rgb]{0,0,0}\frac{1}{2}}\Big{[}\int\frac{d^{3}q}{(2\pi)^{3}}(\frac{1}{q^{2}\,q^{2}}+\frac{-p^{2}}{(q+p)^{2}q^{4}})+\int\frac{d^{3}q}{(2\pi)^{3}}(\frac{-k^{2}}{(q+k)^{2}q^{4}}+\frac{k^{2}\,p^{2}}{(q+k)^{2}(q+p)^{2}\,q^{4}})\Big{]}$
(C.91)
$\displaystyle={\color[rgb]{0,0,0}\frac{1}{2}}\Big{[}\int\frac{d^{3}q}{(2\pi)^{3}}\frac{q^{2}+2(q\cdot
p)}{(q+p)^{2}q^{4}}+\int\frac{d^{3}q}{(2\pi)^{3}}\frac{-k^{2}q^{2}-2k^{2}(q\cdot
p)}{(q+k)^{2}(q+p)^{2}\,q^{4}}\Big{]}$ (C.92)
$\displaystyle={\color[rgb]{0,0,0}\frac{1}{2}}\Big{[}\frac{1}{8|p|}-\frac{p\cdot
p}{8p^{3}}+\frac{-k^{2}}{8}\frac{1}{|p||k||p-k|}+(-2k^{2})(-1)\frac{p^{2}(p\cdot
k)+k^{2}(p\cdot p)}{16k^{3}p^{3}|k-p|}\Big{]}$ (C.93)
$\displaystyle={\color[rgb]{0,0,0}\frac{1}{2}}\frac{p\cdot
k}{8|k|\,|p|\,|k-p|},\qquad\;$ (C.94)

where we have used the “dot $p_{5}$” formulas in Appendix C.1.2, with
$p_{5}=p$ The remaining integrals over $q$ are straightforwardly performed
using formulas we have already given.

Next, we recall the following list of SO(3) non-invariant integrals that we
evaluated in Appendix C.1.2:

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\bm{k\cdot
q}}{(q+k)^{2}q^{4}}=R_{1}(\bm{k};k),\;\;\;\;p_{5}=(0,\bm{k}),$ (C.95)
$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{k_{0}\,q_{0}}{(q+k)^{2}q^{4}}=R_{1}(k_{0};k),\;\;\;\;p_{5}=(k_{0},0,0),$
(C.96) $\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(\bm{k\cdot
q})}{(q+k)^{2}(q+p)^{2}\,q^{4}}=N(\bm{k};p,k),\int\frac{d^{3}q}{(2\pi)^{3}}\frac{k_{0}\,q_{0}}{(q+k)^{2}(q+p)^{2}\,q^{4}}=N(k_{0};p,k),\qquad\;\;\;$
(C.97) $\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(\bm{p\cdot
q})}{(q+k)^{2}\,(q+p)^{2}\,q^{2}}=L(\bm{p};,k,p),\int\frac{d^{3}q}{(2\pi)^{3}}\frac{p_{0}q_{0}}{(q+k)^{2}\,(q+p)^{2}\,q^{2}}=L(p_{0};,k,p),\qquad\;\;\;$
(C.98)

and

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(\bm{k\cdot
q})^{2}}{(q+k)^{2}\,q^{4}}=R_{2}(\bm{k},\bm{k};k),$ (C.99)
$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\bm{q}^{2}}{(q+k)^{2}\,q^{4}}$
(C.100)
$\displaystyle=\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(q_{x}+iq_{y})(q_{x}-iq_{y})}{(q+k)^{2}\,q^{4}}=\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\Big{(}q\cdot(0,1,i)\Big{)}\,\Big{(}q\cdot(0,1,-i)\Big{)}}{(q+k)^{2}\,q^{4}}$
(C.101)
$\displaystyle=R_{2}((0,1,i),(0,1,-i);k)=\frac{2k^{2}+\bm{k}^{2}}{32\,k^{3}},$
(C.102)
$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{q_{0}^{2}}{(q+k)^{2}\,q^{4}}=\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\big{(}q\cdot(1,0,0)\;\big{)}^{2}}{(q+k)^{2}\,q^{4}}$
(C.103)
$\displaystyle=R_{2}((1,0,0),(1,0,0);k)=\frac{k^{2}+k_{0}^{2}}{32k^{3}}.$
(C.104)

Finally, we’ll need:

$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{(\bm{k\cdot
q})^{2}}{(q+k)^{2}(q+p)^{2}\,q^{4}}=R_{3}(\bm{k};k,p),$ (C.107)
$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{q_{0}^{2}}{(q+k)^{2}(q+p)^{2}\,q^{4}}=R_{3}((1,0,0);k,p),$
$\displaystyle\int\frac{d^{3}q}{(2\pi)^{3}}\frac{\bm{q}^{2}}{(q+k)^{2}(q+p)^{2}\,q^{4}}$
$\displaystyle=\int\frac{d^{3}q}{(2\pi)^{3}}\frac{q^{2}-q_{0}^{2}}{(q+k)^{2}(q+p)^{2}\,q^{4}}=\frac{1}{8|k|\,|p|\,|k-p|}-R_{3}((1,0,0);k,p).\qquad\;\;$
(C.108)

Plugging these in, we find the $q$ integral equals

$\displaystyle
I_{bc}\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\text{Tr}[\ldots]_{\mu\nu}(\frac{\delta_{\mu\nu}}{2})D^{\text{Sym}}_{\alpha\beta}(k)$
$\displaystyle=(f_{A}+2f_{B}+f_{C}\,\bm{k^{2}})[\frac{J_{2}(p,k)}{2}+(-p\cdot
k)\frac{J_{3}(p,k)}{2}]+[2f_{B}\bm{k\cdot p}+2f_{C}\bm{k^{2}}\bm{k\cdot
p}+2f_{A}k_{0}p_{0}]J_{3}(p,k)$
$\displaystyle+\frac{f_{A}\,k_{0}^{2}+f_{B}\bm{k}^{2}+f_{C}\bm{k^{4}}}{8k^{3}}+\big{[}2f_{B}p^{2}N(\bm{k};p,k)+2f_{C}\bm{k^{2}}p^{2}\,N(\bm{k};p,k)+2f_{A}\,p^{2}\,N(k_{0};p,k)\;\big{]}$
$\displaystyle+\big{[}2f_{B}\,L(\bm{p};k,p)+2f_{A}\,L(p_{0};k,p)+2f_{C}(\bm{k\cdot
p})\,L(\bm{k};k,p)\big{]}$
$\displaystyle+(-1)\big{[}f_{B}\frac{2k^{2}+\bm{k^{2}}}{16k^{3}}+f_{C}\bm{k^{2}}\frac{k^{2}+\bm{k^{2}}}{16k^{3}}+f_{A}\frac{k^{2}+k_{0}^{2}}{16k^{3}}\big{]}$
$\displaystyle+2p^{2}\,\Big{(}f_{B}\,\,[J_{3}(k,p)-R_{3}((1,0,0);k,p)\;]+f_{C}\,R_{3}(\bm{k};k,p)+f_{A}\,R_{3}((1,0,0);k,p)\Big{)}$
(C.109)

Note that in the definition of $I_{bc}$ above, we do not include the factor
$2$ which counts the contribution from diagrams $b$ and $c$.

###  C.6 Symmetric component of $\Pi^{(a)}$

The symmetric component of $\Pi^{(a)}$ is

$\displaystyle\Pi^{(a)}_{(\text{S})}(p)=\frac{1}{2}\delta_{\mu\nu}\Pi^{(a)}_{\mu\nu}(p)$
(C.110)
$\displaystyle=\frac{\delta_{\mu\nu}}{2}\times\frac{(-1)g^{4}}{N_{f}}\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}q}{(2\pi)^{3}}\text{Tr}[\gamma_{\mu}G(q)\gamma_{\alpha}G(q+k)\gamma_{\nu}G(q+k+p)\gamma_{\beta}G(p+q)]\,D^{\text{Sym}}_{\alpha\beta}(k)$
(C.111)
$\displaystyle=\frac{(-1)g^{4}}{N_{f}}\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}q}{(2\pi)^{3}}$
$\displaystyle\Bigg{(}\delta_{\alpha\beta}\big{[}-4(k\cdot p)(k\cdot
q)-4(k\cdot q)^{2}-2(k\cdot q)(p\cdot q)+3k^{2}(p\cdot q)-p^{2}\,(k\cdot
q)-q^{2}(k\cdot p)$ $\displaystyle-2q^{2}(k\cdot
q)+3k^{2}\,q^{2}\,-2q^{2}(p\cdot
q)-p^{2}\,q^{2}-q^{4}\,\big{]}+k_{\alpha}k_{\beta}[-4q^{2}-4(p\cdot q)]$
$\displaystyle+q_{\alpha}q_{\beta}[2p^{2}-4k^{2}]+p_{\alpha}p_{\beta}[2q^{2}+2(k\cdot
q)]$ $\displaystyle+p_{\alpha}q_{\beta}[2(k\cdot
q)-3k^{2}+2q^{2}]+p_{\beta}q_{\alpha}[-2(k\cdot p)-2(k\cdot q)-k^{2}-4(p\cdot
q)-2q^{2}]$ $\displaystyle+k_{\alpha}q_{\beta}[4(k\cdot p)+4(k\cdot
q)+p^{2}]+q_{\alpha}k_{\beta}[4(k\cdot
q)+p^{2}]+p_{\alpha}k_{\beta}[q^{2}+4(k\cdot q)]+k_{\alpha}p_{\beta}[-2(p\cdot
q)-q^{2}]\Bigg{)}\qquad\;\;$
$\displaystyle\times\frac{1}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}\,D^{\text{Sym}}_{\alpha\beta}(k).$
(C.112)

Above, we have used the parameterization of the symmetric part of the gauge
field propagator in Eq. (C.80).

We rewrite the first part of the integrand using partial fractions:

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)\delta_{\alpha\beta}\Big{(}\frac{2(k\cdot
p)\,k^{2}-k^{4}-\frac{3}{2}k^{2}p^{2}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}+\frac{2k^{2}}{q^{2}\,(q+k+p)^{2}(q+k)^{2}}$
(C.113) $\displaystyle+\frac{{\color[rgb]{0,0,0}-}2(k\cdot
p)+2k^{2}-(q+k)^{2}}{q^{2}\,(q+k+p)^{2}(q+p)^{2}}+\frac{\frac{-1}{2}}{(q+k+p)^{2}(q+k)^{2}}+\frac{\frac{3}{2}}{(q+k+p)^{2}(q+p)^{2}}+\frac{\frac{-1}{2}}{(q+k+p)^{2}q^{2}}$
(C.114) $\displaystyle+\frac{(k\cdot
p)-\frac{q^{2}}{2}}{(q+k+p)^{2}(q+k)^{2}(q+p)^{2}}\Big{)}.$ (C.115)

Integrating this over $q$ gives:

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)\delta_{\alpha\beta}\Big{(}[2(k\cdot
p)\,k^{2}-k^{4}-\frac{3}{2}k^{2}p^{2}]\,J_{4}(k,p)+2k^{2}\,J_{3}(k+p,k)$
(C.116) $\displaystyle+[{\color[rgb]{0,0,0}-}2(k\cdot
p)+2k^{2}]\,J_{3}(k+p,p)-I(p,-k)+\frac{-1}{2}\,J_{2}(k+p,k)+\frac{3}{2}\,J_{2}(k+p,p)+\frac{-1}{2}\,J_{2}(k+p,0)$
(C.117) $\displaystyle+(k\cdot
p)\,J_{3}(k,k-p)+\frac{-1}{2}\,{\color[rgb]{0,0,0}I(k,p)}\Big{)}$ (C.118)
$\displaystyle=D^{\rm
sym}_{\alpha\beta}(k)\delta_{\alpha\beta}\Big{(}[-k^{4}-\frac{3}{2}k^{2}p^{2}]\,J_{4}(k,p)+\frac{|k|}{4p|k-p|}-\frac{|k|}{4p|k+p|}+\frac{-1}{4p}+\frac{1}{16|k-p|}+$
(C.119) $\displaystyle\frac{1}{16|k+p|}+\frac{|k|}{2p\,|k+p|}\Big{)}$ (C.120)
$\displaystyle=D^{\rm
sym}_{\alpha\beta}(k)\delta_{\alpha\beta}\Big{(}[-k^{4}-\frac{3}{2}k^{2}p^{2}]\,J_{4}(k,p)+0+\frac{-1}{4p}+\frac{1}{16|k-p|}+\frac{1}{16|k+p|}+\frac{|k|}{2p\,|k+p|}\Big{)}.$
(C.121)

Next, we use partial fractions to re-express:

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)k_{\alpha}k_{\beta}\frac{-4q^{2}-4(p\cdot
q)}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ $\displaystyle=(D^{\rm
sym}_{\alpha\beta}(k)k_{\alpha}k_{\beta})\Big{(}\frac{2p^{2}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}-\frac{2}{(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
(C.122) $\displaystyle-\frac{2}{(q+k)^{2}(q+k+p)^{2}q^{2}}\Big{)}.$ (C.123)

Integrating this over $q$ gives:

$\displaystyle(D^{\rm
sym}_{\alpha\beta}(k)k_{\alpha}k_{\beta})\Big{(}2p^{2}\,J_{4}(k,p)-2\,J_{3}(p,p-k)-2\,J_{3}(k,k+p)\Big{)}$
(C.124) $\displaystyle=(D^{\rm
sym}_{\alpha\beta}(k)k_{\alpha}k_{\beta})\Big{(}\frac{p}{4|k|\,|k-p|(k\cdot
p)}+\frac{-p}{4|k|\,|k+p|(k\cdot p)}+\frac{-1}{4|k|\,|p|\,|k-p|}$ (C.125)
$\displaystyle+\frac{-1}{4|k|\,|p|\,|k+p|}\Big{)}.$ (C.126)

The next term is

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)q_{\alpha}q_{\beta}\frac{(2p^{2}-4k^{2})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
(C.127) $\displaystyle=(2p^{2}-4k^{2})\frac{f_{C}(\bm{k\cdot
q})^{2}+f_{A}q_{0}^{2}+f_{B}\,\bm{q}^{2}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
(C.128) $\displaystyle=(2p^{2}-4k^{2})\frac{f_{C}(\bm{k\cdot
q})^{2}+(f_{A}-f_{B})q_{0}^{2}+f_{B}\,q^{2}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}.$
(C.129)

Integrating this over $q$ gives:

$\displaystyle(2p^{2}-4k^{2})\Big{(}\,f_{C}\,M_{2}(\bm{k};k,p)+(f_{A}-f_{B})\,M_{2}((1,0,0);k,p)+f_{B}\,J_{3}(k-p,k)\Big{)}.$
(C.130)

Moving on to the next term:

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)p_{\alpha}p_{\beta}\frac{2q^{2}+2(k\cdot
q)}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ $\displaystyle=D^{\rm
sym}_{\alpha\beta}(k)p_{\alpha}p_{\beta}\Big{(}\frac{-k^{2}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}+\frac{1}{(q+k+p)^{2}(q+k)^{2}(q+p)^{2}}$
(C.131) $\displaystyle+\frac{1}{(q+k+p)^{2}(q+p)^{2}\,q^{2}}\Big{)}.$ (C.132)

Integrating over $q$ gives

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)p_{\alpha}p_{\beta}\Big{(}\,-k^{2}\,J_{4}(k,p)+J_{3}(k,k-p)+J_{3}(p,k+p)\Big{)}$
(C.133) $\displaystyle=[f_{C}\bm{(k\cdot
p)}^{2}+f_{A}\,p_{0}^{2}+f_{B}\,\bm{p^{2}}]\;\Big{(}\,\frac{-|k|}{8|p|\,|k-p|\,(k\cdot
p)}+\frac{|k|}{8|p|\,|k+p|\,(k\cdot p)}+\frac{1}{8|k|\,|p|\,|k-p|}$ (C.134)
$\displaystyle+\frac{1}{8|k|\,|p|\,|k+p|}\Big{)}.$ (C.135)

Next, we consider (arranging the terms by their power of $q$):

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)\frac{p_{\alpha}q_{\beta}[\ldots]+q_{\alpha}p_{\beta}[\ldots]}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
$\displaystyle=\frac{-4f_{C}(\bm{k\cdot p})\,(p\cdot q)\,(\bm{k\cdot
q})-4f_{B}\,(p\cdot q)\,(\bm{p\cdot q})-4f_{A}\,(p\cdot
q)\,(p_{0}q_{0})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.136)
$\displaystyle+\frac{[-2f_{B}(k\cdot p)-4f_{B}\,k^{2}](\bm{p\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ $\displaystyle+\frac{[-2f_{C}(k\cdot
p)(\bm{k\cdot p})-4f_{C}\,k^{2}\,(\bm{k\cdot p})]\,(\bm{k\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}+\frac{[-2f_{A}(k\cdot
p)-4f_{A}k^{2}](p_{0}\,q_{0})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}.$ (C.137)

Integrating over $q$ gives

$\displaystyle-4f_{C}(\bm{k\cdot
p})\,F_{1}-4f_{B}\,F_{2}-4f_{A}\,F_{3}+[-2f_{B}(k\cdot
p)-4f_{B}\,k^{2}]\,M(p;k)$ $\displaystyle+[-2f_{C}(k\cdot p)(\bm{k\cdot
p})-4f_{C}\,k^{2}\,(\bm{k\cdot p})]\,M(k;p)$ $\displaystyle+[-2f_{A}(k\cdot
p)-4f_{A}k^{2}]\big{[}\frac{1}{2}J_{3}(k,k+p)-\frac{1}{2}J_{3}(k-p,k)-\frac{p^{2}}{2}J_{4}(k,p)-M(p;k)\big{]},$
(C.138)

where $F_{1}$, $F_{2}$, and $F_{3}$ are defined below:

$\displaystyle F_{1}\,$ $\displaystyle\equiv\int_{q}\frac{(p\cdot
q)\,(\bm{k\cdot q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.143)
$\displaystyle=\int_{q}\frac{1}{2}\frac{[(q+p)^{2}-p^{2}-q^{2}](\bm{k\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
$\displaystyle=\frac{1}{2}\int_{q}\Big{(}\frac{(\bm{k\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}}-\frac{p^{2}(\bm{k\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ $\displaystyle-\frac{(\bm{k\cdot
q})}{(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}L(\bm{k};k,k+p)-p^{2}\,M(k;p)-L(\bm{k};p,p-k)+\bm{k}^{2}\,J_{3}(p,p-k)\Big{)},$
$\displaystyle F_{2}\,$ $\displaystyle\equiv\int_{q}\frac{(p\cdot
q)\,(\bm{p\cdot q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.148)
$\displaystyle=\frac{1}{2}\int_{q}\frac{[(q+p)^{2}-p^{2}-q^{2}]\,(\bm{p\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
$\displaystyle=\frac{1}{2}\Big{(}\int_{q}\frac{\,(\bm{p\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}}-\int_{q}\frac{p^{2}\,(\bm{p\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}-$
$\displaystyle\int_{q}\frac{\,(\bm{p\cdot
q})}{(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}L(\bm{p};k,k+p)-p^{2}\,M(p;k)-L(\bm{p};k,k-p)+\bm{p}^{2}\,J_{3}(k,k-p)\Big{)},$
$\displaystyle F_{3}\,$ $\displaystyle\equiv\int_{q}\frac{(p\cdot
q)\,(p_{0}q_{0})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.152)
$\displaystyle=\frac{1}{2}\Big{(}\int_{q}\frac{\,(p_{0}q_{0})}{q^{2}(q+k)^{2}(q+k+p)^{2}}-\int_{q}\frac{p^{2}\,(p_{0}q_{0})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
$\displaystyle-\int_{q}\frac{\,(p_{0}q_{0})}{(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}L(\vec{p}_{0};k,k+p)-\int_{q}\frac{p^{2}\,(p\cdot
q-\bm{p\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}-L(\vec{p}_{0};k,k-p)+p_{0}^{2}\,J_{3}(k,k-p)\Big{)}$
$\displaystyle=\frac{1}{2}\Big{(}L(\vec{p}_{0};k,k+p)-\big{[}\frac{p^{2}}{2}J_{3}(k,k+p)-\frac{p^{4}}{2}\,J_{4}(k,p)-\frac{p^{2}}{2}J_{3}(k,k-p)-p^{2}\,M(p;k)\big{]}$
$\displaystyle\qquad-L(\vec{p}_{0};k,k-p)+p_{0}^{2}\,J_{3}(k,k-p)\Big{)}.$

Note that $M(p;k)$ is not symmetric with respect to the exchange of its
arguments.

Next, we consider

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)\frac{k_{\alpha}q_{\beta}[\ldots]+q_{\alpha}k_{\beta}[\ldots]}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}=\frac{8(k\cdot
q)\big{[}f_{B}(\bm{k\cdot q})+f_{C}\,(\bm{k^{2}})(\bm{k\cdot
q})+f_{A}\,k_{0}\,q_{0}\big{]}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
$\displaystyle+\frac{(\bm{k\cdot q})\big{[}4f_{B}(k\cdot
p)+4f_{C}\,(\bm{k^{2}})(k\cdot
p)+2f_{B}\,p^{2}+2f_{C}(\bm{k^{2}})\,p^{2}\big{]}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
(C.153) $\displaystyle+\frac{[4f_{A}(k\cdot
p)+2f_{A}\,p^{2}]\,k_{0}q_{0}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}.$ (C.154)

Integrating over $q$ gives

$\displaystyle[8f_{B}+8f_{C}\,\bm{k^{2}}]\,F_{4}+8f_{A}\,[M_{2}(k;k,p)-F_{4}]$
$\displaystyle+\big{[}4f_{B}(k\cdot p)+4f_{C}\,(\bm{k^{2}})(k\cdot
p)+2f_{B}\,p^{2}+2f_{C}(\bm{k^{2}})\,p^{2}\big{]}\,M(k;p)$
$\displaystyle+[4f_{A}(k\cdot
p)+2f_{A}\,p^{2}]\,[\frac{1}{2}J_{3}(k+p,p)-\frac{k^{2}}{2}J_{4}(k,p)-\frac{1}{2}J_{3}(k,k-p)-M(k,p)],$
(C.155)

where, using (C.24),

$\displaystyle F_{4}\,$ $\displaystyle\equiv\int_{q}\frac{(k\cdot
q)(\bm{k\cdot q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.158)
$\displaystyle=\frac{1}{2}\int_{q}\frac{[(q+k)^{2}-k^{2}-q^{2}](\bm{k\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
$\displaystyle=\frac{1}{2}\Big{(}L(\bm{k},p,k+p)-k^{2}\,M(k;p)-L(\bm{k},p,p-k)+\bm{k}^{2}\frac{1}{8|p|\,|p-k|\,|k|}\Big{)}.$

We also note the integral relation,

$\displaystyle F_{5}\,$ $\displaystyle\equiv\int_{q}\frac{(k\cdot
q)\,k_{0}\,q_{0}\big{]}}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.161)
$\displaystyle=\int_{q}\frac{(k\cdot q)\,(k\cdot q-\bm{k\cdot
q})}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$
$\displaystyle=M_{2}(k;k,p)-F_{4}.\qquad\qquad$

where the definition in Eq. (C.38) for $M_{2}$ was used. In fact,
$M_{2}(k;k,p)$ can be simplified further in terms of $J_{2},J_{3},J_{4},I$.

Finally, we consider

$\displaystyle D^{\rm
sym}_{\alpha\beta}(k)\frac{+p_{\alpha}k_{\beta}[\ldots]+k_{\alpha}p_{\beta}[\ldots]}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}=\frac{(k\cdot
q)\;[4f_{B}(\bm{k\cdot p})+4f_{C}\bm{k}^{2}(\bm{k\cdot
p})+4f_{A}\,k_{0}p_{0}]}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}$ (C.162)
$\displaystyle+\frac{(p\cdot q)\;[-2f_{B}(\bm{k\cdot
p})-2f_{C}\bm{k}^{2}(\bm{k\cdot
p})-2f_{A}\,k_{0}p_{0}]}{q^{2}(q+k)^{2}(q+k+p)^{2}(q+p)^{2}}.$ (C.163)

Integrating over $q$ gives

$\displaystyle[4f_{B}(\bm{k\cdot p})+4f_{C}\bm{k}^{2}(\bm{k\cdot
p})+4f_{A}\,k_{0}p_{0}]\,\times\frac{1}{2}[J_{3}(k+p,p)-k^{2}J_{4}(p,k)-J_{3}(k,k-p)]$
$\displaystyle+[-2f_{B}(\bm{k\cdot p})-2f_{C}\bm{k}^{2}(\bm{k\cdot
p})-2f_{A}\,k_{0}p_{0}]\times\frac{1}{2}[J_{3}(k+p,k)-p^{2}J_{4}(p,k)-$
(C.164) $\displaystyle J_{3}(k,k-p)]$ (C.165)

Putting these results together produces $I_{a}$ (schematically):

$\displaystyle
I_{a}\equiv\int\frac{d^{3}q}{(2\pi)^{3}}\text{Tr}[...]_{\mu\nu}(\frac{\delta_{\mu\nu}}{2})D^{\text{Sym}}_{\alpha\beta}(k).$
(C.166)

The result is a lengthy expression that we do not write out here.

###  C.7 Combining the symmetric components of $\Pi^{(a)}$ and $\Pi^{(b+c)}$

We add together (C.166) and (C.109) to find (including multiplying (C.109) by
a factor of 2) the symmetric component:

$\displaystyle\Pi^{(a+b+c)}_{(\text{S})}(p)=\frac{1}{2}\delta_{\mu\nu}\Pi^{(a+b+c)}_{\mu\nu}(p)=\frac{(-1)g^{4}}{N_{f}}\int\frac{d^{3}k}{(2\pi)^{3}}(2\,I_{bc}+I_{a})$
$\displaystyle=\frac{(-1)g^{4}}{N_{f}}\int_{-1}^{1}d\cos\theta\int_{0}^{\Lambda}dk\frac{2\pi
k^{2}}{(2\pi)^{3}}(2\,I_{bc}+I_{a}).$ (C.167)

As before, we take the external momentum $p$ to lie along the $k_{\tau}$
direction so that $k\cdot p=|k||p|\cos\theta$. This choice allows us to set
all $\bm{k\cdot p}=0$ in $2\,I_{bc}+I_{a}$ expression. We plug in the
propagator in (C.80), perform the $dk$ radial direction integral with cut off
$\Lambda$, and perform a $\frac{1}{\Lambda}$ expansion to find:

$\displaystyle\Pi^{(a+b+c)}_{(\text{S})}(p)=-\frac{g^{4}}{N_{f}}|p|\int_{-1}^{1}d\cos\theta\Big{(}f^{w_{x}}_{0}(\cos\theta)+f^{w_{x}}_{L}(\cos\theta)\,\log[\frac{\Lambda}{|p|}]+f^{w_{x}}_{1}(\cos\theta)\,\Lambda\Big{)}.\quad\quad$
(C.168)

$f^{w_{x}}_{L},~{}f^{w_{x}}_{1}$ are odd functions of $\cos\theta$ so they
both vanish after performing the angular integration. Symmetrizing
$f^{w_{x}}_{0}(\cos\theta)$ to remove any antisymmetric part, we find

$\displaystyle\frac{f^{w_{x}}_{0}(z)+f^{w_{x}}_{0}(-z)}{2}$ (C.169)
$\displaystyle\equiv\frac{\Big{(}-5\,C_{Y}\,z(-1+z^{2})+z\sqrt{1-z^{2}}(9+z^{2})+6(-1+z^{2})(C_{Y}-C_{Y}\,z^{2}+2\sqrt{1-z^{2}})\text{arctanh}[z]\Big{)}}{192\pi^{2}\,z\,[\;A_{Y}\sqrt{1-z^{2}}+B_{Y}(1-z^{2})\;\;]},$

where $z=\cos\theta$. Thus, we have

$\displaystyle\sigma^{\text{2loop-
gauge}}_{xx}=\frac{i}{\omega}\Pi^{(a+b+c)}_{(\text{S})}(\bm{p}=0,|p_{0}|\to
i\omega)=\frac{(-1)g^{4}}{N_{f}}\int_{-1}^{1}dz\frac{f^{w_{x}}_{0}(z)+f^{w_{x}}_{0}(-z)}{2}.$
(C.170)

This agrees with Eq. (III.14) in the main text.

As a consistency check, we note that our result agrees with [35, 36] when
$w_{x}=0$:

$\displaystyle\frac{f^{w_{x}=0}_{0}(z)+f^{w_{x}=0}_{0}(-z)}{2}=\frac{g_{X}^{2}}{\kappa^{2}+g_{X}^{4}}\frac{1}{192\pi^{2}\,z}\Big{(}9z+z^{3}+12(z^{2}-1)\text{arctanh}[z]\Big{)}$
(C.171)

and so

$\displaystyle\int_{-1}^{1}d\cos\theta
f^{w_{x}=0}_{0}(\cos\theta)\Big{|}_{\kappa=0,g^{2}_{X}=\frac{1}{16},g=1}\approx
0.00893191=\frac{1}{16}\times 0.14291062.$ (C.172)

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