# Anomalous Hall effect in type-I Weyl metals beyond the noncrossing
approximation

Jia-Xing Zhang1 Wei Chen1,2<EMAIL_ADDRESS>1National Laboratory of
Solid State Microstructures and School of Physics, Nanjing University,
Nanjing, China  2Collaborative Innovation Center of Advanced Microstructures,
Nanjing University, Nanjing, China

###### Abstract

We study the anomalous Hall effect (AHE) in tilted Weyl metals with Gaussian
disorder due to the crossed $X$ and $\Psi$ diagrams in this work. The
importance of such diagrams to the AHE has been demonstrated recently in 2D
massive Dirac model and Rashba ferromagnets. It was shown that the inclusion
of such diagrams dramatically changes the total AHE in such systems. In this
work, we show that the contributions from the $X$ and $\Psi$ diagrams to the
AHE in tilted Weyl metals are of the same order of the non-crossing diagram we
studied in a previous work, but with opposite sign. The total contribution of
the $X$ and $\Psi$ diagrams cancels the majority of the contribution from the
non-crossing diagram in tilted Weyl metals, similar to the 2D massive Dirac
model. We also discussed the difference of the contributions from the crossed
diagrams between 2D massive Dirac model and the tilted Weyl metals. At last,
we discussed the experimental relevance of observing the AHE due to the $X$
and $\Psi$ diagrams in type-I Weyl metal ${\rm Co_{3}Sn_{2}S_{2}}$.

##  I I. Introduction

The anomalous Hall effect (AHE) has been a topic of interest since it’s first
observed in ferromagnetic iron by Edwin Hall in 1881 Hall1881 . It is
analogous to an usual Hall effect but without the need of an external magnetic
field Luttinger1954 ; Haldane1988 . The transverse motion in the anomalous
Hall systems originates from the spin-orbit interaction and to have a net
transverse current, the time reversal symmetry (TRS) has to be broken in the
system Luttinger1954 ; Haldane1988 ; Sinitsyn2007 ; Sinitsyn2008 ; Yang2011 .
In insulator or semiconductor, the anomalous Hall conductivity is quantized
and insensitive to impurity scatterings. In metals, however, the impurity
scatterings affect the AHE significantly, and the AHE in such cases can be
divided to the intrinsic contribution, which is due to the non-trivial
topology of the electronic band structure and remains in the clean limit, and
the extrinsic contribution, which is due to the impurity scatterings
Sinitsyn2007 . The anomalous Hall current can be obtained either from the
quantum Kubo-Streda (QKS) formula Streda1982  or a semi-classical Boltzmann
equation (SBE) approach Sinitsyn2007 ; Sinitsyn2008 . The former approach is
more systematic whereas the latter is physically more transparent.

It is well-known that the Feynman diagrams with crossed impurity lines result
in a longitudinal conductivity which is smaller than the non-crossing diagram
by a factor of $1/\epsilon_{F}\tau$ so the crossed diagrams are usually
neglected in computing the longitudinal conductivity. For a long time, the
crossed diagrams were also ignored for the AHE and only the Feynman diagrams
with non-crossing impurity scattering lines are considered for the AHE
Sinitsyn2007 ; MacDonald2006 ; Sinitsyn2008 . The anomalous Hall conductivity
from the non-crossing diagrams is independent of the impurity scattering rate
or strength, i.e., $\sigma^{a}_{H}\sim\tau^{0}$. However, it was demonstrated
recently that the diagrams with two crossed impurity lines, namely the so-
called X and $\Psi$ diagrams, may also contribute to the anomalous Hall
conductivity with the same order of magnitude as the non-crossing diagram
Ado2015 ; Ado2016 ; Ado2017 . The contributions of such crossed diagrams come
from the skew scatterings on pairs of closely located impurities with distance
of the order of the Fermi wavelength. The account of such crossed diagrams in
a number of anomalous Hall systems changes the total AHE in the systems
dramatically Ado2016 ; Ado2015 ; Ado2017 ; Levchenko2016 . For example, the
inclusion of the X and $\Psi$ diagrams in two-dimensional (2D) Rashba
ferromagnetic metal results in a non-vanishing AHE instead of the vanishing
result under the non-crossing approximation (NCA) Inoue2006 ; Sinova2007 ;
Ado2016 . In 2D massive Dirac model, the X and $\Psi$ diagrams almost cancel
out the NCA contribution at high energy Ado2015 ; Ado2017 . It was also shown
that the same crossed diagrams play an important role for the AHE on the
surface of topological Kondo insulator Levchenko2016 , for the Kerr effect in
chiral p-wave superconductors Levchenko2017 , and for the spin Hall effect in
the presence of strong impurities Ferreira2016 .

The above cases show that the crossed diagrams play an important role for a
complete study of the AHE in a general case. For the reason, we study the
contribution of the crossed diagrams, namely the X and $\Psi$ diagram to the
AHE in tilted Weyl metals with breaking TRS and weak Gaussian disorder in this
work Wan2011 ; Vishwanath2018 . Diagrams with more crossed lines have smaller
contribution in $1/\epsilon_{F}\tau$. For untilted Weyl metals it has been
shown that the impurity scatterings have little effect on the AHE only if the
Fermi energy is not very far from the Weyl nodes Burkov2014 . This is because
the low energy effective Hamiltonian of a single Weyl node of the untilted
Weyl metal gains an emergent TRS and the AHE due to the impurity scatterings
vanish. For tilted Weyl metals, the tilting breaks the TRS of the effective
Hamiltonian of a single Weyl node Pesin2017 ; Zyuzin2017  and the impurity
scatterings have significant effects on the AHE in such system Fu2021 ;
Chen2022 . In a previous paper, we have studied the disorder induced AHE in
the tilted Weyl metals due to the non-crossing diagrams and obtained both the
intrinsic and extrinsic contribution for such diagrams from the quantum Kubo-
Streda formula Chen2022 . We also separated the two different extrinsic
contributions, namely the side jump and skew scattering contribution from the
non-crossing diagrams in this system. The study of the crossed diagrams for
the tilted Weyl metals in this work is an important supplement of the skew
scattering contribution to the AHE in such system.

We show that the contribution from both the $X$ and $\Psi$ diagrams for tilted
Weyl metals is of the same order of the contribution from the NCA diagram,
i.e., $\sim\tau^{0}$. This is different from the 2D massive Dirac model for
which the contribution from the $\Psi$ diagram vanishes for Gaussian disorder
Ado2015 ; Ado2017 . On the other hand, our calculation shows that the total
contribution of the $X$ and $\Psi$ diagram cancels a majority part of the
contribution from the NCA diagram for tilted Weyl metals. This is similar to
the 2D massive Dirac model. However, the inclusion of the $X$ and $\Psi$
diagram does not change the dependence of the anomalous Hall conductivity on
the Fermi energy whereas in 2D massive Dirac model, the crossed diagram
changes the total anomalous Hall conductivity from $\sigma_{xy}\sim
m/\epsilon_{F}$ for NCA diagram to $\sigma_{xy}\sim(m/\epsilon_{F})^{3}$
Ado2017 .

We also discussed the experimental relevance to observe the effects of the $X$
and $\Psi$ diagrams in tilted Weyl metals in experiments, such as ${\rm
Co_{3}Sn_{2}S_{2}}$ Felser2018 ; Ding2019 ; Wang2018 . We point out that the
condition to observe the contributions of the crossed diagrams is much more
demanding than the non-crossing diagram. Moreover, since the intrinsic AHE
from the Chern-Simons term is much higher than the AHE from both the non-
crossing and crossed diagrams in ${\rm Co_{3}Sn_{2}S_{2}}$, the effects of the
disorder on the AHE is not very distinguishable in experiments in this system.
We propose that one can observe the effects of the disorder on the AHE by
measuring the anomalous Nernst effect (ANE) in such system because the
contributions of the disorder to the ANE and AHE are proportional to each
other but the Chern-Simons term has no contribution to the ANE.

This paper is organized as follows. In Sec. II, we present the model and the
calculation of the anomalous Hall effect due to the crossed $X$ and $\Psi$
diagrams in tilted Weyl metals, and compare the AHE from the crossed diagrams
with the non-crossing diagram, as well as with other systems, such as 2D
massive Dirac model. In Sec.III, we have a brief conclusion of the
experimental relevance of observing the effects of $X$ and $\Psi$ diagrams. In
Sec. IV, we have a summary of this work.

##  II II. AHE in tilted Weyl metals due to X and $\Psi$ diagrams

The low energy physics of a Type-I Weyl metal with breaking TRS can be
described by an effective low energy Hamiltonian of two independent Weyl nodes
and a topological Chern-Simons term Burkov2015 ; Chen2022 . The Chern-Simons
term results in an AHE proportional to the distance of the two Weyl nodes and
is not affected by the impurity scatterings. We will then focus on the low
energy effective Hamiltonian of the Weyl nodes in the following, which is

$H=\sum_{\chi}(\chi v\bm{\sigma}\cdot\mathbf{p}+\mathbf{u_{\chi}\cdot p}),$
(1)

where $\chi=\pm 1$ is the chirality of the two Weyl nodes, $\bm{\sigma}$ are
the Pauli matrices and $\mathbf{u}_{\chi}$ is a tilting velocity with
$u_{\chi}<v$ for type-I Weyl metals we consider in this work. Here we assume
the tilting $\mathbf{u}_{+}=-\mathbf{u}_{-}=\mathbf{u}$, i.e., the tilting is
opposite for the two valleys. For this case, the AHE in the two Weyl nodes
adds up instead of cancel out. The Hamiltonian $H_{\chi}$ for each single
valley results in two tilting linear bands $\epsilon_{\pm}=\pm
vp+\mathbf{u_{\chi}\cdot p}$. The tilting term breaks the TRS of a single Weyl
node so the AHE from each valley does not vanish. The term $\chi
v\mathbf{\bm{\sigma}\cdot p}$ breaks the global TRS so the total AHE of the
two valleys is non-zero.

Figure 1: (a)The Feynman diagram of the response function
$\Pi^{\mathbf{I}}_{\alpha\beta}$ under the non-crossing approximation (NCA) in
the spin basis. The thick solid lines are Green’s function in the spin basis
under the Born approximation, and the solid square represents the current
vertex $\Gamma_{\alpha}$ renormalized by the ladder diagram under the NCA.
(b)The X diagram with two crossed impurity lines. (c) and (d): The $\Psi$
diagram with two crossed impurity lines. (e)The recursion equation satisfied
by the renormalized current vertex $\Gamma_{\alpha}$.

We consider weak Gaussian disorder (white noise) with random potential
$V({\mathbf{r}})=V_{0}\sum_{a}\delta({\mathbf{r}-r_{a}})$ and correlation
$\langle
V({\mathbf{r}})V({\mathbf{r}^{\prime}})\rangle=\gamma\delta({\mathbf{r}}-{\mathbf{r}^{\prime}})$,
where $\gamma=n_{i}V_{0}^{2}$ and $n_{i}$ is the impurity density. We assume
that the mean free path of the electrons is much larger than the Fermi
wavelength, i.e., $k_{F}l\gg 1$ or $\epsilon_{F}\tau\gg 1$. The anomalous Hall
conductivity may be written in two parts, i.e., $\sigma_{H}^{\rm I}$ and
$\sigma_{H}^{\rm II}$, in the Kubo-Streda formulaStreda1982 ; Sinitsyn2007 .
Formally $\sigma_{H}^{\rm I}$ takes into account the contribution on the Fermi
surface, and $\sigma_{H}^{\rm II}$ includes contribution from the whole Fermi
sea. Since $\sigma_{H}^{\rm II}$ is not sensitive to impurity scatterings and
its contribution in the clean limit has been studied in previous works for
tilted Weyl metalsPesin2017 ; Zyuzin2017 , we only need to study the
$\sigma_{H}^{\rm I}$ part in this work. The leading order contribution to the
response function $\Pi^{\mathbf{I}}_{\alpha\beta}$ includes the diagrams in
Fig.1, where Fig.1a is the diagram under the NCA and has been studied in our
previous work Chen2022 . The NCA diagram includes both the intrinsic and
extrinsic contributions, and both contributions to the AHE are independent of
the scattering rate $1/\tau$ in the leading order, i.e., $\sim\tau^{0}$. For
the crossed diagrams in Fig. 1(b)-(d), previous works Ado2016 ; Ado2017 ;
Levchenko2016  have shown that for Gaussian disorder, the leading order AHE
from these diagrams for 2D Rashba ferromagnets and massive Dirac models is of
the same order as the non-crossing diagram in Fig.1a, , i.e., $\sim\tau^{0}$.
In the following, we study the contribution of the crossed $X$ and $\Psi$
diagrams for tilted Weyl metals and compare the leading order contribution of
these diagrams with the non-crossing diagram in Fig.1a. Diagrams with more
crossed impurity lines have smaller contribution in $1/\tau$ and so are
negligible.

We assume that the impurity potential is diagonal for both the spin and valley
index, so the two valleys decouple and one can compute the AHE in each valley
separately. The leading order contribution to the AHE from the NCA diagram of
the tilted Weyl metals has been worked out in our previous work Chen2022 . The
total dc anomalous Hall conductivity from the non-crossing diagram for the two
Weyl nodes is $\sigma^{\rm NCA}_{xy}=4e^{2}\epsilon_{F}u/3\pi^{2}v^{2}$ in the
leading order of $u/v$ for $\mathbf{u}$ in the $z$ direction. As a comparison,
we compute the anomalous Hall conductivity in the dc limit due to the crossed
$X$ and $\Psi$ diagram in the tilted Weyl metals in the leading order of $u/v$
in the following.

We consider a uniform electric field ${\mathbf{E}}=-\partial_{t}{\mathbf{A}}$
applied to the system. In the linear response regime
$j^{\rm\mathbf{I}}_{\alpha}=\Pi^{\rm\mathbf{I}}_{\alpha\beta}A^{\beta}$, where
$A^{\alpha}=(0,\mathbf{A})$. The response functions
$\Pi^{\mathbf{I}}_{\alpha\beta}$ in the dc limit for the X and $\Psi$ diagrams
for a single Weyl node (e.g., $\chi=1$) are respectively

$\displaystyle\Pi_{\alpha\beta}^{X}=\gamma^{2}\omega\sum_{\mathbf{p}_{1},...,\mathbf{p}_{4}}\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\delta(\mathbf{p}_{1}+\mathbf{p}_{2}-\mathbf{p}_{3}-\mathbf{p}_{4})$
$\displaystyle\rm
Tr[{\Gamma}_{\alpha}G^{R}(\mathbf{p}_{1})G^{R}(\mathbf{p}_{3})G^{R}(\mathbf{p}_{2}){\Gamma}_{\beta}G^{A}(\mathbf{p}_{2})G^{A}(\mathbf{p}_{4})G^{A}(\mathbf{p}_{1})],$

and

$\displaystyle\Pi_{\alpha\beta}^{\Psi}=\gamma^{2}\omega\sum_{\mathbf{p}_{1},...,\mathbf{p}_{4}}\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\delta(\mathbf{p}_{1}+\mathbf{p}_{4}-\mathbf{p}_{2}-\mathbf{p}_{3})$
$\displaystyle{\rm
Tr}[G^{A}(\mathbf{p}_{1}){\Gamma}_{\alpha}G^{R}(\mathbf{p}_{1})G^{R}(\mathbf{p}_{3})G^{R}(\mathbf{p}_{4})G^{R}(\mathbf{p}_{2}){\Gamma}_{\beta}G^{A}(\mathbf{p}_{2})$
$\displaystyle+G^{A}(\mathbf{p}_{1}){\Gamma}_{\alpha}G^{R}(\mathbf{p}_{1})G^{R}(\mathbf{p}_{2}){\Gamma}_{\beta}G^{A}(\mathbf{p}_{2})G^{A}(\mathbf{p}_{4})G^{A}(\mathbf{p}_{3})],$

where $G^{R/A}$ is the retarded/advanced Green’s function (GF) of the tilted
Weyl metals, and $\Gamma_{\alpha}$ is the current vertex renormalized by the
non-crossing ladder diagram Chen2022 . We have omitted the argument $\epsilon$
in $G^{R/A}(\epsilon,\mathbf{p})$ in the above equations for brevity.

The impurity averaged retarded/advanced GF in a single valley (e.g. with
$\chi=1$) under the first Born approximation is Chen2022

$G^{R/A}(\epsilon,\mathbf{p})=(\epsilon-v\mathbf{\bm{\sigma}\cdot
p}-\mathbf{u\cdot p}-\Sigma^{R/A})^{-1},$ (4)

where the self-energy due to the impurity scatterings is
$\Sigma^{R/A}=\mp\frac{i}{2\tau}[1+\mathbf{\Delta}(\mathbf{u})\cdot{\bm{\sigma}}]$
with $1/\tau=\pi\gamma g(\epsilon_{F})$,
$g(\epsilon_{F})=\int\frac{d^{3}p}{(2\pi)^{3}}\delta(\mathbf{u\cdot
p}+vp-\epsilon_{F})=\frac{\epsilon^{2}_{F}v}{2\pi^{2}(v^{2}-u^{2})^{2}}$ being
the density of states at the Fermi energy $\epsilon_{F}>0$ and
${\mathbf{\Delta}}(\mathbf{u})=-{\mathbf{u}}/v$.

For the calculation in this work, it is convenient to write the $G^{R/A}$ in
Eq.(4) as

$G^{R/A}(\epsilon,{\mathbf{p}})=\frac{(\epsilon\pm\frac{i}{2\tau}-\bm{u}\cdot\bm{p})\sigma^{0}+v\bm{p}\cdot\bm{\sigma}\mp\frac{i}{2\tau}(\bm{\Delta}\cdot\bm{\sigma})}{(\epsilon-\epsilon_{p}^{+}\pm\frac{i}{2\tau^{+}})(\epsilon-\epsilon_{p}^{-}\pm\frac{i}{2\tau^{-}})},$
(5)

with
$1/\tau^{\pm}=\frac{1}{\tau}(1\pm\frac{\mathbf{p}\cdot\mathbf{\Delta}}{p}).$

The renormalized current vertex $\Gamma_{\alpha}$ in Eq.(II) and (II) has been
worked out in our previous work Chen2022 . The bare current vertex for the
tilted Weyl metals is
$\hat{j}_{\alpha}=e(v\sigma_{\alpha}+u_{\alpha}\sigma_{0})$ (we define
$u_{0}\equiv 0$). By expressing $\hat{j}_{\alpha}$ and $\Gamma_{\alpha}$ with
the Pauli matrices as $\hat{j}_{\alpha}={\cal J}_{\alpha\beta}\sigma_{\beta}$
and
$\hat{\Gamma}_{\alpha}=\Gamma_{\alpha\beta}\sigma_{\beta},\alpha,\beta=0,x,y,z$,
one can solve the coefficient of the renormalized current vertex as
$\Gamma_{\alpha\beta}={\cal J}_{\alpha\gamma}{\cal D}_{\gamma\beta}$, where
the summation over the repeated index $\gamma$ is implied as usual and ${\cal
D}=(1-\gamma{\cal I})^{-1}$ is the $4\times 4$ diffusion matrix with the
polarization operator ${\cal I}$ defined as

${\cal
I}_{\alpha\beta}=\frac{1}{2}\int\frac{d{\mathbf{p}}}{(2\pi)^{3}}\rm{Tr}[\sigma_{\alpha}G^{R}(\epsilon+\omega,\mathbf{p}+\mathbf{q})\sigma_{\beta}G^{A}(\epsilon,\mathbf{p})].$
(6)

In our previous work Chen2022 , we have shown that the renormalized current
vertex $\Gamma_{\alpha\beta}=ev{\cal D}_{\alpha\beta}$, i.e., the tilting term
$u_{\alpha}\sigma_{0}$ in the bare current vertex has no contribution to the
AHE and the main effect of the tilting is to produce an anisotropy of the
Fermi surface. We have also worked out the ${\cal I}$ matrix and ${\cal D}$
matrix for the tilted Weyl metals in the previous work Chen2022 , so we will
just apply the results of such matrices for the study in this work.

Denoting the integrand of ${\cal I}_{\alpha\beta}$ in the dc limit as
$I_{\alpha\beta}(\mathbf{p})=\frac{1}{2}\rm{Tr}[\sigma_{\alpha}G^{R}(\epsilon,\mathbf{p})\sigma_{\beta}G^{A}(\epsilon,\mathbf{p})]$,
one gets
$G^{A}\hat{\Gamma}_{\alpha}G^{R}=\Gamma_{\alpha\beta}I_{\beta\gamma}\sigma_{\gamma},\
G^{R}\hat{\Gamma}_{\alpha}G^{A}=\sigma_{\gamma}I_{\gamma\beta}\Gamma_{\alpha\beta}$.
The response functions for the X and $\Psi$ diagrams in Fig.1(b)-(d) can then
be written as

$\displaystyle\Pi_{\alpha\beta}^{X}$ $\displaystyle=$ $\displaystyle
e^{2}\gamma^{2}v^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\sum_{\mathbf{p}_{1},...,\mathbf{p}_{4}}\delta(\mathbf{p}_{1}+\mathbf{p}_{2}-\mathbf{p}_{3}-\mathbf{p}_{4}){\cal
D}_{\alpha\xi}I_{\xi\mu}(\mathbf{p}_{1})F_{\mu\nu}(\mathbf{p}_{3},\mathbf{p}_{4})I_{\nu\gamma}(\mathbf{p}_{2}){\cal
D}^{T}_{\gamma\beta},$ (7) $\displaystyle\Pi_{\alpha\beta}^{\Psi}$
$\displaystyle=$ $\displaystyle
e^{2}\gamma^{2}v^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\sum_{\mathbf{p}_{1},...,\mathbf{p}_{4}}\delta(\mathbf{p}_{1}+\mathbf{p}_{4}-\mathbf{p}_{2}-\mathbf{p}_{3}){\cal
D}_{\alpha\xi}I_{\xi\mu}(\mathbf{p}_{1})M_{\mu\nu}(\mathbf{p}_{3},\mathbf{p}_{4})I_{\nu\gamma}(\mathbf{p}_{2}){\cal
D}^{T}_{\gamma\beta},$ (8)

where we have defined

$\displaystyle F_{\mu\nu}(\mathbf{p}_{3},\mathbf{p}_{4})$
$\displaystyle\equiv$ $\displaystyle{\rm
Tr}[\sigma_{\mu}G^{R}(\epsilon,\mathbf{p}_{3})\sigma_{\nu}G^{A}(\epsilon,\mathbf{p}_{4})],$
(9) $\displaystyle M_{\mu\nu}(\mathbf{p}_{3},\mathbf{p}_{4})$
$\displaystyle\equiv$ $\displaystyle{\rm
Tr}[\sigma_{\mu}\sigma_{\nu}G^{A}(\epsilon,\mathbf{p}_{4})G^{A}(\epsilon,\mathbf{p}_{3})$
(10) $\displaystyle\ \ \
+\sigma_{\nu}\sigma_{\mu}G^{R}(\epsilon,\mathbf{p}_{3})G^{R}(\epsilon,\mathbf{p}_{4})].$

The AHE due to the X and $\Psi$ diagram corresponds to the anti-symmetric part
of the response function $\Pi_{\alpha\beta}^{X}$ and
$\Pi_{\alpha\beta}^{\Psi}$. In the following, we study the AHE in tilted Weyl
metals due to the two diagrams respectively.

###  II.1 A. AHE due to the X diagram

In this subsection, we study the AHE due to the $X$ diagram in tilted Weyl
metals. To do this, we first compute the anti-symmetric part of the response
function $\Pi^{X}_{\alpha\beta}$ in Eq.(7).

For the matrices ${\cal D},I$ and $F$ in Eq.(7), the symmetric parts of these
matrices are ${\cal D}_{0}\sim\tau^{0},I^{s}\sim\tau,F^{s}\sim\tau^{0}$, and
the anti-symmetric parts ${\cal
D}^{a}\sim\tau^{-1},I^{a}\sim\tau^{0},F^{a}\sim\tau^{0}$. In the leading order
of $1/\epsilon_{F}\tau$, the anti-symmetric part of $\Pi_{\alpha\beta}^{X}$ is
then

$\displaystyle\Pi_{\alpha\beta}^{X,a}=e^{2}\gamma^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\sum_{\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{Q}}$
$\displaystyle{\cal
D}_{0,\alpha\gamma}I^{s}_{\gamma\mu}(\mathbf{p}_{1})F^{a}_{\mu\nu}(\mathbf{p}_{1}-\mathbf{Q},\mathbf{p}_{2}+\mathbf{Q})I^{s}_{\nu\eta}(\mathbf{p}_{2}){\cal
D}^{T}_{0,\eta\beta},$

where
$\mathbf{Q}\equiv\mathbf{p}_{1}-\mathbf{p}_{3}=\mathbf{p}_{4}-\mathbf{p}_{2}$.

The vertex correction factor ${\cal D}_{0}$ and ${\cal D}^{T}_{0}$ on the two
ends of $\Pi^{X,a}_{\alpha\beta}$ are constant matrices as a function of $u$
and $v$, as given in our previous work Chen2022 , and when multiplied with the
remaining part of the response function, it results in an extra total factor
$\tilde{\alpha}^{2}\approx 9/4+{\cal O}(u^{2}/v^{2})$ only if the remaining
part is an anti-symmetric matrix of the linear order of $u_{i},i=1,2,3$, which
is the case for both the X and $\Psi$ diagrams. For convenience, we will then
drop the ${\cal D}_{0}$ factor in Eq.(II.1) in the following calculation and
add the vertex correction factor $\tilde{\alpha}^{2}$ at the end.

Since the symmetric part of the $I$ matrix in the dc limit is

$I^{s}_{\alpha\beta}(\mathbf{p})\approx\pi\tau^{+}\delta(\epsilon-\bm{u}\cdot\bm{p}-vp)\frac{1}{p^{2}}\times
p_{\alpha}p_{\beta},$ (12)

the integration over the momentum $\mathbf{p}_{1}$ and $\mathbf{p}_{2}$ is
bound to the Fermi surface due to the $\delta$ function in $I^{s}$.

The anti-symmetric part of the $F$ matrix for the X diagram is
$F^{a}_{\mu\nu}=N_{\mu\nu}(F^{a})/D(F)$, where

$\displaystyle N_{\mu\nu}(F^{a})=2iv^{2}\\{\epsilon^{0\mu\nu
k}[(p_{1}p_{2k}-p_{2}p_{1k})+(p_{1}+p_{2})Q_{k}$ $\displaystyle\ \ \ \ \ \ \ \
+\frac{\mathbf{u}\cdot\mathbf{Q}}{v}(p_{1k}+p_{2k})]-\epsilon^{\mu\nu
lk}(\mathbf{p}_{1}-\mathbf{Q})_{l}(\mathbf{p}_{2}+\mathbf{Q})_{k}\\},$
$\displaystyle
D(F)=[\epsilon-\mathbf{u\cdot}(\mathbf{p}_{1}-\mathbf{Q})-v|\mathbf{p}_{1}-\mathbf{Q}|+\frac{i}{2\tau_{3}^{+}}]$
$\displaystyle\ \ \ \ \ \ \ \
\times[\epsilon-\mathbf{u\cdot}(\mathbf{p}_{1}-\mathbf{Q})+v|\mathbf{p}_{1}-\mathbf{Q}|+\frac{i}{2\tau_{3}^{-}}]$
$\displaystyle\ \ \ \ \ \ \ \
\times[\epsilon-\mathbf{u\cdot}(\mathbf{p}_{2}+\mathbf{Q})-v|\mathbf{p}_{2}+\mathbf{Q}|-\frac{i}{2\tau_{4}^{+}}]$
$\displaystyle\ \ \ \ \ \ \ \ \
\times[\epsilon-\mathbf{u\cdot}(\mathbf{p}_{2}+\mathbf{Q})+v|\mathbf{p}_{2}+\mathbf{Q}|-\frac{i}{2\tau_{4}^{-}}],$
(13)

and
$\frac{1}{\tau_{i}^{\pm}}\equiv\frac{1}{\tau}(1\pm\delta_{i}),\delta_{i}\equiv\frac{\mathbf{p}_{i}\cdot\mathbf{\Delta}}{p_{i}}$,
$\mu,\nu=0,1,2,3$, and $l,k=1,2,3$. In $N_{\mu\nu}(F^{a})$, we have only kept
the leading order in $1/\tau$.

We assume $\mathbf{u}$ in the $z$ direction for simplicity and
$\bm{Q}=Q(\sin\alpha\cos\beta,\sin\alpha\sin\beta,\cos\alpha)$. Rotate the
$z$-axis to the direction of $\bm{Q}$ by the transformation

$\left(\begin{array}[]{c}\hat{\mathbf{x}^{\prime}}\\\
\hat{\mathbf{y}^{\prime}}\\\
\hat{\mathbf{z}^{\prime}}\end{array}\right)=\left(\begin{array}[]{ccc}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\\
-\sin\beta&\cos\beta&0\\\
\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{array}\right)\
\left(\begin{array}[]{c}\hat{\mathbf{x}}\\\ \hat{\mathbf{y}}\\\
\hat{\mathbf{z}}\end{array}\right),$

where $(\hat{\mathbf{x}},\hat{\mathbf{y}},\hat{\mathbf{z}})$ and
$(\mathbf{x}^{\prime},\mathbf{y}^{\prime},\mathbf{z}^{\prime})$ are the bases
of the old and new frames respectively. The coordinates of
$\mathbf{p}_{i},i=1,2$ in the old and new frames are denoted as
$(p_{ix},p_{iy},p_{iz})$ and
$(p^{\prime}_{ix},p^{\prime}_{iy},p^{\prime}_{iz})$ respectively. Assuming in
the rotated frame $\mathbf{p}_{i}=p\
(\sin\theta_{i}\cos\phi_{i}\cdot\hat{\mathbf{x}^{\prime}}+\sin\theta_{i}\sin\phi_{i}\cdot\hat{\mathbf{y}^{\prime}}+\cos\theta_{i}\cdot\hat{\mathbf{z}^{\prime}})$
for $i=1,2$, we then have
$\mathbf{p}_{1}\cdot\mathbf{Q}=p_{1}Q\cos\theta_{1},\mathbf{p}_{2}\cdot\mathbf{Q}=p_{2}Q\cos\theta_{2},\mathbf{u}\cdot\mathbf{Q}=uQ\cos\alpha$.

The coordinates $p_{i\alpha},i=1,2$ in the old frame may be expressed as

$\displaystyle\left\\{\begin{array}[]{ll}p_{i,x}=&p_{i}(\cos\phi_{i}\sin\theta_{i}\cos\alpha\cos\beta-\sin\theta_{i}\sin\phi_{i}\sin\beta+\cos\theta_{i}\sin\alpha\cos\beta),\\\
p_{i,y}=&p_{i}(\cos\phi_{i}\sin\theta_{i}\cos\alpha\sin\beta+\sin\theta_{i}\sin\phi_{i}\cos\beta+\cos\theta_{i}\sin\alpha\sin\beta),\\\
p_{i,z}=&p_{i}(-\sin\theta_{i}\cos\phi\sin\alpha+\cos\theta_{i}\cos\alpha).\\\
\end{array}\right.$ (17)

From the $\delta$ function in $I^{s}$, one can get
$p_{i}=\frac{\epsilon}{v+u\hat{z}_{i}},i=1,2$, where

$\displaystyle\hat{z}_{i}$ $\displaystyle=$ $\displaystyle
p_{iz}/p_{i}=-\sin\theta_{i}\cos\phi_{i}\sin\alpha+\cos\theta_{i}\cos\alpha.$
(18)

Applying the $\delta$ function in $I^{s}$ to replace $\epsilon$ by
$\mathbf{p}_{1}$ and $\mathbf{p}_{2}$ in $D(F)$, we get

$\displaystyle\frac{1}{D(F)}=$
$\displaystyle[(v^{2}-u^{2}\cos^{2}\alpha)Q^{2}-2v^{2}p_{1}Q(\cos\theta_{1}+\frac{u}{v}\cos\alpha)-\frac{i}{\tau}(vp_{1}+uQ\cos\alpha+v\delta_{3}|\mathbf{p}_{1}-\mathbf{Q}|)]^{-1}$
(19)
$\displaystyle\times[(v^{2}-u^{2}\cos^{2}\alpha)Q^{2}+2v^{2}p_{2}Q(\cos\theta_{2}+\frac{u}{v}\cos\alpha)+\frac{i}{\tau}(vp_{2}-uQ\cos\alpha+v\delta_{4}|\mathbf{p}_{2}+\mathbf{Q}|)]^{-1}.$

The AHE due to the $X$ and $\Psi$ diagram is finite only when $\mathbf{u}$ is
non-zero. In this work, we only compute the AHE in the tilted Weyl metals in
the leading order of $\mathbf{u}$ for simplicity. For the reason, we expand
$1/D(F)$ in terms of $u/v$ and keep only the terms up to the linear order of
$u$. We then get

$\displaystyle\frac{1}{D(F)}$ $\displaystyle\approx$
$\displaystyle[v^{2}Q^{2}-2v^{2}\bm{p}_{1}\cdot\bm{Q}-\frac{i}{\tau}vp_{1}-2vp_{1}\bm{u}\cdot\bm{Q}]^{-1}[v^{2}Q^{2}+2v^{2}\bm{p}_{2}\cdot\bm{Q}+\frac{i}{\tau}vp_{2}+2vp_{2}\bm{u}\cdot\bm{Q}]^{-1}$
(20) $\displaystyle\approx$
$\displaystyle\frac{1+\frac{u}{v}\hat{z}_{1}}{v^{2}Q^{2}-2v\epsilon
Q(\cos\theta_{1}+\frac{u}{v}\cos\alpha)-\frac{i}{\tau}\epsilon+vuQ^{2}\hat{z}_{1}}\frac{1+\frac{u}{v}\hat{z}_{2}}{v^{2}Q^{2}+2v\epsilon
Q(\cos\theta_{2}+\frac{u}{v}\cos\alpha)+\frac{i}{\tau}\epsilon+vuQ^{2}\hat{z}_{2}}$
$\displaystyle\approx$
$\displaystyle(1+\frac{u}{v}\hat{z}_{1})(1+\frac{u}{v}\hat{z}_{2})\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{1}-\frac{i}{\tau}\epsilon}(1-\frac{vuQ^{2}\hat{z}_{1}-2\epsilon
uQ\cos\alpha}{v^{2}Q^{2}-2v\epsilon Q\cos\theta_{1}-\frac{i}{\tau}\epsilon})$
$\displaystyle\times\frac{1}{v^{2}Q^{2}+2v\epsilon
Q\cos\theta_{2}+\frac{i}{\tau}\epsilon}(1-\frac{vuQ^{2}\hat{z}_{2}+2\epsilon
uQ\cos\alpha}{v^{2}Q^{2}+2v\epsilon Q\cos\theta_{2}+\frac{i}{\tau}\epsilon}),$

where we have neglected the linear order of $u$ terms $\sim
iuQ\cos\alpha/\tau$ and $i\delta_{3}/\tau,i\delta_{4}/\tau$ since they contain
an extra small factor $1/\tau$.

Putting $I^{s}$ and $F^{a}$ together and neglecting the vertex correction at
the two ends of the $X$ diagram at the moment, we get the anti-symmetric part
of the response function $\Pi_{xy}^{X}$ as

$\displaystyle\Pi_{\alpha\beta}^{X,a}$ $\displaystyle=$
$\displaystyle\pi^{2}e^{2}\gamma^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\int_{0}^{\infty}\frac{dp_{1}}{8\pi^{3}}p^{2}_{1}\int_{0}^{\infty}\frac{dp_{2}}{8\pi^{3}}p^{2}_{2}\int_{0}^{\infty}\frac{dQ}{8\pi^{3}}Q^{2}\int_{0}^{\pi}\sin\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin\theta_{2}d\theta_{2}\int_{0}^{\pi}\sin\alpha
d\alpha$ (21)
$\displaystyle\int_{0}^{2\pi}d\phi_{1}\int_{0}^{2\pi}d\phi_{2}\int_{0}^{2\pi}d\beta\frac{4iv^{2}p_{1\alpha}p_{2\beta}(p_{1}+p_{2})(\bm{p}_{1}\times\bm{p}_{2})\cdot\bm{Q}}{p^{2}_{1}\
p^{2}_{2}\
D(F)}\underset{i=1,2}{\mathbf{\Pi}}\tau_{i}^{+}\delta(\epsilon-\bm{u}\cdot\bm{p}_{i}-vp_{i}).$

The scalar factor
$(\bm{p}_{1}\times\bm{p}_{2})\cdot\bm{Q}=p_{1}p_{2}Q\sin\theta_{1}\sin\theta_{2}\sin(\phi_{2}-\phi_{1})$
in the rotated frame, and
$\tau_{i}^{+}=\tau/(1-\frac{u}{v}\hat{z}_{i}),i=1,2$. For the integrand in
Eq.(21), only the factor $p_{1\alpha}p_{2\beta}$ includes the angle $\beta$
and one can easily integrate out this angle. For $\mathbf{u}$ in the $z$
direction, if the electric field $\mathbf{E}$ is also in the $z$ direction,
$\Pi_{\alpha z}^{X,a}=0$ after the integration over the angle $\beta$ for
$\alpha\neq z$. For the reason, we only need to consider the case when
$\mathbf{E}$ is perpendicular to $\mathbf{u}$. Assuming the electric field
$\mathbf{E}$ in the $y$ direction, $\Pi_{zy}^{X,a}$ is zero with the
integration over $\beta$. We then only need to compute the non-vanishing
component $\Pi_{xy}^{X,a}$.

Since

$\displaystyle\int_{0}^{2\pi}d\beta p_{1x}p_{2y}=\pi
p_{1}p_{2}[\cos\alpha\sin\theta_{1}\sin\theta_{2}\sin(\phi_{2}-\phi_{1})$
$\displaystyle\ \
+\sin\alpha(\cos\theta_{1}\sin\theta_{2}\sin\phi_{2}-\cos\theta_{2}\sin\theta_{1}\sin\phi_{1})],$
(22)

the response function $\Pi_{xy}^{X,a}$ for the $X$ diagram becomes

$\displaystyle\Pi_{xy}^{X,a}$ $\displaystyle=$ $\displaystyle
e^{2}v^{2}\gamma^{2}\tau^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\frac{1}{(2\pi)^{9}}4iv^{2}\pi^{3}$ (23)
$\displaystyle\times$
$\displaystyle\int_{0}^{\infty}Q^{2}dQ\int_{0}^{\pi}\sin\alpha
d\alpha\int_{0}^{\pi}\sin\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin\theta_{2}d\theta_{2}\int_{0}^{2\pi}d\phi_{1}\int_{0}^{2\pi}d\phi_{2}\
Q\sin\theta_{1}\sin\theta_{2}\sin(\phi_{2}-\phi_{1})$ $\displaystyle\times$
$\displaystyle[\cos\alpha\sin\theta_{1}\sin\theta_{2}\sin(\phi_{2}-\phi_{1})+\sin\alpha(\cos\theta_{1}\sin\theta_{2}\sin\phi_{2}-\cos\theta_{2}\sin\theta_{1}\sin\phi_{1})]$
$\displaystyle\times$
$\displaystyle\frac{v}{v-u\hat{z}_{1}}\frac{v}{v-u\hat{z}_{2}}(\frac{\epsilon}{v+u\hat{z}_{1}}+\frac{\epsilon}{v+u\hat{z}_{2}})\frac{\epsilon^{2}}{(v+u\hat{z}_{1})^{3}}\frac{\epsilon^{2}}{(v+u\hat{z}_{2})^{3}}\frac{1}{D(F)}.$

It is easy to check that at $u=0$, the response function in Eq.(23) vanishes.
Expanding Eq.(23) to the linear order of $u$ and combing $1/D(F)$ in Eq.(20) ,
we get

$\displaystyle\Pi_{xy}^{X,a}(u)$ $\displaystyle=$ $\displaystyle
e^{2}v^{2}\gamma^{2}\tau^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\times
4iv^{2}\pi^{3}\frac{1}{(2\pi)^{9}}\times\frac{\epsilon^{5}}{v^{7}}$ (24)
$\displaystyle\times$
$\displaystyle\int_{0}^{\infty}Q^{2}dQ\int_{0}^{\pi}\sin\alpha
d\alpha\int_{0}^{\pi}\sin\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin\theta_{2}d\theta_{2}\int_{0}^{2\pi}d\phi_{1}\int_{0}^{2\pi}d\phi_{2}\times
Q\sin\theta_{1}\sin\theta_{2}\sin(\phi_{2}-\phi_{1})$ $\displaystyle\times$
$\displaystyle[\cos\alpha\sin\theta_{1}\sin\theta_{2}\sin(\phi_{2}-\phi_{1})+\sin\alpha(\cos\theta_{1}\sin\theta_{2}\sin\phi_{2}-\cos\theta_{2}\sin\theta_{1}\sin\phi_{1})]$
$\displaystyle\times$ $\displaystyle\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{1}-\frac{i}{\tau}\epsilon}\frac{1}{v^{2}Q^{2}+2v\epsilon
Q\cos\theta_{2}+\frac{i}{\tau}\epsilon}$ $\displaystyle\times$
$\displaystyle[-3\frac{u}{v}(\hat{z}_{1}+\hat{z}_{2})-2\frac{vuQ^{2}\hat{z}_{1}-2\epsilon
uQ\cos\alpha}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{1}-\frac{i}{\tau}\epsilon}-2\frac{vuQ^{2}\hat{z}_{2}+2\epsilon
uQ\cos\alpha}{v^{2}Q^{2}+2v\epsilon Q\cos\theta_{2}+\frac{i}{\tau}\epsilon}].$

The angular integration over $\phi_{1},\phi_{2}$ and $\alpha$ can be easily
done in the above equation and the contributions from the terms with the
factor $\hat{z}_{1}$ and $\hat{z}_{2}$ vanish after these angular integration.
The response function for the $X$ diagram after the integration over
$\phi_{1},\phi_{2},\alpha$ and $\epsilon$ becomes

$\displaystyle\Pi_{xy}^{X,a}(u)$ $\displaystyle=$
$\displaystyle-\frac{\omega}{12\pi^{3}}e^{2}v^{3}\epsilon_{F}^{2}u\times\int_{0}^{\infty}Q^{4}dQ\int_{0}^{\pi}\sin\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin\theta_{2}d\theta_{2}\sin^{2}\theta_{1}\sin^{2}\theta_{2}\frac{1}{v^{2}Q^{2}-2v\epsilon_{F}Q\cos\theta_{1}-\frac{i}{\tau}\epsilon_{F}}$
(25)
$\displaystyle\times[\frac{1}{v^{2}Q^{2}-2v\epsilon_{F}Q\cos\theta_{1}-\frac{i}{\tau}\epsilon_{F}}-\frac{1}{v^{2}Q^{2}+2v\epsilon_{F}Q\cos\theta_{2}+\frac{i}{\tau}\epsilon_{F}}]\frac{1}{v^{2}Q^{2}+2v\epsilon_{F}Q\cos\theta_{2}+\frac{i}{\tau}\epsilon_{F}}$
$\displaystyle=$
$\displaystyle-i\omega\frac{u}{6\pi^{3}}e^{2}v^{3}\epsilon_{F}^{2}\times\text{Im}\int_{0}^{\infty}dQS^{X}(Q),$

where

$S^{X}(Q)\equiv
Q^{4}\int_{0}^{\pi}\sin\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin\theta_{2}d\theta_{2}\sin^{2}\theta_{1}\sin^{2}\theta_{2}\times\frac{1}{(v^{2}Q^{2}+2v\epsilon_{F}Q\cos\theta_{1}-\frac{i}{\tau}\epsilon_{F})^{2}}\times\frac{1}{v^{2}Q^{2}+2v\epsilon_{F}Q\cos\theta_{2}+\frac{i}{\tau}\epsilon_{F}}.$
(26)

The anomalous Hall conductivity from the $X$ diagram is
$\sigma^{X}_{xy}=\Pi^{X,a}_{xy}/i\omega$, which is then completely real and
dissipationless. The integration in Eq.(25) can be done by a change of
variable $x=\cos\theta_{1},y=\cos\theta_{2}$, as shown in the Appendix. We get
in the leading order of $1/\tau$

$I^{X}\equiv{\rm Im}\int_{0}^{\infty}dQ\
S^{X}(Q)\approx\frac{\pi}{\epsilon_{F}v^{5}}.$ (27)

The leading order response function without vertex correction for the $X$
diagram from a single valley is

$\Pi_{xy}^{X,a}(u)\approx-i\omega\frac{e^{2}\epsilon_{F}u}{6\pi^{2}v^{2}}.$
(28)

The vertex correction adds a factor of $9/4$ in the leading order of $u/v$ to
the response function $\Pi_{xy}^{X}(u)$. For tilted Weyl metals with two
valleys, the response function doubles. We then obtain the leading order
anomalous Hall conductivity of the tilted Weyl metals due to the $X$ diagram
as

$\sigma^{X}_{xy}\approx-\frac{3e^{2}\epsilon_{F}u}{4\pi^{2}v^{2}},$ (29)

which is of the same order of the leading order contribution from the NCA
diagram $\sigma^{\rm NCA}_{xy}=\frac{4e^{2}\epsilon_{F}u}{3\pi^{2}v^{2}}$, but
with opposite sign.

###  II.2 B. AHE due to the $\Psi$ diagram

In this subsection, we study the AHE in tilted Weyl metals due to the $\Psi$
diagram. To do this, we compute the anti-symmetric part of the response
function $\Pi^{\Psi}_{\alpha\beta}$ in Eq.(8).

For the matrices ${\cal D},I$ and $M$ in Eq.(8), the symmetric parts of these
matrices are ${\cal D}_{0}\sim\tau^{0},I^{s}\sim\tau,M^{s}\sim\tau^{0}$, and
the anti-symmetric parts ${\cal
D}^{a}\sim\tau^{-1},I^{a}\sim\tau^{0},M^{a}\sim\tau^{0}$. In the leading order
of $1/\epsilon_{F}\tau$, the anti-symmetric part of $\Pi_{\alpha\beta}^{\Psi}$
is then

$\displaystyle\Pi_{\alpha\beta}^{\Psi,a}=e^{2}\gamma^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\sum_{\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{Q}}{\cal
D}_{0,\alpha\gamma}I^{s}_{\gamma\mu}(\mathbf{p}_{1})M^{a}_{\mu\nu}(\mathbf{Q}-\mathbf{p}_{2},\mathbf{Q}-\mathbf{p}_{1})I^{s}_{\nu\eta}(\mathbf{p}_{2}){\cal
D}^{T}_{0,\eta\beta},$ (30)

where
$\mathbf{Q}\equiv\mathbf{p}_{1}+\mathbf{p}_{4}=\mathbf{p}_{2}+\mathbf{p}_{3}$,
$I^{s}$ is given in Eq.(12) and $M^{a}_{\mu\nu}$ is the anti-symmetric part of

$M_{\mu\nu}\equiv{\rm
Tr}[\sigma_{\mu}\sigma_{\nu}G^{A}(\mathbf{p}_{4})G^{A}(\mathbf{p}_{3})]+{\rm
Tr}[\sigma_{\nu}\sigma_{\mu}G^{R}(\mathbf{p}_{3})G^{R}(\mathbf{p}_{4})].$ (31)

We denote the $G^{A}G^{A}$ term in Eq.(31) as $M^{A}$ and the $G^{R}G^{R}$
term as $M^{R}$. Since $M^{R}=(M^{A})^{*}$, the $M$ matrix is then $M=2\rm Re\
M^{A}$. The anti-symmetric part of the $M^{A}$ matrix in the leading order of
$1/\tau$ is $M^{A,a}_{\mu\nu}=N_{\mu\nu}(M^{A,a})/D(M^{A})$, where

$\displaystyle N_{\mu\nu}(M^{A,a})$ $\displaystyle=$ $\displaystyle\
2iv\epsilon^{0\mu\nu
k}\\{[2(\epsilon-\bm{u}\cdot\bm{Q})+\bm{u}\cdot(\bm{p}_{1}+\bm{p}_{2})]Q_{k}-[\epsilon-\bm{u}\cdot(\bm{Q}-\bm{p}_{2})]p_{1k}-[\epsilon-\bm{u}\cdot(\bm{Q}-\bm{p}_{1})]p_{2k}\\}$
(32) $\displaystyle\
-2v^{2}[(Q-p_{1})_{\mu}(Q-p_{2})_{\nu}-(Q-p_{1})_{\nu}(Q-p_{2})_{\mu}],$
$\displaystyle D(M^{A})$ $\displaystyle=$
$\displaystyle[\epsilon-\bm{u}\cdot(\bm{Q}-\bm{p}_{1})-v|\bm{Q}-\bm{p}_{1}|-\frac{i}{2\tau^{+}_{4}}][\epsilon-\bm{u}\cdot(\bm{Q}-\bm{p}_{1})+v|\bm{Q}-\bm{p}_{1}|-\frac{i}{2\tau^{-}_{4}}]$
(33) $\displaystyle\times$
$\displaystyle[\epsilon-\bm{u}\cdot(\bm{Q}-\bm{p}_{2})-v|\bm{Q}-\bm{p}_{2}|-\frac{i}{2\tau^{+}_{3}}][\epsilon-\bm{u}\cdot(\bm{Q}-\bm{p}_{2})+v|\bm{Q}-\bm{p}_{2}|-\frac{i}{2\tau^{-}_{3}}].$

In the above equation,
$\frac{1}{\tau_{i}^{\pm}}=\frac{1}{\tau}(1\pm\delta_{i}),\delta_{i}=\frac{\mathbf{p}_{i}\cdot\mathbf{\Delta}}{p_{i}}$
as defined before.

For $\mathbf{u}$ in the $z$ direction and
$\mathbf{Q}=Q(\sin\alpha\cos\beta,\sin\alpha\sin\beta,\cos\alpha)$, after the
same rotation of the $z$-axis to the direction of $\mathbf{Q}$ as for the $X$
diagram, and applying the $\delta$ function in $I^{s}$, we obtain

$D(M^{A})\approx[v^{2}Q^{2}-2v^{2}\mathbf{p_{1}\cdot
Q}+\frac{i}{\tau}vp_{1}-2vp_{1}\bm{u}\cdot(2\bm{p}_{1}-\bm{Q})][v^{2}Q^{2}-2v^{2}\mathbf{p_{2}\cdot
Q}+\frac{i}{\tau}vp_{2}-2vp_{2}\bm{u}\cdot(2\bm{p}_{2}-\bm{Q})],$ (34)

where we have neglected the second order $\mathbf{u}$ terms as well as the
$\mathbf{u}/\tau$ terms.

Putting $I^{s}$ and $M^{a}$ together and neglecting the vertex correction at
the two ends of the $\Psi$ diagram at the moment, we get the anti-symmetric
part of the response function for the $\Psi$ diagram as

$\displaystyle\Pi_{\alpha\beta}^{\Psi,a}$ $\displaystyle=$
$\displaystyle\pi^{2}e^{2}\gamma^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}$ (35)
$\displaystyle\sum_{\mathbf{p}_{1},\mathbf{p}_{2},\mathbf{Q}}\left[\frac{2iv\epsilon^{0\mu\nu
k}p_{1,\alpha}p_{1,\mu}p_{2,\nu}p_{2,\beta}[2(\epsilon-\bm{u}\cdot\bm{Q})+\bm{u}\cdot(\bm{p}_{1}+\bm{p}_{2})]Q_{k}}{p^{2}_{1}\
p^{2}_{2}\ \rm
D(M^{A})}+c.c\right]\underset{i=1,2}{\mathbf{\Pi}}\tau_{i}^{+}\delta(\epsilon-\bm{u}\cdot\bm{p}_{i}-vp_{i})$
$\displaystyle=$ $\displaystyle
e^{2}v^{2}\gamma^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}\times
2iv\pi^{2}\frac{1}{(2\pi)^{9}}\int_{0}^{\infty}Q^{2}dQ\int_{0}^{\pi}\sin\alpha
d\alpha\int_{0}^{2\pi}d\beta p_{1\alpha}p_{2\beta}$
$\displaystyle\times\int_{0}^{\infty}dp_{1}\int_{0}^{\infty}dp_{2}\int_{0}^{\pi}\sin\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin\theta_{2}d\theta_{2}\int_{0}^{2\pi}d\phi_{1}\int_{0}^{2\pi}d\phi_{2}\
\tau_{1}^{+}\tau_{2}^{+}(\bm{p}_{1}\times\bm{p}_{2})\cdot\bm{Q}$
$\displaystyle\times[2\epsilon+\mathbf{u}\cdot(\bm{p}_{1}+\bm{p}_{2}-2\bm{Q})]\times\delta(\epsilon-
vp_{1}-\bm{u}\cdot\bm{p}_{1})\delta(\epsilon-
vp_{2}-\bm{u}\cdot\bm{p}_{2})\times[\frac{1}{D(M^{A})}-{\rm c.c}].$

At $\mathbf{u}=0$, the response function $\Pi_{\alpha\beta}^{\Psi,a}$
vanishes. We then expand $\Pi_{\alpha\beta}^{\Psi,a}$ to the linear order of
$\mathbf{u}$ and neglect the higher order contributions. Similar to the $X$
diagram, for $\mathbf{u}$ in the $z$ direction and $\mathbf{E}$ in the $y$
direction, $\Pi^{\Psi,a}_{zy}=0$ and we only need to consider
$\Pi^{\Psi,a}_{xy}$ for the AHE. Keeping only the linear order of $u$ and
integrating out the angle $\beta$, we get

$\displaystyle\Pi_{xy}^{\Psi,a}(u)$
$\displaystyle=e^{2}v^{2}\gamma^{2}\tau^{2}\omega\int\frac{d\epsilon}{2\pi
i}\frac{dn_{F}(\epsilon)}{d\epsilon}4iv\pi^{3}\frac{1}{(2\pi)^{9}}\
\frac{\epsilon^{5}}{v^{6}}$
$\displaystyle\times\int_{0}^{\infty}Q^{2}dQ\int_{0}^{\pi}\sin\alpha
d\alpha\int_{0}^{\pi}\sin\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin\theta_{2}d\theta_{2}\int_{0}^{2\pi}d\phi_{1}\int_{0}^{2\pi}d\phi_{2}\
Q\sin\theta_{1}\sin\theta_{2}\sin(\phi_{2}-\phi_{1})$
$\displaystyle\times[\cos\alpha\sin\theta_{1}\sin\theta_{2}\sin(\phi_{2}-\phi_{1})+\sin\alpha(\cos\theta_{1}\sin\theta_{2}\sin\phi_{2}-\cos\theta_{2}\sin\theta_{1}\sin\phi_{1})]$
$\displaystyle\times\\{\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{1}+\frac{i}{\tau}\epsilon}\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{2}+\frac{i}{\tau}\epsilon}[-\frac{u}{\epsilon}Q\cos\alpha+\frac{u}{2v}(\hat{z}_{1}+\hat{z}_{2})$
$\displaystyle-\frac{2uv\hat{z}_{1}(Q^{2}-\frac{\epsilon}{v}Q\cos\theta_{1}-2\frac{\epsilon^{2}}{v^{2}})+2\epsilon
uQ\cos\alpha}{v^{2}Q^{2}-2\epsilon
vQ\cos\theta_{1}+\frac{i}{\tau}\epsilon}-\frac{2uv\hat{z}_{2}(Q^{2}-\frac{\epsilon}{v}Q\cos\theta_{2}-2\frac{\epsilon^{2}}{v^{2}})+2\epsilon
uQ\cos\alpha}{v^{2}Q^{2}-2\epsilon
vQ\cos\theta_{2}+\frac{i}{\tau}\epsilon}]-{\rm c.c}\\}.$ (36)

After the integration over $\phi_{1},\phi_{2}$ and $\alpha$, we get the anti-
symmetric part of $\Pi_{xy}^{\Psi}$ as

$\displaystyle\Pi_{xy}^{\Psi,a}(u)$ $\displaystyle=$
$\displaystyle-\frac{\omega}{12\pi^{3}}e^{2}v^{3}\epsilon^{2}u\int_{0}^{\infty}Q^{4}dQ\int_{0}^{\pi}\sin\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin\theta_{2}d\theta_{2}\sin^{2}\theta_{1}\sin^{2}\theta_{2}$
(37) $\displaystyle\times\frac{1}{2}\\{\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{1}+\frac{i}{\tau}\epsilon}\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{2}+\frac{i}{\tau}\epsilon}$
$\displaystyle\times[-\frac{1}{2\epsilon^{2}}-(\frac{1}{v^{2}Q^{2}-2\epsilon
vQ\cos\theta_{1}+\frac{i}{\tau}\epsilon}+\frac{1}{v^{2}Q^{2}-2\epsilon
vQ\cos\theta_{2}+\frac{i}{\tau}\epsilon})]-{\rm c.c}\\}$ $\displaystyle=$
$\displaystyle-i\omega\frac{u}{6\pi^{3}}e^{2}v^{3}\epsilon^{2}\times\rm
Im\int_{0}^{\infty}dQS^{\Psi}(Q),$

for which we separate $S^{\Psi}(Q)$ to two parts as

$\displaystyle S^{\Psi}(Q)$ $\displaystyle=$ $\displaystyle
S^{\Psi,1}(Q)+S^{\Psi,2}(Q),$ $\displaystyle S^{\Psi,1}(Q)$ $\displaystyle=$
$\displaystyle-\frac{Q^{4}}{4\epsilon^{2}}\times\int_{0}^{\pi}\sin^{3}\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin^{3}\theta_{2}d\theta_{2}\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{1}+\frac{i}{\tau}\epsilon}\times\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{2}+\frac{i}{\tau}\epsilon},$ (38) $\displaystyle S^{\Psi,2}(Q)$
$\displaystyle=$
$\displaystyle-Q^{4}\times\int_{0}^{\pi}\sin^{3}\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin^{3}\theta_{2}d\theta_{2}\frac{1}{(v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{1}+\frac{i}{\tau}\epsilon)^{2}}\times\frac{1}{v^{2}Q^{2}-2v\epsilon
Q\cos\theta_{2}+\frac{i}{\tau}\epsilon}.$ (39)

As shown in the Appendix, in the leading order of $1/\tau$, the integration
over $S^{\Psi,1}(Q)$ and $S^{\Psi,2}(Q)$ gives

$I^{\Psi,1}\equiv{\rm Im}\int_{0}^{\infty}dQ\
S^{\Psi,1}(Q)\approx\frac{17+16\ln 2}{105}\frac{\pi}{\epsilon_{F}v^{5}},$ (40)

and

$I^{\Psi,2}\equiv{\rm Im}\int_{0}^{\infty}dQ\ S^{\Psi,2}(Q)\approx\frac{1+8\ln
2}{15}\frac{\pi}{\epsilon_{F}v^{5}}.$ (41)

The antisymmetric part of the response function for the $\Psi$ diagram for a
single valley without vertex correction is

$\Pi_{xy}^{\Psi,a}(u)=\Pi_{xy}^{\Psi,1,a}(u)+\Pi_{xy}^{\Psi,2,a}(u)\approx-\frac{4+12\ln
2}{105}i\omega\frac{e^{2}\epsilon_{F}u}{\pi^{2}v^{2}}.$ (42)

Adding the vertex correction factor $9/4$ in the leading order of $u$, and
taking into account the two valleys of the tilted Weyl metals, we get the
total anomalous Hall conductivity due to the $\Psi$ diagram in the leading
order of $u$ as

$\sigma^{\Psi}_{xy}\approx-\frac{6+18\ln
2}{35}\frac{e^{2}\epsilon_{F}u}{\pi^{2}v^{2}}\approx-0.53\frac{e^{2}\epsilon_{F}u}{\pi^{2}v^{2}}.$
(43)

The contribution from the $\Psi$ diagram is also of the same order of the
contribution from the NCA diagram $\sigma^{\rm
NCA}_{xy}=\frac{4e^{2}\epsilon_{F}u}{3\pi^{2}v^{2}}$, but with opposite sign.
This is different from the 2D massive Dirac model, for which the anomalous
Hall conductivity from the $\Psi$ diagram vanishes for Gaussian disorder
Ado2017 .

###  II.3 C. Comparison between the crossed and non-crossing diagrams

Figure 2: The different contributions to the anomalous Hall conductivity from
the non-crossing and crossed diagrams for the tilted Weyl metals. The Fermi
surface contributions from the non-crossing diagram in the plot come from Ref.
Chen2022  and are exact, whereas the results for the crossed diagrams are kept
in the linear order of $u/v$. The intrinsic contribution from the Fermi sea
$\sigma^{\rm II}_{H}$ comes from Ref.Pesin2017 . The black solid line
represents the total Fermi surface contribution from the non-crossing diagram,
and the blue solid line represents the total contribution from both the Fermi
surface and Fermi sea, including both non-crossing and crossed diagrams.

The total contributions of the $X$ and $\Psi$ diagrams to the AHE for the
tilted Weyl metals with two valleys are

$\sigma^{X+\Psi}_{xy}=\sigma^{X}_{xy}+\sigma^{\Psi}_{xy}\approx-1.28\frac{e^{2}\epsilon_{F}u}{\pi^{2}v^{2}}.$
(44)

As a comparison, we plot the different contributions to the anomalous Hall
conductivity of the tilted Weyl metals due to both the non-crossing and
crossed diagrams in Fig.2. The anomalous Hall conductivity from the non-
crossing diagram was obtained in our previous work Chen2022  and includes
three different mechanisms: intrinsic, side jump and skew scattering. The
total anomalous Hall conductivity from the non-crossing diagram in the leading
order of $u/v$ is

$\sigma^{\rm NCA}_{xy}\approx\frac{4e^{2}\epsilon_{F}u}{3\pi^{2}v^{2}}.$ (45)

This contribution includes the intrinsic part $\sigma^{\rm
I}_{int}=\frac{e^{2}\epsilon_{F}u}{3\pi^{2}v^{2}}$ from the Fermi surface and
$\sigma^{\rm II}_{int}=-\frac{e^{2}\epsilon_{F}u}{6\pi^{2}v^{2}}$ from the
Fermi sea. The remaining part is the extrinsic contribution due to impurity
scatterings, including the side jump and skew scattering contribution as shown
in Fig.2.

From Eq.(44) and Eq.(45), we see that the inclusion of the $X$ and $\Psi$
diagram cancels most of the contribution from the non-crossing diagram in the
leading order of $u/v$, as shown in Fig.2. This is similar to the case of the
2D massive Dirac model at large energy with Gaussian disorder. However, for
the 2D massive Dirac model, the inclusion of the $X$ and $\Psi$ diagram
changes the dependence of the total anomalous Hall conductivity on the energy
from $\sigma^{\rm NCA}_{xy}\sim m/\epsilon_{F}$ to $\sigma^{\rm
total}_{xy}\sim(m/\epsilon_{F})^{3}$, which greatly reduces the total
anomalous Hall conductivity in the metallic regime $m/\epsilon_{F}\ll 1$.
Whereas for tilted Weyl metals, the contributions of the $X$ and $\Psi$
diagrams have the same dependence on the Fermi energy as the non-crossing
diagram and the cancellation is due to the opposite signs but close values of
the coefficients of the two contributions.

##  III III. Discussions

The expansion of the response functions of the $X$ and $\Psi$ diagrams to the
second order of $\mathbf{u}$ reveals that the contributions to the AHE from
these diagrams vanish in the second order of $\mathbf{u}$ for the tilted Weyl
metals. The next leading order correction to the AHE we obtained for the $X$
and $\Psi$ diagram is then $\sim(u/v)^{3}$. The same is true for the
contribution to the AHE from the non-crossing diagram Chen2022 . For the
type-I Weyl metals with not very large $u/v$, the anomalous Hall conductivity
in the linear order of $u$ we obtained in this work is then accurate enough.

The contributions to the AHE from the $X$ and $\Psi$ diagrams do not depend on
the disorder strength and scattering rate, and have the same dependence on the
Fermi energy and the tilting of the Weyl metals as the NCA diagram in the
leading order. This makes it hard to distinguish the contributions from the
two types of diagrams in experiments. However, the AHE from the crossed
diagrams originates from the skew scatterings over pairs of closely located
impurities with distances of the order of electron Fermi wavelength Ado2015 ;
Ado2016 ; Ado2017 ; Levchenko2016 . This is very demanding on the disorder
condition. To validate the self-average over the impurities in the response
function, every sub-system of the size of the phase coherence length
$l_{\varphi}$ (which is much smaller than the sample size) needs to contain at
least one pair of such rare impurities. At not very low temperatures, the
phase coherence length $l_{\varphi}$ is very short. This indicates that the
density of the rare impurity pairs needs to be very high to observe the AHE
from the crossed diagrams. Considering the random distribution of the
disorder, it requires a much higher density of the disorder to observe the AHE
from the crossed diagrams than from the non-crossing diagram. This might be
the reason why the anomalous Hall conductivity obtained in experiment in
quasi-2D Dirac materials ${\rm Fe_{3}Sn_{2}}$ is very close to the theoretical
result including only the disorder effects from the NCA diagram Ye2018 ;
Fu2021 .

For the recently studied type-I Weyl metal ${\rm Co_{3}Sn_{2}S_{2}}$ in
experiments Felser2018 ; Ding2019 , the topological Chern-Simons term Chen2022
; Burkov2014 ; Burkov2015  gives an extra anomalous Hall conductivity
$\sigma_{H}\sim\frac{e^{2}}{2\pi^{2}}{\cal K}$ which is proportional to the
distance ${\cal K}$ between the two Weyl nodes. This contribution is
independent of the impurity scatterings and constitutes part of the intrinsic
AHE. For ${\rm Co_{3}Sn_{2}S_{2}}$, the AHE from the Chern-Simons term is one
order of magnitude greater than both the contribution from the non-crossing
diagram and the crossed diagrams of the low energy effective Hamiltonian
Fu2021  so it dominates the total AHE in this system. This makes it hard to
distinguish the contribution of the disorder in experiments, either due to the
NCA diagram or the crossed diagrams. In Ref. Felser2018 , all the AHE measured
in ${\rm Co_{3}Sn_{2}S_{2}}$ was attributed to the intrinsic one. In Ref.
Ding2019 , the authors measured the AHE in both the clean and dirty samples,
but the difference of the anomalous Hall conductivity in the two samples is
only about $10\%$ of the clean case. To better observe the effects of the
disorder and the interplay of the non-crossing and crossed diagrams in
experiments, one may increase the Fermi level of ${\rm Co_{3}Sn_{2}S_{2}}$ by
doping so to enhance the weight of the contribution to the AHE due to both the
non-crossing and crossed diagrams since the anomalous Hall conductivity due to
the Chern-Simons term does not depend on the Fermi energy.

Another way to observe the interplay between the non-crossing and crossed
diagrams in ${\rm Co_{3}Sn_{2}S_{2}}$ in experiments is by the measurement of
the anomalous Nernst effect (ANE) Ding2019 ; Sakai2018 ; Guin2019  in such
system. The ANE only comes from the scatterings on the Fermi surface so the
Chern-Simons term has no contribution to the ANE. The ANE is proportional to
the Fermi surface contribution of the AHE, i,e., $\sigma^{I}_{H}$ we studied
in Ref. Chen2022  and this work, with the ratio $\sim k_{B}T/\epsilon_{F}$
Fu2021 . By measuring the ANE in different disorder conditions, one can tell
whether and when the crossed diagrams play a role in both the ANE and AHE in
the system. Indeed, in Ref. Ding2019 , the ANE in the disordered sample is
about three times of the clean sample, which makes the effects of the disorder
much more discernible in the ANE than in the AHE. The large enhancement of the
ANE in this experiment seems to indicate that the NCA diagram dominates the
contribution in the measured disordered sample according to our calculation in
Ref. Chen2022  for the Gaussian disorder. However, the real system may include
more complicated disorder which may affect the AHE and ANE significantly. For
example, it was shown in Ref.Ado2017  that for the 2D massive Dirac model with
smooth disorder, the anomalous Hall conductivity is enhanced by the $X$ and
$\Psi$ diagrams instead of being canceled as in the case of Gaussian disorder.
A more complete theoretical study including the smooth disorder is then needed
to tell whether the crossed diagrams play a role in such experiments. We will
leave this for a future study since the study of the tilted Weyl metals with
smooth disorder is more complicated than the 2D massive Dirac model due to the
increased dimensionality. On the other hand, to observe the AHE or ANE due to
the crossed diagrams, more experiments with varying disorder conditions may
also need to be carried out in the future.

##  IV IV. Summary

To sum up, we studied the AHE due to the crossed $X$ and $\Psi$ diagrams in
Type-I Weyl metals with Gaussian disorder. We show that similar to the 2D
massive Dirac model, the contributions from the crossed diagrams cancel a
majority part of the contribution from the non-crossing diagram of the low
energy effective Hamiltonian. However, the condition to observe the effect of
the crossed diagrams is more demanding than the non-crossing diagram.
Moreover, since the AHE from the Chern-Simons term is much greater than the
AHE due to the impurity scatterings in real type-I Weyl metals ${\rm
Co_{3}Sn_{2}S_{2}}$, it is hard to distinguish the AHE due to the impurity
scatterings in this system. To observe the contributions or interplay of the
non-crossing and crossed diagrams in this system, one can measure the ANE
effect instead of the AHE because the Chern-Simons term has no contribution to
the ANE and the disorder induced ANE is proportional to that of the AHE .

##  V Acknowledgement

This work is supported by the National Natural Science Foundation of China
under Grant No. 11974166.

##  VI Appendix

In this appendix, we show the details of the integration in Eq.(27) and (40)
-(41) for the $X$ and $\Psi$ diagram.

We first present the integration

$\displaystyle I^{X}$ $\displaystyle=$ $\displaystyle{\rm
Im}\int_{0}^{\infty}dQS^{X}(Q)$ $\displaystyle=$ $\displaystyle{\rm
Im}\int_{0}^{\infty}Q^{4}dQ\
\int_{0}^{\pi}\sin^{3}\theta_{1}d\theta_{1}\int_{0}^{\pi}\sin^{3}\theta_{2}d\theta_{2}\frac{1}{(v^{2}Q^{2}+2v\epsilon_{F}Q\cos\theta_{1}-\frac{i}{\tau}\epsilon_{F})^{2}}\frac{1}{v^{2}Q^{2}+2v\epsilon_{F}Q\cos\theta_{2}+\frac{i}{\tau}\epsilon_{F}}.$

By a change of the variable $x=\cos\theta_{1},y=\cos\theta_{2}$ and denoting
$\epsilon_{F}$ as $\epsilon$ for brevity, we get

$\displaystyle I^{X}$ $\displaystyle=$
$\displaystyle\int_{0}^{\infty}Q^{4}dQ\int_{-1}^{1}dx\int_{-1}^{1}dy(1-x^{2})(1-y^{2})$
$\displaystyle\times\\{-\frac{\frac{\epsilon}{\tau}}{(v^{2}Q^{2}+2v\epsilon
Qy)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}\frac{(v^{2}Q^{2}+2v\epsilon
Qx)^{2}-\frac{\epsilon^{2}}{\tau^{2}}}{[(v^{2}Q^{2}+2v\epsilon
Qx)^{2}+\frac{\epsilon^{2}}{\tau^{2}}]^{2}}+\frac{2\frac{\epsilon}{\tau}(v^{2}Q^{2}+2v\epsilon
Qy)}{(v^{2}Q^{2}+2v\epsilon
Qy)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}\frac{v^{2}Q^{2}+2v\epsilon
Qx}{[(v^{2}Q^{2}+2v\epsilon Qx)^{2}+\frac{\epsilon^{2}}{\tau^{2}}]^{2}}\\}.$

We denote the first and second term in the above equation as $I^{X,1}$ and
$I^{X,2}$ respectively and calculate them separately in the following. With
the variable substitution $v^{2}Q^{2}+2v\epsilon Qy=t,v^{2}Q^{2}+2v\epsilon
Qx=s$ and the relationship

$\frac{\frac{\epsilon}{\tau}}{(v^{2}Q^{2}+2v\epsilon
Qy)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}\approx\pi\delta(v^{2}Q^{2}+2v\epsilon
Qy),$ (47)

we get

$\displaystyle I^{X,1}$ $\displaystyle=$
$\displaystyle-\pi\int_{0}^{\infty}\frac{Q^{4}}{(2\epsilon
vQ)^{2}}dQ\int_{v^{2}Q^{2}-2v\epsilon Q}^{v^{2}Q^{2}+2v\epsilon
Q}ds[1-(\frac{s-v^{2}Q^{2}}{2v\epsilon
Q})^{2}]\frac{s^{2}-\frac{\epsilon^{2}}{\tau^{2}}}{(s^{2}+\frac{\epsilon^{2}}{\tau^{2}})^{2}}\int_{v^{2}Q^{2}-2v\epsilon
Q}^{v^{2}Q^{2}+2v\epsilon Q}dt[1-(\frac{t-v^{2}Q^{2}}{2v\epsilon
Q})^{2}]\delta(t),$ (48) $\displaystyle I^{X,2}$ $\displaystyle=$
$\displaystyle\frac{2\epsilon}{\tau}\int_{0}^{\infty}\frac{Q^{4}}{(2\epsilon
vQ)^{2}}dQ\int_{v^{2}Q^{2}-2v\epsilon Q}^{v^{2}Q^{2}+2v\epsilon
Q}ds[1-(\frac{s-v^{2}Q^{2}}{2v\epsilon
Q})^{2}]\frac{s}{(s^{2}+c^{2})^{2}}\int_{v^{2}Q^{2}-2v\epsilon
Q}^{v^{2}Q^{2}+2v\epsilon Q}dt[1-(\frac{t-v^{2}Q^{2}}{2v\epsilon
Q})^{2}]\frac{t}{t^{2}+c^{2}}.$

We first do the integration of $I^{X,1}$. The integration over $s$ and $t$ for
$I^{X,1}$ in Eq.(48) can be carried out separately at first. To get a non-
vanishing integration over the $\delta(t)$ factor in Eq.(48), $Q$ must be
limited to $0<Q<2\epsilon/v$. Denoting $c\equiv\epsilon/\tau$ and integrating
out $s$ and $t$, $I^{X,1}$ becomes

$\displaystyle I^{X,1}$ $\displaystyle=$
$\displaystyle\int_{0}^{2\epsilon/v}dQ\frac{-\pi Q^{4}}{(2\epsilon
vQ)^{2}}[1-\frac{v^{2}Q^{2}}{4\epsilon^{2}}]$
$\displaystyle\times\\{-(1-\frac{v^{2}Q^{2}}{4\epsilon^{2}})\frac{s}{s^{2}+c^{2}}+\frac{1}{2\epsilon^{2}}[c\pi\delta(s)+\frac{1}{2}\ln(s^{2}+c^{2})]-\frac{1}{4\epsilon^{2}v^{2}Q^{2}}[s+cs\delta(s)-2c\arctan(\frac{s}{c})]\\}|_{s=v^{2}Q^{2}-2v\epsilon
Q}^{v^{2}Q^{2}+2v\epsilon Q}.$

Neglecting the terms small in $1/\tau$, we get

$\displaystyle I^{X,1}$ $\displaystyle=$
$\displaystyle\int_{0}^{2\epsilon/v}dQ\frac{-\pi Q^{4}}{(2\epsilon
vQ)^{2}}[1-\frac{v^{2}Q^{2}}{4\epsilon^{2}}]\times\\{-(1-\frac{v^{2}Q^{2}}{4\epsilon^{2}})\frac{s}{s^{2}+c^{2}}+\frac{1}{4\epsilon^{2}}\ln(s^{2}+c^{2})-\frac{1}{4\epsilon^{2}v^{2}Q^{2}}s\\}|_{s=v^{2}Q^{2}-2v\epsilon
Q}^{\ \ v^{2}Q^{2}+2v\epsilon Q}$ (51) $\displaystyle=$
$\displaystyle\frac{\pi}{2\epsilon v^{5}}[1-\frac{1}{15}(1+8\ln 2)].$

Similarly, after the integration over $s$ and $t$, $I^{X,2}$ becomes

$\displaystyle I^{X,2}$ $\displaystyle=$ $\displaystyle
2\int_{0}^{\infty}\frac{Q^{4}}{(2\epsilon
vQ)^{2}}dQ\left.[(1-\frac{v^{2}Q^{2}}{4\epsilon^{2}})\frac{1}{2}\ln(t^{2}+c^{2})+\frac{1}{2\epsilon^{2}}t-\frac{1}{8\epsilon^{2}v^{2}Q^{2}}t^{2}]\right|_{t=v^{2}Q^{2}-2v\epsilon
Q}^{\ \ v^{2}Q^{2}+2v\epsilon Q}$ (52)
$\displaystyle\times\left.[-(1-\frac{v^{2}Q^{2}}{4\epsilon^{2}})\frac{\pi}{2}\delta(s)+\frac{1}{4\epsilon^{2}}\arctan(\frac{s}{c})]\right|_{s=v^{2}Q^{2}-2v\epsilon
Q}^{\ \ v^{2}Q^{2}+2v\epsilon Q},$

where we have omitted the terms proportional to $c$ or $s\delta(s)$. The
integration over the terms with $\delta(s)$ in Eq.(52) is zero because at
$s=0$, $Q=0$ or $1-\frac{v^{2}Q^{2}}{4\epsilon^{2}}=0$, and the terms with
$\delta(s)$ in the integrand become zero. For the reason, we only need to
consider the terms with $\frac{1}{4\epsilon^{2}}\arctan(\frac{s}{c})$ after
the integration of $s$. Since $c$ is small,
$\arctan(\frac{s}{c})|_{s=v^{2}Q^{2}-2v\epsilon Q}^{\ \ v^{2}Q^{2}+2v\epsilon
Q}$ is non-zero only when $v^{2}Q^{2}-2v\epsilon Q<0<v^{2}Q^{2}+2v\epsilon Q$,
i.e., $0<Q<2\epsilon/v$. In this regime,

$\arctan(\frac{s}{c})|_{s=v^{2}Q^{2}-2v\epsilon Q}^{\ \ v^{2}Q^{2}+2v\epsilon
Q}\approx\pi,$ (53)

and

$\displaystyle I^{X,2}$
$\displaystyle=\frac{\pi}{2\epsilon^{2}}\int_{0}^{2\epsilon/v}\frac{Q^{4}}{(2\epsilon
vQ)^{2}}dQ\times[\frac{1}{2}(1-\frac{v^{2}Q^{2}}{4\epsilon^{2}})\ln\frac{(v^{2}Q^{2}+2v\epsilon
Q)^{2}+c^{2}}{(v^{2}Q^{2}-2v\epsilon Q)^{2}+c^{2}}+\frac{vQ}{\epsilon}]$
$\displaystyle=\frac{\pi}{2\epsilon v^{5}}[1+\frac{1}{15}(1+8\ln 2)].$

Adding $I^{X,1}$ and $I^{X,2}$ together, we get $I^{X}=\pi/\epsilon_{F}v^{5}$
as in Eq.(27).

We next compute $I^{\Psi}\equiv{\rm Im}\int_{0}^{\infty}dQS^{\Psi}(Q)$ for the
$\Psi$ diagram. As shown in the main text, we divide $I^{\Psi}$ to two parts
$I^{\Psi,1}$ and $I^{\Psi,2}$ and compute them separately.

From Eq.(38), we get

$\displaystyle I^{\Psi,1}$ $\displaystyle=$ $\displaystyle{\rm
Im}\int_{0}^{\infty}dQS^{\Psi,1}(Q)$ (54) $\displaystyle=$
$\displaystyle\int_{0}^{\infty}\frac{Q^{4}}{4\epsilon^{2}}dQ\int_{-1}^{1}dx\int_{-1}^{1}dy(1-x^{2})(1-y^{2})$
$\displaystyle\times[\frac{\frac{\epsilon}{\tau}}{(v^{2}Q^{2}-2v\epsilon
Qy)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}\frac{(v^{2}Q^{2}-2v\epsilon
Qx)}{(v^{2}Q^{2}-2v\epsilon
Qx)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}+\frac{\frac{\epsilon}{\tau}}{(v^{2}Q^{2}-2v\epsilon
Qx)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}\frac{v^{2}Q^{2}-2v\epsilon
Qy}{(v^{2}Q^{2}-2v\epsilon Qy)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}]$
$\displaystyle=$
$\displaystyle\frac{\pi}{2\epsilon^{2}}\int_{0}^{\infty}\frac{Q^{4}}{(2\epsilon
vQ)^{2}}dQ\int_{v^{2}Q^{2}-2v\epsilon Q}^{v^{2}Q^{2}+2v\epsilon
Q}ds[1-(\frac{s-v^{2}Q^{2}}{2v\epsilon
Q})^{2}]\frac{s}{s^{2}+\frac{\epsilon^{2}}{\tau^{2}}}\int_{v^{2}Q^{2}-2v\epsilon
Q}^{v^{2}Q^{2}+2v\epsilon Q}dt[1-(\frac{t-v^{2}Q^{2}}{2v\epsilon
Q})^{2}]\delta(t)$ $\displaystyle=$
$\displaystyle\frac{\pi}{2\epsilon^{2}}\int_{0}^{2\epsilon/v}dQ\frac{Q^{4}}{(2\epsilon
vQ)^{2}}(1-\frac{v^{2}Q^{2}}{4\epsilon^{2}})[(1-\frac{v^{2}Q^{2}}{4\epsilon^{2}})\frac{1}{2}\ln\frac{(v^{2}Q^{2}+2v\epsilon
Q)^{2}+c^{2}}{(v^{2}Q^{2}-2v\epsilon Q)^{2}+c^{2}}+\frac{vQ}{\epsilon}]$
$\displaystyle=$ $\displaystyle\frac{\pi}{\epsilon v^{5}}\frac{17+16\ln
2}{105}.$

Similarly, we get

$\displaystyle I^{\Psi,2}$ $\displaystyle=$
$\displaystyle-\int_{0}^{\infty}Q^{4}dQ\times{\rm
Im}\int_{-1}^{1}dx\int_{-1}^{1}dy(1-x^{2})(1-y^{2})[\frac{1}{(v^{2}Q^{2}-2v\epsilon
Qx+\frac{i}{\tau}\epsilon)^{2}}\times\frac{1}{v^{2}Q^{2}-2v\epsilon
Qy+\frac{i}{\tau}\epsilon}]$ (55) $\displaystyle=$
$\displaystyle\int_{0}^{\infty}Q^{4}dQ\int_{-1}^{1}dx\int_{-1}^{1}dy(1-x^{2})(1-y^{2})$
$\displaystyle\times[\frac{\frac{\epsilon}{\tau}}{(v^{2}Q^{2}-2v\epsilon
Qy)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}\frac{(v^{2}Q^{2}-2v\epsilon
Qx)^{2}-\frac{\epsilon^{2}}{\tau^{2}}}{[(v^{2}Q^{2}-2v\epsilon
Qx)^{2}+\frac{\epsilon^{2}}{\tau^{2}}]^{2}}+\frac{2\frac{\epsilon}{\tau}(v^{2}Q^{2}-2v\epsilon
Qy)}{(v^{2}Q^{2}-2v\epsilon
Qy)^{2}+\frac{\epsilon^{2}}{\tau^{2}}}\frac{v^{2}Q^{2}-2v\epsilon
Qx}{[(v^{2}Q^{2}-2v\epsilon Qx)^{2}+\frac{\epsilon^{2}}{\tau^{2}}]^{2}}]$
$\displaystyle=$ $\displaystyle-I^{X,1}+I^{X,2}$ $\displaystyle=$
$\displaystyle\frac{\pi}{\epsilon v^{5}}\frac{1}{15}(1+8\ln 2).$

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