# Disorder Induced Anomalous Hall Effect in Type-I Weyl Metals:
Connection between the Kubo-Streda Formula in the Spin and Chiral basis

Jia-Xing Zhang1 Zhi-Yuan Wang1 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 weak
Gaussian disorder under the Kubo-Streda formalism in this work. To separate
the three different contributions, namely the intrinsic, side jump and skew
scattering contributions, it is usually considered necessary to go to the
eigenstate (chiral) basis of the Kubo-Streda formula. However, it is more
straightforward to compute the total Hall current in the spin basis. For this
reason, we develop a systematic and transparent scheme to separate the three
different contributions in the spin basis for relativistic systems by building
a one-to-one correspondence between the Feynman diagrams of the different
mechanisms in the chiral basis and the products of the symmetric and
antisymmetric parts of the polarization operator in the spin basis. We
obtained the three contributions of the AHE in tilted Weyl metals by this
scheme and found that the side jump contribution exceeds both the intrinsic
and skew scattering contributions for the low-energy effective Hamiltonian. We
compared the anomalous Hall current obtained from our scheme with the results
from the semiclassical Boltzmann equation approach under the relaxation time
approximation and found that the results from the two approaches agree with
each other in the leading order of the tilting velocity.

##  I I. Introduction

The anomalous Hall effect (AHE) has attracted great interest since the
starting work by Karplus and Luttinger Luttinger1954 . Although it is first
discussed in ferromagnetic metals, the AHE has been discovered in various
systems in the following decades, such as magnetic topological insulators
Qi2008 ; Yu2010 ; Chang2013 ; Chang2015 , and Moire materials Xie2020 ;
Eom2000 ; Nomura2006 ; Sondhi1993 ; Yang1994 ; Young2012 , Dirac and Weyl
metals Wan2011 ; Burkov2014 ; Pesin2017 . All these systems include two
ingredients: (i) some sort of spin- or pseudospin-orbit interaction and (ii)
time reversal symmetry (TRS) breaking. The spin-orbit interaction results in a
transverse motion of the electrons perpendicular to the electric field
applied, and the time reversal symmetry breaking is necessary to avoid the
cancellation of the transverse flux in opposite directions Sinitsyn2008 ;
Yang2011 . The AHE in such systems can be divided to the intrinsic
contribution, which is the AHE obtained in the clean limit and is related to
the topology of the electronic band structure Niu1999 ; Niu2002 ; Onoda2002 ,
and the extrinsic contribution which is due to impurity scatterings
Sinitsyn2008 ; Sinitsyn2007 ; Inoue2006 ; Onoda2006 ; Ado2016 . The extrinsic
Hall current can be further divided to the side jump and skew scattering
contribution, according to whether the impurity scattering is symmetric or
asymmetric. In a ferromagnetic semiconductor, the AHE is dominated by the
intrinsic contribution and determined by the Chern number of the filled bands
Onoda2006 . In ferromagnetic metals, however, the extrinsic contribution due
to impurity scatterings can be significant and even cancel out Inoue2006  or
exceed the intrinsic contribution depending on the impurity scattering rate
and strength Onoda2006 .

In this work we study the AHE in a type-I Weyl metal with breaking time
reversal symmetry. For simplicity, we consider the case with the minimum
number of two Weyl nodes Vishwanath2018 . The low energy effective Hamiltonian
of a typical type-I Weyl metal with tilting has the simplest form of spin-
orbit interaction ${\cal
H}_{\chi}=v\chi{\bm{\sigma}}\cdot{\mathbf{k}}+\mathbf{u_{\chi}\cdot k}$ with a
linear dispersion in each Weyl node. The anomalous Hall effect in a untilted
Weyl metal, i.e., $\mathbf{u}=0$, with dilute impurities was studied in Ref.
Burkov2014 . Different from two-dimensional (2D) ferromagnetic metals, the
anomalous Hall effect due to impurity scatterings in the untilted Weyl metal
vanishes only if the Fermi energy is within the linear dispersion regime of
the system. This is because the low-energy Hamiltonian of the untilted Weyl
metals ${\cal H}_{\chi}=v\chi{\bm{\sigma}}\cdot{\mathbf{k}}$ has an emergent
TRS at each single Weyl node. For this reason, the AHE in such a system then
comes completely from the topological Chern-Simons term and is proportional to
the distance between the two Weyl nodes Burkov2014 ; Burkov2015 ; Qi2013 .

The main focus of this work is then to study the impurity scattering induced
AHE in tilted Weyl metals with dilute Gaussian disorder, which is nonvanishing
since the tilting breaks the TRS in a single Weyl node Pesin2017 . We obtain
the intrinsic and extrinsic contribution from the Kubo-Streda formula in the
spin basis Streda1982 . However, it is usually considered hard to separate the
side jump and the skew scattering contribution in the spin basis, since a
single Feynman diagram in the spin basis contains both contributions
Sinitsyn2007 . To separate the two types of extrinsic contributions,
physicists usually turn to the semiclassical Boltzmann equation (SBE) approach
Sinitsyn2006 ; Sinitsyn2007 ; Sinitsyn2008 ; Loss2003 ; Mott1929 ; Smit1955
since it’s physically more transparent. The key ingredient resulting in the
anomalous Hall current is the band mixing by the current vertex and/or
impurity scatterings in the system Luttinger1955 ; Luttinger1958 ;
Sinitsyn2008 . From the SBE approach, one can separate the symmetric and anti-
symmetric impurity scatterings which result in the side-jump and skew
scattering contributions respectively. Based on the SBE approach, Sinitsyn et
al. further figured out the rigorous Feynman diagrams in the band eigenstate
basis (also called the chiral basis) in the Kubo-Streda formalism
corresponding to the side-jump and skew scattering contributions Sinitsyn2007
.

Yet since the Hamiltonian of the anomalous Hall system is usually written in
the spin basis due to the spin-orbit interaction, it is cumbersome to compute
the diagrams corresponding to the side jump and skew scattering contributions
one by one in the chiral basis. Instead, it is straightforward to compute the
total AHE in the spin basis. An efficient and transparent method to separate
the side jump and skew scattering contributions in the spin basis is then
valuable in practice.

In this work we show that by separating the different parts of the
polarization matrix ${\cal I}$ in the spin basis, namely the symmetric part
${\cal I}^{s}$, the intrinsic antisymmetric part ${\cal I}^{a}_{int}$ in the
clean limit and the antisymmetric part ${\cal I}^{a}_{im}$ due to impurity
scatterings, and using these parts as building blocks, one can build up a
rigorous one-to-one correspondence between the Feynman diagrams of the side
jump or skew scattering contribution in the chiral basis and the product of
matrices ${\cal I}^{s},{\cal I}^{a}_{int},{\cal I}^{a}_{im}$ in the spin basis
for Weyl and Dirac systems, as shown in Fig.4. From this correspondence, one
can easily separate the intrinsic, side jump and skew scattering contributions
in the spin basis. By this scheme, we separate the three contributions to the
AHE in tilted Weyl metals and found that the side jump contribution exceeds
both the intrinsic and skew scattering contribution in tilted Weyl metals.

For the isotropic 2D massive Dirac systems, Sinitysn et al. shows that the AHE
due to the three mechanisms obtained from the Kubo-Streda formula matches
completely with the results obtained from the SBE approach Sinitsyn2007 .
However, for anisotropic systems, there is concern that the commonly used SBE
approach under the relaxation time approximation (RTA) in Refs. Sinitsyn2007 ;
Loss2003  may not be reliable since the solution of the SBE in these works
assumes that the relaxation time defined in the solution is independent of the
direction of the incident electrons Sinova2009 ; Sinova2009-2 . This is true
for isotropic systems but not for anisotropic systems. It is then interesting
to compare the AHE obtained from the SBE approach and the result obtained from
the quantum Kubo-Streda formula for tilted Weyl metals, whose Fermi surface is
anisotropic. We found that the RTA is still valid in the leading order of the
tilting velocity $u/v$ [see Eq.(1) for definition] for tilted Weyl metals. As
a result, the AHE obtained from the two approaches agree well with each other
in the leading order of the tilting velocity for all the three mechanisms and
the deviation only comes from higher orders of tilting velocity.

The structure of this paper is as follows. In Sec.II, we compute the AHE of
the tilted Weyl metals using the quantum Kubo-Streda formula in the spin
basis, and separate the intrinsic and extrinsic contributions. We also
separate the different parts ${\cal I}^{s},{\cal I}^{a}_{int},{\cal
I}^{a}_{im}$ of the polarization operator ${\cal I}$ in the spin basis. In
Sec. III.A, we build up a one-to-one correspondence between the Feynman
diagrams in the chiral basis and the matrices defined above in the spin basis
for Weyl and Dirac systems. In Sec.III.B, we use this scheme to separate the
side jump and skew scattering contributions in the tilted Weyl metals. In
Sec.III.C, we compare the three contributions to the AHE we obtained from the
Kubo-Streda formula with the results obtained from the commonly-used SBE
approach. In Sec.IV, we have a comparison of the three different approaches in
studying the AHE, followed by a brief discussion of the third and fourth order
crossed diagrams. We have a brief summary of this paper in Sec. V.

##  II II. Anomalous Hall current from the Kubo-streda formula in the spin
basis

We start with the effective low energy Hamiltonian of a type-I Weyl metal with
breaking TRS

$H=\sum_{\chi}(\chi v\bm{\sigma}\cdot\mathbf{k}+\mathbf{u_{\chi}\cdot k}),$
(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
$u_{\chi}<v$, i.e., we only consider the type-I Weyl metals. The Hamiltonian
$H_{\chi}$ for each single valley results in two tilting linear bands
$\epsilon_{\pm}=\pm vk+\mathbf{u_{\chi}\cdot k}$. Here we assume the tilting
$\mathbf{u}_{+}=-\mathbf{u}_{-}=\mathbf{u}$, i.e., the tilting is opposite for
the two valleys. The tilting term then breaks the TRS of a single Weyl node
but not the global TRS of the whole system, whereas the term $\chi
v\mathbf{\bm{\sigma}\cdot k}$ only breaks the global TRS, but not the TRS of a
single valley. We will see below that if the tilting term is the same for the
two valleys with $\chi=\pm$, the AHE in the two valleys cancels each other.

When the Weyl semi-metal or metal is coupled to an electromagnetic (EM) field
$A^{\alpha}=(A_{0},{\mathbf{A}})$, other than the coupling of the low energy
effective Hamiltonian ]Eq.(1)] to the EM field by the Peierls substitution of
the four-momentum $p^{\alpha}\to p^{\alpha}+eA^{\alpha}$, an extra topological
Chern-Simons term describing the chiral anomaly of the response of the Weyl
semimetal ormetal to the EM field should be included in the action Burkov2015
; Qi2013 ,

$S_{\theta}=-\frac{e^{2}}{8\pi^{2}}\int
dtd^{3}r\partial_{\mu}\theta\epsilon^{\mu\nu\alpha\beta}A_{\nu}\partial_{\alpha}A_{\beta},$
(2)

where $\epsilon^{\mu\nu\alpha\beta}$ is a Levi-Civita antisymmetric tensor,
$\theta=\mathbf{b\cdot r}$ and $\mathbf{b}$ is the separation between the two
Weyl nodes in the momentum space. Here we assume the energy difference between
the two Weyl nodes is zero. The Chern-Simons term results in an anomalous Hall
effect with a transverse current perpendicular to both $\mathbf{b}$ and the
applied electric field,
${\mathbf{j}}=\frac{e^{2}}{2\pi^{2}}{\mathbf{b}}\times{\mathbf{E}}$ Burkov2015
; Qi2013 . This part turns out to be the only intrinsic contribution to the
AHE of the untilted Weyl metal or semi-metal when the doping is not very high.
Moreover, the AHE in untilted Weyl semi-metal/metal is also insensitive to the
impurity scatterings Burkov2014 . For the tilted Weyl metal or semi-metal, the
tilting does not affect the Chern- Simons term since the tilting does not
break the chiral symmetry. However, the AHE for the low energy effective
Hamiltonian Eq.(1) with tilting is no longer vanishing, and is very sensitive
to the impurity scatterings. The main purpose of this work is then to study
the AHE of the low energy effective Hamiltonian [Eq.(1)] for tilted Weyl
metals.

The anomalous Hall current for the low energy effective Hamiltonian can be
written as two parts $j_{H}=j^{\mathbf{I}}_{H}+j_{H}^{\mathbf{II}}$ by the
Kubo-Streda formalism, where $j^{\mathbf{I}}_{H}$ is a contribution from the
Fermi surface and $j^{\mathbf{II}}_{H}$ is a contribution from the Fermi sea
Sinitsyn2007 ; Streda1982 . The latter is insensitive to impurity scatterings
Niu2005 ; Burkov2014  and its contribution in a clean tilted Weyl metal was
studied in Ref Pesin2017 . The contribution $j^{\mathbf{I}}_{H}$ from the
Fermi surface may, however, be significantly affected by impurity scatterings.
Its clean limit has been studied in Ref. Zyuzin2017 . In this work we then
focus on the AHE from the Fermi surface of the tilted Weyl metals due to
impurity scatterings, which remains poorly studied in this system.

We consider dilute non-magnetic impurities with 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 $\gamma=n_{i}V_{0}^{2}$ with $n_{i}$ the impurity density. We assume the
mean free path of the electrons is much larger than the Fermi wavelength,
i.e., $k_{F}l\gg 1$ in this work and we mainly focus on the contribution from
the non-crossing diagram in Fig.1 when computing the AHE. It was shown in Ref.
Ado2016 ; Ado2017  that the crossed diagrams become important for near
impurities with distance of the order of electron Fermi wavelength. However,
due to the intricacy of these diagrams, we will study them in a separate
paper.

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 and add up the contribution together at the end. In the following,
we then focus on the AHE in a single Weyl node.

The impurity averaged retarded Green’s function (GF) in a single valley (e.g.
with $\chi=1$ without loss of generality) under the first Born approximation
is

$G^{R}(\epsilon,\mathbf{k})=(\epsilon-v\mathbf{\bm{\sigma}\cdot
k}-\mathbf{u\cdot k}-\Sigma^{R})^{-1},$ (3)

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

$\Delta_{i}(\mathbf{u})=\frac{1}{g(\epsilon_{F})}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{vk_{i}}{\epsilon_{F}-\mathbf{u\cdot
k}}\delta(\mathbf{u\cdot k}+vk-\epsilon_{F}),$ (4)

where $i=x,y,z$. At $\mathbf{u}=0,\bm{\Delta}(\mathbf{u})=0$ since the
integrand in Eq.(4) is odd. Whereas at finite tilting the Fermi surface is
asymmetric in the momentum space and $\Delta_{i}=-u_{i}/v$. Here
$|\bm{\Delta}|$ describes the difference of the scattering rates of the spin
species parallel or anti-parallel to the tilting $\mathbf{u}$. This is obvious
when one chooses the direction of $\mathbf{u}$ in the $z$ direction so that
only $\Delta_{z}=-u/v$ is non-vanishing. The difference of the impurity
scattering rates of the two spin species turns out to play a key role in the
impurity scattering induced AHE as we will show below.

We consider a uniform electric field ${\mathbf{E}}=-\partial_{t}{\mathbf{A}}$
applied to the system. In the linear response regime $j^{\rm
I}_{\alpha}=\Pi^{\rm I}_{\alpha\beta}A^{\beta}$, where
$A^{\alpha}=(0,\mathbf{A})$, and the leading order contribution of the
response function $\Pi^{\rm I}_{\alpha\beta}$ from the Fermi surface at small
frequency under the non-crossing approximation (NCA) can be expressed by the
Kubo-Streda formula as Streda1982 ; Burkov2014

$\displaystyle\Pi^{\mathbf{I}}_{\alpha\beta}(\omega,{\mathbf{q}})=e^{2}\omega$
$\displaystyle\int\frac{d\epsilon}{2\pi
i}\frac{d{\mathbf{k}}}{(2\pi)^{3}}\partial_{\epsilon}n_{F}(\epsilon)\rm{Tr}[\hat{\Gamma}_{\alpha}G^{R}(\epsilon+\omega,\mathbf{k}+\mathbf{q})\hat{j}_{\beta}G^{A}(\epsilon,\mathbf{k})],$

where $\hat{j}_{\alpha}=u_{\alpha}\sigma_{0}+\chi
v\sigma_{\alpha},\alpha=0,x,y,z$ is the bare current vertex for Hamiltonian
Eq.(1), (we define $u_{0}=0$), $G^{R/A}$ is the retarded (R) or advanced (A)
Green’s function under the first Born approximation, and
$\hat{\Gamma}_{\alpha}$ is the renormalized current vertex due to impurity
scatterings as shown in Fig.1b. Note that the linear response function
$\Pi^{\mathbf{I}}_{\alpha\beta}$ also contains the $G^{R}G^{R}$ and
$G^{A}G^{A}$ terms Sinitsyn2007 , but these terms are smaller in a factor of
$1/k_{F}l$ than the $G^{R}G^{A}$ term in the limit $k_{F}l\gg 1$ so we neglect
it in Eq.(II).

The renormalized current vertex $\hat{\Gamma}_{\alpha}$ satisfies the
recursion equation

$\hat{\Gamma}_{\alpha}(\omega,\mathbf{q})=\hat{j}_{\alpha}+\int\frac{d^{3}k}{(2\pi)^{3}}\gamma
G^{A}(0,\mathbf{k})\hat{\Gamma}_{\alpha}G^{R}(\omega,\mathbf{k+q}).$ (6)

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. (b)The square block is the renormalized current
vertex with the NCA.

The above equation can be solved by expressing the current vertex in the Pauli
matrix basis as $\hat{j}_{\alpha}={\cal
J}_{\alpha\beta}\sigma_{\beta},\alpha,\beta=0,x,y,z$, and the renormalized
current vertex as $\hat{\Gamma}_{\alpha}=\Gamma_{\alpha\beta}\sigma_{\beta}$,
where the summation over the repeated index is implied as usual. For
Hamiltonians with the coefficients ${\cal J}_{\alpha\beta}$ and
$\Gamma_{\alpha\beta}$ independent of the momentum, such as the Weyl and Dirac
systems, the coefficients of the renormalized current vertex can be solved,
with the relationship
$\rm{Tr}[\sigma_{\alpha}\sigma_{\beta}]=2\delta_{\alpha\beta}$, as
$\Gamma_{\alpha\beta}={\cal J}_{\alpha\gamma}D_{\gamma\beta}$, and ${\cal
D}=(1-\gamma{\cal I})^{-1}$ is the $4\times 4$ diffusion matrix with the
polarization operator ${\cal I}$ defined as

${\cal
I}_{\alpha\beta}=\frac{1}{2}\int\frac{d{\mathbf{k}}}{(2\pi)^{3}}\rm{Tr}[\sigma_{\alpha}G^{R}(\epsilon+\omega,\mathbf{k}+\mathbf{q})\sigma_{\beta}G^{A}(\epsilon,\mathbf{k})],$
(7)

and $\alpha,\beta=0,x,y,z$. Note that for the Hamiltonians with non-
relativistic kinetic energy term ${\mathbf{k}}^{2}/2m$, such as 2D Rashba
ferromagnets Ado2016 , ${\cal J}_{\alpha\beta}$ is dependent on the momentum
and the solution of renormalized current vertex is not so simple. We focus on
the case with ${\cal J}_{\alpha\beta}$ independent of the momentum in this
work, such as the relativistic systems, and our study in this work is
applicable only for such systems.

The $4\times 4$ response function matrix $\Pi^{\rm I}(\omega,\mathbf{q})$ with
vertex correction can then be expressed as the following matrix product

$\tilde{\Pi}^{\rm I}(\omega,\mathbf{q})={\cal J}{\cal D}{\cal I}{\cal J}^{T},$
(8)

where $\tilde{\Pi}^{\rm I}\equiv-\frac{\pi i}{\omega e^{2}}\Pi^{\rm I}$, and
${\cal J}$ is the $4\times 4$ matrix with elements ${\cal J}_{\alpha\beta}$,
i.e., the coefficients of the bare current vertex in the Pauli matrix basis as
defined above. The superscript $T$ means the transposition of the
corresponding matrix. The energy $\epsilon$ in $\cal I$ is bounded to the
Fermi energy $\epsilon_{F}$ due to the factor
$\partial_{\epsilon}n_{F}(\epsilon)$ in Eq.(II).

We are interested in the anomalous Hall conductivity in the dc limit of the
diffusive regime, i.e., $ql\ll 1,\omega\tau\ll 1$. However, under the first
Born approximation, the matrix ${\cal D}^{-1}=1-\gamma{\cal I}$ in the exact
dc limit $\omega=0,\bf{q}=0$ has determinant zero and is not invertible. For
the reason, we work at small finite frequency $\omega$ (but keep ${\bf q}$
zero for simplicity) and take the dc limit $\omega\to 0$ at the end of the
calculation.

We separate the ${\cal I}$ matrix to the symmetric and anti-symmetric part as
${\cal I}(\omega,\mathbf{q}\to 0)={\cal I}^{s}(\omega)+{\cal I}^{a}(\omega)$.
Here ${\cal I}^{s}(\omega)$ and ${\cal I}^{a}(\omega)$ in the diffusive limit
$\omega\tau\ll 1$ can be obtained by expanding the small frequency. In the
linear order of the frequency $\omega$, the symmetric part ${\cal
I}^{s}(\omega)$ and the anti-symmetric part ${\cal I}^{a}(\omega)$ are
respectively

$\gamma{\cal
I}^{s}(\omega)=\frac{1}{c_{0}}\left(\begin{array}[]{cccc}c_{0}+\frac{u^{2}}{v}c_{1}+i\omega\tau
b_{0}&c_{1}(\omega)u_{x}&c_{1}(\omega)u_{y}&c_{1}(\omega)u_{z}\\\
c_{1}(\omega)u_{x}&c_{2}(\omega)+c_{3}(\omega)u^{2}_{x}&c_{3}(\omega)u_{x}u_{y}&c_{3}(\omega)u_{x}u_{z}\\\
c_{1}(\omega)u_{y}&c_{3}(\omega)u_{x}u_{y}&c_{2}(\omega)+c_{3}(\omega)u^{2}_{y}&c_{3}(\omega)u_{y}u_{z}\\\
c_{1}(\omega)u_{z}&c_{3}(\omega)u_{x}u_{z}&c_{3}(\omega)u_{y}u_{z}&c_{2}(\omega)+c_{3}(\omega)u^{2}_{z}\end{array}\right),\\\
$ (9)

and

$\displaystyle\gamma{\cal I}^{a}(\omega)$ $\displaystyle=$
$\displaystyle\gamma{\cal I}^{a}_{int}(\omega)+\gamma{\cal
I}^{a}_{im}(\omega),$ (10) $\displaystyle\gamma{\cal
I}_{int,\alpha\beta}^{a}(\omega)$ $\displaystyle=$
$\displaystyle\frac{1}{c_{0}\epsilon_{F}\tau}f(\omega)\epsilon^{\alpha\beta\gamma}u^{\gamma},$
(11) $\displaystyle\gamma{\cal I}_{im,\alpha\beta}^{a}(\omega)$
$\displaystyle=$
$\displaystyle\frac{1}{c_{0}\epsilon_{F}\tau}g(\omega)\epsilon^{\alpha\beta\gamma}\Delta^{\gamma}.$
(12)

In the above expressions, $u^{\gamma}=(0,u_{x},u_{y},u_{z})$ and
$\Delta^{\gamma}=(0,\Delta_{x},\Delta_{y},\Delta_{z})$. The matrix ${\cal
I}_{int,\alpha\beta}^{a}(\omega)$ is the anti-symmetric part of ${\cal I}$ in
the clean limit $\tau\to\infty$, and ${\cal I}_{im,\alpha\beta}^{a}$ is the
anti-symmetric part due to impurity scatterings. The parameters
$c_{i}(\omega)=c_{i}+i\omega\tau b_{i},i=1,2,3$, where $c_{i}$ is the value of
$c_{i}(\omega)$ in the dc limit and $b_{i}$ is the coefficient of the linear
order frequency term. The exact value of $c_{i}$ can be obtained by directly
computing the ${\cal I}$ matrix in the dc limit as in the Appendix A, whereas
the coefficients $b_{i}$ can be obtained by expanding ${\cal
I}(\omega,\mathbf{q}\to 0)$ to the linear order of $\omega$. For the anomalous
Hall conductivity in the dc limit, it turns out that only the values in the dc
limit of $c_{i}(\omega)$ matter at the end and the coefficients $b_{i}$ do not
enter the final results of the dc anomalous Hall conductivity. The same is
true for $f(\omega)$ and $g(\omega)$. For the reason, we only give these
parameters in the dc limit in the following:

$\displaystyle c_{0}=\frac{v^{4}}{(v^{2}-u^{2})^{2}},$ (13) $\displaystyle
b_{0}=\frac{1}{4}[3+5c_{0}-3\frac{u^{2}}{v^{2}}c_{0}+3a(u)\frac{u^{2}}{v^{2}}],$
(14) $\displaystyle c_{1}(\omega\to 0)\equiv
c_{1}=\frac{1}{4v}[a(u)+\frac{u^{2}-3v^{2}}{v^{2}}c_{0}],$ (15) $\displaystyle
c_{2}(\omega\to 0)\equiv
c_{2}=\frac{1}{8}[3+\frac{2u^{2}}{v^{2}-u^{2}}-\frac{v^{2}-u^{2}}{v^{2}}a(u)],$
(16) $\displaystyle c_{3}(\omega\to 0)\equiv
c_{3}=\frac{1}{8u^{2}}[3a(u)-1+4\frac{u^{2}}{v^{2}}c_{0}-\frac{u^{2}}{v^{2}}a(u)],$
$\displaystyle f(\omega\to 0)\equiv
f(u)=\frac{1}{2v}[\frac{v^{2}}{v^{2}-u^{2}}-a(u)],$ (17) $\displaystyle
g(\omega\to 0)\equiv g(u)=-\frac{1}{4}[1+a(u)+\frac{u^{2}}{v^{2}}a(u)],$ (18)
$\displaystyle a(u)\equiv\frac{v^{2}}{u^{2}}(\frac{v}{2u}{\rm
ln}\frac{v+u}{v-u}-1)\approx\frac{1}{3}+\frac{u^{2}}{5v^{2}}+{\cal
O}(\frac{u^{4}}{v^{4}}).$

In the above calculation, when the chirality $\chi$ changes sign but
$\mathbf{u}$ does not, both $f(u)$ and $\mathbf{\Delta}$ changes sign. When
$\mathbf{u}$ changes sign but $\chi$ does not, $f(u)$ does not change sign,
but $\mathbf{\Delta}$ does. The parameter $g(u)$ does not change sign when
$\chi$ or ${\mathbf{u}}$ changes sign. So the sign of the anti-symmetric part
${\cal I}^{a}$ is determined by the product of the sign of $\chi$ and
$\mathbf{u}_{\chi}$ in each valley. In this work, $\chi$ is opposite in the
two valleys since the total chiral charge of the two valleys has to be zero
Vishwanath2018 . The tilting $\mathbf{u}_{\chi}$ then must have opposite signs
in the two valleys to get non-vanishing total AHE as we will see below.

From the $\cal{I}$ matrix, we can obtain the diffusion matrix $\cal{D}$ and
the renormalized current vertex $\hat{\Gamma}$ and finally get the Hall
current through Eq.(8). The main contributions to the AHE in a tilted Weyl
metal are presented as follows.

AHE without vertex correction. In this case, ${\cal D}=1$, the response
function

$\Pi^{\rm I}(\omega,\mathbf{q})=-\frac{\omega}{\pi i}e^{2}{\cal J}{\cal
I}(\omega,\mathbf{q}){\cal J}^{T}.$ (20)

For tilted Weyl metals, ${\cal
J}_{\alpha\beta}=v\delta_{\alpha\beta}+u_{\alpha}\delta_{\beta 0}$. It is easy
to check that in the dc limit the tilting part $u_{\alpha}\delta_{\beta,0}$ in
the bare current vertex of the tilted Weyl metals has no contribution to the
AHE in the linear response and $\Pi^{\rm
I}_{\alpha\beta}(\omega,\mathbf{q})=-\frac{\omega}{\pi i}e^{2}v^{2}{\cal
I}_{\alpha\beta}(\omega,\mathbf{q})$ without vertex correction for the tilted
Weyl metals. The main effect of the tilting is then just to produce an
anisotropy in the Fermi surface.

For isotropic systems, the spatial block of the symmetric part ${\cal I}^{s}$,
i.e., ${\cal I}^{s}_{ij},i,j=x,y,z,$ is diagonal and the diagonal element
corresponds to the longitudinal conductivity that is proportional to $\tau$
Loss2003 . For anisotropic systems, ${\cal I}^{s}_{ij},i,j=x,y,z,$ may contain
off-diagonal elements, as is the case for the tilted Weyl metals in Eq.(9).
This off-diagonal part corresponds to a normal Hall current due to anisotropy
Loss2003 . The anti-symmetric part ${\cal I}^{a}$ of the ${\cal I}$ matrix
comes from TRS breaking and corresponds to an anomalous Hall current. This
part is smaller than the symmetric part ${\cal I}^{s}$ by a factor of
$1/\epsilon_{F}\tau$, consistent with the fact that the anomalous Hall
conductivity is usually smaller than the longitudinal conductivity by
$1/\epsilon_{F}\tau$ Sinitsyn2007 .

The anti-symmetric part ${\cal I}^{a}$ of ${\cal I}(\omega,\mathbf{q})$ in the
limit of $\omega\to 0,\mathbf{q}\to 0$ results in a dc anomalous Hall current
as

$\mathbf{j}_{H}=\frac{e^{2}\epsilon_{F}}{2\pi^{2}v}[f(u)(\mathbf{u\times
E})+g(u)(\mathbf{\Delta\times E})],$ (21)

where $f(u)$ and $g(u)$ are given in Eqs.(17) and (18), and
$\Delta_{i}=-\frac{u_{i}}{v},i=1,2,3$.

Equation (21) above contains two terms. The first term, which we denote as
$\mathbf{j}^{int}_{H}$, corresponds to the clean and dc limit of the anti-
symmetric tensor ${\cal I}$, i.e., ${\cal I}^{a}_{int}(\omega\to 0)$ in
Eq.(11), and together with the contribution $\mathbf{j}^{\rm II}_{H}$ from the
Fermi sea constitute the intrinsic contribution of the low energy effective
Hamiltonian. Note that the contribution corresponding to ${\cal
I}^{a}_{int}(\omega\to 0)$ vanishes in the Weyl metals without tilting, as can
be seen from Eq.(7) where the integration over $\mathbf{k}$ and $-\mathbf{k}$
for the anti-symmetric tensor ${\cal I}^{a}_{\alpha\beta}$ cancels out at
$\omega,\mathbf{q}\to 0$ for this case.

The second term in Eq.(21) corresponds to ${\cal I}^{a}_{im}(\omega\to 0)$ in
Eq.(12). This contribution is due to an unbalanced impurity scattering of the
spin species parallel and anti-parallel to the tilting direction $\mathbf{u}$.
It is proportional to the difference of the impurity scattering rate of the
two spin species, but is independent of the total impurity scattering rate
$1/\tau$. This contribution involves a second order impurity scattering
process $|V_{0}|^{2}$ and constitutes part of the side jump contribution as we
will show in more details in the next section.

Vertex correction. We next consider the AHE in tilted Weyl metals with vertex
correction. Since the anti-symmetric part ${\cal I}^{a}$ is smaller than the
symmetric part ${\cal I}^{s}$ by a factor of $1/\epsilon_{F}\tau$, we can
expand the diffusion matrix $\cal{D}$ treating $\gamma{\cal I}^{a}$ as a small
parameter,

${\cal D}=(1-\gamma{\cal I}^{s}-\gamma{\cal
I}^{a})^{-1}=\sum_{n=0}^{\infty}({\cal D}_{0}\gamma{\cal I}^{a})^{n}{\cal
D}_{0},$ (22)

where $\gamma{\cal I}^{a}\sim 1/\epsilon_{F}\tau$, and ${\cal
D}_{0}\equiv(1-\gamma{\cal I}^{s})^{-1}$ is a constant symmetric matrix
$\sim\tau^{0}$ and satisfying ${\cal D}_{0}=1+{\cal D}_{0}\gamma{\cal I}^{s}$.
The matrix ${\cal D}_{0}$ gives the leading order vertex correction of the
current operator.

For the vertex correction coming from the anti-symmetric part $\gamma{\cal
I}^{a}$ in the ${\cal D}$ matrix, one only needs to keep the lowest order of
this contribution since $\gamma{\cal I}^{a}$ is smaller than $\gamma{\cal
I}^{s}$ by $1/\epsilon_{F}\tau$. For tilted Weyl metals, the response function
$\tilde{\Pi}^{\rm I}={\cal J}{\cal D}{\cal I}{\cal J}^{T}=v^{2}{\cal D}{\cal
I}$. Upon expansion, we get the anti-symmetric part for tilted Weyl metals as

$\displaystyle\tilde{\Pi}^{\rm I}_{a}/v^{2}$ $\displaystyle=$
$\displaystyle[{\cal D}_{0}+{\cal D}^{2}_{0}(\gamma{\cal I}^{a})^{2}{\cal
D}_{0}+...]{\cal I}^{a},$ (23) $\displaystyle+[{\cal D}_{0}(\gamma{\cal
I}^{a}){\cal D}_{0}+{\cal D}^{3}_{0}(\gamma{\cal I}^{a})^{3}{\cal
D}_{0}+...]{\cal I}^{s},$ $\displaystyle=$ $\displaystyle{\cal D}_{0}{\cal
I}^{a}+{\cal D}_{0}(\gamma{\cal I}^{a}){\cal D}_{0}{\cal I}^{s}+...$

The leading order contribution contains two terms, i.e., ${\cal D}_{0}{\cal
I}^{a}$ and ${\cal D}_{0}(\gamma{\cal I}^{a}){\cal D}_{0}{\cal I}^{s}$. Both
of the terms are independent of $\tau$. The addition of the two terms is equal
to ${\cal D}_{0}{\cal I}^{a}{\cal D}_{0}$ for tilted Weyl metals, and ${\cal
J}{\cal D}_{0}{\cal I}^{a}{\cal D}_{0}{\cal J}^{T}$ for general anomalous Hall
systems with ${\cal J}$ independent of the momentum.

To get the matrix ${\cal D}_{0}$, one needs to invert the matrix
$1-\gamma{\cal I}^{s}$. However, in the dc limit $\omega=0,{\mathbf{q}}=0$,
the matrix $1-\gamma{\cal I}^{s}$ has determinant zero due to charge
conservation Burkov2014  and is not invertible. This can be verified by
checking the determinant $\rm Det[1-\gamma{\cal I}^{s}(\omega\to
0)]=-\frac{u^{2}}{v}\frac{c_{1}}{c_{0}^{4}}(c_{0}-c_{2})^{2}(c_{0}-c_{2}-c_{3}u^{2}+vc_{1})$,
for which the factor $c_{0}-c_{2}-c_{3}u^{2}+vc_{1}$ equals to zero exactly.
However, at finite frequency, the matrix $1-\gamma{\cal I}^{s}$ becomes
invertible. From Eq.(9), we get

$\displaystyle\mathcal{D}_{0}(\omega)=\frac{c_{0}}{[c_{0}-c_{2}(\omega)]\mu(\omega)}\left(\begin{array}[]{cccc}c_{0}-c_{2}(\omega)-c_{3}(\omega)u^{2}&c_{1}(\omega)u_{x}&c_{1}(\omega)u_{y}&c_{1}(\omega)u_{z}\\\
c_{1}(\omega)u_{x}&\mu(\omega)+\lambda(\omega)u_{x}^{2}&\lambda(\omega)u_{x}u_{y}&\lambda(\omega)u_{x}u_{z}\\\
c_{1}(\omega)u_{y}&\lambda(\omega)u_{x}u_{y}&\mu(\omega)+\lambda(\omega)u_{y}^{2}&\lambda(\omega)u_{y}u_{z}\\\
c_{1}(\omega)u_{z}&\lambda(\omega)u_{x}u_{z}&\lambda(\omega)u_{y}u_{z}&\mu(\omega)+\lambda(\omega)u_{z}^{2}\\\
\end{array}\right),$ (28)

where

$\displaystyle\mu(\omega)=[(-\frac{u^{2}}{v}c_{1}-i\omega\tau
b_{0})(c_{0}-c_{2}(\omega)-c_{3}(\omega)u^{2})-c_{1}^{2}(\omega)u^{2}]/[c_{0}-c_{2}(\omega)],$
(29)
$\displaystyle\lambda(\omega)=[c_{1}^{2}(\omega)-c_{3}(\omega)(\frac{u^{2}}{v}c_{1}+i\omega\tau
b_{0})]/[c_{0}-c_{2}(\omega)].$ (30)

One can check that in the leading order of $\omega\tau$, the parameter
$\mu(\omega)\propto i\omega\tau$, resulting in a diffusion pole at $\omega\to
0$ in the diffusion matrix ${\cal D}_{0}$. It is interesting to get the
leading order renormalized current vertices from ${\cal D}_{0}$ for the tilted
Weyl metals, which are

$\displaystyle\Gamma_{0}=\Gamma_{0\beta}\sigma_{\beta}=\frac{vc_{0}}{[c_{0}-c_{2}(\omega)]\mu(\omega)}[(c_{0}-c_{2}(\omega)-c_{3}(\omega)u^{2})\sigma_{0}+c_{1}(\omega)\bm{u}\cdot\bm{\sigma}],$
(31)
$\displaystyle\Gamma_{i}=\Gamma_{i\beta}\sigma_{\beta}=\frac{c_{0}}{c_{0}-c_{2}(\omega)}[\frac{c_{0}-c_{2}(\omega)-c_{3}(\omega)u^{2}+vc_{1}(\omega)}{\mu(\omega)}u_{i}\sigma_{0}+v\sigma_{i}+\frac{c_{1}(\omega)+v\lambda(\omega)}{\mu(\omega)}u_{i}(\bm{u}\cdot\bm{\sigma})],i=1,2,3.$
(32)

It is clear from Eq.(31) that the renormalized charge component $\Gamma_{0}$
has a diffusion pole at $\omega\to 0$ due to the charge conservation. However,
since both $c_{0}-c_{2}(\omega)-c_{3}(\omega)u^{2}+vc_{1}(\omega)$ and
$c_{1}(\omega)+v\lambda(\omega)$ in Eq.(32) vanish at $\omega\to 0$ and are
proportional to $i\omega\tau$ in the leading order of the frequency, the
renormalized current vertices $\Gamma_{i},i=1,2,3$ have no diffusion pole.
This is because the current operators $\hat{j}_{i},i=1,2,3$ do not commute
with the Hamiltonian and are not conserved.

From the ${\cal D}_{0}$ and ${\cal I}^{a}$ matrix, we get the leading order
contribution of $\tilde{\Pi}^{\rm I}_{a}(\omega)$ for tilted Weyl metals as

$\displaystyle\tilde{\Pi}^{\rm I}_{a}(\omega)$ $\displaystyle=$ $\displaystyle
v^{2}{\cal D}_{0}(\omega){\cal I}^{a}(\omega){\cal D}_{0}(\omega)$ (33)
$\displaystyle=$
$\displaystyle[\frac{c_{0}}{c_{0}-c_{2}(\omega)}]^{2}v^{2}\mathcal{I}^{a}(\omega).$

Taking the limit $\omega\to 0$, we get the total dc anomalous Hall current for
tilted Weyl metals in a single valley as

${\mathbf{j}}^{\rm
I}_{H}=\frac{e^{2}\epsilon_{F}}{2\pi^{2}v}F(u)[f(u)(\mathbf{u\times
E})+g(u)(\mathbf{\Delta\times E})],$ (34)

where $F(u)=\frac{c^{2}_{0}}{(c_{0}-c_{2})^{2}}$ with $c_{0},c_{2}$ given in
Eqs.(13) and (16), and $f(u)$ and $g(u)$ given in Eqs.(17) and (18). At
${\mathbf{u}}=0,F(u)=\frac{c^{2}_{0}}{(c_{0}-c_{2})^{2}}=9/4$, and at finite
$\mathbf{u}$, the value of $F(u)$ is plotted in Fig.2.

Figure 2: The factor $F(u)$ for the vertex correction as a function of $u/v$.

From Eq.(23), the next leading order vertex correction to the Hall current is
$\sim{\cal D}^{3}_{0}(\gamma{\cal I}^{a})^{2}{\cal I}^{a}+{\cal
D}^{4}_{0}(\gamma{\cal I}^{a})^{3}{\cal I}^{s}\sim(1/\epsilon_{F}\tau)^{2}$,
which is much smaller than the leading order correction in the case
$\epsilon_{F}\tau\gg 1$. For this reason, we only need to keep the leading
order contribution of the AHE which is of the order of $\tau^{0}$.

The total anomalous Hall current in Eq.(34) includes both the intrinsic and
extrinsic contribution. The intrinsic one corresponds to the first term of
Eq.(21). The total contribution subtracting the intrinsic one is then the
extrinsic contribution due to impurity scatterings, which includes the side-
jump and skew scattering contribution. In the next section we show how to
separate these two contributions from the total Hall current in Eq.(34).

##  III III. Separation of the intrinsic, side-jump and skew scattering
contributions in the spin basis

###  III.1 A. General Formalism

The Hall current in Eq.(34) obtained from the Kubo-Streda formula in the spin
basis is rigorous but not physically transparent. It took physicists a long
time to understand the microscopic mechanism of both the intrinsic and
extrinsic contribution of the AHE. With the efforts of the authors in Refs.
Sinitsyn2006 ; Smit1955 ; Berger1970 ; Luttinger1955 ; Luttinger1958 ;
Sinitsyn2007 ; Sinitsyn2008  and others, a semi-classical Boltzmann equation
approach was built to explain the anomalous Hall current. The key ingredient
is the band mixing by the current vertex and/or the impurity potential which
induces a transverse current perpendicular to the electric field. Based on the
semi-classical explanation of the three different mechanisms of the anomalous
Hall currents, i.e., the intrinsic, side-jump and skew scattering, Sinitsyn et
al. figured out the Feynman diagrams corresponding to the three different
mechanisms in the band eigenstate basis (shown in Fig.4) Sinitsyn2007 . The
intrinsic contribution is due to the topological structure of the electron
band and exists even without impurity scatterings. The side-jump contribution
is due to the transverse displacement of the electrons by impurity scatterings
Berger1970 . This scattering is symmetric. Whereas the skew scattering is due
to asymmetric impurity scatterings.

Though it is physically more transparent to separate the different
contributions in the eigenstate or chiral basis, as done in most previous
works Sinitsyn2007 ; Yang2011 ; Sinova2010 , the Hamiltonians of the anomalous
Hall systems are usually given in the spin basis due to the spin-orbit
interaction. For this reason, it is more convenient to calculate the total
anomalous Hall current in the spin basis as we did in the last section.
However, a transparent method to separate the different extrinsic
contributions in the spin basis is lacking and considered to be a difficult
task since a single diagram in the spin basis contains contributions from
different mechanisms Sinitsyn2007 ; Yang2011 .

In this section, we show an efficient and transparent scheme to separate the
different contributions of the AHE in the spin basis for Dirac and Weyl
systems. By separating the symmetric and anti-symmetric part of the $\cal{I}$
matrix and $\cal{D}$ matrix as we did in the last section and using these
matrices as building blocks, we build a one-to-one correspondence between the
Feynman diagrams of the different contributions given in the chiral basis in
Ref Sinitsyn2007  and the matrix products of ${\cal I}^{a}_{int},{\cal
I}^{a}_{im}$ and ${\cal D}_{0}$ we obtained in the spin basis. The result is
shown in Fig.4. For simplicity, we build up this correspondence for tilted
Weyl metals at first and it is easy to generalize the result to other Weyl and
Dirac systems.

For tilted Weyl metals, the tilting part $eu_{\alpha}\sigma_{0}$ in the bare
current vertex has no contribution to the linear response. The ${\cal I}$
matrix remains the same by replacing $\sigma_{\alpha}$ by
$\hat{j}_{\alpha}/v$. The integrand of the ${\cal I}$ matrix in Eq. (7) can
then be replaced by $I_{\alpha\beta}={\rm
Tr}[\hat{j}_{\alpha}G^{R}\hat{j}_{\beta}G^{A}]$ for tilted Weyl metals. To get
better physical understanding, we use the latter notation to expand
$I_{\alpha\beta}$ in this section. The result can be easily generalized to
other systems with different current vertices independent of the momentum.

The integrand $I_{\alpha\beta}$ is the same in the spin and chiral bases. We
expand it in the chiral basis as

$\displaystyle I_{\alpha\beta}=$
$\displaystyle\langle+|\hat{j}_{\alpha}\sum_{s1}|s_{1}\rangle\langle
s_{1}|G^{R}\sum_{s2}|s_{2}\rangle\langle
s_{2}|\hat{j}_{\beta}\sum_{s3}|s_{3}\rangle\langle s_{3}|G^{A}|+\rangle$
$\displaystyle+\langle-|\hat{j}_{\alpha}\sum_{s4}|s_{4}\rangle\langle
s_{4}|G^{R}\sum_{s5}|s_{5}\rangle\langle
s_{5}|\hat{j}_{\beta}\sum_{s6}|s_{6}\rangle\langle s_{6}|G^{A}|-\rangle,$

where $s_{1,...,6}=\pm$ represent the upper and lower eigenbands respectively.
The Green’s function under the Born approximation can be written as

$G^{R/A}=G_{0}^{R/A}+G_{0}^{R/A}\Sigma G^{R/A},$ (36)

where $G_{0}^{R/A}$ is the bare GF without impurity scatterings, and $\Sigma$
is the self-energy due to impurity scatterings given in the last section. In
the eigenstate basis, $G_{0}^{R/A}$ is diagonal, but $\Sigma$ contains both
diagonal and off-diagonal elements, i.e., $\Sigma$ may cause band mixing. For
the reason, $G^{R/A}$ in the Born approximation also contains off-diagonal
elements in the band eigenstate basis. However, its off-diagonal elements are
small in $1/\tau$. For this reason, we only need to keep at most one band off-
diagonal element of $G^{R}$ or $G^{A}$ in each term in Eq.(III.1). The leading
order terms are then

$\displaystyle I_{\alpha\beta}$ $\displaystyle=$
$\displaystyle\langle+|\hat{j}_{\alpha}|+\rangle\langle+|G^{R}|+\rangle\langle+|\hat{j}_{\beta}|+\rangle\langle+|G^{A}|+\rangle$
$\displaystyle+$
$\displaystyle\langle-|\hat{j}_{\alpha}|-\rangle\langle-|G^{R}|-\rangle\langle-|\hat{j}_{\beta}|-\rangle\langle-|G^{A}|-\rangle$
$\displaystyle+$
$\displaystyle\langle+|\hat{j}_{\alpha}|-\rangle\langle-|G^{R}|-\rangle\langle-|\hat{j}_{\beta}|+\rangle\langle+|G^{A}|+\rangle$
$\displaystyle+$
$\displaystyle\langle-|\hat{j}_{\alpha}|+\rangle\langle+|G^{R}|+\rangle\langle+|\hat{j}_{\beta}|-\rangle\langle-|G^{A}|-\rangle$
$\displaystyle+$
$\displaystyle\langle+|\hat{j}_{\alpha}|-\rangle\langle-|G^{R}|+\rangle\langle+|\hat{j}_{\beta}|+\rangle\langle+|G^{A}|+\rangle$
$\displaystyle+$
$\displaystyle\langle+|\hat{j}_{\alpha}|+\rangle\langle+|G^{R}|-\rangle\langle-|\hat{j}_{\beta}|+\rangle\langle+|G^{A}|+\rangle$
$\displaystyle+$
$\displaystyle\langle+|\hat{j}_{\alpha}|+\rangle\langle+|G^{R}|+\rangle\langle+|\hat{j}_{\beta}|-\rangle\langle-|G^{A}|+\rangle$
$\displaystyle+$
$\displaystyle\langle-|\hat{j}_{\alpha}|+\rangle\langle+|G^{R}|+\rangle\langle+|\hat{j}_{\beta}|+\rangle\langle+|G^{A}|-\rangle$
$\displaystyle+$
$\displaystyle\langle-|\hat{j}_{\alpha}|+\rangle\langle+|G^{R}|-\rangle\langle-|\hat{j}_{\beta}|-\rangle\langle-|G^{A}|-\rangle$
$\displaystyle+$
$\displaystyle\langle-|\hat{j}_{\alpha}|-\rangle\langle-|G^{R}|+\rangle\langle+|\hat{j}_{\beta}|-\rangle\langle-|G^{A}|-\rangle$
$\displaystyle+$
$\displaystyle\langle+|\hat{j}_{\alpha}|-\rangle\langle-|G^{R}|-\rangle\langle-|\hat{j}_{\beta}|-\rangle\langle-|G^{A}|+\rangle$
$\displaystyle+$
$\displaystyle\langle-|\hat{j}_{\alpha}|-\rangle\langle-|G^{R}|-\rangle\langle-|\hat{j}_{\beta}|+\rangle\langle+|G^{A}|-\rangle.$
(37)

We separate these terms into three groups:

(i) The first two terms correspond to intraband processes, and both current
vertices are band diagonal without mixing the two bands. As shown in the
Appendix B, these two terms are the dominant symmetric part of the $I$ matrix.
We denote them as $I^{s}$. Among the two terms in $I^{s}$, the first term with
intraband processes in the upper band dominates the contribution to the
integration, since the lower band is far below the Fermi surface and the
scattering processes by current vertex are forbidden. For the reason, the
symmetric part ${\cal I}^{s}$ of the ${\cal I}$ matrix corresponds to the
diagram in Fig. 3(a) in the chiral basis.

(ii) The next two terms involve only band diagonal elements of $G^{R}$ and
$G^{A}$, but both current vertices mix the two bands. For these two terms, we
further separate the part in the clean limit and the remaining part with
disorder scattering by Eq.(36). It is easy to show that the part in the clean
limit dominant (see the Appendix B). This part corresponds to ${\cal
I}^{a}_{int}$ we obtained in the spin basis since it is the only nonvanishing
antisymmetric term in the clean limit in the band eigenstate basis. It then
gives the intrinsic contribution to the AHE from the Fermi surface and
corresponds to the diagram in Fig. 3(c) in the band basis.

(iii)The remaining eight terms in $I_{\alpha\beta}$ are also antisymmetric in
the leading order and are nonvanishing only with impurity scatterings. The
four terms with $\langle-|G^{R/A}|-\rangle$ in the lower band have an extra
smallness compared to the other four terms and so are negligible. For the
remaining four terms, one can expand the band off-diagonal Green’s function
elements $\langle\pm|G^{R/A}|\mp\rangle$ by Eq.(36). Since
$\langle\pm|G_{0}^{R/A}|\mp\rangle=0$,
$\langle\pm|G^{R/A}|\mp\rangle=\langle\pm|G_{0}^{R/A}\Sigma
G^{R/A}|\mp\rangle\approx\langle\pm|G_{0}^{R/A}\Sigma G_{0}^{R/A}|\mp\rangle$,
where the last replacement of $G^{R/A}$ by $G_{0}^{R/A}$ in this element does
not affect the integration of $I^{\alpha\beta}$ in the leading order. The
leading order contribution of this part then corresponds to the four diagrams
in Fig.3d and the matrix ${\cal I}^{a}_{im}$ we obtained in the spin basis.

Note that for all the diagrams in Figs.3(c) and Fig.3(d), replacing the thin
line representing $G^{R/A}_{0}$ by the thick line representing $G^{R/A}$ does
not change the leading order contribution of the AHE. For clarity of physics,
we only keep the minimum number of thick lines in the leading order in this
paper. The same applies to the diagrams in Fig.4.

Figure 3: Correspondence of the symmetric and anti-symmetric part of the
${\cal I}$ matrix in the spin basis to the Feynman diagrams in the band
eigenstate basis for tilted Weyl metals. The thin solid lines represent the
bare GF $G_{0}$ in the eigenstate basis and the thick solid lines represent
the GF under the first Born approximation in the eigenstate basis. The dashed
lines represent impurity scatterings. The diagram in (a) corresponds to the
symmetric part ${\cal I}^{s}$ of the ${\cal I}$ matrix. The solid square in
(b) represents the renormalized current vertex in the chiral basis. It
corresponds to the renormalization by the symmetric part ${\cal D}_{0}$ of the
diffusion matrix ${\cal D}$ in the spin basis. The matrix ${\cal D}_{0}$
satisfies the recursion relationship ${\cal D}_{0}=1+{\cal D}_{0}\gamma{\cal
I}^{s}$ and causes an intraband vertex correction of the current vertex. The
two diagrams in (c) represent the intrinsic antisymmetric part ${\cal
I}^{a}_{int}$ of the ${\cal I}$ matrix. The four diagrams in (d) represent the
antisymmetric part of the ${\cal I}$ matrix due to impurity scatterings ${\cal
I}^{a}_{im}$.  Figure 4: The correspondence of the Feynman diagrams in the
eigenstate (chiral) basis for the intrinsic, side jump and skew scattering
contributions to the matrix product of ${\cal I}^{s},{\cal D}_{0},{\cal
I}^{a}_{int},{\cal I}^{a}_{im}$ obtained in the spin basis for tilted Weyl
metals.

The symmetric part ${\cal I}^{s}$ does not contribute to the Hall current
directly since it involves only intraband scattering. However, it produces a
band-diagonal vertex correction of the current operator through the symmetric
part of the diffusion matrix, i.e., ${\cal D}_{0}$ that we defined in the last
section. The renormalized band-diagonal current vertex is labeled as
$\Upsilon^{++}$ in the band eigenstate basis in Ref Sinitsyn2007 , and
satisfies the recursion relationship shown in Fig. 3b.

The antisymmetric part of diffusion operator ${\cal D}$ is smaller than the
symmetric part by a factor $1/\epsilon_{F}\tau$ so the vertex correction due
to this part does not need a resummation to infinite order. This part
generates a transverse component of the current vertex and one only needs to
keep its lowest order as we did in the last section.

The Feynman diagrams of the intrinsic, side-jump and skew scattering
contribution to the AHE in the leading order of $1/\epsilon_{F}\tau$ in the
band eigenstate basis are given in Ref Sinitsyn2007 , also shown in Fig.4.
With the correspondence in Fig. 3 between the diagrams in the band eigenstate
basis and the matrices in the spin basis, we can easily write down the
matrices for each diagram of the three different mechanisms for tilted Weyl
metals. The results are also presented in Fig.4. From this table, we can
conveniently separate the intrinsic, side jump and skew scattering
contribution in the spin basis for tilted Weyl metals.

The intrinsic contribution corresponds to the ${\cal I}$ matrix in the clean
limit, i.e., ${\cal I}^{a}_{int}$. The side jump contribution involves only
second order impurity scatterings, i.e., only one disorder line. There are
eight diagrams in the eigenstate (chiral) basis for this type, as shown in
Fig.4 Footnote0 . Their total contribution to the AHE is then

$\displaystyle\tilde{\Pi}^{sj}$ $\displaystyle=$ $\displaystyle\gamma{\cal
D}_{0}{\cal I}^{s}{\cal I}^{a}_{int}+{\cal I}^{a}_{int}\gamma{\cal D}_{0}{\cal
I}^{s}+\frac{{\cal D}_{0}{\cal I}^{a}_{im}+{\cal I}^{a}_{im}{\cal D}_{0}}{2}$
(38) $\displaystyle=$ $\displaystyle\frac{{\cal D}_{0}{\cal I}^{a}+{\cal
I}^{a}{\cal D}_{0}}{2}-{\cal I}^{a}_{int}$ $\displaystyle+$
$\displaystyle\frac{({\cal D}_{0}-1){\cal I}^{a}_{int}+{\cal
I}^{a}_{int}({\cal D}_{0}-1)}{2}.$

where $\tilde{\Pi}^{sj}\equiv-\frac{\pi i}{\omega e^{2}v^{2}}\Pi^{sj}$ is the
rescaled response function as in the last section. One may wonder how to
equate the matrix $\gamma{\cal D}_{0}{\cal I}_{s}{\cal I}^{a}_{int}$, i.e.,
$({\cal D}_{0}-1){\cal I}^{a}_{int}$ with the corresponding diagrams on the
left in Fig.4 since ${\cal D}_{0}-1$ only connects to upper band legs whereas
the vertices of ${\cal I}^{a}_{int}$ mix the upper and lower bands. This is
easy to understand by noting that $({\cal D}_{0}-1){\cal I}^{a}_{int}$ is the
leading order contribution of $({\cal D}-1){\cal I}^{a}_{int}$ and the
disorder line of $({\cal D}-1)$ includes all possible inter-band and intra-
band scatterings by disorder. Among all the diagrams corresponding to $({\cal
D}-1){\cal I}^{a}_{int}$, the leading order contribution only corresponds to
the diagrams on the left of Fig.4.

The second line of Eq.(38) equals the leading order contribution with only one
vertex correction ${\cal D}_{0}$ in Eq.(23), subtracting the intrinsic
contribution ${\cal I}^{a}_{int}$. Other than this term, the side jump
contribution contains another term $\frac{({\cal D}_{0}-1){\cal
I}^{a}_{int}+{\cal I}^{a}_{int}({\cal D}_{0}-1)}{2}$, which comes from the
second term in Eq.(23). The remaining part of the second term in Eq.(23) is
the contribution from skew scattering as we show below.

The skew scattering contribution corresponds to the six diagrams in Fig.4.
Each diagram involves two disorder lines, i.e., the fourth order impurity
scattering processes. We can read the total contribution to the AHE from the
skew scattering diagrams as

$\displaystyle\tilde{\Pi}^{sk}$ $\displaystyle=$ $\displaystyle\gamma{\cal
D}_{0}{\cal I}^{s}{\cal I}^{a}_{int}\gamma{\cal D}_{0}{\cal I}^{s}$ (39)
$\displaystyle+\frac{{\cal D}_{0}{\cal I}^{a}_{im}\gamma{\cal D}_{0}{\cal
I}^{s}+\gamma{\cal D}_{0}{\cal I}^{s}{\cal I}^{a}_{im}{\cal D}_{0}}{2},$
$\displaystyle=$ $\displaystyle\frac{({\cal D}_{0}-1){\cal I}^{a}{\cal
D}_{0}+{\cal D}_{0}{\cal I}^{a}({\cal D}_{0}-1)}{2}$
$\displaystyle-\frac{({\cal D}_{0}-1){\cal I}^{a}_{int}+{\cal
I}^{a}_{int}({\cal D}_{0}-1)}{2}.$

The first term of Eq.(39) corresponds to the antisymmetric part of the term
${\cal D}_{0}{\cal I}^{a}{\cal D}_{0}\gamma{\cal I}_{s}$ in Eq.(23), which
contains two vertex corrections ${\cal D}_{0}$. However, the antisymmetric
part of ${\cal D}_{0}{\cal I}^{a}{\cal D}_{0}\gamma{\cal I}_{s}$ is not fully
the skew scattering contribution. It contains part of the side jump
contribution as subtracted in the last line.

It is easy to check that the addition of the intrinsic contribution, the side
jump contribution in Eq.(38) and the skew scattering contribution in Eq.(39)
is equal to the total Hall response $\tilde{\Pi}^{I}_{a}={\cal D}_{0}{\cal
I}^{a}{\cal D}_{0}$ for tilted Weyl metals obtained in the spin basis in
Eq.(23).

The above results can be easily generalized to other anomalous Hall systems
without nonrelativistic kinetic energy term, such as a 2D massive Dirac model
Sinitsyn2007 . For a general anomalous Hall system with ${\cal J}$ independent
of the momentum, one only needs to multiply the matrix ${\cal J}$ and ${\cal
J}^{T}$ on the two ends of the response function matrix, as well as the side
jump contribution in Eq.(38) and skew scattering contribution in Eq.(39). The
total contribution to the Hall response is $\tilde{\Pi}^{I}_{a}={\cal J}{\cal
D}_{0}{\cal I}^{a}{\cal D}_{0}{\cal J}^{T}$.

###  III.2 B. The vertex correction factor ${\cal D}_{0}$

We have a brief discussion of the vertex correction of the current operator in
Fig.3b in this section. The renormalized current vertex due to this vertex
correction is related to the bare current vertex by
$\Gamma_{\alpha\beta}\sigma_{\beta}={\cal J}_{\alpha\gamma}{\cal
D}_{0,\gamma\beta}\sigma_{\beta}$ in the spin basis as we show in the last
section. Here ${\cal D}_{0}$ is the symmetric part of the diffusion matrix
${\cal D}$ in the dc limit.

As discussed in the last section, this vertex correction corresponds to only
the impurity scatterings within the upper band in the ladder diagram Fig.3(b)
in the chiral basis. In the chiral basis, the renormalized band-diagonal
current vertex $\Upsilon^{++}_{\alpha}$ is obtained from the recursion
equation as Sinitsyn2007

$\Upsilon^{++}_{\alpha}({\mathbf{k}})=j^{++}_{\alpha}({\mathbf{k}})+\int\frac{d^{3}{\mathbf{k}^{\prime}}}{(2\pi)^{3}}G^{A+}|V^{++}_{{\mathbf{k}^{\prime}}\mathbf{k}}|^{2}G^{R+}\Upsilon^{++}_{\alpha}({\mathbf{k}^{\prime}}),$
(40)

where
$j^{++}_{\alpha}(\mathbf{k})=\langle\mathbf{k},+|\hat{j}_{\alpha}|\mathbf{k},+\rangle$,
$G^{R/A+}=\langle{\mathbf{k}^{\prime}},+|G^{R/A}|{\mathbf{k}^{\prime}},+\rangle$,
$V_{{\mathbf{k}^{\prime}}{\mathbf{k}}}^{++}=\langle{\mathbf{k}^{\prime}},+|V|{\mathbf{k}},+\rangle$.
For the case with isotropic Fermi surface and isotropic scattering potential,
the scatterings only depend on the angle between ${\mathbf{k}^{\prime}}$ and
${\mathbf{k}}$, and the above equation can be solved exactly by the ansatz

$\Upsilon^{++}_{\alpha}({\mathbf{k}})=\tilde{\alpha}j^{++}_{\alpha}({\mathbf{k}}),$
(41)

where $\tilde{\alpha}$ is a constant depending on the angle between
$\mathbf{k}$ and $\mathbf{k}^{\prime}$ but not the direction of $\mathbf{k}$.
For untilted Weyl metals, as well as many other isotropic systems, such as
graphene and the 2D massive Dirac model, the band-diagonal current vertex in
the chiral basis is $j^{++}_{\alpha}\sim k_{\alpha}$ for $\alpha=x,y,z$. For
these cases, $\tilde{\alpha}$ can be solved as
$\tilde{\alpha}=\tau^{tr}/\tau^{+}$, where $\tau^{tr}$ and $\tau^{+}$ are the
transport and ordinary scattering times of the upper band respectively and can
be expressed as

$1/\tau^{tr}=2\pi\int\frac{d^{3}{\mathbf{k}^{\prime}}}{(2\pi)^{3}}|V^{++}_{{\mathbf{k}^{\prime}}\mathbf{k}}|^{2}(1-\cos({\mathbf{k}^{\prime}}\cdot{\mathbf{k}}))\delta(\epsilon_{\mathbf{k}}-\epsilon^{+}_{\mathbf{k}^{\prime}}),$
(42)
$1/\tau^{+}=2\pi\int\frac{d^{3}{\mathbf{k}^{\prime}}}{(2\pi)^{3}}|V^{++}_{{\mathbf{k}^{\prime}}\mathbf{k}}|^{2}\delta(\epsilon_{\mathbf{k}}-\epsilon^{+}_{\mathbf{k}^{\prime}}).$
(43)

For these isotropic systems, the vertex correction obtained from the above
procedure in the chiral basis should be equivalent to that obtained from
${\cal D}_{0}$ in the spin basis. Indeed, for the spin-orbit interacting
systems with isotropic Fermi surface and isotropic impurity scatterings, the
matrix ${\cal D}_{0}$ is completely diagonal if the Hamiltonian also has time
reversal symmetry, e.g., graphene Chen2012 ; Ando1998 ; Falko2008 , a helical
metal on the surface of topological insulators Burkov2010 , or a single node
of untilted Weyl metals Burkov2014 . And for the isotropic systems with
breaking time reversal symmetry, such as 2D massive Dirac model Sinitsyn2007 ,
${\cal D}_{0}$ is diagonal only for the spatial block with $i,j=x,y,z$. For
all these cases, the vertex correction factor ${\cal D}_{0}$ for the current
vertex $\hat{j}_{i},i=x,y$ can be replaced by the diagonal $(i,i)$ element of
${\cal D}_{0}$, and we have checked that this element is equal to
$\tilde{\alpha}$ obtained from Eqs.(41)-(43). The side jump and skew
scattering contribution in Eqs.(38) and (39) are then greatly simplified for
such systems, and one only needs to compute the parameter $\tilde{\alpha}$ and
the anti-symmetric part ${\cal I}_{int}^{a}$ and ${\cal I}^{a}_{im}$ of the
polarization operator to obtain the three different anomalous Hall
contributions.

However, for the systems with anisotropic Fermi surface, such as tilted Weyl
metals, the impurity scatterings depend not only on the angle between
${\mathbf{k}^{\prime}}$ and ${\mathbf{k}}$, but also the direction of
$\mathbf{k}$. This can be seen by computing $1/\tau^{tr}$ for the tilted Weyl
metals in Eq.(88) in the Appendix C. It is easy to check that it depends on
the direction of $\mathbf{k}$ for finite tilting. In this case, the
integration equation (40) for the renormalized current vertex is intricate and
does not have the simple solution as Eq.(41). On the other hand, the band-
diagonal vertex correction matrix ${\cal D}_{0}$ we obtained in the spin basis
for tilted Weyl metals is no longer diagonal in the spatial block since ${\cal
I}^{s}$ is not diagonal, indicating the complication of the vertex correction
for anisotropic systems.

For the untilted Weyl metals, from our calculation in the last section, ${\cal
D}_{0}$ is diagonal and ${\cal D}_{0}(i,i)=3/2$ for $i=x,y,z$, which is the
same as $\tilde{\alpha}=\tau^{tr}/\tau^{+}$ for this system
($1/\tau^{tr}=n_{i}V^{2}_{0}\epsilon^{2}_{F}/3\pi
v^{3},1/\tau^{+}=n_{i}V^{2}_{0}\epsilon^{2}_{F}/2\pi v^{3}$ for untilted Weyl
metals). For the tilted Weyl metals, ${\cal D}_{0}$ contains both diagonal and
off-diagonal elements, and both play a role in the vertex correction as we
will see below.

###  III.3 C. Separation of the intrinsic, side jump and skew scattering
contribution in the tilted Weyl metals

For the tilted Weyl metals, from the formalism in Sec. III.A and the matrices
${\cal D}_{0},{\cal I}^{a}_{int},{\cal I}^{a}_{im}$ we obtained in Sec.II, we
get the response functions for the three different mechanisms from a single
valley as

$\displaystyle\tilde{\Pi}^{int}(\omega)$ $\displaystyle=$
$\displaystyle\mathcal{I}^{a}_{int}(\omega),$ (44)
$\displaystyle\tilde{\Pi}^{sj}(\omega)$ $\displaystyle=$
$\displaystyle\frac{2c_{2}(\omega)}{c_{0}-c_{2}(\omega)}\mathcal{I}^{a}_{int}(\omega)+\frac{c_{0}}{c_{0}-c_{2}(\omega)}\mathcal{I}^{a}_{im}(\omega),$
(45) $\displaystyle\tilde{\Pi}^{sk}(\omega)$ $\displaystyle=$
$\displaystyle[\frac{c_{2}(\omega)}{c_{0}-c_{2}(\omega)}]^{2}\mathcal{I}^{a}_{int}(\omega)+\frac{c_{0}c_{2}(\omega)}{[c_{0}-c_{2}(\omega)]^{2}}\mathcal{I}^{a}_{im}(\omega).$

Taking the limit $\omega\to 0$, we get the response functions in the dc limit
as

$\displaystyle\tilde{\Pi}^{int}_{\alpha\beta}=\frac{\epsilon_{F}}{4\pi
v^{4}}[\frac{v^{2}}{v^{2}-u^{2}}-a(u)]\epsilon^{\alpha\beta\gamma}u_{\gamma},$
(47) $\displaystyle\tilde{\Pi}^{sj}_{\alpha\beta}=\frac{\epsilon_{F}}{8\pi
v^{4}}\frac{c_{0}}{(c_{0}-c_{2})}[1+a(u)+\frac{u^{2}}{v^{2}}a(u)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \
+4\frac{c_{2}}{c_{0}}(\frac{v^{2}}{v^{2}-u^{2}}-a(u))]\epsilon^{\alpha\beta\gamma}u_{\gamma},$
(48) $\displaystyle\tilde{\Pi}^{sk}_{\alpha\beta}=\frac{\epsilon_{F}}{8\pi
v^{4}}\frac{c_{0}c_{2}}{(c_{0}-c_{2})^{2}}[1+a(u)+\frac{u^{2}}{v^{2}}a(u)$
$\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+2\frac{c_{2}}{c_{0}}(\frac{v^{2}}{v^{2}-u^{2}}-a(u))]\epsilon^{\alpha\beta\gamma}u_{\gamma}.$
(49)

From above, we see that the final results of the response functions only
depend on the parameters $c_{0},c_{2}$ and $a(u)$ obtained in the dc limit in
Sec.II.

Figure 5: The Fermi surface contribution of the anomalous Hall conductivity
due to the three mechanisms: intrinsic, side jump and skew scattering,
obtained from the Kubo-Streda formula as a function of $u/v$, when the
electric field $\mathbf{E}$ is in the $y$ direction and $\mathbf{u}$ is in the
$z$ direction.

In the above calculation we noticed that though the tilting ${\mathbf{u}}$
brings correction to both the diagonal and off-diagonal elements of the ${\cal
D}_{0}$ matrix compared to the untilted case, the effects on the Hall currents
due to the off-diagonal elements of ${\cal D}_{0}$ were canceled by the
effects from part of the diagonal elements. The total effect of the matrix
${\cal D}_{0}$ on the Hall current is equivalent to the factor
$c_{0}/(c_{0}-c_{2})\approx 3/2-9u^{2}/10v^{2}$. However, this factor is
different from $\tilde{\alpha}=\tau^{tr}/\tau^{+}$ for tilted Weyl metals,
which is $\tilde{\alpha}\approx 3/2-3u/2v$. This also indicates that for
tilted Weyl metals with anisotropic Fermi surface, the simple solution
$\tilde{\alpha}=\tau^{tr}/\tau^{+}$ for the vertex correction is no longer
accurate.

From the response function in Eqs.(47)-(49), we can get the dc anomalous Hall
conductivity
$\sigma^{i}_{\alpha\beta}=\frac{e^{2}v^{2}}{\pi}\tilde{\Pi}^{i}_{\alpha\beta},i={\rm
int,sj,sk}$. The anomalous Hall currents from the three different mechanisms
are all proportional to $\mathbf{u\times E}$ and the total anomalous Hall
current from one valley is the same as obtained in Eq.(34). In Fig.5, we plot
the anomalous Hall conductivity due to the three different mechanisms from the
Fermi surface. We can see that in all regimes of $u/v$, the side jump
contribution is the largest and the extrinsic contribution is greater than the
intrinsic contribution.

At small $u/v$, we can expand the results in Eqs.(47)-(49) and get the leading
order Hall currents for the two valleys as

$\displaystyle j^{int}_{H}$ $\displaystyle\approx$
$\displaystyle-\frac{1}{3}\frac{e^{2}\epsilon_{F}}{\pi^{2}v^{2}}\mathbf{u\times
E},$ (50) $\displaystyle j^{sj}_{H}$ $\displaystyle\approx$
$\displaystyle-\frac{5}{6}\frac{e^{2}\epsilon_{F}}{\pi^{2}v^{2}}\mathbf{u\times
E},$ (51) $\displaystyle j^{sk}_{H}$ $\displaystyle\approx$
$\displaystyle-\frac{1}{3}\frac{e^{2}\epsilon_{F}}{\pi^{2}v^{2}}\mathbf{u\times
E}.$ (52)

There is another contribution to the intrinsic anomalous Hall current due to
tilting from the Fermi sea. This contribution was calculated in Ref. Pesin2017
. In the dc limit, this Hall current is $j^{int,{\rm
II}}_{H}\approx\frac{1}{6}\frac{e^{2}\epsilon_{F}}{\pi^{2}v^{2}}\mathbf{u\times
E}$ for the two valleys, which is half of the intrinsic Hall current from the
Fermi surface and has opposite sign. The total intrinsic Hall current in the
leading order of $u/v$ due to tilting is then $j^{int,{\rm
tot}}_{H}\approx-\frac{1}{6}\frac{e^{2}\epsilon_{F}}{\pi^{2}v^{2}}\mathbf{u\times
E}$.

###  III.4 D. Comparison with the semiclassical Boltzmann equation approach

The AHE in the tilted Weyl metals due to the three different mechanisms was
calculated from the SBE approach in Ref.Fu2021 . However, in this work, the
authors neglected the anisotropy of the Fermi surface of the system when
computing the transverse coordinate shift, which results in an incorrect side
jump velocity. For this reason, we redid the calculation of the SBE approach
for the AHE of tilted Weyl metals in the Appendix C.

Another issue of the SBE approach is that the commonly used solution of the
SBE approach under the relaxation-time approximation in Refs. Sinitsyn2007 ;
Loss2003  may become unreliable for anisotropic system, as pointed out in
Refs. Sinova2009 ; Sinova2009-2 . The reason is that the solution of the
nonequilibrium distribution function $g_{\mathbf{k}}$ in the SBE approach
assumes that the relaxation times $\tau^{tr}$ and $\tau^{\bot}$ do not depend
on the direction of the momentum $\mathbf{k}$ (see Appendix C). This is true
for isotropic systems but not the case for anisotropic systems. Strictly
speaking, no scalar relaxation time can be attributed to a given $\mathbf{k}$
state for anisotropic systems. The same problem comes up for the solution of
the anomalous distribution function $g^{a}_{\mathbf{k}}$ due to the coordinate
shift. In Ref.Sinova2009  the authors studied the anisotropic
magnetoresistance (AMR) in 2D Rashba ferromagnets with anisotropic magnetic
impurities. It was shown that the exact result of the AMR in the anisotropic
system is significantly different from the result obtained from the SBE under
the commonly used relaxation time approximation. For this reason, it is
interesting to check whether the AHE from the commonly used SBE approach also
deviates significantly from the result obtained from the quantum Kubo-Streda
formula for tilted Weyl metals, whose Fermi surface is anisotropic.

As shown in Appendix C, we get the AHE in the leading order of $u/v$ from the
SBE approach under the relaxation time approximation after correcting the side
jump velocity in Ref.Fu2021  as

$\displaystyle j^{int}_{H}$ $\displaystyle\approx$
$\displaystyle-\frac{1}{6}\frac{e^{2}\epsilon_{F}}{\pi^{2}v^{2}}\mathbf{u\times
E},$ (53) $\displaystyle j^{sj}_{H}$ $\displaystyle\approx$
$\displaystyle-\frac{5e^{2}\epsilon_{F}}{6\pi^{2}v^{2}}\mathbf{u\times E},$
(54) $\displaystyle j^{sk}_{H}$ $\displaystyle\approx$
$\displaystyle-\frac{e^{2}\epsilon_{F}}{3\pi^{2}v^{2}}\mathbf{u\times E}.$
(55)

The intrinsic anomalous Hall current from the SBE approach includes
contribution from both the Fermi sea and the Fermi surface MacDonald2006 .
This part is the same as the total intrinsic anomalous Hall current we
obtained from the quantum Kubo-Streda formula in the last section. Moreover,
the side jump and skew scattering contributions we obtained from the SBE
approach also agree with the result from the quantum Kubo-Streda formula in
the leading order of $u/v$. The reason for this full match between the two
approaches is because the relaxation time for tilted Weyl metals defined in
Eq.(88)-(90) is a constant independent of the momentum in the zeroth order of
tilting and depends on the momentum only at higher orders of $u/v$. For this
reason, the solution of the SBE under the relaxation time approximation is
still valid in the leading order of $u/v$ in tilted Weyl metals, and the AHEs
obtained from the SBE and Kubo-Streda formula agree with each other in the
leading order of $u/v$.

##  IV IV. Discussion

For some of the anisotropic systems, it is still possible to solve the SBE
exactly by introducing momentum dependent relaxation time, as shown in Ref.
Sinova2009 . However, the difficulty of solving the SBE in this way is greatly
enhanced because this approach involves solving two integral equations of the
distribution function and there is no general solution for different models.
On the other hand, one encounters a similar problem for anisotropic systems
when solving the vertex correction of the current operator from the recursion
equation in the chiral basis, as shown in Sec. III.B. For anisotropic systems,
the only convenient approach to get the rigorous anomalous Hall current for
Gaussian disordered systems is then to apply the Kubo-Streda formula in the
spin basis, since one can solve the recursion equation for the vertex
correction in this basis exactly. Our scheme in this work to separate the
contributions from the three different mechanisms in the spin basis of the
Kubo-Streda formula is then especially important for anisotropic systems.

Figure 6: The Feynman diagrams for the response function
$\Pi^{\mathbf{I}}_{\alpha\beta}$ due to the third order impurity scatterings.
The thick solid lines are the Green’s function in the spin basis under the
Born approximation.

Though we mainly focus on the Gaussian disorder in this work, we also have a
brief discussion of the skew scattering contribution due to third order
impurity scatterings because this contribution is inversely proportional to
the impurity density and may become significant when the impurity density is
very dilute. This contribution requires a third order correlation of the
impurity potential and the corresponding Feynman diagrams are shown in Fig.6.
We computed this contribution to the AHE in tilted Weyl metals from the Kubo-
Streda formula, as shown in Appendix D, and found that the result obtained
from the Kubo-Streda formula for this contribution is also the same as the
result obtained from the SBE under the relaxation time approximation in the
leading order of $u/v$.

There has also been an awareness for a long time that the fourth-order crossed
diagrams may give significant contribution to the AHE Sinitsyn2007 ;
Sinitsyn2006 . However, due to the intricacy of the calculation of these
diagrams, only recently Ado et al. computed the contributions of these
diagrams for 2D Rashba ferromagnets and found that they become important for
near impurities with distance comparable to the Fermi wavelength of the
electrons Ado2016 . We will present the study of these diagrams for tilted
Weyl metals in a different paper.

##  V IV. Summary

To sum up, we studied the anomalous Hall effect in disordered type-I Weyl
metals with finite tilting in the Kubo-Streda formalism in the spin basis. We
developed an efficient and transparent scheme in this basis to separate the
Hall current from the three different mechanisms: intrinsic, side jump and
skew scattering. This scheme is applicable for general relativistic systems,
such as Weyl and Dirac models, both isotropic and anisotropic. We compared the
anomalous Hall current for tilted Weyl metals obtained in this way with the
results from the SBE approach and found that in the leading order of the
tilting velocity, the results from the two approaches agree well with each
other. Our scheme is especially important for studying the AHE in anisotropic
systems since both the SBE approach and the Kubo-Streda formula in the chiral
basis encounter difficulty in studying the AHE for such systems.

##  VI Acknowledgements

We thank Rui Wang for helpful discussions. This work is supported by the
National Natural Science Foundation of China under Grant No. 11974166.

##  VII Appendix

###  VII.1 A. Calculation of the ${\cal I}$ matrix

In this appendix, we show the calculation of the anti-symmetric part of the
polarization operator ${\cal I}$. The calculation of the symmetric part is
similar. Since the final result of the dc anomalous Hall conductivity only
depends on the parameters of the ${\cal I}$ matrix in the dc limit, for
simplicity we only show the calculation in the dc limit in this appendix.

The polarization operator in the dc limit is

${\cal I}_{\alpha\beta}(\omega\to 0,\mathbf{q}\to
0)=\frac{1}{2}\int\frac{d^{3}{\mathbf{k}}}{(2\pi)^{3}}\rm
Tr[\sigma^{\alpha}G^{R}(\epsilon+\omega,\mathbf{k}+\mathbf{q})\sigma^{\beta}G^{A}(\epsilon,\mathbf{k})].$
(56)

The Green’s function of the tilted Weyl metals under the first Born
Approximation is

$G^{R/A}(\omega\to 0,{\mathbf{q}\to
0})=\frac{1}{(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk\pm\frac{i}{2\tau^{+}})(\epsilon-\mathbf{u}\cdot\mathbf{k}+vk\pm\frac{i}{2\tau^{-}})}[(\epsilon\pm\frac{i}{2\tau}-\mathbf{u}\cdot\mathbf{k})\sigma^{0}+v\mathbf{k}\cdot\mathbf{\bm{\sigma}}\mp\frac{i}{2\tau}(\mathbf{\Delta}\cdot\mathbf{\bm{\sigma}})],$
(57)

where ${\mathbf{\Delta}}=-{\mathbf{u}}/v$, $\tau$ is given in the main text
and
$1/\tau^{\pm}=\frac{1}{\tau}(1\pm\frac{{\mathbf{k}}\cdot{\mathbf{\Delta}}}{k})$.

The numerator of the integrand of ${\cal I}_{\alpha\beta}$ is

$N=\frac{1}{2}\rm
Tr\\{\sigma^{\alpha}[(\epsilon+\frac{i}{2\tau}-\mathbf{u}\cdot\mathbf{k})\sigma_{0}+v\mathbf{k}\cdot\mathbf{\bm{\sigma}}-\frac{i}{2\tau}(\mathbf{\Delta}\cdot\mathbf{\bm{\sigma}})]\sigma^{\beta}[(\epsilon-\frac{i}{2\tau}-\mathbf{u}\cdot\mathbf{k})\sigma_{0}+v\mathbf{k}\cdot\mathbf{\bm{\sigma}}+\frac{i}{2\tau}(\mathbf{\Delta}\cdot\mathbf{\bm{\sigma}})]\\}.$
(58)

We separate its antisymmetric part into two parts:

$(1):\ \frac{1}{2}v\ \rm
Tr[(\epsilon+\frac{i}{2\tau}-\mathbf{u}\cdot\mathbf{\bm{k}})\sigma^{\alpha}\sigma^{\beta}(\mathbf{k}\cdot\mathbf{\bm{\sigma}})+(\epsilon-\frac{i}{2\tau}-\mathbf{u}\cdot\mathbf{k})\sigma^{\alpha}(\mathbf{k}\cdot\mathbf{\bm{\sigma}})\sigma^{\beta}],$
(59)

$(2):\ \frac{1}{2}(\frac{i}{2\tau})\ \rm
Tr[(\epsilon+\frac{i}{2\tau}-\mathbf{u}\cdot\mathbf{k})\sigma^{\alpha}\sigma^{\beta}(\mathbf{\Delta}\cdot\mathbf{\bm{\sigma}})-(\epsilon-\frac{i}{2\tau}-\mathbf{u}\cdot\mathbf{k})\sigma^{\alpha}(\mathbf{\Delta}\cdot\mathbf{\bm{\sigma}})\sigma^{\beta}].$
(60)

Note that the term $\rm
Tr[\sigma^{\alpha}(\mathbf{k}\cdot\mathbf{\bm{\sigma}})\sigma^{\beta}(\mathbf{\Delta}\cdot\mathbf{\bm{\sigma}})-\sigma^{\alpha}(\mathbf{\Delta}\cdot\mathbf{\bm{\sigma}})\sigma^{\beta}(\mathbf{k}\cdot\mathbf{\bm{\sigma}})]$
also contains an anti-symmetric part, but the integration over this anti-
symmetric part vanishes. Since ${\rm
Tr}(\sigma^{\alpha}\sigma^{\beta}\sigma^{\gamma})=2i\epsilon^{\alpha\beta\gamma}$,
parts (1) and (2) can be simplified as

$(1)=\frac{1}{2}v\times\frac{i}{\tau}k^{\gamma}Tr(\sigma^{\alpha}\sigma^{\beta}\sigma^{\gamma})=-\frac{1}{\tau}\epsilon^{\alpha\beta\gamma}vk^{\gamma},$
(61)
$(2)=(\frac{i}{2\tau})(\epsilon-\mathbf{u}\cdot\mathbf{k})\Delta^{\gamma}Tr(\sigma^{\alpha}\sigma^{\beta}\sigma^{\gamma})=-\frac{1}{\tau}(\epsilon-\mathbf{u}\cdot\mathbf{k})\epsilon^{\alpha\beta\gamma}\Delta^{\gamma}.$
(62)

The denominator of the integrand of ${\cal I}$ is

$\displaystyle D$ $\displaystyle=$
$\displaystyle\frac{1}{(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk+\frac{i}{2\tau^{+}})(\epsilon-\mathbf{u}\cdot\mathbf{k}+vk+\frac{i}{2\tau^{-}})}\frac{1}{(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk-\frac{i}{2\tau^{+}})(\epsilon-\mathbf{u}\cdot\mathbf{k}+vk-\frac{i}{2\tau^{-}})}$
(63) $\displaystyle=$
$\displaystyle\frac{1}{(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk)^{2}+(\frac{1}{2\tau^{+}})^{2}}\frac{1}{(\epsilon-\mathbf{u}\cdot\mathbf{k}+vk)^{2}+(\frac{1}{2\tau^{-}})^{2}}.$

For $\epsilon_{F}\tau\gg 1$, $1/\tau^{-}$ is negligible,

$D\approx
2\pi\tau^{+}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk)\times\frac{1}{4v^{2}k^{2}}.$
(64)

The total anti-symmetric part of the ${\cal I}$ matrix is then

$\displaystyle\mathcal{I}^{a}$
$\displaystyle=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{4v^{2}k^{2}}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk)\times
2\pi\tau^{+}(-\frac{1}{\tau})\epsilon^{\alpha\beta\gamma}[vk^{\gamma}+(\epsilon-\mathbf{u}\cdot\mathbf{k})\Delta^{\gamma}]$
$\displaystyle=-\frac{1}{(2\pi)^{2}}\epsilon^{\alpha\beta\gamma}\int d^{3}k\
\frac{1}{4vk^{2}}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk)\frac{\tau^{+}}{\tau}(k^{\gamma}+k\Delta^{\gamma})$
$\displaystyle=-\frac{1}{(2\pi)^{2}}\epsilon^{\alpha\beta\gamma}\int d^{3}k\
\frac{1}{4vk^{2}}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk)\frac{1}{1+(\mathbf{k}\cdot\mathbf{\Delta}/k)}(k^{\gamma}+k\Delta^{\gamma})$
(65)

To do the above integration, we rotate the $z$-axis to the direction of the
vector $\mathbf{u}$. Suppose
$\mathbf{u}=u(\sin\alpha\cos\beta\cdot\hat{\mathbf{x}}+\sin\alpha\sin\beta\cdot\hat{\mathbf{y}}+\cos\alpha\cdot\hat{\mathbf{z}})$
in the old coordinate. The rotation to the new coordinate is

$\left(\begin{array}[]{c}\hat{\mathbf{x}^{\prime}}\\\
\hat{\mathbf{y}^{\prime}}\\\
\hat{\mathbf{z}^{\prime}}\end{array}\right)=\left(\begin{array}[]{ccc}\cos\alpha\cos\beta&\cos\alpha\sin\beta&-\sin\alpha\\\
-\sin\beta&\cos\beta&0\\\
\sin\alpha\cos\beta&\sin\alpha\sin\beta&\cos\alpha\end{array}\right)\
\left(\begin{array}[]{c}\hat{\mathbf{x}}\\\ \hat{\mathbf{y}}\\\
\hat{\mathbf{z}}\end{array}\right).$ (66)

Now we have $\mathbf{u}=u\hat{\mathbf{z}^{\prime}}$. Suppose $\mathbf{k}=k\
(\sin\theta\cos\phi\cdot\hat{\mathbf{x}^{\prime}}+\sin\theta\sin\phi\cdot\hat{\mathbf{y}^{\prime}}+\cos\theta\cdot\hat{\mathbf{z}^{\prime}}),\
\mathbf{u}\cdot\mathbf{k}=uk\cos\theta$ in the new coordinate. In the old
coordinate,

$\left\\{\begin{array}[]{ll}k_{x}=&k(\cos\alpha\cos\phi\sin\theta\cos\beta-\sin\theta\sin\phi\sin\beta+\cos\theta\sin\alpha\cos\beta),\\\
k_{y}=&k(\cos\alpha\cos\phi\sin\theta\sin\beta+\sin\theta\sin\phi\cos\beta+\cos\theta\sin\alpha\sin\beta),\\\
k_{z}=&k(-\sin\theta\cos\phi\sin\alpha+\cos\theta\cos\alpha).\\\
\end{array}\right.\\\ $ (67)

We then have $\int_{0}^{2\pi}d\phi\ k^{\gamma}=2\pi
k\cos\theta\frac{u^{\gamma}}{u},\ \delta(\epsilon-\mathbf{u\cdot
k}-vk)=\frac{1}{u\cos\theta+v}\delta(k-\frac{\epsilon}{u\cos\theta+v})$.

The intrinsic part of ${\cal I}^{a}$ is obtained by setting $\tau\to\infty$
and $\Delta=0$. We then get

$\displaystyle\mathcal{I}_{int}^{a}$
$\displaystyle=-\lim_{\tau\rightarrow\infty}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{4v^{2}k^{2}}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk)\times
2\pi\frac{\tau^{+}}{\tau}\epsilon^{\alpha\beta\gamma}vk^{\gamma}$
$\displaystyle=-\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{4v^{2}k^{2}}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk)\times
2\pi\epsilon^{\alpha\beta\gamma}vk^{\gamma}$
$\displaystyle=-\frac{\epsilon^{\alpha\beta\gamma}}{2\pi}\frac{u^{\gamma}}{u}\int_{0}^{\infty}k^{2}dk\int_{0}^{\pi}\sin\theta
d\theta
k\cos\theta\frac{1}{u\cos\theta+v}\delta(k-\frac{\epsilon}{u\cos\theta+v})\frac{1}{4vk^{2}}$
$\displaystyle=\frac{\epsilon_{F}}{4v^{4}\pi}\epsilon^{\alpha\beta\gamma}u^{\gamma}[\frac{v^{2}}{v^{2}-u^{2}}-a(u)],$
(68)

where
$a(u)=\frac{v^{3}}{2u^{3}}\ln\frac{v+u}{v-u}-\frac{v^{2}}{u^{2}}=\frac{1}{3}+{\cal
O}(u^{2}/v^{2})$.

The remaining part of ${\cal I}^{a}$ is ${\cal I}^{a}_{im}$, i.e.,

$\displaystyle\mathcal{I}_{im}^{a}$ $\displaystyle=$
$\displaystyle\mathcal{I}^{a}-\mathcal{I}_{int}^{a}=-\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{4v^{2}k^{2}}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}-vk)\times
2\pi\epsilon^{\alpha\beta\gamma}[(\frac{\tau^{+}}{\tau}-1)vk^{\gamma}+\frac{\tau^{+}}{\tau}(\epsilon-\mathbf{u}\cdot\mathbf{k})\Delta^{\gamma}]$
(69) $\displaystyle=$ $\displaystyle-\frac{\epsilon}{8\pi
v^{3}}\epsilon^{\alpha\beta\gamma}\Delta_{\gamma}[1+a(u)+\frac{u^{2}}{v^{2}}a(u)].$

We can compute the symmetric part of the ${\cal I}$ matrix similarly. The
resulting ${\cal I}^{s}$ matrix is shown in the main text.

###  VII.2 B. Connection between the ${\cal I}$ matrix in the spin basis and
diagrams in the chiral basis

In this appendix, we show the correspondence between the symmetric and anti-
symmetric parts of the ${\cal I}$ matrix in the spin basis and the Feynman
diagrams in the chiral basis.

The integrand $I_{\alpha\beta}$ of the ${\cal I}$ matrix in the band
eigenstate basis is expanded as in Eq.(III.1) in the main text. We denote the
elements of the GF in the eigenstate basis under the Born approximation as
$\langle+|G|+\rangle=G^{+},\langle-|G|-\rangle=G^{-},\langle+|G|-\rangle=G^{+-},\langle-|G|+\rangle=G^{-+}$,
and the same notation for $\hat{j}_{\alpha}$.

The first two terms of $I_{\alpha\beta}$ in Eq.(III.1) are

$I^{s}_{\alpha\beta}=j^{++}_{\alpha}G^{R+}j^{++}_{\beta}G^{A+}+j^{--}_{\alpha}G^{R-}j^{--}_{\beta}G^{A-}.$
(70)

It is obvious that $I^{s}_{\alpha\beta}$ is symmetric under the exchange of
$\alpha$ and $\beta$. Since $G^{R+}G^{A+}\approx
2\pi\tau^{+}\delta(\epsilon-\epsilon_{+})$ and $G^{R-}G^{A-}\approx
2\pi\tau^{-}\delta(\epsilon-\epsilon_{-})$, for Fermi energy $\epsilon>0$,
$G^{R-}G^{A-}\sim 0$. The dominant contribution of ${\cal
I}_{\alpha\beta}^{s}$ then only contains the first term with upper band
scattering. The symmetric matrix ${\cal I}_{s}$ then corresponds to the
diagram in Fig.3(a).

The next two terms in Eq.(III.1) are
$I^{(2)}_{\alpha\beta}=j^{+-}_{\alpha}G^{R-}j^{-+}_{\beta}G^{A+}+j^{-+}_{\alpha}G^{R+}j^{+-}_{\beta}G^{A-}$.
We show below that its dominant part is anti-symmetric. Since
$j^{+-}_{\alpha}=(j^{-+}_{\alpha})^{*},G^{R-}G^{A+}=(G^{R+}G^{A-})^{*}$, one
can write

$G^{R-}G^{A+}=X+iY,\ G^{R+}G^{A-}=X-iY,\ j^{+-}_{\alpha}=\rm Re\
j^{+-}_{\alpha}+i\ Im\ j^{+-}_{\alpha},\ j^{-+}_{\beta}=\ Re\
j^{-+}_{\beta}+i\ \rm Im\ j^{-+}_{\beta},$ (71)

and

$\displaystyle\ \ \
j^{+-}_{\alpha}G^{R-}j^{-+}_{\beta}G^{A+}+j^{-+}_{\alpha}G^{R+}j^{+-}_{\beta}G^{A-}$
$\displaystyle=(\rm Re\ j^{+-}_{\alpha}+i\ Im\ j^{+-}_{\alpha})(Re\
j^{-+}_{\beta}+i\ Im\ j^{-+}_{\beta})(X+iY)+(Re\ j^{+-}_{\alpha}-i\ Im\
j^{+-}_{\alpha})(Re\ j^{-+}_{\beta}-i\ Im\ j^{-+}_{\beta})(X-iY)$
$\displaystyle=2X(\rm Re\ j_{\alpha}^{+-}Re\ j^{-+}_{\beta}-Im\
j_{\alpha}^{+-}Im\ j^{-+}_{\beta})-2Y(Im\ j_{\alpha}^{+-}Re\
j^{-+}_{\beta}+Re\ j_{\alpha}^{+-}Im\ j^{-+}_{\beta}).$ (72)

Since $\rm Re\ j^{-+}_{\alpha}=Re\ j^{+-}_{\alpha},Im\ j^{-+}_{\alpha}=-Im\
j^{+-}_{\alpha}$, the first term in Eq.(VII.2) is symmetric and the second
term is anti-symmetric. However, since $X=Re[G^{R+}G^{A-}]$ is smaller than
$G^{R+}G^{A+}$ by a factor $1/\epsilon_{F}\tau$, the symmetric part of this
equation is negligible compared to the symmetric part in ${\cal I}^{s}$. We
then only need to keep the anti-symmetric part $-2{\rm Im}[G^{R+}G^{A-}]({\rm
Im}\ j_{\alpha}^{+-}{\rm Re}\ j^{-+}_{\beta}+{\rm Re}\ j_{\alpha}^{+-}{\rm
Im}\ j^{-+}_{\beta})$ of $I^{(2)}_{\alpha\beta}$.

In $I^{(2)}_{\alpha\beta}$, we can expand
$G^{R/A}=G_{0}^{R/A}+G_{0}^{R/A}\Sigma^{R/A}G^{R/A}$. Since the self-energy
$\Sigma$ introduces an extra small parameter $1/\tau$ which cannot be
compensated by another $\tau$, the dominant contribution of
$I^{(2)}_{\alpha\beta}$ is equal to the integration of
$j^{+-}_{\alpha}G_{0}^{R-}j^{-+}_{\beta}G_{0}^{A+}+j^{-+}_{\alpha}G_{0}^{R+}j^{+-}_{\beta}G_{0}^{A-}$.
Since this is the only nonvanishing antisymmetric part in the clean limit, it
is equal to ${\cal I}^{a}_{int}$ in the spin basis, and corresponds to the two
diagrams in the chiral basis in Fig.3c in the main text.

Among the remaining eight terms in $I_{\alpha\beta}$, the four terms with
$G^{A-}$ are smaller in $1/\epsilon_{F}\tau$ and so are neglected. The
remaining four terms contain $G^{+-}$ or $G^{-+}$ and $G^{+}$. One can expand
$G^{+-}=G^{+-}_{0}+G_{0}^{+}\Sigma G^{-}\approx G^{+-}_{0}+G_{0}^{+}\Sigma
G_{0}^{-}$. It is easy to check that the replacement in the last equation does
not change the dominant contribution of the integration over
$I_{\alpha\beta}$. Since $G^{+-}_{0}=0$ in the eigenstate basis, and the
dominant contribution can have only one $G^{-}$ line Sinitsyn2007 , the
dominant four terms with $G^{+-}$ or $G^{-+}$ correspond to the four diagrams
in Fig.3(d).

The integrand of the four diagrams in Fig.3(d) can be written as

$\displaystyle I^{(3)}_{\alpha\beta}$
$\displaystyle=j_{\alpha}^{++}j_{\beta}^{-+}G_{0}^{R+}\Sigma^{R}G_{0}^{R-}G^{A+}+j_{\beta}^{++}j_{\alpha}^{-+}G_{0}^{A+}\Sigma^{A}G_{0}^{A-}G^{R+}$
(73) $\displaystyle+\
j_{\alpha}^{++}j_{\beta}^{+-}G^{R+}G_{0}^{A-}\Sigma^{A}G_{0}^{A+}+j_{\beta}^{++}j_{\alpha}^{+-}G^{A+}G_{0}^{R-}\Sigma^{R}G_{0}^{R+}.$

Since
$G_{0}^{R+}\Sigma^{R}G_{0}^{R-}G^{A+}=(G_{0}^{A+}\Sigma^{A}G_{0}^{A-}G^{R+})^{*},G^{R+}G_{0}^{A-}\Sigma^{A}G_{0}^{A+}=(G^{A+}G_{0}^{R-}\Sigma^{R}G_{0}^{R+})^{*}$,
the symmetric part of $I^{(3)}_{\alpha\beta}$ is

$I^{3s}_{\alpha\beta}=(j_{\alpha}^{++}j_{\beta}^{-+}+j_{\beta}^{++}j_{\alpha}^{-+}){\rm
Re}(G_{0}^{R+}\Sigma^{R}G_{0}^{R-}G^{A+})+(j_{\alpha}^{++}j_{\beta}^{+-}+j_{\beta}^{++}j_{\alpha}^{+-}){\rm
Re}(G^{R+}G_{0}^{A-}\Sigma^{A}G_{0}^{A+}),$ (74)

and the antisymmetric part is

$I^{3a}_{\alpha\beta}=i(j_{\alpha}^{++}j_{\beta}^{-+}-j_{\beta}^{++}j_{\alpha}^{-+}){\rm
Im}(G_{0}^{R+}\Sigma^{R}G_{0}^{R-}G^{A+})+i(j_{\alpha}^{++}j_{\beta}^{+-}-j_{\beta}^{++}j_{\alpha}^{+-}){\rm
Im}(G^{R+}G_{0}^{A-}\Sigma^{A}G_{0}^{A+}).$ (75)

However, the symmetric part $I^{3s}_{\alpha\beta}$ is smaller than the
symmetric part $I^{s}_{\alpha\beta}$ by a factor $1/\epsilon_{F}\tau$. So we
only need to keep the antisymmetric part of $I^{(3)}_{\alpha\beta}$, which is
equal to $I^{a}_{im}$ obtained in the spin basis since this antisymmetric part
is nonvanishing only with impurity scattering.

###  VII.3 C.The intrinsic, side jump and skew scattering contributions from
the SBE approach under the relaxation time approximation

In this appendix, we redo the calculation of the intrinsic, side jump and skew
scattering contributions in tilted Weyl metals from the SBE approach under the
relaxation time approximation (RTA) in Ref.Fu2021 .

The Hamiltonian of the three-dimensional (3D) Weyl metal in each valley in our
main text is $H_{\chi}=\chi
v_{F}\mathbf{\bm{\sigma}}\cdot\mathbf{k}+\mathbf{u}_{\chi}\cdot\mathbf{k}$,
with $\chi=\pm 1$ and ${\mathbf{u}}_{+}=-{\mathbf{u}}_{-}$. As we show in the
main text, in this case the contributions to the AHE in the two valleys add up
instead of cancel out. In the following, we only need to compute the different
contributions to the AHE for a single valley and double the result at the end
for two valleys.

The two eigenstates of $H_{\chi}$ with $\chi=+1$ are

$\left|u^{+}_{k}\rangle\right.=\left(\begin{array}[]{c}\cos(\theta/2)\\\
\sin(\theta/2)e^{i\phi}\end{array}\right),\left|u^{-}_{k}\rangle\right.=\left(\begin{array}[]{c}\sin(\theta/2)\\\
-\cos(\theta/2)e^{i\phi}\end{array}\right),$ (76)

with $\cos\theta=k_{z}/k,\tan\phi=k_{y}/k_{x}$.

Intrinsic contribution. The intrinsic Hall conductivity for a single valley is

$\sigma^{int}_{xy}=e^{2}\int_{k<k_{F}}\frac{d^{3}k}{(2\pi)^{3}}\mathbf{\Omega}^{+}(k)=\frac{e^{2}\epsilon_{F}}{12\pi^{2}v^{2}}u,$
(77)

where the Berry curvature of the upper band is
$\Omega^{+}(k)=-\frac{{\mathbf{k}}}{2k^{3}}$. The total intrinsic contribution
from the two valleys doubles and is
$\sigma^{int}_{xy}=\frac{e^{2}\epsilon_{F}}{6\pi^{2}v^{2}}u$, consistent with
the value in Ref Fu2021 .

Side jump contribution. There are two different mechanisms for the side jump
contribution. One is directly due to the transverse coordinate shift

$\delta\mathbf{r}_{\mathbf{k},\mathbf{k}^{\prime}}=\langle
u_{k^{\prime}}^{+}|i\frac{\partial}{\partial\mathbf{k}^{\prime}}u_{k}^{+}\rangle-\langle
u_{k}^{+}|i\frac{\partial}{\partial\mathbf{k}}u_{k}^{+}\rangle-(\frac{\partial}{\partial\mathbf{k}^{\prime}}+\frac{\partial}{\partial\mathbf{k}})\arg(\langle
u^{+}_{k^{\prime}}|u^{+}_{k}\rangle).$ (78)

The other is due to the anomalous distribution function $g^{a}_{k}$ resulting
from the coordinate shift $\delta\mathbf{r}_{k,k^{\prime}}$. The two
contributions correspond to symmetric diagrams in the chiral basis and so
should have the same value Sinitsyn2007 . Both contributions come from the
second order symmetric impurity scattering $\omega^{(2)}_{k,k^{\prime}}$. In
the following, we compute the two contributions respectively from the SBE
approach.

From Eqs.(76) and (78), we get

$\delta\mathbf{r}_{\mathbf{k},\mathbf{k}^{\prime}}=\frac{1}{4}\frac{k^{\prime}+k}{k^{\prime
2}k^{2}}({\mathbf{k}^{\prime}}\times{\mathbf{k}})\times\frac{1}{|\langle
u^{+}_{k^{\prime}}|u^{+}_{k}\rangle|^{2}}.$ (79)

For systems with isotropic Fermi surface, $k=k^{\prime}$, Eq.(79) reduces to
$\delta\mathbf{r}_{k,k^{\prime}}=\frac{\mathbf{\Omega}^{+}(k)\times(\mathbf{k}^{\prime}-\mathbf{k})}{|\langle
u_{k^{\prime}}^{+}|u^{+}_{k}\rangle|^{2}}$, where
$\Omega^{+}(k)=-\frac{{\mathbf{k}}}{2k^{3}}$ is the Berry curvature of the
upper band of the tilted Weyl metals. This is the form used in Ref.Fu2021  to
compute the side jump contribution in tilted Weyl metals. However, the
anisotropy of the Fermi surface plays an important role for tilted Weyl metals
and the coordinate shift we obtained in Eq.(79) after taking into account this
anisotropy gives a different side-jump velocity from the isotropic formula.

The side-jump velocity is

$\mathbf{v}^{sj}_{\mathbf{k}}=\sum_{k^{\prime}}\omega^{(2)}_{\mathbf{k},\mathbf{k}^{\prime}}\delta\mathbf{r}_{\mathbf{k},\mathbf{k}^{\prime}},$
(80)

where the second order symmetric scattering rate
$\omega^{(2)}_{\mathbf{k},\mathbf{k}^{\prime}}=2\pi|V_{\mathbf{k},\mathbf{k}^{\prime}}|^{2}\delta(\epsilon_{k}-\epsilon_{k^{\prime}})$
with $|V_{\mathbf{k},\mathbf{k}^{\prime}}|^{2}=n_{i}V_{0}^{2}|\langle
u^{+}_{k^{\prime}}|u^{+}_{k}\rangle|^{2}$. The side jump velocity is then

$\mathbf{v}_{\mathbf{k}}^{sj}=-\frac{1}{4}\frac{\mathbf{k}}{k^{2}}\times 2\pi
n_{i}V_{0}^{2}\sum_{k^{\prime}}\delta(\epsilon_{k}-\epsilon_{k^{\prime}})\frac{k+k^{\prime}}{k^{\prime
2}}\mathbf{k}^{\prime}.$ (81)

For simplicity, we assume the tilting in the $z$ direction, i.e.
${\mathbf{u}}=(0,0,u)$. We then get

$\displaystyle\mathbf{v}_{\mathbf{k}}^{sj}$ $\displaystyle=$
$\displaystyle-\frac{1}{4}\frac{\mathbf{k}\times\hat{z}}{k^{2}}\times 2\pi
n_{i}V_{0}^{2}\frac{1}{(2\pi)^{3}}\int_{0}^{\infty}dk^{\prime}\int_{0}^{\pi}\sin\theta^{\prime}d\theta^{\prime}\int_{0}^{2\pi}d\phi^{\prime}\delta(\epsilon_{k}-\epsilon_{k^{\prime}})(k+k^{\prime})k^{\prime}_{z}$
(82) $\displaystyle\approx$
$\displaystyle\frac{5n_{i}V_{0}^{2}}{12\pi}\frac{u}{v^{2}}(\mathbf{k}\times\hat{z}).$
(83)

In the last line, we have only kept the leading order of $u/v$. The above
result does not depend on whether $\mathbf{u}$ is in the z direction, so
$\mathbf{v}_{k}^{sj}\sim\mathbf{k}\times\mathbf{u}$. Note that neglecting the
anisotropy of the Fermi surface of tilted Weyl metals at the calculation of
$\delta\mathbf{r}_{k,k^{\prime}}$ will result in an incorrect results
$\mathbf{v}_{k}^{sj}=\frac{2n_{i}V_{0}^{2}u}{3\pi
v^{2}}(\mathbf{k}\times\mathbf{u})$.

We separate the non-equilibrium distribution function as
$f(\epsilon_{\mathbf{k}})=f_{0}(\epsilon_{\mathbf{k}})+g_{\mathbf{k}}+g^{a}_{\mathbf{k}}$,
where $f_{0}$ is the equilibrium distribution, $g_{\mathbf{k}}$ is the usual
non-equilibrium distribution without considering the coordinate shift and
$g^{a}_{\mathbf{k}}$ is the anomalous part due to coordinate shift. The non-
equilibrium parts $g_{\mathbf{k}}$ and $g^{a}_{\mathbf{k}}$ satisfy the
following equations respectively Sinitsyn2007

$\displaystyle e\mathbf{E}\cdot\mathbf{v}_{\mathbf{k}}\frac{\partial
f_{0}}{\partial\epsilon^{+}_{\mathbf{k}}}=-\sum_{k^{\prime}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}(g_{\mathbf{k}}-g_{\mathbf{k}}),$
(84)
$\displaystyle\sum_{k^{\prime}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}(g_{\mathbf{k}}^{a}-g^{a}_{\mathbf{k}}+\frac{-\partial
f_{0}}{\partial\epsilon^{+}_{k}}e\mathbf{E}\cdot\delta\mathbf{r}_{\mathbf{k},\mathbf{k}^{\prime}})=0,$
(85)

where
${\mathbf{v}_{\mathbf{k}}}=\frac{\partial\epsilon_{k}^{+}}{\partial\mathbf{k}}$.

Assuming the electric field $\mathbf{E}$ is in the $xy$ plane,
$g_{\mathbf{k}}$ is then solved by the ansatz solution Loss2003

$g_{\mathbf{k}}=(-\frac{\partial
f_{0}}{\partial\epsilon_{\mathbf{k}}^{+}})e\mathbf{E}\cdot(A{\mathbf{v}_{\mathbf{k}}}+B{\mathbf{v}_{\mathbf{k}}}\times\hat{{\mathbf{u}}}),$
(86)

where $A$ and $B$ are assumed to be constant independent of the direction of
$\mathbf{k}$, and $\hat{{\mathbf{u}}}$ is the unit vector in the direction of
$\mathbf{u}$. The solution for $g_{\mathbf{k}}$ under this assumption is then

$g_{\mathbf{k}}=(-\frac{\partial
f_{0}}{\partial\epsilon_{\mathbf{k}}^{+}})e\mathbf{E}\cdot({\mathbf{v}_{\mathbf{k}}}\tau^{tr}+({\mathbf{v}_{\mathbf{k}}}\times\hat{{\mathbf{u}}})\tau^{tr}\tau_{\perp}^{tr}/\tau^{\perp}),$
(87)

where

$\displaystyle 1/\tau^{tr}$ $\displaystyle=$
$\displaystyle\sum_{k^{\prime}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}(1-\frac{\mathbf{v}_{k}\cdot\mathbf{v}_{k^{\prime}}}{|\mathbf{v}_{k}|^{2}}),$
(88) $\displaystyle 1/\tau^{\bot}$ $\displaystyle=$
$\displaystyle\sum_{k^{\prime}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}\
\frac{\mathbf{v}_{k^{\prime}}\cdot(\mathbf{v}_{k}\times\hat{{\mathbf{u}}})}{|\mathbf{v}_{k}\times\hat{{\mathbf{u}}}|^{2}},$
(89) $\displaystyle 1/\tau_{\perp}^{tr}$ $\displaystyle=$
$\displaystyle\sum_{k^{\prime}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}(1-\frac{(\mathbf{v}_{k^{\prime}}\times\hat{{\mathbf{u}}})\cdot(\mathbf{v}_{k}\times\hat{{\mathbf{u}}})}{|\mathbf{v}_{k}\times\hat{{\mathbf{u}}}|^{2}}).$
(90)

It is easy to check that for tilted Weyl metals, $\tau^{tr},\tau^{\bot}$ and
$\tau_{\perp}^{tr}$ depend on the direction of $\mathbf{k}$, contradictory
with the ansatz. However, in the leading order of $u/v$, they reduce to the
constants in untilted Weyl metals, and
$1/\tau^{tr}\approx\sum_{\mathbf{k}^{\prime}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}[1-\cos(\mathbf{k}\cdot\mathbf{k^{\prime}})]$,
$1/\tau_{\perp}^{tr}=\int\frac{d^{3}k^{\prime}}{(2\pi)^{3}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}(1-\left|\frac{v^{\perp}_{\mathbf{k}^{\prime}}}{v^{\perp}_{\mathbf{k}}}\right|\cos(\phi-\phi^{\prime}))$,
and
$1/\tau^{\bot}=\int\frac{d^{3}k^{\prime}}{(2\pi)^{3}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}\left|\frac{v^{\perp}_{\mathbf{k}^{\prime}}}{v^{\perp}_{\mathbf{k}}}\right|\sin(\phi-\phi^{\prime})$,
where $v^{\perp}_{\mathbf{k}}=\mathbf{v}_{\mathbf{k}}\times\hat{{\mathbf{u}}}$
is a vector perpendicular to both $\mathbf{v}_{\mathbf{k}}$ and
$\hat{{\mathbf{u}}}$.

Assuming $\mathbf{E}$ is in the $y$ direction, the anomalous Hall conductivity
in a single valley due to the coordinate shift is Sinitsyn2007

$\sigma^{\rm
sj,I}_{xy}=e\int\frac{d^{3}k}{(2\pi)^{3}}(g_{\mathbf{k}}/E_{y})v_{{\mathbf{k}},x}^{sj},$
(91)

where $v_{{\mathbf{k}},x}^{sj}$ is the $x$ component of the side jump velocity
$\mathbf{v}_{\mathbf{k}}^{sj}$, and $g_{\mathbf{k}}$ is solved in Eq.(87).

For the symmetric second order impurity scattering
$\omega^{(2)}_{k,k^{\prime}}$, $1/\tau^{tr}\approx
1/\tau^{tr}_{\bot}\approx\frac{\epsilon_{F}^{2}n_{i}V_{0}^{2}}{3\pi v^{3}}$ in
the leading order of $u/v$, and $1/\tau^{\bot}=0$, so the second term in
Eq.(87) has zero contribution to the side jump contribution. From Eqs.(83) and
Eq.(87), we get the side jump contribution in the leading order of $u/v$ due
to transverse coordinate shift for a single valley under the RTA as

$\sigma^{\rm sj,I}_{xy}=\frac{5e^{2}\epsilon_{F}}{24\pi^{2}v^{2}}u_{z}.$ (92)

One can calculate $\sigma^{\rm sj,I}_{yz}$ in the same way and get the side
jump current $j_{H}^{\rm
sj,I}=\frac{5e^{2}\epsilon_{F}}{24\pi^{2}v^{2}}\mathbf{E}\times\mathbf{u}$.

The second part of the side jump contribution comes from the anomalous
distribution function due to coordinate shift. The anomalous distribution
function solved from Eq.(85) under the RTA is

$g_{\mathbf{k}}^{a}=\frac{5n_{i}V_{0}^{2}u}{12\pi v^{2}}\frac{\partial
f_{0}}{\partial\epsilon^{+}_{\mathbf{k}}}[e\mathbf{E}\cdot(\mathbf{k}\times\mathbf{u})]\tau^{tr}_{\bot}.$
(93)

Here $1/\tau^{tr}_{\bot}$ is defined in Eq.(90) in the leading order of $u/v$.

The Hall conductivity in the leading order of $u/v$ due to the anomalous
distribution for a single valley under the RTA is then

$\sigma_{xy}^{\rm
sj,II}=e\int\frac{d^{3}k}{(2\pi)^{3}}(g_{k}^{a}/E_{y})v_{k,x}=\frac{5e^{2}\epsilon_{F}}{24\pi^{2}v^{2}}u_{z},$
(94)

which is equal to the contribution due to coordinate shift. The Hall current
due to this part is also $j_{H}^{\rm
sj,II}=\frac{5e^{2}\epsilon_{F}}{24\pi^{2}v^{2}}\mathbf{E}\times\mathbf{u}$.

It was pointed out in Ref.Sinitsyn2007  that the product of
$g^{a}_{\mathbf{k}}$ and $v^{sj}_{\mathbf{k}}$ may give a finite contribution
to the AHE in anisotropic systems. However, it is easy to check that under the
RTA, this contribution to the Hall current is zero in the leading order of
$u/v$.

The total side jump contribution under the RTA for two valleys is then

$j_{H}^{\rm
sj}=\frac{5e^{2}\epsilon_{F}}{6\pi^{2}v^{2}}\mathbf{E}\times\mathbf{u}.$ (95)

Skew scattering contribution. The skew scattering contribution is due to the
asymmetric fourth-order impurity scatterings. Still assuming the electric
field $\mathbf{E}$ in the $y$ direction, the Hall conductivity from the skew
scattering contribution for a single valley is Sinitsyn2007

$\sigma^{sk}_{xy}=e\sum_{\mathbf{k}}(g_{\mathbf{k}}/E_{y})v_{{\mathbf{k}},x},$
(96)

where the solution of $g_{\mathbf{k}}$ under RTA is still Eq.(87) but with the
contribution of fourth-order scattering
$\omega_{\mathbf{k},\mathbf{k}^{\prime}}^{(4)}=-\frac{2\epsilon_{F}}{3v^{4}}(n_{i}V_{0}^{2})^{2}u\sin\theta\sin\theta^{\prime}\sin(\phi-\phi^{\prime})\delta(\epsilon_{\mathbf{k}}-\epsilon_{\mathbf{k}^{\prime}})$
taken into account. The contribution of
$\omega_{\mathbf{k},\mathbf{k}^{\prime}}^{(4)}$ to $1/\tau^{tr}$ and
$1/\tau^{tr}_{\bot}$ is zero, but
$1/\tau^{\bot}=-\frac{1}{9}\frac{\epsilon^{3}_{F}u}{\pi^{2}v^{7}}(n_{i}V_{0}^{2})^{2}$
is non-zero for
$\omega_{\mathbf{k},\mathbf{k}^{\prime}}=\omega_{\mathbf{k},\mathbf{k}^{\prime}}^{(4)}$.
The skew scattering contribution under RTA is then

$\sigma^{sk}_{xy}=-\int\frac{d^{3}k}{(2\pi)^{3}}\left[(\frac{\partial
f_{0}}{\partial\epsilon_{\mathbf{k}}^{+}})e\mathbf{E}\cdot({\mathbf{v}_{\mathbf{k}}}\times\hat{{\mathbf{u}}})\tau^{tr}\tau_{\perp}^{tr}/\tau^{\perp}\right]v_{\mathbf{k},x}/E_{y},$
(97)

where the contribution to $1/\tau^{tr}$ and $1/\tau^{tr}_{\bot}$ still comes
from the second order impurity scatterings.

Putting all things together, we get the leading order total skew scattering
contribution under RTA for two valleys as

$j_{H}^{\rm
sk}=\frac{e^{2}\epsilon_{F}}{3\pi^{2}v^{2}}\mathbf{E}\times\mathbf{u}.$ (98)

which is the same as the value obtained in Ref. Fu2021 .

The sign of the intrinsic contribution we obtained is opposite to that of Ref.
Fu2021  since we use a different convention in Eq.(77). The signs of the three
contributions we obtained are then the same as the signs we obtained from the
Kubo-Streda formula.

###  VII.4 D.The skew-scattering contribution due to third order impurity
scatterings

We compute the anomalous Hall effect due to third order skew scatterings of
impurities by the Kubo-Streda formula in this appendix. For simplicity, we
only keep the leading order of $u/v$ in this section.

The diagrams corresponding to these processes are shown in Fig.6. We first
consider the response function without the vertex correction at the two ends.
The corresponding response function for the two diagrams is

$\displaystyle\Pi_{\alpha\beta}^{\rm bare}=$ $\displaystyle
e^{2}v^{2}n_{i}V_{1}^{3}\frac{i\omega}{2\pi}\sum_{\mathbf{k}_{1}}\sum_{\mathbf{k}_{2}}\sum_{\mathbf{k}_{3}}\\{{\rm
Tr}[G^{A}(\epsilon,\mathbf{k}_{1})\sigma_{\alpha}G^{R}(\epsilon,\mathbf{k}_{1})G^{R}(\epsilon,\mathbf{k}_{3})G^{R}(\epsilon,\mathbf{k}_{2})\sigma_{\beta}G^{A}(\epsilon,\mathbf{k}_{2})]$
(99) $\displaystyle\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \
+{\rm
Tr}[G^{A}(\epsilon,\mathbf{k}_{1})\sigma_{\alpha}G^{R}(\epsilon,\mathbf{k}_{1})G^{R}(\epsilon,\mathbf{k}_{2})\sigma_{\beta}G^{A}(\epsilon,\mathbf{k}_{2})G^{A}(\epsilon,\mathbf{k}_{3})]\\},$

where the third order distribution $\langle V_{i}^{3}\rangle=V^{3}_{1}$ for
the impurity potential. The energy in the Green’s function is bounded to the
Fermi energy, i.e., $\epsilon=\epsilon_{F}$.

The integrand of Eq.(99) can be related to the integrand of the polarization
matrix as

$G^{A}(\epsilon,\mathbf{k}_{1})\sigma_{\alpha}G^{R}(\epsilon,\mathbf{k}_{1})=I_{\alpha\mu}(\epsilon,\mathbf{k}_{1})\sigma_{\mu},\
G^{R}(\epsilon,\mathbf{k}_{2})\sigma_{\beta}G^{A}(\epsilon,\mathbf{k}_{2})=\sigma_{\nu}I_{\nu\beta}(\epsilon,\mathbf{k}_{2})$
(100)

where $I_{\alpha\mu}(\epsilon,\mathbf{k})=\frac{1}{2}\rm
Tr[\sigma_{\alpha}G^{R}(\epsilon,\mathbf{k})\sigma_{\mu}G^{A}(\epsilon,\mathbf{k})]$.

Defining

$f_{\mu\nu}(\epsilon,\mathbf{k}_{3})={\rm
Tr}[\sigma_{\mu}G^{R}(\epsilon,\mathbf{k}_{3})\sigma_{\nu}]+{\rm
Tr}[\sigma_{\mu}\sigma_{\nu}G^{A}(\epsilon,\mathbf{k}_{3})],$ (101)

the response function in Eq.(99) then becomes

$\displaystyle\Pi_{\alpha\beta}^{\rm
bare}=e^{2}v^{2}n_{i}V_{1}^{3}\frac{i\omega}{2\pi}\sum_{\mathbf{k}_{1}}\sum_{\mathbf{k}_{2}}\sum_{\mathbf{k}_{3}}I_{\alpha\mu}(\mathbf{k}_{1})f_{\mu\nu}(\mathbf{k}_{3})I_{\nu\beta}(\mathbf{k}_{2})=\frac{i\omega}{2\pi}e^{2}v^{2}n_{i}V_{1}^{3}{\cal
I}_{\alpha\mu}{\cal F}_{\mu\nu}{\cal I}_{\nu\beta}$ (102)

where ${\cal I}$ is the polarization matrix defined in the main text and
${\cal F}_{\mu\nu}=\sum_{\mathbf{k}_{3}}f_{\mu\nu}(\epsilon,\mathbf{k}_{3})$.

The AHE corresponds to the anti-symmetric part of $\Pi_{\alpha\beta}$. Since
the anti-symmetric part of ${\cal I}$ is smaller than its symmetric part by
$1/\epsilon_{F}\tau$, yet the symmetric part and anti-symmetric part of ${\cal
F}$ have the same order of magnitude in terms of $1/\epsilon_{F}\tau$, we only
need to keep the symmetric part of ${\cal I}$ and the anti-symmetric part of
${\cal F}$.

The leading order anti-symmetric part of $f_{\mu\nu}(\epsilon,\mathbf{k}_{3})$
is

$\displaystyle f_{\mu\nu}^{a}(\epsilon,\mathbf{k}_{3})$
$\displaystyle=2iv\epsilon_{\mu\nu\gamma}k_{3\gamma}[\frac{1}{(\epsilon-\mathbf{u}\cdot\mathbf{k}_{3}-vk_{3}-\frac{i}{2\tau})(\epsilon-\mathbf{u}\cdot\mathbf{k}_{3}+vk_{3}-\frac{i}{2\tau})}-{\rm
c.c}]$ (103) $\displaystyle\approx-4\pi
v\epsilon_{\mu\nu\gamma}k_{3\gamma}\frac{1}{\epsilon-\mathbf{u}\cdot\mathbf{k}_{3}+vk_{3}}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}_{3}-vk_{3}).$

We assume $\mathbf{u}$ in the $z$ direction, i.e., $\mathbf{u}=(0,0,u)$ for
simplicity in this appendix. The integration over
$f_{\mu\nu}^{a}(\epsilon,\mathbf{k}_{3})$ is then nonzero only for $\gamma=z$
in the above expression. We get

$\displaystyle{\cal
F}_{12}=\sum_{\mathbf{k}_{3}}f_{12}^{a}(\epsilon,\mathbf{k}_{3})$
$\displaystyle=-\frac{4\pi
v}{(2\pi)^{3}}\int_{0}^{\infty}k_{3}^{2}dk_{3}\int_{0}^{\pi}\sin\theta_{3}d\theta_{3}\int_{0}^{2\pi}d\phi_{3}\delta(\epsilon-\mathbf{u}\cdot\mathbf{k}_{3}-vk_{3})\frac{k_{3z}}{\epsilon-\mathbf{u}\cdot\mathbf{k}_{3}+vk_{3}}$
(104) $\displaystyle\approx\frac{\epsilon_{F}^{2}u}{\pi v^{4}}.$

For $\mathbf{E}$ in the $y$ direction, the leading order response function
corresponding to the AHE for the third order impurity scatterings is then

$\Pi^{\rm bare}_{xy}=\frac{i\omega}{2\pi}e^{2}v^{2}n_{i}V_{1}^{3}{\cal
I}_{11}{\cal F}_{12}{\cal
I}_{22}\approx\frac{i\omega}{2\pi}\frac{e^{2}\epsilon_{F}^{6}\tau^{2}u}{36\pi^{3}v^{8}}n_{i}V_{1}^{3}.$
(105)

Adding the vertex correction from the ladder diagram at the two ends of the
diagrams in Fig.6, we get the full response function at the leading order of
$u/v$ as

$\Pi_{xy}^{(3)}=\frac{9}{4}\Pi^{\rm
bare}_{xy}=\frac{9}{4}\frac{i\omega}{2\pi}e^{2}v^{2}n_{i}V_{1}^{3}{\cal
I}_{11}{\cal F}_{12}{\cal
I}_{22}=\frac{i\omega}{2\pi}\frac{e^{2}\epsilon_{F}^{6}\tau^{2}u}{16\pi^{3}v^{8}}n_{i}V_{1}^{3}.$
(106)

The corresponding Hall conductivity for two valleys is

$\sigma_{xy}^{(3)}=2\frac{\Pi^{(3)}_{xy}}{i\omega}=\frac{1}{4}\frac{e^{2}\epsilon_{F}^{2}u}{\pi^{2}v^{2}}\frac{V_{1}^{3}}{n_{i}V_{0}^{4}},$
(107)

where we have used $n_{i}V_{0}^{2}=\frac{2\pi v^{3}}{\epsilon_{F}^{2}\tau}$.

As a comparison, we next compute this contribution by the SBE approach under
the RTA.

The Hall conductivity from the SBE approach is

$\sigma^{(3)}_{xy}=e\sum_{\mathbf{k}}(g_{\mathbf{k}}/E_{y})(\mathbf{v}_{\mathbf{k}})_{x},$
(108)

where $g_{\mathbf{k}}$ is solved in Appendix C but with
$\omega_{\mathbf{k},\mathbf{k}^{\prime}}$ now includes
$\omega_{\mathbf{k},\mathbf{k}^{\prime}}^{(3)}=-\frac{\epsilon^{2}u}{2v^{4}}n_{i}V_{1}^{3}\sin\theta^{\prime}\sin\theta\sin(\phi-\phi^{\prime})\delta(\epsilon_{\mathbf{k}}-\epsilon_{\mathbf{k}^{\prime}})$.

For $\sigma^{(3)}_{xy}$, only $g_{\mathbf{k}}=-\frac{\partial
f_{0}}{\partial\epsilon_{\mathbf{k}}}e\mathbf{E}\cdot(\mathbf{v}_{k}\times\hat{z})(\tau^{tr})^{2}/\tau^{\perp}$
has non-zero contribution for $\mathbf{E}$ in the $y$ direction.

For $\omega^{(3)}_{\mathbf{k},\mathbf{k}^{\prime}}$,

$\frac{1}{\tau^{\perp}}=\sum_{\mathbf{k}^{\prime}}\omega_{\mathbf{k},\mathbf{k}^{\prime}}^{(3)}\frac{\sin\theta^{\prime}}{\sin\theta}\sin(\phi-\phi^{\prime})=-\frac{\epsilon^{2}u}{2v^{4}}n_{i}V_{1}^{3}\times\frac{\epsilon^{2}}{6\pi^{2}v^{3}}+o(u^{2}),$
(109)

The leading order contribution to $1/\tau^{tr}$ still comes from the second
order scattering and $1/\tau^{tr}=\frac{\epsilon^{2}n_{i}V_{0}^{2}}{3\pi
v^{3}}$. We then get the Hall conductivity for two valleys due to the third
order scattering as

$\sigma_{xy}^{(3)}=\frac{1}{4}\frac{e^{2}\epsilon_{F}^{2}u}{\pi^{2}v^{2}}\frac{V_{1}^{3}}{n_{i}V_{0}^{4}}.$
(110)

This result is consistent with Eq.(107) obtained from the Kubo-Streda formula
as well as the result in Ref.Fu2021 .

## References

  * (1) R. Karplus and J. M. Luttinger, Phys. Rev. 95, 1154(1954).
  * (2) X.-L. Qi, T. L. Hughes, and S.-C. Zhang, Phys. Rev. B 78, 195424(2008).
  * (3) R. Yu, W. Zhang, H.-J. Zhang, S.-C. Zhang, X. Dai, and Z. Fang, Science 329, 61(2010).
  * (4) C.-Z. Chang, J. Zhang, X. Feng, J. Shen, Z. Zhang, M. Guo, K. Li, Y. Ou, P. Wei, L.-L. Wang, et al., Science 340, 167(2013).
  * (5) C.-Z. Chang, W. Zhao, D. Y. Kim, H. Zhang, B. A. Assaf, D. Heiman, S.-C. Zhang, C. Liu, M. H. Chan, and J. S. Moodera, Nature Mater. 14, 473(2015).
  * (6) M. Xie, and A. H. MacDonald, Phys. Rev. Lett. 124, 097601(2020).
  * (7) J. Eom, H. Cho, W. Kang, K. Campman, A. Gossard, M. Bichler, and W. Wegscheider, Science 289, 2320(2000).
  * (8) K. Nomura, and A. H. MacDonald, Phys. Rev. Lett. 96, 256602(2006).
  * (9) S. L. Sondhi, A. Karlhede, S. Kivelson, and E. Rezayi, Phys. Rev. B 47, 16419(1993).
  * (10) K. Yang, K. Moon, L. Zheng, A. MacDonald, S. Girvin, D. Yoshioka, and S.-C. Zhang, Phys. Rev. Lett. 72, 732(1994).
  * (11) A. F. Young, C. R. Dean, L. Wang, H. Ren, P. Cadden-Zimansky, K. Watanabe, T. Taniguchi, J. Hone, K. L. Shepard, and P. Kim, Nature Phys. 8, 550(2012).
  * (12) X. Wan, A. M. Turner, A. Vishwanath, and S. Y. Savrasov, Phys. Rev. B 83, 205101(2011).
  * (13) A. A. Burkov, Phys. Rev. Lett. 113, 187202(2014).
  * (14) J. F. Steiner, A.V.Andreev, and D. A. Pesin, Phys. Rev. Lett. 119, 036601(2017).
  * (15) N. A. Sinitsyn, J Phys.: Cond. Matt. 20, 023201(2008).
  * (16) S. Yang, H. Pan, Y. Yao, and Q. Niu, Phys. Rev. B 83, 125122(2011).
  * (17) G. Sundaram and Q. Niu, Phys. Rev. B 59, 14915 (1999).
  * (18) T. Jungwirth, Q. Niu, and A. H. MacDonald, Phys. Rev. Lett. 88, 207208 (2002).
  * (19) M. Onoda and N. Nagaosa, J. Phys. Soc. Jpn. 71, 19 (2002).
  * (20) N. A.Sinitsyn, A.H.MacDonald, T.Jungwirth, V. K. Dugaev, and J. Sinova, Phys. Rev. B 75, 045315(2007).
  * (21) J. Inoue, T. Kato, Y. Ishikawa, H. Itoh, G. Bauer, and L. W. Molenkamp, Phys. Rev. Lett. 97, 046604(2006).
  * (22) S. Onoda, N. Sugimoto, and N. Nagaosa, Phys. Rev. Lett. 97, 126602(2006).
  * (23) I. A. Ado, I. A. Dmitriev, P. M. Ostrovsky, and M. Titov, Phys. Rev. Lett. 117, 046601(2016).
  * (24) N. P. Armitage, E. J. Mele, A. Vishwanath, Rev. Mod. Phys. 90, 015001(2018).
  * (25) A. A. Burkov, J Phys.: Cond Matt. 27, 113201(2015).
  * (26) P. Hoser, and X. Qi, C. R. Physique 14, 857(2013).
  * (27) P. Streda, J. Phys. C 15, L717(1982).
  * (28) N. A. Sinitsyn, Q. Niu, and A. H. MacDonald, Phys. Rev. B 73, 075318(2006).
  * (29) J. Schliemann and D. Loss, Phys. Rev. B 68, 165311(2003).
  * (30) N. F. Mott, Proc. R. Soc. London, Ser. A 124, 425(1929).
  * (31) J. Smit, Physica Amsterdam 21, 877(1955).
  * (32) J. M. Luttinger and W. Kohn, Phys. Rev. 97, 869(1955).
  * (33) J. M. Luttinger, Phys. Rev. 112, 739(1958).
  * (34) N. A. Sinitsyn, Qian Niu, Jairo Sinova, and Kentaro Nomura Phys. Rev. B 72, 045346(2005).
  * (35) Y. Ferreiros, A.A.Zyuzin, and J. H. Bardarson, Phys. Rev. B 96, 115202(2017).
  * (36) L. Berger, Phys. Rev. B 2, 4559(1970).
  * (37) A. A. Kovalev, J. Sinova, and Y. Tserkovnyak, Phys. Rev. Lett. 105, 036601(2010).
  * (38) Among the eight diagrams for side jump contribution, four correspond to the contribution directly from the transverse coordinate shift and the other four correspond to the contribution of the anomalous distribution due to the coordinate shift. The two types have the same contributions to AHE and we call both of them the side jump contribution.
  * (39) I. A. Ado, I. A. Dmitriev, P. M. Ostrovsky, and M. Titov, Phys. Rev. B 96, 235148(2017).
  * (40) A. A. Burkov. A. S. Nunez and A. H. MacDonald, Phys. Rev. B 70, 155308(2004).
  * (41) A. A. Burkov and D. G. Hawthorn, Phys. Rev. Lett. 105, 066802(2010).
  * (42) W. Chen and A. A. Clerk, Phys. Rev. B 86, 125443(2012).
  * (43) K. Kechedzhi, O. Kashuba, and V. I. Falko, Phys. Rev. B, 77, 193403(2008).
  * (44) N. Shon and T. Ando, J. Phys. Soc. Jpn. 67, 2421(1998).
  * (45) N. A. Sinitsyn, J. E. Hill, Hongki Min, Jairo Sinova, and A. H. MacDonald, Phys. Rev. Lett. 97, 106804(2006).
  * (46) M. Papaj, and L. Fu, Phys. Rev. B 103, 075424(2021).
  * (47) K. Vyborny, A. A. Kovalev, J. Sinova, and T. Jungwirth, Phys. Rev. B 79, 045427(2009).
  * (48) A. A. Kovalev, Y. Tserkovnyak, K. Vyborny, and J. Sinova, Phys. Rev. B 79, 195129(2009).