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Semiclassical magnetotransport including the effects of the Berry curvature and Lorentz force

Seungchan Woo1,2    Brett Min3,2    Hongki Min1,2 [email protected] 1 Department of Physics and Astronomy, Seoul National University, Seoul 08826, Korea 2 Center for Theoretical Physics (CTP), Seoul National University, Seoul 08826, Korea 3 Department of Physics, McGill University, Montréal, Québec H3A 2T8, Canada
Abstract

In topological semimetals and insulators, negative longitudinal magnetoresistance and angle-dependent planar Hall effect have been reported arising from the Berry curvature. Using the Boltzmann transport theory, we present a closed-form expression for the nonequilibrium distribution function which includes both the effects of the Berry curvature and Lorentz force. Using this formulation, we obtain analytical expressions for conductivity and resistivity tensors in Weyl semimetals demonstrating a characteristic field dependence arising from the competition between the two effects.

I Introduction

In topological materials with a nonvanishing Berry curvature, the positive magnetoconductance has been observed experimentally in the presence of parallel electric and magnetic fields KIM2013 ; HUANG2015 ; XIONG2015 ; LI2015 ; LI2016a ; ZHANG2016 ; LI2016b ; ZHANG2016 ; ZHANG2017 ; WANG2012 ; HE2013 ; WANG2015 ; WIEDMANN2016 ; ARNOLD2016 ; BREUNIG2017 ; ASSAF2017 . This is a unique feature to topological materials that does not occur in conventional magnetotransport which only takes classical Lorentz force effect into account.

The prevailing explanation for the positive longitudinal magnetoconductance is the so-called chiral anomaly. In 1983, Nielsen and Ninomiya suggested the chiral anomaly in Weyl fermions under a strong magnetic field regime where the chiral zeroth Landau level creates a one-dimensional conducting channel that pumps electrons from one Weyl node to another NIELSEN1983 . In 2013, Son and Spivak discussed chiral anomaly in Weyl semimetals under weak external magnetic field using the semi-classical Boltzmann approach SON2013 . They argued the positive magnetoconductivity that scales quadratically in magnetic field is due to topological charge pumping. One can expect possible detection of chiral anomaly between the valleys, given that the intervalley scattering is negligible compared to the intravalley scattering SEKINE2017 . However, chiral anomaly cannot be responsible for observed positive magnetoconductivity in topological insulators (TIs) where chiral charges are not well defined WANG2012 ; HE2013 ; WANG2015 ; WIEDMANN2016 ; ARNOLD2016 ; BREUNIG2017 ; ASSAF2017 . In topological materials such as TIs DAI2017 and Weyl semimetals (WSMs) KIM2014 , it is suggested that the anomalous velocity induced by the non-trivial Berry curvature alone can generate an additional contribution to the conductivity that grows with the magnetic field.

To understand magnetotransport properties quantitatively, it is important to consider both the effects of the anomalous velocity due to the non-trivial Berry curvature and the classical Lorentz force effect. Most of the previous studies KIM2014 ; DAI2017 ; NANDY2017 have focused on the Berry curvature effect, while only a few took the Lorentz force into considerations in describing magnetotransport behaviors IMRAN2018 ; JOHANSSON2019 . In this paper, we revisit semiclassical treatment of magnetotransport in topological materials to shed more light on the origin of the observed positive magnetoconductivity. We present a general semiclassical formula for conductivity which fully incorporates the Berry curvature and the Lorentz force. From the Boltzmann transport equation, we obtain a closed-form expression for the nonequilibrium distribution function by solving the corresponding self-consistent equation. We then apply our formula to WSMs and express the magnetoconductivity in terms of dimensionless parameters characterizing magnetic fields associated with the Lorentz force and the Berry curvature, respectively.

II Semiclassical Boltzmann magnetotransport theory for topological materials

The semiclassical Boltzmann transport equation governs the time evolution of a non-equilibrium distribution function f=f(𝐫,𝐤,t)f=f(\mathbf{r},\mathbf{k},t) at position 𝐫\mathbf{r} and momentum 𝐤\mathbf{k}. It states that the time evolution of ff equals the probability rate of electrons being scattered in and out of the distribution, called the collision term (ft)coll\Big{(}\frac{\partial f}{\partial t}\Big{)}_{\text{coll}}:

dfdt=(ft)coll.\frac{df}{dt}=\Big{(}\frac{\partial f}{\partial t}\Big{)}_{\text{coll}}. (1)

In a homogeneous sample with a steady external perturbation, there are no position 𝐫\mathbf{r} nor time tt dependence in the distribution function ff. Then, dfdt\frac{df}{dt} simplifies to

dfdt=𝐤˙f𝐤𝐤,\frac{df}{dt}=\dot{\mathbf{k}}\cdot\frac{\partial f_{\mathbf{k}}}{\partial\mathbf{k}}, (2)

where we have included a subscript 𝐤\mathbf{k} to the non-equilibrium distribution function ff to indicate that it is only a function of 𝐤\mathbf{k}. In a simple relaxation time approximation, the collision integral is replaced by the ratio between the deviation from the equilibrium Fermi-Dirac distribution f𝐤(0)f^{(0)}_{\mathbf{k}} and the average time τ\tau between successive collisions. Hence,

𝐤˙f𝐤𝐤=g𝐤τ,\dot{\mathbf{k}}\cdot\frac{\partial f_{\mathbf{k}}}{\partial\mathbf{k}}=-\frac{g_{\mathbf{k}}}{\tau}, (3)

where g𝐤=f𝐤f𝐤(0)g_{\mathbf{k}}=f_{\mathbf{k}}-f^{(0)}_{\mathbf{k}}. Here we assume a constant transport relaxation time τ\tau in momentum 𝐤\mathbf{k} and magnetic field 𝐁\mathbf{B} for a given chemical potential. Note that when we fully consider the collision integral in the system with a nontrivial Berry curvature, the transport relaxation time τ\tau may show a field dependent anisotropy induced by the coupling between the Berry curvature and magnetic field PARK2021 . However, in weak magnetic field regime, we can assume τ\tau as a constant. Furthermore, when short-range scattering is dominant or charged impurities are fully screened, we can assume that τ\tau does not depend on 𝐤\mathbf{k} XIAO2005 ; Ashcroft1976 ; Ziman1960 .

Here we provide a closed-form expression for magnetoconductivity in the presence of both a non-trivial Berry curvature and the Lorentz force within the semiclassical Boltzmann approach. The semiclassical equations of motion for electrons with a charge q=eq=-e in the presence of the Berry curvature are given by XIAO2010

𝐫˙\displaystyle\dot{\mathbf{r}} =ϵ𝐦(𝐤)𝐤𝐤˙×𝛀𝐤,\displaystyle=\frac{\partial\epsilon_{\mathbf{m}}(\mathbf{k})}{\hbar\partial\mathbf{k}}-\dot{\mathbf{k}}\times\mathbf{\Omega}_{\mathbf{k}}, (4a)
𝐤˙\displaystyle\hbar\dot{\mathbf{k}} =q𝐄+qc𝐫˙×𝐁,\displaystyle=q\mathbf{E}+\frac{q}{c}\dot{\mathbf{r}}\times\mathbf{B}, (4b)

where ϵ𝐦(𝐤)=ϵ0(𝐤)𝐦𝐤𝐁\epsilon_{\mathbf{m}}(\mathbf{k})=\epsilon_{0}(\mathbf{k})-\mathbf{m}_{\mathbf{k}}\cdot\mathbf{B}, ϵ0(𝐤)\epsilon_{0}(\mathbf{k}) is the unperturbed band energy, 𝐦𝐤\bf{m}_{\mathbf{k}} is an orbital magnetic moment that couple to the magnetic field and 𝛀𝐤\mathbf{\Omega}_{\mathbf{k}} is the Berry curvature. It has been reported that disorder affects not only the carrier distribution but also the semiclassical equations of motion, generating a correction to the velocity proportional to the disorder strength ATENCIA2022 . In this work, we neglect this correction assuming a weak disorder potential for simplicity. In the presence of a magnetic field, 𝐫˙\dot{\mathbf{r}} and 𝐤˙\dot{\mathbf{k}} in Eq. (4) are coupled through the Lorentz force. Combining these two equations of motion, we have

𝐫˙\displaystyle\dot{\mathbf{r}} =D𝐤1[𝐯𝐤q𝐄×𝛀𝐤qc(Ω𝐤𝐯𝐤)𝐁],\displaystyle=D^{-1}_{\mathbf{k}}[\mathbf{v}_{\mathbf{k}}-\frac{q}{\hbar}\mathbf{E}\times\mathbf{\Omega}_{\mathbf{k}}-\frac{q}{\hbar c}({\Omega}_{\mathbf{k}}\cdot\mathbf{v}_{\mathbf{k}})\mathbf{B}], (5a)
𝐤˙\displaystyle\hbar\dot{\mathbf{k}} =D𝐤1[q𝐄+qc𝐯𝐤×𝐁q2c(𝐄𝐁)𝛀𝐤],\displaystyle=D^{-1}_{\mathbf{k}}[q\mathbf{E}+\frac{q}{c}\mathbf{v}_{\mathbf{k}}\times\mathbf{B}-\frac{q^{2}}{\hbar c}(\mathbf{E}\cdot\mathbf{B})\mathbf{\Omega}_{\mathbf{k}}], (5b)

where 𝐯𝐤=1ϵ𝐦(𝐤)𝐤\mathbf{v}_{\mathbf{k}}=\frac{1}{\hbar}\frac{\partial\epsilon_{\mathbf{m}}(\mathbf{k})}{\partial\mathbf{k}} and D𝐤=1qc(𝛀𝐤𝐁)D_{\mathbf{k}}=1-\frac{q}{\hbar c}(\mathbf{\Omega}_{\mathbf{k}}\cdot\mathbf{B}) represents the modified density of states in the phase space due to the Berry curvature effect XIAO2005 . The first term in the square bracket in Eq. (5b) corresponds to electric force due to an electric field 𝐄\mathbf{E} and the second term represents Lorentz force due to a magnetic field 𝐁\mathbf{B}. The last term is anomalous electromagnetic force due to the Berry curvature which leads to positive magnetoconductivity in topological materials.

Replacing f𝐤=f𝐤0+g𝐤f_{\mathbf{k}}=f^{0}_{\mathbf{k}}+g_{\mathbf{k}} according to the relaxation time approximation in Eq. (3), we have

𝐤˙1(f𝐤(0)+g𝐤)𝐤=g𝐤τ.\displaystyle\hbar\dot{\mathbf{k}}\cdot\frac{1}{\hbar}\frac{\partial(f^{(0)}_{\mathbf{k}}+g_{\mathbf{k}})}{\partial\mathbf{k}}=-\frac{g_{\mathbf{k}}}{\tau}. (6)

Plugging Eq. (5) into Eq. (6), we get

D𝐤1[\displaystyle D^{-1}_{\mathbf{k}}\Big{[} q𝐄𝐯𝐤f𝐤(0)ϵ𝐤+qc(𝐯𝐤×𝐁g𝐤𝐤)\displaystyle q\mathbf{E}\cdot\mathbf{v}_{\mathbf{k}}\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}+\frac{q}{\hbar c}\left(\mathbf{v}_{\mathbf{k}}\times\mathbf{B}\cdot\frac{\partial g_{\mathbf{k}}}{\partial\mathbf{k}}\right) (7)
q2c(𝐄𝐁)(Ω𝐤𝐯𝐤)f𝐤(0)ϵ𝐤]=g𝐤τ.\displaystyle-\frac{q^{2}}{\hbar c}(\mathbf{E}\cdot\mathbf{B})(\Omega_{\mathbf{k}}\cdot\mathbf{v}_{\mathbf{k}})\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{]}=-\frac{g_{\mathbf{k}}}{\tau}.

In previous studies DAI2017 ; KIM2014 ; NANDY2017 of magnetoconductivity and planar Hall conductivity in topological materials, Lorentz force effect has been often neglected. Therefore, in order to better understand positive longitudinal magnetoconductivity and angle-dependent planar Hall conductivity in topological materials, we take all the terms in Eq. (7) into consideration.

We can rewrite Eq. (7) in the following form:

q𝐄𝐯~𝐤f𝐤(0)ϵ𝐤+qc(𝐯~𝐤×𝐁)g𝐤𝐤=g𝐤τ,q\mathbf{E}\cdot\tilde{\mathbf{v}}_{\mathbf{k}}\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}+\frac{q}{\hbar c}(\tilde{\mathbf{v}}_{\mathbf{k}}\times\mathbf{B})\cdot\frac{\partial g_{\mathbf{k}}}{\partial\mathbf{k}}=-\frac{g_{\mathbf{k}}}{\tau}, (8)

where

𝐯~𝐤=D𝐤1[𝐯𝐤qc(𝐯𝐤𝛀𝐤)𝐁].\tilde{\mathbf{v}}_{\mathbf{k}}={D_{\mathbf{k}}}^{-1}\Big{[}\mathbf{v}_{\mathbf{k}}-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})\mathbf{B}\Big{]}. (9)

Figure 1 shows a schematic picture of 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}} in WSMs in the presence of a magnetic field 𝐁\mathbf{B}. As shown in Fig. 1, 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}} is greater in magnitude than v𝐤v_{\mathbf{k}} in all 𝐤\mathbf{k} except when 𝐯𝐤\mathbf{v}_{\mathbf{k}} is parallel to 𝐁\mathbf{B} giving 𝐯~𝐤=𝐯𝐤\tilde{\mathbf{v}}_{\mathbf{k}}=\mathbf{v}_{\mathbf{k}}. In general, it is this effect due to Berry curvature that leads to enhanced magnetoconductivity in topological materials.

Refer to caption
Figure 1: Schematic picture comparing 𝐯𝐤\mathbf{v}_{\mathbf{k}} (blue) and 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}} (red) near the Weyl nodes described by the Hamiltonian H=χvF𝐤𝝈H=\chi\hbar v_{\rm F}\mathbf{k}\cdot\bm{\sigma} for χ=±1\chi=\pm 1 in the presence of a constant magnetic field 𝐁=Bxx^\mathbf{B}=B_{x}\hat{x}. Here the length of an arrow represents the velocity normalized by vFv_{\rm F}.

Our goal now is to solve for g𝐤g_{\mathbf{k}} in Eq. (8). We assume that a small deviation from the equilibrium distribution takes the form of g𝐤=𝐯~𝐤𝐆g_{\mathbf{k}}=\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\mathbf{G} where 𝐆\mathbf{G} is some arbitrary vector that is independent of 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}}. Upon plugging g𝐤=𝐯~𝐤𝐆g_{\mathbf{k}}=\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\mathbf{G} into Eq. (8), we obtain

qτf𝐤(0)ϵ𝐤𝐄𝐯~𝐤+qτc(𝐯~𝐤×𝐁)𝕄1𝐆+𝐯~𝐤𝐆=0,q\tau\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\mathbf{E}\cdot\tilde{\mathbf{v}}_{\mathbf{k}}+\frac{q\tau}{c}(\tilde{\mathbf{v}}_{\mathbf{k}}\times\mathbf{B})\cdot\mathds{M}^{-1}\mathbf{G}+\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\mathbf{G}=0, (10)

where 𝕄ij1=1v~𝐤,jki\mathds{M}^{-1}_{ij}=\frac{1}{\hbar}\frac{\partial\tilde{\rm v}_{\mathbf{k},j}}{\partial k_{i}} is the inverse mass tensor. Factoring 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}} out,

g𝐤=\displaystyle g_{\mathbf{k}}= 𝐯~𝐤[qτ(f𝐤(0)ϵ𝐤)𝐄qτc(𝐁×𝕄1𝐆)].\displaystyle\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\bigg{[}q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\mathbf{E}-\frac{q\tau}{c}(\mathbf{B}\times\mathds{M}^{-1}\mathbf{G})\bigg{]}. (11)

From g𝐤=𝐯~𝐤𝐆g_{\mathbf{k}}=\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\mathbf{G} and Eq. (11), we obtain a self-consistent form of 𝐆\mathbf{G} as

𝐆=𝐆0qτc(𝐁×𝕄1𝐆),\mathbf{G}=\mathbf{G}_{0}-\frac{q\tau}{c}(\mathbf{B}\times\mathds{M}^{-1}\mathbf{G}), (12)

where

𝐆0=qτ(f𝐤(0)ϵ𝐤)𝐄.\mathbf{G}_{0}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\mathbf{E}. (13)

The solution for Eq. (12) can be obtained as

𝐆=qτ(f𝐤(0)ϵ𝐤)(𝟙+qτc𝔽𝕄1)1𝐄,\mathbf{G}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\Big{(}\mathds{1}+\frac{q\tau}{c}\mathds{F}\mathds{M}^{-1}\Big{)}^{-1}\mathbf{E}, (14)

where 𝔽ij=iϵijkBk\mathds{F}_{ij}=\sum_{i}\epsilon_{ijk}B_{k} is the magnetic field strength tensor (see Appendix A for a detailed derivation). Thus, we obtain

g𝐤=qτ(f𝐤(0)ϵ𝐤)𝐯~𝐤[(𝟙+qτc𝔽𝕄1)1𝐄].g_{\mathbf{k}}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\Big{[}\Big{(}\mathds{1}+\frac{q\tau}{c}\mathds{F}\mathds{M}^{-1}\Big{)}^{-1}\mathbf{E}\Big{]}. (15)

The current is therefore computed as

Jα=qddk(2π)dD𝐤f𝐤vα,\displaystyle J_{\alpha}=q\int\frac{d^{d}k}{(2\pi)^{d}}D_{\mathbf{k}}f_{\mathbf{k}}v_{\alpha}, (16)

where α\alpha is the direction in which we measure the current. Let us denote the current associated with f𝐤(0)f^{(0)}_{\mathbf{k}} as Jαint=qddk(2π)dD𝐤f𝐤(0)vαJ^{\text{int}}_{\alpha}=q\int\frac{d^{d}k}{(2\pi)^{d}}D_{\mathbf{k}}f^{(0)}_{\mathbf{k}}{{v}}_{\alpha} and that associated with g𝐤g_{\mathbf{k}} as Jαext=qddk(2π)dD𝐤g𝐤vαJ^{\text{ext}}_{\alpha}=q\int\frac{d^{d}k}{(2\pi)^{d}}D_{\mathbf{k}}g_{\mathbf{k}}{{v}}_{\alpha}. From now on, we will only focus on the extrinsic contribution JαextJ^{\text{ext}}_{\alpha} arising from scatterings.

Plugging Eq. (15) into JαextJ^{\text{ext}}_{\alpha} and using Jαext=βσαβEβJ_{\alpha}^{\text{ext}}=\sum_{\beta}\sigma_{\alpha\beta}E_{\beta} relation, we finally arrive at

σαβ=q2\displaystyle\sigma_{\alpha\beta}=q^{2}\int ddk(2π)dD𝐤τ(f𝐤(0)ϵ𝐤)v~α\displaystyle\frac{d^{d}k}{(2\pi)^{d}}D_{\mathbf{k}}\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\tilde{v}_{\alpha} (17)
×(𝐯~𝐤{[𝟙+qτc𝔽𝕄1]1β^}),\displaystyle\times\Big{(}\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\Big{\{}\Big{[}\mathds{1}+\frac{q\tau}{c}\mathds{F}\mathds{M}^{-1}\Big{]}^{-1}\hat{\beta}\Big{\}}\Big{)},

where β\beta is the direction of an electric field.

Now let us assume that the mobility tensor eτ𝕄1e\tau\mathds{M}^{-1} is set to a constant μ\mu for simplicity. Then 𝐆\mathbf{G} in Eq. (14) becomes (see Appendix A)

𝐆=qτ(f𝐤(0)ϵ𝐤)𝐄+μc𝐄×𝐁+μ2c2(𝐄𝐁)𝐁1+μ2c2|𝐁|2.\mathbf{G}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\frac{\mathbf{E}+\frac{\mu}{c}\mathbf{E}\times\mathbf{B}+\frac{\mu^{2}}{c^{2}}(\mathbf{E}\cdot\mathbf{B})\mathbf{B}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}. (18)

Note that the obtained 𝐆\mathbf{G} is consistent with the assumption that 𝐆\mathbf{G} is independent of 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}}. Then g𝐤g_{\mathbf{k}} is given by

g𝐤=qτ(f𝐤(0)ϵ𝐤)𝐯~𝐤𝐄+μc𝐄×𝐁+μ2c2(𝐄𝐁)𝐁1+μ2c2|𝐁|2.g_{\mathbf{k}}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\frac{\mathbf{E}+\frac{\mu}{c}\mathbf{E}\times\mathbf{B}+\frac{\mu^{2}}{c^{2}}(\mathbf{E}\cdot\mathbf{B})\mathbf{B}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}. (19)

Using Eq. (19), we finally obtain the following form for magnetoconductivity

σαβ=\displaystyle\sigma_{\alpha\beta}= q2ddk(2π)dD𝐤τ(f𝐤(0)ϵ𝐤)1+μ2c2|𝐁|2\displaystyle q^{2}\int\frac{d^{d}k}{(2\pi)^{d}}\frac{D_{\mathbf{k}}\tau(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}
×[v~αv~βμcv~α(𝐯~𝐤×𝐁)β+μ2c2(𝐯~𝐤𝐁)v~αBβ].\displaystyle\times\Big{[}\tilde{v}_{\alpha}\tilde{v}_{\beta}-\frac{\mu}{c}\tilde{v}_{\alpha}(\tilde{\mathbf{v}}_{\mathbf{k}}\times\mathbf{B})_{\beta}+\frac{\mu^{2}}{c^{2}}(\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\mathbf{B})\tilde{v}_{\alpha}B_{\beta}\Big{]}.

The above form is a general expression of magnetoconductivity for topological materials including the Berry curvature and Lorentz force within the semiclassical regime under the assumption that 𝐆\mathbf{G} in g𝐤=𝐯~𝐤𝐆g_{\mathbf{k}}=\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\mathbf{G} is independent of 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}} and the mobility tensor is a constant.

III Magnetotransport in Weyl semimetals

In this section, we study the magnetotransport properties of WSMs in three dimensions using a closed-form expression for magnetoconductivity Eq. (II) discussed in the previous section. For simplicity, we consider a single Weyl node described by the Hamiltonian H=χvF𝐤𝝈H=\chi\hbar v_{\rm F}\mathbf{k}\cdot\bm{\sigma} which has isotropic linear dispersion, where χ=±1\chi=\pm 1 are for the different chiralities of Weyl fermions and 𝝈\bm{\sigma} are the Pauli matrices.

III.1 Longitudinal magnetoconductivity

To investigate longitudinal magnetoconductivity σxx(𝐁)\sigma_{xx}(\mathbf{B}) in WSMs, without loss of generality, we set the electric and magnetic field orientations as 𝐄=Exx^\mathbf{E}=E_{x}\hat{x} and 𝐁=Bxx^+Byy^\mathbf{B}=B_{x}\hat{x}+B_{y}\hat{y}, respectively. Organizing terms in Eq. (II) in powers of μ\mu and using Eq. (9),

σxx=q2d3k(2π)3τ(f𝐤(0)ϵ𝐤)D𝐤11+μ2c2|𝐁|2[Σ𝐤(0)+Σ𝐤(1)+Σ𝐤(2)],\sigma_{xx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})D^{-1}_{\mathbf{k}}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}\Sigma_{\mathbf{k}}^{(0)}+\Sigma_{\mathbf{k}}^{(1)}+\Sigma_{\mathbf{k}}^{(2)}\Big{]}, (21)

where Σ𝐤(i)\Sigma_{\mathbf{k}}^{(i)} is a sum of the terms that include iith order of μ\mu. Due to the D𝐤1D^{-1}_{\mathbf{k}} term in Eq. (21), it is difficult to obtain an analytic expression for σxx\sigma_{xx} incorporating the full density of states correction. Therefore, to obtain a simple closed form result, we first assume D𝐤1D^{-1}_{\mathbf{k}} as 11 in Eq. (21). We will discuss the correction beyond this approximation later. Here, Σ𝐤(i)\Sigma_{\mathbf{k}}^{(i)} are defined as

Σ𝐤(0)=\displaystyle\Sigma_{\mathbf{k}}^{(0)}= vx22qcvx(𝐯𝐤𝛀𝐤)|𝐁|cosΓ\displaystyle~{}v^{2}_{x}-2\frac{q}{\hbar c}v_{x}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|\cos\Gamma
+q2(c)2(𝐯𝐤𝛀𝐤)2|𝐁|2cos2Γ,\displaystyle+\frac{q^{2}}{(\hbar c)^{2}}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{2}\cos^{2}\Gamma, (22a)
Σ𝐤(1)=\displaystyle\Sigma_{\mathbf{k}}^{(1)}= μc[vx(𝐯𝐤×𝐁)x\displaystyle~{}\frac{\mu}{c}\Big{[}-v_{x}(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}
+qc(𝐯𝐤×𝐁)x(𝐯𝐤𝛀𝐤)|𝐁|cosΓ],\displaystyle~{}+\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|\cos\Gamma\Big{]}, (22b)
Σ𝐤(2)=\displaystyle\Sigma_{\mathbf{k}}^{(2)}= μ2c2[(𝐯𝐤𝐁)vx|𝐁|cosΓ\displaystyle~{}\frac{\mu^{2}}{c^{2}}\Big{[}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})v_{x}|\mathbf{B}|\cos\Gamma
qc(𝐯𝐤𝛀𝐤)|𝐁|3vxcosΓ\displaystyle~{}-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{3}v_{x}\cos\Gamma
qc(𝐯𝐤𝐁)(𝐯𝐤𝛀𝐤)|𝐁|2cos2Γ\displaystyle~{}-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{2}\cos^{2}\Gamma
+q2(c)2(𝐯𝐤𝛀𝐤)2|𝐁|4cos2Γ],\displaystyle~{}+\frac{q^{2}}{(\hbar c)^{2}}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{4}\cos^{2}\Gamma\Big{]}, (22c)

where Γ\Gamma is the angle between 𝐄\bf E and 𝐁\bf B. The first term vx2v^{2}_{x} in Eq. (22) gives a well known longitudinal conductivity in the absence of magnetic field: σxx(𝐁=0)=q2N0Dσ0\sigma_{xx}(\mathbf{B}=0)=q^{2}N_{0}D\equiv\sigma_{0} where N0{N_{0}} is the density of states at the Fermi energy and D=vF2τ/dD=v_{\rm F}^{2}\tau/d is the diffusion constant with d=3d=3.

Collecting terms that would give us non-zero contribution after momentum integral, Eq. (21) can be rewritten as

σxx=q2d3k(2π)3τ(f(0)ϵ𝐤)(1+μ2c2|𝐁|2)\displaystyle\sigma_{xx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau(-\frac{\partial f^{(0)}}{\partial\epsilon_{\mathbf{k}}})}{(1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2})} (23)
×[vx2+q2(c)2(𝐯𝐤𝛀𝐤)2|𝐁|2cos2Γ\displaystyle\times\bigg{[}v^{2}_{x}+\frac{q^{2}}{(\hbar c)^{2}}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{2}\cos^{2}\Gamma
+μ2c2vx2|𝐁|2cos2Γ+μ2c2q2(c)2(𝐯𝐤𝛀𝐤)2|𝐁|4cos2Γ].\displaystyle+\frac{\mu^{2}}{c^{2}}v^{2}_{x}|\mathbf{B}|^{2}\cos^{2}\Gamma+\frac{\mu^{2}}{c^{2}}\frac{q^{2}}{(\hbar c)^{2}}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{4}\cos^{2}\Gamma\bigg{]}.

The Berry curvature in an isotropic WSM is 𝛀𝐤=χ𝐤2|𝐤|3\mathbf{\Omega}_{\mathbf{k}}=\chi\frac{\mathbf{k}}{2|\mathbf{k}|^{3}}. Therefore, 𝐯𝐤𝛀𝐤\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}} becomes χvF/(2kF2)\chi{v_{\rm F}}/({2k_{\rm F}^{2}}) where vFv_{\rm F} and kFk_{\rm F} are the Fermi velocity and Fermi wave vector, respectively. Note that all the surviving terms are even functions of magnetic field |𝐁||\mathbf{B}| and independent of χ\chi. Here we emphasize that there are two kinds of magnetic field effects: the Lorentz force and anomalous velocity effect due to the Berry curvature. The terms that are related to the Lorentz force comes with a μc|𝐁|\frac{\mu}{c}|\mathbf{B}| factor. On the other hand, the terms that are related to the Berry curvature comes with a qc|𝛀𝐤||𝐁|=q2(c)212kF2\frac{q}{\hbar c}|\mathbf{\Omega}_{\mathbf{k}}||\mathbf{B}|=\frac{q^{2}}{(\hbar c)^{2}}\frac{1}{2k_{\rm F}^{2}} factor.

Since the terms coupled with the magnetic field are proportional to either μc|𝐁|\frac{\mu}{c}|\mathbf{B}| or qc|𝛀𝐤||𝐁|\frac{q}{\hbar c}|\mathbf{\Omega}_{\mathbf{k}}||\mathbf{B}|, we introduce the following dimensionless parameters:

bμ\displaystyle b_{\mu} \displaystyle\equiv μc|𝐁|qτcvFkF|𝐁|,\displaystyle\frac{\mu}{c}|\mathbf{B}|\equiv\frac{q\tau}{\hbar c}\frac{v_{\rm F}}{k_{\rm F}}|\mathbf{B}|, (24a)
bBC\displaystyle b_{\rm BC} \displaystyle\equiv qc|𝛀𝐤||𝐁|qc12kF2|𝐁|.\displaystyle~{}\frac{q}{\hbar c}|\mathbf{\Omega}_{\mathbf{k}}||\mathbf{B}|~{}\equiv\frac{q}{\hbar c}\frac{1}{2k_{\rm F}^{2}}|\mathbf{B}|. (24b)

Note that these two dimensionless parameters are related with each other as bμ/bBC=2kFlb_{\mu}/b_{\rm BC}=2k_{\rm F}l, where l=vFτl=v_{\rm F}\tau is the mean-free path. This gives us an important insight that the Lorentz force effect cannot be simply neglected when studying magnetotransport in topological materials within the semiclassical Boltzmann approach which is valid in kFl1k_{\rm F}l\gg 1 regime.

Carrying out the momentum integral in Eq. (23), we obtain a longitudinal magnetoconductivity in WSMs for an arbitrary external magnetic field under the assumption that D𝐤1=1D^{-1}_{\mathbf{k}}=1 (see Appendix B.1):

σxx(𝐁)=σ0[11+bμ2+(bμ21+bμ2+3bBC2)cos2Γ].\sigma_{xx}(\mathbf{B})=\sigma_{0}\Big{[}\frac{1}{1+b^{2}_{\mu}}+\Big{(}{\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}+3b^{2}_{\rm BC}}\Big{)}\cos^{2}{\Gamma}\Big{]}. (25)

Note that σxx(𝐁)\sigma_{xx}(\mathbf{B}) in Eq. (25) is independent of χ\chi, as bBCb_{\text{BC}} comes with a power of two. Thus, the contribution from multiple degenerate Weyl nodes enters as a degeneracy factor when internode scatterings are neglected. Most of previous works for longitudinal magnetoconductivity in WSMs using Boltzmann transport theory reported σxx(Γ)σ0+Δσcos2Γ\sigma_{xx}(\Gamma)\sim\sigma_{0}+\Delta\sigma\cos^{2}{\Gamma} where Δσ\Delta\sigma is the additional positive magnetoconductivity that scales quadratically in magnetic field KIM2014 ; DAI2017 ; NANDY2017 ; IMRAN2018 . Instead, Eq. (25) shows a non-monotonic behavior of magnetoconductivity in magnetic field due to the competition between the Berry curvature and Lorentz force depending on kFlk_{\rm F}l, as shown in Fig. 2. This result is consistent with the previous calculation for WSMs obtained by expanding the non-equilibrium distribution function in Fourier harmonics IMRAN2018 .

Refer to caption
Figure 2: Longitudinal conductivity as a function of bμb_{\mu} at selected angles between 𝐄\mathbf{E} and 𝐁\mathbf{B} in WSMs for kFl=4{k}_{\rm F}l=4.

We now come back to the approximation we made: D𝐤1=1D^{-1}_{\mathbf{k}}=1. Note that for weak magnetic fields, we could Taylor expand D𝐤1D^{-1}_{\mathbf{k}} as

D𝐤1=1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)2+.D^{-1}_{\mathbf{k}}=1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}+\cdots. (26)

We emphasize that earlier studies of magnetotransport also took the D𝐤D_{\mathbf{k}} correction into account but incompletely. For instance, Kim et alKIM2014 only took even terms in Taylor expanded D𝐤1D^{-1}_{\mathbf{k}} in Eq. (26) which gave rise to an additional positive correction to magnetoconductivity described in Eq. (25). However, numerical calculations of the full longitudinal magnetoconductivity [Eq. (21)] show a reduced value compared to Eq. (25). The reduced magnetoconductivity is due to terms that are odd orders in |𝐁||\mathbf{B}| in Eq. (21) coupling with odd orders in |𝐁||\mathbf{B}| terms in the Taylor expanded D𝐤1D^{-1}_{\mathbf{k}} in Eq. (26). As a result, the pairs of odd terms in |𝐁||\mathbf{B}| give rise to non-vanishing even terms in |𝐁||\mathbf{B}| which additionally give negative corrections to the magnetoconductivity.

By incorporating first three terms in Eq. (26), we obtain the longitudinal magnetoconductivity in WSMs up to quadratic order in bBC2b_{\rm BC}^{2} as

σxx(𝐁)\displaystyle\sigma_{xx}(\mathbf{B}) σ01+bμ2\displaystyle\approx\frac{\sigma_{0}}{1+b_{\mu}^{2}}
×{1+15bBC2+[75bBC2+bμ2(1+85bBC2)]cos2Γ}.\displaystyle\times\Big{\{}1+\frac{1}{5}b^{2}_{\rm BC}+\Big{[}\frac{7}{5}b^{2}_{\rm BC}+b_{\mu}^{2}\Big{(}1+\frac{8}{5}b^{2}_{\rm BC}\Big{)}\Big{]}\cos^{2}{\Gamma}\Big{\}}.

See Appendix B.1 for detailed derivations. This result is well matched with the previous work which focused on specific angles between applied electric and magnetic fields IMRAN2018 .

III.2 Planar Hall conductivity

To investigate the planar Hall conductivity in WSMs, we again set the electric and magnetic field orientations as 𝐄=Exx^\mathbf{E}=E_{x}\hat{x} and 𝐁=Bxx^+Byy^\mathbf{B}=B_{x}\hat{x}+B_{y}\hat{y}, respectively. Neglecting terms that would give zero contribution to conductivity, Eq. (II) gives the following form of the planar Hall conductivity σyx(𝐁)\sigma_{yx}(\mathbf{B}) assuming D𝐤1=1D^{-1}_{\mathbf{k}}=1 (see Appendix B.2):

σyx(𝐁)=σ0[(bμ21+bμ2+3bBC2)sin(Γ)cos(Γ)].\sigma_{yx}(\mathbf{B})=\sigma_{0}\Big{[}\Big{(}{\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}+3b^{2}_{\rm BC}}\Big{)}\sin{\Gamma}\cos{\Gamma}\Big{]}. (28)

Note that σyx(𝐁)\sigma_{yx}(\mathbf{B}) in Eq. (28) is independent of χ\chi. Equation (28) is the analytic form of the planar Hall conductivity in WSMs for an arbitrary angle of the applied magnetic field |𝐁||\mathbf{B}|. The angle dependence of the planar Hall conductivity is well matched with the previous works σyx(Γ)sin(Γ)cos(Γ)\sigma_{yx}(\Gamma)\sim\sin{\Gamma}\cos{\Gamma} BURKOV2017 ; NANDY2017 ; LI2018 ; LIANG2019 ; LIU2019 ; MA2019 ; LI2020 but the field dependence shows an additional contribution from the Lorentz force in addition to a quadratic field dependence due to the Berry curvature. As shown in Fig 3, the planar Hall conductivity shows different |𝐁||\mathbf{B}| dependence at different kFlk_{\rm F}l regimes. For large kFlk_{\rm F}l, the planar Hall conductivity roughly increases with |𝐁||\mathbf{B}| quadratically at low |𝐁||\mathbf{B}| field and saturates at high |𝐁||\mathbf{B}| field. For low kFlk_{\rm F}l regime, the planar Hall conductivity shows |𝐁|2|\mathbf{B}|^{2} dependence with no sign of saturation in a broad range of magnetic field as reported in the previous studies NANDY2017 .

Refer to caption
Figure 3: Planar Hall conductivity as a function of bμb_{\mu} in WSMs at Γ=π4\Gamma=\frac{\pi}{4} for kFl=0.5,2,5k_{\rm F}l=0.5,2,5.

We now come back to the approximation we made: D𝐤1=1D^{-1}_{\mathbf{k}}=1. Again, including the Taylor expanded D𝐤1D^{-1}_{\mathbf{k}} in Eq. (26), we obtain the planar Hall magnetoconductivity in WSMs as

σyx(𝐁)σ0[bμ21+bμ2(1+15bBC2)+75bBC2]sin(Γ)cos(Γ).\sigma_{yx}(\mathbf{B})\approx\sigma_{0}\Big{[}\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}\Big{(}1+\frac{1}{5}b^{2}_{\rm BC}\Big{)}+\frac{7}{5}b^{2}_{\rm BC}\Big{]}\sin{\Gamma}\cos{\Gamma}. (29)

See Appendix B.2 for detailed derivations.

III.3 Hall conductivity

To investigate the Hall conductivity in WSMs, we set the electric and magnetic field orientations as 𝐄=Exx^\mathbf{E}=E_{x}\hat{x} and 𝐁=Byy^\mathbf{B}=B_{y}\hat{y}, respectively. Then Eq. (II) gives the following form of Hall conductivity σzx(𝐁)\sigma_{zx}(\mathbf{B}) assuming D𝐤1=1D^{-1}_{\mathbf{k}}=1 (see Appendix B 3):

σzx(𝐁)=σ0(bμ1+bμ2).\sigma_{zx}(\mathbf{B})=\sigma_{0}\Big{(}\frac{b_{\mu}}{1+b^{2}_{\mu}}\Big{)}. (30)

Note that this result is identical to the conventional magnetotransport result. This implies that there is no anomalous velocity effect in the Hall conductivity.

We now come back to the approximation we made: D𝐤1=1D^{-1}_{\mathbf{k}}=1. Again, including the Taylor expanded D𝐤1D^{-1}_{\mathbf{k}} in Eq. (26), we can obtain the Hall magnetoconductivity in WSMs as

σzx(𝐁)σ0bμ1+bμ2(1+15bBC2).\sigma_{zx}(\mathbf{B})\approx\sigma_{0}\frac{b_{\mu}}{1+b^{2}_{\mu}}\left(1+\frac{1}{5}b^{2}_{\text{BC}}\right). (31)

See Appendix B.3 for detailed derivations.

III.4 Conductivity and resistivity tensors

Combining the previous results of magnetoconductivity in WSMs, we obtain the following conductivity tensor under 𝐁=Bxx^+Byy^\mathbf{B}=B_{x}\hat{x}+B_{y}\hat{y} with D𝐤1=1D^{-1}_{\mathbf{k}}=1:

𝝈(𝐁)=σ0[11+bμ2+(bμ21+bμ2+3bBC2)cos2Γ(bμ21+bμ2+3bBC2)sin(Γ)cos(Γ)bμ1+bμ2sin(Γ)(bμ21+bμ2+3bBC2)sin(Γ)cos(Γ)11+bμ2+(bμ21+bμ2+3bBC2)sin2Γbμ1+bμ2cos(Γ)bμ1+bμ2sin(Γ)bμ1+bμ2cos(Γ)11+bμ2],\displaystyle{\bm{\sigma}(\mathbf{B})}=\sigma_{0}\begin{bmatrix}\frac{1}{1+b^{2}_{\mu}}+({\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}+3b^{2}_{\rm BC}})\cos^{2}{\Gamma}&({\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}+3b^{2}_{\rm BC}})\sin{\Gamma}\cos{\Gamma}&-\frac{b_{\mu}}{1+b^{2}_{\mu}}\sin{\Gamma}\\ ({\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}+3b^{2}_{\rm BC}})\sin{\Gamma}\cos{\Gamma}&\frac{1}{1+b^{2}_{\mu}}+({\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}+3b^{2}_{\rm BC}})\sin^{2}{\Gamma}&\frac{b_{\mu}}{1+b^{2}_{\mu}}\cos{\Gamma}\\ \frac{b_{\mu}}{1+b^{2}_{\mu}}\sin{\Gamma}&-\frac{b_{\mu}}{1+b^{2}_{\mu}}\cos{\Gamma}&\frac{1}{1+b^{2}_{\mu}}\\ \end{bmatrix}, (32)

where Γ\Gamma is the angle between applied electric and magnetic fields. Since the resistivity tensor 𝝆(𝐁)\bm{\rho}(\mathbf{B}) is the inverse of the conductivity tensor 𝝈(𝐁)\bm{\sigma}(\mathbf{B}), we obtain 𝝆(𝐁)\bm{\rho}(\mathbf{B}) in the following form:

𝝆(𝐁)=ρ0\displaystyle\bm{\rho}(\mathbf{B})=\rho_{0} [1+3bBC2sin2Γ1+3bBC23bBC2sin(Γ)cos(Γ)1+3bBC2bμsin(Γ)3bBC2sin(Γ)cos(Γ)1+3bBC21+3bBC2cos2Γ1+3bBC2bμcos(Γ)bμsin(Γ)bμcos(Γ)1],\displaystyle\begin{bmatrix}\frac{1+3b^{2}_{\rm BC}\sin^{2}{\Gamma}}{1+3b^{2}_{\rm BC}}&-\frac{3b^{2}_{\rm BC}\sin{\Gamma}\cos{\Gamma}}{1+3b^{2}_{\rm BC}}&b_{\mu}\sin{\Gamma}\\ -\frac{3b^{2}_{\rm BC}\sin{\Gamma}\cos{\Gamma}}{1+3b^{2}_{\rm BC}}&\frac{1+3b^{2}_{\rm BC}\cos^{2}{\Gamma}}{1+3b^{2}_{\rm BC}}&-b_{\mu}\cos{\Gamma}\\ -b_{\mu}\sin{\Gamma}&b_{\mu}\cos{\Gamma}&1\\ \end{bmatrix}, (33)

where ρ0=ρxx(𝐁=0)=σ01\rho_{0}=\rho_{xx}(\mathbf{B}=0)=\sigma_{0}^{-1}. Note that although both the Lorentz force and the Berry curvature induced terms are present in σij(𝐁)\sigma_{ij}(\mathbf{B}), the Lorentz force induced terms do not appear in the resistivity ρij(𝐁)\rho_{ij}(\mathbf{B}) where i,j=x,yi,j=x,y. This result agrees with the previous one reporting that ρxx=ρΔρcos2Γ\rho_{xx}=\rho_{\parallel}-\Delta\rho\cos^{2}{\Gamma}, ρxy=Δρsin(Γ)cos(Γ)\rho_{xy}=-\Delta\rho\sin{\Gamma}\cos{\Gamma} where ρ\rho_{\parallel} (ρ\rho_{\perp}) is the resistivity in longitudinal (transverse) magnetic field and Δρ=ρρ=3bBC2/(1+3bBC2)\Delta\rho=\rho_{\perp}-\rho_{\parallel}=3b^{2}_{\rm BC}/(1+3b^{2}_{\rm BC}) is the resistivity anisotropy JAN1957 ; BURKOV2017 . For the conductivity and resistivity tensors including the Taylor expanded D𝐤1D^{-1}_{\mathbf{k}}, see Appendix B.5.

IV Conclusion

In this work, we presented a closed-form expression for the magnetoconductivity using the semiclassical magnetotransport theory that fully incorporates the Berry curvature and the Lorentz force effects. We then applied this formula to WSMs and obtained analytic expressions for the longitudinal, planar Hall and Hall conductivities in terms of dimensionless parameters bμb_{\mu} and bBCb_{\rm BC} which are normalized magnetic fields associated with the Lorentz force and the Berry curvature, respectively. From these results, we showed a non-monotonic field dependence in the longitudinal and planar Hall conductivities depending on kFlk_{\rm F}l. Furthermore, we clearly demonstrated that although the Lorentz force effect is manifested in the planar Hall conductivity, its contribution vanishes in the planar Hall resistivity.

Acknowledgements.
This work was supported by the National Research Foundation of Korea (NRF) grant funded by the Korea government (MSIT) (Grant No. 2018R1A2B6007837) and Creative-Pioneering Researchers Program through Seoul National University (SNU).

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Appendix A Derivation of the nonequilibrium distribution function

Here we go through a detailed derivation of the nonequilibrium distribution function. Let us start with Eq. (12) in the main text:

𝐆\displaystyle\mathbf{G} =𝐆0qτc(𝐁×𝕄1𝐆),\displaystyle=\mathbf{G}_{0}-\frac{q\tau}{c}(\mathbf{B}\times\mathds{M}^{-1}\mathbf{G}), (34)

where 𝕄ij1=1v~𝐤,jki\mathds{M}^{-1}_{ij}=\frac{1}{\hbar}\frac{\partial\tilde{\rm v}_{\mathbf{k},j}}{\partial k_{i}} is the inverse mass tensor and

𝐆0=qτ(f𝐤(0)ϵ𝐤)𝐄.\mathbf{G}_{0}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\mathbf{E}. (35)

To solve 𝐆\mathbf{G} in Eq. (34), note that Eq. (34) is given by the following self-consistent form for a vector 𝐱\mathbf{x}:

𝐱=𝐚+(M𝐱)×𝐛,\mathbf{x}=\mathbf{a}+(M\mathbf{x})\times\mathbf{b}, (36)

where 𝐚\mathbf{a} and 𝐛\mathbf{b} are vectors and MM is a matrix. Then Eq. (36) can be rewritten as

xi=ai+jkϵijk(M𝐱)jbk=ai+jkFijMjkxk,\displaystyle x_{i}=a_{i}+\sum_{jk}\epsilon_{ijk}(M\mathbf{x})_{j}b_{k}=a_{i}+\sum_{jk}F_{ij}M_{jk}x_{k}, (37)

where Fij=kϵijkbkF_{ij}=\sum_{k}\epsilon_{ijk}b_{k}. Thus, we have

𝐱=𝐚+FM𝐱=(1FM)1𝐚.\mathbf{x}=\mathbf{a}+FM\mathbf{x}=(1-FM)^{-1}\mathbf{a}. (38)

Using the above result Eq. (38), we obtain 𝐆\mathbf{G} as the following form:

𝐆=qτ(f𝐤(0)ϵ𝐤)(𝟙+qτc𝔽𝕄1)1𝐄,\mathbf{G}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\Big{(}\mathds{1}+\frac{q\tau}{c}\mathds{F}\mathds{M}^{-1}\Big{)}^{-1}\mathbf{E}, (39)

where 𝔽ij=iϵijkBk\mathds{F}_{ij}=\sum_{i}\epsilon_{ijk}B_{k} is the magnetic field strength tensor. Finally, the nonequilibrium distribution function g𝐤=𝐯~𝐤𝐆g_{\mathbf{k}}=\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\mathbf{G} is given by

g𝐤=qτ(f𝐤(0)ϵ𝐤)𝐯~𝐤[(𝟙+qτc𝔽𝕄1)1𝐄].g_{\mathbf{k}}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\Big{[}\Big{(}\mathds{1}+\frac{q\tau}{c}\mathds{F}\mathds{M}^{-1}\Big{)}^{-1}\mathbf{E}\Big{]}. (40)

The mobility tensor is defined as

μ~ij=eτ𝕄ij1,\tilde{\mu}_{ij}=e\tau\mathds{M}^{-1}_{ij}, (41)

which is in general a non-diagonal matrix. For a system with an isotropic energy dispersion in the absence of a magnetic field, the mobility tensor is given by a scalar multiple of an identity matrix. For simplicity, we assume that the mobility tensor is set to a constant μ\mu. Then, we can rewrite Eq. (34) as

𝐆\displaystyle\mathbf{G} =𝐆0μc𝐁×𝐆.\displaystyle=\mathbf{G}_{0}-\frac{\mu}{c}\mathbf{B}\times\mathbf{G}. (42)

Note that Eq. (42) is given by the following self-consistent form for a vector 𝐱\mathbf{x}:

𝐱=𝐚+𝐱×𝐛,\mathbf{x}=\mathbf{a}+\mathbf{x}\times\mathbf{b}, (43)

where 𝐚\mathbf{a} and 𝐛\mathbf{b} are vectors. Then Eq. (43) can be rewritten as

xi\displaystyle x_{i} =\displaystyle= ai+jkϵijkxjbk\displaystyle a_{i}+\sum_{jk}\epsilon_{ijk}x_{j}b_{k} (44)
=\displaystyle= ai+jkϵijk(aj+lmϵjlmxlbm)bk\displaystyle a_{i}+\sum_{jk}\epsilon_{ijk}\left(a_{j}+\sum_{lm}\epsilon_{jlm}x_{l}b_{m}\right)b_{k}
=\displaystyle= ai+jkϵijkajbk+jklm(δimδklδilδkm)xlbmbk\displaystyle a_{i}+\sum_{jk}\epsilon_{ijk}a_{j}b_{k}+\sum_{jklm}(\delta_{im}\delta_{kl}-\delta_{il}\delta_{km})x_{l}b_{m}b_{k}
=\displaystyle= ai+(𝐚×𝐛)i+(𝐱𝐛)bib2xi\displaystyle a_{i}+(\mathbf{a}\times\mathbf{b})_{i}+(\mathbf{x}\cdot\mathbf{b})b_{i}-b^{2}x_{i}
=\displaystyle= 11+b2[ai+(𝐚×𝐛)i+(𝐚𝐛)bi].\displaystyle\frac{1}{1+b^{2}}\left[a_{i}+(\mathbf{a}\times\mathbf{b})_{i}+(\mathbf{a}\cdot\mathbf{b})b_{i}\right].

Here, we used jϵijkϵmjl=δimδklδilδkm\sum_{j}\epsilon_{ijk}\epsilon_{mjl}=\delta_{im}\delta_{kl}-\delta_{il}\delta_{km} and 𝐱𝐛=(𝐚+𝐱×𝐛)𝐛=𝐚𝐛\mathbf{x}\cdot\mathbf{b}=(\mathbf{a}+\mathbf{x}\times\mathbf{b})\cdot\mathbf{b}=\mathbf{a}\cdot\mathbf{b}. Thus, we have

𝐱=11+b2[𝐚+(𝐚×𝐛)+(𝐚𝐛)𝐛].\mathbf{x}=\frac{1}{1+b^{2}}\left[\mathbf{a}+(\mathbf{a}\times\mathbf{b})+(\mathbf{a}\cdot\mathbf{b})\mathbf{b}\right]. (45)

Using the result of Eq. (45), we therefore obtain 𝐆\mathbf{G} as the following form:

𝐆=qτ(f𝐤(0)ϵ𝐤)𝐄+μc𝐄×𝐁+μ2c2(𝐄𝐁)𝐁1+μ2c2|𝐁|2.\displaystyle\mathbf{G}=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\frac{\mathbf{E}+\frac{\mu}{c}\mathbf{E}\times\mathbf{B}+\frac{\mu^{2}}{c^{2}}(\mathbf{E}\cdot\mathbf{B})\mathbf{B}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}. (46)

Note that Eq. (39) is reduced to Eq. (46) when the mobility tensor is given by μ~ij=μδij\tilde{\mu}_{ij}=\mu\delta_{ij}.

Finally, the nonequilibrium distribution function g𝐤=𝐯~𝐤𝐆g_{\mathbf{k}}=\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\mathbf{G} is given by

g=qτ(f𝐤(0)ϵ𝐤)𝐯~𝐤𝐄+μc𝐄×𝐁+μ2c2(𝐄𝐁)𝐁1+μ2c2|𝐁|2.\displaystyle g=q\tau\Big{(}-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}}\Big{)}\tilde{\mathbf{v}}_{\mathbf{k}}\cdot\frac{\mathbf{E}+\frac{\mu}{c}\mathbf{E}\times\mathbf{B}+\frac{\mu^{2}}{c^{2}}(\mathbf{E}\cdot\mathbf{B})\mathbf{B}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}. (47)

Appendix B Magnetoconductivity of Weyl semimetals

In this section, we derive the magnetoconductivity of Weyl semimetals including the Taylor expanded D𝐤1D_{\mathbf{k}}^{-1}. Here we set the magnetic field orientation as 𝐁=Bxx^+Byy^\mathbf{B}=B_{x}\hat{x}+B_{y}\hat{y}.

B.1 Longitudinal magnetoconductivity

Here we go through a detailed derivation of the longitudinal magnetoconductivity. Let us start with Eq. (21) in the main text:

σxx=q2d3k(2π)3τ(f𝐤(0)ϵ𝐤)D𝐤11+μ2c2|𝐁|2[Σ𝐤(0)+Σ𝐤(1)+Σ𝐤(2)]σxx(0)+σxx(1)+σxx(2),\sigma_{xx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})D^{-1}_{\mathbf{k}}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}\Sigma_{\mathbf{k}}^{(0)}+\Sigma_{\mathbf{k}}^{(1)}+\Sigma_{\mathbf{k}}^{(2)}\Big{]}\equiv\sigma^{(0)}_{xx}+\sigma^{(1)}_{xx}+\sigma^{(2)}_{xx}, (48)

where Σ𝐤(i)\Sigma^{(i)}_{\mathbf{k}} is a sum of the terms that include iith order of μ\mu described in Eq. (22) in the main text which corresponds to conductivity of σxx(i)\sigma^{(i)}_{xx} respectively. Inserting the Taylor expanded density of states correction

D𝐤1=1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)2+D^{-1}_{\mathbf{k}}=1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}+\cdots (49)

to Eq. (48) and focusing on Σ𝐤(0)\Sigma^{(0)}_{\mathbf{k}}, σxx(0)\sigma^{(0)}_{xx} yields

σxx(0)=q2d3k(2π)3τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2[vx22qcvx(𝐯𝐤𝛀𝐤)|𝐁|cosΓ+q2(c)2(𝐯𝐤𝛀𝐤)2|𝐁|2cos2Γ],\displaystyle\sigma^{(0)}_{xx}=q^{2}\!\!\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}v^{2}_{x}-\frac{2q}{\hbar c}v_{x}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|\cos\Gamma+\frac{q^{2}}{(\hbar c)^{2}}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{2}\cos^{2}\Gamma\Big{]}, (50)

where we replaced (f𝐤(0)ϵ𝐤)=δ(kkF)vF(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})=\frac{\delta(k-k_{\rm F})}{\hbar v_{\rm F}} for zero temperature. Expanding and throwing away terms that will give zero contribution after the momentum integral due to odd order in kik_{i} (i=x,y,z)(i=x,y,z), we are left with the following angular integral after switching to spherical coordinates then integrating kk out:

σxx(0)=q202π𝑑ϕ0π𝑑θkF2sinθ(2π)3τI(0)(θ,ϕ)vF(1+bμ2),\sigma^{\text{(0)}}_{xx}=q^{2}\int^{2\pi}_{0}d\phi\int^{\pi}_{0}d\theta\frac{k_{\rm F}^{2}\sin\theta}{(2\pi)^{3}}\frac{\tau I^{(0)}(\theta,\phi)}{\hbar v_{\rm F}(1+b^{2}_{\mu})}, (51)

where

I(0)(θ,ϕ)=vF2[\displaystyle I^{(0)}(\theta,\phi)=v_{\rm F}^{2}\Big{[} sin2θcos2ϕ+bBC2cos2Γ2bBC2cos2Γsin2θcos2ϕ\displaystyle\sin^{2}\theta\cos^{2}\phi+b^{2}_{\text{BC}}\cos^{2}\Gamma-2b^{2}_{\text{BC}}\cos^{2}\Gamma\sin^{2}\theta\cos^{2}\phi (52)
+\displaystyle+ bBC2(cos2Γsin4θcos4ϕ+sin2Γsin4θcos2ϕsin2ϕ)\displaystyle b^{2}_{\text{BC}}(\cos^{2}\Gamma\sin^{4}\theta\cos^{4}\phi+\sin^{2}\Gamma\sin^{4}\theta\cos^{2}\phi\sin^{2}\phi)
+\displaystyle+ bBC4(cos4Γsin2θcos2ϕ+cos2Γsin2Γsin2θsin2ϕ)].\displaystyle b^{4}_{\text{BC}}(\cos^{4}\Gamma\sin^{2}\theta\cos^{2}\phi+\cos^{2}\Gamma\sin^{2}\Gamma\sin^{2}\theta\sin^{2}\phi)\Big{]}.

Finally, carrying out the angular integral leads to

σxx(0)=σ01+bμ2[1+bBC25(8cos2Γ+sin2Γ)+bBC4cos2Γ],\sigma^{\text{(0)}}_{xx}=\frac{\sigma_{0}}{1+b^{2}_{\mu}}\left[1+\frac{b^{2}_{\text{BC}}}{5}\left(8\cos^{2}\Gamma+\sin^{2}\Gamma\right)+b^{4}_{\text{BC}}\cos^{2}\Gamma\right], (53)

where σ0=q2(kF22π2vF)(τvF23)=q2N0D\sigma_{0}=q^{2}(\frac{k_{\rm F}^{2}}{2\pi^{2}\hbar v_{\rm F}})(\frac{\tau v_{\rm F}^{2}}{3})=q^{2}N_{0}D is the longitudinal magnetoconductivity in the absence of magnetic field.

Focusing now on Σ𝐤(1)\Sigma^{(1)}_{\mathbf{k}}, σxx(1)\sigma^{(1)}_{xx} yields

σxx(1)=q2d3k(2π)3τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2μc[vx(𝐯𝐤×𝐁)x+qc(𝐯𝐤×𝐁)x(𝐯𝐤𝛀𝐤)|𝐁|cosΓ].\sigma^{(1)}_{xx}=q^{2}\!\!\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\frac{\mu}{c}\Big{[}-v_{x}(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}+\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|\cos\Gamma\Big{]}. (54)

The above whole expression vanishes after the momentum integral, because (𝐯𝐤×𝐁)x=vyBzvzBy=vzBy(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}=v_{y}B_{z}-v_{z}B_{y}=-v_{z}B_{y} as Bz=0B_{z}=0. This will lead to a single order in vzv_{z} in every term in the integrand therefore result in zero after integration.

Finally, focusing on Σ𝐤(2)\Sigma^{(2)}_{\mathbf{k}}, σxx(2)\sigma^{(2)}_{xx} yields

σxx(2)=q2d3k(2π)3\displaystyle\sigma^{(2)}_{xx}=q^{2}\!\!\int\frac{d^{3}k}{(2\pi)^{3}}\!\!\!\!\! τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2[(𝐯𝐤𝐁)vx|𝐁|cosΓqc(𝐯𝐤𝛀𝐤)|𝐁|3vxcosΓ\displaystyle\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})v_{x}|\mathbf{B}|\cos\Gamma-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{3}v_{x}\cos\Gamma (55)
\displaystyle- qc(𝐯𝐤𝐁)(𝐯𝐤𝛀𝐤)|𝐁|2cos2Γq2(c)2(𝐯𝐤𝛀𝐤)2|𝐁|4cos2Γ].\displaystyle\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{2}\cos^{2}\Gamma-\frac{q^{2}}{(\hbar c)^{2}}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{4}\cos^{2}\Gamma\Big{]}.

Simplifying the product of the Taylor expanded D𝐤1D^{-1}_{\mathbf{k}} and terms in the square bracket while again, keeping only the non-zero contribution, we are left with the following angular integral after switching to spherical coordinates then integrating kk out:

σxx(2)=q202π𝑑ϕ0π𝑑θkF2sinθ(2π)3τI(2)(θ,ϕ)vF(1+bμ2),\sigma^{(2)}_{xx}=q^{2}\int^{2\pi}_{0}d\phi\int^{\pi}_{0}d\theta\frac{k_{\rm F}^{2}\sin\theta}{(2\pi)^{3}}\frac{\tau I^{(2)}(\theta,\phi)}{\hbar v_{\rm F}(1+b^{2}_{\mu})}, (56)

where

I(2)(θ,ϕ)=vF2bμ2[\displaystyle I^{(2)}(\theta,\phi)=v^{2}_{\rm F}b^{2}_{\mu}\Big{[} sin2θcos2ϕcos2Γ+bBC2(cos2Γcos2Γsin2θcos2ϕsin2θcos2ϕcos4Γ\displaystyle\sin^{2}\theta\cos^{2}\phi\cos^{2}\Gamma+b^{2}_{\text{BC}}\big{(}\cos^{2}\Gamma-\cos^{2}\Gamma\sin^{2}\theta\cos^{2}\phi-\sin^{2}\theta\cos^{2}\phi\cos^{4}\Gamma (57)
\displaystyle- sin2θsin2ϕsin2Γcos2Γ+sin4θcos4ϕcos4Γ+3sin4θcos2ϕsin2ϕcos2Γsin2Γ)\displaystyle\sin^{2}\theta\sin^{2}\phi\sin^{2}\Gamma\cos^{2}\Gamma+\sin^{4}\theta\cos^{4}\phi\cos^{4}\Gamma+3\sin^{4}\theta\cos^{2}\phi\sin^{2}\phi\cos^{2}\Gamma\sin^{2}\Gamma\big{)}
+\displaystyle+ bBC4(sin2θcos2ϕcos4Γ+sin2θsin2ϕsin2Γcos2Γ)].\displaystyle b^{4}_{\text{BC}}\left(\sin^{2}\theta\cos^{2}\phi\cos^{4}\Gamma+\sin^{2}\theta\sin^{2}\phi\sin^{2}\Gamma\cos^{2}\Gamma\right)\Big{]}.

Finally, carrying out the angular integral leads to

σxx(2)=σ0bμ21+bμ2(cos2Γ+85bBC2cos2Γ+bBC4cos2Γ).\sigma^{(2)}_{xx}=\sigma_{0}\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}\left(\cos^{2}\Gamma+\frac{8}{5}b^{2}_{\text{BC}}\cos^{2}\Gamma+b^{4}_{\text{BC}}\cos^{2}\Gamma\right). (58)

Adding up the σxx(i)\sigma^{(i)}_{xx}s, we have

σxx=\displaystyle\sigma_{xx}= σ01+bμ2[1+bBC25(8cos2Γ+sin2Γ)+bBC4cos2Γ+bμ2(cos2Γ+85bBC2cos2Γ+bBC4cos2Γ)].\displaystyle\frac{\sigma_{0}}{1+b^{2}_{\mu}}\left[1+\frac{b^{2}_{\text{BC}}}{5}\left(8\cos^{2}\Gamma+\sin^{2}\Gamma\right)+b^{4}_{\text{BC}}\cos^{2}\Gamma+b^{2}_{\mu}\left(\cos^{2}\Gamma+\frac{8}{5}b^{2}_{\text{BC}}\cos^{2}\Gamma+b^{4}_{\text{BC}}\cos^{2}\Gamma\right)\right]. (59)

B.2 Planar Hall conductivity

Here we go through a detailed derivation of the planar Hall conductivity. Starting with Eq. (20) in the main text for d=3d=3, we again express it in powers of μ\mu while using the definition for 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}}. We then obtain,

σyx=q2d3k(2π)3τ(f𝐤(0)ϵ𝐤)D𝐤11+μ2c2|𝐁|2[Σ𝐤(0)+Σ𝐤(1)+Σ𝐤(2)]σyx(0)+σyx(1)+σyx(2),\sigma_{yx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})D^{-1}_{\mathbf{k}}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}\Sigma_{\mathbf{k}}^{(0)}+\Sigma_{\mathbf{k}}^{(1)}+\Sigma_{\mathbf{k}}^{(2)}\Big{]}\equiv\sigma^{(0)}_{yx}+\sigma^{(1)}_{yx}+\sigma^{(2)}_{yx}, (60)

where

Σ𝐤(0)=\displaystyle\Sigma_{\mathbf{k}}^{(0)}= vyvxqcvy(𝐯𝐤𝛀𝐤)|𝐁|cosΓqcvx(𝐯𝐤𝛀𝐤)|𝐁|sinΓ+(qc)2(𝐯𝐤𝛀𝐤)2|𝐁|2cosΓsinΓ,\displaystyle~{}v_{y}v_{x}-\frac{q}{\hbar c}v_{y}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|\cos\Gamma-\frac{q}{\hbar c}v_{x}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|\sin\Gamma+\left(\frac{q}{\hbar c}\right)^{2}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{2}\cos\Gamma\sin\Gamma, (61a)
Σ𝐤(1)=\displaystyle\Sigma_{\mathbf{k}}^{(1)}= μc[vy(𝐯𝐤×𝐁)x+qc(𝐯𝐤𝛀𝐤)(𝐯𝐤×𝐁)x|𝐁|sinΓ],\displaystyle~{}\frac{\mu}{c}\left[-v_{y}(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}+\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}|\mathbf{B}|\sin\Gamma\right], (61b)
Σ𝐤(2)=\displaystyle\Sigma_{\mathbf{k}}^{(2)}= μ2c2[(𝐯𝐤𝐁)vy|𝐁|cosΓqc(𝐯𝐤𝛀𝐤)|𝐁|3vycosΓqc(𝐯𝐤𝐁)(𝐯𝐤𝛀𝐤)|𝐁|2cosΓsinΓ\displaystyle~{}\frac{\mu^{2}}{c^{2}}\Big{[}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})v_{y}|\mathbf{B}|\cos\Gamma-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{3}v_{y}\cos\Gamma-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{2}\cos\Gamma\sin\Gamma
+\displaystyle+ (qc)2(𝐯𝐤𝛀𝐤)2|𝐁|4cosΓsinΓ],\displaystyle\left(\frac{q}{\hbar c}\right)^{2}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{4}\cos\Gamma\sin\Gamma\Big{]}, (61c)

each corresponds to planar Hall conductivity of σyx(i)\sigma^{(i)}_{yx} respectively. Inserting the Taylor expanded density of states correction

D𝐤1=1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)2+D^{-1}_{\mathbf{k}}=1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}+\cdots (62)

to Eq. (60) and focusing on Σ𝐤(0)\Sigma^{(0)}_{\mathbf{k}}, σyx(0)\sigma^{(0)}_{yx} yields

σyx(0)=q2d3k(2π)3τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2\displaystyle\sigma^{(0)}_{yx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}} [vyvxqcvy(𝐯𝐤𝛀𝐤)|𝐁|cosΓqcvx(𝐯𝐤𝛀𝐤)|𝐁|sinΓ\displaystyle\!\!\!\!\!\!\Big{[}v_{y}v_{x}-\frac{q}{\hbar c}v_{y}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|\cos\Gamma-\frac{q}{\hbar c}v_{x}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|\sin\Gamma (63)
+\displaystyle+ (qc)2(𝐯𝐤𝛀𝐤)2|𝐁|2cosΓsinΓ],\displaystyle\left(\frac{q}{\hbar c}\right)^{2}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{2}\cos\Gamma\sin\Gamma\Big{]},

where we replaced (f𝐤(0)ϵ𝐤)=δ(kkF)vF(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})=\frac{\delta(k-k_{\rm F})}{\hbar v_{\rm F}} for zero temperature. Expanding and throwing away terms that will give zero contribution after the momentum integral due to odd order in kik_{i} (i=x,y,z)(i=x,y,z), we are left with the following angular integral after switching to spherical coordinates then integrating kk out:

σyx(0)=q202π𝑑ϕ0π𝑑θkF2sinθ(2π)3τI(0)(θ,ϕ)vF(1+bμ2),\sigma^{(0)}_{yx}=q^{2}\int^{2\pi}_{0}d\phi\int^{\pi}_{0}d\theta\frac{k_{\rm F}^{2}\sin\theta}{(2\pi)^{3}}\frac{\tau I^{(0)}(\theta,\phi)}{\hbar v_{\rm F}(1+b^{2}_{\mu})}, (64)

where

I(0)(θ,ϕ)=vF2[\displaystyle I^{(0)}(\theta,\phi)=v_{\rm F}^{2}\Big{[} bBC2(cosΓsinΓsin2θcos2ϕcosΓsinΓ\displaystyle b^{2}_{\text{BC}}\big{(}\cos\Gamma\sin\Gamma-\sin^{2}\theta\cos^{2}\phi\cos\Gamma\sin\Gamma (65)
\displaystyle- sin2θsin2ϕcosΓsinΓ+2sin4θcos2ϕsin2ϕcosΓsinΓ)\displaystyle\sin^{2}\theta\sin^{2}\phi\cos\Gamma\sin\Gamma+2\sin^{4}\theta\cos^{2}\phi\sin^{2}\phi\cos\Gamma\sin\Gamma\big{)}
+\displaystyle+ bBC4(sin2θcos2ϕcos3ΓsinΓ+sin2θsin2ϕcosΓsin3Γ)].\displaystyle b^{4}_{\text{BC}}\big{(}\sin^{2}\theta\cos^{2}\phi\cos^{3}\Gamma\sin\Gamma+\sin^{2}\theta\sin^{2}\phi\cos\Gamma\sin^{3}\Gamma\big{)}\Big{]}.

Finally, carrying out the angular integral leads to

σyx(0)=σ01+bμ2(75bBC2+bBC4)cosΓsinΓ.\sigma^{(0)}_{yx}=\frac{\sigma_{0}}{1+b^{2}_{\mu}}\left(\frac{7}{5}b^{2}_{\text{BC}}+b^{4}_{\text{BC}}\right)\cos\Gamma\sin\Gamma. (66)

Focusing now on Σ𝐤(1)\Sigma^{(1)}_{\mathbf{k}}, σyx(1)\sigma^{(1)}_{yx} yields

σyx(1)=q2d3k(2π)3τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2μc[vy(𝐯𝐤×𝐁)x+qc(𝐯𝐤𝛀𝐤)(𝐯𝐤×𝐁)x|𝐁|sinΓ].\sigma^{(1)}_{yx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\frac{\mu}{c}\left[-v_{y}(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}+\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}|\mathbf{B}|\sin\Gamma\right]. (67)

The above whole expression vanishes after the momentum integral, because (𝐯𝐤×𝐁)x=vyBzvzBy=vzBy(\mathbf{v}_{\mathbf{k}}\times\mathbf{B})_{x}=v_{y}B_{z}-v_{z}B_{y}=-v_{z}B_{y} as Bz=0B_{z}=0. This will lead to a single order in vzv_{z} in every term in the integrand therefore result in zero after integration.

Finally, focusing on Σ𝐤(2)\Sigma^{(2)}_{\mathbf{k}}, σyx(2)\sigma^{(2)}_{yx} yields

σyx(2)=q2d3k(2π)3τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2μ2c2[(𝐯𝐤𝐁)vy|𝐁|cosΓqc(𝐯𝐤𝛀𝐤)|𝐁|3vycosΓ\displaystyle\sigma^{(2)}_{yx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\frac{\mu^{2}}{c^{2}}\Big{[}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})v_{y}|\mathbf{B}|\cos\Gamma-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{3}v_{y}\cos\Gamma (68)
qc(𝐯𝐤𝐁)(𝐯𝐤𝛀𝐤)|𝐁|2cosΓsinΓ+(qc)2(𝐯𝐤𝛀𝐤)2|𝐁|4cosΓsinΓ].\displaystyle-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{2}\cos\Gamma\sin\Gamma+\left(\frac{q}{\hbar c}\right)^{2}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}|\mathbf{B}|^{4}\cos\Gamma\sin\Gamma\Big{]}.

Simplifying the product of the Taylor expanded D𝐤1D^{-1}_{\mathbf{k}} and terms in the square bracket while again, keeping only the non-zero contribution, we are left with the following angular integral after switching to spherical coordinates then integrating kk out:

σyx(2)=q202π𝑑ϕ0π𝑑θkF2sinθ(2π)3τI(2)(θ,ϕ)vF(1+bμ2),\sigma^{(2)}_{yx}=q^{2}\int^{2\pi}_{0}d\phi\int^{\pi}_{0}d\theta\frac{k_{\rm F}^{2}\sin\theta}{(2\pi)^{3}}\frac{\tau I^{(2)}(\theta,\phi)}{\hbar v_{\rm F}(1+b^{-2}_{\mu})}, (69)

where

I(2)(θ,ϕ)=vF2bμ2\displaystyle I^{(2)}(\theta,\phi)=v_{\rm F}^{2}b^{2}_{\mu} [\displaystyle\Big{[} sin3θsin2ϕcosΓsinΓ+bBC2(sinθcosΓsinΓsin3θcos2ϕcos3ΓsinΓsin3θsin2ϕsinΓcosΓ\displaystyle\sin^{3}\theta\sin^{2}\phi\cos\Gamma\sin\Gamma+b^{2}_{\text{BC}}\big{(}\sin\theta\cos\Gamma\sin\Gamma-\sin^{3}\theta\cos^{2}\phi\cos^{3}\Gamma\sin\Gamma-\sin^{3}\theta\sin^{2}\phi\sin\Gamma\cos\Gamma (70)
\displaystyle- sin3θsin2ϕcosΓsin3Γ+sin5θcos2ϕsin2ϕcos3ΓsinΓ+2sin5θcos2ϕsin2ϕcos3ΓsinΓ\displaystyle\sin^{3}\theta\sin^{2}\phi\cos\Gamma\sin^{3}\Gamma+\sin^{5}\theta\cos^{2}\phi\sin^{2}\phi\cos^{3}\Gamma\sin\Gamma+2\sin^{5}\theta\cos^{2}\phi\sin^{2}\phi\cos^{3}\Gamma\sin\Gamma
+\displaystyle+ sin5θsin4ϕcosΓsin3Γ)+bBC4(sin3θcos2ϕcos3ΓsinΓ+sin3θsin2ϕcosΓsin3Γ)].\displaystyle\sin^{5}\theta\sin^{4}\phi\cos\Gamma\sin^{3}\Gamma\big{)}+b^{4}_{\text{BC}}\big{(}\sin^{3}\theta\cos^{2}\phi\cos^{3}\Gamma\sin\Gamma+\sin^{3}\theta\sin^{2}\phi\cos\Gamma\sin^{3}\Gamma\big{)}\Big{]}.

Finally, carrying out the angular integral leads to

σyx(2)=σ0bμ21+bμ2(1+85bBC2+bBC4)cosΓsinΓ.\sigma^{(2)}_{yx}=\sigma_{0}\frac{b^{2}_{\mu}}{1+b^{2}_{\mu}}\Big{(}1+\frac{8}{5}b^{2}_{\text{BC}}+b^{4}_{\text{BC}}\Big{)}\cos\Gamma\sin\Gamma. (71)

Adding up the σyx(i)\sigma^{(i)}_{yx}s, we have

σyx=σ01+bμ2[(75bBC2+bBC4)cosΓsinΓ+bμ2(1+85bBC2+bBC4)cosΓsinΓ].\sigma_{yx}=\frac{\sigma_{0}}{1+b^{2}_{\mu}}\left[\left(\frac{7}{5}b^{2}_{\text{BC}}+b^{4}_{\text{BC}}\right)\cos\Gamma\sin\Gamma+b^{2}_{\mu}\Big{(}1+\frac{8}{5}b^{2}_{\text{BC}}+b^{4}_{\text{BC}}\Big{)}\cos\Gamma\sin\Gamma\right]. (72)

B.3 Conductivity σzx\sigma_{zx}

Here we go through a detailed derivation of σzx\sigma_{zx}. Starting with Eq. (20) in the main text for d=3d=3, we again express it in powers of μ\mu while using the definition for v~𝐤\tilde{v}_{\mathbf{k}}. We then obtain

σzx=q2d3k(2π)3τ(f𝐤(0)ϵ𝐤)D𝐤11+μ2c2|𝐁|2[Σ𝐤(0)+Σ𝐤(1)+Σ𝐤(2)]σzx(0)+σzx(1)+σzx(2),\sigma_{zx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})D^{-1}_{\mathbf{k}}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}\Sigma_{\mathbf{k}}^{(0)}+\Sigma_{\mathbf{k}}^{(1)}+\Sigma_{\mathbf{k}}^{(2)}\Big{]}\equiv\sigma^{(0)}_{zx}+\sigma^{(1)}_{zx}+\sigma^{(2)}_{zx}, (73)

where

Σ𝐤(0)=\displaystyle\Sigma_{\mathbf{k}}^{(0)}= vzvxqc(𝐯𝐤𝛀𝐤)vz|𝐁|cosΓ,\displaystyle v_{z}v_{x}-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})v_{z}|\mathbf{B}|\cos\Gamma, (74a)
Σ𝐤(1)=\displaystyle\Sigma_{\mathbf{k}}^{(1)}= μcvz(𝐯𝐤×𝐁)x,\displaystyle~{}-\frac{\mu}{c}v_{z}\left(\mathbf{v}_{\mathbf{k}}\times\mathbf{B}\right)_{x}, (74b)
Σ𝐤(2)=\displaystyle\Sigma_{\mathbf{k}}^{(2)}= μ2c2[(𝐯𝐤𝐁)qc(𝐯𝐤𝛀𝐤)|𝐁|2]vz|𝐁|cosΓ,\displaystyle~{}\frac{\mu^{2}}{c^{2}}\left[(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{2}\right]v_{z}|\mathbf{B}|\cos\Gamma, (74c)

each corresponds to Hall conductivity of σzx(i)\sigma^{(i)}_{zx} respectively. Inserting the Taylor expanded density of states correction to Eq. (73) and focusing on Σ𝐤(0)\Sigma^{(0)}_{\mathbf{k}}, σzx(0)\sigma^{(0)}_{zx} yields

σzx(0)=q2d3k(2π)3τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2[vzvxqc(𝐯𝐤𝛀𝐤)vz|𝐁|cosΓ],\displaystyle\sigma^{(0)}_{zx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}v_{z}v_{x}-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})v_{z}|\mathbf{B}|\cos\Gamma\Big{]}, (75)

where we replaced (f𝐤(0)ϵ𝐤)=δ(kkF)vF(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})=\frac{\delta(k-k_{\rm F})}{\hbar v_{\rm F}} for zero temperature. The above whole expression in Eq. (75) vanishes after the momentum integral.

Focusing now on Σ𝐤(1)\Sigma^{(1)}_{\mathbf{k}}, σzx(1)\sigma^{(1)}_{zx} yields

σzx(1)=q2d3k(2π)3τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2[μcvz(𝐯𝐤×𝐁)x].\displaystyle\sigma^{(1)}_{zx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}-\frac{\mu}{c}v_{z}\left(\mathbf{v}_{\mathbf{k}}\times\mathbf{B}\right)_{x}\Big{]}. (76)

Expanding and throwing away terms that will give zero contribution after the momentum integral due to odd order in kik_{i} (i=x,y,z)(i=x,y,z), we are left with the following angular integral after switching to spherical coordinates then integrating kk out:

σzx(2)=q202π𝑑ϕ0π𝑑θkF2sinθ(2π)3τbμI(1)(θ,ϕ)vF(1+bμ2),\sigma^{(2)}_{zx}=q^{2}\int^{2\pi}_{0}d\phi\int^{\pi}_{0}d\theta\frac{k_{\rm F}^{2}\sin\theta}{(2\pi)^{3}}\frac{\tau b_{\mu}I^{(1)}(\theta,\phi)}{\hbar v_{\rm F}(1+b^{2}_{\mu})}, (77)

where

I(1)(θ,ϕ)=vF2[cos2θsinθsinΓ+bBC2(cos2θsin3θcos2ϕcos2ΓsinΓ+cos2θsin3θsin2ϕsin3Γ)].I^{(1)}(\theta,\phi)=v_{\rm F}^{2}\Big{[}\cos^{2}\theta\sin\theta\sin\Gamma+b^{2}_{\text{BC}}\big{(}\cos^{2}\theta\sin^{3}\theta\cos^{2}\phi\cos^{2}\Gamma\sin\Gamma+\cos^{2}\theta\sin^{3}\theta\sin^{2}\phi\sin^{3}\Gamma\big{)}\Big{]}. (78)

Finally, carrying out the angular integral leads to

σzx(1)=σ0bμ1+bμ2(1+15bBC2)sinΓ.\sigma^{(1)}_{zx}=\sigma_{0}\frac{b_{\mu}}{1+b^{2}_{\mu}}\left(1+\frac{1}{5}b^{2}_{\text{BC}}\right)\sin\Gamma. (79)

Finally, focusing on Σ𝐤(2)\Sigma^{(2)}_{\mathbf{k}}, σzx(2)\sigma^{(2)}_{zx} yields

σzx(0)=q2d3k(2π)3τδ(kkF)vF1+qc(𝐁𝛀𝐤)+(qc)2(𝐁𝛀𝐤)21+μ2c2|𝐁|2μ2c2[(𝐯𝐤𝐁)qc(𝐯𝐤𝛀𝐤)|𝐁|2]vz|𝐁|cosΓ.\sigma^{(0)}_{zx}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau\delta(k-k_{\rm F})}{\hbar v_{\rm F}}\frac{1+\frac{q}{\hbar c}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})+(\frac{q}{\hbar c})^{2}(\mathbf{B}\cdot\mathbf{\Omega}_{\mathbf{k}})^{2}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\frac{\mu^{2}}{c^{2}}\left[(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{B})-\frac{q}{\hbar c}(\mathbf{v}_{\mathbf{k}}\cdot\mathbf{\Omega}_{\mathbf{k}})|\mathbf{B}|^{2}\right]v_{z}|\mathbf{B}|\cos\Gamma. (80)

The above expression vanishes after the momentum integral.

Adding up the σzx(i)\sigma^{(i)}_{zx}s, we have

σzx=σ0bμ1+bμ2(1+15bBC2)sinΓ.\sigma_{zx}=\sigma_{0}\frac{b_{\mu}}{1+b^{2}_{\mu}}\left(1+\frac{1}{5}b^{2}_{\text{BC}}\right)\sin\Gamma. (81)

Note that σzx\sigma_{zx} at Γ=π2\Gamma={\pi\over 2} in Eq. (81) corresponds to the Hall conductivity.

B.4 Conductivity σzz\sigma_{zz}

Here we go through a detailed derivation of the σzz\sigma_{zz}. Starting with Eq. (20) in the main text for d=3d=3, we again express it in powers of μ\mu while using the definition for 𝐯~𝐤\tilde{\mathbf{v}}_{\mathbf{k}}. We then obtain

σzz=q2d3k(2π)3τ(f𝐤(0)ϵ𝐤)D𝐤11+μ2c2|𝐁|2[Σ𝐤(0)+Σ𝐤(1)+Σ𝐤(2)]σzz(0)+σzz(1)+σzz(2),\sigma_{zz}=q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\tau(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})D^{-1}_{\mathbf{k}}}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}\Big{[}\Sigma_{\mathbf{k}}^{(0)}+\Sigma_{\mathbf{k}}^{(1)}+\Sigma_{\mathbf{k}}^{(2)}\Big{]}\equiv\sigma^{(0)}_{zz}+\sigma^{(1)}_{zz}+\sigma^{(2)}_{zz}, (82)

where

Σ𝐤(0)=\displaystyle\Sigma_{\mathbf{k}}^{(0)}= vz2,\displaystyle v^{2}_{z}, (83a)
Σ𝐤(1)=\displaystyle\Sigma_{\mathbf{k}}^{(1)}= μcvz(𝐯𝐤×𝐁)z,\displaystyle~{}-\frac{\mu}{c}v_{z}\left(\mathbf{v}_{\mathbf{k}}\times\mathbf{B}\right)_{z}, (83b)
Σ𝐤(2)=\displaystyle\Sigma_{\mathbf{k}}^{(2)}= 0,\displaystyle~{}0, (83c)

as Bz=0B_{z}=0. Note that after the momentum integral, Σ𝐤(1)\Sigma^{(1)}_{\mathbf{k}} will vanish due to odd order in vzv_{z}. The only remaining contribution is from Σ𝐤(0)\Sigma^{(0)}_{\mathbf{k}} which gives

σzz(0)=\displaystyle\sigma^{(0)}_{zz}= q2d3k(2π)3D𝐤1τ(f𝐤(0)ϵ𝐤)1+μ2c2|𝐁|2vz2\displaystyle q^{2}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{D^{-1}_{\mathbf{k}}\tau(-\frac{\partial f^{(0)}_{\mathbf{k}}}{\partial\epsilon_{\mathbf{k}}})}{1+\frac{\mu^{2}}{c^{2}}|\mathbf{B}|^{2}}v^{2}_{z} (84)
=\displaystyle= q2dΩvF(2π)3τvF2kF21+bμ2(cos2θ+bBC2cos2θsin2θcos2ϕcos2Γ+bBC2cos2θsin2θsin2ϕsin2Γ)sinθ\displaystyle q^{2}\int\frac{d\Omega}{\hbar v_{\rm F}(2\pi)^{3}}\frac{\tau v_{\rm F}^{2}k_{\rm F}^{2}}{1+b^{2}_{\mu}}\Big{(}\cos^{2}\theta+b^{2}_{\text{BC}}\cos^{2}\theta\sin^{2}\theta\cos^{2}\phi\cos^{2}\Gamma+b^{2}_{\text{BC}}\cos^{2}\theta\sin^{2}\theta\sin^{2}\phi\sin^{2}\Gamma\Big{)}\sin\theta
=\displaystyle= σ01+bμ2(1+15bBC2).\displaystyle\frac{\sigma_{0}}{1+b^{2}_{\mu}}\left(1+\frac{1}{5}b^{2}_{\text{BC}}\right).

B.5 Conductivity and resistivity tensors

Finally, we have the following conductivity tensor including the Taylor expanded D𝐤1D_{\mathbf{k}}^{-1}:

𝝈(𝐁)\displaystyle{\bm{\sigma}(\mathbf{B})} =\displaystyle= σ01+bμ2\displaystyle\frac{\sigma_{0}}{1+b^{2}_{\mu}}
×\displaystyle\times [1+15bBC2+[75bBC2+bμ2(1+85bBC2)]cos2Γ[bμ2(1+15bBC2)+75(1+bμ2)bBC2]sin(Γ)cos(Γ)bμ(1+15bBC2)sinΓ[bμ2(1+15bBC2)+75(1+bμ2)bBC2]sin(Γ)cos(Γ)1+15bBC2+[75bBC2+bμ2(1+85bBC2)]sin2Γbμ(1+15bBC2)cosΓbμ(1+15bBC2)sinΓbμ(1+15bBC2)cosΓ1+15bBC2].\displaystyle\begin{bmatrix}1+\frac{1}{5}b^{2}_{\rm BC}+\Big{[}\frac{7}{5}b^{2}_{\rm BC}+b_{\mu}^{2}\Big{(}1+\frac{8}{5}b^{2}_{\rm BC}\Big{)}\Big{]}\cos^{2}{\Gamma}&\Big{[}{b^{2}_{\mu}}(1+\frac{1}{5}b^{2}_{\rm BC})+\frac{7}{5}(1+b^{2}_{\mu})b^{2}_{\rm BC}\Big{]}\sin{\Gamma}\cos{\Gamma}&-b_{\mu}\left(1+\frac{1}{5}b^{2}_{\text{BC}}\right)\sin\Gamma\\ \Big{[}{b^{2}_{\mu}}(1+\frac{1}{5}b^{2}_{\rm BC})+\frac{7}{5}(1+b^{2}_{\mu})b^{2}_{\rm BC}\Big{]}\sin{\Gamma}\cos{\Gamma}&1+\frac{1}{5}b^{2}_{\rm BC}+\Big{[}\frac{7}{5}b^{2}_{\rm BC}+b_{\mu}^{2}\Big{(}1+\frac{8}{5}b^{2}_{\rm BC}\Big{)}\Big{]}\sin^{2}{\Gamma}&b_{\mu}\left(1+\frac{1}{5}b^{2}_{\text{BC}}\right)\cos\Gamma\\ b_{\mu}\left(1+\frac{1}{5}b^{2}_{\text{BC}}\right)\sin\Gamma&-b_{\mu}\left(1+\frac{1}{5}b^{2}_{\text{BC}}\right)\cos\Gamma&1+\frac{1}{5}b^{2}_{\text{BC}}\\ \end{bmatrix}.

The resistivity tensor 𝝆(𝐁)\bm{\rho}(\mathbf{B}) is then obtained by the inverse of the conductivity tensor 𝝈(𝐁)\bm{\sigma}(\mathbf{B}) as

𝝆(𝐁)\displaystyle{\bm{\rho}(\mathbf{B})} =\displaystyle= ρ0[50+10bBC2+70bBC2sin2Γ50+90bBC2+16bBC435bBC2cosΓsinΓ25+45bBC2+8bBC45bμsinΓ5+bBC235bBC2cosΓsinΓ25+45bBC2+8bBC450+10bBC2+70bBC2cos2Γ50+90bBC2+16bBC45bμcosΓ5+bBC25bμsinΓ5+bBC25bμcosΓ5+bBC255+bBC2].\displaystyle\rho_{0}\begin{bmatrix}\frac{50+10b^{2}_{\text{BC}}+70b^{2}_{\text{BC}}\sin^{2}\Gamma}{50+90b^{2}_{\text{BC}}+16b^{4}_{\text{BC}}}&-\frac{35b^{2}_{\text{BC}}\cos\Gamma\sin\Gamma}{25+45b^{2}_{\text{BC}}+8b^{4}_{\text{BC}}}&\frac{5b_{\mu}\sin\Gamma}{5+b^{2}_{\text{BC}}}\\ -\frac{35b^{2}_{\text{BC}}\cos\Gamma\sin\Gamma}{25+45b^{2}_{\text{BC}}+8b^{4}_{\text{BC}}}&\frac{50+10b^{2}_{\text{BC}}+70b^{2}_{\text{BC}}\cos^{2}\Gamma}{50+90b^{2}_{\text{BC}}+16b^{4}_{\text{BC}}}&-\frac{5b_{\mu}\cos\Gamma}{5+b^{2}_{\text{BC}}}\\ -\frac{5b_{\mu}\sin\Gamma}{5+b^{2}_{\text{BC}}}&\frac{5b_{\mu}\cos\Gamma}{5+b^{2}_{\text{BC}}}&\frac{5}{5+b^{2}_{\text{BC}}}\\ \end{bmatrix}. (86)

Note that although both the Lorentz force and the Berry curvature induced terms are present in σij(𝐁)\sigma_{ij}(\mathbf{B}), the Lorentz force induced terms do not appear in the resistivity ρij(𝐁)\rho_{ij}(\mathbf{B}) where i,j=x,yi,j=x,y even if we include the Taylor expanded D𝐤1D^{-1}_{\mathbf{k}}.