# Phonon helicity induced by electronic Berry curvature in Dirac materials

Lun-Hui Hu  Department of Physics, the Pennsylvania State University,
University Park, PA 16802  Jiabin Yu  Condensed Matter Theory Center,
Department of Physics, University of Maryland, College Park, Maryland 20742,
USA  Department of Physics, the Pennsylvania State University, University
Park, PA 16802  Ion Garate  Département de Physique, Institut Quantique and
Regroupement Québécois sur les Matériaux de Pointe, Université de Sherbrooke,
Sherbrooke, Québec, Canada J1K 2R1  Chao-Xing Liu<EMAIL_ADDRESS>Department of
Physics, the Pennsylvania State University, University Park, PA 16802

###### Abstract

In two-dimensional insulators with time-reversal (TR) symmetry, a nonzero
local Berry curvature of low-energy massive Dirac fermions can give rise to
nontrivial spin and charge responses, even though the integral of the Berry
curvature over all occupied states is zero. In this work, we present a new
effect induced by the electronic Berry curvature. By studying electron-phonon
interactions in BaMnSb2, a prototype two-dimensional Dirac material possessing
two TR-related massive Dirac cones, we find that the nonzero local Berry
curvature of electrons can induce a phonon angular momentum. The direction of
this phonon angular momentum is locked to the phonon propagation direction,
and thus we refer to it as “phonon helicity”, in a way that is reminiscent of
electron helicity in spin-orbit-coupled electronic systems. We discuss
possible experimental probes of such phonon helicity.

Introduction - Berry phase and Berry curvature play a vital role in various
branches of physics[1], owing to their deep connection to gauge field theories
and differential geometry[2, 3]. In condensed matter systems with broken TR
symmetry, the integration of the electronic Berry curvature over all the
occupied states in the first Brillouin zone underpins a number of Hall-related
phenomena, including the quantum Hall and quantum anomalous Hall effect [4, 5,
6, 7], (intrinsic) anomalous Hall effect [8] and orbital magnetic moments [9,
10]. Similarly, the Berry phase and curvature of nonelectronic quasiparticles
such as phonons and magnons lead to a thermal Hall effect in insulating
crystals with broken TR[11, 12, 13, 14, 15, 16]. In a TR invariant system, the
integral of Berry curvature over all the occupied states vanishes.
Nevertheless, nonvanishing Berry curvature can still appear locally in the
Brillouin zone, inducing a variety of electronic phenomena, such as the spin
[17] and valley Hall effects [18, 19, 20], nonlinear Hall effect [21, 22, 23,
24, 25], piezo-electromagnetic response [26, 27, 28, 29, 30, 31, 32] and
quantized circular photo-galvanic effect in Weyl semimetals [33, 34, 35, 36].

In certain two-dimensional (2D) insulators with TR symmetry, the low-energy
electronic excitations are 2D massive Dirac fermions. In these systems, the
gapped Dirac cones appear in pairs at two TR-related momenta (called two
valleys below) with opposite local Berry curvatures due to TR symmetry.
Although the total Hall conductivity vanishes, as proposed in Refs. [31] and
[32], a nonzero Hall current can be induced by dynamical strain. Specifically,
the dynamical strain plays the role of an artificial gauge field, dubbed
“pseudo-gauge field” in literature[37]. One may wonder whether there are more
ways (other than dynamical strain) to induce nonvanishing charge and spin
response [17] for the local Berry curvature in TR invariant gapped systems.
Furthermore, one may ask whether the electronic local Berry curvature can
induce nontrivial behaviors in quasiparticles other than electrons.

In this work, we study the phonon dynamics in a prototype of 2D Dirac
material, BaMnSb2. By combining symmetry analysis and deformation potential
theory, we find that the electron-phonon (e-ph) coupling in this material has
the same form as the coupling between Dirac fermions and a $\mathrm{U}(1)$
gauge field, meaning that phonons can also act as pseudo-gauge fields and
thereby induce nonvanishing Hall currents [31, 32]. The electronic local Berry
curvature can induce an out-of-plane phonon angular momentum (PAM), whose sign
reverses for opposite in-plane phonon momenta. We refer to it as ”phonon
helicity”, because it resembles the helical spin texture of spin-orbit-coupled
bands in the momentum space for electronic systems. Specifically, the nonzero
electronic local Berry curvature first introduces a self-energy correction to
the phonon Green’s function, which is an odd function of the phonon momentum
q, and then this self-energy correction makes the lattice vibration
elliptical, eventually giving rise to nonzero PAM. This Berry-curvature
contribution to phonon dynamics can be probed through measuring the total
phonon angular momentum with a temperature gradient or through optical
measurements of spatial dispersion of dielectric function.

Model Hamiltonian– We take BaMnSb2 as a prototype model system. BaMnSb2 is a
layered material [38] with alternating Ba-Sb layers and Mn-Sb layers stacked
along the $(001)$ direction. Density-functional theory (DFT) calculations in
Refs. [39] and [38] suggest that the electronic bands near the Fermi energy
mainly come from the $p$-orbitals of the Sb atoms in the Ba-Sb layers, while
the Mn-Sb layers serve as insulating barriers that prevent the tunneling of
electrons along the $(001)$ direction. Consequently, the bands near the Fermi
energy are almost non-dispersive along the $(001)$ direction, and thus BaMnSb2
is a quasi-2D material. The unit cell in each Ba-Sb layer contains two Sb
atoms, labeled as Sb1 and Sb2 in Fig. 1(a), both forming a square lattice.
Distortion shifts Sb2 atoms away from the center of the squares formed by Sb1
atoms, resulting in a series of zig-zag chains. Due to the zig-zag distortion,
the point group of BaMnSb2 is $C_{2v}$, spanned by a two-fold rotation along
the $x$-axis ($C_{2x}$) and $z$-directional ($\sigma_{v}(xy)$) mirror planes.
Magnetic moments in the Mn-Sb layers have little influence on the low-energy
bands near the Fermi energy in the Ba-Sb layer, and thus the TR symmetry is
taken into account.

We first describe the low-energy electronic properties. DFT calculations [38]
indicate that the low-energy bands can be captured by two 2D massive Dirac
cones at two momenta $K_{\pm}=(\pi/a,\pm k_{y_{0}})$, noted in Fig. 1(d). The
energy band sequence around $K_{\pm}$ is schematically shown in Fig. 1(e), in
which four low-energy bands at each momentum come from the two atomic orbitals
($p_{x,y}$ orbitals) and two spin states of Sb1 atoms. The low-energy
Hamiltonian is $\mathcal{H}_{s}=\int
d^{2}k\Psi_{s,\mathbf{k}}^{\dagger}h_{s}(\mathbf{k})\Psi_{s,\mathbf{k}}$ with

$\displaystyle
h_{s}(\mathbf{k})=sv_{0}(k_{y}\tau_{3}+k_{x}\tau_{1})+sm_{0}\tau_{2},$ (1)

where $v_{0}$ is the velocity of Dirac electrons and $m_{0}$ is the Dirac mass
tuned by the zig-zag distortion. Here $s=\pm$ labels two valleys, and at each
valley, the basis is labeled by
$\Psi^{\dagger}_{s}(\mathbf{k})=(c_{s,p_{x},\mathbf{k}}^{\dagger},c_{s,p_{y},\mathbf{k}}^{\dagger})$,
and $\tau_{i=0,1,2,3}$ are Pauli matrices for the orbital index. The spin-
valley locking allows us to only consider electrons with one specific spin
direction around each valley at low energies [38], as shown in Fig. 1(e).
Within this low-energy subspace, $\sigma_{v}(xy)$ corresponds to an identity
matrix and does not change the 2D momentum. Thus, we only consider $C_{2x}$
and TR ($\mathcal{T}$) operations henceforth. The Hamiltonian (1) transforms
as $C_{2x}h_{s}(k_{x},k_{y})C_{2x}^{-1}=h_{-s}(k_{x},-k_{y})$ and
$\mathcal{T}h_{s}(\mathbf{k})\mathcal{T}^{-1}=h_{-s}(-\mathbf{k})$, where the
symmetry representations are $C_{2x}=\tau_{3}$ and
$\mathcal{T}=\tau_{0}\mathcal{K}$ ($\mathcal{K}$ stands for complex
conjugation). Within one valley, we only have the combined symmetry
$C_{2x}\mathcal{T}=\tau_{3}\mathcal{K}$.

For the phonons, we focus on the in-plane vibrational modes, which are
decoupled from the out-of-plane modes due to the opposite parities under
$\sigma_{v}(xy)$. There are two Sb atoms in one unit cell and each atom can
vibrate in the $x$ or $y$ direction. Thus, we have in total four phonon modes,
two acoustic and two optical, which are classified according to the $C_{2v}$
point group, as shown in table I of Sec. I.A in Supplemental Material (SM)
[40]. Specifically, the lattice vibrations along the $x$ and $y$ directions
are decoupled at ${\bf q}=0$ due to the opposite signs under $C_{2x}$; the
vibration patterns of these four modes are schematically shown in Fig. 1(b)
and (c). The displacement vectors for these modes are labelled by
$\vec{u}_{A_{1}}^{a},\vec{u}_{B_{1}}^{a},\vec{u}_{A_{1}}^{o},\vec{u}_{B_{1}}^{o}$,
where $a/o$ are for acoustic and optical phonon modes and $A_{1},B_{1}$ are
the corresponding irreducible representations
($C_{2x}\vec{u}_{A_{1}}^{a/o}=\vec{u}_{A_{1}}^{a/o}$ and
$C_{2x}\vec{u}_{B_{1}}^{a/o}=-\vec{u}_{B_{1}}^{a/o}$). These four phonon modes
can also be obtained from the dynamical matrix for this system, as discussed
in Sec. I.B of SM.

Figure 1:  (a) Lattice structure of BaMnSb2. The green and orange dots
represent two Sb atoms in each unit. Two acoustic phonons and two optical
phonons are shown in (b) and (c), respectively. (d) First Brillouin zone. The
two red dots label the location of two valleys $K_{\pm}$. (e) Low energy
electron bands at Dirac cones $K_{\pm}$. The up/down-arrow represents spin
polarization direction, and our model only focuses on the bands within the box
with the dashed lines.

The e-ph coupling takes the general form

$\displaystyle\mathcal{H}_{e-ph}^{\eta}=\sum_{\mathbf{k},\mathbf{q}}\sum_{\lambda,s,\alpha,\beta}\mathbf{g}^{\eta,\lambda,s}_{\alpha,\beta}(\mathbf{q})Q_{\eta,\lambda,\mathbf{q}}c_{s,\alpha,\mathbf{k}}^{\dagger}c_{s,\beta,\mathbf{k}-\mathbf{q}},$
(2)

where
$Q_{\eta,\lambda,\mathbf{q}}=(b_{\mathbf{q},\lambda}+b^{\dagger}_{-\mathbf{q},\lambda})/\sqrt{2M\omega_{\lambda}}$
labels the phonon displacement operator for the phonon mode
$|\eta,\lambda\rangle$ ($\eta$ for acoustic/optical and $\lambda$ for
$A_{1}$/$B_{1}$); $M$ is the mass of Sb atoms; $b_{\mathbf{q},\lambda}$ is a
bosonic operator that annihilates a phonon mode $\lambda$ with momentum
$\mathbf{q}$; $\alpha,\beta$ represent $p$-orbitals and
$\mathbf{g}^{\eta,\lambda,s}_{\alpha,\beta}(\mathbf{q})$ is the element of the
e-ph coupling vertex, which satisfies the Hermitian condition
$\mathbf{g}^{\eta,\lambda,s}(\mathbf{q})=[\mathbf{g}^{\eta,\lambda,s}(-\mathbf{q})]^{\dagger}$.
The detailed form of $\mathbf{g}^{\eta,\lambda,s}$ can be derived from the
deformation theory combined with the ${\bf k\cdot p}$ theory, as shown in
Secs. II and III of SM [40]. Here we only focus on long-wavelength optical
phonons (dropping the index $\eta$ below) and keep the spin-conserving intra-
valley scattering process. Thus, we obtain

$\mathbf{g}^{A_{1},s}=g_{0}\tau_{0}+sg_{2}\tau_{2}+g_{3}\tau_{3};\,\,\mathbf{g}^{B_{1},s}=g_{1}\tau_{1},$
(3)

to the lowest order in ${\bf q}$, where $g_{i}$’s are real and material-
dependent parameters[41, 42]. The e-ph coupling vertex matrices satisfy the
relations
$C_{2x}\mathcal{T}\mathbf{g}^{A_{1},s}\mathcal{T}^{-1}C_{2x}^{-1}=\mathbf{g}^{A_{1},s}$
and
$C_{2x}\mathcal{T}\mathbf{g}^{B_{1},s}\mathcal{T}^{-1}C_{2x}^{-1}=-\mathbf{g}^{B_{1},s}$.
Comparing Eq. (3) with Eq. (1) (see Sec. II.C of SM [40]), one can see that
$g_{0}$, $g_{1}$ and $g_{3}$ terms lead to a pseudo-gauge field

$\displaystyle{A}_{pse,\mu}^{s}=(g_{0}Q_{A_{1}},s\tfrac{g_{1}}{v_{0}}Q_{B_{1}},s\tfrac{g_{3}}{v_{0}}Q_{A_{1}})_{\mu},$
(4)

where $e=1$ is assumed and $\mu=0,1,2$. The $g_{2}$ term acts as a correction
to Dirac mass. Taking into account both valleys, we note that
$\tfrac{g_{1}}{v_{0}}Q_{B_{1}}$ and $\tfrac{g_{3}}{v_{0}}Q_{A_{1}}$ couple to
electrons oppositely at two valleys, while the coupling between
$g_{0}Q_{A_{1}}$ and electrons is the same for the two valleys. Similar
considerations apply to acoustic phonons as well (cf. Sec. III of SM [40]).

Figure 2:  The off-diagonal phonon self-energy $\Sigma_{AB}$ induced by the
electronic Berry curvature. We plot the real and imaginary parts of
$\Sigma_{AB}$ (in meV) as a function of $\omega/m_{0}$ and $q_{y}/q_{0}$ in
(a) and (b), respectively. (c) shows Im$[\Sigma_{AB}]$ as a function of
$q_{y}$ for the frequency values $\omega/m_{0}=0,0.4,0.8,1.2$. The dashed line
gives the pure contribution from the Berry curvature. The inset shows the
partial derivative of $\Sigma_{AB}$. The parameter values are $m_{0}=25$ meV,
$v_{0}=100$ nm$\cdot$meV, $\omega_{A_{1}}=20$ meV, $\omega_{B_{1}}=30$ meV,
and $g_{0}=80\sqrt{20}$ meV$\cdot$nm-1, $g_{1}=80\sqrt{30}$ meV$\cdot$nm-1.
Also, $q_{0}=m_{0}/v_{0}$ and
$\Sigma_{0}=g_{0}g_{1}\mathcal{A}N_{3}^{+}/(v_{0}M\sqrt{\omega_{A}\omega_{B}})$.

Phonon self-energy from electronic Berry curvature – Next, we study the
optical phonon Green’s function and the optical phonon spectrum renormalized
by both electron-electron (e-e) and e-ph interactions. The bare phonon Green’s
function is $D_{0}(q,i\omega_{n})=\text{Diag}\\{D_{A_{1}},D_{B_{1}}\\}$, where
$D_{\lambda}=2\omega_{\lambda}/((i\omega_{n})^{2}-\omega_{\lambda}^{2})$ and
$\omega_{\lambda}$ is the bare frequency for the $\lambda$ mode. Here,
$\omega_{n}=2\pi n/\beta$ is the Matsubara frequency and $\beta=1/k_{B}T$.
Following standard formalism [42], the full phonon Green’s function $D({\bf
q},i\omega_{n})$ can be obtained from $D^{-1}({\bf
q},i\omega_{n})=D_{0}^{-1}({\bf q},i\omega_{n})-\Sigma({\bf q},i\omega_{n})$,
where the phonon self-energy $\Sigma=\Sigma^{p}+\Sigma^{e}$ contains both the
correction from the bare e-ph interactions ($\Sigma^{p}$) and the additional
correction due to the screened e-e Coulomb interaction ($\Sigma^{e}$). When
the Fermi level $\mu$ is inside the bulk electronic gap ($|\mu|<2|m_{0}|$) and
when $|\omega|<2|m_{0}|$, the $\Sigma^{e}$ term only renormalizes the bare
dielectric constant $\epsilon_{0}$, the bare phonon energy $\omega_{\lambda}$
and the e-ph coupling strength $g_{i}$ (see Sec. IV of the SM [40]).
Hereafter, we only consider the contribution from the bare e-ph interaction
($\Sigma^{p}$) in Eq. (2), assuming that $\omega_{\lambda}$ and $g_{i}^{o}$
have been renormalized by the e-e interaction. Thus, the superscript $p$ is
hereafter omitted.

The diagonal components of the phonon self-energies are then found to be
$\Sigma_{AA}=(\Pi_{00}+\Pi_{33}+\Pi_{03}+\Pi_{30})/(2M\omega_{A})$ and
$\Sigma_{BB}=\Pi_{11}/(2M\omega_{B})$, where

$\displaystyle\Pi_{ij}(\mathbf{q},i\omega_{n})=\frac{g_{i}g_{j}}{\beta}\sum_{s,\mathbf{k},iq_{m}}\text{tr}[\tau_{i}G_{s}(\mathbf{k},iq_{m})\tau_{j}G_{s}(\mathbf{k}^{\prime},i\omega_{n}^{\prime})],$
(5)

$\mathbf{k}^{\prime}=\mathbf{k}+\mathbf{q}$,
$i\omega_{n}^{\prime}=i\omega_{n}+iq_{m}$, $q_{m}=(2m+1)\pi/\beta$, and
$G_{s}(\mathbf{k},iq_{m})=[iq_{m}-h_{s}(\mathbf{k})+\mu]^{-1}$ is the
Matsubara Green’s function for free electrons at valley $K_{s}$. The diagonal
self-energies only provide corrections to the dispersion of the
$\lambda$-phonon, and thus can be included by redefining
$\omega_{\lambda}(\mathbf{q})$.

A more interesting physical effect emerges from the off-diagonal self-energy
$\Sigma_{AB}=(\Pi_{01}+\Pi_{31})/(2M\sqrt{\omega_{A}\omega_{B}})$, where the
contribution from the $g_{2}$ term has been neglected (see Sec. V of SM [40]
for a detailed justification). $\Sigma_{AB}$ hybridizes A and B phonons and
has a deep connection to the Berry curvature of Dirac cones. To see that, we
first consider $\omega=0$ and treat ${\bf q}$ as a perturbation, so that
$\Sigma_{AB}({\bf q},\omega=0)=\Sigma_{AB}(0)+{\bf q}\cdot(\partial_{\bf
q}\Sigma_{AB})_{{\bf q}=0}+\mathcal{O}(q^{2})$. As shown in Sec. IV.C of SM
[40], TR symmetry requires
$\Sigma_{AB}(\mathbf{q},\omega)=\Sigma_{AB}^{\ast}(-\mathbf{q},\omega)$, while
the $C_{2x}$ symmetry leads to
$\Sigma_{AB}(q_{x},q_{y},\omega)=-\Sigma_{AB}(q_{x},-q_{y},\omega)$. Combining
these two symmetry constraints together with the perturbation expansion, we
find $\Sigma_{AB}({\bf q},\omega=0)\approx
iq_{y}(\partial_{q_{y}}\text{Im}[\Sigma_{AB}])_{{\bf q}=0}$. We notice
$\Sigma_{AB}(0)=0$ since the $A_{1}$ and $B_{1}$ modes belong to different
irreducible representations, but the $q_{y}$-dependent term is still allowed.
Remarkably, a direct calculation gives

$\displaystyle\left(\partial_{q_{y}}\text{Im}[\Sigma_{AB}(q_{y},0)]\right)_{q_{y}=0}=\frac{g_{0}g_{1}\mathcal{A}(N_{3}^{+}-N_{3}^{-})}{2v_{0}M\sqrt{\omega_{A}\omega_{B}}},$
(6)

where $\mathcal{A}$ is the sample area and the coefficient
$N_{3}^{s}=\frac{1}{6}\epsilon^{\alpha\beta\gamma}\text{tr}\int_{-\infty}^{\infty}d\omega^{\prime}\int\frac{d^{2}k}{(2\pi)^{2}}G_{s}^{-1}\partial_{k_{\alpha}}G_{s}G_{s}^{-1}\partial_{k_{\beta}}G_{s}G_{s}^{-1}\partial_{k_{\gamma}}G_{s}$
is the Thouless-Kohmoto-Nightingale-den Nijs formula for the integral of the
Berry curvature around the $s$-valley (i.e., the valley Chern number; see Sec.
IV.D of SM [40]). TR requires $N_{3}^{+}+N_{3}^{-}=0$. Thus, to linear order
in ${\bf q}$, $\Sigma_{AB}({\bf
q})=ig_{0}g_{1}\mathcal{A}N_{3}^{+}q_{y}/(v_{0}M\sqrt{\omega_{A}\omega_{B}})\triangleq
iq_{y}\Sigma_{0}$, and the local Berry curvature around one valley enters into
the off-diagonal phonon self-energy. This is a key result of the present work.
Alternatively (see Sec. VI of SM [40]), Eq. (6) can be obtained from a Chern-
Simons term that appears in the effective action for phonons after integrating
out the electrons. Such derivation confirms that $\Sigma_{AB}$ results from
the valley Chern numbers.

The above analytical expression is derived in the limit $\omega\to 0$ and
${\bf q}\to 0$, while the $\lambda$-optical phonon has a finite frequency
$\omega_{\lambda}$. To see the finite frequency behavior of $\Sigma_{AB}$, we
numerically evaluate it according to Eq. (5), and show the real and imaginary
parts of $\Sigma_{AB}$ in Fig. 2(a) and (b), respectively. When the frequency
is larger than electron energy gap ($\omega>2|m_{0}|$), the interband
scattering of electrons from the conduction to valence bands can give rise to
a large correction to the phonon self-energy. This is reflected by the red
color lines for both $\text{Re}[\Sigma_{AB}]$ and $\text{Im}[\Sigma_{AB}]$ in
Fig. 2(a) and (b), corresponding to the dynamical Kohn anomaly [43] occurring
at $q=\sqrt{(\omega-m_{0})^{2}-m_{0}^{2}}$ for $\mu=0$ (see Sec. VII of SM
[40]). Below electron energy gap ($\omega<2|m_{0}|$), $\text{Re}[\Sigma_{AB}]$
is exactly zero, while $\text{Im}[\Sigma_{AB}]$ is still non-zero. In Fig.
2(c), we plot $\text{Im}[\Sigma_{AB}]$ as a function of $q_{y}$ for
$\omega/m_{0}=0,0.4,0.8,1.2$ separately, from which a linear-dependence on
$q_{y}$ appears in the long-wavelength limit. The pure Berry curvature
contribution extracted from Eq. (6) is indicated by the gray dashed line,
which shows a good coincidence with numerical results for a small $q_{y}$. The
inset in Fig. 2(c) shows $(\partial_{q_{y}}\text{Im}[\Sigma_{AB}])/\Sigma_{0}$
at low $q_{y}$, and only a small derivation from unity is found for finite
$q_{y}$ and $\omega$. Therefore, our numerical simulations demonstrate that
the Berry curvature contribution to the off-diagonal self-energy is dominant
when the phonon frequency $\omega$ is well below electron energy gap. We
notice that in a related material compound, SrMnSb2, the optical phonon
frequency has been measured to be $\sim 15\,{\rm meV}$ [44], well below the
expected electronic gap of $\sim 50$ meV.

Figure 3:  (a) and (b) show the spectrum and angular momentum of optical
phonons with $m_{0}=25$ meV. The phonon dispersions are labeled by black
dashed lines, while the color represents ${\cal P}$ in Eq. (7). (c) shows
Im[$\Sigma_{AB}$] (black line) and the phonon circular polarization (red and
blue lines) as a function of $m_{0}$ with $q_{y}=m_{0}/2v_{0}$ and $\omega=10$
meV. The phonon circular polarization has the same sign as ${\cal P}$. The
inset is the schematics for the elliptical vibration of phonon modes.
Parameters: $v_{0}=100$ meV$\cdot$nm, $\omega_{A_{1}}=20$ meV,
$\omega_{B_{1}}=30$ meV, $g_{0}=80\sqrt{20}$ meV$\cdot$nm-1,
$g_{1}=80\sqrt{30}$ meV$\cdot$nm-1. Also, $q_{0}=m_{0}/v_{0}$ and
$\Sigma_{0}=g_{0}g_{1}\mathcal{A}N_{3}^{+}/(v_{0}M\sqrt{\omega_{A}\omega_{B}})$.

Phonon helicity – Next, we explore the influence of $\Sigma_{AB}$ in the
phonon dynamics by studying the full phonon Green’s function $D({\bf
q},\omega)$. The phonon dispersion can be directly extracted from the poles of
$D({\bf q},\omega)$, and is depicted by the black dashed lines in Fig. 3(a-b)
for optical phonons. Since the $A_{1}$ and $B_{1}$ phonon modes describe
lattice vibration along the $x$ and $y$ directions, respectively, and the off-
diagonal term $\Sigma_{AB}$ that couples these two modes is purely imaginary,
we expect that $\Sigma_{AB}$ can make phonon modes elliptically polarized at
finite $q_{y}$, as schematically depicted in the inset of Fig. 3(c). To
quantify this, we define the momentum-resolved PAM as

$\displaystyle\mathcal{P}(\mathbf{q},\omega)=\text{Tr}\left[\left(\hat{P}_{+}-\hat{P}_{-}\right)\hat{A}(\mathbf{q},\omega)\right],$
(7)

where
$\hat{A}(\mathbf{q},\omega)=i[D(\mathbf{q},\omega)-D^{\dagger}(\mathbf{q},\omega)]/(2\pi)$
is the phonon spectral function and the projection operators
$\hat{P}_{\pm}=|\pm\rangle\langle\pm|$ project the phonon modes to the left
and right circular polarized basis $|\pm\rangle=(\vec{u}_{o,A_{1}}\pm
i\,\vec{u}_{o,B_{1}})/\sqrt{2}$. The total PAM discussed in literature [45,
46, 47, 48] is the thermal average of $\mathcal{P}(\mathbf{q},\omega)$ over
all the phonon modes. We highlight two features of ${\cal P}$: (1)
$\mathcal{P}({\bf q},\omega)=0$ when ${\bf q}||\hat{\bf x}$ but
$\mathcal{P}({\bf q},\omega)\neq 0$ when ${\bf q}||\hat{\bf y}$ (see Fig.
3(a)); (2) $\mathcal{P}(q_{y},\omega)=-\mathcal{P}(-q_{y},\omega)$, as
required by TR symmetry (see Fig. 3(b)). The latter feature suggests the
helical nature of each phonon band and gives rise to phonon helicity. Both
features can be understood from the symmetry property of $\Sigma_{AB}$: (1)
$\Sigma_{AB}$ vanishes along the $q_{x}$ direction; (2) $\Sigma_{AB}\propto
q_{y}$, so that its sign reverses under $q_{y}\to-q_{y}$.

More interestingly, since the local Berry curvature around one valley changes
sign under a band inversion, we anticipate that the phonon helicity will also
reverse its sign. To test this, we tune the Dirac mass $m_{0}$ in Eq. (1), and
evaluate $\Sigma_{AB}$ and $\mathcal{P}$ for the lower branch in Fig. 3(b) as
a function of $m_{0}$ at the momentum $q_{y}=m_{0}/2v_{0}$ in Fig. 3(c). One
can clearly see that both $\Sigma_{AB}$ and $\mathcal{P}$ reverse signs across
the phase transition point at $m_{0}=0$. Thus, we conclude that an electronic
topological phase transition will leave its fingerprint on PAM $\mathcal{P}$.

We also notice that along the $q_{y}$ direction, the $B_{1}$ mode is
longitudinal while the $A_{1}$ mode is transverse. Therefore, a nonzero
$\Sigma_{AB}$ mixes the longitudinal and transverse modes of the in-plane
phonons to form elliptical vibrational modes. The longitudinal-transverse
mixing has been discussed previously in the context of the phonon Hall
viscosity [49, 50, 51]. However, there is an essential difference: our system
does not break TR while the phonon Hall viscosity breaks TR. Consequently, in
Ref. [49], the PAM is an even function of ${\bf q}$, as reflected by the
$q^{2}$-dependent coupling term between the longitudinal and transverse modes,
while in our system, the PAM is an odd function of ${\bf q}$, thereby
generating phonon helicity.

Discussion and Conclusion–. We have demonstrated that electronic Berry
curvature in a 2D TR-invariant Dirac material can induce an off-diagonal self-
energy correction to the optical phonon Green’s function, giving rise to
elliptical polarization of phonon modes. Our theory can also be directly
applied to acoustic phonons. However, the e-ph coupling linearly depends on
${\bf q}$ for acoustic phonons, and thus the off-diagonal phonon self-energy
has a $q_{y}^{3}$ dependence, as discussed in Sec. VI of SM [40].

Our study is qualitatively different from recent works investigating the
influence of the chiral anomaly on sound velocity and attenuation in
Weyl/Dirac semimetals[52, 53, 54, 55, 56]. For one thing, we focus on
electrical insulators. Despite of several early works studying the Berry phase
effect on phonon dynamics [14, 15, 16], the influence of electronic band
topology on the PAM and helicity have not been mentioned before. Furthermore,
although we take BaMnSb2 as an example, the proposed mechanism for generating
phonon helicity might be adapted to other Dirac materials. While a careful
treatment is required to verify this for each material, possible candidates
include HgTe/CdTe and InAs/GaSb quantum wells[57, 58, 59, 60],
(LaO)2(SbSe2)2[61], boron nitride[62, 63], 1T’-WTe2 [64, 65, 66], and
others[67, 68, 69, 70].

The Berry curvature contribution to phonon helicity can be detected in the
following ways. Due to TR symmetry, the equilibrium phonon occupation number
is symmetric under $q_{y}\to-q_{y}$; the total PAM is accordingly zero.
However, away from equilibrium, a temperature gradient applied along the $y$
axis breaks the $q_{y}\to-q_{y}$ symmetry and, combined with phonon helicity,
leads to a nonzero total PAM. This mechanism is analogous to the Edelstein
effect in electronic systems and has been discussed in Ref. [47]. For the
optical phonon that can directly couple to light, the elliptical polarization
will be reflected in the dielectric function and thus might be probed through
optical measurements, as discussed in Sec. VIII of SM [40].

Acknowledgments–. We thank Zhen Bi, Shengxi Huang, Jainendra Jain, Ziqiang
Mao, Yinming Shao, and Shuang Zhang for helpful discussions. This work at Penn
State is primarily supported by a DOE grant (DE-SC0019064). L.-H. H. also
acknowledges the support of the Office of Naval Research (Grant 361 No.
N00014-18-1-2793). I. Garate acknowledges financial support from the Canada
First Research Excellence Fund (CFREF) and the Natural Sciences and
Engineering Research Council of Canada (NSERC). J. Yu is supported by the
Laboratory for Physical Sciences.

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Supplementary material for “Phonon helicity induced by electronic Berry
curvature in Dirac materials”

##  Appendix A Long-wavelength phonon modes for the quasi-2D Sb layers

In this section, we study the phonon modes of BaMnSb2 with zigzag distortion
[38], with a focus on the the effective quasi-2D Sb layer (see Fig. 4(a)).
First, we classify the phonon modes according to the $C_{2v}$ point group.
This group has three symmetry operators: $C_{2x}$ is the two-fold rotation
along the $x$-axis, while $\sigma_{v}(xy)$ and $\sigma_{v}(xz)$ are two mirror
reflections about the $xy$ and $xz$ planes, respectively:

$\displaystyle C_{2v}=\\{C_{2x},\sigma_{v}(xy),\sigma_{v}(xz)\\}.$ (8)

There are two Sb atoms in each unit cell (labeled by green and orange filled
circles in Fig. 4) and each atom can move along $x$, $y$ and $z$ directions.
Thus, the total number of degrees of freedom is six, giving rise to six phonon
modes. We characterize the long-wavelength phonon modes by the displacements
of the two atoms away from their equilibrium positions,
$(u_{1,x},u_{1,y},u_{1,z},u_{2,x},u_{2,y},u_{2,z})$.

Table 1:  The phonon modes for the quasi-2D Sb layers, including three acoustic and three phonon modes. The corresponding lattice vibrations at $\Gamma$ point are classified by the $C_{2v}$ point group. Phonon modes | $E$ | $C_{2x}$ | $\sigma_{v}(xy)$ | $\sigma_{v}(xz)$ | Irrep.
---|---|---|---|---|---
Acoustic | $(1,0,0,1,0,0)$ | $+1$ | $+1$ | $+1$ | $+1$ | $A_{1}$
$(0,1,0,0,1,0)$ | $+1$ | $-1$ | $+1$ | $-1$ | $B_{1}$
$(0,0,1,0,0,1)$ | $+1$ | $-1$ | $-1$ | $+1$ | $B_{2}$
Optical | $(1,0,0,-1,0,0)$ | $+1$ | $+1$ | $+1$ | $+1$ | $A_{1}$
$(0,1,0,0,-1,0)$ | $+1$ | $-1$ | $+1$ | $-1$ | $B_{1}$
$(0,0,1,0,0,-1)$ | $+1$ | $-1$ | $-1$ | $+1$ | $B_{2}$

The results are summarized in Table 1, from which we learn that the lattice
vibration along the $z$-axis is odd under mirror reflection $\sigma_{v}(xy)$,
while that along the $x$ or $y$ directions is even. Therefore, the
$z$-direction mode can not couple to the electronic states around Fermi
energy, which are even under $\sigma_{v}(xy)$. As a result, we will only focus
on the in-plane phonon modes.

###  A.1 Symmetry-based classification of in-plane phonons

Once again, we only consider the effective 2D lattice structure for Sb atoms
with the zigzag distorted phase of BaMnSb2 (see Fig. 4(a)). To study the
lattice vibration in the $x-y$ plane, we ignore the mirror $\sigma_{v}(xy)$,
which acts as an identity for the in-plane phonon modes. In addition, $C_{2x}$
and $\sigma_{v}(xz)$ act in the same way on the in-plane phonon modes;
hereafter only $C_{2x}$ will be taken into account for the symmetry analysis
for in-plane phonon modes. Then, effectively the point group symmetry is
reduced to $C_{2}$. There are in total four degree of freedom for a unit cell,
i.e., one Sb with $(u_{1,x},u_{1,y})$ and the other Sb with
$(u_{2,x},u_{2,y})$. Under the $C_{2x}$ rotation,

$\displaystyle C_{2x}:\begin{cases}(u_{1,x},u_{1,y})\to(u_{1,x},-u_{1,y}),\\\
(u_{2,x},u_{2,y})\to(u_{2,x},-u_{2,y}).\end{cases}$ (9)

Therefore, the reducible characters for the in-plane phonon modes are
$\chi(E)=4$ and $\chi(C_{2x})=0$. Based on the $C_{2}$ point group, the are
two $A_{1}$ and two $B_{1}$ irreducible modes [71], which are shown in Table.
1. Here $A_{1}$ and $B_{1}$ are two 1D irreducible representations (irrep.) in
Mulliken symbols, which is taken to label the phonon modes. Thus, there are
two $A_{1}$-type phonon modes as well as two $B_{1}$-type phonon modes. Next,
we will solve the dynamical ($D$-) matrix to show the phonon modes explicitly.

Figure 4: In (a), it shows the lattice structure for Sb layer. In (b), the
acoustic phonon mode. In (c), the optical phonon mode. The black arrow
indicates the relative vibration direction.

###  A.2 Characterization of in-plane phonons from the dynamical matrix

In this subsection, we determine the frequencies and displacement vectors
corresponding to in-plane phonon modes. The representative Sb atoms are
labeled by integer numbers shown in Fig. 4(a). We treat the interatomic forces
in the harmonic approximation, and we only consider the spring constants
between nearest and next-nearest atoms:

$\displaystyle\begin{split}&\text{Spring constant between
}\text{Sb}_{0}\;\text{and}\;\text{Sb}_{1,2,3,4}:\gamma_{0},\\\ &\text{Spring
constant between }\text{Sb}_{0}\;\text{and}\;\text{Sb}_{6,7}:\gamma_{1},\\\
&\text{Spring constant between
}\text{Sb}_{0}\;\text{and}\;\text{Sb}_{5,8}:\gamma_{2}.\end{split}$ (10)

Here, we have neglected the forces between neighboring Sb planes, which
generally lead to vibrations along the $z$-axis. Next, we calculate the force
constant directly. For simplicity, we take lattice constant $a=b$, and assume
$c=\delta a\times\frac{a}{2}$ with $-1<\delta a<1$; $\delta a=0$ corresponds
to the absence of zigzag distortion. Since we have two Sb atoms in one unit
cell, we consider them separately.

####  A.2.1 Force constants for the first Sb atom

First, let us choose the position of the Sb0 atom as the origin of
coordinates. The normalized coordinates for the nearest and the next-nearest
atoms are

$\displaystyle\begin{split}&\text{Sb}_{1}:(1,0),\;\text{Sb}_{2}:(0,1),\;\text{Sb}_{3}:(-1,0),\;\text{Sb}_{4}:(0,-1),\\\
&\text{Sb}_{5}:(\delta a+1,1)/A,\;\text{Sb}_{6}:(\delta
a-1,1)/B,\;\text{Sb}_{7}:(\delta a-1,-1)/B,\;\text{Sb}_{8}:(\delta
a+1,-1)/A,\end{split}$ (11)

with normalization factors $A=\sqrt{1+(\delta a+1)^{2}}$ and
$B=\sqrt{1+(\delta a-1)^{2}}$. Let us briefly review how to calculate the
force constants, using Sb0 as an example. The potential energy associated to
the displacement of this atom is

$\displaystyle E\approx
E_{0}+\sum_{i}\sum_{\alpha,\beta=\\{x,y\\}}\phi_{\alpha,\beta}(0,i){u}_{0,\alpha}{u}_{i,\beta},$
(12)

where $E_{0}$ is a constant and the index $i$ represents the nearby atoms:
$i=5,6,7,8$ for the nearest neighbors and $i=1,2,3,4$ for the next-nearest
neighbors (see Fig. 4(a)). The force constant is defined as the second
derivative of the potential energy evaluated at the equilibrium position,

$\displaystyle\phi_{\alpha,\beta}(0,i)=\frac{\partial^{2}E}{\partial
u_{0,\alpha}\partial u_{i,\beta}}=-\frac{\partial F_{\alpha}}{\partial
u_{i,\beta}},$ (13)

where $F_{\alpha}=\sum_{i}f_{\alpha}(0,i)$ is the total force acting on the
Sb0 atom along the $\alpha$-direction. With the spring constants $\gamma(0,i)$
between Sb0 and its neighbors Sbi given in Eq. (10), we have

$\displaystyle
F_{\alpha}=\sum_{i}\gamma(0,i)e_{\alpha}(i)\left(\sum_{\beta}e_{\beta}(i)u_{i,\beta}\right),$
(14)

where $\vec{e}(i)=(e_{x}(i),e_{y}(i))$ is the unit vector along ${\bf R}_{i}$
(the straight line connecting Sb0 and Sbi), obtained from Eq. (11). Therefore,
we have the force constant

$\displaystyle\phi_{\alpha\beta}(0,i)=-\gamma(0,i)e_{\alpha}(i)e_{\beta}(i).$
(15)

In this way, we calculate all the force constants between Sb0 and its
neighbors (Sb5,6,7,8 and Sb1,2,3,4). For the nearest neighbors, we get

$\begin{aligned} \phi_{xx}(0,5)&=\phi_{xx}(0,8)=-\gamma_{2}\left(\frac{\delta
a+1}{A}\right)^{2},\\\
\phi_{yy}(0,5)&=\phi_{yy}(0,8)=-\gamma_{2}\left(\frac{1}{A}\right)^{2},\\\
\phi_{xy}(0,5)&=-\phi_{xy}(0,8)=-\gamma_{2}\frac{\delta
a+1}{A^{2}},\end{aligned}\text{ }\begin{aligned}
\phi_{xx}(0,6)&=\phi_{xx}(0,7)=-\gamma_{1}\left(\frac{\delta
a-1}{B}\right)^{2},\\\
\phi_{yy}(0,6)&=\phi_{yy}(0,7)=-\gamma_{1}\left(\frac{1}{B}\right)^{2},\\\
\phi_{xy}(0,6)&=-\phi_{xy}(0,7)=-\gamma_{1}\frac{\delta
a-1}{B^{2}}.\end{aligned}$

The force constants for the next-nearest neighbors are

$\displaystyle\begin{split}&\phi_{xx}(0,1)=\phi_{xx}(0,3)=\phi_{yy}(0,2)=\phi_{yy}(0,4)=-\gamma_{0},\end{split}$
(16)

all others being zero. Moreover, the potential energy of the crystal and the
force on a given atom should be invariant under a rigid body displacement of
the whole crystal. This implies the constraint

$\displaystyle\phi_{\alpha\beta}(0,0)=-\sum_{i=1}^{8}\phi_{\alpha\beta}(0,i).$
(17)

Therefore, we have

$\displaystyle\begin{split}\phi_{xx}(0,0)&=-\sum_{i=1}^{8}\left[\phi_{xx}(0,i)\right]=2\gamma_{0}+2\gamma_{2}\left(\frac{\delta
a+1}{A}\right)^{2}+2\gamma_{1}\left(\frac{\delta a-1}{B}\right)^{2},\\\
\phi_{yy}(0,0)&=-\sum_{i=1}^{8}\left[\phi_{yy}(0,i)\right]=2\gamma_{0}+2\gamma_{2}\left(\frac{1}{A}\right)^{2}+2\gamma_{1}\left(\frac{1}{B}\right)^{2},\\\
\phi_{xy}(0,0)&=\phi_{yx}(0,0)=0.\end{split}$ (18)

####  A.2.2 Force constants for the second Sb atom

Next, let us consider the second Sb atom and choose the position of the Sb6 as
the origin of coordinates (see Fig. 4(a)). Repeating the above calculations
for the force constants, we get

$\begin{aligned} \phi_{xx}(6,11)&=\phi_{xx}(6,3)=-\gamma_{2}\left(\frac{\delta
a+1}{A}\right)^{2},\\\
\phi_{yy}(6,11)&=\phi_{yy}(6,3)=-\gamma_{2}\left(\frac{1}{A}\right)^{2},\\\
\phi_{xy}(6,11)&=-\phi_{xy}(6,3)=-\gamma_{2}\frac{\delta a+1}{A^{2}},\\\
\end{aligned}\begin{aligned}
\phi_{xx}(6,2)&=\phi_{xx}(6,0)=-\gamma_{1}\left(\frac{\delta
a-1}{B}\right)^{2},\\\
\phi_{yy}(6,2)&=\phi_{yy}(6,0)=-\gamma_{1}\left(\frac{1}{B}\right)^{2},\\\
\phi_{xy}(6,2)&=-\phi_{xy}(6,0)=-\gamma_{1}\frac{\delta a-1}{B^{2}}.\\\
\end{aligned}$

Also,

$\begin{split}\phi_{xx}(6,5)&=\phi_{xx}(6,9)=\phi_{yy}(6,7)=\phi_{yy}(6,10)=-\gamma_{0},\\\
\phi_{xx}(6,6)&=-\sum_{j}\left[\phi_{xx}(6,j)\right]=2\gamma_{0}+2\gamma_{2}\left(\frac{\delta
a+1}{A}\right)^{2}+2\gamma_{1}\left(\frac{\delta a-1}{B}\right)^{2},\\\
\phi_{yy}(6,6)&=-\sum_{j}\left[\phi_{yy}(6,j)\right]=2\gamma_{0}+2\gamma_{2}\left(\frac{1}{A}\right)^{2}+2\gamma_{1}\left(\frac{1}{B}\right)^{2},\end{split}$
(19)

Here, $j$ is summed over nearest neighbors ($0,2,3,11$) and next-nearest
neighbors ($5,7,9,10$). All other force constants are zero.

####  A.2.3 The dynamical $D$-matrix

We are now ready to calculate the dynamical matrix ($D$-matrix) on the basis

$\displaystyle\\{u_{1,x},u_{1,y},u_{2,x},u_{2,y}\\}.$ (20)

The dynamical $D$-matrix is defined from the equation of motion in the
momentum space for the lattice vibrations,

$\displaystyle\omega^{2}u_{n,\alpha}=\sum_{m,\beta}{\cal
D}_{\alpha\beta}(m,n)u_{m,\beta},$ (21)

where $m,n\in\\{1,2\\}$ labels the two Sb atoms in one unit cell and the
momentum label ${\bf q}=(q_{x},q_{y})$ is implicit in ${\cal
D}_{\alpha\beta}(m,n)$. The relation between the matrix elements of the
$D$-matrix and the force constants of the preceding subsection is given by
(cf. Fig. 4(a))

$\displaystyle{\cal
D}_{\alpha\beta}(m,n)=\frac{1}{\sqrt{M_{m}M_{n}}}\sum_{i}\phi_{\alpha\beta}(m,i)e^{-i{\bf
q}\cdot{\bf R}_{i}},$ (22)

where $M_{n}=M_{m}=M$ is the mass of Sb atoms. The precise meaning of the
above summation is as follows:

$\displaystyle\begin{cases}m=1,n=1:\;\sum_{i}\phi_{\alpha\beta}(m,i)e^{-i{\bf
q}\cdot{\bf R}_{i}}\to\sum_{i=\\{0,1,2,3,4\\}}\phi_{\alpha\beta}(0,i)e^{-i{\bf
q}\cdot{\bf R}_{i}},\\\ m=2,n=2:\;\sum_{i}\phi_{\alpha\beta}(m,i)e^{-i{\bf
q}\cdot{\bf
R}_{i}}\to\sum_{i=\\{6,5,7,9,10\\}}\phi_{\alpha\beta}(6,i)e^{-i{\bf
q}\cdot{\bf R}_{i}},\\\ m=1,n=2:\;\sum_{i}\phi_{\alpha\beta}(m,i)e^{-i{\bf
q}\cdot{\bf R}_{i}}\to\sum_{i=\\{5,6,7,8\\}}\phi_{\alpha\beta}(0,i)e^{-i{\bf
q}\cdot{\bf R}_{i}},\\\ m=2,n=1:\;\sum_{i}\phi_{\alpha\beta}(m,i)e^{-i{\bf
q}\cdot{\bf R}_{i}}\to\sum_{i=\\{0,2,3,11\\}}\phi_{\alpha\beta}(6,i)e^{-i{\bf
q}\cdot{\bf R}_{i}}.\end{cases}$ (23)

After a straightforward calculation, the elements of the dynamical $D$-matrix
for the first Sb atom in one unit cell are given by

$\displaystyle\begin{split}{\cal
D}_{xx}(1,1)&=\frac{1}{M}\left[\phi_{xx}(0,0)+\phi_{xx}(0,1)e^{iq_{x}a}+\phi_{xx}(0,3)e^{-iq_{x}a}\right]\\\
&=\frac{2}{M}\left[\gamma_{2}\left(\frac{\delta
a+1}{A}\right)^{2}+\gamma_{1}\left(\frac{\delta
a-1}{B}\right)^{2}+\gamma_{0}(1-\cos(q_{x}a))\right],\\\ {\cal
D}_{yy}(1,1)&=\frac{1}{M}\left[\phi_{yy}(0,0)+\phi_{yy}(0,2)e^{iq_{y}a}+\phi_{yy}(0,4)e^{-iq_{y}a}\right]\\\
&=\frac{2}{M}\left[\gamma_{2}\left(\frac{1}{A}\right)^{2}+\gamma_{1}\left(\frac{1}{B}\right)^{2}+\gamma_{0}(1-\cos(q_{y}a))\right],\\\
{\cal D}_{xy}(1,1)&={\cal D}_{yx}(1,1)=0.\end{split}$ (24)

For the second Sb atom in one unit cell, we have

$\displaystyle{\cal D}_{xx}(2,2)={\cal D}_{xx}(1,1),{\cal D}_{yy}(2,2)={\cal
D}_{yy}(1,1),\text{ and }{\cal D}_{xy}(2,2)={\cal D}_{yx}(2,2)=0.$ (25)

Concerning the off-diagonal terms between these two Sb atoms, we have

$\displaystyle{\cal D}_{xx}(1,2)$
$\displaystyle=\frac{1}{M}\left[\phi_{xx}(0,5)e^{i\frac{a}{2}\left(q_{x}(\delta
a+1)+q_{y}\right)}+\phi_{xx}(0,8)e^{i\frac{a}{2}\left(q_{x}(\delta
a+1)-q_{y}\right)}+\phi_{xx}(0,6)e^{i\frac{a}{2}\left(q_{x}(\delta
a-1)+q_{y}\right)}+\phi_{xx}(0,7)e^{i\frac{a}{2}\left(q_{x}(\delta
a-1)-q_{y}\right)}\right],$
$\displaystyle=\frac{2}{M}\cos\left(q_{y}\frac{a}{2}\right)\left[-\gamma_{2}\left(\frac{\delta
a+1}{A}\right)^{2}e^{iq_{x}\frac{a}{2}(\delta
a+1)}-\gamma_{1}\left(\frac{\delta
a-1}{B}\right)^{2}e^{iq_{x}\frac{a}{2}(\delta a-1)}\right],$
$\displaystyle{\cal D}_{yy}(1,2)$
$\displaystyle=\frac{2}{M}\cos\left(q_{y}\frac{a}{2}\right)\left[-\gamma_{2}\left(\frac{1}{A}\right)^{2}e^{iq_{x}\frac{a}{2}(\delta
a+1)}-\gamma_{1}\left(\frac{1}{B}\right)^{2}e^{iq_{x}\frac{a}{2}(\delta
a-1)}\right],$ (26) $\displaystyle{\cal D}_{xy}(1,2)$
$\displaystyle=i\frac{2}{M}\sin\left(q_{y}\frac{a}{2}\right)\left[-\gamma_{2}\frac{\delta
a+1}{A^{2}}e^{iq_{x}\frac{a}{2}(\delta a+1)}-\gamma_{1}\frac{\delta
a-1}{B^{2}}e^{iq_{x}\frac{a}{2}(\delta a-1)}\right],$ $\displaystyle{\cal
D}_{yx}(1,2)$ $\displaystyle={\cal D}_{xy}(1,2).$

Likewise,
$\mathcal{D}_{\tau\tau^{\prime}}(2,1)=\mathcal{D}_{\tau\tau^{\prime}}^{\ast}(1,2)$.
Therefore, the four-by-four $D$-matrix is given by

$\displaystyle\mathcal{D}=\left[\begin{array}[]{cccc}{\cal
D}_{xx}(1,1)&0&{\cal D}_{xx}(1,2)&{\cal D}_{xy}(1,2)\\\ 0&{\cal
D}_{yy}(1,1)&{\cal D}_{xy}(1,2)&{\cal D}_{yy}(1,2)\\\ {\cal
D}_{xx}^{\ast}(1,2)&{\cal D}_{xy}^{\ast}(1,2)&{\cal D}_{xx}(2,2)&0\\\ {\cal
D}_{xy}^{\ast}(1,2)&{\cal D}_{yy}^{\ast}(1,2)&0&{\cal
D}_{yy}(2,2)\end{array}\right].$ (31)

The vibration spectrum is the solution of $|\mathcal{D}({\bf
q})-\omega^{2}|=0$. To gain analytical understanding, hereafter we consider
the vibrational modes at the $\Gamma$ point ($q_{x}=q_{y}=0$), which has the
full lattice symmetry. Then, the $D$-matrix is

$\displaystyle\mathcal{D}({\bf
q}=0)=\left[\begin{array}[]{cccc}D_{1}&0&D_{3}&0\\\ 0&D_{2}&0&D_{4}\\\
D_{3}&0&D_{1}&0\\\ 0&D_{4}&0&D_{2}\end{array}\right],$ (36)

where $D_{1}=\frac{2}{M}\left[\gamma_{2}\left(\frac{\delta
a+1}{A}\right)^{2}+\gamma_{1}\left(\frac{\delta a-1}{B}\right)^{2}\right]$,
$D_{2}=\frac{2}{M}\left[\gamma_{2}\left(\frac{1}{A}\right)^{2}+\gamma_{1}\left(\frac{1}{B}\right)^{2}\right]$,
$D_{3}=-D_{1}$, and $D_{4}=-D_{2}$. By diagonalizing ${\cal D}({\bf 0})$, we
obtain the frequencies and displacement vectors for the four in-plane
vibrational modes:

$\displaystyle\begin{split}\omega_{a,A_{1}}&=0\text{ and
}\vec{u}_{a,A_{1}}=(1,0,1,0)^{T}/\sqrt{2},\\\ \omega_{a,B_{1}}&=0\text{ and
}\vec{u}_{a,B_{1}}=(0,1,0,1)^{T}/\sqrt{2},\\\ \omega_{o,A_{1}}&=2D_{1}\text{
and }\vec{u}_{o,A_{1}}=(-1,0,1,0)^{T}/\sqrt{2},\\\
\omega_{o,B_{1}}&=2D_{2}\text{ and
}\vec{u}_{o,B_{1}}=(0,-1,0,1)^{T}/\sqrt{2}.\end{split}$ (37)

For acoustic phonons, ${\bf q}=0$ corresponds to a rigid translation of the
crystal (hence $\omega=0$) and its phonon modes are shown in Fig. 4(b). For
optical phonons, the two Sb atoms in one unit cell move $180^{\circ}$ out of
phase, shown in Fig. 4(c). As discussed next, this is consistent with the
symmetry analysis.

###  A.3 Symmetry construction of the dynamical matrix at $q=0$

In the preceding subsection, we have constructed the $D$-matrix by using the
force constant approach with harmonic approximation. In this subsection, we
recover the same for of the $D$-matrix using symmetry arguments. We begin by
rewriting the general form of dynamical $D$-matrix in the
$\\{u_{1,x},u_{1,y},u_{2,x},u_{2,y}\\}$ basis as

$\displaystyle{\cal D}=\left(\begin{array}[]{cccc}{\cal D}_{xx}(1,1)&{\cal
D}_{xy}(1,1)&{\cal D}_{xx}(1,2)&{\cal D}_{xy}(1,2)\\\ {\cal D}_{yx}(1,1)&{\cal
D}_{yy}(1,1)&{\cal D}_{yx}(1,2)&{\cal D}_{yy}(1,2)\\\ {\cal D}_{xx}(2,1)&{\cal
D}_{xy}(2,1)&{\cal D}_{xx}(2,2)&{\cal D}_{xy}(2,2)\\\ {\cal D}_{yx}(2,1)&{\cal
D}_{yy}(2,1)&{\cal D}_{yx}(2,2)&{\cal D}_{yy}(2,2)\end{array}\right).$ (42)

Let us consider the symmetry constraints due to the symmetry operators
$\\{C_{2x},\sigma_{v}(xz)\\}$, following Ref. [72]. The transformation of the
dynamical matrix under crystal symmetry operations is given by

$\boldsymbol{\Gamma}({\bf q};\\{R|{\bf v}(R)+{\bf x}(l)\\}){\cal D}({\bf
q})\boldsymbol{\Gamma}^{-1}({\bf q};\\{R|{\bf v}(R)+{\bf x}(l)\\})={\cal
D}(R{\bf q}),$ (43)

where the representation matrix $\boldsymbol{\Gamma}$ of the symmetry operator
$\\{R|{\bf v}(R)+{\bf x}(l)\\}$ is given by

$\Gamma_{\alpha\beta}(m,n|{\bf q};\\{R|{\bf v}(R)+{\bf
x}\\})=R_{\alpha\beta}\delta(m,F_{o}(n;R))e^{i{\bf q}\cdot[\\{R|{\bf
v}(R)+{\bf x}(l)\\}^{-1}{\bf x}(m)-{\bf x}(n)]}.$ (44)

Here, $m,n\in\\{{1,2}\\}$ denote the two Sb atoms in a unit cell,
$\alpha,\beta\in\\{x,y\\}$, $R$ is a real $2\times 2$ orthogonal matrix
representation of one of the proper or improper rotations of the point group,
${\bf v}(R)$ is a vector which is smaller than any primitive translation
vector of the crystal (in symmorphic space groups such as ours, ${\bf
v}(R)={\bf 0}$ for all $R$), ${\bf x}(l)$ is the position of the origin of
$l-$th unit cell, ${\bf x}(m)$ is the position of an atom $m$ relative to the
origin of the unit cell, and $\delta(m,F_{o}(n;R))$ indicates which atom $m$
is brought into the position of $n$ under the symmetry operation.

Hereafter, we focus on $R\in\\{C_{2x},\sigma_{v}(xz)\\}$ and ${\bf q}=0$.
Thus,

$\displaystyle\Gamma_{\alpha\beta}(m,n|{\bf 0};\\{R|{\bf v}(R)+{\bf
x}(l)\\})=R_{\alpha\beta}\delta(m,n),$ (45)

where $R=\text{Diag}[1,-1]$. Therefore,
$\boldsymbol{\Gamma}=\text{Diag}[1,-1,1,-1]$ and, according to Eq. (43), we
arrive at

$\displaystyle{\cal D}=\left(\begin{array}[]{cccc}{\cal D}_{xx}(1,1)&0&{\cal
D}_{xx}(1,2)&0\\\ 0&{\cal D}_{yy}(1,1)&0&{\cal D}_{yy}(1,2)\\\ {\cal
D}_{xx}(2,1)&0&{\cal D}_{xx}(2,2)&0\\\ 0&{\cal D}_{yy}(2,1)&0&{\cal
D}_{yy}(2,2)\end{array}\right).$ (50)

The dynamical matrix must be Hermitian, ${\cal D}^{\dagger}({\bf q})={\cal
D}({\bf q})$. In addition, time-reversal symmetry requires ${\cal D}({\bf
q})={\cal D}^{\ast}(-{\bf q})$. As a result, the $D$-matrix becomes

$\displaystyle{\cal D}=\left(\begin{array}[]{cccc}{\cal D}_{xx}(1,1)&0&{\cal
D}_{xx}(1,2)&0\\\ 0&{\cal D}_{yy}(1,1)&0&{\cal D}_{yy}(1,2)\\\ {\cal
D}_{xx}(1,2)&0&{\cal D}_{xx}(2,2)&0\\\ 0&{\cal D}_{yy}(1,2)&0&{\cal
D}_{yy}(2,2)\end{array}\right),$ (55)

where all the matrix elements are real. One more constraint on ${\cal D}$
comes from the fact that the lattice energy does not change for an overall
translation of the whole lattice, i.e. the acoustic phonon should have zero
energy for ${\bf q}=0$. The overall translation of the whole lattice along the
$x$-direction is described by the vector
$(u_{1,x},u_{1,y},u_{2,x},u_{2,y})=(1,0,1,0)$. Thus, we expect

${\cal D}(1,0,1,0)^{T}=0.$ (56)

Similarly, the overall translation of the whole lattice along the $y$
direction gives rise to

${\cal D}(0,1,0,1)^{T}=0.$ (57)

These two equations result in

$\displaystyle{\cal D}_{xx}(1,1)={\cal D}_{xx}(2,2)=-{\cal D}_{xx}(1,2),\text{
and }{\cal D}_{yy}(1,1)={\cal D}_{yy}(2,2)=-{\cal D}_{yy}(1,2),$ (58)

which makes the form of the $D$-matrix identical to the one obtained above
from the force constant model (see Eq. (36)).

##  Appendix B Electron-phonon interaction Hamiltonian for optical phonons

In this section, we derive the low-energy effective Hamiltonian for the
interaction between electrons and optical phonons in BaMnSb2. The main
objective is to obtain Eq. (5) of the main text. We begin by briefly reviewing
key aspects of low-energy electrons [38], which are massive Dirac fermions
located at ${\bf K}_{\pm}=(\pi/a,\pm k_{y_{0}})$:

  * 1.

Only the $p_{x}$ and $p_{y}$ orbitals of one atom in a unit cell contribute to
the states near the Fermi energy. Thus, the basis of electronic states is
given by

$\displaystyle
C_{\mathbf{k}}^{\dagger}=\left(c_{p_{x},\uparrow}^{\dagger},c_{p_{x},\downarrow}^{\dagger},c_{p_{y},\uparrow}^{\dagger},c_{p_{y},\downarrow}^{\dagger}\right).$
(59)

  * 2.

The transformation of the electronic states under symmetry operations is
described by

$\displaystyle\begin{split}C_{2x}C_{\mathbf{K}_{+}}^{\dagger}C_{2x}^{-1}&=C_{C_{2x}\mathbf{K}_{+}}^{\dagger}(-i\tau_{3}\sigma_{1}),\\\
\mathcal{T}C_{\mathbf{K}_{+}}^{\dagger}\mathcal{T}^{-1}&=C_{-\mathbf{K}_{+}}^{\dagger}(i\tau_{0}\sigma_{2}),\\\
\mathcal{P}C_{\mathbf{K}_{+}}^{\dagger}\mathcal{P}^{-1}&=C_{\mathbf{K}_{+}}^{\dagger}(i\tau_{3}\sigma_{3}),\end{split}$
(60)

where $\tau_{i}$ and $\sigma_{i}$ are Pauli matrices acting in orbital and
spin subspaces, respectively. Also, $\mathcal{T}$ is the time-reversal
operator. Both $C_{2x}$ and $\mathcal{T}$ interchange the valleys ${\bf
K_{+}}$ and ${\bf K_{-}}$. Thus, on each valley, only the combined symmetry
$\mathcal{P}=C_{2x}\mathcal{T}$ remains.

We now derive the electron-phonon (e-ph) Hamiltonian following Ref. [73,74].
The displacement of atoms away from their equilibrium positions gives rise to
a deformation potential $\delta U$, which couples to the electron density as

$\displaystyle\mathcal{H}_{e-ph}$ $\displaystyle=\int
d^{2}r\,\Psi^{\dagger}(\mathbf{r})\Psi(\mathbf{r})\delta U(\mathbf{r}),$ (61)
$\displaystyle\delta U(\mathbf{r})$ $\displaystyle=\sum_{{\bf
l},s}\left[U(\mathbf{r}-\mathbf{R}_{\mathbf{l},s}-\mathbf{Q}_{\mathbf{l},s})-U(\mathbf{r}-\mathbf{R}_{\mathbf{l},s})\right]\simeq\sum_{{\bf
l},s}\mathbf{Q}_{\mathbf{l},s}\cdot\frac{\partial
U(\mathbf{r}-\mathbf{R}_{\mathbf{l},s})}{\partial\mathbf{R}_{\mathbf{l},s}},$
(62)

where
$U(\mathbf{r})=\sum_{\mathbf{l},s}U_{s}(\mathbf{r}-\mathbf{R}_{\mathbf{l},s})$
is the periodic lattice potential, ${\bf l}$ is the position of the $l-$th
unit cell, $s$ labels atoms inside the unit cell,
$\mathbf{R}_{\mathbf{l},s}=\mathbf{l}+\mathbf{r}_{s}$ is the position of atom
$s$ in unit cell $l$,

$\displaystyle\mathbf{Q}_{\mathbf{l},s}=\sum_{\mathbf{q},\lambda}\sqrt{\frac{\hbar}{2M_{s}N\omega_{\mathbf{q},\lambda}}}e^{i\mathbf{q}\cdot(\mathbf{l}+\mathbf{r}_{s})}\mathbf{P}_{{\bf
q},s}^{\lambda}\left(b_{\mathbf{q},\lambda}+b^{\dagger}_{-\mathbf{q},\lambda}\right)$
(63)

is the displacement operator for atom $s$ in unit cell $l$,
$b_{\mathbf{q},\lambda}$ is an operator that annihilates a phonon mode
$\lambda$ with momentum $\mathbf{q}$, $\omega_{\lambda}(\mathbf{q})$ is the
phonon frequency, $N$ is the number of unit cells in the crystal, $M_{s}$ is
the mass of atom $s$ and ${\bf P}_{{\bf q},s}^{\lambda}$ is the polarization
vector corresponding to atom $s$ in phonon mode $\lambda$ with momentum
$\mathbf{q}$. In addition, the fermion field operator $\Psi(\mathbf{r})$ can
be expanded into a sum over Bloch wave functions,
$\Psi(\mathbf{r})=\frac{1}{\sqrt{\mathcal{A}}}\sum_{\mathbf{k}}\sum_{\tau\sigma,{\bf
k}}e^{i\mathbf{k}\cdot\mathbf{r}}u_{\tau\sigma}(\mathbf{r})c_{\sigma\tau}(\mathbf{k})$,
where $\mathcal{A}$ is the area of the 2D sample. Here, $u_{\tau\sigma,{\bf
k}}(\mathbf{r})$ is the periodic part of the Bloch wave function, which
satisfies $u_{\tau\sigma,{\bf k}}(\mathbf{r})=u_{\tau\sigma,{\bf
k}}(\mathbf{r}+\mathbf{l})$ and is calculated by assuming that the atoms are
in their equilibrium positions.

The low-energy electrons are contained in the vicinity of two valleys
$K_{\pm}$, which are well-separated in momentum space. Then, we truncate the
momentum sum in the fermion field operator $\Psi(\mathbf{r})$ around the
vicinity of the two valleys,

$\displaystyle\Psi(\mathbf{r})=\frac{1}{\sqrt{\mathcal{A}}}\sum_{|\mathbf{k}|<\Lambda}\sum_{\kappa}e^{i\mathbf{K}_{\kappa}\cdot\mathbf{r}}\sum_{\tau\sigma}e^{i\mathbf{k}\cdot\mathbf{r}}u_{\tau\sigma,\kappa}(\mathbf{r})c_{\sigma\tau,\kappa}(\mathbf{k})+\mathcal{O}\left(\frac{k}{\Lambda}\right),$
(64)

where $\kappa=\pm$ label the two valleys, now ${\bf k}$ is the momentum
measured from a valley and $\Lambda$ is a cutoff under which the linear
dispersion of Dirac electrons is valid. Based on this approximation, we can
derive the e-ph coupling Hamiltonian at the two valleys separately.

Neglecting intervalley electron-phonon scattering and performing some
straightforward manipulations (details can be found in Ref. [73,74]), the e-ph
Hamiltonian becomes

$\displaystyle\mathcal{H}_{e-ph}$ $\displaystyle=\sum_{{\bf k},{\bf
q}}\sum_{\lambda,\kappa}\sum_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}g^{\lambda,\kappa}_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}(\mathbf{q})Q_{\lambda}(\mathbf{q})c_{\sigma\tau,\kappa}^{\dagger}({\bf
k})c_{\sigma^{\prime}\tau^{\prime},\kappa}(\mathbf{k}-\mathbf{q}),$ (65)
$\displaystyle
g^{\lambda,\kappa}_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}(\mathbf{q})$
$\displaystyle=\frac{1}{\sqrt{N}\mathcal{A}}\sum_{s}\int
d^{2}r\,u_{\sigma\tau,\kappa}^{\ast}(\mathbf{r})u_{\sigma^{\prime}\tau^{\prime},\kappa}(\mathbf{r})e^{-i\mathbf{q}\cdot\mathbf{r}}\times\left[\mathbf{P}^{\lambda}_{{\bf
q},s}\cdot\frac{\partial
U(\mathbf{r}-\mathbf{r}_{s})}{\partial\mathbf{r}_{s}}\right],$ (66)

where $\lambda=\\{A_{1},B_{1}\\}$ labels the two optical in-plane phonon modes
and
$Q_{\lambda}(\mathbf{q})=\sqrt{\frac{\hbar}{2M\omega_{\lambda}(\mathbf{q})}}\left(b_{\mathbf{q},\lambda}+b^{\dagger}_{-\mathbf{q},\lambda}\right)$
is the displacement operator for mode $\lambda$ at momentum ${\bf q}$. Note
that $\hbar=1$ is assumed in the main text. Below, we only consider the e-ph
coupling for long-wavelength phonons. Moreover, the Hermiticity of the e-ph
coupling Hamiltonian requires that
$\mathbf{g}^{\lambda,\kappa}(-\mathbf{q})=\left[\mathbf{g}^{\lambda,\kappa}(\mathbf{q})\right]^{\dagger}$,
where $\mathbf{g}^{\lambda,\kappa}({\bf q})$ is a $4\times 4$ matrix whose
matrix elements are
$g^{\lambda,\kappa}_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}(\mathbf{q})$.

###  B.1 Symmetry analysis for the electron-phonon coupling vertex

In this subsection, we focus on the symmetry analysis for the e-ph coupling
Hamiltonian in Eq. (65). For notational simplicity, we relabel the two atoms
in a unit cell as Sb1 and Sb2 (this is not to be confused with the labels in
Fig. 4(a)). We restrict ourselves to long-wavelength optical phonons, for
which the e-ph coupling matrix $\mathbf{g}^{\lambda,\kappa}(\mathbf{q})$ is
independent of $\mathbf{q}$ to leading order in ${\bf q}$. This approximation
relies on $\mathbf{g}^{\lambda,\kappa}({\bf q})$ being an analytic function of
${\bf q}$ at $q\to 0$. Analyticity is expected in our problem, due to the non-
degeneracy of low-energy electronic and phononic states. Moreover, the matrix
elements of $\mathbf{g}^{\lambda,\kappa}({\bf 0})$ are real due to the
Hermiticity condition.

In the low-energy effective theory for electrons, the spin degree of freedom
is locked to the valley degree of freedom. As a result, the e-ph coupling for
long-wavelength phonons is spin-conserving. We will furthermore assume that
the e-ph coupling is spin-independent. Then, we drop the spin index and work
with the spinless basis
$C_{\mathbf{k}}^{\dagger}=(c_{p_{x}}^{\dagger},c_{p_{y}}^{\dagger})$. At the
same time, the four-by-four matrix $\mathbf{g}^{\lambda,\kappa}({\bf 0})$ is
reduced to a two-by-two matrix, with elements given by

$\displaystyle
g^{\lambda,\kappa}_{\tau,\tau^{\prime}}(0)=\frac{1}{\sqrt{N}\mathcal{A}}\sum_{s}\int
d^{2}\mathbf{r}\;u_{\tau,\kappa}^{\ast}(\mathbf{r})u_{\tau^{\prime},\kappa}(\mathbf{r})\times\left[\mathbf{P}^{\lambda}_{{\bf
q},s}\cdot\frac{\partial
U(\mathbf{r}-\mathbf{r}_{s})}{\partial\mathbf{r}_{s}}\right].$ (67)

Next, we determine the matrix structure of $\mathbf{g}^{\lambda,\kappa}({\bf
0})$ for $A_{1}$ and $B_{1}$ in-plane optical phonons, using symmetry
arguments. For brevity, we hereafter drop the ${\bf q}=0$ indicator from
$\mathbf{g}^{\lambda,\kappa}$. We start with the e-ph Hamiltonian for
electrons at the $K_{+}$ valley,

$\displaystyle\mathcal{H}_{e-ph}^{+}=C_{\mathbf{K_{+}}}^{\dagger}\left[\mathbf{g}^{\lambda,+}Q_{\lambda}(\mathbf{q}\to
0)\right]C_{\mathbf{K_{+}}}.$ (68)

The invariance of $\mathcal{H}_{e-ph}^{+}$ under the combined symmetry
operator $\mathcal{P}=C_{2x}\mathcal{T}$ implies
$\mathcal{P}\mathcal{H}_{e-ph}^{+}\mathcal{P}^{-1}={\cal H}_{e-ph}^{+}$, where

$\displaystyle\mathcal{P}\mathcal{H}_{e-ph}^{+}\mathcal{P}^{-1}=C_{\mathbf{K_{+}}}^{\dagger}\left[(i\tau_{3})(x_{\lambda}\mathbf{g}^{\lambda,+}Q_{\lambda}(\mathbf{q}\to
0))^{\ast}(-i\tau_{3})\right]C_{\mathbf{K_{+}}}$ (69)

and the coefficient $x_{\lambda}$ (which will be either $+1$ or $-1$,
depending on $\lambda$) is defined via
$\mathcal{P}Q_{\lambda}\mathcal{P}^{-1}=x_{\lambda}Q_{\lambda}$. Therefore, we
have
$\tau_{3}(\mathbf{g}^{\lambda,+})^{\ast}\tau_{3}=x_{\lambda}\mathbf{g}^{\lambda,+}$.

For the $\lambda=A_{1}$ phonon mode, the Sb1 atom moves along the $x$
direction. Hence, $x_{A_{1}}=+1$. We do not need to consider the Sb2 atom in
the unit cell, since there are no electron states of Sb2 near the Fermi
surface. Therefore, we have
$\tau_{3}(\mathbf{g}^{A_{1},+})^{\ast}\tau_{3}=\mathbf{g}^{A_{1},+}$, which
leads to

$\displaystyle\mathbf{g}^{A_{1},+}=g_{0}^{o}\tau_{0}+g_{2}^{o}\tau_{2}+g_{3}^{o}\tau_{3}.$
(70)

In the $K_{-}$ valley, time-reversal symmetry imposes

$\mathbf{g}^{{A_{1}},-}=g_{0}^{o}\tau_{0}-g_{2}^{o}\tau_{2}+g_{3}^{o}\tau_{3}.$
(71)

For $\lambda=B_{1}$, the Sb1 atom moves along the $y$ direction. Then,
$x_{B_{1}}=-1$ and
$\tau_{3}(\mathbf{g}^{B_{1},+})^{\ast}\tau_{3}=-\mathbf{g}^{B_{1},+}$, which
leads to

$\mathbf{g}^{B_{1},+}=g_{1}^{o}\tau_{1}.$ (72)

For the $K_{-}$ valley, time-reversal symmetry imposes

$\mathbf{g}^{B_{1},-}=g_{1}^{o}\tau_{1}.$ (73)

In summary,

$\displaystyle\begin{split}\mathbf{g}^{o,A_{1},s}&=g_{0}^{o}\tau_{0}+sg_{2}^{o}\tau_{2}+g_{3}^{o}\tau_{3},\\\
\mathbf{g}^{o,B_{1},s}&=g_{1}^{o}\tau_{1},\end{split}$ (74)

where $g_{0,1,2,3}^{o}$ are material-dependent real parameters. This coincides
with Eq. (5) of the main text.

###  B.2 Analysis by deformation potential theory

In this subsection, we present an alternative derivation of the e-ph
$\mathbf{g}^{\lambda,\kappa}$ matrix. The starting point is Eq. (67). Again,
all the low-energy electronic states are coming from the Sb1 atom, and the
e-ph coupling is assumed to be independent of spin. Therefore, the expression
for the e-ph coupling at the $K_{+}$ valley is given by

$\displaystyle g^{\lambda,+}_{\tau,\tau^{\prime}}(\mathbf{q})$
$\displaystyle=\frac{1}{\sqrt{N}\mathcal{A}}\int
d^{2}r\,u_{\tau,+}^{\ast}(\mathbf{r})u_{\tau^{\prime},+}(\mathbf{r})e^{-i\mathbf{q}\cdot\mathbf{r}}\times\left[\mathbf{P}^{\lambda}_{\bf
q}\cdot\frac{\partial
U(\mathbf{r}-\mathbf{r}_{1})}{\partial\mathbf{r}_{1}}\right],$ (75)

where $\mathbf{r}=(x,y)$, $\mathbf{r}_{1}=(x_{1},y_{1})$, $\mathbf{P}^{A}_{\bf
q}=(1,0)$ and $\mathbf{P}^{B}_{\bf q}=(0,1)$ (when $q\to 0$). Then, we study
the constraints due to the combined symmetry operator $\mathcal{P}$. Firstly,
it gives rise to $\mathcal{P}:\,(x,y)\to(x,-y)$. And the Bloch wave basis are
transfered as $\mathcal{P}:\,u_{p_{x}}(x,y)\to u_{p_{x}}^{\ast}(x,-y)$ and
$u_{p_{y}}(x,y)\to-u_{p_{y}}^{\ast}(x,-y)$. As for the lattice potential,
$\mathcal{P}:\,U(x-x_{1},y-y_{1})\to U(x-x_{1},-y+y_{1})$. Therefore,

  * (1)

For the $\lambda=A_{1}$ phonon, all the element of $g^{A_{1},+}$ matrix in the
orbital subspace,

$\displaystyle
g^{A_{1},+}_{\tau,\tau^{\prime}}=\frac{1}{\sqrt{N}\mathcal{A}}\int
d^{2}r\,\frac{\partial U(\mathbf{r}-\mathbf{r}_{1})}{\partial
x_{1}}u_{\tau,+}^{\ast}(x,y)u_{\tau^{\prime},+}(x,y),$ (76)

where $\mathbf{r}_{1}=(x_{1},0)$ because all the Sb1 atoms move along the
$x$-direction. Under the combined $\mathcal{P}$ symmetry operator, we have

$\displaystyle\begin{split}g^{A_{1},+}_{p_{x},p_{x}}&=\frac{1}{\sqrt{N}\mathcal{A}}\int
d^{2}r\,\frac{\partial U(\mathbf{r}-\mathbf{r}_{1})}{\partial
x_{1}}u_{p_{x},+}^{\ast}(x,y)u_{p_{x},+}(x,y),\\\
&\to\frac{1}{\sqrt{N}\mathcal{A}}\int d^{2}r\,\frac{\partial
U(x-x_{1},-y)}{\partial
x_{1}}u_{p_{x},+}(x,-y)u_{p_{x},+}^{\ast}(x,-y)=g^{A_{1},+}_{p_{x},p_{x}},\end{split}$
(77)

where we apply $P$ to everything inside the integral of the first line, and we
replace $y\to-y$ for the second line. Similarly, one can easily check that

$\displaystyle g^{A_{1},+}_{p_{y},p_{y}}\to g^{A_{1},+}_{p_{y},p_{y}},\quad
g^{A_{1},+}_{p_{x},p_{y}}\to-g^{A_{1},+}_{p_{y},p_{x}}.$ (78)

From the above results, we have

$\displaystyle\left(\begin{array}[]{cc}g^{A_{1},+}_{p_{x},p_{x}}&g^{A_{1},+}_{p_{x},p_{y}}\\\
g^{A_{1},+}_{p_{y},p_{x}}&g^{A_{1},+}_{p_{y},p_{y}}\end{array}\right)=\left(\begin{array}[]{cc}g^{A_{1},+}_{p_{x},p_{x}}&-g^{A_{1},+}_{p_{y},p_{x}}\\\
-g^{A_{1},+}_{p_{x},p_{y}}&g^{A_{1},+}_{p_{y},p_{y}}\end{array}\right)\Rightarrow\left(\begin{array}[]{cc}g^{A_{1},+}_{p_{x},p_{x}}&g^{A_{1},+}_{p_{x},p_{y}}\\\
g^{A_{1},+}_{p_{y},p_{x}}&g^{A_{1},+}_{p_{y},p_{y}}\end{array}\right)=g_{0}^{o}\tau_{0}+g_{2}^{o}\tau_{2}+g_{3}^{o}\tau_{3}.$
(85)

For the $K_{-}$ valley, the time-reversal symmetry leads to
$\mathbf{g}^{{A_{1}},-}=g_{0}^{o}\tau_{0}-g_{2}^{o}\tau_{2}+g_{3}^{o}\tau_{3}$.

  * (2)

For the $\lambda=B_{1}$ phonon, the matrix elements of the
$\mathbf{g}^{B_{1},+}$ matrix in the orbital subspace read

$\displaystyle
g^{B_{1},+}_{\tau,\tau^{\prime}}=\frac{1}{\sqrt{N}\mathcal{A}}\int
d^{2}\mathbf{r}\frac{\partial U(\mathbf{r}-\mathbf{r}_{1})}{\partial
y_{1}}u_{\tau,+}^{\ast}(\mathbf{r})u_{\tau^{\prime},+}(\mathbf{r}).$ (86)

Please note that $\frac{\partial U(\mathbf{r}-\mathbf{r}_{1})}{\partial
y_{1}}$ also changes sign under the $\mathcal{P}$ symmetry operator. In the
same way, we have

$\displaystyle g^{B_{1},+}_{p_{x},p_{x}}\to-g^{B_{1},+}_{p_{x},p_{x}},\quad
g^{B_{1},+}_{p_{y},p_{y}}\to-g^{B_{1},+}_{p_{y},p_{y}},\quad
g^{B_{1},+}_{p_{x},p_{y}}\to g^{B_{1},+}_{p_{y},p_{x}}.$ (87)

Similar relations apply for the $K_{-}$ valley. Then,
$\mathbf{g}^{B_{1},+}=\mathbf{g}^{B_{1},-}=g_{1}^{o}\tau_{1}$.

In sum, the two methods of this and the preceding subsection give consistent
results to construct the e-ph Hamiltonian. As a result, the full Hamiltonian
can be written as follows,

$\displaystyle\text{Electron Hamiltonian: }h_{\pm}(\mathbf{k})=\pm
v_{0}(k_{y}\tau_{3}+k_{x}\tau_{1})\pm m_{0}\tau_{2},$ (88)
$\displaystyle\text{e-ph Hamiltonian at $K_{+}$: }{\cal
H}^{+}_{e-ph}=\sum_{\mathbf{k},\mathbf{q}}C^{\dagger}_{\mathbf{k}}\left[\left(g_{0}^{o}\tau_{0}+g_{2}^{o}\tau_{2}+g_{3}^{o}\tau_{3}\right)Q^{A}_{\mathbf{q}}+g_{1}^{o}\tau_{1}Q^{B}_{\mathbf{q}}\right]C_{\mathbf{k}+\mathbf{q}},$
(89) $\displaystyle\text{e-ph Hamiltonian at $K_{-}$: }{\cal
H}^{-}_{e-ph}=\sum_{\mathbf{k},\mathbf{q}}C^{\dagger}_{\mathbf{k}}\left[\left(g_{0}^{o}\tau_{0}-g_{2}^{o}\tau_{2}+g_{3}^{o}\tau_{3}\right)Q^{A}_{\mathbf{q}}+g_{1}^{o}\tau_{1}Q^{B}_{\mathbf{q}}\right]C_{\mathbf{k}+\mathbf{q}},$
(90)

where ${\bf k}$ is the momentum measured from a valley and the $g$-parameters
depend on the material details.

###  B.3 Phonon-induced pseudo-gauge fields for Dirac electrons

In this section, we show that the e-ph coupling leads to electronic pseudo-
gauge fields. This can be seen by rewriting the relevant Hamiltonians in real
space.

$\displaystyle H$ $\displaystyle=\sum_{s}\int
d^{2}\mathbf{r}\,c^{\dagger}_{s}(\mathbf{r})\left\\{sv_{0}\left[(-i\partial_{y}+A_{y})\tau_{3}+(-i\partial_{x}+A_{x})\tau_{1}\right]+sm_{0}\tau_{2}+A_{0}\tau_{0}\right\\}c_{s}(\mathbf{r}),$
(91) $\displaystyle H_{e-ph}$ $\displaystyle=\sum_{s}\int
d^{2}\mathbf{r}\,c^{\dagger}_{s}(\mathbf{r})\left\\{\left(g_{0}^{o}\tau_{0}+sg_{2}^{o}\tau_{2}+g_{3}^{o}\tau_{3}\right)Q_{A_{1}}(\mathbf{r})+g_{1}^{o}\tau_{1}Q_{B_{1}}(\mathbf{r})\right\\}c_{s}(\mathbf{r})\
,$ (92)

where we have included the $U(1)$ gauge field $A_{\mu}$, the matrix is
$(-,+,+)$, the unit system is chosen such that $\hbar=c=1$, we have absorbed
the elementary charge $e$ into $A_{\mu}$, and we have used

$\displaystyle\begin{split}&c^{\dagger}_{s,\boldsymbol{k}}=\frac{1}{\sqrt{\mathcal{A}}}\int
d^{2}re^{\mathrm{i}\boldsymbol{k}\cdot\boldsymbol{r}}c^{\dagger}_{s}(\mathbf{r})\\\
&Q_{\lambda,\boldsymbol{k}}=\frac{1}{\mathcal{A}}\int
d^{2}re^{\mathrm{i}\boldsymbol{k}\cdot\boldsymbol{r}}Q_{\lambda}(\mathbf{r}).\end{split}$
(93)

Here the unit for the above operators are:
$[c^{\dagger}_{s}(\mathbf{r})]=[nm^{-1}]=[meV]$,
$[c^{\dagger}_{s,\boldsymbol{k}}]=[1]$,
$[Q_{\lambda,\boldsymbol{k}}]=[nm]=[meV^{-1}]$, and
$[Q_{\lambda}(\mathbf{r})]=[nm]=[meV^{-1}]$. Notice that $[nm]=[meV^{-1}]$ in
the case of $\hbar=c=1$. Adding $H_{e-ph}$ to $H$ gives

$\displaystyle H_{tot}=\sum_{s}\int
d^{2}\mathbf{r}\,c^{\dagger}_{s}(\mathbf{r})\left\\{\widetilde{A}_{s,0}\tau_{0}+sv_{0}\left[(-i\partial_{y}+\widetilde{A}_{s,y})\tau_{3}+(-i\partial_{x}+\widetilde{A}_{s,x})\tau_{1}\right]+s(m_{0}+g_{2}^{o}Q_{A_{1}}(\mathbf{r}))\tau_{2}\right\\}c_{s}(\mathbf{r})\
,$ (94)

where $\widetilde{A}_{s,0}=A_{0}+g_{0}^{o}Q_{A_{1}}(\mathbf{r})$,
$\widetilde{A}_{s,x}=A_{x}+s\frac{1}{v_{0}}g_{1}^{o}Q_{B_{1}}(\mathbf{r})$,
and
$\widetilde{A}_{s,y}=A_{y}+s\frac{1}{v_{0}}g_{3}^{o}Q_{A_{1}}(\mathbf{r})$. As
a result, we can define the pseudo-gauge field at $K_{s}$ as

$\displaystyle{A}_{pse,\mu}^{s}(\mathbf{r})=\left(g_{0}^{o}Q_{A_{1}}(\mathbf{r}),s\frac{1}{v_{0}}g_{1}^{o}Q_{B_{1}}(\mathbf{r}),s\frac{1}{v_{0}}g_{3}^{o}Q_{A_{1}}(\mathbf{r})\right)_{\mu},$
(95)

where $\mu=0,1,2$ and the pseudo-gauge field couples to the electrons in the
same way as the $U(1)$ gauge field at each valley. The unit of
${A}_{pse,\mu}^{s}(\mathbf{r})$ is [meV], since $[v_{0}]=[meV\cdot nm]=[1]$
and $[g_{0,1,3}^{o}]=[meV\cdot nm^{-1}]=[meV^{2}]$. In Eq. (4) of the main
text, we show the above pseudo-gauge field in the momentum space.

##  Appendix C Electron-phonon interaction Hamiltonian for acoustic phonons

In this section, we derive the electron-acoustic phonon coupling Hamiltonian
from the deformation potential theory. In principle, very similar expressions
as the ones obtained for the optical phonons could be used. In practice, those
expressions do not clearly show how the e-ph coupling for acoustic phonons
vanishes in the $q\to 0$ limit. Consequently, we follow another approach,
discussed in Ref. [mahan_book,abrikosov_book] to derive the leading-order
electron-phonon coupling.

Since the acoustic phonon frequency vanishes at $q\to 0$, Ref.
[abrikosov_book] suggests to use the following operator instead of the
operator $Q_{\lambda}(q)$ [abrikosov_book]

$\displaystyle\tilde{Q}_{\lambda}(\mathbf{q})=\omega_{\lambda}(\mathbf{q})Q_{\lambda}(\mathbf{q})=\sqrt{\frac{\hbar\omega_{\lambda}(\mathbf{q})}{2M}}\left(b_{\mathbf{q},\lambda}+b^{\dagger}_{-\mathbf{q},\lambda}\right).$
(96)

In terms of $\tilde{Q}_{\lambda}(\mathbf{q})$, Ref. [abrikosov_book] further
suggests that the coupling between long-wavelength acoustic phonons and
electrons at valley is

$\displaystyle{\cal
H}_{e-ph}=\sum_{\mathbf{k},\mathbf{q}}\sum_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}\sum_{\lambda,\kappa}\tilde{g}_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}^{\lambda,\kappa}\tilde{Q}_{\lambda}(\mathbf{q})c_{\sigma\tau,\kappa}^{\dagger}(\mathbf{k})c_{\sigma^{\prime}\tau^{\prime},\kappa}(\mathbf{k}-\mathbf{q}),$
(97)

where $\tilde{g}_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}^{\lambda,\kappa}$
are all independent of the phonon momentum $\mathbf{q}$, to leading order. Now
we transform Eq. (97) back to the e-ph coupling in terms of $Q_{\lambda}(q)$

$\mathcal{H}_{e-ph}=\sum_{{\bf k},{\bf
q}}\sum_{\lambda,\kappa}\sum_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}g^{\lambda,\kappa}_{\sigma\tau,\sigma^{\prime}\tau^{\prime}}(\mathbf{q})Q_{\lambda}(\mathbf{q})c_{\sigma\tau,\kappa}^{\dagger}({\bf
k})c_{\sigma^{\prime}\tau^{\prime},\kappa}(\mathbf{k}-\mathbf{q})\ ,$ (98)

where Eq. (97) suggests that the matrix $g^{\lambda,\kappa}$ must be of linear
order of the $|\mathbf{q}|$. Clearly, although the e-ph coupling matrix is
$q-$independent for long-wavelength optical phonons, it is linear in ${\bf q}$
for acoustic phonons. Nevertheless, Eq. (97) does not take the direction of
$\mathbf{q}$ into consideration, though the order of $\mathbf{q}$ is correct.
The direction of $\mathbf{q}$ does matter here, since the distortion effect
breaks the four-fold rotational symmetry. To include the direction of
$\mathbf{q}$, we perform the symmetry analysis for $g^{\lambda,\kappa}$, while
only keeping the linear order of $\mathbf{q}$. After taking the combined
symmetry $\mathcal{P}=C_{2x}\mathcal{T}$ and the Hermitian condition into
account, all the matrix elements are purely imaginary. Then, we arrive at

$\displaystyle\begin{split}\mathbf{g}^{A_{1},s}&=iq_{x}(g_{0}^{a}\tau_{0}+sg_{2}^{a}\tau_{2}+g_{3}^{a}\tau_{3})+iq_{y}(g_{1}^{a}\tau_{1}),\\\
\mathbf{g}^{B_{1},s}&=iq_{y}(g_{0}^{a}\tau_{0}+sg_{2}^{a}\tau_{2}+g_{3}^{a}\tau_{3})+iq_{x}(g_{1}^{a}\tau_{1}),\end{split}$
(99)

where $s$ is the valley index. We emphasize that the pseudo-gauge field given
by the acoustic phonons has the same form as the strain tensor, as reflected
by linear $\bf q$ dependence of the e-ph couplings[75].

##  Appendix D Green’s function for two coupled in-plane phonon modes

In this section, we discuss the phonon Green’s function in the presence of
electron-phonon (e-ph) and electron-electron (e-e) interactions. The
diagrammatic form of the Green’s function is shown in Fig. 5, which follows
Ref. [43] (we have reproduced the same results using a functional integral
approach, in which electrons are integrated out with the aid of a Hubbard-
Stratonovich transformation in order to obtain an effective action for
phonons). We will focus on the two optical in-plane phonons labeled by
$\lambda=\\{A_{1},B_{1}\\}$. Then, the Matsubara Green’s function for the
phonons is defined as

$\displaystyle D_{\lambda\lambda^{\prime}}(\mathbf{q},\tau)=-\langle
T_{\tau}A_{\mathbf{q},\lambda}(\tau)A_{-\mathbf{q},\lambda^{\prime}}(0)\rangle,$
(100)

where
$A_{\mathbf{q},\lambda}=b_{\mathbf{q},\lambda}+b^{\dagger}_{-\mathbf{q},\lambda}$,
$\tau$ is the imaginary time and $T_{\tau}$ denotes the imaginary-time ordered
product. Note that the displacement operator is
$Q_{\lambda}(\mathbf{q})=\sqrt{\frac{\hbar}{2M\omega_{\lambda}}}A_{\mathbf{q},\lambda}$.
In $(\mathbf{q},\omega)$ space, we use

$\displaystyle
D_{\lambda\lambda^{\prime}}(\mathbf{q},iq_{m})=\int_{0}^{\beta}e^{iq_{m}\tau}D_{\lambda\lambda^{\prime}}(\mathbf{q},\tau)\,d\tau,$
(101)

where $q_{m}=2\pi m/\beta$ with $\beta=1/k_{B}T$ the inverse of temperature
and integer $m$. Then, the bare (non-interacting) phonon Green’s function is
given by

$\displaystyle
D_{0}(\mathbf{q},iq_{m})=\text{Diag}\\{D_{A_{1}A_{1}},D_{B_{1}B_{1}}\\},$
(102)

where
$D_{\lambda\lambda}=2\omega_{\lambda}/((iq_{m})^{2}-\omega_{\lambda}^{2})$ and
$\omega_{\lambda}$ is the bare frequency for the $\lambda$-phonon mode. In the
presence of electron-phonon interactions, the full Green’s function is no
longer diagonal in the $\\{A_{1},B_{1}\\}$ basis. In our calculation, we first
ignore the $g_{2}$-term in the e-ph coupling Hamiltonian because it only
renormalizes the Dirac electron’s mass; we discuss its effect to the off-
diagonal phonon self-energy later (see Sec. LABEL:section-g2-self-energy). The
full phonon Green’s function is given by

$\displaystyle
D(\mathbf{q},iq_{m})=\left[1-D_{0}(\mathbf{q},iq_{m})\Sigma(\mathbf{q},iq_{m})\right]^{-1}D_{0}(\mathbf{q},iq_{m}),$
(103)

where
$\Sigma(\mathbf{q},iq_{m})=\Sigma^{p}(\mathbf{q},iq_{m})+\Sigma^{e}(\mathbf{q},iq_{m})$
is the phonon self-energy due to bare e-ph interactions ($\Sigma^{p}$) and the
screened e-e interaction ($\Sigma^{e}$). The phonon self-energy can be
represented as a $2\times 2$ matrix in $\\{A_{1},B_{1}\\}$ space. Its diagonal
entries are

$\displaystyle\begin{split}\Sigma_{AA}^{p}&=\frac{\hbar}{2M\omega_{A}}(\Pi_{00}^{p}+\Pi_{33}^{p}+\Pi_{03}^{p}+\Pi_{30}^{p})\text{
and }\Sigma_{BB}^{p}=\frac{\hbar}{2M\omega_{B}}\Pi_{11}^{p},\\\
\Sigma_{AA}^{e}&=\frac{\hbar}{2M\omega_{A}}(\Pi_{00}^{e}+\Pi_{33}^{e}+\Pi_{03}^{e}+\Pi_{30}^{e})\text{
and }\Sigma_{BB}^{e}=\frac{\hbar}{2M\omega_{B}}\Pi_{11}^{e}.\end{split}$ (104)

Likewise, the off-diagonal self-energies are

$\displaystyle\begin{split}\Sigma_{AB}^{p}&=\frac{\hbar}{2M\sqrt{\omega_{A}\omega_{B}}}(\Pi_{01}^{p}+\Pi_{31}^{p}),\\\
\Sigma_{AB}^{e}&=\frac{\hbar}{2M\sqrt{\omega_{A}\omega_{B}}}(\Pi_{01}^{e}+\Pi_{31}^{e}),\end{split}$
(105)

with $\Sigma_{BA}({\bf q},\omega)=\Sigma_{AB}(-{\bf q},-\omega)$. Again, we
will take $\hbar=1$ convention in the following calculations for simplicity.
Here, the components $\Pi^{p}_{ij}({\bf q},iq_{m})$ with $i,j=0,1,3$ are given
by

$\displaystyle\Pi_{ij}^{p}(\mathbf{q},iq_{m})=\frac{g_{i}^{o}g_{j}^{o}}{\beta}\sum_{s,\mathbf{k},\omega_{n}}\text{tr}[\tau_{i}G_{s}(\mathbf{k},i\omega_{n})\tau_{j}G_{s}(\mathbf{k}^{\prime},i\omega_{n}+iq_{m})],$
(106)

where $\mathbf{k}^{\prime}=\mathbf{k}+\mathbf{q}$, $s$ labels the two valleys,
$G_{s}(\mathbf{k},i\omega_{n})=[-i\omega_{n}-\mu+h_{s}(\mathbf{k})]^{-1}$ is
the bare Matsubara Green’s function of Dirac electrons at valley $K_{s}$,
$\omega_{n}=(2n+1)/\beta$ is a fermionic Matsubara frequency and
$q_{m}=2m/\beta$ is a bosonic Matsubara frequency. The retarded phonon self-
energy $\Sigma({\bf q},\omega)$ is obtained by analytical continuation
$iq_{m}\to\omega+i0^{+}$.

Similarly,

$\displaystyle\Pi_{ij}^{e}(\mathbf{q},iq_{m})=\frac{g_{i}^{o}g_{j}^{o}}{\beta^{2}\mathcal{V}}\left(\sum_{s,\mathbf{k},\omega_{n}}\text{tr}[\tau_{i}G_{s}(\mathbf{k},i\omega_{n})\tau_{0}G_{s}(\mathbf{k}^{\prime},i\omega_{n}+iq_{m})]V_{q}^{RPA}\sum_{s,\mathbf{k},\omega_{l}^{\prime}}\text{tr}[\tau_{0}G_{s}(\mathbf{k},i\omega_{l}^{\prime})\tau_{j}G_{s}(\mathbf{k}^{\prime},i\omega_{l}^{\prime}+iq_{m})]\right),$
(107)

where $V_{q}^{RPA}=V_{q}/(1-V_{q}\Pi_{0}^{e})$ is the RPA-screened e-e
interaction, with $V_{q}=e^{2}/(\epsilon_{\infty}q^{2})$ the “bare” Coulomb
potential ($\epsilon_{\infty}$ denotes the screening from high-energy
electronic bands that are not included in Eq. (88)). $\mathcal{V}$ is the 3D
volume. Also, the electronic polarization function $\Pi_{0}^{e}$ is defined as

$\displaystyle\Pi^{e}_{0}(\mathbf{q},iq_{m})=\int_{0}^{2\pi/a_{z}}\frac{dk_{z}}{2\pi}\left\\{\frac{1}{\beta\mathcal{A}}\sum_{s,\mathbf{k},\omega_{n}}\text{Tr}[\tau_{0}G(\mathbf{k},\omega_{n})\tau_{0}G(\mathbf{k}+\mathbf{q},\omega_{n}+q_{m})]\right\\}.$
(108)

which will be calculated in the next subsection. Here $\mathcal{A}$ is the
area of the 2D Sb layer, the electric polarization is defined in 3D with a sum
over $\mathbf{k}$ in 2D and one integral along $k_{z}$, and $a_{z}$ is the
z-direction lattice constant. Since the BaMnSb is just a layered material, and
we assume the inter-layer coupling is zero.

The units for the quantities in the above expressions are summarized as
$[Q]=[nm],[c_{\mathbf{k}}]=[1],[V]=[nm^{3}],[a_{z}]=[nm],[\frac{\hbar}{2M\omega_{A}}]=[nm^{2}],[g]=[meV\cdot
nm^{-1}],[G]=[meV^{-1}],[\beta]=[meV^{-1}],[V_{q}]=[meV\cdot
nm^{3}],[\sum_{k}]=[1],[\sum_{\omega}]=[1]$, then
$[\Pi^{p}]=[\Pi^{e}]=[meV\cdot nm^{-2}]$, $[\Sigma]=[meV]$, and
$[\Pi^{e}_{0}]=[meV^{-1}\cdot nm^{-3}]$. Moreover, $[2M\omega_{A}]=[nm^{-2}]$
for $\hbar=e=1$ (this convention is used in the main text). In the following
sections, we take $[nm]=[meV^{-1}]$ to double-check the units for simplicity
by taking $[v_{0}]=[1]$ ($v_{0}$ is Fermi velocity of Dirac fermions).

Figure 5:  (a) Dyson equation for the phonon Green’s function (GF). (b)
Equation for the phonon self-energy.

###  D.1 The electronic polarization function for the insulating phase

In this subsection, we calculate the contribution of low-energy electrons in
BaMnSb2 to the electronic polarization function $\Pi_{0}^{e}({\bf q},iq_{m})$.
We assume throughout that the chemical potential is inside the electron
bandgap. We calculate the contributions from the $K_{+}$ and $K_{-}$ valleys
separately, and add them in the end. The corresponding electronic Green’s
function at valley $\mathbf{K}_{s}$ (a $2\times 2$ matrix) is

$\displaystyle
G_{s}(\mathbf{k},i\omega_{n})=\left(-i\omega_{n}-\mu+h_{s}(\mathbf{k})\right)^{-1}=\frac{P_{+}(\mathbf{k})}{-i\omega_{n}-\mu+E_{\mathbf{k}}}+\frac{P_{-}(\mathbf{k})}{-i\omega_{n}-\mu-
E_{\mathbf{k}}},$ (109)

where $P_{\pm}(\mathbf{k})=\frac{1}{2}[1\pm h_{s}({\bf k})/E_{\mathbf{k}}]$
are projector operators and
$E_{\mathbf{k}}=\sqrt{v_{0}^{2}(k_{x}^{2}+k_{y}^{2})+m^{2}}$. We begin with
contributions from electrons at $K_{+}$ valley to the electric polarization
function,

$\displaystyle\frac{1}{\beta}\sum_{\mathbf{k}}\sum_{\omega_{n}}\text{Tr}[\tau_{0}G(\mathbf{k},\omega_{n})\tau_{0}G(\mathbf{k}+\mathbf{q},\omega_{n}+q_{m})].$
(110)

First, we recognize that

$\displaystyle\tau_{0}G(\mathbf{k},\omega_{n})\tau_{0}G(\mathbf{k}+\mathbf{q},\omega_{n}+q_{m})=\sum_{\alpha\beta}f_{\alpha\beta}(\mathbf{k},\mathbf{q},\omega_{n},q_{m})P_{\alpha}(\mathbf{k})P_{\beta}(\mathbf{k}+\mathbf{q}),$
(111)

where $\alpha,\beta=\\{+,-\\}$ and
$f_{\alpha\beta}(\mathbf{k},\mathbf{q},\omega_{n},q_{m})=\frac{1}{-i\omega_{n}-\mu+\alpha
E_{\mathbf{k}}}\frac{1}{-i(\omega_{n}+q_{m})-\mu+\beta
E_{\mathbf{k}+\mathbf{q}}}$. Then, the summation over Matsubara frequencies
gives

$\displaystyle\begin{split}k_{B}T\sum_{\omega_{n}}f_{\alpha\beta}(\mathbf{k},\mathbf{q},\omega_{n},q_{m})&=\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha E_{\mathbf{k}}},\end{split}$ (112)

where $n_{F}(x)=1/(1+e^{x/k_{B}T})$ is the Fermi distribution function so that
$n_{F}(x>0)=0$ and $n_{F}(x<0)=1$ at zero temperature. For the trace part, we
have

$\displaystyle\text{Tr}[P_{\alpha}(\mathbf{k})P_{\beta}(\mathbf{k}+\mathbf{q})]$
$\displaystyle=\frac{1}{2}+\frac{1}{2}\alpha\beta\left[\frac{E_{\mathbf{k}}}{E_{\mathbf{k}+\mathbf{q}}}+\frac{v_{0}^{2}(k_{x}q_{x}+k_{y}q_{y})}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right].$
(113)

The contribution from electrons at the $K_{-}$ valley can be computed in the
same way, simply by replacing the projection $P_{\alpha}$ with $P_{-\alpha}$.
Therefore, summing over the contribution from both valleys, we obtain

$\displaystyle\Pi^{e}_{0}(\mathbf{q},iq_{m})=\frac{1}{a_{z}\mathcal{A}}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left\\{1+\alpha\beta\left[\frac{E_{\mathbf{k}}}{E_{\mathbf{k}+\mathbf{q}}}+\frac{v_{0}^{2}(k_{x}q_{x}+k_{y}q_{y})}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right]\right\\}.$
(114)

In the $m=0$ limit, the term inside the curly brackets tends to
$1+\alpha\beta\cos(\theta_{\mathbf{k},\mathbf{k}^{\prime}})$, where
$\theta_{\mathbf{k},\mathbf{k}^{\prime}}$ is the angle between $\mathbf{k}$
and $\mathbf{k}^{\prime}$. In that limit, our expression for $\Pi^{e}_{0}$
reproduces that of graphene [76].

In the zero-temperature limit and when the chemical potential is inside the
bandgap, only interband transitions $\alpha\neq\beta$ contribute to
$\Pi^{e}_{0}$. In the static limit, these contributions amount to a
renormalization of $\epsilon_{\infty}$ in the screened Coulomb interaction.
More precisely, we obtain[77]

$\displaystyle\Pi^{e}_{0}(\mathbf{q},\omega=0)\approx-q^{2}/(6a_{z}\pi|m_{0}|),$
(115)

here $a_{z}$ is the z direction lattice constant. which is valid when we take
$v_{0}q\ll m_{0}$ limit. Thus, the contribution from Dirac fermions to the
static dielectric function is

$\displaystyle\epsilon(\mathbf{q})=1-V_{q}\Pi_{0}^{e}(\mathbf{q},\omega=0)\approx
1+e^{2}/(6a_{z}\pi\epsilon_{\infty}|m_{0}|)\triangleq\epsilon_{RPA},$ (116)

where $V_{q}=e^{2}/(\epsilon_{\infty}q^{2})$ is the Coulomb potential. We
check the unit of $[V_{q}]=[meV^{-2}]$ and
$[\Pi_{0}^{e}(\mathbf{q},iq_{m})]=[meV^{2}]$, so that $\epsilon_{RPA}$ is a
dimensionless parameter. Thus, Eq. (116) gives rise to the the RPA-screened
Coulomb potential for the insulating electron system.

###  D.2 The diagonal phonon self-energies $\Sigma_{AA}$ and $\Sigma_{BB}$

In this subsection, we discuss the diagonal phonon self-energies,
$\Sigma_{\lambda\lambda}$, for $\lambda=\\{A_{1},B_{1}\\}$. These quantities
were defined in Eq. (104). We first calculate all the necessary elements of
$\Pi^{p}_{ij}$, with $i,j=0,1,3$, following Eq. (106). The calculation is
similar to that of the polarization function described above. The explicit
expressions are as follows:

$\displaystyle\Pi_{00}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=(g_{0}^{o})^{2}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{k+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[1+\alpha\beta\left[\frac{E_{\mathbf{k}}}{E_{\mathbf{k}+\mathbf{q}}}+\frac{v_{0}^{2}(k_{x}q_{x}+k_{y}q_{y})}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right]\right],$
(117) $\displaystyle\Pi_{33}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=(g_{3}^{o})^{2}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[1+\alpha\beta\frac{v_{0}^{2}(k_{y}^{2}-k_{x}^{2})+v_{0}^{2}(k_{y}q_{y}-k_{x}q_{x})-m_{0}^{2}}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right],$
(118) $\displaystyle\Pi_{03}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=g_{0}^{o}g_{3}^{o}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[i\alpha\beta\frac{v_{0}q_{x}m_{0}}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right]=-\Pi_{30}^{p}(\mathbf{q},iq_{m}),$
(119) $\displaystyle\Pi_{11}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=(g_{1}^{o})^{2}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[1-\alpha\beta\frac{v_{0}^{2}(k_{y}^{2}-k_{x}^{2})+v_{0}^{2}(k_{y}q_{y}-k_{x}q_{x})+m_{0}^{2}}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right],$
(120) $\displaystyle\Pi_{01}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=g_{0}^{o}g_{1}^{o}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[-i\alpha\beta\frac{v_{0}q_{y}m_{0}}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right]=-\Pi_{10}^{p}(\mathbf{q},iq_{m}).$
(121)

In the insulating regime ($\mu$ inside the electron band gap), only interband
transitions ($\alpha\neq\beta$) contribute at zero temperature. For latter
reference, when $|\omega|<2|m_{0}|$ (where $\omega$ is a real frequency), the
retarded response functions $\Pi_{ii}^{p}({\bf q},\omega)$ are purely real
($i=0,1,3$), while the retarded response functions $\Pi^{p}_{01}({\bf
q},\omega)$ and $\Pi^{p}_{03}({\bf q},\omega)$ are purely imaginary.

####  D.2.1 Diagonal self-energies in the absence of electron-electron
interactions

We first consider only the e-ph interaction. Since
$\Pi_{03}^{p}+\Pi_{30}^{p}=0$, the diagonal self-energies are given by
$\Sigma_{AA}^{p}=\frac{1}{2M\omega_{A}}[\Pi_{00}^{p}+\Pi_{33}^{p}]$ and
$\Sigma_{BB}^{p}=\frac{1}{2M\omega_{B}}\Pi_{11}^{p}$. Before calculations, we
double-check the unit of $[\Sigma_{AA,AB,BB}]=[meV]$, since $[M]=[meV]$,
$[\omega_{A,B}]=[meV]$ and $[\Pi^{p}]=[meV^{3}]$. In the insulating regime and
if $|\omega|<2|m_{0}|$, these self-energies are purely real at zero
temperature.

####  D.2.2 Contribution of electron-electron interactions to the diagonal
self-energies

Now, we incorporate the e-e interactions, which contribute to the diagonal
self-energies as follows:

$\displaystyle\Sigma_{AA}^{e}(\mathbf{q},iq_{m})$
$\displaystyle=\frac{1}{2M\omega_{A}}\left[\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{00}^{p}(\mathbf{q},iq_{m})V_{q}^{RPA}\Pi_{00}^{p}(\mathbf{q},iq_{m})+\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{30}^{p}(\mathbf{q},iq_{m})V_{q}\Pi_{03}^{p}(\mathbf{q},iq_{m})\right],$
(122) $\displaystyle\Sigma_{BB}^{e}(\mathbf{q},iq_{m})$
$\displaystyle=\frac{1}{2M\omega_{B}}\left[\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{10}^{p}(\mathbf{q},iq_{m})V_{q}^{RPA}\Pi_{01}^{p}(\mathbf{q},iq_{m})\right].$
(123)

In the insulating regime and if $|\omega|<2|m_{0}|$, these self-energies are
purely real at zero temperature.

###  D.3 The off-diagonal phonon self-energy $\Sigma_{AB}$

Next, let us consider the off-diagonal self-energy defined in Eq. (105), which
couples the two optical phonons at a finite phonon momentum and generates the
phonon helicity. To compute this self-energy, we need the following
quantities:

$\displaystyle\Pi_{01}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=g_{0}^{o}g_{1}^{o}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[-i\alpha\beta\frac{v_{0}q_{y}m_{0}}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right]=-\Pi_{10}^{p}(\mathbf{q},iq_{m}),$
(124) $\displaystyle\Pi_{31}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=g_{3}^{o}g_{1}^{o}\sum_{\mathbf{k}}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\times\left[\alpha\beta\frac{v_{0}^{2}(k_{x}(k_{y}+q_{y})+k_{y}(k_{x}+q_{x})}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right].$
(125)

####  D.3.1 Off-diagonal self-energy in the absence of electron-electron
interactions

Hereafter, we focus on $\Sigma_{AB}^{p}$ computed for ${\bf q}=(q_{x},0)$ or
${\bf q}=(0,q_{y})$. These special cases will allow us to gain some analytical
understanding on the phonon helicity. The trace part for
$\Pi_{31}^{p}(q,iq_{m})$ is an odd function of $k_{x}$ or $k_{y}$, and thus

$\displaystyle\Pi_{31}^{p}(q_{x},q_{y}=0,iq_{m})$
$\displaystyle=g_{3}^{o}g_{1}^{o}\sum_{\mathbf{k},\omega_{n}}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\times\left[\alpha\beta\frac{v_{0}^{2}(2k_{x}+q_{x})}{E_{k}E_{k+q}}\right]\times
k_{y}=0,$ (126) $\displaystyle\Pi_{31}^{p}(q_{x}=0,q_{y},iq_{m})$
$\displaystyle=g_{3}^{o}g_{1}^{o}\sum_{\mathbf{k},\omega_{n}}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\times\left[\alpha\beta\frac{v_{0}^{2}(2k_{y}+q_{y})}{E_{k}E_{k+q}}\right]\times
k_{x}=0.$ (127)

It follows that

$\displaystyle\begin{split}\Sigma_{AB}^{p}(q_{x},q_{y}=0,iq_{m})&=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\left[\Pi_{31}^{p}(q_{x},q_{y}=0,iq_{m})+\Pi_{01}^{p}(q_{x},q_{y}=0,iq_{m})\right]\\\
&=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\Pi_{01}^{p}(q_{x},q_{y}=0,iq_{m}),\\\
\Sigma_{AB}^{p}(q_{x}=0,q_{y},iq_{m})&=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\left[\Pi_{31}^{p}(q_{x}=0,q_{y},iq_{m})+\Pi_{01}^{p}(q_{x}=0,q_{y},iq_{m})\right]\\\
&=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\Pi_{01}^{p}(q_{x}=0,q_{y},iq_{m}),\end{split}$
(128)

####  D.3.2 Contribution of electron-electron interactions to the off-diagonal
self-energy

Next, we consider the additional correction to the off-diagonal phonon coming
from e-e interactions, which is given by

$\displaystyle\Sigma_{AB}^{e}(\mathbf{q},iq_{m})=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\left[\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{00}^{p}(\mathbf{q},iq_{m})V_{q}^{RPA}\Pi_{01}^{p}(\mathbf{q},iq_{m})+\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{30}^{p}(\mathbf{q},iq_{m})V_{q}^{RPA}\Pi_{01}^{p}(\mathbf{q},iq_{m})\right].$
(129)

Let us consider these two terms one-by-one. From Eqs. (119) and (121), we
obtain

$\displaystyle\Pi_{30}^{p}(\mathbf{q},iq_{m})\sim q_{x}\text{ and
}\Pi_{01}^{p}(\mathbf{q},iq_{m})\sim q_{y},$ (130)

to linear order in ${\bf q}$. As a result, one of these two terms must vanish
when ${\bf q}$ is along the $x$ or $y$ axis. Then, it simplifies the off-
diagonal self energy as

$\displaystyle\Sigma_{AB}^{e}(\mathbf{q},iq_{m})=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\left[\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{00}^{p}(\mathbf{q},iq_{m})V_{q}^{RPA}\Pi_{01}^{p}(\mathbf{q},iq_{m})\right],$
(131)

with ${\bf q}=(q_{x},0)$ or ${\bf q}=(0,q_{y})$.

Adding the results from Eqs. (128) and (131), the total off-diagonal self-
energy becomes

$\displaystyle\begin{split}\Sigma_{AB}(\mathbf{q},iq_{m})&=\left[1+\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{00}^{p}(\mathbf{q},iq_{m})V_{q}^{RPA}\right]\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\Pi_{01}^{p}(\mathbf{q},iq_{m}),\\\
&=\left[1+\Pi_{0}^{e}(\mathbf{q},iq_{m})V_{q}^{RPA}\right]\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\Pi_{01}^{p}(\mathbf{q},iq_{m})\\\
&=\frac{1}{\epsilon(\mathbf{q})}\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\Pi_{01}^{p}(\mathbf{q},iq_{m}),\end{split}$
(132)

where the RPA dielectric function $\epsilon(\mathbf{q})$ is given by Eq.
(116). In other words, the influence of e-e interactions in the off-diagonal
phonon energy amounts to renormalizing the e-ph coupling parameters as

$\displaystyle
g_{0}^{o}g_{1}^{o}\to\frac{1}{\epsilon(\mathbf{q})}g_{0}^{o}g_{1}^{o}\approx\frac{g_{0}^{o}g_{1}^{o}}{\epsilon_{RPA}}.$
(133)

In sum, after renormalizing the e-ph couplings, we have

$\displaystyle\Sigma_{AB}(\mathbf{q},iq_{m})=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}[\Pi_{01}^{p}(\mathbf{q},iq_{m})],$
(134)

with ${\bf q}=(q_{x},0)$ or ${\bf q}=(0,q_{y})$. This is the result used in
the main text to determine the phonon helicity.

####  D.3.3 Symmetry constraints on the off-diagonal self-energy

Now, let us focus on the symmetry constraints on the off-diagonal self-energy.
The symmetry operators include the time-reversal symmetry and the two-fold
rotation symmetry. In this subsection, we work with real frequencies.

  * (1.)

Time-reversal symmetry $\mathcal{T}$ imposes the following constraints on the
electron Green’s function:

$\displaystyle\mathcal{T}G_{\pm}(k_{x},k_{y},\omega)\mathcal{T}^{-1}=G_{\mp}^{\ast}(-k_{x},-k_{y},\omega).$
(135)

It follows that

$\displaystyle\begin{split}\Pi_{i1}^{p}(\mathbf{q},\omega)&=\sum_{s}x_{i}(s)g_{i}^{o}g_{1}^{o}\frac{1}{\beta}\sum_{\mathbf{k},\omega^{\prime}}\text{tr}[\tau_{i}G_{s}(\mathbf{k},\omega^{\prime})\tau_{1}G_{s}(\mathbf{k}^{\prime},\omega^{\prime\prime})],\\\
&=\left(\sum_{s}x_{i}(s)g_{i}^{o}g_{1}^{o}\frac{1}{\beta}\sum_{\mathbf{k},\omega^{\prime}}\text{tr}[\tau_{i}^{\ast}G_{s}^{\ast}(\mathbf{k},\omega^{\prime})\tau_{1}^{\ast}G_{s}^{\ast}(\mathbf{k}^{\prime},\omega^{\prime\prime})]\right)^{\ast},\\\
&=\left(\sum_{s}x_{i}(s)g_{i}^{o}g_{1}^{o}\frac{1}{\beta}\sum_{-\mathbf{k},\omega^{\prime}}\text{tr}[\tau_{i}^{\ast}G_{-s}(-\mathbf{k},\omega^{\prime})\tau_{1}G_{-s}(-\mathbf{k}^{\prime},\omega^{\prime\prime})]\right)^{\ast}=\left(\Pi_{i1}^{p}(-\mathbf{q},\omega)\right)^{\ast},\end{split}$
(136)

where $x(s)=s\delta_{i,2}+\delta_{i,0}+\delta_{i,3}$ with $i=0,2,3$, we
replace $s\to-s$ in the last line, and we use
$x(-s)\tau_{i}^{\ast}=x(s)\tau_{i}$. Therefore, we have

$\displaystyle\Sigma_{AB}(\mathbf{q},\omega)=\Sigma_{AB}^{\ast}(-\mathbf{q},\omega).$
(137)

  * (2.)

The two-fold rotation symmetry $C_{2x}$ imposes the following constraint on
the Green’s function:

$\displaystyle
C_{2x}G_{\pm}(k_{x},k_{y},\omega)C_{2x}^{-1}=G_{\mp}(k_{x},-k_{y},\omega).$
(138)

Along the lines of the analysis above,

$\displaystyle\begin{split}\Pi_{i1}^{p}(q_{x},q_{y},\omega)&=\sum_{s}x_{i}(s)g_{i}^{o}g_{1}^{o}\frac{1}{\beta}\sum_{\mathbf{k},\omega^{\prime}}\text{tr}[\tau_{i}G_{s}(\mathbf{k},\omega^{\prime})\tau_{1}G_{s}(\mathbf{k}^{\prime},\omega^{\prime\prime})],\\\
&=\sum_{s}x_{i}(s)g_{i}^{o}g_{1}^{o}\frac{1}{\beta}\sum_{\mathbf{k},\omega^{\prime}}\text{tr}[C_{2x}\tau_{i}C_{2x}^{-1}C_{2x}G_{s}(\mathbf{k},\omega^{\prime})C_{2x}^{-1}C_{2x}\tau_{1}C_{2x}^{-1}C_{2x}G_{s}(\mathbf{k}^{\prime},\omega^{\prime\prime})C_{2x}^{-1}],\\\
&=-\sum_{s}x_{i}(s)g_{i}^{o}g_{1}^{o}\frac{1}{\beta}\sum_{\mathbf{k},\omega^{\prime}}\text{tr}[C_{2x}\tau_{i}C_{2x}^{-1}G_{-s}(k_{x},-k_{y},\omega^{\prime})\tau_{1}G_{-s}(k_{x}^{\prime},-k_{y}^{\prime},\omega^{\prime\prime})],\\\
&=-\Pi_{i1}^{p}(q_{x},-q_{y},\omega),\end{split}$ (139)

where we used $x_{i}(s)C_{2x}\tau_{i}C_{2x}^{-1}=x_{i}(-s)\tau_{i}$ with
$i=0,2,3$. Then we arrive at

$\displaystyle\Sigma_{AB}(q_{x},q_{y},\omega)=-\Sigma_{AB}(q_{x},-q_{y},\omega).$
(140)

Based on the above two symmetries, we find $\Sigma_{AB}({\bf
q},\omega=0)\approx iq_{y}(\partial_{q_{y}}\text{Im}[\Sigma_{AB}])_{{\bf
q}=0}$ as shown in the main text.

###  D.4 Berry curvature contribution to the off-diagonal self-energy in the
$\omega=0$ limit

In this section, we prove that the off-diagonal self-energy
$\Sigma_{AB}(q_{y},\omega)$ is proportional to TKNN invariant in the $q_{y}\to
0$ and $\omega=0$ limit. As discussed in the last subsection,
$\Sigma_{AB}(q_{y},\omega=0)$ is linearly proportional to $q_{y}$ due to the
TR symmetry and $C_{2x}$ rotational symmetry. Next, we demonstrate that the
coefficient of the linear term is given by

$\displaystyle
a_{0}\frac{\partial}{\partial_{q_{y}}}\text{Im}[\Sigma_{AB}(q_{y},\omega=0)]|_{q_{y}=0}=N_{3},$
(141)

where
$a_{0}=\frac{2Mv_{0}\sqrt{\omega_{A}\omega_{B}}}{g_{0}g_{1}\mathcal{A}}$, and
the right hand side is given by

$\displaystyle
N_{3}=\frac{\epsilon^{\alpha\beta\gamma}}{6}\text{tr}\int_{-\infty}^{\infty}d\omega^{\prime}\int\frac{d^{2}\mathbf{k}}{(2\pi)^{2}}G^{-1}\partial_{k_{\alpha}}GG^{-1}\partial_{k_{\beta}}GG^{-1}\partial_{k_{\gamma}}G$
(142)

is the TKNN invariant and
$\alpha,\beta,\gamma\in\\{i\omega^{\prime},k_{x},k_{y}\\}$. Without loss of
generality, we set $\hbar=1$. To prove the above equation, let us begin with
$a_{0}\Sigma_{AB}(q_{y})$ with a symmetric form

$\displaystyle\frac{1}{\beta\mathcal{A}}\sum_{\mathbf{k}}\sum_{\omega_{n}}\text{Tr}[\tau_{0}G(\mathbf{k}-\frac{\bf
q}{2},\omega_{n})\tau_{x}G(\mathbf{k}+\frac{\bf q}{2},\omega_{n}+q_{m})].$
(143)

After taking the partial differential with the $q_{y}\to 0$ limit, we find
that

$\displaystyle\partial_{q_{y}}G(\mathbf{k}\pm\mathbf{\frac{q}{2}},\omega_{n})|_{q_{y}=0}=\pm\frac{1}{2}\partial_{k_{y}}G(\mathbf{k},\omega_{n}).$
(144)

Therefore, the left hand side of Eq. (141) becomes

$\displaystyle\frac{1}{4}\text{tr}\int_{-\infty}^{\infty}d\omega^{\prime}\int\frac{d^{2}k}{(2\pi)^{2}}\left(-\partial_{k_{y}}G\tau_{x}G+G\tau_{x}\partial_{k_{y}}G\right).$
(145)

For a single Dirac Hamiltonian of the type $h({\bf
k})=v_{0}(k_{x}\tau_{x}+k_{y}\tau_{y})+m\tau_{z}$, with velocity $v_{0}$, we
have

$\displaystyle\frac{1}{v_{0}}\partial_{k_{x}}G^{-1}=-\tau_{x},\;\partial_{i\omega}G^{-1}=\tau_{0}.$
(146)

Substituting Eq. (146) into Eq. (145), we find

$\displaystyle
a_{0}\frac{\partial}{\partial_{q_{y}}}\text{Im}[\Sigma_{AB}(q_{y},\omega=0)]|_{q_{y}=0}=\frac{1}{2}\text{tr}\int_{-\infty}^{\infty}d\omega^{\prime}\int\frac{d^{2}k}{(2\pi)^{2}}\Big{(}-(\partial_{k_{y}}G)(\partial_{k_{x}}G^{-1})G(\partial_{i\omega^{\prime}}G^{-1})+G(\partial_{k_{x}}G^{-1})(\partial_{k_{y}}G)(\partial_{i\omega^{\prime}}G^{-1})\Big{)}.$
(147)

By using the identity

$\displaystyle G\partial_{\alpha}G^{-1}=-(\partial_{\alpha}G)G^{-1}$ (148)

with $\alpha=\\{i\omega^{\prime},k_{x},k_{y}\\}$, we can simplify the Eq.
(147) as

$\displaystyle\frac{1}{2}\text{tr}\int_{-\infty}^{\infty}d\omega^{\prime}\int\frac{d^{2}k}{(2\pi)^{2}}\Big{(}(G^{-1}\partial_{k_{x}}G)(G^{-1}\partial_{k_{y}}G)(G^{-1}\partial_{i\omega^{\prime}}G)-(G^{-1}\partial_{k_{y}}G)(G^{-1}\partial_{k_{x}}G)(G^{-1}\partial_{i\omega^{\prime}}G)\Big{)}=N_{3}.$
(149)

where the cyclic permutation for the trace is used. This concludes the proof
of Eq. (141).

##  Appendix E The additional self-energy correction due to the $g_{2}^{o}$
e-ph coupling vertex

Up until now, we have studied the contribution to the phonon self-energy
originating from $g_{0}^{o}$, $g_{1}^{o}$ and $g_{3}^{o}$; the corresponding
electron-phonon interactions behave as pseudo-gauge fields for electrons. In
doing so, we have ignored the effect of the $g_{2}^{o}$ term, which
renormalizes the Dirac mass of the electrons. One may ask how this term will
influence our main results concerning the electronic Berry curvature
contribution to the phonon helicity. To address this question, we write the
correction from the $g_{2}^{o}$ e-ph coupling to the off-diagonal phonon self-
energy as

$\displaystyle\Delta\Sigma_{AB}=\Delta\Sigma^{p}_{AB}+\Delta\Sigma^{e}_{AB},$
(150)

where
$\Delta\Sigma^{p}_{AB}=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\Pi_{21}^{p}$,
$\Delta\Sigma^{e}_{AB}=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{20}^{p}\times
V_{q}^{RPA}\times\Pi_{01}^{p}$ and

$\displaystyle\Pi_{01}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=g_{0}^{o}g_{1}^{o}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[-i\alpha\beta\frac{v_{0}q_{y}m_{0}}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right],$
(151) $\displaystyle\Pi_{21}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=g_{2}^{o}g_{1}^{o}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[i\alpha\frac{v_{0}k_{y}}{E_{\mathbf{k}}}-i\beta\frac{v_{0}(k_{y}+q_{y})}{E_{\mathbf{k}+\mathbf{q}}}\right],$
(152) $\displaystyle\Pi_{20}^{p}(\mathbf{q},iq_{m})$
$\displaystyle=g_{2}^{o}g_{0}^{o}\sum_{\mathbf{k}}\sum_{\alpha\beta}\frac{n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta E_{\mathbf{k}+\mathbf{q}})}{-iq_{m}+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}\left[\alpha\frac{m}{E_{\mathbf{k}}}+\beta\frac{m}{E_{\mathbf{k}+\mathbf{q}}}\right].$
(153)

###  E.1 The self-energy correction $\Delta\Sigma^{p}_{AB}$ in the absence of
electron-electron interactions

In this subsection, let us first study the self-energy correction
$\Delta\Sigma^{p}_{AB}=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\Pi_{21}^{p}$.
Because we focus on the insulating phase at zero temperature, we can set
$\mu=0$. Accordingly,

$\displaystyle
n_{F}(-E_{\mathbf{k}})=n_{F}(-E_{\mathbf{k}+\mathbf{q}})=1,\text{ and
}n_{F}(E_{\mathbf{k}})=n_{F}(E_{\mathbf{k}+\mathbf{q}})=0,$ (154)

which leads to

$\displaystyle\Pi_{21}^{p}(\mathbf{q},\omega)\approx g_{2}^{o}g_{1}^{o}\times
i\omega\sum_{\mathbf{k}}\frac{2}{(E_{\mathbf{k}}+E_{\mathbf{k}+\mathbf{q}})^{2}}\left[\frac{v_{0}k_{y}}{E_{\mathbf{k}}}+\frac{v_{0}(k_{y}+q_{y})}{E_{\mathbf{k}+\mathbf{q}}}\right].$
(155)

Here, we have used $iq_{m}\to\omega+i0^{+}$ and we have kept only the leading
order term in $\omega$. We check that $\Pi_{21}^{p}$ satisfies the symmetry
requirements with respect to time-reversal and the two-fold rotation $C_{2x}$,

$\displaystyle\Pi_{21}^{p}(q_{x},q_{y}=0,\omega)=0,\quad\Pi_{21}^{p}(q_{x}=0,q_{y},\omega)=-\Pi_{21}^{p}(q_{x}=0,-q_{y},\omega),\quad\Pi_{21}^{p}(q_{x},q_{y},\omega)=\left(\Pi_{21}^{p}(-q_{x},-q_{y},\omega)\right)^{\ast}.$
(156)

To compare with the $\Pi_{01}^{p}\propto iq_{y}$ term, we evaluate
$\Pi_{21}^{p}$ to leading order in $q_{y}$ and we focus on the results along
the $q_{y}$-axis, i.e., $q_{x}=0$. Performing a Taylor expansion on $q_{y}$
and keeping terms up to linear order, we arrive at

$\displaystyle\begin{split}\Pi_{21}^{p}(q_{x}=0,q_{y},\omega)&\approx
g_{2}^{o}g_{1}^{o}\times
i\omega\sum_{\mathbf{k}}\frac{1}{2E_{\mathbf{k}}^{2}}\left(1-\frac{v_{0}^{2}k_{y}q_{y}}{E_{\mathbf{k}}^{2}}\right)\left[\frac{v_{0}k_{y}}{E_{\mathbf{k}}}+\frac{v_{0}q_{y}}{E_{\mathbf{k}}}+\frac{v_{0}k_{y}}{E_{\mathbf{k}}}\left(1-\frac{v_{0}^{2}k_{y}q_{y}}{E_{\mathbf{k}}^{2}}\right)\right]\\\
&\approx g_{2}^{o}g_{1}^{o}\times
i\omega\sum_{\mathbf{k}}\left[\frac{v_{0}q_{y}}{2E_{\mathbf{k}}^{3}}+\frac{v_{0}k_{y}}{2E_{\mathbf{k}}^{3}}\times\left(2-3\frac{v_{0}^{2}k_{y}q_{y}}{E_{\mathbf{k}}^{2}}\right)\right]=g_{2}^{o}g_{1}^{o}\times
i\omega\sum_{\mathbf{k}}\left[\frac{v_{0}q_{y}}{2E_{\mathbf{k}}^{3}}-\frac{v_{0}k_{y}}{2E_{\mathbf{k}}^{3}}\frac{3v_{0}^{2}k_{y}q_{y}}{E_{\mathbf{k}}^{2}}\right]\\\
&=g_{2}^{o}g_{1}^{o}\times i\omega
v_{0}q_{y}\sum_{\mathbf{k}}\left[\frac{1}{2E_{\mathbf{k}}^{3}}-\frac{3v_{0}^{2}k_{y}^{2}}{2E_{\mathbf{k}}^{5}}\right]=g_{2}^{o}g_{1}^{o}\times
i\omega
v_{0}q_{y}\left(\frac{1}{2|m|v_{0}^{2}}-\frac{1}{2}\frac{3v_{0}^{2}}{2}\frac{2}{3|m|v_{0}^{4}}\right)=0.\end{split}$
(157)

This indicates that the leading order term must be $q_{y}^{3}$, namely,

$\displaystyle\Pi_{21}^{p}(q_{x}=0,q_{y},\omega)\propto
g_{2}^{o}g_{1}^{o}\times(i\omega q_{y}^{3}).$ (158)

In sum, in the long-wavelength regime, the contribution of the $g_{2}$-term to
the off-diagonal phonon self-energy is small compared to that from
$\Pi_{01}^{p}$.

###  E.2 Contribution of electron-electron interactions to the self-energy
correction $\Delta\Sigma^{e}_{AB}$

In this subsection, we study the self-energy correction
$\Delta\Sigma^{e}_{AB}=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\frac{1}{(g_{0}^{o})^{2}\mathcal{V}}\Pi_{20}^{p}\times
V_{q}^{RPA}\times\Pi_{01}^{p}$, where $\Pi_{01}^{p}\sim iq_{y}$. To do so, we
only need to evaluate $\Pi_{20}^{p}$. Setting $q_{x}=0$, a straightforward
calculation gives

$\displaystyle\Pi_{20}^{p}(q_{x}=0,q_{y},\omega)$ $\displaystyle\approx
g_{2}^{o}g_{0}^{o}\times
m\omega\sum_{\mathbf{k}}\frac{2}{(E_{\mathbf{k}}+E_{\mathbf{k}+\mathbf{q}})^{2}}\left[\frac{1}{E_{\mathbf{k}}}-\frac{1}{E_{\mathbf{k}+\mathbf{q}}}\right],$
(159)

By inspection, we notice that $\Pi_{20}^{p}(q_{x}=0,q_{y},\omega)$ is at least
of order $q_{y}^{2}$. Performing a Taylor expansion to order $q_{y}^{2}$, we
arrive at

$\displaystyle\begin{split}\Pi_{20}^{p}(q_{x}=0,q_{y},\omega)&\approx
g_{2}^{o}g_{0}^{o}\times
m\omega\sum_{\mathbf{k}}\frac{1}{2E_{\mathbf{k}}^{3}}\left[1-\frac{v_{0}^{2}k_{y}q_{y}}{E_{\mathbf{k}}^{2}}\right]\left[\frac{v_{0}^{2}k_{y}q_{y}}{E_{\mathbf{k}}^{2}}+\frac{v_{0}^{2}q_{y}^{2}}{2{E_{\mathbf{k}}^{2}}}\left(1-\frac{3v_{0}^{2}k_{y}^{2}}{E_{\mathbf{k}}^{2}}\right)\right],\\\
&\approx g_{2}^{o}g_{0}^{o}\times
m\omega\sum_{\mathbf{k}}\frac{1}{2E_{\mathbf{k}}^{3}}\left[\frac{v_{0}^{2}k_{y}q_{y}}{E_{\mathbf{k}}^{2}}+\frac{v_{0}^{2}q_{y}^{2}}{2E_{\mathbf{k}}^{2}}\left(1-\frac{5v_{0}^{2}k_{y}^{2}}{E_{\mathbf{k}}^{2}}\right)\right],\\\
&\approx g_{2}^{o}g_{0}^{o}\times
m\omega\left(\frac{1}{4}v_{0}^{2}q_{y}^{2}\right)\times\left(0+\frac{1}{3|m|^{3}v_{0}^{2}}-5v_{0}^{2}\frac{1}{2}\frac{2}{15|m|^{3}v_{0}^{4}}\right)=0.\end{split}$
(160)

This indicates that the leading order must be $q_{y}^{4}$, namely,

$\displaystyle\Pi_{20}^{p}(q_{x}=0,q_{y},\omega)\sim g_{2}^{o}g_{0}^{o}\times
m\omega q_{y}^{4}.$ (161)

Consequently,

$\displaystyle\Delta\Sigma^{e}_{AB}=\left(\frac{1}{g_{0}^{o}}\right)^{2}\Pi_{20}^{p}\times
V_{q}^{RPA}\times\Pi_{01}^{p}\sim g_{2}^{o}g_{1}^{o}\times(i\omega
q_{y}^{3}).$ (162)

In sum, the full contribution of the $g_{2}^{o}$ e-ph coupling to the off-
diagonal phonon self-energy scales as

$\displaystyle\Delta\Sigma_{AB}=\Delta\Sigma^{p}_{AB}+\Delta\Sigma^{e}_{AB}\sim\Delta\Sigma^{p}_{AB}\sim
g_{2}^{o}g_{1}^{o}\times(i\omega q_{y}^{3}).$ (163)

For long-wavelength phonons, this correction can be neglected compared to the
$\Sigma_{AB}\propto iq_{y}$ contribution obtained from $g_{0}^{o}$ and
$g_{1}^{o}$.

##  Appendix F Field-theoretic derivation for the off-diagonal coupling
between phonons

In this section, we show an alternative derivation for phonon hybridization
from the field theory. Here we use the real time, which is related to the
imaginary-time formalism in other sections by a Wick rotation.

We begin by integrating out the 2D Dirac fermions [comtet2000aspects] in Eq.
(94). This gives the following Chern-Simons term in the effective action:

$\displaystyle\mathcal{S}_{eff}=\sum_{s}\frac{\sigma_{s}}{2}\int
dt\,d^{2}\mathbf{r}\,\epsilon^{\mu\nu\rho}\widetilde{A}_{s,\mu}\partial_{\nu}\widetilde{A}_{s,\rho},$
(164)

where the Einstein summation convention is adopted for the Greek letters
($\mu,\nu,\rho$), $s$ is the valley index,

$\sigma_{s}=\frac{s}{4\pi}\text{sgn}(m_{0})$ (165)

is the Hall conductance, and we have assumed zero temperature with the
chemical potential inside the mass gap ($\hbar=e=1$ units are used). So we
have $[\mathcal{S}_{eff}]=[1]$ due to
$[t]=[meV^{-1}],[r]=[meV^{-1}],[\widetilde{A}_{s,\mu}]=[meV]$. The effective
action can be split into three parts

$S_{eff}=S_{eff}^{A}+S_{eff}^{A-pse}+S_{eff}^{pse}\ ,$ (166)

where $S_{eff}^{A}$ is the part solely for the $U(1)$ gauge field $A$,
$S_{eff}^{A-pse}$ labels the interaction between the $U(1)$ gauge field and
the pseudo-gauge field, and $S_{eff}^{pse}$ is solely for the pseudo-gauge
field.

Next, we apply the above equation to BaMnSb2. For $S_{eff}^{pse}$, we can use
the form of the pseudo-gauge field Eq. (95) to derive the coupling between
$A_{1}$ and $B_{1}$ optical phonons, resulting in

$S_{eff}^{pse}=\mathcal{A}T\sum_{q}\frac{\text{sgn}(m_{0})}{4\pi}\frac{g_{0}g_{1}}{v_{0}}q_{y}\boldsymbol{Q}^{T}_{q}\left(\begin{matrix}0&\mathrm{i}\\\
-\mathrm{i}&0\end{matrix}\right)\boldsymbol{Q}_{-q}+\text{diagonal part}\ ,$
(167)

where $\boldsymbol{Q}_{q}=(Q_{A_{1},q},Q_{B_{1},q})^{T}$ with
$q=(\mathrm{i}\omega,\boldsymbol{q})$, and $T$ is redefined as the time
interval in this section. Check the unit:
$[\mathcal{A}]=[meV^{-2}],[T]=[meV^{-1}],[g_{0,1}]=[meV^{2}],[v_{0}]=[1],[q_{y}]=[meV],[\boldsymbol{Q}_{q}]=[meV^{-1}]$,
so that $[S_{eff}^{pse}]=[1]$. The diagonal parts will renormalize the phonon
frequency as discussed in Sec. (IV.B). Following the standard techniques in
Ref. [79], we take functional derivations of the partition function to
calculate the total phonon Green’s function. Here we are using the real-time
Green’s function, which is done by converting Eq. (100) and Eq. (103) to the
real time. Then, the off-diagonal term plays the role of self-energy
corrections, which is given by

$\Sigma_{AB}=\mathcal{A}\frac{1}{2\pi}\frac{g_{0}g_{1}}{v_{0}}\frac{m_{0}}{|m_{0}|}q_{y}\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\
,$ (168)

which coincides with the results in Sec. (IV.D) and Eq. (6) in the main text,
as $N_{3}^{+}-N_{3}^{-}=\frac{1}{2\pi}m_{0}/|m_{0}|$. Nevertheless, the
expression in Eq. (168) does not include the screening of electron-phonon
couplings due to electron-electron interactions; our objective here is to
provide a simple and intuitive reason for why the electronic Chern number
alters phonon properties and leads to phonon helicity. Screening effects of
Sec. IV can be recovered in the present formalism by keeping higher order
terms in electromagnetic gauge fields in the perturbative expansion of the
phonon effective action. In the Sec. V above, we have shown the effect of
electron-phonon interaction by analyzing all the self-energy corrections due
to e-e and e-ph interactions.

On the other hand, $S_{eff}^{A-pse}$ reads

$S_{eff}^{A-pse}=\frac{\text{sgn}(m_{0})}{2\pi}\int
d^{3}x(\frac{g_{1}}{v_{0}}Q_{B_{1}}E_{y}-\frac{g_{3}}{v_{0}}Q_{A_{1}}E_{x})=Q\sum_{q}\boldsymbol{E}_{q}^{T}\boldsymbol{Q}_{-q}+...\
,$ (169)

where $Q=\mathcal{V}T\frac{\text{sgn}(m_{0})(g_{1}-g_{3})}{4\pi v_{0}}$, and
“…” corresponds to the anisotropic part. In this work, we neglect the
anisotropic part, and then the full effective action for optical phonons,
including $\mathcal{S}_{eff}^{A-pse}$ and $\mathcal{S}_{eff}^{pse}$, reads

$\displaystyle\mathcal{S}_{ph}$
$\displaystyle=\frac{1}{2}\sum_{\omega,\mathbf{q}}\mathbf{Q}_{q}^{T}\mathcal{M}(\mathbf{q},\omega)\mathbf{Q}_{-q}+\sum_{\omega,\mathbf{q}}Q\mathbf{E}_{q}^{T}\boldsymbol{Q}_{-q},$
(170) $\displaystyle\mathcal{M}(\mathbf{q},\omega)$
$\displaystyle=\left[\begin{array}[]{cc}\omega^{2}-\omega_{A}^{2}&\sqrt{\omega_{A}\omega_{B}}\Sigma_{AB}\\\
-\sqrt{\omega_{A}\omega_{B}}\Sigma_{AB}&\omega^{2}-\omega_{B}^{2}\end{array}\right],$
(173)

where $\omega_{A}^{2}$ and $\omega_{B}^{2}$ include the diagonal phonon self-
energies’ corrections, and $\Sigma_{BA}({\bf q},\omega)=\Sigma_{AB}(-{\bf
q},-\omega)$ is used.

##  Appendix G Phonon helicity for acoustic phonons

The general electron-acoustic phonon coupling vertex is given by

$\displaystyle\begin{split}\mathbf{g}^{a,A_{1},s}&=iq_{x}(g_{0}^{a}\tau_{0}+sg_{2}^{a}\tau_{2}+g_{3}^{a}\tau_{3})+iq_{y}(g_{1}^{a}\tau_{1}),\\\
\mathbf{g}^{a,B_{1},s}&=iq_{y}(g_{0}^{a}\tau_{0}+sg_{2}^{a}\tau_{2}+g_{3}^{a}\tau_{3})+iq_{x}(g_{1}^{a}\tau_{1}).\end{split}$
(174)

The off-diagonal self-energy is similar to that discussed for optical phonons
and is given by

$\displaystyle\Sigma_{AB}(\mathbf{q},iq_{m})=\frac{1}{2M\sqrt{\omega_{A}^{a}\omega_{B}^{a}}}[\Pi_{01}^{p}(\mathbf{q},iq_{m})],$
(175)

since we only focus on the $q_{y}$ and $q_{x}$ directions in the discussion of
phonon dynamics. Different from the optical phonon with a constant frequency
$\omega_{A}$ and $\omega_{B}$ for the limit ${\bf q}\rightarrow 0$, the
acoustic phonons possess a linear dispersion, given by
$\omega_{\lambda}^{a}=v_{\lambda}|q|$ for the $\lambda$-mode in the limit
${\bf q}\rightarrow 0$.

Figure 6:  (a) and (b) show the spectrum and circular polarization of optical
phonons with $m_{0}=25$ meV. The phonon dispersions are labeled by black
dashed lines, while the color represents the polarization. Parameters:
$v_{0}=100$ meV$\cdot$nm, $\mu=0$, $v_{A_{1}}=15$ meV$\cdot$nm, $v_{B_{1}}=20$
meV$\cdot$nm, $g_{0}^{a}=80\sqrt{15}$ meV, $g_{1}^{a}=80\sqrt{20}$ meV. Here
$q_{0}=m_{0}/v_{0}$ and
$\Sigma_{0}=g_{0}g_{1}\mathcal{A}N_{3}^{+}/(v_{0}M\sqrt{\omega_{A}\omega_{B}})$.

All the off diagonal self-energy results obtained from the electron-optical
phonons coupling can be directly applied to the electron-acoustic phonon
coupling. The only difference is that $\Sigma_{AB}$ has $q_{y}^{3}$ dependence
for acoustic phonons, instead of a linear $q_{y}$ dependence for optical
phonons. The numerical results are shown in Fig. 6, where the black lines
depict the phonon dispersion including the correction from the off-diagonal
self-energy. The colors in Fig. 6 show non-vanishing circular phonon
polarization or phonon angular momentum at a finite $q_{y}$. Similar to the
optical phonons, we find helical structure of the phonon angular momentum as a
function of the momentum ${\bf q}$ for acoustic phonons.

##  Appendix H The dynamical Kohn anomaly

In this section, we analytically analyze the singularity and kinks for the
off-diagonal phonon self-energy
$\Sigma_{AB}(\mathbf{q},\omega)\sim\Pi_{01}^{p}(\mathbf{q},\omega)$ at finite
$\omega$ and ${\bf q}$. Using the identity

$\displaystyle\frac{1}{x\pm i\eta}=\mathcal{P}\frac{1}{x}\mp i\pi\delta(x),$
(176)

where $\mathcal{P}$ means the principal value, we obtain

$\displaystyle\frac{1}{-\omega-i\eta+\beta E_{\mathbf{k}+\mathbf{q}}-\alpha
E_{\mathbf{k}}}=\mathcal{P}\frac{1}{-\omega+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha E_{\mathbf{k}}}+i\pi\delta(-\omega+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha E_{\mathbf{k}}).$ (177)

Thus, for the off-diagonal self-energy
$\Sigma_{AB}(\mathbf{q},\omega)=\frac{1}{2M\sqrt{\omega_{A}\omega_{B}}}\Pi_{01}^{p}(\mathbf{q},\omega)$,
and the real component of $\Pi_{01}^{p}(\mathbf{q},\omega)$ is given by

$\displaystyle\text{Re}[\Pi_{01}^{p}(\mathbf{q},\omega)]=g_{0}g_{1}\sum_{k}\sum_{\alpha\beta}\left(n_{F}(\mu-\alpha
E_{\mathbf{k}})-n_{F}(\mu-\beta
E_{\mathbf{k}+\mathbf{q}})\right)\left[\alpha\beta\frac{\pi
v_{0}q_{y}m_{0}}{E_{\mathbf{k}}E_{\mathbf{k}+\mathbf{q}}}\right]\delta(-\omega+\beta
E_{\mathbf{k}+\mathbf{q}}-\alpha E_{\mathbf{k}}).$ (178)

For simplicity, we concentrate on the insulating regime
($0<=\mu<|m_{0}|<\omega/2$) at zero temperature. With
$n_{F}(\mu+E_{\mathbf{k}})=n_{F}(\mu+E_{\mathbf{k}^{\prime}})=0$ and
$n_{F}(\mu-E_{\mathbf{k}})=n_{F}(\mu-E_{\mathbf{k}^{\prime}})=1$
($\mathbf{k}^{\prime}=\mathbf{k}+\mathbf{q}$), we obtain

$\displaystyle\frac{\text{Re}[\Pi_{01}^{p}(\mathbf{q},\omega)]}{g_{0}g_{1}\pi
v_{0}m_{0}q_{y}}$
$\displaystyle=\sum_{\mathbf{k}}\frac{1}{E_{\mathbf{k}}E_{\mathbf{k}^{\prime}}}\left(\delta(-\omega-
E_{\mathbf{k}^{\prime}}-E_{\mathbf{k}})-\delta(-\omega+E_{\mathbf{k}^{\prime}}+E_{\mathbf{k}})\right)=-\sum_{\mathbf{k}}\frac{\delta(-\omega+E_{\mathbf{k}^{\prime}}+E_{\mathbf{k}})}{E_{\mathbf{k}}E_{\mathbf{k}^{\prime}}}.$
(179)

In polar coordinates, we have

$\displaystyle\sum_{k}=\mathcal{A}\int\frac{d^{2}k}{(2\pi)^{2}}=\mathcal{A}\int_{0}^{\infty}kdk\int_{0}^{2\pi}d\theta,$
(180)

where $\mathcal{A}$ is the area of the crystal in the $xy$ plane. Next, we
choose ${\bf q}=(0,q)$ with $q\geq 0$, so that $\theta$-dependence only
appears for the terms involving
$k^{\prime}=\sqrt{k^{2}+q^{2}+2kq\cos(\theta-\pi/2)}=\sqrt{k^{2}+q^{2}+2kq\sin\theta}$.
Then, making a change of variables, we have

$\displaystyle\begin{split}&\int_{0}^{2\pi}d\theta\to
2\int_{-\pi/2}^{\pi/2}d\theta\to\int_{|k-q|}^{k+q}dk^{\prime}\frac{4k^{\prime}}{\sqrt{(2kq)^{2}-((k^{\prime})^{2}-k^{2}-q^{2})^{2}}}.\end{split}$
(181)

After this replacement, we obtain

$\displaystyle\frac{\text{Re}[\Pi_{01}^{p}(\mathbf{q},\omega)]}{g_{0}g_{1}\pi
v_{0}m_{0}q_{y}}=-\mathcal{A}\int_{0}^{\infty}dk\int_{|k-q|}^{k+q}dk^{\prime}g(k,k^{\prime})\delta(-\omega+E_{\mathbf{k}^{\prime}}+E_{\mathbf{k}}),$
(182)

where
$g(k,k^{\prime})=\frac{4kk^{\prime}}{\sqrt{(2kq)^{2}-((k^{\prime})^{2}-k^{2}-q^{2})^{2}}}$
is a continuous function and $E_{k}=\sqrt{k^{2}+m_{0}^{2}}$ with $v_{0}\equiv
1$. With some algebra, this expression can be further simplified as

$\displaystyle\frac{\text{Re}[\Pi_{01}^{p}(\mathbf{q},\omega)]}{g_{0}g_{1}\pi
v_{0}m_{0}q_{y}}=-\mathcal{A}\int_{0}^{+\infty}dk\,g(k,k_{0})\theta(k_{0}-|k-q|)\theta(k+q-k_{0}),$
(183)

where $\theta(...)$ is the step function,
$k_{0}=\sqrt{(\omega-\sqrt{k^{2}+m_{0}^{2}})^{2}-m_{0}^{2}}$. The integrand
contains two step functions, which confine the range of the integration. Fix
all these parameters here except $\omega$, we try to find if the range of
integration changes by tunning $\omega$, which indicates a singularity/kink
for this integration.

  * (1.)

For the step function $\theta(k+q-k_{0})$, the $k+q-k_{0}$ increase
monotonically by $k$. And $\left(k+q-k_{0}\right)_{k=0}=q-k_{0}$ can changes
sign, once it happens, the range of the integration will change.

  * (2.)

For the step function $\theta(k_{0}-|k-q|)$, $k_{0}-|k-q|$ is a monotonically
increasing function if $k<q$ and a monotonically decreasing function for
$k>q$. Also, $\left(k_{0}-|k-q|\right)_{k=0}=k_{0}-q$ could change sign.

Therefore, we find that $(k+q-k_{0})_{k=0}=0$ gives rise to the
singularity/kink, therefore, the kink locates at $q_{0}=\sqrt{(\omega-
m_{0})^{2}-m_{0}^{2}}$, shown in the main text.

##  Appendix I Spatial dispersion of the dielectric function

In this section, we study the influence of the phonon helicity of optical
phonons on the dielectric response of the material. From the action for purely
phonon modes (see Sec. (VI)), the equation of motion for the phonon dynamics
can be obtained my minimizing the effective action (170),

$\displaystyle\mathbf{Q}_{\mathbf{q}}\mathcal{M}(\mathbf{q},\omega)+Q\mathbf{E}(\mathbf{q},\omega)=0.$
(184)

The optical phonon can produce a macroscopic polarization $\mathbf{P}$, given
by $\mathbf{P}(\omega,q_{y})=Q\mathbf{\Psi}_{q_{y}}(\omega)=\chi\mathbf{E}$
with the electric susceptibility $\chi$ to be a two-by-two matrix. The complex
dielectric function matrix is then given by

$\displaystyle\epsilon=\epsilon_{0}I_{2\times 2}+\chi=\epsilon_{0}I_{2\times
2}+\frac{Q^{2}}{N(q_{y},\omega)}\left[\begin{array}[]{cc}\omega^{2}-\omega_{B}^{2}&C\\\
-C&\omega^{2}-\omega_{A}^{2}\end{array}\right],$ (187)

where
$N(q_{y},\omega)=(\omega^{2}-\omega_{A}^{2})(\omega^{2}-\omega_{B}^{2})-\omega_{A}\omega_{B}\Sigma_{AB}(q_{y},\omega)$
and $C=\sqrt{\omega_{A}\omega_{B}}\Sigma_{AB}$. Note that $C$ is purely
imaginary. Since the off-diagonal term is an odd function in $q_{y}$, one can
check the dielectric function matrix satisfies the Onsager relation due to the
time-reversal symmetry. The off-diagonal component of the dielectric constant
is of particular interest and is given by
$\epsilon_{xy}=i\sqrt{\omega_{A}\omega_{B}}\Sigma_{AB}(q_{y},\omega)/N(q_{y},\omega)\sim
i\sqrt{\omega_{A}\omega_{B}}g_{0}g_{1}m_{0}q_{y}/(2\pi
v_{0}|m_{0}|(\omega^{2}-\omega_{A}^{2})(\omega^{2}-\omega_{B}^{2}))$ up to the
lowest order in ${\bf q}$. Along $q_{x}$ axis, the off-diagonal self-energy
vanishes, indicating that the off-diagonal term of the dielectric function
also becomes zero. Therefore, one can see that the Berry curvature
contribution to the optical phonon self-energy will also enter into the
spatial dispersion (${\bf q}$ dependence) of the off-diagonal dielectric
function through the optical phonon-light coupling. Since this term will have
a resonance at the optical phonon frequency, it can be easily distinguished
from other contribution through tuning the light frequency. However, it may
not be easy to measure this effect in optical absorption/reflectivity, because
it vanishes when $q\to 0$ and the photon wave vectors corresponding to optical
phonon frequencies are small.