# Angular Momentum of Phonons and Einstein-de Haas Effect

Lifa Zhang  Department of Physics, The University of Texas at Austin, Austin,
Texas 78712, USA  Qian Niu  Department of Physics, The University of Texas at
Austin, Austin, Texas 78712, USA  International Center for Quantum Materials,
Peking University, Beijing 100871, China

(27 Jan 2014)

###### Abstract

We study angular momentum of phonons in a magnetic crystal. In the presence of
a spin-phonon interaction, we obtain a nonzero angular momentum of phonons,
which is an odd function of magnetization. At zero temperature, phonon has a
zero-point angular momentum besides a zero-point energy. With increasing
temperature, the total phonon angular momentum diminishes and approaches to
zero in the classical limit. The nonzero phonon angular momentum can have a
significant impact on the Einstein-de Haas effect. To obtain the change of
angular momentum of electrons, the change of phonon angular momentum needs to
be subtracted from the opposite change of lattice angular momentum.
Furthermore, the finding of phonon angular momentum gives a potential method
to study the spin-phonon interaction. Possible experiments on phonon angular
momentum are also discussed.

###### pacs:

63.20.-e 63.20.kk 75.70.Ak

The Einstein-de Haas effect einde ; frenkel79 , a phenomenon of mechanical
rotation induced by a magnetization change, was originally designed to prove
the existence of Ampere’s molecular currents; but subsequent experiments
stewart18  showed that the magnetic moment of an atom is dominated by spin
while contribution from orbital motion to the magnetic moment is almost
absent. The Einstein-de Haas experiment together with the Barnett experiment
barn15 ; barn35  (a change of magnetization resulting from a mechanical
rotation) has provided an effective method of measuring the gyromagnetic ratio
for various materials davi53 ; kitt49 ; reck69 . The accuracy of gyromagnetic
ratio is crucial to determining of orbital and spin contribution in total
magnetization stoh95 ; ulme04 ; lidb09 ; boeg10 ; pere11 ; niem12 ; dzya11 .

Due to conservation of total angular momentum of the whole system in the
Einstein-de Haas effect, the change of angular momentum of electrons
(including both spin and orbital parts) has taken to be equal in magnitude but
opposite in sign to the change of lattice angular momentum, which corresponds
to mechanical rotation. However, the mechanical rotation only reflects angular
momentum of the rigid-body lattice where atoms are assumed in the
corresponding equilibrium positions; while phonons, which come from atomic
vibrations around equilibrium positions, are assumed to have no macroscopic
angular momentum. Recently, a remarkable phenomenon of phonon Hall effect was
observed in a paramagnetic insulator phee1 ; phee2 , which is indeed a
surprise since phonons as neutral quasi-particles cannot directly couple to
magnetic field via Lorentz force. The following theoretical studies phet1 ;
phet2  showed that through Raman spin-phonon interaction the magnetic field
can have an effective force to distort phonon transport, and thus drive a
circulating heat flow qin11 . Therefore a natural question arises: can such
circulating phonons have nontrivial angular momentum and emergent macroscopic
effects?

In this Letter, we study angular momentum of phonons in a magnetic crystal in
a microscopic picture. It is found that the Raman spin-phonon interaction
induces a nonzero phonon angular momentum, which is an odd function of
magnetization. In addition to a zero-point energy, phonon has a zero-point
angular momentum at zero temperature. Such zero-point phonon angular momentum
is offset by that of excited phonon modes such that the total angular momentum
of phonons vanishes in the classical limit. Phonon angular momentum can not be
ignored in total angular momentum especially in magnetic materials with large
magnetization and spin-phonon interaction. Revisiting the Einstein-de Haas
effect we find that phonon angular momentum needs to be subtracted in
calculating angular momentum of electrons. With this correction, spin and
orbital angular momentum can be precisely determined. Besides the Einstein-de
Haas effect, nontrivial phonon angular momentum can be applied to the study of
spin-phonon interaction, thermal Hall effect, and other topics related to
phonons.

Angular momentum of phonons – The lattice angular momentum related to
mechanical rotation only reflects the rigid-body motion of the lattice.
However, angular momentum of phonons has never been considered. In a
microscopic picture, we can define angular momentum of phonons as

$\bm{J}^{\rm ph}=\sum_{l\alpha}\bm{u}_{l\alpha}\times\dot{\bm{u}}_{l\alpha}.$
(1)

Here $\bm{u}_{l\alpha}$ is a displacement vector of the $\alpha$-th atom in
the $l$-th unit cell, multiplied by square root of mass. Along $z$ direction,
$J^{\rm
ph}_{z}=\sum_{l\alpha}(u_{l\alpha}^{x}\dot{u}_{l\alpha}^{y}-u_{l\alpha}^{y}\dot{u}_{l\alpha}^{x})$.
One can present the displacement in the second quantization form as
$u_{l}=\sum_{k}\epsilon_{k}e^{i({\bf
R}_{l}\cdot{\bm{k}}-\omega_{k}t)}\sqrt{\frac{\hbar}{2\omega_{k}N}}\;a_{k}+{\rm
h.c.}$, with $k=(\bm{k},\sigma)$ specifying a wave vector $\bm{k}$ and a
branch $\sigma$, where $\epsilon_{k}$ is a displacement polarization vector.
Then the phonon angular momentum can be written as supp

$\displaystyle J_{z}^{\rm ph}$ $\displaystyle=$
$\displaystyle\frac{\hbar}{2}\sum_{k,k^{\prime}}\epsilon_{k}^{\dagger}M\epsilon_{k^{\prime}}\left(\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}}+\sqrt{\frac{\omega_{k^{\prime}}}{\omega_{k}}}\right)a_{k}^{\dagger}a_{k^{\prime}}\delta_{{\bm{k}},{\bm{k}}^{\prime}}e^{i(\omega_{k}-\omega_{k^{\prime}})t}$
(2)
$\displaystyle+\frac{\hbar}{2}\sum_{k}\epsilon_{k}^{\dagger}M\epsilon_{k}.$

Here $M=\left(\begin{smallmatrix}0&-i\\\ i&0\end{smallmatrix}\right)\otimes
I_{n\times n}$, and $n$ is the number of atoms in one unit cell. In
equilibrium, the angular momentum of phonons reduces to supp :

$J^{\rm
ph}_{z}=\sum_{\sigma,\bm{k}}l_{\bm{k},\sigma}^{z}\,[f(\omega_{\bm{k},\sigma})+\frac{1}{2}],\;\;l_{\bm{k},\sigma}^{z}=(\epsilon^{\dagger}_{\bm{k},\sigma}\,M\,\epsilon_{\bm{k},\sigma})\hbar,$
(3)

where $f(\omega_{k})=\frac{1}{e^{\hbar\omega_{k}/k_{B}T}-1}$ is the Bose-
Einstein distribution. In Eq.(S60), we do summation over all wave vector
points and all phonon branches ($\omega\geq 0$). Here $l_{\bm{k},\sigma}^{z}$
is the phonon angular momentum of branch $\sigma$ at wave vector $\bm{k}$,
which is real and proportional to $\hbar$. At zero temperature, the total
phonon angular momentum is $J^{\rm
ph}_{z}(T=0)=\sum_{\sigma,\bm{k}}\frac{1}{2}\,l_{\bm{k},\sigma}^{z}$, which
means that each mode of $(\bm{k},\sigma)$ has a zero-point angular momentum
$\frac{1}{2}l_{\bm{k},\sigma}^{z}=\frac{\hbar}{2}(\epsilon^{\dagger}_{\bm{k},\sigma}\,M\,\epsilon_{\bm{k},\sigma})$
besides a zero-point energy of $\hbar\omega_{\bm{k},\sigma}/2$.

For an ionic crystal lattice in a uniform external magnetic field, the
Hamiltonian reads in a compact form phet1 ; phet2 ; holz72 ; wang09 :

$H=\frac{1}{2}(p-{\tilde{A}}u)^{T}(p-{\tilde{A}}u)+\frac{1}{2}u^{T}Ku,$ (4)

where $u$ is a column vector of displacements from lattice equilibrium
positions, multiplied by square root of mass; $p$ is a conjugate momentum
vector, and $K$ is a force constant matrix. The cross term $u^{T}{\tilde{A}}p$
can be interpreted as a Raman spin-phonon interaction ray67 ; iose95 . The
superscript $T$ stands for the matrix transpose. ${\tilde{A}}$, an
antisymmetric real matrix qin12 , has a dimension of $N\,d\times N\,d$ where
$N$ is the number of total sites and $d$ is the dimension of lattice
vibrations; in a proper approximation it can be block diagonal with elements
$\Lambda_{\alpha}=\left(\begin{smallmatrix}0&\lambda_{\alpha}\\\
-\lambda_{\alpha}&0\\\ \end{smallmatrix}\right)$ with respect to the
$\alpha$-th ionic site, where we only consider two-dimensional ($x$ and $y$
directions) motion of the lattice ($d=2$). Here $\lambda_{\alpha}$ has a
dimension of frequency, and is proportional to the spin-phonon interaction and
magnetization, which is assumed to be proportional to magnetic field for a
paramagnetic material. The magnetic field is applied along $z$ direction. The
polarization vector $\epsilon$ satisfies
$\bigl{[}(-i\omega+A)^{2}+D\bigr{]}\epsilon=0,$ where $D({\bf
k})=-A^{2}+\sum_{l^{\prime}}K_{ll^{\prime}}e^{i({\bf R}_{l^{\prime}}-{\bf
R}_{l})\cdot{\bf k}}$ is the dynamic matrix and $A$ is block diagonal with the
element of $\Lambda_{\alpha}$, and has a dimension of $2n\times 2n$ where $n$
is the number of sites per unit cell.

In absence of spin-phonon interaction, the system reduces to a trivial phonon
system $H=\frac{1}{2}p^{T}p+\frac{1}{2}u^{T}Ku$. Solving the simple eigenvalue
problem as
$D(\bm{k})\epsilon_{\bm{k},\sigma}=\omega_{\bm{k},\sigma}^{2}\epsilon_{\bm{k},\sigma}$
with $D^{T}({\bf k})=D^{*}({\bf k})=D(-{\bf k})$, one can have
$\omega_{-\bm{k},\sigma}=\omega_{\bm{k},\sigma},\epsilon_{-\bm{k},\sigma}=\epsilon_{\bm{k},\sigma}^{*}$,
then we obtain $l_{-\bm{k},\sigma}^{z}=-l_{\bm{k},\sigma}^{z}$ and $J^{\rm
ph}_{z}=0$ supp . Thus for a phonon system without a spin-phonon interaction,
the total angular momentum of phonons is zero.

Figure 1:  (Color online) (a) The phonon angular momentum $J_{z}^{\rm ph}$ of
one unit cell as a function of $\lambda$ at temperature $T=0$ K for different
lattice symmetries. (b) The contour plot of the phonon angular momentum
$J_{z}^{\rm ph}$ of one unit cell as a function of $\lambda$ and temperature
$T$. (c) The phonon angular momentum $J_{z}^{\rm ph}$ of one unit cell from
different phonon bands as a function of temperature $T$ at $\lambda=$ 1 THz,
where the arrow denotes the Debye temperature of the model ($T_{D}=358$ K).
(d) The phonon angular momentum $J_{z}^{\rm ph}$ of one unit cell from
different phonon bands as a function of $\lambda$ at $T=0$ K. The phonon
angular momenta in (b)-(d) are calculated for a honeycomb lattice. All the
phonon angular momenta are in the unit of $\hbar$.

For a phonon system with a spin-phonon interaction,
$\epsilon_{-\bm{k},\sigma}=\epsilon_{\bm{k},-\sigma}^{*}\neq\epsilon_{\bm{k},\sigma}^{*}$,
and then $l_{-\bm{k},\sigma}^{z}\neq-l_{\bm{k},\sigma}^{z}$, thus one can get
a nonzero phonon angular momentum, which is shown in Fig. 1. We calculate
phonon angular momentum for lattices with the following parameters: the
longitudinal spring constant is $K_{L}=0.144\,$eV/(uÅ2) and the transverse one
is $K_{T}=K_{L}/4$; the unit cell lattice vectors are $(a,0),\,(0,a)$ for a
square lattice and $(a,0),\,(a/2,a\sqrt{3}/2)$ for other lattices with
$a=1\,$Å. We take $\lambda_{\alpha}=\lambda$ for the model calculation
comment2 . Figure 1(a) shows that honeycomb and kagome lattices have larger
phonon angular momenta than those of triangle and square lattices, which means
that lattices with more sites per unit cell can have a larger phonon angular
momentum. We can understand this trend by observing that optical bands are
more important in contributing to the phonon angular momentum than the
acoustic ones. In Fig. 1(c) and (d) we plot the phonon angular momentum
contributing from different bands in a honeycomb lattice. It is shown that the
phonon angular momentum from acoustic bands almost vanishes at low
temperatures (see Fig. 1(c)) and if $\lambda$ is not large (see Fig. 1(d)),
thus the optical bands dominate the contribution to the total phonon angular
momentum. With more sites per unit cell more optical bands present, thus
phonon angular momentum will be larger.

By using the relations
$\epsilon_{-\bm{k},\sigma}^{*}(-A)=\epsilon_{\bm{k},\sigma}(A)$,
$\omega_{-\bm{k},\sigma}(-A)=\omega_{\bm{k},\sigma}(A)$, $M^{T}=-M$, we can
obtain $J^{\rm ph}_{z}(-\lambda)=-J^{\rm ph}_{z}(\lambda)$. Since $\lambda$ is
proportional to magnetization, the total angular momentum of phonon will
change sign when magnetization changes sign. As shown in Fig. 1(a), (b) and
(d), the total angular momentum of phonons per unit cell increases as
$\lambda$ increases; but the increase rate will decrease.

Angular Momentum in the Classical Limit – At the high temperature limit, from
Eq. (S60) we have supp :

$J^{\rm
ph}_{z}(T\rightarrow\infty)=\sum_{\sigma>0,\bm{k}}[(\frac{k_{B}T}{\hbar\omega_{\bm{k},\sigma}}+\frac{\hbar\omega_{\bm{k},\sigma}}{12k_{B}T})l_{\bm{k},\sigma}^{z}].$
(5)

It seems that the phonon angular momentum would be linear with temperature at
the high temperature limit. However, the first term vanishes due to the fact
of
$\sum_{\sigma>0,\bm{k}}\frac{\epsilon^{\dagger}_{\bm{k},\sigma}\,M\,\epsilon_{\bm{k},\sigma}}{\omega_{\bm{k},\sigma}}=0$
supp . Therefore at a high temperature total phonon angular momentum is
proportional to $1/T$ and tends to zero as

$J^{\rm
ph}_{z}(T\rightarrow\infty)=\sum_{\sigma>0,\bm{k}}\frac{\hbar\omega_{\bm{k},\sigma}}{12k_{B}T}l_{\bm{k},\sigma}^{z}\rightarrow
0.$ (6)

The phonon angular momentum per unit cell changing with temperature is shown
in Fig. 1(b) and (c). Whatever a magnetic field is applied, the phonon angular
momentum per unit cell decreases with increasing temperature and tends to zero
at the high temperature limit ($T\gg T_{D}$). With increasing temperature more
modes are exited, the angular momentum of which has the direction opposite to
that of zero-point angular momentum; at the high temperature limit, the phonon
angular momentum of all the excited modes exactly cancels out the zero-point
angular momentum ($\sum
l_{\bm{k},\sigma}^{z}\,f(\omega_{\bm{k},\sigma},T\rightarrow\infty)=-\sum\frac{1}{2}\,l_{\bm{k},\sigma}^{z}$
). We can understand the absent phonon angular momentum in the classical limit
as follows. At high temperatures, classical statistical mechanics is
applicable to calculate phonon angular momentum. Summation over quantum states
becomes a phase-space integral with respect to $p$ and $u$. One can do a
change of variable to make the kinetic energy in the Hamiltonian Eq. (4) into
a usual form $p^{2}/2$, thus removing the effect of ${\tilde{A}}u$; for such a
pure harmonic system, the angular momentum of phonons is zero as discussed
above. Furthermore, the Bohr van Leeuwen theorem states that in classical
mechanics the thermal average of the magnetization is always zero bvl , which
also makes the angular momentum of phonons vanish at the classical limit.
Therefore, the phonon angular momentum is meaningful only in low-temperature
quantum systems.

Revisit the Einstein-de Haas Effect – The Einstein-de Haas effect einde
showed a mechanical rotation of a freely suspended body caused by the change
in its magnetization. In their experiment einde , Einstein and de Haas
employed a resonance method in which the magnetic field was periodic and tuned
to be the natural frequency of the rod and its suspension, which provided
measurements for the ratio between the change in magnetization and the one in
total angular momentum. Traditionally the total angular momentum is assumed as
$\bm{J}^{\rm tot}=\bm{J}^{\rm lat}+\bm{J}^{\rm spin}+\bm{J}^{\rm orb}$, thus
due to conservation of angular momentum, one obtains $\Delta\bm{J}^{\rm
lat}=-(\Delta\bm{J}^{\rm spin}+\Delta\bm{J}^{\rm orb})$ which is determined by
the mechanical rotation of the sample kitt49 . However, from a microscopic
point of view, the angular momentum of all atoms in the sample can be written
as

$\bm{J}^{\rm
atom}=\sum_{l\alpha}(\bm{R}_{l\alpha}+\bm{u}_{l\alpha})\times(\dot{\bm{R}}_{l\alpha}+\dot{\bm{u}}_{l\alpha}),$
(7)

where $\bm{R}_{l\alpha}$ is the equilibrium position of the $\alpha$-th atom
in the $l$-th unit cell, multiplied by square root of its mass. The angular
momentum of lattice is

$\bm{J}^{\rm lat}=\sum_{l\alpha}\bm{R}_{l\alpha}\times\dot{\bm{R}}_{l\alpha},$
(8)

which really reflects the mechanical rotation of rigid-body motion of the
sample. In equilibrium, the cross terms related with $\bm{u}$ or
$\dot{\bm{u}}$ are zero, then $\bm{J}^{\rm atom}=\bm{J}^{\rm lat}+\bm{J}^{\rm
ph}$. Thus the total angular momentum should be

$\bm{J}^{\rm tot}=\bm{J}^{\rm lat}+\bm{J}^{\rm ph}+\bm{J}^{\rm
spin}+\bm{J}^{\rm orb}.$ (9)

The global conservation of angular momentum does not explain how the angular
momentum is actually transferred from individual electrons or atoms to the
whole rigid body; the Raman type spin-phonon interaction can be ubiquitous and
plays an essential role. According to the discussion in the above section, we
know that in the presence of the spin-phonon interaction, the phonon band
structure is nontrivial and gives nonzero angular momentum $\bm{J}^{\rm ph}$.
Based on conservation of total angular momentum we obtain

$\Delta\bm{J}^{\rm spin}+\Delta\bm{J}^{\rm orb}=-\Delta\bm{J}^{\rm
lat}-\Delta\bm{J}^{\rm ph}.$ (10)

Therefore to obtain the change of angular momentum of electrons, one needs to
subtract the contribution of phonon from the opposite change of lattice
angular momentum. On the other hand one can measure the total magnetization
change as

$\Delta M=\Delta M^{\rm spin}+\Delta M^{\rm orb}.$ (11)

Combining Eq. (10), Eq. (11) together with the facts of $\Delta M^{\rm
orb}=\frac{e}{2m}\Delta J^{\rm orb}$ and $\Delta M^{\rm
spin}=\frac{e}{m}\Delta J^{\rm spin}$, one can easily determine $\Delta M^{\rm
spin}$ and $\Delta M^{\rm orbit}$.

The phonon can make a significant contribution to total angular momentum,
while the magnitude of phonon angular momentum depends on the value of
$\lambda$. The parameter $\lambda$ can be obtained from phonon dispersion
relation since our calculation shows that in the presence of spin-phonon
interaction degenerate phonon modes split at $\Gamma$ point with a gap of
$2\lambda$. By means of Raman scattering experiments, literatures split1 ;
split2  show that the phonon splitting ranges up to about 26 cm-1 in
paramagnetic CeF3 at $T=1.9$ K and $B=6$ T, thus $\lambda$ can be about $0.39$
THz and phonon angular momentum per unit cell is about 0.02 $\hbar$. One also
can estimate the parameter $\lambda$ from phonon Hall effect. For a
paramagnetic terbium gallium garnet Tb3Ga5O12, the parameter $\lambda$ is
estimated as $\lambda=0.1$ cm${}^{-1}\simeq 3$ GHz at $B=1$ T and $T=5.45$ K
phet1 , thus in such material phonon angular momentum per unit cell is about
$1.6\times 10^{-4}\hbar$, which is relatively small. However, one can observe
a much larger phonon angular momentum when magnetization is saturated in this
paramagnetic material since the parameter $\lambda$ is proportional to
magnetization. In the phonon Hall effect experiment phee1 ; phee2 , the
paramagnetic insulator was chosen to manifest the phonon contribution in the
thermal transport where the contribution from electron and magnon can be
neglected. However, the spin-phonon interaction is widely present in various
magnetic materials spimat1 ; spimat2 ; spimat3 ; spimat4 . Ferromagnetic
materials have very large magnetization, thus one can expect a large phonon
angular momentum. One also can observe evident phonon angular momentum in
materials with strong spin-phonon interaction by using Raman spectroscopy,
such as La2NiMnO6 iliev07 , Sr2CoO4 pand13  and cupric oxide chen95 .

Thus for materials with strong spin-phonon interaction together with large
magnetization, the zero-point angular momentum of phonons can be significant.
According to previous studies, in some ferromagnetic materials the calculated
orbital magnetic moment is only a few percent of the total magnetic moment,
that is, the orbital angular momentum is also around a few percent of $\hbar$
reck69 , thus the phonon angular momentum can not be ignored. With improvement
of experimental technique in past decades, accuracy of measurement has been
much enhanced thus phonon angular momentum should be measurable.

Possible experiment to separate phonon angular momentum – One can do
experiments on a ferromagnetic insulator with saturation magnetization, where
electron transport can be ignored. Due to the property of phonon angular
momentum – it decreases with increasing temperature and vanishes in the
classical limit, one can measure the change of lattice angular momentum at low
and high temperatures to separate the phonon angular momentum from the others.
Here the temperature scale should be the Debye temperature which divides the
quantum and classical regions. On the other hand, in order to avoid the
involvement of magnons, we need to do experiments at temperatures low compared
to the Curie temperature. This demands that the Curie temperature must be much
higher than the Debye temperature. Thus the angular momentum of magnons almost
keeps constant while that of phonons changes dramatically with changing
temperature. Fortunately, this can be satisfied by many ferromagnetic
materials where their Curie temperature are around 1000 K while their Debye
temperatures are less than 500 K kitt04 .

Besides application to measurement of gyromagnetic ratio, nontrivial phonon
angular momentum provides us a possible efficient route to study spin-phonon
interaction in magnetic materials. On the other hand, to separate the
contribution from phonons and magnons to thermal Hall effect in ferromagnetic
materials is an open problem, phonon angular momentum can give a way to obtain
the phonon contribution.

Acknowledgements – We thank Junren Shi, Yang Gao, Zhenhua Qiao, Xiao Li, Ran
Cheng for helpful discussions. We acknowledge support from DOE-DMSE (DE-
FG03-02ER45958), NBRPC (2012CB-921300), NSFC (91121004), and the Welch
Foundation (F-1255).

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##  I Supplementary information for “Angular Momentum of Phonons and the
Einstein-de Hass Effect”

###  I.1 Derivation of Eq.(2) and (3)

The angular momentum of phonons is

$\displaystyle J_{z}^{{\rm{ph}}}$ $\displaystyle=$
$\displaystyle\sum\limits_{l\alpha}{(u_{l\alpha}^{x}\dot{u}_{l\alpha}^{y}-u_{l\alpha}^{y}\dot{u}_{l\alpha}^{x})}$
(S7) $\displaystyle=$
$\displaystyle\sum\limits_{l\alpha}{\left({\begin{array}[]{*{20}c}{u_{l\alpha}^{x}}\\\
{u_{l\alpha}^{y}}\\\
\end{array}}\right)^{T}\left({\begin{array}[]{*{20}c}0&1\\\ {-1}&0\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{\dot{u}_{l\alpha}^{x}}\\\
{\dot{u}_{l\alpha}^{y}}\\\ \end{array}}\right).}$

For a unit cell with two atoms $n=2$, that is, $\alpha=1,2$, the angular
momentum of phonons can be written as

$J_{z}^{{\rm{ph}}}\;=\sum\limits_{l}{\left({\begin{array}[]{*{20}c}{u_{l1}^{x}}\\\
{u_{l1}^{y}}\\\ {u_{l2}^{x}}\\\ {u_{l2}^{y}}\\\
\end{array}}\right)^{T}\left({\begin{array}[]{*{20}c}0&1&{}\hfil&{}\hfil\\\
{-1}&0&{}\hfil&{}\hfil\\\ {}\hfil&{}\hfil&0&1\\\ {}\hfil&{}\hfil&{-1}&0\\\
\end{array}}\right)}\left({\begin{array}[]{*{20}c}{\dot{u}_{l1}^{x}}\\\
{\dot{u}_{l1}^{y}}\\\ {\dot{u}_{l2}^{x}}\\\ {\dot{u}_{l2}^{y}}\\\
\end{array}}\right).$ (S9)

Using the
$u_{l}=\left(\begin{smallmatrix}u_{l1}^{x}\;u_{l1}^{y}\;u_{l2}^{x}\;u_{l2}^{y}\end{smallmatrix}\right)^{T}$,
we obtain

$J_{z}^{{\rm{ph}}}=\sum\limits_{l}u_{l}^{T}iM{\dot{u}_{l}}.$ (S10)

where $M=\left(\begin{smallmatrix}0&-i\\\ i&0\end{smallmatrix}\right)\otimes
I_{n\times n}$. By using the second quantization for $u_{l}$ as

$u_{l}=\sum_{k}\epsilon_{k}e^{i({\bf
R}_{l}\cdot{\bm{k}}-\omega_{k}t)}\sqrt{\frac{\hbar}{2\omega_{k}N}}\;a_{k}+{\rm
h.c.},$

we obtain

$J_{z}^{{\rm{ph}}}=\frac{\hbar}{2N}\sum_{l}\sum_{k,k^{\prime}}\left(\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}}\epsilon_{k}^{\dagger}M\epsilon_{k^{\prime}}a_{k}^{\dagger}a_{k^{\prime}}+\sqrt{\frac{\omega_{k^{\prime}}}{\omega_{k}}}\epsilon_{k^{\prime}}^{T}(-M)\epsilon_{k}^{*}a_{k^{\prime}}a_{k}^{\dagger}\right)e^{i(\bm{k}^{\prime}-\bm{k}){\bm{R}_{l}}}e^{i(\omega_{k}-\omega_{k^{\prime}})t}.$
(S11)

Here we ignore the $a\,a$ and $a^{\dagger}\,a^{\dagger}$ terms since they vary
rapidly with time and have no contribution in equilibrium.

Since
$\epsilon_{k^{\prime}}^{T}(-M)\epsilon_{k}^{*}=\epsilon_{k}^{\dagger}M\epsilon_{k^{\prime}}$
and
$\frac{1}{N}\sum_{l}e^{i(\bm{k}^{\prime}-\bm{k}){\bm{R}_{l}}}=\delta_{{\bm{k}},{\bm{k}}^{\prime}}$,
then

$J_{z}^{{\rm{ph}}}=\frac{\hbar}{2}\sum_{k,k^{\prime}}\epsilon_{k}^{\dagger}M\epsilon_{k^{\prime}}\left(\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}}a_{k}^{\dagger}a_{k^{\prime}}+\sqrt{\frac{\omega_{k^{\prime}}}{\omega_{k}}}a_{k^{\prime}}a_{k}^{\dagger}\right)\delta_{{\bm{k}},{\bm{k}}^{\prime}}e^{i(\omega_{k}-\omega_{k^{\prime}})t}.$
(S12)

Due to the communication relation
$[a_{\bm{k},\sigma^{\prime}},a_{\bm{k},\sigma}^{\dagger}]=\delta_{\sigma,\sigma^{\prime}}$,
we obtain Eq.(2) in the main text, that is

$J_{z}^{{\rm{ph}}}=\frac{\hbar}{2}\sum_{k,k^{\prime}}\epsilon_{k}^{\dagger}M\epsilon_{k^{\prime}}\left(\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}}+\sqrt{\frac{\omega_{k^{\prime}}}{\omega_{k}}}\right)a_{k}^{\dagger}a_{k^{\prime}}\delta_{{\bm{k}},{\bm{k}}^{\prime}}e^{i(\omega_{k}-\omega_{k^{\prime}})t}+\frac{\hbar}{2}\sum_{k}\epsilon_{k}^{\dagger}M\epsilon_{k}.$
(S13)

In equilibrium we know $\langle
a_{\bm{k},\sigma^{\prime}}^{\dagger}a_{\bm{k},\sigma}\rangle=f(\omega_{k})\delta_{\sigma,\sigma^{\prime}}$,
then we obtain

$J_{z}^{{\rm{ph}}}=\hbar\sum_{k}\epsilon_{k}^{\dagger}M\epsilon_{k}f(\omega_{k})+\frac{\hbar}{2}\sum_{k}\epsilon_{k}^{\dagger}M\epsilon_{k},$
(S14)

which is the Eq.(3) in the main text.

###  I.2 Proof of the zero angular momentum of trivial phonon system without
spin-phonon interaction

For the trivial phonon system $H=\frac{1}{2}p^{T}p+\frac{1}{2}u^{T}Ku$, one
has an eigenvalue problem as

$D(\bm{k})\epsilon(\bm{k},\sigma)=\omega_{\bm{k},\sigma}^{2}\epsilon(\bm{k},\sigma).$
(S15)

Due to $D^{\dagger}=D$, then

$\epsilon^{\dagger}(\bm{k},\sigma)D(\bm{k})=\omega_{\bm{k},\sigma}^{2}\epsilon^{\dagger}(\bm{k},\sigma).$
(S16)

From Eq.S15, for wave vector $-\bm{k}$, one has

$D(-\bm{k})\epsilon(-\bm{k},\sigma)=\omega_{-\bm{k},\sigma}^{2}\epsilon(-\bm{k},\sigma),$

then

$\epsilon^{T}(-\bm{k},\sigma)D^{T}(-\bm{k})=\omega_{-\bm{k},\sigma}^{2}\epsilon^{T}(-\bm{k},\sigma).$

And since $D^{T}(-\bm{k})=D^{T}(\bm{k})$, then

$\epsilon^{T}(-\bm{k},\sigma)D(\bm{k})=\omega_{-\bm{k},\sigma}^{2}\epsilon^{T}(\bm{k},\sigma).$
(S17)

From Eq. (S16) and Eq. (S17), we can have

$\displaystyle\omega_{-\bm{k},\sigma}$ $\displaystyle=$
$\displaystyle\omega_{\bm{k},\sigma}$ (S18)
$\displaystyle\epsilon(-\bm{k},\sigma)$ $\displaystyle=$
$\displaystyle\epsilon^{*}(\bm{k},\sigma).$ (S19)

Therefore,

$l_{-\bm{k},\sigma}^{z}=(\epsilon^{\dagger}_{-\bm{k},\sigma}\,M\,\epsilon_{-\bm{k},\sigma})\hbar=(\epsilon^{T}_{\bm{k},\sigma}\,M\,\epsilon^{*}_{\bm{k},\sigma})\hbar=(\epsilon^{\dagger}_{\bm{k},\sigma}\,M^{T}\,\epsilon_{\bm{k},\sigma})\hbar,$
(S20)

because of $M^{T}=-M$, then

$l_{-\bm{k},\sigma}^{z}=-l_{\bm{k},\sigma}^{z}.$ (S21)

And $f(\omega_{-\bm{k},\sigma})=f(\omega_{\bm{k},\sigma})$, thus

$J^{\rm
ph}_{z}=\sum_{\sigma,\bm{k}}l_{\bm{k},\sigma}^{z}\,[f(\omega_{\bm{k},\sigma})+\frac{1}{2}]=0,$

where the summation is over all the $\bm{k}$ points in the first Brillouin
zone, and all the branches ($\omega_{\sigma}\geq 0$) are included.

###  I.3 Another Proof of Eq. (3) for systems with a spin-phonon interaction

In the presence of the Raman type spin phonon interaction, as stated in the
main text, the Hamiltonian is

$H=\frac{1}{2}(p-{\tilde{A}}u)^{T}(p-{\tilde{A}}u)+\frac{1}{2}u^{T}Ku.$ (S22)

The polarization vector $\epsilon$ satisfies

$\bigl{[}(-i\omega+A)^{2}+D\bigr{]}\epsilon=0,$ (S23)

where $D({\bf k})=-A^{2}+\sum_{l^{\prime}}K_{ll^{\prime}}e^{i({\bf
R}_{l^{\prime}}-{\bf R}_{l})\cdot{\bf k}}$ denotes the dynamic matrix and $A$
is block diagonal with elements $\Lambda_{\alpha}$. Here, we use the short-
hand notation $k=({\bf k},\sigma)$ to specify both the wavevector and the
phonon branch. $D,K_{l,l^{\prime}},$ and $A$ are all $nd\times nd$ matrices,
where $n$ is the number of particles in one unit cell and $d$ is the dimension
of the vibration.

Equation (S23) is not a standard eigenvalue problem. However, we can describe
the system by a polarization vector as $x=(\mu,\epsilon)^{T}$, where $\mu$ and
$\epsilon$ are associated with the momenta and coordinates, respectively.
Using Bloch’s theorem, one has $i\frac{\partial}{{\partial t}}x=H_{\rm
eff}x,\;H_{\rm eff}=i\left(\begin{smallmatrix}-A&-D\\\
I_{nd}&-A\end{smallmatrix}\right),$ where the $I_{nd}$ is the $nd\times nd$
identity matrix. Therefore, one obtains the eigenvalue problem of the equation
of motion $H_{\rm
eff}\,x_{k}=\omega_{k}\,x_{k},\;\;\;\;\tilde{x}_{k}^{T}\,H_{\rm
eff}=\omega_{k}\,\tilde{x}_{k}^{T},$ where $x_{k}=(\mu_{k},\epsilon_{k})^{T}$
is the right eigenvector, and the left eigenvector is chosen as
${\tilde{x}}_{k}^{T}=(\epsilon^{\dagger}_{k},-\mu^{\dagger}_{k})/(-2i\omega_{k})$,
in such choice the second quantization of the Hamiltonian is satisfied. The
orthonormal condition holds between the left and right eigenvectors, as
${\tilde{x}_{\sigma,{\bf k}}}^{T}\;x_{\sigma^{\prime},{\bf
k}}=\delta_{\sigma\sigma^{\prime}}$. We also have the completeness relation as
$\sum_{\sigma}x_{\sigma,{\bf k}}\otimes{\tilde{x}_{\sigma,{\bf
k}}}^{T}=I_{2nd}$. The normalization of the eigenmodes is equivalent to
$\epsilon_{k}^{\dagger}\,\epsilon_{k}+\frac{i}{\omega_{k}}\epsilon_{k}^{\dagger}\,A\,\epsilon_{k}=1$.

From the eigenvalue problem Eq. (S23), we know that the completed set contains
branches of negative frequency. The short-hand notation $k=({\bf k},\sigma)$
can include negative branches, $-k$ means $(-{\bf k},-\sigma)$. In order to
simplify the notation, for all the branches, we define
$a_{-k}=a_{k}^{\dagger}$. The time dependence of the operators is given by:
$a_{k}(t)=a_{k}e^{-i\omega_{k}t},a_{k}^{\dagger}(t)=a_{k}^{\dagger}e^{i\omega_{k}t}.$

Inserting the equation of motion $\dot{u}_{l}=p_{l}-Au_{l}$, that is

$\left({\begin{array}[]{*{20}c}{\dot{u}_{l\alpha}^{x}}\\\
{\dot{u}_{l\alpha}^{y}}\\\
\end{array}}\right)=\left({\begin{array}[]{*{20}c}{p_{l\alpha}^{x}}\\\
{p_{l\alpha}^{y}}\\\
\end{array}}\right)-\left({\begin{array}[]{*{20}c}0&\lambda_{l\alpha}\\\
{-\lambda_{l\alpha}}&0\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{u_{l\alpha}^{x}}\\\
{u_{l\alpha}^{y}}\\\ \end{array}}\right),$ (S24)

Eq. LABEL:eq-s1 becomes

$\displaystyle J_{z}^{{\rm{ph}}}$ $\displaystyle=$
$\displaystyle\sum\limits_{l\alpha}{\left({\begin{array}[]{*{20}c}{u_{l\alpha}^{x}}\\\
{u_{l\alpha}^{y}}\\\
\end{array}}\right)^{T}\left({\begin{array}[]{*{20}c}0&1\\\ {-1}&0\\\
\end{array}}\right)}\left({\begin{array}[]{*{20}c}{p_{l\alpha}^{x}}\\\
{p_{l\alpha}^{y}}\\\
\end{array}}\right)+\left({\begin{array}[]{*{20}c}{u_{l\alpha}^{x}}\\\
{u_{l\alpha}^{y}}\\\
\end{array}}\right)^{T}\left({\begin{array}[]{*{20}c}\lambda_{l\alpha}&0\\\
0&\lambda_{l\alpha}\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{u_{l\alpha}^{x}}\\\
{u_{l\alpha}^{y}}\\\ \end{array}}\right)$ (S37) $\displaystyle=$
$\displaystyle\sum\limits_{l\alpha}{\left({\begin{array}[]{*{20}c}{u_{l\alpha}^{x}}\\\
{u_{l\alpha}^{y}}\\\ {-p_{l\alpha}^{x}}\\\ {-p_{l\alpha}^{y}}\\\
\end{array}}\right)^{T}}\left({\begin{array}[]{*{20}c}0&1&\lambda_{l\alpha}&0\\\
{-1}&0&0&\lambda_{l\alpha}\\\ 0&0&0&0\\\ 0&0&0&0\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{p_{l\alpha}^{x}}\\\
{p_{l\alpha}^{y}}\\\ {u_{l\alpha}^{x}}\\\ {u_{l\alpha}^{y}}\\\
\end{array}}\right).$ (S50)

For a unit cell with two atoms $n=2$, the angular momentum of phonons can be
written as

$J_{z}^{{\rm{ph}}}\;=\sum\limits_{l}{\left({\begin{array}[]{*{20}c}{u_{l1}^{x}}\\\
{u_{l1}^{y}}\\\ {u_{l2}^{x}}\\\ {u_{l2}^{y}}\\\ {-p_{l1}^{x}}\\\
{-p_{l1}^{y}}\\\ {-p_{l2}^{x}}\\\ {-p_{l2}^{y}}\\\
\end{array}}\right)^{T}\left({\begin{array}[]{*{20}c}0&1&{}\hfil&{}\hfil&\lambda_{l1}&0&{}\hfil&{}\hfil\\\
{-1}&0&{}\hfil&{}\hfil&0&\lambda_{l1}&{}\hfil&{}\hfil\\\
{}\hfil&{}\hfil&0&1&{}\hfil&{}\hfil&\lambda_{l2}&0\\\
{}\hfil&{}\hfil&{-1}&0&{}\hfil&{}\hfil&0&\lambda_{l2}\\\
{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil\\\
{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil\\\
{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil\\\
{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil&{}\hfil\\\
\end{array}}\right)}\left({\begin{array}[]{*{20}c}{p_{l1}^{x}}\\\
{p_{l1}^{y}}\\\ {p_{l2}^{x}}\\\ {p_{l2}^{y}}\\\ {u_{l1}^{x}}\\\
{u_{l1}^{y}}\\\ {u_{l2}^{x}}\\\ {u_{l2}^{y}}\\\ \end{array}}\right).$ (S52)

Using the $\chi_{l}=\left(\begin{smallmatrix}p_{l}\\\
u_{l}\end{smallmatrix}\right)$ and
$\tilde{\chi}_{l}=\left(\begin{smallmatrix}u_{l}\\\
-p_{l}\end{smallmatrix}\right)$, where $u_{l},p_{l}$ are column vectors of
displacements and conjugate momenta for the $l-$th unit, if $n=2$, then
$u_{l}=\left({\begin{array}[]{*{20}c}{u_{l1}^{x}}&{u_{l1}^{y}}&{u_{l2}^{x}}&{u_{l2}^{y}}\\\
\end{array}}\right)^{T}$, similar for $p_{l}$. Then we obtain Eq. (3) in the
main text as

$J^{\rm
ph}_{z}=\sum_{l}\tilde{\chi}^{T}_{l}\left(\begin{array}[]{cc}iM&-iMA\\\
0&0\end{array}\right)\chi_{l}.$ (S53)

By the second quantization
$\chi_{l}=\sqrt{\frac{\hbar}{N}}\sum_{k}x_{k}e^{i{\bf R}_{l}\cdot{\bf
k}}\sqrt{\frac{1}{2|\omega_{k}|}}\;a_{k};\;\;\tilde{\chi}_{l}=\sqrt{\frac{\hbar}{N}}\sum_{k}\tilde{x}_{k}e^{-i{\bf
R}_{l}\cdot{\bf
k}}(-2i\omega_{k})\sqrt{\frac{1}{2|\omega_{k}|}}\;a_{k}^{\dagger}$, and
$x_{k}=\left(\begin{smallmatrix}\mu_{k}\\\
\epsilon_{k}\end{smallmatrix}\right)$, $k=({\bf k},\sigma)$,
$\tilde{x}_{k}^{T}=\frac{1}{-2i\omega_{k}}\left(\begin{smallmatrix}\epsilon^{\dagger}_{k}\;&-\mu^{\dagger}_{k}\end{smallmatrix}\right)$,
we obtain

$J_{z}^{{\rm{ph}}}\;=\sum\limits_{l,k,k^{\prime}}{e^{i{\bm{R}}_{l}(\bm{k}-\bm{k}^{\prime})}}\frac{\hbar}{{N}}\frac{1}{{2\sqrt{|\omega_{k^{\prime}}||\omega_{k}|}}}\left({\begin{array}[]{*{20}c}{\varepsilon_{k^{\prime}}^{\dagger}}&{-\mu_{k^{\prime}}^{\dagger}}\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}iM&{-iMA}\\\ 0&0\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{\mu_{k}}\\\
{\varepsilon_{k}}\\\ \end{array}}\right)a_{k^{\prime}}^{\dagger}a_{k}.$ (S54)

By using the fact of
$\sum\limits_{l}{e^{i{\bm{R}}_{l}(\bm{k}-\bm{k}^{\prime})}}=N\delta_{\bm{k},\bm{k}^{\prime}}$,
we obtain

$J_{z}^{{\rm{ph}}}\;=\sum\limits_{\bm{k},\sigma,\sigma^{\prime}}{\frac{\hbar}{{2\sqrt{|\omega_{\bm{k},\sigma}||\omega_{\bm{k},\sigma^{\prime}}|}}}\left({\begin{array}[]{*{20}c}{\varepsilon_{\bm{k},\sigma^{\prime}}^{\dagger}}&{-\mu_{\bm{k},\sigma^{\prime}}^{\dagger}}\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}iM&{-iMA}\\\ 0&0\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{\mu_{\bm{k},\sigma}}\\\
{\varepsilon_{\bm{k},\sigma}}\\\
\end{array}}\right)a_{\bm{k},\sigma^{\prime}}^{\dagger}a_{\bm{k},\sigma}}.$
(S55)

In equilibrium, we know that $\langle
a_{\bm{k},\sigma^{\prime}}^{\dagger}a_{\bm{k},\sigma}\rangle=f(\omega_{k}){\rm
sign}(\sigma)\delta_{\sigma,\sigma^{\prime}}$, then

$\langle
J_{z}^{{\rm{ph}}}\rangle\;=\sum\limits_{k}{\frac{\hbar}{{2|\omega_{k}|}}}\left({\begin{array}[]{*{20}c}{\varepsilon_{k}^{\dagger}}&{-\mu_{k}^{\dagger}}\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}iM&{-iMA}\\\ 0&0\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{\mu_{k}}\\\
{\varepsilon_{k}}\\\ \end{array}}\right)\langle a_{k}^{\dagger}a_{k}\rangle.$
(S56)

And using the relation between the momentum and displacement polarization
vectors $\mu_{k}=-i\omega_{k}\epsilon_{k}+A\epsilon_{k}$, we get

$\left({\begin{array}[]{*{20}c}{\varepsilon_{k}^{\dagger}}&{-\mu_{k}^{\dagger}}\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}iM&{-iMA}\\\ 0&0\\\
\end{array}}\right)\left({\begin{array}[]{*{20}c}{\mu_{k}}\\\
{\varepsilon_{k}}\\\
\end{array}}\right)=\omega\varepsilon_{k}^{\dagger}M\varepsilon_{k}.$ (S57)

Then the phonon angular momentum in equilibrium is

$J^{\rm
ph}_{z}=\frac{\hbar}{2}\sum_{k}\epsilon^{\dagger}_{k}\,M\,\epsilon_{k}\,f(\omega_{k}),$
(S58)

by using $\omega_{k}{\rm sign}(\sigma)=|\omega_{k}|$.

Figure S1:  (Color online) (a) The angular momentum $l_{\bm{k},\sigma}^{z}$ of
bands $\sigma=1$ \- 4 as a function of $k_{x}$. (b)-(d) The sum of the angular
momentum of four bands
$\sum_{\sigma>0}l_{\bm{k},\sigma}^{z}\,[f(\omega_{\bm{k},\sigma})+\frac{1}{2}]$
at different temperatures. The parameters are the same with those in the main
text.

We have $\epsilon_{-\bm{k},-\sigma}=\epsilon^{*}_{\bm{k},\sigma}$, then

$(\epsilon^{\dagger}_{-\bm{k},-\sigma}\,M\,\epsilon_{-\bm{k},-\sigma})\hbar=(\epsilon^{T}_{\bm{k},\sigma}\,M\,\epsilon^{*}_{\bm{k},\sigma})\hbar=(\epsilon^{\dagger}_{\bm{k},\sigma}\,M^{T}\,\epsilon_{\bm{k},\sigma})\hbar=-(\epsilon^{\dagger}_{\bm{k},\sigma}\,M\,\epsilon_{\bm{k},\sigma})\hbar,$
(S59)

but $f(\omega_{-\bm{k},-\sigma})\neq f(\omega_{\bm{k},\sigma})$, thus phonon
angular momentum at $-\bm{k},-\sigma$ cannot cancel out that at
$\bm{k},\sigma$, and we can get a nonzero phonon angular momentum. Here,
$k=({\bm{k}},\sigma)$ includes all the positive $\sigma>0$ and negative
$\sigma<0$ branches. For the negative branches, that is $\sigma<0,\omega<0$,
using $\epsilon_{-k}^{*}=\epsilon_{k};\;\omega_{-k}=-\omega_{k}$ , then

$\frac{\hbar}{2}\sum\limits_{\bm{k},\sigma<0}{\varepsilon_{\bm{k},\sigma}^{\dagger}M\varepsilon_{\bm{k},\sigma}f(\omega_{\bm{k},\sigma})}=\frac{\hbar}{2}\sum\limits_{\bm{k}^{\prime},\sigma^{\prime}>0}{\varepsilon_{-\bm{k}^{\prime},-\sigma^{\prime}}^{\dagger}M\varepsilon_{-\bm{k}^{\prime},-\sigma^{\prime}}f(\omega_{-\bm{k}^{\prime},-\sigma^{\prime}})}=\frac{\hbar}{2}\sum\limits_{\bm{k}^{\prime},\sigma^{\prime}>0}{\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}^{T}M\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}^{*}f(-\omega_{\bm{k}^{\prime},\sigma^{\prime}})}.$

And
$\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}^{T}M\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}^{*}=\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}^{\dagger}M^{T}\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}$,
$M^{T}=-M$,
$f(-\omega_{\bm{k}^{\prime},\sigma^{\prime}}=-(1+f(\omega_{\bm{k}^{\prime},\sigma^{\prime}})$,
then

$\frac{\hbar}{2}\sum\limits_{\bm{k},\sigma<0}{\varepsilon_{\bm{k},\sigma}^{\dagger}M\varepsilon_{\bm{k},\sigma}f(\omega_{\bm{k},\sigma})}=\frac{\hbar}{2}\sum\limits_{\bm{k}^{\prime},\sigma^{\prime}>0}{\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}^{\dagger}M\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}(f(\omega_{\bm{k}^{\prime},\sigma^{\prime}})}+1).$

Therefore we obtain the angular momentum of phonons as:

$J^{\rm
ph}_{z}=\sum_{\sigma>0,\bm{k}}l_{\bm{k},\sigma}^{z}\,[f(\omega_{\bm{k},\sigma})+\frac{1}{2}],\;\;l_{\bm{k},\sigma}^{z}=(\epsilon^{\dagger}_{\bm{k},\sigma}\,M\,\epsilon_{\bm{k},\sigma})\hbar.$
(S60)

Here due to
$\epsilon_{-\bm{k},\sigma}=\epsilon_{\bm{k},-\sigma}^{*}\neq\epsilon_{\bm{k},\sigma}^{*}$,
we cannot obtain $l_{-\bm{k},\sigma}^{z}=-l_{\bm{k},\sigma}^{z}$, thus the
phonon angular momentum at the wave vector $-\bm{k}$ cannot cancel out that at
$\bm{k}$, and we can obtain a nonzero angular momentum of phonons.

We show the phonon angular momentum as a function of $k_{x}$ ($k_{y}=0$) in
Fig. S1. Due to the 6-fold symmetry of the honeycomb lattice, the angular
momentum of phonons is an even function of $k_{x}$, as shown in Fig. S1. For
band 1 the angular momenta of all the wave-vector points are negative, while
they are positive for band 4; for both band 2 and 3, the phonon angular
momentum (by doing the summation over all the four bands) at $\bm{\Gamma}$
point is opposite to that at $\bm{K}$ point. The total angular momentum of
four branches has a maximum at $\bm{\Gamma}$ point, and it decreases as
$k_{x}$ increasing. The phonon angular momentum arrives at its minimum at the
point $\bm{K}$ ($k=4\pi/3$) with a positive value around $0.08\hbar$. The
phonons at every $\bm{k}$ points in the Brillouin zone has nonzero phonon
angular momentum. The difference of the total phonon angular momentum between
those at $\bm{\Gamma}$ and $\bm{K}$ will be smoothed by the temperature, which
can be seen in Fig. S1 (b)(c)(d). When $T=0$ K, the phonon angular momentum
$\bm{\Gamma}$ is around 20 times of that at $\bm{K}$ (Fig. S1 (b)), and the
ratio decreases to about 2 when $T=30K$ (see Fig. S1 (c)). When $T=300$ K and
the phonon angular momentum will be almost the same for all the wave-vector
points, as shown in Fig. S1 (d).

###  I.4  phonon angular momentum as a odd function of $\lambda$

$\displaystyle J_{z}^{{\rm{ph}}}(-\lambda)$ $\displaystyle=$
$\displaystyle\hbar\sum\limits_{\bm{k},\sigma>0}{\varepsilon_{\bm{k},\sigma}^{\dagger}(-\lambda)M\varepsilon_{\bm{k},\sigma}(-\lambda)(f(\omega_{\bm{k},\sigma}(-\lambda))}+1)$
(S61) $\displaystyle\Downarrow$
$\displaystyle\varepsilon_{-\bm{k},\sigma}^{*}(-\lambda)=\varepsilon_{\bm{k},\sigma}(\lambda)\quad\omega_{-\bm{k},\sigma}(-\lambda)=\omega_{\bm{k},\sigma}(\lambda)$
$\displaystyle=$
$\displaystyle\hbar\sum\limits_{\bm{k},\sigma>0}{\varepsilon_{-\bm{k},\sigma}^{T}(\lambda)M\varepsilon_{-\bm{k},\sigma}^{*}(\lambda)(f(\omega_{-\bm{k},\sigma}(\lambda))}+1)$
$\displaystyle\Downarrow$
$\displaystyle\varepsilon_{-\bm{k},\sigma}^{T}(\lambda)M\varepsilon_{-\bm{k},\sigma}^{*}(\lambda)=\varepsilon_{-\bm{k},\sigma}^{\dagger}(\lambda)M^{T}\varepsilon_{-\bm{k},\sigma}(\lambda)=-\varepsilon_{-\bm{k},\sigma}^{\dagger}(\lambda)M\varepsilon_{-\bm{k},\sigma}(\lambda)$
$\displaystyle=$
$\displaystyle-\hbar\sum\limits_{\bm{k}^{\prime},\sigma>0}{\varepsilon_{\bm{k}^{\prime},\sigma}^{\dagger}(\lambda)M\varepsilon_{\bm{k}^{\prime},\sigma}(\lambda)(f(\omega_{\bm{k}^{\prime},\sigma}(\lambda))}+1)$
$\displaystyle=$ $\displaystyle-J_{z}^{{\rm{ph}}}(\lambda).$

###  I.5  phonon angular momentum in the classical limit

At a high temperature, we can expand the Einstein-Bose distribution in Taylor
series as

$f(x)=\frac{1}{{e^{x}-1}}\simeq\frac{1}{x}-\frac{1}{2}+\frac{1}{{12}}x+O(x^{2}),\quad
x=\frac{{\hbar\omega}}{{k_{B}T}}.$ (S62)

Inserting it to Eq. (5) in the main text, we have

$J^{\rm
ph}_{z}(T\rightarrow\infty)=\sum_{\sigma>0,\bm{k}}[(\frac{k_{B}T}{\hbar\omega}+\frac{\hbar\omega}{12k_{B}T})l_{\bm{k},\sigma}^{z}],$
(S63)

The term linear to $T$ is zero, that is,
$\sum_{\sigma>0,\bm{k}}\frac{\epsilon^{\dagger}_{\bm{k},\sigma}\,M\,\epsilon_{\bm{k},\sigma}}{\omega_{\bm{k},\sigma}}=0$,
we prove it as follows.

Since we have the completeness relation as

$\sum_{\sigma}x_{\bm{k},\sigma}\otimes{\tilde{x}_{\bm{k},\sigma}}^{T}=I_{4n\times
4n},$ (S64)

with $x_{k}=\left(\begin{smallmatrix}\mu_{k}\\\
\epsilon_{k}\end{smallmatrix}\right)$,
$\tilde{x}_{k}^{T}=\frac{1}{-2i\omega_{k}}\left(\begin{smallmatrix}\epsilon^{\dagger}_{k}\;&-\mu^{\dagger}_{k}\end{smallmatrix}\right)$.
Thus, the off-diagonal block

$\sum_{\sigma}\epsilon_{\bm{k},\sigma}\otimes{\tilde{\epsilon}_{\bm{k},\sigma}}^{\dagger}/(-2i\omega_{\bm{k},\sigma})=O_{2n\times
2n},$ (S65)

where $O_{2n\times 2n}$ is a zero matrix. Then we have, for arbitrary $i$ and
$j$,

$\sum_{\sigma}\frac{\epsilon_{i}(\bm{k},\sigma)\epsilon_{j}^{*}(\bm{k},\sigma)}{\omega_{\bm{k},\sigma}}=0.$
(S66)

Similar as that in Section II, we can prove

$\sum\limits_{\bm{k},\sigma<0}{\frac{{\varepsilon_{\bm{k},\sigma}^{\dagger}M\varepsilon_{\bm{k},\sigma}}}{{\omega_{\bm{k},\sigma}}}}=\sum\limits_{\bm{k}^{\prime},\sigma^{\prime}>0}{\frac{{\varepsilon_{-\bm{k}^{\prime},-\sigma^{\prime}}^{\dagger}M\varepsilon_{-\bm{k}^{\prime},-\sigma^{\prime}}}}{{\omega_{\bm{k}^{\prime},-\sigma^{\prime}}}}}=\sum\limits_{\bm{k}^{\prime},\sigma^{\prime}>0}\frac{{\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}^{T}M\varepsilon_{\bm{k}^{\prime},\sigma^{\prime}}^{*}}}{{-\omega_{\bm{k}^{\prime},\sigma^{\prime}}}}=\sum\limits_{\bm{k},\sigma>0}{\frac{{\varepsilon_{\bm{k},\sigma}^{\dagger}M\varepsilon_{\bm{k},\sigma}}}{{\omega_{\bm{k},\sigma}}}}.$

Then

$\sum\limits_{\bm{k},\sigma>0}{\frac{{\varepsilon_{\bm{k},\sigma}^{\dagger}M\varepsilon_{\bm{k},\sigma}}}{{\omega_{\bm{k},\sigma}}}}=\frac{1}{2}\sum\limits_{\bm{k},\sigma}{\frac{{\varepsilon_{\bm{k},\sigma}^{\dagger}M\varepsilon_{\bm{k},\sigma}}}{{\omega_{\bm{k},\sigma}}}}=\frac{1}{2}\sum\limits_{\scriptstyle\bm{k},\sigma\hfill\atop\scriptstyle
j,i\hfill}{\frac{{\varepsilon_{j}^{*}(\bm{k},\sigma)M_{ji}\varepsilon_{i}(\bm{k},\sigma)}}{{\omega_{\bm{k},\sigma}}}}=\frac{1}{2}\sum\limits_{\scriptstyle\bm{k}\hfill\atop\scriptstyle
j,i\hfill}{M_{ji}\sum\limits_{\sigma}{\frac{{\varepsilon_{j}^{*}(\bm{k},\sigma)\varepsilon_{i}(\bm{k},\sigma)}}{{\omega_{\bm{k},\sigma}}}}=}0.$
(S67)