Chiral phonons in two-dimensional materials

The Einstein–de Haas effect [9, 19] is a phenomenon of mechanical rotation that is induced by the magnetization change. It is Einstein's only experiment during his life, which collaborated with de Haas. As shown in figure 1, the experiment was originally designed to prove the existence of Ampere's molecular currents. Einstein considered that the electron rotation around the nucleus and the electron spin formed the molecule electric current, which was the origin of the magnetism of matter. The ratio of atomic magnetic moment ($M$)

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle M= {e \over 2\pi r/v} \pi r^2= {1\over 2}evr \nonumber \end{align} \tag{ 1 }$$

to the total $J$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle J=mvr \nonumber \end{align} \tag{ 2 }$$

denotes the gyromagnetic ratio

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle {M \over J}= g{e\over 2m}, \nonumber \end{align} \tag{ 3 }$$

where $g$ is the Lande factor. The published results of $g$ in the Einstein–de Haas experiment was 1.02, but later experiments showed that $g\approx2$ which indirectly confirmed the electron spin found in the Stern–Gerlach experiment. In the ferromagnet, the atomic magnetic moment mainly comes from the spin instead of the orbital moment. According to the relation of $M$ with respect to $J$ and the conservation of $J$, we have

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta M^{orb}= {e \over 2m } \Delta J^{orb} \nonumber \end{align} \tag{ 4 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta M^{spin}= {e \over m } \Delta J^{spin} \nonumber \end{align} \tag{ 5 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta M= \Delta M^{spin} + \Delta M^{orb} \nonumber \end{align} \tag{ 6 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta J= \Delta J^{spin} +\Delta J^{orb}= -\Delta J^{lattice} \nonumber \end{align} \tag{ 7 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta M= g{e \over 2m } \Delta J. \nonumber \end{align} \tag{ 8 }$$

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The Einstein–de Haas effect provided an effective way to measure the $g$ of different materials. The accuracy of $g$ is key to determine the contribution of orbit and spin to the total $M$. The macroscopic visible mechanic rotation just reflects the motion of rigid body. The $J$ of phonon ($J^{\,ph}$), which is zero in the traditional sense, was not taken into account in the Einstein–de Haas effect. If phonon has nonzero $J$, the total $J$ should consider the contribution of phonon so that the $g$ need to be corrected.

Since the discovery of classic Hall effect of electrons, Leduc [20] proposed that the heat conductor of metal should have similar behavior. The transverse heat current arose once the temperature gradient was exerted, which was the thermal Hall effect or the Righi-Leduc effect. Different from the case of metal where the electron are main carriers, the phonons are the carrier of the heat current in the electric insulating crystal. The phonon Hall effect (PHE) was subsequently found by Strohm et al [6] and subsequently reproduced by Inyushkin and Taldenkov [22]. As shown in figure 2, when the magnetic field is exerted in the paramagnetic insulating thin film with longitudinal temperature gradient, we can observe the transverse temperature difference in the direction perpendicular to the longitudinal temperature gradient. The PHE is a exciting finding because phonon is a neutral pasi-particle which can not directly interacts with the magnetic field and thus suffers from the Lorentz force. This creative finding provides a new degree to manipulate the phonon by the magnetic field.

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Since the discover of PHE, some theories [23, 24] such as the spin–phonon interaction were proposed to understand the PHE. The magnetic field, which can not directly interacts with the phonon, only induces the polarization of the paramagnetic ion. The coupling between the paramagnetic ion and the phonon induces the PHE, which is also called the Raman spin–phonon interaction. For phonon Hall effect, the topological nature was investigated and topological phase transition was found [7], and a recent review on topological phonons is also reported [25]. The magnetic field exerts an effective force to distort the transport of phonon so that a transverse heat current emerges [26]. Thus, a natural question is whether the phonon current has nonzero angular momentum with some new macroscopic effects?

The lattice angular momentum corresponding to the mechanical rotation of rigid-body motion of the sample is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle J_{l\alpha}=\sum\limits_{l\alpha}R_{l\alpha}\times \dot{R}_{l\alpha}, \nonumber \end{align} \tag{ 9 }$$

it only reflects the rigid-body motion. In a microscopic picture, similar to the lattice angular momentum (rigid-body part), the phonon angular momentum can be defined as [8]

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle J^{\,ph}=\sum\limits_{l\alpha}u_{l\alpha}\times \dot{u}_{l\alpha}. \nonumber \end{align} \tag{ 10 }$$

In the equation, $\mu_{l\alpha}$ is a displacement vector which belong to the $\alpha$th atom in the $l$th unit cell, here we're going to do the mass normalization by multiplying the square root of the mass, and the purpose is to simplify the angular momentum. Along $z$ direction, $J_{z}^{\,ph}=\sum_{l\alpha}(u_{l\alpha}^{x}\dot{u}_{l\alpha}^{y}-u_{l\alpha}^{y}\dot{u}_{l\alpha}^{x})$. In the second quantization form, the displacement can be given as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle u_{l}=\sum^{k}\varepsilon_{k}e^{i(R_{l}.k-\omega_{k}t)}\sqrt{\frac{\hbar}{2\omega_{k}N}a_{k}}+H.c., \nonumber \end{align} \tag{ 11 }$$

with $k=(k,\sigma)$, $k$ and $\sigma$ are the wave vector and branch, respectively. $\varepsilon_{k}$ is a displacement polarization vector. Then the total phonon angular momentum can be written as [8]

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle & J_{z}^{\,ph}=\frac{\hbar}{2}\sum_{k,k'}\epsilon_{k}^{\dagger}M\epsilon_{k^{\prime}}(\sqrt{\frac{\omega_{k}}{\omega_{k^{\prime}}}} + \sqrt{\frac{\omega_{k^{\prime}}}{\omega_{k}}})a_{k}^{\dagger}\nonumber \\ &a_{k^{\prime}}\delta_{k,k^{\prime}}e^{i(\omega_{k}-\omega_{k^{\prime}})t} +\frac{\hbar}{2} \sum_{k}\epsilon_{k}^{\dagger}M\epsilon_{k}. \nonumber \end{align} \tag{ 12 }$$

Here

$$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle M=\left( \begin{array}{@{}cc@{}} 0 & -i \nonumber \\ i & 0 \nonumber \\ \end{array} \right)\otimes I_{n\times n}, \nonumber \end{align*}$$

$n$ is the quantity of atoms in a unit cell. In equilibrium, the total phonon angular momentum changes into [8],

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle \label{13} J_{z}^{\,ph}=\sum\limits_{\sigma ,k} l_{k,\sigma}^{z}[\,f(\omega_{k,\sigma})+\frac{1}{2}], l_{k,\sigma}^{z}=(\epsilon_{k,\sigma}^{\dagger}M\epsilon_{k,\sigma})\hbar, \nonumber \end{align} \tag{ 13 }$$

$f(\omega_{k})= 1/(e^{\hbar\omega_{k}/k_{B}T}-1)$ is the Bose–Einstein distribution. In equation (13), all the wave vectors and all phonon branches are summed $(\omega\geqslant 0)$. Here, $l_{k,\sigma}^{z}$ is the phonon angular momentum with wave vector $k$ and branch $\sigma$, it is real and in proportion to $\hbar$. At zero temperature, the form of the total phonon angular momentum is $J_{Z}^{\,ph}(T=0)=\sum_{\sigma,k}\frac{1}{2}l_{k,\sigma}^{z}$, it means that every phonon mode $(k,\sigma)$ has a zero-point energy of $\hbar\omega_{k,\sigma}/2$ and has a nonzero angular momentum(zero-point angular momentum) $\newcommand{\re}{{\rm Re}} \newcommand{\e}{{\rm e}} \renewcommand{\dag}{\dagger}\frac{1}{2}l_{k,\sigma}^{z}=(\frac{\hbar}{2})(\epsilon_{k,\sigma}^{\dagger}M\epsilon_{k,\sigma})$.

Recently, the helicity-resolved Raman scattering has been experimentally observed in transition-metal dichalcogenide (TMD) atomic layers [27]. The authors found that the helicity of incident photons can be completely reversed by the Brillouin-zone-center $(\Gamma)$ phonons. Such a finding implies that not only the valley phonons but also the $\Gamma$ phonons which involved in the intravalley scattering of electrons can have chirality. Therefore, in the Brillouin zone high-symmetry points of the hexagonal lattices, investigating chirality of the phonons is highly desirable.

The two-dimensional honeycomb AB lattice, which has two sublattices (A and B) in each unit cell, can be used as a simplified model to demonstrate the general features of chiral phonons in monolayer materials, such as gapped graphene with isotopic doping [28] or staggered sublattice potential [29], hexagonal boron nitride [30]. By calculating the eigenvectors, the sublattice vibrations at valleys can be plotted, as shown in figure 3(a). We can see that the whole vibrations at the valleys are circularly polarized. The valley phonon modes from band 1 to band 4 are shown in the figure. Among it, two sublattices of band 1 and band 4 do circular motion in the opposite direction; two sublattices of band 2 and band 3 are different from the above, one is still, the other is a circular motion. The direction of the circular motion will be reversed from $K$ to $K'$.

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The phonon chirality can be characterized by the polarization of phonons along the $z$ direction. The phonon eigenvectors can be written as [10] $\newcommand{\e}{{\rm e}} \epsilon=(x_{1}y_{1}x_{2}y_{2})^{T}$. A new basis of $|R \rangle$ and $|L \rangle$) can be used to represent the right-handed circular polarization and left-handed circular polarization.

$$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle |R_1\rangle&\equiv \frac{1}{\sqrt{2}}(1 \ i \ 0\ ... \ ... \ 0)^T,\nonumber \\ |L_1\rangle&\equiv \frac{1}{\sqrt{2}}(1 \ -i\ 0 \ ... \ ... \ 0)^T,\nonumber \\ |R_n\rangle&\equiv \frac{1}{\sqrt{2}}(0 \ \ ... \ ... \ 0 \ 1 \ i)^T,\nonumber \\ |L_n\rangle&\equiv \frac{1}{\sqrt{2}}(0 \ \ ... \ ... \ 0 \ 1 \ -i)^T. \nonumber \end{align*}$$

Using above basis, the phonon eigenvector $\newcommand{\e}{{\rm e}} \epsilon$ can be represented as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \epsilon=\sum_{\alpha=1}^{n}\epsilon_{R_{\alpha}}|R_{\alpha}\rangle+\epsilon_{L_{\alpha}}|L_{\alpha}\rangle , \nonumber \end{align} \tag{ 14 }$$

where $\newcommand{\e}{{\rm e}} \epsilon_{R_{\alpha}}=\langle R_{\alpha}|\epsilon\rangle=\frac{1}{\sqrt{2}}(x_{\alpha}-iy_{\alpha})$, $\newcommand{\e}{{\rm e}} \epsilon_{L_{\alpha}}=\langle L_{\alpha}|\epsilon\rangle=$$\frac{1}{\sqrt{2}}(x_{\alpha}+iy_{\alpha})$. The phonon circular polarization operator along the $z$ direction is defined as

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \hat{S}^{z}\equiv \sum(|R_{\alpha}\rangle\langle R_{\alpha}|-|L_{\alpha}\rangle\langle L_{\alpha}|), \nonumber \end{align} \tag{ 15 }$$

phonon circular polarization can be expressed as

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle s_{ph}^{z}=\epsilon^{\dagger}\hat{S}^{z}\epsilon\hbar = \sum(|\epsilon_{R\alpha}|^{2}-|\epsilon_{L\alpha}|^{2})\hbar. \nonumber \end{align} \tag{ 16 }$$

Here, the hexagonal lattices has two atoms in a unit cell, so the value of $n$ is 2. Since $\newcommand{\e}{{\rm e}} \sum_{\alpha}|\epsilon_{R_{\alpha}}|^{2}+|\epsilon_{L_{\alpha}}|^{2}=1$, the value of the phonon circular polarization is between $\pm\hbar$. Along the $z$ direction, the $s_{ph}^{z}$ has the same form with the phonon angular momentum $j_{k,\sigma}^{z}$ [8]. We use $s_{\alpha}^{z}$ to represent the contribution of each sublattice of the unit cell to the phonon circular polarization. As shown in figure 3(a), at valley $K$, for band 2, $s_{A}^{z}=0, s_{B}^{z}=-\hbar$; for band 3, $s_{A}^{z}=\hbar, s_{B}^{z}=0$; for band 1 and band 4, the magnitudes of sublattices A and B are different with opposite circular vibrations. Therefore,the circular polarization of phonon $s_{ph}^{z}$ which is at the valleys can be nonzero. As mentioned above, introducing a staggered sublattice on-site potential or isotope doping can lead to the similar chiral phonons at the valleys. At the Brillouin zone center $\Gamma$, we can see doubly degenerate acoustic modes and optical modes, both of which are not circularly polarized. Through superposing the degenerated modes the circularly polarized phonon modes can always be obtained.

At the high-symmetry points $\Gamma,K,K^{\prime}$ in a honeycomb lattice,the phonons are invariant to the threefold discrete rotations in the direction perpendicular to the two-dimensional honeycomb lattice plane (figure 3(b)). The eigenvector has a phase change [10]

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Re(\frac{2\pi}{3},z)u_{k}=e^{-i(2\pi/3)l_{ph}^{k}}u_{k}, u_{k}=\varepsilon_{k}e^{i(R_{l} \cdotp k-\omega t)}. \nonumber \end{align} \tag{ 17 }$$

Here $\Re$ is the rotation operator, $l_{ph}^{k}$ in the equation is defined as the pseudoangular momentum (PAM) and has values of 1, $-1$, or 0, $\mu_{k}$ is the phonon wave function. The phonon wave function is consisted of two parts, the local (intracell) part $\varepsilon_{k}$ and the nonlocal (intercell) part $e^{iR_{l}.k}$. Thus, under the threefold rotation, we can get the spin phonon PAM ($l^{s}$) for the intracell part and the orbital phonon PAM ($l^{o}$) for the intercell part. The total phonon PAM

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle l_{ph}^{k}=l^{s}+l^{o}. \nonumber \end{align} \tag{ 18 }$$

For the left-handed or right-handed polarization of A and B atoms, it can be represented by the spin PAM $l_{R}^{s}=1$, $l_{L}^{s}=-1$.

The orbital PAM is obtained by the change of the phase $e^{iR_{l}.k}$ under the threefold rotation. As shown in figure 3(b), at the valley $K$ or $K^{\prime}$, the orbital PAM of the sublattices A and B is opposite and the value is 1 or $-1$. At the $\Gamma$ point, the sublattices A and B have no phase change, so the orbital PAM is zero. The spin PAM is obtained by the phase change of $\varepsilon_{k}$. At the valley $K$, $l_{A}^{s}=-1, l_{B}^{s}=1$ for bands 1 and 4; For band 2, $l_{A}^{s}$  =  0, $l_{B}^{s}=-1$; for band 3, $l_{A}^{s}=1$, $l_{B}^{s}=0$. At the valley $K^{\prime}$, the value of spin PAM is opposite to the valley $K$.

As the phonon wave function $\mu_{k}$ is the eigenstate of the rotation operator $\Re$, if both sublattices are vibrating, the phonon PAM equals to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle l_{ph}=l_{A}^{s}+l_{A}^{o}=l_{B}^{s}+l_{B}^{o}. \nonumber \end{align} \tag{ 19 }$$

The values of phonon PAM are listed in figure 3(c). We can see that the orbital PAM and spin PAM of A and B can verify above equation. If one of the sublattices does not vibrate, the phonon PAM is determined by the other vibrational sublattice. At the $\Gamma$ point, the orbital PAM is zero, so the above equation changes into

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \label{20} l_{ph}^{k}=l_{A}^{s}=l_{B}^{s}. \nonumber \end{align} \tag{ 20 }$$

As mentioned in the end of above section, the sublattices can do circular vibration by superposing the doubly degenerate modes. Therefore, the value of the $l_{ph}^{k}$ is 1 or $-1$.

In the unit cell, each sublattice can be selected as the center of the threefold rotation with a zero orbital phonon PAM. Thus, the spin PAM is equal to the phonon PAM according to equation (20). That is to say the value of spin PAM is $\pm1$ or 0. Therefore, all the phonon modes at valleys (figure 3(a)) do circular motion, otherwise the phonon modes are still.

In a uniform external magnetic field, the Hamiltonian for ionic crystals is written as [23, 24]

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle H=\frac{1}{2}(\,p-\tilde{A}u)^{T}(\,p-\tilde{A}u)+\frac{1}{2}u^{T}Ku, \nonumber \end{align} \tag{ 21 }$$

$u$ is a column vector which from lattice equilibrium positions of displacements, the quality normalization has been completed by multiplying square root of mass, $K$ is a force constant matrix, $p$ is a conjugate momentum vector, the $\mu^{T}\tilde{A}p$ term reflects the Raman spin–phonon interaction [31, 32], $\tilde{A}$ is an antisymmetric real matrix [33], the dimension of the matrix is $Nd\times Nd$, $d$ is the dimension of the lattice vibrations and $N$ is the number of total sizes. $\tilde{A}$ can be block diagonal, and the element is

$$\begin{align*} \newcommand{\e}{{\rm e}} \displaystyle \Lambda_{\alpha}=\left( \begin{array}{@{}cc@{}} 0 & \lambda_{\alpha} \nonumber \\ -\lambda_{\alpha} & 0 \nonumber \\ \end{array} \right), \nonumber \end{align*}$$

where we only consider two-dimensional motion in the honeycomb lattice, so $d=2$($x$ and $y$ directions). $\lambda_{\alpha}$ has the dimension of frequency, and it is proportional to the magnetization and the spin–phonon interaction.

When there is no spin–phonon interaction on the system, the Hamiltonian is reduced to

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle H=\frac{1}{2}p^{T}p+\frac{1}{2}\mu^{T}K\mu. \nonumber \end{align} \tag{ 22 }$$

Use $D^{T}(k)=D^{*}(k)=D(-k)$ to solve the eigenvalue equation $D(k)\varepsilon_{k,\sigma}=\omega^{2}_{k,\sigma}\varepsilon_{k,\sigma}$, we can get $\omega_{-k,\sigma}=\omega_{k,\sigma}$, $\varepsilon_{-k,\sigma}=\varepsilon^{*}_{k,\sigma}$ and $l_{-k,\sigma}^{z}=-l_{k,\sigma}^{z}$. The total phonon angular momentum $J_{z}^{\,ph}=0$¹ if we sum over all phonon modes. However, when the system has a spin–phonon interaction, $l_{-k,\sigma}^{z}\neq-l_{k,\sigma}^{z}$, the total phonon angular momentum $J_{z}^{\,ph}\neq0$, which is shown in figure 4.

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With spin–phonon interaction, the phonon angular momenta at zero temperature for different lattices are shown in figure 4(a). The phonon angular momenta of honeycomb and kagome lattices are larger than the triangle and square lattices. The honeycomb and kagome lattices have more sites in one unit cell than the triangle and square lattices, which means the more sites of lattices, the more phonon angular momentum they have. This trend can be understood from the fact that the contribution of optical bands to phonon angular momentum is greater than that of acoustic bands. As shown in figures 4(c) and (d), if the value of $\lambda$ is small,the phonon angular momentum of the acoustic bands almost vanishes at low temperature, so the optical bands do the main contribution to the total phonon angular momentum. Therefore, the more sites of per unit cell, the more optical bands it has, thus the phonon angular momentum will be bigger. As shown in figure 4(a), the total angular momentum $J_{z}^{\,ph}$ increased with the increase of $\lambda$. As shown in figure 4(b), since $\lambda$ is proportional to the magnetization, the sign of phonon angular momentum also changes when the sign of magnetization changes $J_{z}^{\,ph}(-\lambda)=-J_{z}^{\,ph}(\lambda)$. This can explain the phenomenon that the transverse temperature difference reverses when the magnetic field is reversed in the phonon hall effect.

At the high temperature limit, because of

$$\begin{align} \renewcommand{\dag}{\dagger} \newcommand{\e}{{\rm e}} \newcommand{\re}{{\rm Re}} \displaystyle \sum_{\sigma>0 ,k}(\varepsilon_{k,\sigma}^{\dagger}M\varepsilon_{k,\sigma}/\omega_{k,\sigma})=0, \nonumber \end{align} \tag{ 23 }$$

the phonon angular momentum goes to zero. Whether or not the magnetic field is applied, the phonon angular momentum of each unit cell decreases with the increase of temperature and goes to zero at the high temperature limit (the temperature is far greater than the Debye temperature). As the temperature increases, more phonon modes are stimulated. The angular momentums of these modes have opposite direction to the zero-point angular momentum. In the high temperature limit, all stimulated modes are offset by zero-point angular momentum. The vanish of phonon angular momentum can be understood as follows: at high temperature, the classical statistical mechanics is applied to the calculation of phonon angular momentum. A phase-space integral with respect to $p$ and $u$ can be obtained by the summation over quantum states, it can change to a pure harmonic system by variable substitution. As discussed earlier, the phonon angular momentum is zero. Furthermore, the Bohr-van Leeuwen theorem illuminates that the average magnetization in the classical mechanics is always zero [34], which makes the phonon angular momentum vanish in the classical limit. Therefore, only in the low-temperature quantum systems that the phonon angular momentum is meaningful.

The Einstein–de Haas effect shows that a freely suspended body has mechanical rotation due to its change of magnetization. Einstein and De Haas adopted the resonance method in the experiment. The periodic magnetic field is tuned to be the intrinsic frequency of the suspended objects. Their experiment provide a way to measure the ratio between the change of magnetization and the change of total angular momentum. It is traditionally believed that the total angular momentum of the system is equal to the sum of the lattice angular momentum, spin angular momentum and orbital angular momentum

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle J^{tot}=J^{lat}+J^{spin}+J^{orb}. \nonumber \end{align} \tag{ 24 }$$

Due to the conservation of angular momentum, we known that the change of the lattice angular momentum is equal to the sum of the change of the spin angular momentum and the orbital angular momentum

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta J^{lat}=-(\Delta J^{spin}+\Delta J^{orb}). \nonumber \end{align} \tag{ 25 }$$

The change of the lattice angular momentum is determined by the mechanical rotation of the sample [35]. But from the microscopic point of view, the angular momentums of all atoms in the sample are equal to the sum of the lattice angular momentum and the phonon angular momentum in equilibrium. So the total angular momentum will have one more phonon angular momentum [8].

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle J^{tot}=J^{lat}+J^{spin}+J^{orb}+J^{\,ph}. \nonumber \end{align} \tag{ 26 }$$

As discussed in the previous section, when there is a spin–phonon interaction, the total phonon angular momentum is not zero. According to the conservation of total angular momentum

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta J^{lat}+\Delta J^{\,ph}=-(\Delta J^{spin}+\Delta J^{orb}). \nonumber \end{align} \tag{ 27 }$$

On the other hand, the total magnetization change can be measured using

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta M=\Delta M^{spin}+\Delta M^{orb}, \nonumber \end{align} \tag{ 28 }$$

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta M=g\frac{e}{2m}\Delta J. \nonumber \end{align} \tag{ 29 }$$

Obviously the $g\frac{e}{2m}$ now needs to be corrected. Since the spin–phonon interaction is ubiquitous in the magnetic lattices, the phonon angular momentum is very important.

Phonons have important contribution to total angular momentum, and the magnitude of phonon angular momentum depends on the value of $\lambda$. The calculation shows that in the presence of spin–phonon interaction, the degenerate phonon modes split at $\Gamma$ point and the gap is 2$\lambda$, so the parameter $\lambda$ can be obtained from the phonon dispersion relation. This kind of split of the degenerate acoustic modes can be measured by the inelastic neutron scattering or Raman scattering. Because the phonon angular momentum proportional to the magnetization, larger phonon angular momentum can be observed when magnetization in paramagnetic materials is saturated. In the phonon hall effect experiment, the paramagnetic insulator is used to highlight the contribution of phonons in the thermal transport, while the contribution of the electrons and the magnons can be ignored. However, spin–phonon interaction extensively exist in various magnetic materials [36]. A very large magnetization do Ferromagnetic materials have, so we can observe the obvious phonon angular momentum if the spin–phonon interaction is strong.

For the materials with large magnetization and strong spin–phonon interaction, the zero-point phonon angular momentum is important. According to previous studies, the orbital magnetic moment calculated in some ferromagnetic materials are only a small part of the total magnetic moment, which means that the orbital angular momentum is approximately $\hbar$ times a small value [37]. The phonon angular momentum is comparable to the size of the electron orbital angular momentum, so the phonon angular momentum cannot be ignored.

In ferromagnetic insulators, the transport of electrons are negligible. Because the phonon angular momentum decreases with the rising temperature and it tends to zero under the condition of the classical limit, the change of lattice angular momentum can be measured under the low temperature ($T_{low}$) and the high temperature ($T_{high}$), so that the the phonon angular momentum can be separated. Here, the dividing line between the low temperature and the high temperature should be the Debye temperature ($T_{Debye}$), which is the critical temperature divides the quantum and classical regions. On the other hand, in order to avoid the participation of the magnon, the high temperature needs to be lower than the Curie temperature, which makes the Curie temperature to be much higher than Debye temperature. So the angular momentum of magnon is almost constant, but the phonon angular momentum changes significantly with the temperature. Fortunately, many ferromagnetic materials can satisfy $T_{low}<T_{Debye}$, $T_{Debye}<T_{high}<T_{Curie}$, their Curie temperature is about 1000 K, and their Debye temperature is lower than 500 K [38].

Except for the application in measuring the gyromagnetic ratio, the nonzero phonon angular momentum also provides an effective method to study the spin–phonon interaction of magnetic materials. In ferromagnetic materials, there is an open question to separate the contributions to the thermal hall effect from phonons and magnons, a method to obtain the phonon contribution is gived by the phonon angular momentum.

Due to the breaking of spatial inversion symmetry, the separated valleys $K$ and $K'$ in momentum space are no longer equivalent, this kind of nonequivalence property can be described by the Berry curvature or orbital magnetic moment. The two valleys constitute another degree of freedom for the electrons besides charge and spin. Therefore, the construction of the valley degree of freedom depends on the spatial inversion symmetry breaking of the material systems. The emergence of valley degree of freedom leads to the the valleytronics.

There are many graphene-like hexagonal lattice materials, such as boron nitride, TMD, gallium selenide, etc. Unlike graphene, the A and B sublattice structures in these materials are composed of different atoms, which destroys the spatial inversion symmetry. The existence of valley degree of freedom in these materials create conditions for the study of valleytronics.

On the $AB$ double-layer graphene materials, we can apply a electric field perpendicular to the two-dimensional plane to break the spatial inversion symmetry of graphene, which make the Berry curvature and the orbital magnetic moment of $K$ and $K'$ have opposite direction with the same magnitude [39]. The Berry curvature and orbital magnetic moment which are associated with the valleys lead some novel effects carried such as the valley hall effect and the optical transition selection rules [39]. As shown in figure 5, because of the inverse Berry curvature, the electrons in $K$ and $K'$ valleys move in the direction perpendicular to the applied electric field, and they deflect transversely with opposite direction, which is called the valley Hall effect. As shown in figure 6, only the left-handed(right-handed) circularly polarized photos are involved in the interband transition of $K'(K)$ valley electrons. This is just the optical transition selection rules which are relevant to valleys, i.e. the circular polarization dichroism of valleys. All of these provide possible physical methods to regulate a particular valley state.

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The scattering of the electrons in the intravalley involves a polarized photoexcitation and photoluminescence, and the phonons of the brillouin-zone-corner(valley) are also involved in intervalley electron scattering [5]. In view of the deterministic chirality of valley electrons, it is natural to ask two questions: do the valley phonons have definite chirality? what role do the chiral phonons play in the intervalley scattering of electrons? It was later observed in the TMDs that the phonons of the Brillouin-zone-center$(\Gamma)$ could completely reverse the polarization of the incident photons, i.e. the optical polarity inversion phenomena. This finding imply that, except for the phonons in the valleys, the $\Gamma$ phonons which involved in the intralley scattering of electrons also have the chirality [27]. Therefore, it is necessary to study the chiral phonons at the high-symmetry points of the brillouin zone.

4.2.1. The helicity-resolved Raman scattering in TMD

The moment of the electrons along the direction $z$ is 0, the state will not change under the threefold rotation, the orbits of the sublattices A and B determine the PAM. If we assume that the orbit of sublattice A corresponds to the valence band and the orbit of sublattice B corresponds to the conduction band, the pseudoangular momentum $l_{c(\nu)}=\pm\tau,\tau=\pm 1$. Therefore, in the gapped graphene, electrons absorb the right-handed or left-handed circularly polarizated photons to have interband transition. Because of the conservation of the PAM, the selection rule can be obtained

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle l_{c}-l_{\nu}=l_{photon}=\pm\tau, \nonumber \end{align} \tag{ 30 }$$

the difference between the conduction bands angular momentum and the valence bands angular momentum is equal to the photon angular momentum, as is shown in figure 7.

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The electron in the conduction band is scattered in the intravalley by a phonon at $\Gamma$, then through emitting a photon it can combine with a hole in the valence band. First order Raman scattering is named for this process. Because the PAM is conserved in this process, the selection rule is

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta l_{photon}=l_{ph}. \nonumber \end{align} \tag{ 31 }$$

At $\Gamma$ point, doubly degenerate optical modes are Raman active, has PAM of 1, $-1$, so it is expected to have a helicity-resolved Raman $G$ peak in the honeycomb lattice (gapped graphene or a boron nitride monolayer). In the whole process, the incident right-handed (left-handed) photon emits a left-handed (right-handed) phonon or absorbs a right-handed (left-handed) phonon, then turns to the left-handed (right-handed) photon. As shown in figure 8 [10], the helicity-resolved Raman scattering in layered TMD is explained by this selection rule.

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4.2.2. Intervalley scattering of electrons

The intervalley scattering of electrons can be achieved by emitting a circularly polarized valley phonon $(l_{ph}=\pm1)$. Because phonons have PAM under a threefold rotational symmetry, under the condition that both momentum and energy are conserved, the selection rule

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle l_{c(\nu)}(K)-l_{c(\nu)}(K')=\pm1 \nonumber \end{align} \tag{ 32 }$$

can be obtained. The difference between the conduction band (valence band) angular momentum at $K$ and $K'$ point is equal to the PAM of the valley phonon. It is well known that the doubly resonance peak $D$ in the graphene Raman spectrum reflects the intervalley scattering of phonons which adjacent to the $K$ point [42, 43]. Therefore, according to the Raman spectrum, a valley phonon with definite PAM and frequency can be observed [10].

4.2.3. Chiral phonons in monolayer ${MoS_{2}}$

There is a band-gap $\Delta=1.65$ eV in the spin–orbit coupling MoS$_{2}$ (the single layer is converted into a direct band-gap) and a spin splitting $\lambda_{k,\nu}=150$ meV for the highest valence band at $K$ valley. For the lowest conductance band, there is a $3$ meV splitting, which can be neglected [44]. By absorbing or transmitting a photon, a valley center phonon is involved in the intervalley electron scattering at valley centers, and the selection rules and energy conservation are

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \Delta l_{el}=\pm l_{ph}\pm l_{photon}, \lambda_{K,\nu}=\pm \hbar\omega_{ph}\pm \hbar\omega_{photon}. \nonumber \end{align} \tag{ 33 }$$

The change of electron angular momentum is equal to the algebraic sum of the phonon angular momentum and photon angular momentum, and the spin splitting of the valence band is equal to the algebraic sum of the phonon energy and photon energy, where $+$ means emission and  −  means absorbtion.

For monolayer MoS$_{2}$, at valleys, the PAM of the conduction bands are $1, -1$, and the PAM of the valence band is $0$ [45]. The right-handed polarized photon excites a pair of excitons (conduction electron and valence band hole) at valley $K$, the right-handed photon has a band-gap energy. Because the stimulated electrons are in the center of the valleys, and the phonons have energy, so by emitting a phonon, they cannot be scattered to another valley center. However, since the valence band has a large spin splitting, through absorbing a stimulated circular polarized photon the hole can be scattered to another valley, and a chiral valley phonon can be emitted, the spin of electron is fixed here.

As shown in figure 9(a), there is a resonance peak with energy $\lambda_{k,\nu}+\hbar\omega_{ph}$ can be observed by the stimulated right-handed light which scanning on the sample. According to the selection rule, the chiral phonon emitted at $K$ point has the PAM of $-1$. The resonance peak is in $164.4$ meV (due to the small phonon polarization of 48.0 meV, the peak of 98.0 meV is not evident). For stimulated left-handed photons, only a peak of $190$ meV is observed, which is corresponding to the phonon mode of $40$ meV. The other two modes do not participate because they are odd under the mirror operation since in the scattering process the whole system keeps even under a mirror operation relative to the $x$-$y$ plane. Different valleys are electrons and holes located in, which after being scattered by phonons and photons, respectly. This is consistent with the recent discovery that the low energy excitons have large momentum [46]. Similarly, a left-handed photon with 1.65 eV excits a hole of $K'$ and emits a chiral phonon of $K'$. So by a stimulated photon, we can obtain a chiral phonon at a definite valley. By irradiating samples using the two-step polarized light, there created a large quantity of valley phonons which has definite frequencies [10].

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4.2.4. Chiral phonons in other 2D systems

Chiral phonons not only exit in monolayer MoS$_{2}$, but also exist in many other 2D systems. In graphene/hexagonal boron nitride (G/h-BN) heterostructure, the broken inversion symmetry and the interlayer interaction of G/h-BN open the band gap of ZO1(K) and ZO3(K), LO1(K) and LO3(K), shown in figures 10(a)–(c). The nonzero phonon polarization at the first-Brillouinzone corners (valleys) is shown in figure 10(d). The phonon chirality can be characterized by the polarization of phonons [10], which means the chiral phonons occur at valleys. Furthermore, the vertical stress is effective to adjust the band gap while holding the phonon chirality of G/h-BN heterostructure, which is favorable for the ultrafast infrared spectroscopy or Raman measurement [47]. In addition, chiral phonons also be found in center-stacked bilayer triangle lattices [48]. In this model, there are two atoms in one unit cell and considers three dimensional motion—six phonon bands can be obtained. Because the two sublattices have different masses, the spatial inversion symmetry is broken. The band gap is opened with nonzero phonon polarization, which is shown in figure 11. In bilayer lattice structure, the interaction between layers (interlayer coupling) can not be ignored and the mass of sublattice in different layers must be considered, too. By adjusting these parameters, they find that the phonon chirality remain robust with changing interlayer coupling and mass ratio of sublattices [48].

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In all above 2D systems, chiral phonons exist at Brillouin-zone corners (valleys). But the nondegenerate chiral phonons also can be observed near to ${\boldsymbol \Gamma}$ point [49]. As shown in figure 12(a), through the periodic atomic mass doping of sublattice $m_{c}$, the $\sqrt{3}\times\sqrt{3}$ superlattices are tailored on a honeycomb AB lattice. Considering two dimensional motion and the model has six atoms in one unit cell, there are twelve phonon branches, shown in figure 12(b). Moreover, the nondegenerate chiral phonons of two valleys are folded to $\Gamma$ point, thus we can obtain nonzero phonon polarization at Brillouin-zone center, see figures 12(c) and (d). Since the nonzero phonon polarization occur at $\Gamma$ point where the LO/TO splitting can not be ignored, the relation between LO/TO splitting and phonon polarization has also been discussed in this article [49].

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4.3.1. Circularly polarized infrared absorption or emission

Because the phonons of different bands at valleys have different PAM, we can observe the circular polarization infrared spectra in the intervalley scattering of the valley phonons (phonon bands).The nonzero ionic magnetic moment is caused by the circular polarization vibrations of the sublattices at the valleys, and it can couple to photons directly. So through polarized infrared absorption or emission, the valley phonons can be observed. As discussed earlier, the valley phonons can be excited during the electron or hole intervalley scattering process. As shown in figure 13, assuming the electrons can excite the phonons of band $\varepsilon_{1}$ at the valley $K'$ massively, through a photon with energy $\varepsilon_{2}-\varepsilon_{1}$ stimulation which is left-circularly-polarized infrared or a photon with energy $\varepsilon_{3}-\varepsilon_{1}$ simulation which is right-handed infrared, the corresponding left-handed or right-handed photoluminescence can be observed.

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If a lot of valley phonons at band $\varepsilon_{2}$ are created, through the infrared stimulation, we can observe the left-handed photoluminescence with energy $\varepsilon_{3}-\varepsilon_{2}$ ($\varepsilon_{2}-\varepsilon_{1}$), or the right-handed photoluminescence with energy $\varepsilon_{4}-\varepsilon_{2}$. Therefore, the resonance peak of infrared spectrum can distinguish the valley phonons. Figure 13 shows the bands and corresponding PAM of phonons at valley $K'$, the blue ellipses mark the valley phonons which are excited during the intervalley electron scattering, the green lines reflect all possible absorption or emission of polarized infrared lights [10].

4.3.2. Valley phonon hall effect

Because of the opposite Berry curvature, electrons in different valleys flow to opposite transverse edges under an in-plane electric field, which is called the electronic valley hall effect [39, 50]. Recently, in monolayer MoS$_{2}$ transistors [51] and in graphene superlattices [52] the valley hall effect of electrons has been observed experimentally. As discussed above, phonons with definite frequencies at the valleys can be produced massively. So, analogy to the electronic valley hall effect, if the Berry curvature of valley phonons is nonzero, the valley phonon hall effect can be observed when there is an in-plane gradient strain field, this provides another method to observe valley phonons.

Due to breaking of the spatial inversion symmetry, the nonzero phonon Berry curvature is observed in the valley, as shown in figure 14(a). At the valleys, band 2 and band 3 have large Berry curvature, while band 1 and band 4 has small Berry curvature. Because of the nonzero Berry curvature, when applying a strain gradient $E_{strain}$ along the $x$ direction, the phonons excited at different valleys will move along different transverse direction since

$$\begin{align} \newcommand{\e}{{\rm e}} \displaystyle \nu_{atom}\propto -E_{strain}\times\Omega, \nonumber \end{align} \tag{ 34 }$$

which is analogous to the electrons. As shown in figure 14(b), if the polarization of the photon is reversed, the direction of the transverse phonon current is also reversed. Because the phonons accumulated on one edge, we can observe the transverse temperature difference. If the circular polarization of the stimulated photon is reversed, the temperature difference changes sign. The phenomenon of phonon hall effect can be obtained in the paramagnetic insulator, and the transport of phonons can be distorted by the magnetic field so that the transverse temperature difference can be observed, which has led to many researches in this field [23]. At nonmagnetic system with broken inversion symmetry, the transverse valley phonon hall effect induced by Berry curvature will bring new applications.

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