Two-dimensional rotational symmetry SO $^{\perp }(2)$

If the density matrices $\hat{\rho}$ and $\hat{\breve{\rho}}$ possess the SO $^{\perp }(2)$ symmetry, it follows from Eqs. (5) and (6) that all densities of type $\rho$ should be SO $^{\perp }(2)$ scalars. Now it is convenient to decompose densities of type $\bm{s}$ into components perpendicular and parallel to the symmetry axis:

$$\bm{s}(\bm{r},\bm{r}')=\bm{s}_{\perp}(\bm{r},\bm{r}') +\bm{s}_z(\bm{r},\bm{r}') .$$ (30)

The perpendicular components $\bm{s}_{\perp}(\bm{r},\bm{r}')$ are SO $^{\perp }(2)$ vectors and the parallel components $\bm{s}_z(\bm{r},\bm{r}')$ are SO $^{\perp }(2)$ invariants.

The following SO $^{\perp }(2)$ scalars can be constructed from the perpendicular and parallel components of the position vectors $\bm{r}$ and $\bm{r}'$: $z,\ z',$ the $z$-coordinates of $\bm{r}_z$ and $\bm{r}'_z$, respectively, $r^2_{\perp}=\bm{r}_{\perp}\cdot\bm{r}_{\perp},\ \bm{r}_{\perp}\cdot\bm{r}'_{\perp},\ r^{\prime2}_{\perp}=\bm{r}'_{\perp}\cdot\bm{r}'_{\perp}$, and $\bm{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp})$ and $\bm{r}'_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp})=(z'/z)\bm{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp})$. As we see the two latter scalars are dependent on each other because $\bm{r}_z\parallel\bm{r}'_z$. The square $(\bm{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp}))^2 =z^2(r^2_{\perp}r'^2_{\perp}-(\bm{r}_{\perp}\cdot\bm{r}'_{\perp})^2)$ is dependent on other scalars, namely on $z^2,\ r_{\perp}^2,\ r_{\perp}^{\prime 2}$ and $\bm{r}_{\perp}\cdot\bm{r}'_{\perp}$. Hence, the nonlocal scalar densities have the form

$$\rho (\bm{r}, \bm{r}')=\varrho_0(z,z',r^2_{\perp},\bm{r}_{\perp}\... ...rp},r^{\prime 2}_{\perp}, \bm{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp}) ),$$ (31)

where $\varrho_0$ is an arbitrary scalar function, linear in the last argument. There are three vectors invariant under SO $^{\perp }(2)$ and parallel to the symmetry axis, namely $\bm{r}_z$, $\bm{r}'_z$, and $\bm{r}_{\perp}\times\bm{r}'_{\perp}$. Consequently, the z-component of $\bm{s}(\bm{r}$, $\bm{r}')$ is of the form

$$\bm{s}_z(\bm{r}, \bm{r}') = \varrho_z(z,z',r^2_{\perp},\bm{r}_{\perp}\cdot\bm{r}'_{\perp},r^{\prime 2}_{\perp}, \bm{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp}))\bm{r}_z$$

$$+ \varrho_z'(z,z',r^2_{\perp},\bm{r}_{\perp}\cdot\bm{r}'_{\perp},r^{\prime 2}_{\perp}, \bm{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp}))\bm{r}'_z$$

$$+ \varphi_z(z,z',r^2_{\perp},\bm{r}_{\perp}\cdot\bm{r}'_{\perp},r^{... ...dot(\bm{r}_{\perp}\times\bm{r}'_{\perp}))(\bm{r}_{\perp}\times\bm{r}'_{\perp}),$$ (32)

where $\varrho_z$, $\varrho_z'$, and $\varphi_z$ are scalar functions. There exist four independent SO $^{\perp }(2)$ vectors that lie in the plane perpendicular to the symmetry axis: $\bm{r}_{\perp}$, $\bm{r}'_{\perp}$, $\bm{r}_z\times\bm{r}_{\perp}$ and $\bm{r}'_z\times\bm{r}'_{\perp}$. Hence, the SO $^{\perp }(2)$ vector component of the nonlocal spin density takes the form:

$$\bm{s}_{\perp}(\bm{r}, \bm{r}') = \varrho_{\perp}(z,z',r^2_{\perp},\bm{r}_{\perp}\cdot\bm{r}'_{\per... ...e 2}_{\perp}, \bm{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp}))\bm{r}_{\perp}$$

$$+\varrho_{\perp}'(z,z',r^2_{\perp},\bm{r}_{\perp}\cdot\bm{r}'_{\p... ... 2}_{\perp}, \bm{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp}))\bm{r}'_{\perp}$$

$$+\varphi_{\perp}(z,z',r^2_{\perp},\bm{r}_{\perp}\cdot\bm{r}'_{\pe... ...m{r}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp}))(\bm{r}_z\times\bm{r}_{\perp})$$

$$+\varphi_{\perp}'(z,z',r^2_{\perp},\bm{r}_{\perp}\cdot\bm{r}'_{\p... ...}_z\cdot(\bm{r}_{\perp}\times\bm{r}'_{\perp}))(\bm{r}'_z\times\bm{r}'_{\perp}),$$ (33)

with $\varrho_{\perp}$, $\varrho_{\perp}'$, $\varphi_{\perp}$, and $\varphi_{\perp}'$ being scalar functions. General forms of the local zero-order densities are obtained from Eqs. (31), (32), and (33) by putting $\bm{r}_z =\bm{r}'_z$ and $\bm{r}_{\perp}=\bm{r}'_{\perp}$. This gives:

$$\rho (\bm{r}) = \rho_0(z,r_{\perp}),$$ (34)

$$\bm{s}_z(\bm{r}) = \rho_z(z,r_{\perp})\bm{r}_z,$$ (35)

$$\bm{s}_{\perp}(\bm{r}) = \rho_{\perp}(z,r_{\perp})\bm{r}_{\perp}+\phi_{\perp}(z,r_{\perp})(\bm{r}_z\times\bm{r}_{\perp}),$$ (36)

where $\rho_0$, $\rho_z$, $\rho_{\perp}$, and $\phi_{\perp}$ are arbitrary scalar functions of $z$ and $r_{\perp}=\sqrt{r^2_{\perp}}$. The general form of the remaining scalar densities $\tau (\bm{r})$ and $J(\bm{r})$ is the same as that of Eq. (34).

The components of the differential operator,

$$\bm{\nabla}=\bm{\nabla}_z+\bm{\nabla}_{\perp} ,$$ (37)

have the same transformation properties under SO $^{\perp }(2)$ rotations as the corresponding components of the position vector (29). Hence, the densities $\bm{j}(\bm{r})$, $\bm{J}(\bm{r})$, $\bm{T}(\bm{r})$, and $\bm{F}(\bm{r})$ have the components parallel to the $z$-axis that are invariant under SO $^{\perp }(2)$ and the SO $^{\perp }(2)$ vector components that are perpendicular to the symmetry axis. Therefore, they all take general forms given by Eqs. (35) and (36).

It follows from the definitions (12), (15), and (37) that the components $(\underline{\mathsf{J}})_{az}$ ($a=x,\ y$) of the symmetric traceless spin-current density form the SO $^{\perp }(2)$ vector while the components $(\underline{\mathsf{J}})_{ab}$ ( $a,\ b=x,\ y$) form the SO $^{\perp }(2)$ symmetric traceless tensor. The following four symmetric traceless tensors can be formed with vectors $\bm{r}_z$ and $\bm{r}_{\perp}$:

$$\underline{\bm{r}_z\otimes\bm{r}_{\perp}} = \tfrac{1}{2}(\bm{r}_z\otimes\bm{r}_{\perp}+\bm{r}_{\perp}\otimes\bm{r}_z),$$ (38)

$$\underline{\bm{r}_z\otimes(\bm{r}_z\times\bm{r}_{\perp}}) = \tfrac{1}{2}(\bm{r}_z\otimes(\bm{r}_z\times\bm{r}_{\perp})+(\bm{r}_z\times\bm{r}_{\perp})\otimes\bm{r}_z),$$ (39)

$$\underline{\bm{r}_{\perp}\otimes\bm{r}_{\perp}} = \bm{r}_{\perp}\otimes\bm{r}_{\perp}-\tfrac{1}{2}r^2_{\perp}\mathsf{1}_{\perp},$$ (40)

$$\underline{\bm{r}_{\perp}\otimes(\bm{r}_z\times\bm{r}_{\perp}}) = \tfrac{1}{2}(\bm{r}_{\perp}\otimes(\bm{r}_z\times\bm{r}_{\perp})+(\bm{r}_z\times\bm{r}_{\perp})\otimes\bm{r}_{\perp}),$$ (41)

where $(\mathsf{1}_{\perp})_{ab}=(1-\delta_{az})\delta_{ab}$ ( $a,\ b=x,\ y,\ z$) is the unit tensor in the plane perpendicular to the symmetry axis. Two tensors (38) and (39) transform under SO $^{\perp }(2)$ like vectors perpendicular to the $z$-axis. The tensors (40) and (41) are the SO $^{\perp }(2)$ tensors. Consequently, the general form of the symmetric traceless spin-current density is:

$$\underline{\mathsf{J}}(\bm{r}) = \theta_{z\perp}(z,r_{\perp})\underline{\bm{r}_z\otimes\bm{r}_{\pe... ...{zz\perp}(z,r_{\perp})\underline{\bm{r}_z\otimes(\bm{r}_z\times\bm{r}_{\perp})}$$

$$+ \theta_{\perp\perp}(z,r_{\perp})\underline{\bm{r}_{\perp}\otimes\... ...p}(z,r_{\perp})\underline{\bm{r}_{\perp}\otimes(\bm{r}_z\times\bm{r}_{\perp})},$$ (42)

where $\theta_{z\perp}$, $\theta_{zz\perp}$, $\theta_{\perp\perp}$, and $\theta_{\perp z\perp}$ are scalar functions. We note in passing that in the case of the axial symmetry, all local densities formally look like the spherical symmetric nonlocal fields (20), (21), and (22) with $\bm{r}=\bm{r}_z$ and $\bm{r}'=\bm{r}_{\perp}$, provided that $\bm{r}_z\cdot\bm{r}_{\perp}=0$.