Does Spin-Orbital Coupling (SOC) have no effects on orbitals with different $l$? SOC term is:
$$H_{\mathrm{SOC}}=\alpha \vec{L} \cdot \vec{S}=\frac{\alpha}{2}\left(L^{+} \sigma^{+}+L^{-} \sigma^{-}+L^{z} \sigma^{z}\right)$$
from spin Orbital Coupling matrix in p-orbital basis, we know that the matrix elements between orbitals with different $l$ will vanish, e.g.
$$\langle s|H_{SOC} |p_{x/y/z}\rangle=0$$
since $L^{z/+/-}$ cannot change $l$. But in practice, some reference in research will gives SOC gap between orbitals with different $l$. I am confused of it.
Example:
Following figure comes from this paper,

we can find that there exists a SOC gap between $p$ and $d$ orbitals with different $l$ in the system with $C_4$ symmetry.However, according to the analysis of Hamiltonian elements, $\langle p|H_{SOC} |d\rangle$ always vanishes.
 A: Thanks for sharing this interesting work. Let me try to explain as far as I understand myself.
I think your understanding of SOC is incomplete; you forget about the importance of the orbital hybridization. For example, let us go back to one of the simplest situation: the SOC in graphene (or i.e. Kane-Mele model). In this situation, a Dirac cone is formed at the $K$ point by the $p_z$ orbitals, where the SOC is opening a small energy gap at this Dirac point. So actually, if you think about this, this involves some interaction between two $p_z$ states? Now, in the Kane-Mele model, this SOC effect is considered as some next-nearest-neighbor (NNN) hopping process with following steps:
\begin{equation}
p_{z\uparrow}^{A}\xrightarrow{SOC}p_{+\downarrow}^{A}\xrightarrow{V}s_{\downarrow}^{B}\xrightarrow{V}p_{+\downarrow}^{A}\xrightarrow{SOC}p_{z\uparrow}^{A}
\end{equation}
with $p_\pm=p_x\pm ip_y$, and with $A,B$ representing the two sublattices of honeycomb structure. Now, in this process, you see that the true SOC interaction is happening on-site (because after all SOC is an atomic effect), while this SOC interaction is accompanied with a spin-flip, where the electrons undergo a virtual hopping (SOC coupling) from the $\pi$ band ($p_z$) to $\sigma$ band ($p_\pm$), which is indeed allowed by your $H_{SOC}$. Because of the strong $sp_2$ hybridization, the electron can virtually hop to the $s$-orbital at the neighboring site and repeat the process in reverse. The resulting SOC effect we observe in graphene is thus actually only allowed by this second-order virtual hopping process (two times a virtual hopping $V$). This is actually illustrating the importance of all the bands that are not lying close to the Fermi level. In this case the $p_\pm$ orbitals hybridized with the $s$-orbitals; they allow this process to happen. Now a similar situation is also happening in many other topological insulators, where there is $s$-$p$ hybridization and band crossing of them, which subsequently results in the SOC lifting these crossing degeneracies. I expect a similar situation is happening in your example. I hope that makes sense to you?
Important to mention here, is the fact that due to the second-order nature of this process, the SOC effects becomes very small (e.g. SOC-induced gap in graphene few meV).
