I've been reading about the Kane-Mele model and noticed that it does not include nearest-neighbor spin-orbit coupling, with explanations often pointing to mirror symmetry as the reason. Could someone explain why mirror symmetry in the model leads to the exclusion of nearest-neighbor spin-orbit coupling? I assume, in the Kane-mele model, the basis is the $p_z$ orbitals which are perpendicular to the 2D plane. Considering this, I think the spin-orbit coupling matrix elements shall not differ much between nearest-neighbor and next-nearest-neighour.


  • $\begingroup$ Does physics.stackexchange.com/questions/101892/… provide any useful insight to you? I agree that one should be able to answer this just using symmetries, but I am stuck on that. $\endgroup$ Nov 16, 2023 at 17:37
  • $\begingroup$ @NandagopalManoj Thanks for pointing out the question and answer. The answer merely says the SOC is zero for nearest neighbor coupling without mentioning the reason. $\endgroup$
    – Frank
    Nov 19, 2023 at 6:06
  • $\begingroup$ The intuition is that, for the NNN case, the spin orbit coupling is induced by the ion in the other Sublattice sitting between the two sites that the electron is hopping on. The $L \cdot S$ term (where $L$ is measured with respect to the intermediary ion) is what gives you the Kane-Mele term. For NN hopping, this term cancels out due to reflection symmetry along the line joining the two sites. Does that help? $\endgroup$ Nov 19, 2023 at 8:15
  • $\begingroup$ @NandagopalManoj Why does the reflection symmetry cause the $L\cdot S$ term to vanish? Writing the matrix element of $L\cdot S$ between two nearest neighbor sites ($\langle p_z^{(A)}|L\cdot S|p_z^{(B)}\rangle$), I still don't understand why the coupling is zero... $\endgroup$
    – Frank
    Nov 19, 2023 at 17:48
  • 2
    $\begingroup$ @NandagopalManoj please consider translating your comments into a proper response $\endgroup$
    – Mauricio
    Nov 22, 2023 at 16:02

1 Answer 1


Spin orbit coupling is a relativistic correction to the energy levels of an electron orbiting a nucleus. It tells us that there is an energy contribution that is proportional to $L\cdot S$, where $L$ is the angular momentum relative to the nucleus and $S$ is the spin of the electron. The intuitive picture for the Kane-Mele term goes like this:

Imagine the hopping of an electron from site $A$ to site $B$. We will view this in a path integral type picture, where the electron takes all possible paths from site $A$ to site $B$ and sums the corresponding amplitudes. The majority of the total amplitude comes from paths close to the saddle point path -- close to the straight line connecting $A$ and $B$.

As the electron follows a particular path, it interacts with all the nuclei in the honeycomb lattice of graphene. In particular, let us focus on the spin-orbit $L\cdot S$ terms coming from all the nuclei.

Nearest neighbor case

Now, let us assume $A$ and $B$ are nearest neighbors. We will keep in mind that the lattice has a reflection symmetry where the mirror lies on the line joining $A$ and $B$. Imagine a spin $\uparrow$ particle takes a spacetime path $\mathcal{P}$ from $A$ to $B$. Let this path have an action associated to it $S_\uparrow[\mathcal{P}]$. The total probability amplitude for hopping of a spin $\uparrow$ particle is given by the path integral $$ \Psi_\uparrow(A\to B) = \sum_{\mathcal{P}} e^{iS_\uparrow[\mathcal{P}]}$$ Now consider the opposite spin and the reflected path $\mathcal{P}'$. Reflection symmetry tells us that the SOC contribution from an ion $C$ in the lattice to is $S_\uparrow[\mathcal{P}]$ is exactly equal to the SOC contribution of $C'$ (the reflected counterpart of $C$) to $S_\downarrow[\mathcal{P}']$ -- to see this just draw an arbitrary path from $A$ to $B$ and pick an arbitrary ion $C$ and look at $L\cdot S$ at an instance in the path; repeat for opposite spin with $\mathcal{P}'$ and $C'$. The non-SOC contribution to the action should also be same for $\mathcal{P}$ and $\mathcal{P}'$ (regardless of spin) due to reflection symmetry. Therefore, taking into account the SOC contribution from all ions $C$, \begin{align} \Psi_\downarrow(A\to B) &= \sum_{\mathcal{P}} e^{iS_\downarrow[\mathcal{P}]} \\ &= \sum_{\mathcal{P}'} e^{iS_\downarrow[\mathcal{P}']} \\ &= \sum_{\mathcal{P}} e^{iS_\uparrow[\mathcal{P}]} \\ &= \Psi_\uparrow(A\to B) \end{align} that is, the hopping amplitude for both spins must be exactly the same. The Kane-Mele term is not allowed.

Next-nearest neighbor case

If $A$ and $B$ are next-nearest neighbors, the symmetry that we take advantage of (with mirror axis on the line joining $A$ and $B$) is not present anymore. The two spins can have different hopping amplitudes. This allows for the Kane-Mele term. It is also intruitively clear that there is one ion $C$ that will have the maximal SOC contribution, and there is no equivalent ion $C'$ to cancel this contribution.


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