Dear Qingtian,
First, I would like to precise that I am not a specialist of Graphene or spin hall effect (neither Kwant) so you need to take the
solution I give you with precaution and you need to do tests knowing the expected results.
The
amazing thing with Kwant is that the process of defining the Hamiltonan
becomes simpler once you precise the Tight
binding model.
binding model.
So, when you have the lattice expression of the Hamiltonian forget the x and y directions and work as in a graph (sites
neighbors and links).
The starting point to your problem is may be the simpler case of electrons with spin in honeycomb lattice (without SO interaction).
You should understand that the state | n,m, sigma> =|n,m>
|sigma> ,where is the the tensorial product. Your hamiltonian becomes H=h
1 . h is the spinless hamiltonian in graphene and 1 is the 2x2
identity matrix. So now, with the definition of the tensorial product (matricial version) you can understand why we need to change in the spinless hamiltonian all
the elements (sites and hopping) by hij*1 (hij---->identity *hij) to obtain the whole Hamiltonian with spins.
With the interaction presented in the model of Kane & Mele, you need to do the same and understand how is the Hamiltonian is
written as a Tensorial product.
you will understand that in the Hamiltonian
with the second nearest neighbors you need to change the elements hij by
matrices
as follows:
as follows:
hij ------> 1j*t2*Vij*Sigma_z
you need just to be careful with the hopping Vij which are site-dependent .
For this you need to just define the Vij in the clockwise: Vij=+1 by choosing the directions for the next nearest neighbors as
follows: (1,0) (-1,-1) (0,-1)
we do this for the two sublattices a, b.
The other anticlockwise hoppings Vij are directly insured by hermeticity.
The same procedure is done for the leads.
The results at the end seems to me at least "not bizarre" but you need to check this with specialists.
The program following these remarks is included with this mail.
Regards
Adel Abbout