[Kwant] continuous vs discrete leads

[Patrik Arvoy]

Dear kwant users,

If I want to attach a continuous lead instead of its discrete counterpart
to a discrete system, I get the error message 'Use site_family(1, 2)
instead of site_family((1, 2))!'.

The discrete lead is according to the following

    prim_vecs=tinyarray.array([(a,0.,0.),(0.,a,0.),(0.,0.,a)])

    offset1=tinyarray.array((-0.7, 0.0, 0.0))
    lat1=kwant.lattice.Monatomic(prim_vecs, offset1, norbs=no)


    syst[lat1(0, 0, -1)] = e1*sigma_0
    syst[amorphous_lat(0), lat1(0, 0, -1)] = tl*sigma_0
    syst[amorphous_lat(N), lat1(0, 0, -1)] = tl*sigma_0

    sym = kwant.TranslationalSymmetry([0, 0, -a])
    dn_lead = kwant.Builder(sym, conservation_law=-sigma_z)

    if discrete:
        sym = kwant.TranslationalSymmetry([0, 0, -a])
        dn_lead = kwant.Builder(sym, conservation_law=-sigma_z)

        dn_lead[lat1(0, 0, -2)] = e1*sigma_0
        dn_lead[lat1.neighbors()] = t0*sigma_0
        syst.attach_lead(dn_lead)

Now if I replace the 'if discrete:' part with the following

    if continuous:
        t00=0.0
        Leadham =
"t00*sigma_0*k_x**2+t00*sigma_0*k_y**2-t0*sigma_0*k_z**2+(2*t0+e1)*sigma_0"
        template = kwant.continuum.discretize(Leadham grid_spacing=a)
        dn_lead.fill(template, lat1, lat1(0, 0, -2))
        syst.attach_lead(dn_lead)

I have to add k_x and k_y with zero coefficient (t00) to solve the symmetry
complain (is there a neater way?)

where is the error message ('Use site_family(1, 2) instead of
site_family((1, 2))!') originated from?

Thanks
Patrik