Hi, Sorry, as far as I know momentum operators can only be present in form of polynomial expression.
You could define Hamiltonian using Taylor expression with some cutoff but this may lead to many and long hoppings. Best, Rafał On Sat, 11 Jan 2020, 09:51 Barış Özgüler, <ozgu...@wisc.edu> wrote: > Hi all, > > I tried to use kwant.continuum.discretize for my Hamiltonian which > includes sqrt(k_x * k_x + k_y * k_y). Not surprisingly, I get the error > "TypeError: can't convert expression to float". Is there a way to > discretize the term including "sqrt" using kwant.continuum.discretize or > another discretizer of Kwant? > > Thank you, > > Barış >
