0
$\begingroup$

I have introduced onsite disorder in Graphene. How will I write the momentum space Hamiltonian for such a system?(System size is 25 x 25 unit cells) If I don't add onsite disorder then we can write the momentum space Hamiltonian(k space) easily. The tight binding Hamiltonian in terms of creation and annihilation operator reads as, $$H = -t \sum_{<i,j>} c_i^{\dagger} c_j + \sum_{i} w_i c_i^{\dagger} c_i$$ where,the first quantity on r.h.s is the nearest neighbor hopping term, t is the nearest neighbor hopping strength, $c_i^\dagger$ is the creation operator, $c_i$ is the annihilation operator. The second quantity is for onsite disorder where, $w_i$ is the randomly generated onsite energy at site $i$.

The important thing here is that the second quantity on r.h.s breaks the translation symmetry.

$\endgroup$
0
$\begingroup$

In momentum-space, the hopping term will be diagonal in momenta and the on-site disorder term will mix/scatter different momenta. (In other words, momentum is no longer conserved when disorder is present.) With periodic boundary conditions, one natural way of expressing the disorder term is $$ \sum_i w_i c_i^\dagger c_i = \sum_{k,k'} W_{k,k'} c_k^\dagger c_{k'}, $$ where the scattering matrix is $$ W_{k,k'} = \frac{1}{N} \sum_{j=1}^N w_j e^{i(2\pi/N)j(k-k')}. $$

$\endgroup$
3
  • $\begingroup$ Ok. then, is there any way to get the single particle eigen state(real space) for such a system ? $\endgroup$ – Hope Ful Apr 12 '20 at 20:37
  • $\begingroup$ @HopeFul The standard approach is to express the Hamiltonian as a matrix, and diagonalize it numerically. $\endgroup$ – Anyon Apr 12 '20 at 23:20
  • $\begingroup$ Can you kindly clarify, how will I write this matrix, I am struggling to do it for several days. Complete mathematical steps you can share $\endgroup$ – Hope Ful Apr 13 '20 at 9:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.