# How can I identify momentum eigenstates in a tight binding model with degenerate energy eigenstates?

Summary:

I numerically diagonalize a tight binding Hamiltonian to get energy eigenvectors, some of which are degenerate. However, the numerically diagonalized degenerate eigenvectors are not necessarily eigenvectors of momentum as well. How can I numerically get eigenvectors of both momentum and the Hamiltonian?

More Details:

I am using a single orbital tight binding model in graphene, and applying a magnetic field so that Landau Levels appear. Specifically, I am using the tight binding model

$$H = t \sum_{r, r' \in NN}a^{\dagger}(r)b(r')e^{i\theta_{r, r'}} + h.c.$$

where $$\theta_{r,r'} = q \int_r^{r'}A(r) \cdot dr$$. A is the vector potential, and q is the particle charge. $$a^\dagger, a, b^\dagger, b$$ are the creation and annihilation operators on the A and B sublattices of graphene. Picking the Landau Gauge, $$A = -B y \hat x$$, and $$\theta$$ becomes $$\theta_{r, r'} = \frac{1}{2} q B (x'-x)(y'+y)$$

I want to find exact solutions to this Hamiltonian, so I numerically diagonalize the tight binding model on a lattice with periodic boundary conditions. As expected, I get Landau levels. Of course, each Landau level has high degeneracy. Therefore when I numerically diagonalize the tight binding Hamiltonian, I get many eigenvectors with the same eigenvalue.

Translational symmetry in the x-direction is maintained, so $$k_x$$ should be a good quantum number. Additionally, translation by $$\Delta_y = \frac{4\pi}{qB}$$ leave the Hamiltonian invariant as well, so $$k_y$$ should be a good quantum number as well, although the Bruillon zone for valid $$k_y$$ is reduced from the case where there is no magnetic field.

When I numerically diagonalize the tight binding Hamiltonian, I get eigenvectors of the Hamiltonian, but not necessarily eigenvectors with quantum numbers $$k_x$$ and $$k_y$$. Is there a way for me to numerically find simultaneous eigenvectors of $$p_x$$, $$p_y$$, and $$H$$?

Alternatively, is there a way to express the momentum operator $$p_x$$ and $$p_y$$ in the tight binding basis? If I can express the operators $$p_x$$ and $$p_y$$ in the tight binding basis, I can easily find simultaneous eigenvectors of $$p_x$$, $$p_y$$, and $$H$$ numerically.

I am aware that the low-energy graphene Hamiltonian can be expanded around the K and K' points, and exact solutions for the landau levels as a function of $$k_x$$ can be obtained. However, I need full, exact solutions to the complete tight binding Hamiltonian so this is not helpful for me.

Recall that when you diagonalize your Hamiltonian, that corresponds to changing basis into that of $$a_E, a^\dagger_E$$, which are the annihilation and creation operators of particles with energy $$E$$, meaning your Hamiltonian becomes
$$H = \sum_E E\ (a^\dagger_E a_E + b^\dagger_E b_E).$$
Similarly you can find your momentum operator as $$p_a = \sum_{p} p\ a^\dagger_p a_p$$, and similarly for your $$b$$ particles. If you then Fourier transform these operators from momentum to position you will have your momentum operator in the tight binding basis.