Current operator in continuum model of graphene

For the graphene hamiltonian with NNN hopping, the wavefunctions are of the form: $(\psi_A ,\psi_B)^T$. The current from A(i) to B(j) site in the lattice model is given by: $$J_{ij}=\mathrm{i}t(c^{\dagger}_ic_j-c^{\dagger}_jc_i)$$ where $t$ is the hopping parameter.

1) How can this operator be generalised to the continuum model? Is it same as the general way in which Dirac current is defined?

2) What does the following convey:(Does it in some sense capture the A to B current?) $\hat{O}\propto\mathrm{i}\langle\psi_B^{\dagger}\psi_A-\psi_A^{\dagger}\psi_B\rangle$

• where did you get this expression for current? The graphene tight-binding Hamiltonian is normally more complicated that this and includes a pseudo-spin quantum number representing sites A and B. Been out of the field for a few years so perhaps I'm just not up-to-date. – Greg Petersen Mar 23 '16 at 4:35
• This link may help:physics.stackexchange.com/questions/70088/… – vbj Mar 23 '16 at 4:49

1 Answer

The general way to find the current opperator is to gauge the U(1) symmetry and take the derivative of the Hamiltonian with respect to the gauge field $a_{ij}$ or $a_\mu$ and then turn off the gauge field:

$$\text{lattice: }J_{ij}=\left.\frac{\partial H}{\partial a_{ij}}\right|_{a_{ij}\to 0},\text{ continuum: }J^\mu=\left.\frac{\partial H}{\partial a_\mu}\right|_{a_\mu\to 0}.$$

On the lattice, after gauging the U(1) symmetry, $H=-\sum_{ij} t (e^{\mathrm{i} a_{ij}} c_{i}^\dagger c_{j}+h.c.)$, so the derivative with respect to $a_{ij}$ gives $J_{ij}= -t(\mathrm{i} c_{i}^\dagger c_{j}+h.c.)$. In the field theory, $H=\int \mathrm{d}^d x\; c^\dagger(\mathrm{i}\partial_\mu+a_\mu)\gamma^\mu c$, so the derivative with respect to $a_\mu$ gives $J^\mu=c^\dagger \gamma^\mu c$. You only need to figure out how to relate the lattice Hamiltonian to the continuum Hamiltonian. A simple method is to Fourier transform to the momentum space, expand around the gapless momentum points, and transform back to the real space.In general, all operators like $\mathrm{i}(\psi_i^\dagger \psi_j-\psi_j^\dagger\psi_i)$ can be interprete as a current.