It's well known that the classical Hamiltonian governing the dynamics of a charged particle in a static magnetic field is
$$ S_{cl}[x,\dot{x}] = \int_0^t dt' \frac{1}{2}m\dot{\vec{x}}^2 + e\vec{A}(x)\cdot \dot{\vec{x}}$$
with corresponding Hamiltonian (after appropriately accounting for primary constraints),
$$H = \frac{(m\dot{\vec{x}} - e\vec{A})^2}{2m} = \frac{p_{c}^2}{2m}$$
Now suppose we want to approximate the continuum theory with a lattice model (lattice constant $a$, say). It's canonical to approximate this by a tight-binding model with Peierls substitution,
$$H' = \sum_{\langle ij \rangle} J a^\dagger_i a_j \exp\left[-iea \vec{A} \cdot (\vec{r}_i - \vec{r}_j)\right] + h.c.$$
Let's use 2D square lattice, with uniform perpendicular magnetic field $\vec{B} = B_0 \hat{z}$. We are at liberty to choose Landau gauge, and set $\vec{A} = B_0 x \hat{y}$. The tight-binding Hamiltonian is easy to diagonalize, giving
$$H = \sum_{\mathbf{q}} \omega(\mathbf{q})a^\dagger_{\mathbf{q}} a_{\mathbf{q}}$$ $$\omega(\mathbf{q}) = 2J \left[\cos(q_x) + \cos(q_y + eB_0)\right] $$
In my train of thought, the Lorentz force should cause there to be something like a Hall effect in the propagator. However, calculating
$$\mathcal{M}_{fi} = \langle x_{\mathbf{r}} | e^{-iHt} | x_{\mathbf{r'}} \rangle = \langle 0| a_{\mathbf{r}} e^{-iHt} a^\dagger_{\mathbf{r'}} |0\rangle$$ $$ = \sum_{\mathbf{k},\mathbf{k'}} \langle 0 | a_k e^{-i \mathbf{k} \cdot \mathbf{r}} \prod_{\mathbf{q}} \left\{ \exp\left[ -i t \omega(\mathbf{q}) a^\dagger_{\mathbf{q}} a_{\mathbf{q}}\right] \right\} e^{i \mathbf{k}' \cdot \mathbf{r'}} a^\dagger_{\mathbf{k}'} |0\rangle$$ $$ =_{\text{single-particle}} \sum_{\mathbf{k},\mathbf{k'}} \langle 0 | a_k e^{-i \mathbf{k} \cdot \mathbf{r}} \exp\left[ -i t \omega(\mathbf{q}) \right] a^\dagger_{\mathbf{k}'} a_{\mathbf{k'}} e^{i \mathbf{k}' \cdot \mathbf{r'}} a^\dagger_{\mathbf{k}'} |0\rangle$$ $$ = \sum_k \exp\left[-i\mathbf{k}(\mathbf{r}-\mathbf{r}') - i t \omega(\mathbf{k}) \right]\langle 0 | 0\rangle $$
This is an issue because the final integral over the Brilloun zone is separable:
$$\mathcal{M}_{fi} = \int_{-\pi/a}^{\pi/a} \exp[-ik_x\Delta x - 2it\cos(k_x)] dk_x \int_{-\pi/a}^{\pi/a} \exp[-ik_y\Delta y - 2it\cos(k_y +eB_0)] dk_y $$
So regardless of what this matrix element evaluates to, there is no asymmetry $y \to -y$ other than a physically irrelevant phase difference.
My best guess of what's going on is that we're missing some important paths from the path integral (namely, the curved ones) and are incapable of approximating them by discrete, square jumps. In principle, a locally curved path will have lower energy cost than a discrete-hop approximation to it, in a situation analogous to the famous pi=4 proof based on successive approximations to a circle. I naively expected there to be some sort of destructive interference brought about by the plaquette fluxes, but this is not the case.
The questions are:
- What is responsible for the discretized tight-binding model's failure to reproduce the classical result that moving charged bodies curve in magnetic fields? Is my explanation above believable?
- The classical Lorentz force (as can be obtained from stationary points of the original action) is fundamentally an effect arising from the coherence of phase sums along worldlines. Is it possible to see this in a lattice model?
