Does electric field break discrete translational symmetry or not?

To solve a problem with an electric field in a lattice, one can look at the following Hamiltonian: $$H=\frac{p^2}{2m}+Ex$$. However, this Hamiltonian does not respect discrete translational symmetry. So in some literature this Hamiltonian is used: $$H=\frac{(p-eA)^2}{2m}$$, where $A$ is time dependent and is $A=Et$. This way, discrete translational symmetry are preserved.

However, I think whether a Hamiltonian has translational symmetry is a physical thing and should not change with gauge. Then the above formalism really puzzles me. Any comments?

The physical statement is that the Hamiltonian is translationally invariant up to gauge transformations. Gauge transformations relate two different descriptions of the same physical system, so a system satisfying this property is translationally invariant from a physical point of view. This property is in fact true for both of the Hamiltonians given in the question. For the first Hamiltonian, we note that a translation $x \to x + a$ is the same as shifting the scalar potential by a constant: $Ex \to Ex + Ea$, and a constant shift in the scalar potential can be expressed in terms of a gauge transformation. Recall that a general gauge transformation for the gauge potential $(\varphi,\textbf{A})$ is $$\varphi \to \varphi + \partial_t f, \quad \textbf{A} \to \textbf{A} - \nabla f$$ for some function $f(\textbf{r},t)$. In this case we set $f = Eat$.
• Can you explicitly show the gauge transformation resulting in $Ex\to Ex+Ea$? – Chong Wang Oct 25 '16 at 3:57