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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?

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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 $f = Eat$.

Now, for practical purposes, it is more convenient to choose a gauge in which the Hamiltonian is translationally invariant exactly, not just up to gauge transformations. This is possible for a uniform electric field, but not in general. In fact, it is easily seen that for a uniform magnetic field there is no such gauge. (Actually, in the case of uniform electric field there's a problem too, in the sense that we have obtained explicit spatial translation symmetry at the cost of breaking explict time translation symmetry.)

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    $\begingroup$ Can you explicitly show the gauge transformation resulting in $Ex\to Ex+Ea$? $\endgroup$ – Chong Wang Oct 25 '16 at 3:57
  • $\begingroup$ @ChongWang I added it above. $\endgroup$ – Dominic Else Oct 27 '16 at 0:37

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