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 $f = Eat$.

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.)

  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.