# Breaking translational invariance on a 1D periodic lattice

I am seeking some clarification on the process of breaking translational symmetry in a bosonic lattice by applying a uniform external magnetic field, which was stated as a fact in this paper: https://arxiv.org/pdf/cond-mat/0608350.pdf. While this paper outlines the approach to a 2D square lattice, I am considering the simpler case of a 1D periodic lattice with equal spacing.

A 1D bosonic lattice with $N$ sites and a lattice spacing of $d$ is described by the Hamiltonian $$\hat H = -t\sum_m (a_m^\dagger a_{m+1} + a_{m+1}^\dagger a_m) - \mu \sum_m a_m^\dagger a_m$$ where $a_m$ is the annihilation operator acting on the $m^{th}$ site, $t$ is the hopping amplitude between nearest neighbours and $\mu$ is the chemical potential. This Hamiltonian is easily diagonalized by a Fourier transform $$a_m = \frac {1}{\sqrt{N}} \sum_k e^{ik(md)}a_k$$ where the sum is taken over all $k$ in the Brillouin zone ($k = \frac{2\pi j}{Nd}$ for $j=0,1,...,N-1$.) Such a transformation yields the diagonalized Hamiltonian $$\hat H = -\sum_k [\mu + 2t \cos(kd)]a_k^\dagger a_k.$$

Now, upon applying some external uniform magnetic field to the lattice, one picks up a Peierls phase $e^{i \phi}$ upon hopping in some direction perpendicular to the applied field. Choosing our gauge to pick up the phase upon hopping in the direction of the 1D lattice, we obtain a modified Hamiltonian $$\hat H = -t\sum_m (e^{-i\phi}a_m^\dagger a_{m+1} + e^{i\phi}a_{m+1}^\dagger a_m) - \mu \sum_m a_m^\dagger a_m.$$ Repeating the same procedure as above, the Hamiltonian is diagonalized as $$\hat H = -\sum_k [\mu + 2t \cos(kd-\phi)] a_k^\dagger a_k.$$ Thus the result of applying an external magnetic field is simply shifting the energy spectrum. I do not see how this breaks translational symmetry as the eigenvectors are seemingly equivalent to the $\phi=0$. Can anybody explain this?

• Where is it claimed that this breaks translational symmetry? – Norbert Schuch Jul 17 '18 at 21:11