1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Where is it claimed that this breaks translational symmetry? $\endgroup$ – Norbert Schuch Jul 17 '18 at 21:11
1
$\begingroup$

In this case 1D is qualitatively different than 2D. Phases in the 1D model with nearest neighbour tunneling always can be "gauged out" by adding a phase to a definition of annichilation(creation) operators (with the exception of periodic boundary conditions when some total phase could be accumulated on the whole ring -- giving so called twisted boundary conditions). So simplification you have chosen to use will not work here.

$\endgroup$

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.