Large gauge transformation in $\mathrm{U}(1)$ flux threading argument

Question

In Oshikawa's flux threading argument for the $\mathrm{U}(1)\times T$ Lieb-Schultz-Mattis (LSM) theorem, the author defined a so-called large gauge transformation $$U=\exp\left(i\frac{2\pi}{L}\sum_{\mathbf{r}} x_\mathbf{r} n_\mathbf{r} \right) \tag{1}$$ where $L$ is the number of sites in the $x$-direction, the sum is over all lattice sites $\mathbf{r}$, $x_\mathbf{r}$ is the $x$-coordinate and $n_\mathbf{r}$ is number operator at $\mathbf{r}$ (note that the argument is independent of particle statistics). Assume periodic boundary condition in $x$. Also assume the ground state is gapped and no spontanous symmetry breaking occurs. First, a $\mathrm{U}(1$) flux of $2\pi$ is adiabatically embeded in the system. The Hamiltonian is changed from $H(\Phi=0)$ to $H(\Phi=2\pi)$, even though the spectrum is unchaged. The author then argued that $$UH(\Phi=2\pi)U^{-1}=H(\Phi=0), \tag{2}$$ i.e. $U$ transforms the twisted Hamiltonian back to itself.

How does one justify eqn. (2) for a generic $\mathrm{U}(1)$ and translation symmetric form of $H(\Phi)$? Also, can someone give a non-trivial example to demonstrate this?

Answer 1

Eq (2) has nothing to do with translation invariance. It also has very little to do with the actual form of the Hamiltonian. Let us just consider a one-dimensional chain and let the site index runs from $j=1$ to $j=L$. We can then introduce background U(1) gauge potential $A_{j,j+1}$. The "flux", or the holonomy $\Phi$ is defined as $\Phi=\sum_{j=1}^L A_{j,j+1}$. Suppose we perform a gauge transformation $e^{i\sum_j \theta_j n_j}$, then

$A_{j,j+1}\rightarrow A_{j,j+1}+\theta_{j}-\theta_{j+1}$

$H(2\pi)$ in Oshikawa's paper is defined in the gauge $A_{j,j+1}=\frac{2\pi}{L}$, under the large gauge transformation $A_{j,j+1}=0$ for $1\leq j<L$, and $A_{L,1}=2\pi\equiv 0$, which gives $H(0)$.

If you need an example, just take the nearest-neighbor hopping model

$ H=-\sum_j e^{i A_{j,j+1}}a_j^\dagger a_{j+1} + \cdots $

where $n_j=a^\dagger_j a_j$.