$\mathbb{Z}_2$ gauge theory and disorder

Question

I am confused about basics of $\mathbb{Z}_2$ (and likely other) gauge theories and plain disorder. Let $$H=H_F + h\,H_{EM}$$ $$H_F = -t\sum_l (c^\dagger_l \sigma^z_{l,l+1} c_{l+1} + h.c.)$$ be (the 'texbook') chain of length $N$ of spinless fermions $c^{(\dagger)}_l$ at sites $l$, coupled via a $\mathbb{Z}_2$ Peierls factor $\sigma^z_{l,l+1}$ to the gauge field. The pure gauge field's Hamiltonian $H_{EM}$ is not relevant and $h$ is some coupling constant. As usual, $H$ should commute with all local generators $G_l = \sigma^x_{l-1,l} (-)^{n_l} \sigma^x_{l,l+1}$, where $n_l=c^\dagger_l c_l$ is the particle number.

Now my problem(?) is, that from the literature, it is never absolutely clear to me, but often it seems implied, that for $h=0$, i.e. when matter decouples from the gauge field, then, $H_F$ represents a translationally invariant free fermion Hamiltonian, which can be solved by Fourier transformation to produce the standard $\cos(k)$ dispersion.

I understand, that the superselection sectors with $G_l=1$ or $G_l=-1$ for all $l$ are special, but for $h=0$ and instead of the $G_l$, one can also just use that all $[H_F,\sigma^z_{l,l+1}]=0$ and classify w.r.t. $2^N$ (mostly) random and static distributions of $\sigma^z_{l,l+1}=\pm 1$. Therefore I would expect, that most of the eigenstates of $H_F$ are Anderson-localized with energies that have nothing to do with a $\cos(k)$-band.

So, would a proper statement be, that only the gauge vacuum returns the free matter of the un-gauged theory at $h=0$ and moreover, that the fermions in almost all of the remaining superselection sectors are localized. ... Or is there some 'voodoo' I am lacking?

Answer 1

Do I undestand correctly that you want to neglect completely the constraints defined by $G_l$ operators? Then you can indeed diagonalize all $\sigma_{l,l+1}^z$. Within a subspace defined by one set of eigenvalues $\sigma_{l,l+1}^z \psi = (-1)^{\alpha_{l, l+1}} \psi$, $\alpha_{l, l+1} \in \{ 0 , 1 \}$ we have that $H_F$ is the free Hamiltonian with some signs included: $H_F = - t \sum_{l} (-1)^{\alpha_{l, l+1}} c_l^{\dagger} c_{l+1} + \mathrm{h.c.}$ Now redefine operators $c_l$ as $c_l = (-1)^{\theta_l} c_l'$ with some $\theta_l \in \{ 0 , 1 \}$. This does not change the form of canonical anticommutation relations. This redefinition amounts to replacement $\alpha_{l, l+1} \to \alpha_{l , l+1} + \theta_l + \theta_{l+1}$ (addition taken modulo two). Choosing suitable $\theta$ you can eliminate all $\alpha$ (except for one in the case of closed chains -- in this case two inequivalent $\alpha$ correspond to periodic and antiperiodic boundary conditions). The final conculsion is that (except for the periodic/antiperiodic distinction in the closed chain case) the spectrum does not depend on $\alpha_{l, l+1}$ at all. So I think the answer is no, there is no localization phenomenon at play here.