Kramers-Wannier Duality in the Transverse Ising Model

Question

This question is about the Kramers-Wannier duality transformation in the (1+1)-d transverse field Ising model. The Hamiltonian is

$$H(J,g)= -J \sum_i \sigma_i^z \sigma_{i+1}^z -g \sum_i \sigma_i^x$$

and I am concerned about the system on a ring with $N$ sites, with either periodic boundary conditions $\sigma_1^z \sigma_N^z$ or anti-periodic boundary conditions $-\sigma_1^z \sigma_N^z$.

How do I define the duality transformation for the system with PBC or APBC? For open boundary conditions with $N$ sites, the answer, as given in the answer to this question here, 

$$\mu_i^z = \prod_{j \leq i}^N \sigma_j^x \text{ for } i = 1,\cdots, N$$

$$ \mu_i^x = \sigma_i^z \sigma_{i+1}^z \text{ for } i = 1,\cdots,N-1$$ and

$$ \mu_N^x = \sigma_N^z $$,

which gives the same anticommutation relations as $\sigma$. 

But using this transformation on the system on a ring means that the boundary term $\sigma_1^z \sigma_N^z$ gets mapped to $\mu_1^x \mu_2^x \cdots \mu_{N-1}^x$, which is ugly and doesn't give a local term in the dual spin system. I know that the answer involves mapping different sectors of the Hilbert space separately (decomposed in boundary conditions + symmetry charge sectors). in fact the exact correspondence is that

1) (PBC, Symmetry = $\prod_i^N \sigma_i^x$ = +1) with Hamiltonian (J,g) gets mapped to (PBC, Symmetry = +1) with Hamiltonian (g, J)

2) (PBC, Symmetry = -1) with H(J,g) gets mapped to (APBC, Symmetry = +1) with H(g,J)

3) (APBC, Symmetry = +1) with H(J,g) gets mapped to (PBC, Symmetry = -1) with H(g,J)

4) (APBC, Symmetry = -1) with H(J,g) gets mapped to (APBC, Symmetry - -1) with H(g,J)

but the duality transformation as defined above for the open boundary conditions model doesn't make this mapping explicit.

The duality transformation in the PBC/APBC case is quite likely awkward (but doable) - but it is precisely this exact mapping which I am interested in. Try as I might, I have not been successful in writing the transformation down. 

Could someone help me make this transformation explicit? Or, if the duality transformation as written down for the OBC case works, how to show that the mapping I've written down between the sectors in the Hilbert space happens?

Answer 1

The ordinary KW duality has a subtlety that the dual Ising symmetry is actually a gauge $\mathbb{Z}/2$ symmetry, not a global one. This introduces some difficulties on a circle since it has topology. I think this turns your nonlocal operator into a gauge transformation.

Let me argue this point about the duality. Consider the ferromagnetic phase of the 1+1d Ising model. This has a doubly degenerate ground state. One usually says it is dual to the paramagnetic phase of another 1+1d Ising model. This statement makes no sense, because the latter has no ground state degeneracy. The proper statement is that the paramagnetic phase of the 1+1d Ising model is dual to the GAUGED ferromagnetic phase of the 1+1d Ising model.

For a discussion of these subtleties and some other fun stuff, I recommend Yoni Bentov's wonderful paper http://arxiv.org/abs/1412.0154