At the critical point, is Kramers-Wannier duality a unitary symmetry of the model?

Question

I have in mind the transverse ising model and its (self-dual) generalizations, such as

$$H_{TI} = \sum_i \sigma^z_{i}\sigma^z_{i+1} + h \sigma^x_{i}$$ and $$H_{SDANNI} = \sum_i (\sigma^z_{i}\sigma^z_{i+1}+\Delta \sigma^z_{i}\sigma^z_{i+2}) + h( \sigma^x_{i}+\Delta \sigma^x_{i}\sigma^x_{i+1})$$

These models are self-dual under a Kramers-Wannier mapping, which is often used to show that $h=1$ is the location of the critical point. There are often some subtleties with this transformation to do with the tails of the mapping and the degeneracy of ground states.

---

At the critical point of $h=1$, does there exist a unitary $U$ implementing the duality such that $U H_{TI} U^\dagger = H_{TI}$ and $U H_{SDANNI} U^\dagger = H_{SDANNI}$?

My motivation to ask this is fairly vague. I keep feeling that these models have an extra symmetry at the critical point of $h=1$ relative to their $h\neq 1$ counterparts, and I suspect it is related to the fact that they are dual to themselves at the critical point. In particular, I want to implement this symmetry in exact diagonalization. My mind says that it's unlikely one can find a unitary operator (or even any linear operator) to implement Kramers-Wannier duality, but my heart hopes that it exists and it's known.

Answer 1

The self-duality of the TFI model can not be a unitary map, because it maps a Hamiltonian with spontaneous Ising symmetry breaking in the ground states ($h<1$) to one in the paramagnetic state with a unique ground state ($1/h>1$), and unitary maps ought to preserve degeneracy. In fact the same argument shows that it can not even be an invertible linear transformation.

Another way to see that KW duality is very different from a symmetry is to think about what happens in a closed system, with periodic boundary condition. You can show that the operator mapping, as written, only works in a sector of the Hilbert space with a given total $\mathbb{Z}_2$ charge. The total charge gets translated to the $\mathbb{Z}_2$ flux of the transformed model.

As Heidar mentioned in the comment, you can implement KW duality as an non-invertible defect line in the (1+1)d theory (most conveniently in the 2D classical Ising model, see https://arxiv.org/abs/1601.07185, or just thinking purely in terms of the CFT, as thoroughly discussed in https://arxiv.org/abs/1802.04445).

Now you may ask whether it is possible to write down the map explicitly in some way, at the level of states. I don't think anyone has done this and it would be interesting to try. There are two things you can do:

1. Use Jordan-Wigner transformation to fermionize the TFI chain, into a chain of Majorana fermions. Then the KW duality becomes a lattice translation in the fermionic representation, which is a unitary symmetry. In principle you can write down how the translation acts on states.

2. In general, it is possible sometimes to represent these dualities in 1+1d using a matrix product operator (MPO) (the general theory is laid out in https://arxiv.org/abs/1511.08090, although in the context of 2D topological phase).

Answer 2

I fully agree with what Meng Cheng has written about the impossibility for duality to be a unitary transformation in the case of periodic boundary conditions. Nevertheless, if one doesn't mind breaking translational and parity symmetries, it is possible to define duality as unitary transformation and find self-dual hamiltonian. Let $|s_1,\ldots,s_N\rangle$ are eigenvectors of $\hat{\sigma}_j^z$ operators. Id est $$ \hat{\sigma}_j^z |s_1,\ldots,s_N\rangle = s_j |s_1,\ldots,s_N\rangle. $$ Let's define operator $\hat{U}$ in the following way: $$ \hat{U}|s_1,\ldots,s_N\rangle = |t_1,\ldots,t_N\rangle, $$ where $t_j = s_{j-1}s_j$ for $j = 2,\ldots,N$ and $t_1 = s_1$. It is not hard to see that the following relations are valid $$ \hat{U}\ \hat{\sigma}_{j-1}^z\hat{\sigma}_j^z\ \hat{U}^\dagger = \hat{\sigma}_j^z,\quad j = 2,\ldots,N, $$ $$ \hat{U}\ \hat{\sigma}_{1}^z\ \hat{U}^\dagger = \hat{\sigma}_{1}^z, $$ $$ \hat{U}\hat{\sigma}_j^x\hat{U}^\dagger = \hat{\sigma}_j^x\hat{\sigma}_{j+1}^x,\quad j = 1,\ldots,N-1, $$ $$ \hat{U}\hat{\sigma}_N^x\hat{U}^\dagger = \hat{\sigma}_N^x. $$ Now, if $$ \hat{H} = \hat{\sigma}_1^z + \sum_{j=2}^N\hat{\sigma}_{j-1}^z\hat{\sigma}_j^z + \sum_{j=1}^N\hat{\sigma}_j^x, $$ then $$ \hat{U}\hat{H}\hat{U}^\dagger = \sum_{j=1}^N\hat{\sigma}_j^z + \sum_{j=1}^{N-1}\hat{\sigma}_j^x\hat{\sigma}_{j+1}^x + \hat{\sigma}_N^x. $$ Transformed hamiltonian is essentially the initial one up to the spin rotation and spatial reflection unitary transformations.

Update. There is no need to break parity symmetry. Hamiltonian $$ \hat{H}_2 = \sum_{j=2}^{N}\hat{\sigma}_{j-1}^z\hat{\sigma}_j^z + \sum_{j=1}^{N-1}\hat{\sigma}_j^x. $$ is self-dual. It is also invariant under $\hat{\sigma}_j^z \to -\hat{\sigma}_j^z$, $\hat{\sigma}_j^x \to \hat{\sigma}_j^x$ transformation.