Jordan-Wigner transformation for lattice models without $U(1)$ symmetry

Question

The Jordan-Wigner transformation is a powerful approach to studying one-dimensional spin models. The following dictionary between spin operators and creation/annihilation operators for fermions allows us to recast spin models in terms of fermions:

$$S_{j}^{+} = c_{j}^{\dagger}\exp\left[i\pi\sum_{k<j}n_{k}\right],\hspace{12pt} S_{j}^{z} = n_{j}-\frac{1}{2}$$

For example, a Heisenberg spin-1/2 chain in a magnetic field becomes

$$H = J\sum_{j}\vec{S}_{j}\cdot\vec{S}_{j+1} - h\sum_{j}S_{j}^{z} \longrightarrow \sum_{j}\left[\frac{J}{2}\left(c_{j}^{\dagger}c_{j+1}+c_{j+1}^{\dagger}c_{j}\right) + J\left(n_{j}-\frac{1}{2}\right)\left(n_{j+1}-\frac{1}{2}\right) -hn_{j}\right].$$

The original spin model has a $U(1)$-symmetry associated with rotations about the magnetic field axis; in the fermion model, this is translated into conservation of total fermion number. There are several consequences of this for, e.g., quantum critical points induced by an applied magnetic field. Moreover, this is why the fermion theory is local in spite of the apparent non-locality in the Jordan-Wigner dictionary.

My question is, what can we say about such a model in which the $U(1)$ symmetry is broken from the beginning?

There are many choices which can accomplish this, but one simple choice is to include a Heisenberg anisotropy. This destroys our freedom to choose $h\propto\hat{z}$, so the spin model will look something like

$$H = J\sum_{j}\left(\vec{S}_{j}\cdot\vec{S}_{j+1} + \lambda S_{j}^{y}S_{j+1}^{y}\right) - \sum_{j}\vec{h}\cdot\vec{S}_{j}$$

The $U(1)$ symmetry is certainly broken now; moreover, if $h$ has components in the x or y directions, the corresponding fermion model appears to have a nonlocal structure, since a single operator $S^{\pm}$ gets turned into a long-range string of fermion operators. Are there worked out examinations of such models in the literature? All examples of the Jordan-Wigner mapping I have seen feature number conservation and spatial locality, but it seems to me violations of these constraints are very common.

Answer 1

The case of your model with the field along $\hat{x}$ is known as the transverse-field XXZ chain, see e.g. Dmitriev et al., One-dimensional anisotropic Heisenberg model in the transverse magnetic field, JETP 95, 538 (2002). The combination of transverse and longitudinal (along $\hat{y}$) was discussed in Dmitriev and Krivnov, Phys. Rev. B 70, 144414 (2004). In the latter case, part of the Jordan-Wigner-transformed Hamiltonian is non-local. The authors avoid that issue by implementing a mean-field theory, and finding the non-local part vanishes in the mean-field approximation. As for number non-conserving terms, in the mean-field theory they can always be handled through a Bogoliubov transformation.

Answer 2

The Jordan-Wigner transformation doesn't require a U(1) symmetry. All it requires is a $\mathbb Z_2$ symmetry.

You can easily see this if you reverse engineer your way from a general fermionic Hamiltonians. Since all terms in the Hamiltonian must have an even number of fermionic creation/annihilation operators, the resulting spin model obtained from a local fermionic Hamiltonian will be a local spin Hamiltonian. It will, however, not have a U(1) symmetry, unless the original fermionic Hamiltonian had a U(1) symmetry (that is, it was particle number preserving). All what remains is a $\mathbb Z_2$ symmetry which originates from the fermionic parity symmetry.

You can see this in your example, where you mention an anisotropy in the Heisenberg interaction. Here, $S^x_1S^x_2$ is mapped to $(a^\dagger_1-a_1)(a^\dagger_2 a_2)$, while $S^y_1S^y_2$ is mapped to $-(a^\dagger_1-a_1)(a^\dagger_2+a_2)$ (the exact signs depend how you put your Jordan-Wigner string). If you add those two with the same weight, you get a particle number preserving hopping Hamiltonian. Otherwise, you get a fermionic Hamiltonian which additionally contains pairing terms, yet is still local.

The story is different if in your example, you don't focus on the anisotropy in the Heisenberg interaction (which you highlight), but instead on the magnetic field: If you align the magnetic field in an arbitrary direction, also the $\mathbb Z_2$ symmetry (flipping the spin in a direction orthogonal to the field, e.g. in the example above $y\to -y$) is potentially broken, and the field is mapped to a non-local fermionic operator. Indeed, combining such a field with an anisotropic Heisenberg interaction likely breaks the $\mathbb Z_2$ altogether, while for an isotropic Heisenberg model with a general field direction, such a mapping still exists (you just have to redefine the Jordan-Wigner transformation such that $a$ and $a^\dagger$ are lowering and raising operators in the preferred basis of the field).

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Since you ask for examples: The most standard example would be the solution of the 1D transverse field Ising model by mapping it to free fermions (which then contain both pairing and hopping terms).