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Toric code is one of the few exactly solvable model in condensed matter, however, like the paper (http://arxiv.org/abs/1104.5485) that uses SU(2) slave fermion to "solve" Kitaev's honeycomb model, is it possible to use the same approach to mean-field toric code? It seems that for each plaquette or star operator in toric code's Hamiltonian, it needs 8 fermions since they are 4-spin interaction terms, which is pretty daunting if we try the mean-field approach. Any comments will be welcomed.

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  • $\begingroup$ Minor comment to the post (v1): In the future please link to abstract pages rather than pdf files, e.g., arxiv.org/abs/1104.5485 $\endgroup$
    – Qmechanic
    Commented Oct 31, 2015 at 15:40

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Yes, there is a similar mean-field approach to toric code. It was introduced in Xiao-Gang Wen's book Section 10.2 and 10.3, known as the Wen plaquette model.

In the toric code model, the qubits are defined on the links, which are not very convenient in terms of the spin liquid language. So the first step is to redefine a smaller square lattice, such that the qubits are on the sites. This illustrated in the following figure as going from the left (toric code model) to the right (Wen plaquette model).

Toric code to Wen plaquette

The toric code model (on the left) is given by the following Hamiltonian

$$H=-\sum_v Q_v-\sum_f B_f,$$

where $v$ stands for the vertex and $f$ stands for the face (plaquette) on the square lattice. The qubits are defined on the link. The stabilizers $Q_v$ and $B_f$ depicted in the figure are given by

$$Q_v=\sigma_1^z\sigma_2^z\sigma_3^z\sigma_4^z, B_f=\sigma_5^x\sigma_6^x\sigma_7^x\sigma_8^x.$$

The same definition is just translated through out the lattice. In the Wen plaquette model (on the right), the qubits are defined on the site. We represent the Pauli operators on a single qubit by drawing lines through the site: $\sigma^z = $, $\sigma^x = $, such that a ring around the plaquette stands for the operator

$O_p = $$=\sigma_1^z\sigma_2^x\sigma_3^z\sigma_4^x.$

It is not hard to figure out that under a basis transformation of the qubits, the toric code Hamiltonian becomes

$$H=-\sum_p O_p,$$

where $p$ stands for the plaquette of the smaller square lattice. The plaquettes in the smaller lattice are naturally partitioned into red and blue plaquettes following the check-board pattern. The operator $O_p$ surrounding a red (blue) plaquette corresponds to the $Q_v$ ($B_f$) operator in the toric code model.

Now every qubit can be considered as a spin-1/2 on each site, and we are ready to run the standard SU(2) projective construction for the spin liquid. We introduce four Majorana spinons on each site (equivalent to two complex spinons), denoted as $\chi_i^0,\chi_i^1,\chi_i^2,\chi_i^3$. They can be arranged in a matrix form as

$$F_i=\sigma^0\chi_i^0+\mathrm{i}\sigma^1\chi_i^1+\mathrm{i}\sigma^2\chi_i^2+\mathrm{i}\sigma^3\chi_i^3.$$

The SU(2) spin and SU(2) gauge transformation simply acts on $F_i$ as left and right SU(2) rotations respectively. Then the onsite spin operator can be fractionalized as

$$\sigma_i^a = \mathrm{Tr}F_i^\dagger \sigma^a F_i, (a=1,2,3),$$

under the gauge constraint that $\chi_i^0\chi_i^1\chi_i^2\chi_i^3=1$. Or explicitly, we have $\sigma_i^z=\mathrm{i}\chi_i^0\chi_i^3-\mathrm{i}\chi_i^1\chi_i^2$ and $\sigma_i^x=\mathrm{i}\chi_i^0\chi_i^1-\mathrm{i}\chi_i^2\chi_i^3$, which can be graphically represented as follows:

enter image description here

Under this fractionalization, the ring operator $O_p$ simply becomes a product of 8 Majorana fermions around the plaquette,

$$O_p=\chi_1\chi_2\chi_3\chi_4\chi_5\chi_6\chi_7\chi_8.$$

The arrangement of the Majorana fermions are shown below.

enter image description here

One can then take the mean field decomposition in the following manner

$$O_p=\langle\chi_1\chi_2\rangle\langle\chi_3\chi_4\rangle\langle\chi_5\chi_6\rangle\langle\chi_7\chi_8\rangle.$$

The corresponding mean field Hamiltonian $H_\text{MF}$ simply describes Majorana fermions dimerized across each link.

$$H_\text{MF}=\sum_{\langle i j \rangle\in\text{x-link}}\mathrm{i}u\chi_i^1\chi_j^3+\sum_{\langle i j \rangle\in\text{y-link}}\mathrm{i}u\chi_i^0\chi_j^2$$

enter image description here

So the fermion spectrum is fully gapped, as each pair of Majorana fermions forms a bounding state independently. Because Majorana fermions are real, the mean field ansatz also breaks the SU(2) gauge structure down to $Z_2$, thus resulting in a $Z_2$ spin liquid, described by the emergent $Z_2$ topological order at low energy.

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  • $\begingroup$ Thanks Everett for the detailed answer. By the way, do you know where can I look into for reference about all the results listed in your last paragraph? $\endgroup$
    – huyichen
    Commented Oct 31, 2015 at 4:45
  • $\begingroup$ @huyichen They are in Xiao-Gang's book Sec. 10.2. $\endgroup$ Commented Oct 31, 2015 at 5:03
  • $\begingroup$ What is the relation of Majorana fermions dimerized and the spectrum being gapped? $\endgroup$
    – huyichen
    Commented Oct 31, 2015 at 20:17
  • $\begingroup$ @huyichen I have added some more figures and words to explain. You can solve the mean-field Hamiltonian to see the spectrum is indeed fully gapped and actually also flat. $\endgroup$ Commented Oct 31, 2015 at 21:57
  • $\begingroup$ Is u in the mean field theory a constant from the other expectation values or should it depend on the position $ij$ of the link? $\endgroup$
    – huyichen
    Commented Nov 1, 2015 at 2:24

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