Let us say that we modify the familiar Wen plaquette model so that only one kind of plaquette operator is in the Hamiltonian which is a product of $\sigma_z$s, what kind of topological order or state is this system in?

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    $\begingroup$ The plaquette term in Wen's model is like $\sigma^z\sigma^x\sigma^z\sigma^x$, and the same term for every plaquette. So what do you mean to keep only one kind of plaquette operator? Also, Wen's plaquette model may not be familiar to many people, so it is better to give a reference. $\endgroup$ – Meng Cheng Nov 11 '15 at 20:16
  • $\begingroup$ As you know, Wen's plaquette model is nothing but the toric code and you can transform it into a version with plauqette operators $\sigma^z$ $\sigma^z$ $\sigma^z$ $\sigma^z$ on even sublattice and $\sigma^x$ $\sigma^x$ $\sigma^x$ $\sigma^x$ on odd sublattice. I am here asking what if you are left with only one on all plaquettes? $\endgroup$ – huyichen Nov 11 '15 at 20:55
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    $\begingroup$ You should probably mention Kitaev's toric code instead of Wen's plaquette model to be more clear. If you only have only $\sigma^z$'s, then you don't have enough stabilizers. So there will be an extensive ground state degeneracy (something like $2^{N_\text{plaq}/2}$), labelled basically by the eigenvalues of $\sigma^x \sigma^x \sigma^x \sigma^x$ on odd plaquettes. It is not a gapped phase at all. $\endgroup$ – Meng Cheng Nov 11 '15 at 21:23
  • $\begingroup$ @MengCheng Could you elaborate on why the phase is not gapped? $\endgroup$ – huyichen Nov 12 '15 at 0:25
  • $\begingroup$ @MengCheng If you replace all $\sigma^x$ products on the (odd) plaquettes by $\sigma^z$ products -- I guess this is what the OP means? -- then you still have the same number of stabilizers. However, it is easy to see that the ground space degeneracy is still extensive (e.g., you can choose all spins in a column to be equal, then any column can have any value.) $\endgroup$ – Norbert Schuch Nov 12 '15 at 17:06

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