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This question concerns the very interesting paper: ''Symmetry protected topological (SPT) orders and the group cohomology of their symmetry group'' by Chen et al., http://arxiv.org/abs/1106.4772

In this paper, in Section II.F. (left column of page 7), it is said that the unitary transformation U that goes from the trivial SPT state to a non-trivial SPT state whose wavefunction is written as a product of group cocycles is local. Indeed, citing the paper, ``Then, using the local unitary transformation $U=\prod_{\vartriangle} \nu_3(1,g_i,g_j,g_k) \prod_{\triangledown} \nu_3^{-1}(1,g_i,g_j,g_k)$, we find that the above ideal ground state wave function is given by $\Phi = U \Phi_0$ and...'' where $\Phi_0$ is the trivial SPT state and $\Phi$ the non-trivial one.

Given the fact that $U$ is symmetry-preserving, my question is: how can U be local, since, by definition, two different SPT phases cannot be linked by a symmetry-preserving local unitary transformation (as stated in the paper)?

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The trivial and non-trivial SPT states both are symmetric under on-site unitary symmetry transformations. The trivial and non-trivial SPT states can be mapped into each other by local unitary transformations (the $U$ in you question). Although such a $U$ is a local unitary transformation, 1. it is not on-site, 2. it is not the on-site unitary symmetry transformations, 3. it is not symmetric under the on-site symmetry transformations.

If two states can be mapped into each other by symmetric local unitary transformations, then the two states have the same SPT order.

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  • $\begingroup$ Thanks very much for your answer. According to your second sentence, this means that, if the $U$ defined above is local, it must be not-symmetric. I don't see how $U$ can be not-symmetric, since the product of cocycles is invariant under symmetry. It therefore seems to me that $U$ is non-local, unlike what is being said in the paper. $\endgroup$ – Thomas May 18 '15 at 7:08
  • $\begingroup$ @Thomas The unitary transformation $U$ is local but not on-site. It is a product of operators over the simplexes (triangles), but not a product of operators on each site, therefore it fits in the rule 1. of Prof. Wen's answer above. Also be careful that $\nu_3(1,g_i,g_j,g_k)\to \nu_3(1,g g_i,g g_j,g g_k)$ is not a symmetric transformation of the cocycle, because $g$ only acts on the spacial simplex but not touching the first element $1$. So whether or not $U$ is symmetric is not obvious at the first glance. $\endgroup$ – Everett You May 18 '15 at 19:37
  • $\begingroup$ Thanks for Everett You's comment which is very clear. I modified my answer to make it more clear. $\endgroup$ – Xiao-Gang Wen May 21 '15 at 15:16
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It seems that you thought $\Phi_0$ is a trivial SPT while $\Phi$ is nontrivial. This does not make sense without defining the symmetry transformation. The fact that $\Phi=U\Phi_0$ means both are product state ($U$ is a honest local unitary transformation). However, symmetry transformation is defined differently in the two states: for $\Phi$ the symmetry action is simply $|g_i\rangle\rightarrow |gg_i\rangle$, while on $\Phi_0$, the symmetry action is much more complicated (pull back the simple one on $\Phi$ by $U$). So both are nontrivial SPT phases if $\nu_3$ is a nontrivial cocycle, but with different definitions of symmetry transformations. In other words, if we start from $\Phi_0$ with "trivial" symmetry transformation $|g_i\rangle\rightarrow |gg_i\rangle$, transforming to $\Phi_0$ by $U$, one still gets a trivial SPT phase.

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There is an important distinction here which I feel the other answers have not addressed. One can check that the unitary $U$ as given in the question is indeed symmetric in the sense that $[U,W(g)] = 0$, where $W(g)$ is the representation of the symmetry. It is also a local unitary, in the sense that one can find a local (possibly time-dependent) Hamiltonian $H(t)$ such that $U$ is the time evolution of $H$, $$U = \mathcal{T} \exp\left(-i\int_0^1 H(t) dt\right).$$ However, one can prove that the Hamiltonian $H(t)$ cannot be chosen to be symmetric, i.e. necessarily $[H(t),W(g)] \neq 0$ for some $t$. Only local unitaries that can be generated by symmetric Hamiltonians have a right to be called symmetric local unitaries.

(If we instead define a local unitary to be a finite-depth quantum circuit, then the corresponding statement is that $U$ cannot be written as a finite-depth quantum circuit in which each layer is individually symmetric.)

A slightly different (but equivalent) way to think about it is that $U$ is only symmetric on a system with no boundaries. If we try to implement it on a system with boundary we find that it must break the symmetry. A truly symmetric local unitary should respect the symmetry regardless of boundary conditions.

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