Let $P_n$ denote the Pauli group on $n$ qubits (think of n as a large number).

Let $G=\left<g_1,...,g_n\right> < P_n$ be some abelian subgroup such that each $g_i$ acts on at most $k\ll n$ qubits. Assume that $g_i$ are independent and $-I\notin G$.

  • Does there always exist some unitary $U$ such that $U^\dagger G U=\left<X_1,...,X_n\right>$?

  • If so, what does $U$ look like? Can we ensure $U$ is "simple" enough? More precisely, can we find such a $U$ when we restrict ourselves to constant depth unitary quantum circuits?


1 Answer 1


Any set of commuting stabilizers can be mapped to any other set of commuting stabilizers by applying a unitary from the Clifford group, see http://arxiv.org/abs/quant-ph/9807006.

The Clifford group is generated by CNOT, Hadamard, and the phase gate $\left(\begin{smallmatrix}1\\&i\end{smallmatrix}\right)$, so your $U$'s indeed have a special form.

However, $U$ can't always be chosen to be local. Consider, e.g., the Toric Code on a sphere: It is given by $n$ local and commuting stabilizers, and its ground state (=the joint $+1$ eigenstate of all stabilizers) has global (topological) entanglement, and thus cannot be converted to a product state with constant depth circuits.

  • $\begingroup$ The thing is: you can NOT express any of the globally entangled ground states of the toric code with n independant AND local constraints: If the code is on $n^2$ qubits, then you can describe it's ground space as the stabilized space of at most $n^2-2$ local constraints. Here it's not the case as the stabilized space is unique since we have n commuting pauli elements, and therefore, there is no global entanglement involved. $\endgroup$ Commented Aug 24, 2015 at 12:27
  • $\begingroup$ Although the toric code itself is irrelevant, the spherical "code" (it's not really a code but it's still well defined) looks like the perfect candidate for my question, as it can be described as the unique stabilized space of n local Pauli constraints (stars and plaquettes on some tessellation of the sphere) $\endgroup$ Commented Aug 24, 2015 at 12:28
  • $\begingroup$ @user3001348 That's why I took a toric code on a sphere -- while it is indeed not a code, it still has topological entanglement. $\endgroup$ Commented Aug 24, 2015 at 12:28
  • $\begingroup$ Oups, I missed the sphere part :) Anyways, how do you prove that this state can't be constructed with some local depth circuit? The only proof I know of involves global entanglement wich requires a degenerate ground space $\endgroup$ Commented Aug 24, 2015 at 12:31
  • $\begingroup$ @user3001348 This might take some thought, but I'm sure it'll work out. After all, the state on any subregion of the sphere is no diffent from a contractible region of the toric code state on the torus and, e.g., there is a topological correction to the entanglement entropy. $\endgroup$ Commented Aug 24, 2015 at 12:33

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