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The Pauli group, $P_n$, is given by $$P_n=\{ \pm 1, \pm i\}\otimes \{ I,\sigma_x,\sigma_y,\sigma_z\}^{\otimes n}$$ Abelian subgroups of this which do not contain the element $(-1)*I$ correspond to a stabilizer group. If there are $r$ generators of one such subgroup, $\mathcal{G}$, then the $+1$ eigenstate has $2^{n-r}$ basis elements.

This then leads to the natural question of whether we have that $r\le n$ and how can it be proved (either way)?

I guess a (valid?) proof would be along the lines of that if $r \gt n$ we would have a bias of fractional dimension - this is not allowed so $r\lt n$. But if one exists I would prefer a proof considering only the group properties and not the space which it acts on.

Quantum spaghettification
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1 Answers1

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Consider a subgroup $G $ of the Pauli group with at least one operator that acts non-trivially on some qubit.

  • Given any qubit $j $, for which the group contains an operator $S_j $ which acts on $j $ non-trivially, there is a Clifford group operator $C_j $ such that $C_j S_j C_j^\dagger =Z_j $, acting on qubit $j $ alone. (Why?)

  • If $G_j = \{ C_j S C_j^\dagger \,\vert\, S \in G \}$ and $G $ is abelian, then $G_j = \langle Z_j \rangle \oplus G'_j$, where $G'_j $ does not act on qubit $j $. (Why?)

  • By induction, we can transform any abelian subgroup on $n $ qubits to a group with at most $n+1$ generators, where up to $n $ of them act on a single qubit with a $Z $ operator. (And what then would the remaining one be?)

From this, we can prove that a stabiliser group on $n $ qubits has at most $n $ generators; and with only a little more work, we can show that a stabiliser group with $r $ generators stabilises a subspace of dimension $2^{n-r} $.

Niel de Beaudrap
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