I am currently reading the beginning of the paper Topological Quantum Memory and I am confused about one part:

The check operators generate an Abelian group, the code’s stabilizer.

When I learned group theory, I learned that if I have a group action $M: G \to G$, where $M$ is the action map and $G_4$ is the group, the stabilizer of the group is the set of all elements of the group that are each sent to itself.

What is the group action here? Or is there another definition of the stabilizer in the context of quantum computing?


1 Answer 1


A group action of a group $G$ on a set $X$ is a map $\phi:\, G\times X \rightarrow X$ such that $$ \phi(e,x) = x, \quad\text{and}\quad \phi(g, \phi(h,x) ) = \phi(gh,x), $$ for all $x\in X$ and $g,h\in G$ ($e\in G$ is the identity element). Usually, one simply writes $g\cdot x \equiv \phi(g,x)$.

Given such a group action, the stabilizer subgroup of an element $x \in X$ is defined as $$ G_x := \left\{ g\in G \; | \; g\cdot x = x \right\}, $$ i.e. the subgroup which fixes $x$. Likewise, for any subset $Y\subset X$, we can define the stabilizer $G_Y$ as the intersection of all $G_x$ for $x\in Y$. That is, $G_Y$ is the subgroup of $Y$ which fixes $S$ point-wise.

Note that you can also reverse the construction and ask: Given a subgroup $H$ of $G$, what is the subset $X_H\subset X$ which is stabilized by $H$? This is exactly the subset of fixed points of $H$.

Here, we consider $X=\mathbb C^{2^n}$ and the unitary group $G=\mathrm U(2^n)$ with its natural action/representation by matrix multiplication. Then, the stabilized set of a subgroup $H<\mathrm U(2^n)$ is always a subspace, namely the common 1-eigenspace of all $h\in H$. For this to be non-empty, it is sufficient that all $h\in H$ commute (and that all $h\in H$ have an eigenvalue 1). Thus, we'll consider Abelian subgroups in the following.

Note that given a subspace $V\subset \mathbb C^{2^n}$, its stabilizer $\mathrm U(2^n)_V \simeq \mathrm U(V^\perp)$ is generally huge. Albeit, you always get away with a much smaller group $\mathcal{S}_V \subset \mathrm U(2^n)_V$ of order $|\mathcal{S}_V| = \dim V$ which nevertheless stabilizes $V$.

In the context of quantum codes, we call $V\subset \mathbb C^{2^n}$ the code space or simply code and want to regard it as an embedding of a $k$-qubit Hilbert space $\mathbb C^{2^k}$ into the $n$-qubit Hilbert space $\mathbb C^{2^n}$. Hence, we have $\dim V = 2^k$, and we can always find a minimal stabilizer subgroup with $|\mathcal{S}_V|=2^k$.

For stabilizer codes, we restrict to $G=\mathcal{P}_n\subset \mathrm U(2^n)$, the $n$-qubit Pauli group, defined as $$ \mathcal{P}_n = \left\{ i^k \sigma_1\otimes\dots\otimes\sigma_n \; | \; k\in\mathbb Z_4, \sigma_j \in \{\mathbb I, X, Y, Z \} \right\}. $$ The stabilizer subgroups of stabilizer codes than have the form $\mathcal S = \langle P_1,\dots,P_k\rangle$, where $P_i\in \mathcal{P}_n$ are commuting and independent generators. Finally, there's the additional condition that $-\mathbb I \notin \mathcal S$ (since the stabilized subspace would otherwise be $\{0\}$). This is equivalent to requiring that this subgroup contains only Hermitian operators.

To see this, note that any element of the Pauli group is either Hermitian or anti-Hermitian. Any Hermitian element $P$ has order two, $P^2 = PP^\dagger = \mathbb I$, as $P$ is also unitary. For an anti-Hermitian element, $(iP)^2 = - \mathbb I$, hence those have order four. This also shows that $\mathcal S$ cannot contain anti-Hermitian elements as otherwise $- \mathbb I \in \mathcal S$. This argumentation also shows that $|\mathcal S| = 2^k$.

  • $\begingroup$ This is a very detailed answer. Thank you very much!! One question: In the last sentence of your post, you mention that "For the Pauli group, this is equivalent to the requirement that all elements of the group are Hermitian." Can you please kindly explain the equivalence? I do not completely understand. $\endgroup$
    – Debbie
    Mar 18, 2022 at 14:35
  • $\begingroup$ @Debbie You're welcome. I have updated my answer. $\endgroup$ Mar 18, 2022 at 14:57
  • 1
    $\begingroup$ Very nice and detailed answer! $\endgroup$
    – Lior
    Mar 18, 2022 at 15:56
  • $\begingroup$ @MarkusHeinrich Thanks again. $\endgroup$
    – Debbie
    Mar 18, 2022 at 17:41
  • $\begingroup$ can't you have a subgroup of non-commutying unitaries with a non-empty common 1-eigenspace? Trivial example being in $\mathbb{C}^3$ the set (and subgroup) of matrices of the form $1\oplus U$ for all $U\in\mathbf{U}(2)$, which contains $1\oplus X$ and $1\oplus Y$ which do not commute, but also stabilises the subspace $\mathbb{C} e_1$ $\endgroup$
    – glS
    Mar 18, 2022 at 19:02

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