# Stabilizer groups#

Among the many subgroups of the Pauli group $\group{P}_n$, there is a particular type of subgroup, called the stabilizer group $S$, which satisfy the following properties:

- All pairs of elements in $S$ commute, i.e. for every $P,Q \in S$, we have $PQ = QP$.
- $-I = - I_0 \otimes I_1 \otimes \cdots \otimes I_{n-1}$ is not a part of $S$.

We will see later on why these properties are useful, but for now, let's try to create some examples of these groups so we can build some intuition.

Example 1: Let $P_1 = Z \otimes Z \otimes I$ and $P_2 = Z \otimes I \otimes Z$. Then, we note that

\begin{align} P_1P_2 = I \otimes Z \otimes Z = P_3,\\ P_1^2 = P_2^2 = P_3^2 = I \otimes I \otimes I. \end{align}This tells us that \begin{equation} S = \set{I \otimes I \otimes I, Z \otimes Z \otimes I, Z \otimes I \otimes Z, I \otimes Z \otimes Z}, \end{equation} is closed. Multiplying these elements together in any way does not produce any operator other than these four. Since, the identity is also part of this set, this is a subgroup of $\group{P}_n$.

#### Task 1 (On paper)#

Produce the multiplication table of the subgroup $S$ defined in example 1 above.

$\times$ | $I$ | $P_1$ | $P_2$ | $P_3$ |
---|---|---|---|---|

$I$ | ||||

$P_1$ | ||||

$P_2$ | ||||

$P_3$ |

Further, we claim that $S$ is a stabilizer group. Obviously, we see that $-I$ is not part of this group, so the second property is trivially satisfied. What about the first one? If you produced the multiplicaton table above, correctly, you will note that indeed every pair commutes.

#### Task 2 (On paper)#

Take $P_1 = X \otimes X \otimes I$ and $P_2 = X \otimes I \otimes X$.

Multiply them together in every combination, till you find a set of elements that is closed. The identity should be part of this set. Hence this is another stabilizer group.

Write down the multiplication table of this group, and verify that every pair of element commutes.

## Generators of a group#

This is an opportune moment to introduce the notion of the *generators* of a group. The generators of a group can be used to construct all other elements of the group. For instance, for Example 1, we note that
\begin{align}
I \otimes I \otimes I &= P_1^2 = P_2^2, \\
Z \otimes Z \otimes I &= P_1, \\
Z \otimes I \otimes Z &= P_2, \\
I \otimes Z \otimes Z &= P_1P_2.
\end{align}
Every element in the group is some product of $P_1$ and $P_2$. Therefore, $P_1$ and $P_2$ are called the generators of the group $S$.

Similarly, in Task 2, you created a group generated by $X \otimes X \otimes I$ and $X \otimes I \otimes X$.

The advantage of identifying the generators for stabilizer groups is that they reduce the storage space for specifying a group. This is because of two properties.

- Property of stabilizer groups: All elements commute.
- Property of Pauli operators: $P^2 = 1$ (ignoring elements of the form $\pm \iu P$.

Let $S$ be a stabilizer group, with generators $\set{g_i}_{i=0}^{m-1}$. The first property means that we can write any element $h$ in $S$ as a product of generators. \begin{equation} h = g_0^{j_0} g_1^{j_1} \cdots g_{m-1}^{j_{m-1}}, \end{equation} where we don't care about the order of the generators in the product. The power of $j_i$ on each generator tells us how many times it appears in the product in order to create $h$.

The second property means that $j_i$ can only be $0$ or $1$.

This means that any element $h$ is completely specified by the bitstring $(j_0, j_1, \dots, j_{m-1})$. Recognizing that there are $2^m$ such strings, the following lemma immediately follows.

**Lemma:** If $S$ has $m$ generators, then it has $2^m$ elements.

To work with stabilizer codes, we will be depending highly on generators of the stabilizer groups.

#### Task 3 (On paper)#

Determine the bitstring representation of all elements in $S$ for

- Example 1
- Task 2

## States stabilized by stabilizer groups#

In stabilizer states, we saw that multiple operators can stabilize the same state.

Since, stabilizer groups are just a set of operators, a very natural question to ask is, what states are stabilized by these operators?

Example 1: Above, we created the stabilizer group \begin{equation} S = \set{I \otimes I \otimes I, Z \otimes Z \otimes I, Z \otimes I \otimes Z, I \otimes Z \otimes Z}. \end{equation} Let's try to construct the state simultaneously stabilized by all these operators.

- The identity stabilizes all states.
- We already know that $Z \otimes Z \otimes I, Z \otimes I \otimes Z$ jointly stabilize $\set{\ket{000},\ket{111}}$.
- It's easy to discover that $I \otimes Z \otimes Z$ has stabilizer states $\set{\ket{000}, \ket{100}, \ket{011}, \ket{111}}$.

Hence, the common intersection is $\set{\ket{000},\ket{111}}$. Therefore,

The group $S$ stabilizes the states $\set{\ket{000},\ket{111}}$, and any of their linear combinations.

We note that $\set{\ket{000},\ket{111}}$ is the basis of the repetition code. The connection between $S$ and repetition code will be fleshed out in the next notebook.

### Too much work#

Actually, we are doing too much work in determining the stabilizer states. Each element of $S$ is specified by a product of two generators $P_1$ and $P_2$.

If $P_1$ and $P_2$ both stabilize a state $\ket{\psi}$, then any $h \in S$ (which is just some product of $P_1$ and $P_2$) will also stabilize $\ket{\psi}$.

If $P_1$ or $P_2$ don't stabilize a state, then that state cannot be included in our final set of states.

Taken together this means that we only need to determine states that are stablized by the all generators to determine the states stabilized by the stabilizer group.

**Lemma:** The basis of states stabilized by a stabilizer group $S$, is the intersection of states stabilized by the generators of $S$.