# Pauli group#

We are going to study the simplest class of quantum error-correcting codes, called the stabilizer codes. To understand the stabilizer codes, we need to build up some theory.

## The Pauli matrices#

Recall the Pauli matrices along with the identity matrix, $\set{I, X, Y, Z}$. These matrices have the property that multiplying them together in any way, gets us back to the same set of matrices up to a factor of $\pm 1$ or $\pm \iu$.

For instance, \begin{equation} XY = \iu Z, \quad YZ = \iu X, \quad ZX = \iu X. \end{equation}

#### Task 1 (On paper)#

Complete the following multiplication table.

$\times$ | $I$ | $X$ | $Y$ | $Z$ |
---|---|---|---|---|

$I$ | $II = I$ | $IX=X$ | $IY=Y$ | $IZ=Z$ |

$X$ | $XI = X$ | $XY=\iu Z$ | ||

$Y$ | $YI = Y$ | |||

$Z$ | $ZI = Z$ |

The first row and colum have been completed for you, as well as one additional element.

### Commutation relations#

There are one additional property of these matrices that we will use over and over again, and that is the *commutation properties*.

Note that for $P, Q \in \set{I, X, Y, Z}$, we have either

- $P$ and $Q$ commute, i.e. $PQ = QP$, or
- $P$ and $Q$ anti-commute, i.e. $PQ = -QP$.

You can verify this by comparing the $(i,j)$ entry in the table above with the $(j,i)$ one. Either the entries are the same ($i$th and $j$th operators commute) or they differ by a minus sign ($i$th and $j$th operators anti-commute).

### The Pauli group $\group{P}_1$#

What we have discovered that the 16 matrices, \begin{equation} \group{P}_1 = \set{\pm I, \pm \iu I, \pm X, \pm \iu X, \pm Y, \pm \iu Y, \pm Z, \pm \iu Z}, \end{equation} form a group. This means the following things:

- Multiplying any two elements of $\group{P}_1$ gives us another element of $\group{P}_1$. You have explicitly checked this above. This property is called closure (under multiplication) - equivalently, we say the set $\group{P}_1$ is closed under multiplication.
- We can multiply elements of $\group{P}_1$ in any order. Meaning $(XZ)Y = X(YZ)$, where the brackets indicate the order of multiplication. This is true because the elements of $\group{P}_1$ are matrices.
- There is an identity element in $\group{P}_1$.
- For every element in $\group{P}_1$, its inverse is also inside $\group{P}_1$. For example, the inverse of $X$ is $X$ itself. The inverse of $\iu X$ is $-\iu X$.

*Question:* Determine the inverse of every element in $\group{P}_1$.

Note that we have 16 elements here, because there are 4 pauli matrices and 4 possible phase factors ($\pm 1, \pm \iu$).

### The Pauli group $\group{P}_2$#

Now, consider the tensor product of two Pauli matrices, I mean elements such as $X \otimes Z$ and $ Y \otimes Z$. You are already familiar with the multiplication of such elements, \begin{equation} (P_1 \otimes P_2)(P_3 \otimes P_4) = P_1P3 \otimes P_2P_4, \end{equation} and $(X \otimes Z)(Y \otimes Z) = (\iu Z) \otimes I = \iu(Z \otimes I)$.

Suppose, we collect all such elements into a set. How many such elements are there? In $P_1\otimes P_2$, the matrix $P_1$ can be one of four Pauli matrices, and so can $P_2$. Additionally, we have four possible phase factors. In total, we will have $4^{2+1}=64$ elements in a set we will call $\group{P}_2$.

Again, we go over the four properties of groups, and make sure $\group{P}_2$ satisfies them.

- We have confirmed that multipling these 64 elements together in any way gives us one of the 64 elements.
- Multiplication order does not matter as before.
- The element $I \otimes I$ is the identity element.
- We can easily construct the inverse of any element, and it will be one among the 64 elements.

*Question:* Determine the inverse of $\iu (X \otimes Y)$.

### The Pauli group $\group{P}_n$#

It should be clear now, how to extend the above analysis to tensor products of $n$ Pauli matrices, $P_1 \otimes P_2 \otimes \cdots \otimes P_n$. By the same arguments as above, there will be $4^{n+1}$ elements in $\group{P}_n$.

All our discussion of the stabilizer codes will occur in the context of the Pauli group.

## Subgroups of the Pauli group#

A *subgroup* is a subset of the elements of the group with two properties:

- The subset is closed under multiplication.
- For every element in the subset, its inverse element is also in the subset.

These two properties directly imply that the identity must be part of the subgroup, so that multiplying an element with its inverse (both in the subgroup) give another element in the subgroup.

We can find many groups of the Pauli group $\group{P}_n$.

A trivial example is $\set{I}$. Yes, the subset which has only one element. It's easy to see that both properties are satisfied.

Here is an example. Consider $\group{P}_n$, which has elements of the form $P = P_1 \otimes P_2 \otimes P_3$. This group has a subgroup, which is defined as follow.

Let $P_2 = I$, so that $Q = P_1 \otimes I \otimes P_3$, and consider all such elements in $\group{P}_3$, i.e., we have the subset \begin{equation} G = \set{ Q \in \group{P}_3 \text{ such that } Q=P_1 \otimes I \otimes P_3}. \end{equation}

It is quite easy to see that

- multiplying any two elements in $G$ yields another element in $G$.
- the inverse of $Q=P_1 \otimes I \otimes P_3$ is just $Q^{-1} = P_1^{-1} \otimes I \otimes P_3^{-1}$, which should also be in $G$.

Therefore $G$ is a subgroup of $\group{P}_3$.

*Question:* Consider the set
\begin{equation}
G = \set{I \otimes I \otimes I, X \otimes I \otimes I, I \otimes Z \otimes I, X \otimes Z \otimes I}.
\end{equation}
Does this form a subgroup of $\group{P}_3$?