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.

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 elements of $\group{P}_1$ satisfy the group axioms:

  • 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.
  • Given three elements $P, Q, R$ in the $\group{P}_1$, we have $(PQ)R = P(QR)$. This is true because the elements of $\group{P}_1$ are matrices. This property is called associativity of multiplication.
  • There is an identity element, $I$, 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 $\group{P_1}$ has 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, meaning elements such as $X \otimes Z$ or $ Y \otimes Z$. You are already familiar with the multiplication of such elements. Given $P = P^{(0)} \otimes P^{(1)}$ and $Q = Q^{(0)} \otimes Q^{(1)}$ in $\group{P}_2$, \begin{equation} PQ = (P^{(0)} \otimes P^{(1)})(Q^{(0)} \otimes Q^{(1)}) = P^{(0)}Q^{(0)} \otimes P^{(1)}Q^{(1)}. \end{equation}

Example: $(X \otimes Z)(Y \otimes Z) = (\iu Z) \otimes I = \iu(Z \otimes I)$.

Suppose, we collect all such elements into a set. How many such elements are there? In $P^{(0)}\otimes P^{(1)}$, the matrix $P^{(0)}$ can be one of four Pauli matrices, and so can $P^{(1)}$. 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 axioms 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 continues to be associative.
  • The element $I \otimes I$ is the identity element.
  • We can easily construct the inverse of any element $P^{(0)} \otimes P^{(1)}$ as ${P^{(0)}}^{-1} \otimes {P^{(1)}}^{-1}$, 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, $P^{(0)} \otimes P^{(1)} \otimes \cdots \otimes P^{(n-1)}$. 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 that itself form a group. Meaning the subset satisfies all four of the axioms of the group.

Example: Consider the $\set{I,X} \subset \group{P}_1$. We can check all four axioms.

  • $IX = X$, $I^2 = I$, $X^2 = I$ are all in the subset, so closure of multiplication is satified.
  • Associativity continues to hold.
  • The identity is in the subset.
  • The inverse of $I$ is $I$ and the inverse of $X$ is $X$. So all elements have their inverse in the subset as well.

Hence, this subset is a subgroup of $\group{P}_1$.

Question: Does $\set{I,\iu X} \subset \group{P}_1$ form a subgroup of $\group{P}_1$?

Example: Consider $\group{P}_3$, which has elements of the form $P = P^{(0)} \otimes P^{(1)} \otimes P^{(2)}$. This group has a subgroup, which is defined as follow.

Let $P^{(1)} = I$, so that $Q = P^{(0)} \otimes I \otimes P^{(2)}$, and consider all such elements in $\group{P}_3$, i.e., we have the subset \begin{equation} \group{H} = \set{ Q \in \group{P}_3 \text{ such that } Q=P^{(0)} \otimes I \otimes P^{(2)}}. \end{equation}

To check that $\group{H}$ is indeed a subgroup, we check all four properties.

  • Closure under multiplication. Its quite clear that multiplying two elements of the form $Q$ will yield another element of the form $Q$, and we included all elements of this form in $\group{H}$, so this property is satisfied.
  • Associativity continues to hold.
  • The identity $I \otimes I \otimes I$ is of the form $Q$, so it is in $\group{H}$.
  • The inverse of $Q$ is just ${P^{(0)}}^{-1} \otimes I \otimes {P^{(2)}}^{-1}$, which is still in the form of $Q$. So all elements of $\group{H}$ also have their inverse inside $\group{H}$.

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

Question: How many elements are there in $G$?

For finite-sized groups, such as the Pauli group, we don't actually have to check all four axioms.

Theorem: Let $\group{H}$ be a nonempty finite subset of a group $\group{G}$. Then $\group{H}$ is a subgroup of $\group{G}$ if $\group{H}$ is closed under multiplication.

So, we only have to check the closure axiom, and if it holds so will the other three axioms. This makes the task of checking if a subset is a subgroup much easier.

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 $\group{P}_3$?