# Painless Paulis

February 18, 2021. Quick tips for avoiding nasty algebra with Pauli matrices.

#### Introduction

In elementary quantum computing, you spend a lot of time doing messy, explicit calculations with $2\times 2$ matrices. Sometimes this is unavoidable, but sometimes, there are shortcuts which let us bypass this mess. Let’s remind ourselves of the basic objects, the Pauli matrices:

$\sigma_1 = X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \\ \end{bmatrix}, \quad \sigma_2 = Y = \begin{bmatrix} 0 & -i \\ i & 0 \\ \end{bmatrix}, \quad \sigma_3 = Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \\ \end{bmatrix}.$

We won’t really need these! Instead, we’ll only be using the algebraic relations

$\sigma_a \sigma_b = \delta_{ab}I + i \epsilon_{abc}\sigma_c,$

where $I$ is the $2\times 2$ identity matrix, $\delta_{ab}$ is $1$ if $a = b$ and $0$ otherwise, and $\epsilon_{abc}$ is $1$ when $abc$ is an even permutation of $123$, $-1$ if it’s odd, and $0$ otherwise. We are also using the Einstein summation notation, where the repeated index $c$ is summed over. As special cases, we have

$X^2 = Z^2 = I, \quad XZ = -XZ. \tag{1} \label{ops}$

This will be our key hack!

One of the most important gates in universal quantum computing is the Hadamard gate, since it is used to generate entanglement from unentangled input states. Our basic observation is that

$H = \frac{1}{\sqrt{2}} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \\ \end{bmatrix} = \frac{1}{\sqrt{2}}(X + Z).$

We can first check that the inverse of $H$ is itself, i.e. $H^2 = I$. This follows from (\ref{ops}), since

\begin{align*} H^2 & = \frac{1}{2}(X + Z)^2 \\ & = \frac{1}{2}(X^2 + XZ + ZX + Z^2) \\ & = \frac{1}{2}(X^2 + XZ - XZ + Z^2) = \frac{1}{2}(X^2 + Z^2) = I. \end{align*}

Another important property of $H$ is that it takes the basis states $|0\rangle$ and $|1\rangle$ to eigenstates of $X$. To see this, we first note from (\ref{ops}) that

$X(X + Z) = X^2 + XZ = Z^2 + XZ = (X + Z)Z.$

Hence, using the fact that $Z|n\rangle = (-1)^n|n\rangle$ for $n = 0, 1$, we get

$XH|n\rangle = \frac{1}{\sqrt{2}}X(X + Z)|n\rangle = \frac{1}{\sqrt{2}}(X + Z)Z|n\rangle = (-1)^nH|n\rangle.$

Thus, $H|n\rangle$ is an eigenstate of $X$ with eigenvalue $(-1)^n$, with $|+\rangle = H|0\rangle$ and $|-\rangle = H|1\rangle$. Since $H^2 = I$, we can run this argument backwards to conclude that $H|\pm\rangle$ return the eigenstates of $Z$. This may seem trivial, but combining these two results implies that the eigenstates of $H$ itself can be written

$|H_\pm\rangle = \frac{1}{\sqrt{2}}(|+\rangle \pm |0\rangle).$

Let’s check:

$H|H_\pm\rangle = \frac{1}{\sqrt{2}}(H|+\rangle \pm H|0\rangle) = \frac{1}{\sqrt{2}}(|0\rangle \pm |+\rangle) = \pm|H_\pm\rangle.$

I think this algebraic approach is nicer than explicitly solving an eigenvalue equation. If we use $H$ to make observations, then it’s also easy to check the probability of observing $|H_\pm\rangle$ for a state written in terms of the computational basis, $|\psi\rangle = \alpha |0\rangle + \beta|1\rangle$. We just use known overlaps:

\begin{align*} \langle H_\pm |\psi \rangle & = \frac{1}{2}\left[\alpha\langle +|0\rangle \pm \langle 0|0\rangle + \beta \langle +|1\rangle \pm \langle 0|1\rangle \right]\\ & = \frac{1}{2}\left[\frac{\alpha}{\sqrt{2}} + \frac{\beta}{\sqrt{2}} \pm 1\right]\\ \Longrightarrow \quad p(H = \pm 1) & = |\langle H_\pm |\psi \rangle|^2 = \frac{1}{4}(\alpha + \beta \pm \sqrt{2})^2. \end{align*}

And if you’re wondering about the overlaps $\langle +|n\rangle$, there’s a shortcut for computing those too:

$\langle + | n\rangle = \langle 0 | H |n\rangle = \frac{1}{\sqrt{2}}\langle 0|(X+Z)|n\rangle = \frac{1}{\sqrt{2}}(\langle 1|n\rangle + (-1)^n\langle 0|n\rangle) = \frac{1}{\sqrt{2}}.$

#### Easy entanglement

So, maybe you don’t want to measure $H$ very often. We do use it all the time to entangle things, with the canonical example being the generation of a maximally entangled Bell pair. This is achieved using Hadamard and a CNOT gate, which acts on computational basis states as

$\text{CNOT}|m\rangle |n\rangle = |m\rangle |m\oplus n\rangle,$

where $\oplus$ is addition modulo $2$, also called XOR.