Pauli gates and the Hadamard gate

The Pauli matrices

$$X=\begin{pmatrix}0&1\\1&0\end{pmatrix},\quad Y=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\quad Z=\begin{pmatrix}1&0\\0&-1\end{pmatrix}.$$

X exchanges $\lvert0\rangle$ and $\lvert1\rangle$. Y sends $\lvert0\rangle$ to $i\lvert1\rangle$ and $\lvert1\rangle$ to $-i\lvert0\rangle$. Z leaves $\lvert0\rangle$ alone and negates $\lvert1\rangle$.

Each Pauli squares to I. On a basis state, Z’s minus may be only a global phase; on a superposition it can change a relative phase.

Hadamard changes the basis

$$H\lvert0\rangle=\lvert+\rangle,\qquad H\lvert1\rangle=\lvert-\rangle.$$

Because $H^2=I$, it also maps $\lvert+\rangle$ back to 0 and $\lvert-\rangle$ back to 1. This is why H before Z readout implements an X-basis measurement.

A worked identity

Trace $HZH\lvert0\rangle$: the first H gives plus, Z gives minus, and the second H gives 1. On $\lvert1\rangle$, the sequence gives 0. Thus $HZH=X$ on a basis, hence on all states by linearity.

Practise

Use the qubit explorer to compare X and Y on $\lvert0\rangle$. Their outcome probabilities agree, but their statevectors differ by a global phase. Then test them on a superposition: agreement on one input is not agreement as operators.

Reference: IBM Quantum — Single systems.