Pauli Matrices

The gates $X$ , $Y$ , and $Z$ that you have applied in the last three lessons each correspond to a $2 \times 2$ matrix. Together they are called the Pauli matrices, named after the physicist Wolfgang Pauli. Knowing their explicit form lets you read off how each gate transforms any state vector by ordinary matrix–vector multiplication.

The three matrices

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

A few patterns worth memorising:

All three are Hermitian ( $M = M^\dagger$ , i.e. $M$ equals its own conjugate transpose) and unitary ( $M M^\dagger = I$ ). As a consequence, every Pauli gate is its own inverse.
$X$ and $Z$ are real, with only $0$ s and $\pm 1$ s. $Y$ is the odd one out — it introduces the imaginary unit $i$ .
The diagonal of $Z$ reads $(+1, -1)$ , recording that $|0\rangle$ and $|1\rangle$ pick up phases $+1$ and $-1$ respectively under $Z$ .

Reading the action from the matrix

Matrix–vector multiplication connects the matrix entries to the gate's action. For $X$ acting on a general qubit $\alpha|0\rangle + \beta|1\rangle$ :

X \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \beta \\ \alpha \end{pmatrix},

so $X$ swaps the two amplitudes — exactly the bit-flip you observed in the simulator.

For $Z$ :

Z \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} \alpha \\ -\beta \end{pmatrix}.

The amplitude of $|0\rangle$ is unchanged while the amplitude of $|1\rangle$ picks up a minus sign.

For $Y$ :

Y \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} \begin{pmatrix} \alpha \\ \beta \end{pmatrix} = \begin{pmatrix} -i\beta \\ i\alpha \end{pmatrix}.

$Y$ both swaps the amplitudes and multiplies each by $\pm i$ — a combination of $X$ and $Z$ twisted by $i$ . In fact one can verify that $Y = iXZ$ .

Determinants as a consistency check

For any $2 \times 2$ matrix $\bigl(\begin{smallmatrix}a & b \\ c & d\end{smallmatrix}\bigr)$ , the determinant is $ad - bc$ . Computing it for each Pauli matrix gives

\det(X) = 0 \cdot 0 - 1 \cdot 1 = -1, \quad \det(Y) = 0 \cdot 0 - (-i)(i) = -((-i)(i)) = -(-i^2) = -(1) = -1, \quad \det(Z) = (1)(-1) - 0 = -1.

All three determinants equal $-1$ . This is not a coincidence. The group $SU(2)$ consists of $2\!\times\!2$ unitary matrices with determinant exactly $+1$ ; the Pauli matrices have $\det = -1$ , so they lie in $U(2)$ but outside $SU(2)$ . On the Bloch sphere each Pauli gate acts as a $180°$ rotation about its namesake axis ( $X$ about $\hat{x}$ , $Y$ about $\hat{y}$ , $Z$ about $\hat{z}$ ) — a proper rotation — so the $\det = -1$ in the $2\!\times\!2$ matrix sense does not indicate a reflection on the Bloch sphere.

Try it

Use the determinant formula to compute $\det(X) + \det(Y) + \det(Z)$ and return the result.

Run your code to see the quantum state.