Vectors, matrices and linear algebra

A state as a column

We represent the basis states with two simple columns:

$$\lvert0\rangle=\begin{pmatrix}1\\0\end{pmatrix},\qquad \lvert1\rangle=\begin{pmatrix}0\\1\end{pmatrix}.$$

A general pure qubit has column $(\alpha,\beta)^T$. Its squared norm is $|\alpha|^2+|\beta|^2=1$.

A gate as a matrix

The X gate exchanges the two amplitudes:

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

To multiply a matrix by a column, take each row’s dot product with the column. For the first row above, $0\alpha+1\beta=\beta$.

The Hadamard gate is

$$H=\frac1{\sqrt2}\begin{pmatrix}1&1\\1&-1\end{pmatrix}.$$

Applying it to $\lvert0\rangle$ gives $(1,1)^T/\sqrt2$. Applying it again gives $(1,0)^T$. The cancellation in the second component is interference in matrix form.

Why unitary matrices?

An ideal closed-system gate $U$ satisfies $U^\dagger U=I$, where the dagger means conjugate transpose. This property preserves inner products and norms, so valid states remain normalised. It also means $U^{-1}=U^\dagger$.

Measurement and noisy evolution need more general descriptions; do not assume every physical operation is a unitary gate on the subsystem you observe.

Practise

Calculate $H\lvert1\rangle$. You should obtain $(1,-1)^T/\sqrt2$. Then verify $X^2=I$ and $H^2=I$ directly. Use the gate playground to compare the numerical outputs with your calculation.

Reference: IBM Quantum — Single systems.