哈达玛门（H 门）的矩阵为
$$H = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix}，\qquad H^\dagger = H$$
矩阵—矢量乘法逐行展开：$H|0\rangle = \frac{1}{\sqrt{2}}\,(1\times 1 + 1\times 0,\; 1\times 1 + (-1)\times 0)^{T} = \frac{1}{\sqrt{2}}\,(1, 1)^{T} = \frac{|0\rangle + |1\rangle}{\sqrt{2}} \equiv |+\rangle$；同理 $H|1\rangle = \frac{1}{\sqrt{2}}\,(1, -1)^{T} = \frac{|0\rangle - |1\rangle}{\sqrt{2}} \equiv |-\rangle$。归一化验证 $|1/\sqrt{2}|^2 + |1/\sqrt{2}|^2 = 1/2 + 1/2 = 1$，印证幺正变换保范数。H 把基矢变成均匀叠加，并把相位写入符号：$|+\rangle$ 与 $|-\rangle$ 的差别只在 $|1\rangle$ 分量的正负号。