Quantum Country exercise 8

This is one in a series of posts working through the exercises in the Quantum Country online introduction to quantum computing and related topics. The exercises in the original document are not numbered; I have added my own numbers for convenience in referring to them.

Exercise 8. Show that the matrices Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} and Z = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} are unitary.

Answer: For Y we have

\begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}^\dagger \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \left( \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}^T \right)^* \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}^* \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}

= \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix} = \begin{bmatrix} -i^2 & 0 \\ 0 & -i^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

For Z we have

\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}^\dagger \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I

So both Y and Z are unitary.

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