## 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.

This entry was posted in quantum country. Bookmark the permalink.