Exercise 2.19#
The Pauli matrices are given by
(1)#\[\begin{split}
I= \begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix}, X = \begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix},Y = \begin{pmatrix}
0 & -i\\ i &0
\end{pmatrix},Z = \begin{pmatrix}
1 & 0\\ 0 &-1
\end{pmatrix}
\end{split}\]
The Hermitian conjugate of Pauli matrices can be directly obtained by
(2)#\[\begin{split}
I^{\dagger}= \begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix}, X^{\dagger} = \begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix},Y^{\dagger} = \begin{pmatrix}
0 & -i\\ i &0
\end{pmatrix},Z^{\dagger} = \begin{pmatrix}
1 & 0\\ 0 &-1
\end{pmatrix}
\end{split}\]
Compare eq. (1) and eq. (2) we could conclude that \(I=I^{\dagger}, X=X^{\dagger}, Y=Y^{\dagger}, Z=Z^{\dagger}\) and \(I, X, Y, Z\) are all Hermitian operators. Meanwhile, we could calculate
(3)#\[\begin{split}
\begin{align}
I^{\dagger}I &= II^{\dagger} = \begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix}\begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix} = \begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix} = I \\
X^{\dagger}X &= XX^{\dagger} = \begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix}\begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix} = \begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix} = I \\
Y^{\dagger}Y &= YY^{\dagger} = \begin{pmatrix}
0 & -i\\ i &0
\end{pmatrix}\begin{pmatrix}
0 & -i\\ i &0
\end{pmatrix} = \begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix} = I \\
Z^{\dagger}Z &= ZZ^{\dagger} = \begin{pmatrix}
1 & 0\\ 0 &-1
\end{pmatrix}\begin{pmatrix}
1 & 0\\ 0 &-1
\end{pmatrix} = \begin{pmatrix}
1 & 0\\ 0 & 1
\end{pmatrix} = I \\
\end{align}
\end{split}\]
Note that we have \(I^{\dagger}I = II^{\dagger}\), \(X^{\dagger}X = XX^{\dagger}\), \(Y^{\dagger}Y = YY^{\dagger}\) and \(Z^{\dagger}Z = ZZ^{\dagger}\) since \(I, X, Y, Z\) are Hermitian operators. From eq. (3) we could conclude that, \(I, X, Y, Z\) are all unitary matrices.