Exercise 4.7¶
The Pauli matrices are given by
(1)¶\[\begin{split}
X = \begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix},Y = \begin{pmatrix}
0 & -i\\ i &0
\end{pmatrix}
\end{split}\]
Then we could check
(2)¶\[\begin{split}
XYX = \begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix}\begin{pmatrix}
0 & -i\\ i &0
\end{pmatrix}\begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix} = \begin{pmatrix}
i & 0\\ 0 & -i
\end{pmatrix}\begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix} = \begin{pmatrix}
0 & i\\ -i & 0
\end{pmatrix} = -Y
\end{split}\]
The rotation operator about the \(\hat{y}\) axis is given by
(3)¶\[\begin{split}
R_y(\theta) = \cos\frac{\theta}{2}\ I - i\sin\frac{\theta}{2}Y = \begin{pmatrix}
\cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2}
\end{pmatrix}
\end{split}\]
Then we could check
(4)¶\[\begin{split}
\begin{align}
XR_y(\theta)X &= \begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix}\begin{pmatrix}
\cos\frac{\theta}{2} & -\sin\frac{\theta}{2} \\ \sin\frac{\theta}{2} & \cos\frac{\theta}{2}
\end{pmatrix}\begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix} \\
&= \begin{pmatrix}
\sin\frac{\theta}{2} & \cos\frac{\theta}{2}\\ \cos\frac{\theta}{2} & -\sin\frac{\theta}{2}
\end{pmatrix}\begin{pmatrix}
0 & 1\\ 1 &0
\end{pmatrix} = \begin{pmatrix}
\cos\frac{\theta}{2} & \sin\frac{\theta}{2}\\ -\sin\frac{\theta}{2} & \cos\frac{\theta}{2}
\end{pmatrix}
\end{align}
\end{split}\]
Also, for \(R_y(-\theta)\) we have
(5)¶\[\begin{split}
R_y(-\theta) = \begin{pmatrix}
\cos\frac{-\theta}{2} & -\sin\frac{-\theta}{2} \\ \sin\frac{-\theta}{2} & \cos\frac{-\theta}{2}
\end{pmatrix} = \begin{pmatrix}
\cos\frac{\theta}{2} & \sin\frac{\theta}{2} \\ -\sin\frac{\theta}{2} & \cos\frac{\theta}{2}
\end{pmatrix}
\end{split}\]
According to eq. (4) and eq. (5), we have
(6)¶\[
XR_y(\theta)X = R_y(-\theta)
\]
A relative simple way to prove \(XR_y(\theta)X = R_y(-\theta)\) is to use the relation \(XYX=-X\) directly. That is
(7)¶\[\begin{split}
\begin{align}
XR_y(\theta)X &= X\left(\cos\frac{\theta}{2} I-i\sin\frac{\theta}{2} Y\right)\\
&= \cos\frac{\theta}{2} XIX - i\sin\frac{\theta}{2} XYX \\
&= \cos\frac{\theta}{2} I + i\sin\frac{\theta}{2}Y \\
&= \cos\frac{\theta}{2} I - i\sin\left(\frac{-\theta}{2}\right)Y = R_y(-\theta)
\end{align}
\end{split}\]