# 布洛赫球面

## 布洛赫球面諸點與純態的對應

 $\alpha = \cos \theta \, e^{i \delta},\quad \beta = \sin \theta \, e^{i (\delta + \phi)} \,$
$\Rightarrow |\psi \rangle = \cos \theta \, e^{i \delta} \, |0 \rangle + \sin \theta \, e^{i (\delta + \phi)} \, |1 \rangle = e^{i \delta}( \cos \theta \, |0 \rangle + \sin \theta \, e^{i \phi} \,|1 \rangle )$


  $|\psi \rangle = \cos \theta \, |0 \rangle + \sin \theta \, e^{i \phi} \,|1 \rangle$


 $0 \leq \theta \leq \frac{\pi}{2} \Rightarrow 0 \leq 2\theta \leq \pi, \quad$
$0 \leq \phi < 2 \pi$


$2\theta \,$$\phi \,$的所有分佈在三維空間$\mathbb{R}^3$中畫出來，就可以得到一個球面，此即布洛赫球面，如同圖1。

 $\begin{matrix} x & = & \sin 2 \theta \times \cos \phi \\ y & = & \sin 2 \theta \times \sin \phi \\ z & = & \cos 2 \theta \end{matrix}$


• $|0 \rangle$$z_+: \, (0,0,1)$
• $|1 \rangle$$z_-: \, (0,0,-1)$

## 習慣差異

 $\begin{matrix} x & = & \sin \theta \times \cos \phi \\ y & = & \sin \theta \times \sin \phi \\ z & = & \cos \theta \end{matrix}$


 $0 \leq \theta \leq \pi ,\quad 0 \leq \phi < 2\pi$


 $|\psi \rangle = \cos \frac{\theta}{2} \, |0 \rangle + \sin \frac{\theta}{2} \, e^{i \phi} \,|1 \rangle$


## 布洛赫球與混合態

$\frac{1}{2}\mathbf{1} = \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
$= \frac{1}{2} \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix} + \frac{1}{2} \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}$$= \frac{1}{2} |0 \rangle\langle 0| + \frac{1}{2} |1 \rangle\langle 1| = \frac{1}{2} z_+ + \frac{1}{2} z_-$
$= \frac{1}{2} \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{pmatrix} + \frac{1}{2} \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} & \frac{1}{2} \end{pmatrix} = \frac{1}{2} x_+ + \frac{1}{2} x_-$
$= \frac{1}{2} \begin{pmatrix} \frac{1}{2} & -\frac{i}{2} \\ \frac{i}{2} & \frac{1}{2} \end{pmatrix} + \frac{1}{2} \begin{pmatrix} \frac{1}{2} & \frac{i}{2} \\ -\frac{i}{2} & \frac{1}{2} \end{pmatrix} = \frac{1}{2} y_+ + \frac{1}{2} y_-$

## 外部連結

Density Operator of a Single Qubit: The Bloch Sphere