Bloch sphere
Basic introduction
Bloch sphere representation of a "single qubit state".
Bloch shpere
If we ahve any qubit state |\psi\rangle \in \mathbb{C}^{2} can be mapped to a point on the bloch sphere. Assume state |\psi\rangle such that
|\psi\rangle = \alpha|0\rangle + \beta|1\rangle, \ \alpha, \beta \ \in \mathbb{C}, \ \alpha^{2}+\beta^{2} = 1
We set
\alpha = \text{cos} \frac{\theta}{2} e^{i\delta}, \ \beta = \text{sin} \frac{\theta}{2} e^{i(\delta+\phi)}
thus,
\begin{array}{ll}
|\psi\rangle & = \text{cos} \frac{\theta}{2} e^{i\delta}|0\rangle + \text{sin}\frac{\theta}{2} e^{i(\delta+\phi)}|1\rangle \\
& = e^{i\delta}(\text{cos}\frac{\theta}{2}|0\rangle + \text{sin}e^{i\phi}\frac{\theta}{2}|1\rangle)
\end{array}
where e^{i\delta} is a global phase, somce it has same effects on |0\rangle and |1\rangle and cannot measure duing experiment, we can ignore it.
For the relative phase, e^{i\phi}, can canont ignore it. We set \text{sin} and \text{cos} a non-negative \mathbb{R}. From the bloch sphere, we can have
\begin{array}{ll}
x = \text{sin}\theta\text{cos}\phi\\
y = \text{sin}\theta\text{sin}\phi\\
z = \text{cos}\theta
\end{array}
where
\begin{array}{ll}
0 \leq \theta \leq \pi\\
0 \leq \phi < 2\pi\\
\end{array}
thus,
|\psi\rangle = \text{cos} \frac{\theta}{2}|0\rangle + \text{sin} \frac{\theta}{2}e^{i\theta}|1\rangle.
Common representations
Z-basis (computational basis)
- |0\rangle: North pole (z = 1)
- |1\rangle: South pole (z = -1)
X-basis
- |+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle) (x axis)
- |-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle) (-x axis)
Y-basis
- |i+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + i|1\rangle) (y axis)
- |i-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - i|1\rangle) (-y axis)