Consider the ket vector $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, where $\alpha$ and $\beta$ are complex-valued probability amplitudes, $|0\rangle$ and $|1\rangle$ are the computational bases.

We can express $\alpha$ and $\beta$ in the exponential form [Wil17, (3.6)]:

$\alpha = r_0e^{j\varphi_0}, \qquad \beta = r_1e^{j\varphi_1},$

we can rewrite $|\psi\rangle$ as

$|\psi\rangle = e^{j\varphi_0}\underbrace{\left\{\cos\left(\frac{\theta}{2}\right)|0\rangle + e^{j(\varphi_1-\varphi_0)}\sin\left(\frac{\theta}{2}\right)|1\rangle\right\}}_{\text{Portion that matters}},$

where $\theta$ is related to $r_0$ and $r_1$ by:

$r_0 \triangleq \cos\left(\frac{\theta}{2}\right), \qquad r_1 \triangleq \sin\left(\frac{\theta}{2}\right).$

Thus, given $|\psi\rangle = \alpha|0\rangle + \beta|1\rangle$, we can rewrite it in the physically equivalent form:

$|\psi\rangle = \cos\left(\frac{\theta}{2}\right)|0\rangle + e^{j\varphi}\sin\left(\frac{\theta}{2}\right)|1\rangle,$

where $\varphi\triangleq\varphi_1-\varphi_0$, $0\leq\varphi\lt2\pi$, $0\leq\theta\leq\pi$. Note:

$\left|\cos\left(\frac{\theta}{2}\right)\right|^2 + \left|e^{j\varphi}\right|^2\left|\sin\left(\frac{\theta}{2}\right)\right|^2 = 1.$

💡 Above, $\theta/2$ is used instead of $\theta$ because for visualisation using a Bloch sphere (more on this later), as $\theta$ ranges from 0 to $\pi$, the values of $\cos$ and $\sin$ are confined within $[0,1]$ to ensure the qubit representation is unique [Wil17, p. 57].

So what is a Bloch sphere?

As explained in Fig. 2, a sphere provides the necessary number of degrees of freedom to represent a ket vector.